# questions

tags: ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-double-star-system-consists-of-two-stars-a-and-b-which-have-time-period-t-a-and-t-b-radius-r-a-and-r-b-and-mass-m-a-and-m-b-choose-the-correct-optio

slug:
a-double-star-system-consists-of-two-stars-a-and-b-which-have-time-period-t-a-and-t-b-radius-r-a-and-r-b-and-mass-m-a-and-m-b-choose-the-correct-optio-82746

question:
<div class="question">A double star system consists of two stars $A$ and $B$ which have time period $T_A$ and $T_B$. Radius $R_A$ and $R_B$ and mass $M_A$ and $M_B$. Choose the correct option.</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>If $T_A&gt;T_B$, then $R_A&gt;R_B$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>If $T_A&gt;T_B$, then $M_A&gt;M_B$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\left(\frac{T_A}{T_B}\right)^2=\left(\frac{R_A}{R_B}\right)^3$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$T_A=T_B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$T_A=T_B$</span> </div>

answer:
<div class="solution">In case of binary star system angular velocity and hence the time period of both the stars are equal.</div>


tags: ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-few-electric-field-lines-for-a-system-of-two-charges-q-1-and-q-2-fixed-at-two-different-points-on-the-x-axis-are-shown-in-the-figure-these-lines-sug

slug:
a-few-electric-field-lines-for-a-system-of-two-charges-q-1-and-q-2-fixed-at-two-different-points-on-the-x-axis-are-shown-in-the-figure-these-lines-sug-34862

question:
<div class="question">A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/R5lzBNcidnS01bSprkXrbMp3l9bUBkdMt6YycbuULb8.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\left|Q_1\right|&gt;\left|Q_2\right|$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\left|Q_1\right| &lt; \left|Q_2\right|$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>at a finite distance to the left of $Q_1$ the electric field is zero<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\left|Q_1\right|&gt;\left|Q_2\right|$<br/>, <br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> </div>

answer:
<div class="solution">From the behaviour of electric lines, we can say that $Q_1$ is positive and $Q_2$ is negative. Further, $\left|Q_1\right|&gt;\left|Q_2\right|$<br/>At some finite distance to the right of $Q_2$, electric field will be zero. Because electric field due to $Q_1$ is towards right (away from $Q_1$ ) and due to $Q_2$ is towards left (towards $Q_2$ ). But since magnitude of $Q_1$ is more, the two fields may cancel each other because distance of that point from $Q_1$ will also be more.<br/>$\therefore$ The correct options are (a) and (d).</div>


tags: ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-glass-tube-of-uniform-internal-radius-r-has-a-valve-separating-the-two-identical-ends-initially-the-valve-is-in-a-tightly-closed-position-end-1-has-

slug:
a-glass-tube-of-uniform-internal-radius-r-has-a-valve-separating-the-two-identical-ends-initially-the-valve-is-in-a-tightly-closed-position-end-1-has-30310

question:
<div class="question">A glass tube of uniform internal radius $(r)$ has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aTOnsM_ko_rZ4DTOZ1tXO4djJRRDKbR9XCHC2FJmEWk.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>air from end 1 flows towards end 2. No change in the volume of the soap bubbles<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>no change occurs<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases<br/></span> </div>

answer:
<div class="solution">$\Delta p_1=\frac{4 T}{r_1}$ and $\Delta p_2=\frac{4 T}{r_2}$<br/>$$<br/>\begin{array}{cc} <br/>&amp; r_1 &lt; r_2 \\<br/>\therefore &amp; \Delta p_1&gt;\Delta p_2<br/>\end{array}<br/>$$<br/>Air will flow from 1 to 2 and volume of bubble at end- 1 will decrease.<br/>Therefore, correct option is (b).</div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-hollow-pipe-of-length-08-m-is-closed-at-one-end-at-its-open-end-a-05-m-long-uniform-string-is-vibrating-in-its-second-harmonic-and-it-resonates-with

slug:
a-hollow-pipe-of-length-08-m-is-closed-at-one-end-at-its-open-end-a-05-m-long-uniform-string-is-vibrating-in-its-second-harmonic-and-it-resonates-with-43352

question:
<div class="question">A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~ms}^{-1}$, the mass of the string is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$5 \mathrm{~g}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$10 \mathrm{~g}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$20 \mathrm{~g}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$40 \mathrm{~g}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$10 \mathrm{~g}$<br/></span> </div>

answer:
<div class="solution">$2\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}$ $\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2}$<br/>( $\mu=$ mass per unit length of wire)<br/>or $\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}$<br/>Solving we get $\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}$<br/>$\therefore$ Mass of string<br/>$$<br/>=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}<br/>$$<br/>$\therefore$ The correct option is (b).</div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2021 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-horizontal-force-f-is-applied-at-the-centre-of-mass-of-a-cylindrical-object-of-mass-m-and-radius-r-perpendicular-to-its-axis-as-shown-in-the-figure-

slug:
a-horizontal-force-f-is-applied-at-the-centre-of-mass-of-a-cylindrical-object-of-mass-m-and-radius-r-perpendicular-to-its-axis-as-shown-in-the-figure-39442

question:
<div class="question"><p>A horizontal force <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math> is applied at the centre of mass of a cylindrical object of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> and radius <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math>, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi></math>. The centre of mass of the object has an acceleration <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math>. The acceleration due to gravity is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi></math>. Given that the object rolls without slipping, which of the following statement(s) is (are) correct?</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c40aaf47-970f-42c6-a45c-5b3be0c39e9a-image.png" style="width: 200px; height: 177px;"/></p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">For the same <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math>, the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> does not depend on whether the cylinder is solid or hollow.</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">For a solid cylinder, the maximum possible value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi>μ</mi><mi>g</mi></math>.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">The magnitude of the frictional force on the object due to the ground is always <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><mi>m</mi><mi>g</mi></math>.</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">For a thin-walled hollow cylinder, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math>.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value">For a solid cylinder, the maximum possible value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi>μ</mi><mi>g</mi></math>., For a thin-walled hollow cylinder, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math>.</span> </div>

answer:
<div class="solution"><p>Let moment of inertia <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>I</mi></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/f4eb0c6e-932c-407c-b3e4-d033856c2fa9-758f5244-8b39-44c0-b24d-ebcc14e0b238-image.png" style="width: 250px; height: 256px;"/></p><p>For pure rolling, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mi>α</mi><mi>R</mi></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>−</mo><mi>f</mi><mo>=</mo><mi>m</mi><mi>a</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>1</mn></mfenced></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mi>R</mi><mo>=</mo><mi>I</mi><mi>α</mi></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mi>R</mi><mo>=</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><mi>R</mi></mfrac></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac></math></p><p>Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>−</mo><mfrac><mrow><mi>I</mi><mi>a</mi></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>m</mi><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mo> </mo><mi>F</mi><mo>=</mo><mfenced><mrow><mi>m</mi><mo>+</mo><mfrac><mi>I</mi><msup><mi>R</mi><mn>2</mn></msup></mfrac></mrow></mfenced><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For the same value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math>, value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> depends on <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math>.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For a solid cylinder<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>I</mi><mo>=</mo><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac></mrow></mfenced></math> (to calculate maximum <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math>)</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mi>μ</mi><mi>m</mi><mi>g</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>μ</mi><mi>m</mi><mi>g</mi><mo>=</mo><mfenced><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac></mfenced><mo>×</mo><mfrac><mi>a</mi><msup><mi>R</mi><mn>2</mn></msup></mfrac></math></p><p>Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mn>2</mn><mi>μ</mi><mi>g</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>f</mi></math> is dependent on applied force.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> For thin-walled hollow cylinder<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>I</mi><mo>=</mo><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow></mfenced></math>,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>=</mo><mfenced><mrow><mi>m</mi><mo>+</mo><mfrac><mrow><mi>m</mi><msup><mi>R</mi><mn>2</mn></msup></mrow><msup><mi>R</mi><mn>2</mn></msup></mfrac></mrow></mfenced><mi>a</mi><mo>=</mo><mn>2</mn><mi>m</mi><mi>a</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mi>a</mi><mo>=</mo><mfrac><mi>F</mi><mrow><mn>2</mn><mi>m</mi></mrow></mfrac></math> </p></div>


tags: ['Mathematics', 'Ellipse', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-hyperbola-having-the-transverse-axis-of-length-2-sin-is-confocal-with-the-ellipse-3-x-2-4-y-2-12-then-its-equation-is

slug:
a-hyperbola-having-the-transverse-axis-of-length-2-sin-is-confocal-with-the-ellipse-3-x-2-4-y-2-12-then-its-equation-is-14840

question:
<div class="question">A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then, its equation is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$<br/></span> </div>

answer:
<div class="solution">The given ellipse is<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\<br/>&amp; \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>$$<br/>\therefore \quad a e=1<br/>$$<br/>Hence the eccentricity $e_1$, of the hyperbola is given by<br/>$$<br/>\begin{array}{rlrl} <br/>&amp; &amp; 1 &amp; =e_1 \sin \theta \\<br/>\Rightarrow &amp; &amp; e_1 &amp; =\operatorname{cosec} \theta \\<br/>\Rightarrow &amp; b^2 &amp; =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta<br/>\end{array}<br/>$$<br/>Hence, the hyperbola is<br/>$$<br/>\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1<br/>$$</div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>
source_db: MarksBatch1_P1.db
title:
a-lamina-is-made-by-removing-a-small-disc-of-diameter-2-r-from-a-bigger-disc-of-uniform-mass-density-and-radius-2-r-as-shown-in-the-figure-the-moment--1

slug:
a-lamina-is-made-by-removing-a-small-disc-of-diameter-2-r-from-a-bigger-disc-of-uniform-mass-density-and-radius-2-r-as-shown-in-the-figure-the-moment-1-30622

question:
<div class="question">A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing though $O$ and $P$ is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/FCGosuZSs9bvX9TAOyMX0Db910xOh_xEi9Gnk8hO45Q.original.fullsize.png"/></div>

answer:
<div class="solution">Let $\sigma$ be the surface mass density.
<br/> <br/>Moment of inertia of the lamina about axes passing through 'O'
<br/> <br/>$\begin{array}{l}
<br/> <br/>I_{O}=\frac{1}{2} \sigma\left[\pi(2 R)^{2}\right] \times(2 R)^{2}- \\
<br/> <br/>=\frac{13}{2} \pi \sigma R^{4} \quad\left[\frac{1}{2}\left(\sigma \pi R^{2}\right)^{2}+\sigma\left(\pi R^{2}\right) \times R^{2}\right]
<br/> <br/>\end{array}$
<br/> <br/>Moment of inertia of the lamina about axes passing through 'P'
<br/> <br/>$\begin{array}{l}
<br/> <br/>I_{P}=8 \pi \sigma \mathrm{R}^{4}+\sigma \pi(2 \mathrm{R})^{2} \times(2 \mathrm{R})^{2} \\
<br/> <br/>{\left[\frac{1}{2} \sigma\left(\pi R^{2}\right) R^{2}+\sigma\left(\pi R^{2}\right)\left(\sqrt{(2 R)^{2}+R^{2}}\right)^{2}\right]}
<br/> <br/>\end{array}$
<br/> <br/>$\begin{array}{l}
<br/> <br/>=24 \pi \sigma \mathrm{R}^{4}-5.5 \sigma \pi \mathrm{R}^{4}=18.5 \pi \sigma \mathrm{R}^{4} \\
<br/> <br/>\therefore \quad \frac{I_{P}}{I_{O}}=\frac{18.5 \pi \sigma R^{4}}{\frac{13}{2} \pi \sigma R^{4}}=\frac{37}{13} \approx 3
<br/> <br/>\end{array}$</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">6</span> </div>
source_db: MarksBatch1_P1.db
title:
a-large-glass-slab-3-5-of-thickness-8-cm-is-placed-over-a-point-source-of-light-on-a-plane-surface-it-is-seen-that-light-emerges-out-of-the-top-surfac

slug:
a-large-glass-slab-3-5-of-thickness-8-cm-is-placed-over-a-point-source-of-light-on-a-plane-surface-it-is-seen-that-light-emerges-out-of-the-top-surfac-12503

question:
<div class="question">A large glass slab $\left(\mu=\frac{5}{3}\right)$ of thickness 8 $\mathrm{cm}$ is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R \mathrm{~cm}$. What is the value of $R$ ?</div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/36cY2ycowgTDhPchNs-SiYyjriRvKBG1N-mYocUGmvo.original.fullsize.png"/><br/><br/>$$<br/>\frac{R}{t}=\tan \theta_C<br/>$$<br/>or $\quad R=t\left(\tan \theta_C\right)$<br/>$$<br/>\begin{aligned}<br/>&amp; \text { But, } \quad \sin \theta_C=\frac{1}{\mu}=\frac{3}{5} \\<br/>&amp; \therefore \quad \tan \theta_C=\frac{3}{4} \\<br/>&amp; \therefore \quad R=\frac{3}{4} t=\frac{3}{4}(8 \mathrm{~cm})=6 \mathrm{~cm} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>Hence the answer is 6 .</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-light-beam-is-travelling-from-region-i-to-region-iv-refer-figure-the-refractive-index-in-region-i-ii-iii-and-iv-are-n-0-2-n-0-6-n-0-and-8-n-0-respec

slug:
a-light-beam-is-travelling-from-region-i-to-region-iv-refer-figure-the-refractive-index-in-region-i-ii-iii-and-iv-are-n-0-2-n-0-6-n-0-and-8-n-0-respec-21573

question:
<div class="question">A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV are $n_0, \frac{n_0}{2}, \frac{n_0}{6}$ and $\frac{n_0}{8}$, respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/XM7-nvO7wvCxwFQBPcgcs6H5x6K1A6HAofcbqGfpC6g.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{3}{4}\right)$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{8}\right)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{4}\right)$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sin ^{-1}\left(\frac{1}{3}\right)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\sin ^{-1}\left(\frac{1}{8}\right)$<br/></span> </div>

answer:
<div class="solution">Critical angle from region III to region IV<br/>$$<br/>\sin \theta_c=\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}<br/>$$<br/>Now applying Snell's law in region I and region III :<br/>$$<br/>n_0 \sin \theta=\frac{n_0}{6} \sin \theta_c<br/>$$<br/>or<br/>$$<br/>\begin{aligned}<br/>\sin \theta &amp; =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8} \\<br/>\theta &amp; =\sin ^{-1}\left(\frac{1}{8}\right)<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (b).</div>


tags: ['Physics', 'Work Power Energy', 'JEE Main']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">8</span> </div>
source_db: MarksBatch1_P1.db
title:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-1

slug:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-1-30944

question:
<div class="question">A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WKlYQVtpL-uxteeE1iYredzRLzJ0_pwkpH4jdbu3QDI.original.fullsize.png"/><br/></div>

answer:
<div class="solution">$$<br/>\text { } \begin{aligned}<br/>\text { Now, } B &amp; =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\<br/>&amp; =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\<br/>a &amp; =\frac{\text { Net pulling force }}{\text { Total mass }} \\<br/>&amp; =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\<br/>s &amp; =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}<br/>\end{aligned}<br/>$$<br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BeQguzyOCEjbb25-f_J4-t2aGHS-oU6VRYL-wWQF-Ys.original.fullsize.png"/><br/><br/><br/>$$<br/>\begin{aligned}<br/>&amp; T-0.36 g=0.36 a=0.36 \frac{g}{3} \\<br/>&amp; \therefore \quad T=0.48 g \\<br/>&amp; \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\<br/>&amp; =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\<br/>&amp; =0.08(10)^2=8 \mathrm{~J} \\<br/>&amp;<br/>\end{aligned}<br/>$$</div>


tags: ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">8</span> </div>
source_db: MarksBatch1_P1.db
title:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms

slug:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-ms-66985


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-light-ray-travelling-in-glass-medium-is-incident-on-glassair-interface-at-an-angle-of-incidence-the-reflected-r-and-transmitted-t-intensities-both-a

slug:
a-light-ray-travelling-in-glass-medium-is-incident-on-glassair-interface-at-an-angle-of-incidence-the-reflected-r-and-transmitted-t-intensities-both-a-61328

question:
<div class="question">A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected $(R)$ and transmitted $(T)$ intensities, both as function of $\theta$, are plotted. The correct sketch is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8-2OElXNEWowMDwk6QM0zKc2EYtOtP1JZE_EFnzTF0I.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/MRxmX1Tor6LkgA33aw9FVk8crkAbKnw_SgDn1-nDFDs.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OMaB27Dwg6Rl0ATVY-Af7D0a69q79QdneMKQ989T0f4.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2SrQ0m4NG6zXfh1kZhFGBFx5biIvPp9IOPH4zj2V_6s.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OMaB27Dwg6Rl0ATVY-Af7D0a69q79QdneMKQ989T0f4.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">After critical angle reflection will be $100 \%$ and transmission is $0 \%$. Options (b) and (c) satisfy this condition. But option (c) is the correct option. Because in option (b) transmission is given $100 \%$ at $\theta=0^{\circ}$, which is not true.<br/>Analysis of Question<br/>From examination point of view question is moderately difficult. Because it takes little time to understand the given graphs clearly.</div>


tags: ['Mathematics', 'Three Dimensional Geometry', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-line-with-positive-direction-cosines-passes-through-the-point-p-2-1-2-and-makes-equal-angles-with-the-coordinate-axes-the-line-meets-the-plane-2-x-y

slug:
a-line-with-positive-direction-cosines-passes-through-the-point-p-2-1-2-and-makes-equal-angles-with-the-coordinate-axes-the-line-meets-the-plane-2-x-y-26191

question:
<div class="question">A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+z=9$ at point $Q$. The length of the line segment $P Q$ equals</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>1<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\sqrt{2}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>2</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\sqrt{3}$<br/></span> </div>

answer:
<div class="solution">$$<br/>\text { Since, } l=m=n=\frac{1}{\sqrt{3}}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ich1orC4Q3FFYIKtKIixDQa3M4swfCBvT6uaMCex8bo.original.fullsize.png"/><br/><br/><br/>$\therefore$ Equation of line are<br/>$$<br/>\begin{gathered}<br/>\frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\<br/>\Rightarrow \quad x-2=y+1=z-2=r \text { (say) }<br/>\end{gathered}<br/>$$<br/>$\therefore$ Any point on the line is<br/>$$<br/>Q \equiv(r+2 r-1, r+2)<br/>$$<br/>$\because Q$ lies on the plane $2 x+y+z=9$<br/>$$<br/>\begin{aligned}<br/>&amp; \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\<br/>&amp; \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\<br/>&amp; \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\<br/>&amp; =\sqrt{3}<br/>\end{aligned}<br/>$$</div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">6</span> </div>
source_db: MarksBatch1_P1.db
title:
a-long-circular-tube-of-length-10-m-and-radius-03-m-carries-a-current-i-along-its-curved-surface-as-shown-a-wire-loop-of-resistance-0005-and-of-radius

slug:
a-long-circular-tube-of-length-10-m-and-radius-03-m-carries-a-current-i-along-its-curved-surface-as-shown-a-wire-loop-of-resistance-0005-and-of-radius-87473

question:
<div class="question">A long circular tube of length $10 \mathrm{~m}$ and radius $0.3 \mathrm{~m}$ carries a current I along its curved surface as shown. A wire loop of resistance $0.005 \Omega$ and of radius $0.1 \mathrm{~m}$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I=I_0 \cos 300 t$, where $I_0$ is constant. If the magnetic moment of the loop is $N \mu_0 I_0 \sin 300 t$, then $N$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CbAZwTPq4oMzOvE4BMMDId0ZYqPEcVqy6u9nDKdL4lA.original.fullsize.png"/><br/></div>

answer:
<div class="solution">Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is<br/>$$<br/>B=\mu_0 n i<br/>$$<br/>Here, $n=$ number of turns per unit length<br/>$\therefore n i=$ current per unit length<br/>In the given problem,<br/><br/>$$<br/>\begin{aligned}<br/>&amp; n i=\frac{I}{L} \\<br/>\therefore \quad B &amp; =\frac{\mu_0 I}{L}<br/>\end{aligned}<br/>$$<br/>Flux passing through the circular coil is<br/>$$<br/>\phi=B S=\left(\frac{\mu_0 I}{L}\right)\left(\pi r^2\right)<br/>$$<br/>Induced emf, $e=-\frac{d \phi}{d t}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \cdot \frac{d I}{d t}$<br/>Induced current,<br/>$$<br/>i=\frac{e}{R}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \frac{d I}{d t}<br/>$$<br/>Magnetic moment $i A=i \pi r^2$<br/>or $\quad M=-\left(\frac{\mu_0 \pi^2 r^4}{L R}\right) \cdot \frac{d I}{d t}$<br/>Given, $\quad I=I_0 \cos 300 t$<br/>$$<br/>\therefore \quad \frac{d I}{d t}=-300 I_0 \sin (300 t)<br/>$$<br/><br/>Substituting in Eq. (i), we get<br/>$$<br/>\begin{aligned}<br/>M &amp; =\left(\frac{300 \pi^2 r^4}{L R}\right) \mu_0 I_0 \sin 300 t \\<br/>\therefore \quad N &amp; =\frac{300 \pi^2 r^4}{L R}<br/>\end{aligned}<br/>$$<br/>Substituting the values, we get<br/>$$<br/>\begin{aligned}<br/>N &amp; =\frac{300(22 / 7)^2(0.1)^4}{(10)(0.005)} \\<br/>&amp; =5.926<br/>\end{aligned}<br/>$$<br/>or $\quad N \simeq 6$<br/>Analysis of Question<br/>Question is difficult to understand inside the examination hall. But 5 to $10 \%$ question in IIT JEE are always difficult. Students should not panic. Because topper of IIT JEE scores approximately $80-90 \%$.</div>


tags: ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-long-hollow-conducting-cylinder-is-kept-coaxially-inside-another-long-hollow-conducting-cylinder-of-larger-radius-both-the-cylinders-are-initially-e

slug:
a-long-hollow-conducting-cylinder-is-kept-coaxially-inside-another-long-hollow-conducting-cylinder-of-larger-radius-both-the-cylinders-are-initially-e-61563

question:
<div class="question">A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>a potential difference appears between the two cylinders when a charge density is given to the inner cylinder<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>a potential difference appears between the two cylinders when a charge density is given to the outer cylinder<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>no potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>no potential difference appears between the two cylinders when same charge density is given to both the cylinders</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>a potential difference appears between the two cylinders when a charge density is given to the inner cylinder<br/></span> </div>

answer:
<div class="solution">There will be an electric field between two cylinders (using Gauss theorem. This electric field will produce a potential difference.<br/>$\therefore$ Answer is (a).</div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-long-insulated-copper-wire-is-closely-wound-as-a-spiral-of-n-turns-the-spiral-has-inner-radius-a-and-outer-radius-b-the-spiral-lies-in-the-xyplane-a

slug:
a-long-insulated-copper-wire-is-closely-wound-as-a-spiral-of-n-turns-the-spiral-has-inner-radius-a-and-outer-radius-b-the-spiral-lies-in-the-xyplane-a-37403

question:
<div class="question">A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the XY-plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WmycTUrrVf7IKN4c5VBT4TB4PBxzkeLl7dO0m0tVDr4.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sUooC5bCiYfSxQLUrgtSEhqOIKhrcCw8OvgH71nhc30.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OkdpbkPeRzs32HIE0PC_5k62_Xe10HO1oaGyCgqX0PU.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/6aA8dsRAmibaeIeEGBWhgL5Hqp3JSZHy07T-JtrFE_4.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GOAVi2UUHeYkKsJUWy0MiygIxqx6Ss8UyFnYWKY_jfk.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sUooC5bCiYfSxQLUrgtSEhqOIKhrcCw8OvgH71nhc30.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">If we take a small strip of $d r$ at distancer from centre, then number of turns in this strip would be<br/>$$<br/>d N=\frac{N}{b-a} d r<br/>$$<br/>Magnetic field due to this element at the centre of the coil will be<br/>$$<br/>\begin{aligned}<br/>&amp; d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r} \\<br/>\therefore \quad &amp; B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}<br/>\end{aligned}<br/>$$<br/>$\therefore$ Correct answer is (a).<br/>Analysis of Question<br/>(i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number $3.245$ of IE Irodov.<br/>(ii) Interestingly the same question was asked in IIT-JEE 2001 also.</div>


tags: ['Physics', 'Electromagnetic Induction', 'JEE Advanced', 'JEE Advanced 2021 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-long-straight-wire-carries-a-current-i-2-ampere-a-semicircular-conducting-rod-is-placed-beside-it-on-two-conducting-parallel-rails-of-negligible-res

slug:
a-long-straight-wire-carries-a-current-i-2-ampere-a-semicircular-conducting-rod-is-placed-beside-it-on-two-conducting-parallel-rails-of-negligible-res-11416

question:
<div class="question"><p>A long straight wire carries a current, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi><mo>=</mo><mn>2</mn></math> ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi>cm</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mo> </mo><mi>cm</mi></math> from the wire.  At time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math>, the rod starts moving on the rails with a speed <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>0</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> (see the figure).</p><p>A resistor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi><mo>=</mo><mn>1.4</mn><mo> </mo><mi mathvariant="normal">Ω</mi></math> and a capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub><mo>=</mo><mn>5.0</mn><mo> </mo><mi>μF</mi></math> are connected in series between the rails. At time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is uncharged. Which of the following statement(s) is (are) correct? [<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>μ</mi><mn>0</mn></msub><mo>=</mo><mn>4</mn><mi>π</mi><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></math> SI units. Take <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ln</mi><mn>2</mn><mo>=</mo><mn>0.7</mn></math>]</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/b7441289-f46c-471e-9ace-bb7533e4ce9e-image.png" style="width: 200px; height: 275px;"/></p></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1.2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>.4</mo><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value">Maximum current through <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1.2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></math> ampere, Maximum charge on capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>0</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup></math> coulomb.</span> </div>

answer:
<div class="solution"><p>Magnetic field due the long wire, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mo>=</mo><mfrac><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>I</mi></mrow><mrow><mn>2</mn><mi>π</mi><mi>x</mi></mrow></mfrac></math></p><p>Motional emf in small element, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">d</mi><mi>ε</mi><mo>=</mo><mi>B</mi><mi>v</mi><mo>d</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub><mfenced><mn>2</mn></mfenced></mrow><mrow><mn>2</mn><mi>π</mi><mi>x</mi></mrow></mfrac><mi mathvariant="normal">d</mi><mi>x</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>ε</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub></mrow><mi>π</mi></mfrac><msubsup><mo>∫</mo><mn>1</mn><mn>4</mn></msubsup><mfrac><mn>1</mn><mi>x</mi></mfrac><mi mathvariant="normal">d</mi><mi>x</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mi>ε</mi><mo>=</mo><mfrac><mrow><mn>3</mn><msub><mi>μ</mi><mn>0</mn></msub></mrow><mi>π</mi></mfrac><mi>ln</mi><mn>4</mn></math><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></math></p><p>Maximum charge on capacitor, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>q</mi><mi>max</mi></msub><mo>=</mo><mi>ε</mi><mi>C</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfenced><mrow><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></mrow></mfenced><mfenced><mrow><mn>5</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>6</mn></mrow></msup></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>84</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>13</mn></mrow></msup><mo> </mo><mi mathvariant="normal">C</mi><mo>=</mo><mn>8.4</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>12</mn></mrow></msup><mo> </mo><mi mathvariant="normal">C</mi></math></p><p>Maximum current in circuit, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>I</mi><mi>max</mi></msub><mo>=</mo><mfrac><mi>ε</mi><mi>R</mi></mfrac><mo>=</mo><mfrac><mrow><mn>16.8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>−</mo><mn>7</mn></mrow></msup></mrow><mn>1.4</mn></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>6</mn></mrow></msup><mo> </mo><mi mathvariant="normal">A</mi></math></p></div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-loop-carrying-current-i-lies-in-the-x-y-plane-as-shown-in-the-figure-the-unit-vector-k-is-coming-out-of-the-plane-of-the-paper-the-magnetic-moment-o

slug:
a-loop-carrying-current-i-lies-in-the-x-y-plane-as-shown-in-the-figure-the-unit-vector-k-is-coming-out-of-the-plane-of-the-paper-the-magnetic-moment-o-12791

question:
<div class="question">A loop carrying current $I$ lies in the $x-y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ZQI4dW5wmYwSBqFfVnVz4R2SgFtX3OlMJYgeeFwFsz8.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$a^{2} I \hat{k}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$(2 \pi+1) a^{2} I \hat{k}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$</span> </div>

answer:
<div class="solution">Magnetic moment of a current carrying loop $\vec{M}=N I \vec{A}$
<br/> <br/>Here $N=1, A=a^{2}+2 \pi\left(\frac{a}{2}\right)^{2}=a^{2}\left[1+\frac{\pi}{2}\right]$
<br/> <br/>From Screw law, direction of $m$ is outward or in +ve z-direction.
<br/> <br/>$\therefore \quad \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/gqv5YWas2u0OrTX0h5hHA_A8CiX4RBcAhWprW1WAXoM.original.fullsize.png"/></div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-1

slug:
a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-1-39438

question:
<div class="question">A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a &lt; x &lt; 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a &lt; x &lt; 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/_N5VIaTqGyv5jcrg44Q0Bg2NBIG0WPxXf5MuRx9KtH4.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mh_9iwlp6wjACA2hwq5AR2fr7YhbhDe2HfcZMMPWs4E.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5DLSkEyTqEIOhN9TjwKWqu3FgulcbE1av2dEBbNN1r0.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uNBd5cdR_7EE6qPM2BbIMVGhSvFPMwO_gXUHVMsYF48.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$<br/>Correct option is (a).</div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov

slug:
a-magnetic-field-b-b-0-j-exists-in-the-region-a-x-2-a-and-b-b-0-j-in-the-region-2-a-x-3-a-where-b-0-is-a-positive-constant-a-positive-point-charge-mov-65056

question:
<div class="question">A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a &lt; x &lt; 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a &lt; x &lt; 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/_N5VIaTqGyv5jcrg44Q0Bg2NBIG0WPxXf5MuRx9KtH4.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mh_9iwlp6wjACA2hwq5AR2fr7YhbhDe2HfcZMMPWs4E.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5DLSkEyTqEIOhN9TjwKWqu3FgulcbE1av2dEBbNN1r0.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uNBd5cdR_7EE6qPM2BbIMVGhSvFPMwO_gXUHVMsYF48.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CuWOOBYq3jS2YpduXSHKvuhX2J8MwJbhYpKeuhCF9vw.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$<br/>Correct option is (a).</div>


tags: ['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-man-walks-a-distance-of-3-units-from-the-origin-towards-the-northeast-n-4-5-e-direction-from-there-he-walks-a-distance-of-4-units-towards-the-northw

slug:
a-man-walks-a-distance-of-3-units-from-the-origin-towards-the-northeast-n-4-5-e-direction-from-there-he-walks-a-distance-of-4-units-towards-the-northw-56920

question:
<div class="question">A man walks a distance of 3 units from the origin towards the North-East $\left(\mathrm{N} 45^{\circ} \mathrm{E}\right)$ direction. From there, he walks a distance of 4 units towards the North-West ( $\mathrm{N} 45^{\circ} \mathrm{W}$ ) direction to reach a point $P$. Then, the position of $P$ in the Argand plane is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$3 e^{i \pi / 4}+4 i$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$(3-4 i) e^{i \pi / 4}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$(4+3 i) e^{i \pi / 4}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$(3+4 i) e^{i \pi / 4}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$(3+4 i) e^{i \pi / 4}$</span> </div>

answer:
<div class="solution">Let $O A=3$, so that the complex number associated with $A$ is $3 e^{i \pi / 4}$. If $z$ is the complex number associated with $P$, then<br/>$$<br/>\begin{array}{rlrl} <br/>&amp; &amp; \frac{z-3 e^{i \pi / 4}}{0-3 e^{i \pi / 4}} &amp; =\frac{4}{3} e^{-i \pi / 2}=-\frac{4 i}{3} \\<br/>\Rightarrow &amp; 3 z-9 e^{i \pi / 4} &amp; =12 i^{i \pi / 4} \\<br/>\Rightarrow &amp; z &amp; z &amp; =(3+4 i) e^{i \pi / 4}<br/>\end{array}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/qlN5mJTHFvHbuCdEp3r2OS8iZQJDWNFAAOodCbaNB6Y.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-massless-rod-b-d-is-suspended-by-two-identical-massless-strings-a-b-and-c-d-of-equal-lengths-a-block-of-mass-m-is-suspended-point-p-such-that-bp-is-

slug:
a-massless-rod-b-d-is-suspended-by-two-identical-massless-strings-a-b-and-c-d-of-equal-lengths-a-block-of-mass-m-is-suspended-point-p-such-that-bp-is-87850

question:
<div class="question">A massless rod $B D$ is suspended by two identical massless strings $A B$ and $C D$ of equal lengths. A block of mass ' $m$ ' is suspended point $P$ such that $B P$ is equal to ' $x$ ', if the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of $x$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kFubdTzCAp7Sxepzvw02bmX6QUSOvEenKMtjcCm52SA.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$1 / 5$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$1 / 4$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$41 / 5$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$31 / 4$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$1 / 5$<br/></span> </div>

answer:
<div class="solution">$f \propto v \propto \sqrt{T}$<br/>Further $\Sigma \tau_p=0$<br/>or<br/>or<br/>$$<br/>f_{A B}=2 f_{C D} \Rightarrow T_{A B}=4 T_{C D}<br/>$$<br/>$$<br/>\begin{aligned}<br/>T_{A B}(x) &amp; =T_{C D}(l-x) \\<br/>4 x &amp; =l-x\left(\text { as } T_{A B}-4 T_{C D}\right) \\<br/>x &amp; =\frac{l}{5}<br/>\end{aligned}<br/>$$</div>


tags: ['Physics', 'Thermal Properties of Matter', 'JEE Main']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">9</span> </div>
source_db: MarksBatch1_P1.db
title:
a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c--1

slug:
a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c-1-32812

question:
<div class="question">A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]</div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfSQoXhjZNvHl8wU_-OPaLyrnErrKwulkUaBdtVyaCo.original.fullsize.png"/><br/><br/><br/>Heat will flow both sides from point $P$.<br/>$$<br/>\begin{aligned}<br/>L_1 \frac{d m_1}{d t} &amp; =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\<br/>&amp; =\frac{400}{(\lambda x) / k A} \\<br/>L_1 \frac{d m_2}{d t} &amp; =\frac{400-100}{(100-\lambda) x / k A}<br/>\end{aligned}<br/>$$<br/>In about two equations,<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\<br/>&amp; L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}<br/>\end{aligned}<br/>$$<br/>Solving these two equations we get $\lambda=9$.</div>


tags: ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">9</span> </div>
source_db: MarksBatch1_P1.db
title:
a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c-

slug:
a-metal-rod-a-b-of-length-10-x-has-its-one-end-a-in-ice-at-0-c-and-the-other-end-b-in-water-at-10-0-c-if-a-point-p-on-the-rod-is-maintained-at-40-0-c-70467

question:
<div class="question">A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]</div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfSQoXhjZNvHl8wU_-OPaLyrnErrKwulkUaBdtVyaCo.original.fullsize.png"/><br/><br/><br/>Heat will flow both sides from point $P$.<br/>$$<br/>\begin{aligned}<br/>L_1 \frac{d m_1}{d t} &amp; =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\<br/>&amp; =\frac{400}{(\lambda x) / k A} \\<br/>L_1 \frac{d m_2}{d t} &amp; =\frac{400-100}{(100-\lambda) x / k A}<br/>\end{aligned}<br/>$$<br/>In about two equations,<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\<br/>&amp; L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}<br/>\end{aligned}<br/>$$<br/>Solving these two equations we get $\lambda=9$.</div>


tags: ['Physics', 'Oscillations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-metal-rod-of-length-l-and-mass-m-is-pivoted-at-one-end-a-thin-disc-of-mass-m-and-radius-r-l-is-attached-at-its-centre-to-the-free-end-of-the-rod-con

slug:
a-metal-rod-of-length-l-and-mass-m-is-pivoted-at-one-end-a-thin-disc-of-mass-m-and-radius-r-l-is-attached-at-its-centre-to-the-free-end-of-the-rod-con-44195

question:
<div class="question">A metal rod of length $L$ and mass $m$ is pivoted at one end. A thin disc of mass $M$ and radius $R( &lt; L)$ is attached at its centre to the free end of the rod. Consider two ways the disc is attached. Case A-the disc is not free to rotate about its centre and case $B$-the disc is free to rotate about its centre. The rod-disc system performs<br/>SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true?<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2Z9bbEPubSFDgbkK7f83upg2RB4bCyUHjrCj_uu9Q6U.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Restoring torque in case $A=$ Restoring torque in case $B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Restoring torque in case $A &lt; $ Restoring torque in case $B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Angular frequency for case $A&gt;$ Angular frequency for case $B$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Angular frequency for case $A &lt; $ Angular frequency for case $B$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Restoring torque in case $A=$ Restoring torque in case $B$<br/>, <br/>Restoring torque in case $A &lt; $ Restoring torque in case $B$</span> </div>

answer:
<div class="solution">$\tau_A=\tau_B=m g \frac{L}{2} \sin \theta+\operatorname{MgL} \sin \theta$<br/>$=$ Restoring torque about point $O$.<br/>In case $A$, moment of inertia will be more. Hence, angular acceleration $(\alpha=\tau / I)$ will be less. Therefore, angular frequency will be less.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/1s5RswcuRs38mBLtlc7w0I0vX4MgAYvlJTzjP3tzLLk.original.fullsize.png"/><br/><br/>$\therefore$ Correct options are (a) and (d).<br/>Analysis of Question<br/>Question is difficult from my point of view. Because this type of SHM is rarely taught in the class and questions of this type are not given in standard books.</div>


tags: ['Physics', 'Current Electricity', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-meter-bridge-is-set-up-as-shown-in-figure-to-determine-an-unknown-resistance-x-using-a-standard-10-resistor-the-galvanometer-shows-null-point-when-t

slug:
a-meter-bridge-is-set-up-as-shown-in-figure-to-determine-an-unknown-resistance-x-using-a-standard-10-resistor-the-galvanometer-shows-null-point-when-t-36255

question:
<div class="question">A meter bridge is set up as shown in figure, to determine an unknown resistance $X$ using a standard $10 \Omega$ resistor. The galvanometer shows null point when tapping key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-5wAQXPr0ogSu-GX8RP1V0KuFgaPpJl2ZFwXoZMVR7U.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$10.2 \Omega$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$10.6 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$10.8 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$11.1 \Omega$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$10.6 \Omega$<br/></span> </div>

answer:
<div class="solution">Using the concept of balanced Wheatstone bridge, we have<br/>$$<br/>\begin{array}{rlrl} <br/>&amp; \frac{P}{Q} &amp; =\frac{R}{S} \\<br/>\therefore &amp; &amp; \frac{X}{(52+1)} &amp; =\frac{10}{(48+2)} \\<br/>\therefore &amp; &amp; X &amp; =\frac{10 \times 53}{50}=10.6 \Omega<br/>\end{array}<br/>$$<br/>$\therefore$ Correct option is (b).<br/>Analysis of Question<br/>Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.</div>


tags: ['Physics', 'Kinetic Theory of Gases', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">$3.16$</span> </div>
source_db: MarksBatch1_P1.db
title:
a-mixture-of-2-moles-of-helium-gas-atomic-mass-4-amu-and-1-mole-of-argon-gas-atomic-mass-40-amu-is-kept-at-300-k-in-a-container-the-ratio-of-the-rms-s

slug:
a-mixture-of-2-moles-of-helium-gas-atomic-mass-4-amu-and-1-mole-of-argon-gas-atomic-mass-40-amu-is-kept-at-300-k-in-a-container-the-ratio-of-the-rms-s-94660

question:
<div class="question">A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass $=40 \mathrm{amu}$ ) is kept at $300 \mathrm{~K}$ in a container. The ratio of the rms speeds $\left(\frac{v_{\mathrm{rms}}(\text { helium })}{v_{\mathrm{rms}}(\operatorname{argon})}\right)$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$0.32$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$0.45$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2.24$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$3.16$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">Using $V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow V_{r m s} \propto \frac{1}{\sqrt{M}}$ $\frac{v_{\text {rms (helium) }}}{v_{\text {rms (argon) }}}=\sqrt{\frac{M_{\text {argon }}}{M_{\text {helium }}}}=\sqrt{\frac{40}{4}}=\sqrt{10} \approx 3.16$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-monoatomic-ideal-gas-undergoes-a-process-in-which-the-ratio-of-p-to-v-at-any-instant-is-constant-and-equals-to-1-what-is-the-molar-heat-capacity-of-

slug:
a-monoatomic-ideal-gas-undergoes-a-process-in-which-the-ratio-of-p-to-v-at-any-instant-is-constant-and-equals-to-1-what-is-the-molar-heat-capacity-of-86103

question:
<div class="question">A monoatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to 1 . What is the molar heat capacity of the gas?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{4 R}{2}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{3 R}{2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{5 R}{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>0</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{4 R}{2}$<br/></span> </div>

answer:
<div class="solution">The molar heat capacity for any process is given by a following expression<br/>$C=C_V+\frac{R}{1-\gamma}$ when $p V^\gamma=$ constant and $C_p / C_V=\gamma$<br/>Here, $\frac{p}{V}=1$ i.e., $p V^{-1}=$ constant<br/>$$<br/>\therefore \quad C=\frac{3}{2} R+\frac{R}{1-(-1)}=\frac{3}{2} R+\frac{R}{2}=\frac{4}{2} R<br/>$$</div>


tags: ['Physics', 'Capacitance', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-1

slug:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-1-22884

question:
<div class="question">A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kXR5joieLkCoSaad8_WTmvKA05ny_GxeO0ptHUUMHLY.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d-3 v t}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> </div>

answer:
<div class="solution">After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.<br/>Now, time constant as function of time<br/>$$<br/>\begin{aligned}<br/>\tau_c &amp; =C R \\<br/>&amp; \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\<br/>&amp; =\frac{6 \varepsilon_0 R}{5 d+3 v t}<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (a).</div>


tags: ['Physics', 'Capacitance', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i

slug:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-3-d-i-97609

question:
<div class="question">A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kXR5joieLkCoSaad8_WTmvKA05ny_GxeO0ptHUUMHLY.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{6 \varepsilon_0 R}{5 d-3 v t}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{6 \varepsilon_0 R}{5 d+3 v t}$<br/></span> </div>

answer:
<div class="solution">After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.<br/>Now, time constant as function of time<br/>$$<br/>\begin{aligned}<br/>\tau_c &amp; =C R \\<br/>&amp; \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\<br/>&amp; =\frac{6 \varepsilon_0 R}{5 d+3 v t}<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (a).</div>


tags: ['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-particle-moves-in-the-x-y-plane-under-the-influence-of-a-force-such-that-its-linear-momentum-is-p-t-a-i-cos-k-t-j-sin-k-t-where-a-and-k-are-constant-1

slug:
a-particle-moves-in-the-x-y-plane-under-the-influence-of-a-force-such-that-its-linear-momentum-is-p-t-a-i-cos-k-t-j-sin-k-t-where-a-and-k-are-constant-1-92214

question:
<div class="question">A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t)=A[\hat{\mathbf{i}} \cos k t-\hat{\mathbf{j}} \sin k t]$, where $A$ and $k$ are constants. The angle between the force and the momentum is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$0^{\circ}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$30^{\circ}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$45^{\circ}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$90^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$90^{\circ}$</span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\<br/>&amp; \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}<br/>\end{aligned}<br/>$$<br/>Since,<br/>F $\cdot \mathbf{p}=0$<br/>$\therefore$ Angle between $\mathbf{F}$ and pshould be $90^{\circ}$.<br/>$\therefore$ Correct option is $(\mathrm{d})$.</div>


tags: ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2018 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">2</span> </div>
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-10-3-kg-and-charge-10-c-is-initially-at-rest-at-time-t-0-the-particle-comes-under-the-influence-of-an-electric-field-e-t-e-0-sin-t--1

slug:
a-particle-of-mass-10-3-kg-and-charge-10-c-is-initially-at-rest-at-time-t-0-the-particle-comes-under-the-influence-of-an-electric-field-e-t-e-0-sin-t-1-14819

question:
<div class="question">A particle, of mass <math><msup><mrow><mn>10</mn></mrow><mrow><mo>-</mo><mn>3</mn></mrow></msup><mi> kg</mi></math> and charge <math><mn>1.0</mn><mi> </mi><mn>C</mn></math>, is initially at rest. At time <math><mi>t</mi><mo>=</mo><mn>0</mn></math>, the particle comes under the influence of an electric field <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>E</mi><mo>→</mo></mover><mfenced separators="|"><mi>t</mi></mfenced><mo>=</mo><msub><mi>E</mi><mn>0</mn></msub><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mi> </mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math> where <math><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mn>1.0</mn><mi> N</mi><mi> </mi><msup><mrow><mn>C</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> and <math><mi>ω</mi><mo>=</mo><msup><mrow><mn>10</mn></mrow><mrow><mn>3</mn></mrow></msup><mi> rad</mi><mi> </mi><msup><mrow><mn>s</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>. Consider the effect of only the electrical force on the particle. Then the maximum speed, in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>m</mn><mo> </mo><msup><mn>s</mn><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>, attained by the particle at subsequent times is ____________.</div>

answer:
<div class="solution"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>3</mn></mrow></msup><mi> kg</mi><mo>,</mo><mi> </mi><mi> </mi><mi>q</mi><mo>=</mo><mn>1 </mn><mn>C</mn><mo>,</mo><mi> </mi><mi>t</mi><mo>=</mo><mn>0</mn></math><br/>Force on particle will be<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>=</mo><mi>q</mi><mi>E</mi><mo>=</mo><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math><br/>At <math><msub><mrow><mi>v</mi></mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></msub><mo>,</mo><mi> </mi><mi>F</mi><mo>=</mo><mn>0</mn></math> <math><mo>⇒</mo><mi>q</mi><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>⁡</mo><mrow><mi>ω</mi><msub> <mi>t</mi> <mi>o</mi> </msub> <mo>=</mo><mn>0</mn></mrow> </mrow></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>ω</mi><msub><mi>t</mi><mi>o</mi></msub><mo>=</mo><mi>π</mi></math><br/><math><mi>F</mi><mo>=</mo><mi>q</mi><msub><mrow><mi>E</mi></mrow><mrow><mn>0</mn></mrow></msub><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>⁡</mo><mrow><mi>ω</mi><mi>t</mi></mrow></mrow></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>d</mn><mi>v</mi></mrow><mrow><mn>d</mn><mi>t</mi></mrow></mfrac><mo>=</mo><mi>q</mi><mfrac><msub><mi>E</mi><mn>0</mn></msub><mi>m</mi></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mo>∫</mo><mn>0</mn><mi>v</mi></msubsup><mn>d</mn><mi>v</mi><mo>=</mo><msubsup><mo>∫</mo><mn>0</mn><mrow><mi>π</mi><mo>∕</mo><mi>ω</mi></mrow></msubsup><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mi>m</mi></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mo> </mo><mn>d</mn><mi>t</mi></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>-</mo><mn>0</mn><mo>=</mo><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><msubsup><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow></mfenced><mn>0</mn><mrow><mi>π</mi><mo>∕</mo><mi>ω</mi></mrow></msubsup><mi> </mi></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>-</mo><mn>0</mn><mo>=</mo><mfrac><mrow><mi>q</mi><msub><mi>E</mi><mn>0</mn></msub></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfenced separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mi>π</mi></mfenced></mrow></mfenced><mo>-</mo><mfenced separators="|"><mrow><mo>-</mo><mi>cos</mi><mfenced><mn>0</mn></mfenced></mrow></mfenced></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><mfrac><mrow><mn>1</mn><mo>×</mo><mn>1</mn></mrow><mrow><mi>m</mi><mi>ω</mi></mrow></mfrac><mo>×</mo><mn>2</mn><mo>=</mo><mn>2</mn><mo> m</mo><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-m-and-charge-q-moving-with-velocity-v-enters-region-ii-normal-to-the-boundary-as-shown-in-the-figure-region-ii-has-a-uniform-magnet

slug:
a-particle-of-mass-m-and-charge-q-moving-with-velocity-v-enters-region-ii-normal-to-the-boundary-as-shown-in-the-figure-region-ii-has-a-uniform-magnet-86730

question:
<div class="question">A particle of mass $m$ and charge $q$, moving with velocity $v$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $l$. Choose the correct choice(s)<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2R-9CzzQsKQiIWeWGqRQfQoFeKBfJB3YsXFFqvgUbGU.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>The particle enters Region III only if its velocity $v&gt;\frac{q l B}{m}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>The particle enters Region III only if its velocity $v &lt; \frac{q I B}{m}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>The particle enters Region III only if its velocity $v&gt;\frac{q l B}{m}$<br/>, <br/>Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$<br/>, <br/>Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I</span> </div>

answer:
<div class="solution">$\overrightarrow{\mathbf{V}} \perp \overrightarrow{\mathbf{B}}$ in region II. Therefore, path of particle is circle in region II. Particle enters in region III if, radius of circular path, $r&gt;1$ or $\frac{m v}{B q}&gt;1$ or $v&gt;\frac{B q l}{m}$<br/>If $v=\frac{B q l}{m}, r=\frac{m v}{B q}=l$, particle will turn back and path length will be maximum as shown in figure in region II. If particle returns to region I, time spent in region II will be $t=\frac{T}{2}=\frac{\pi m}{B q}$, which is independent of $v$.<br/>$\therefore$ correct options are (a), (c) and (d).<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0u3oe3umQiRM7QrzjiMfVEpq5UCFuBEPfFEcz8oRLs4.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2015 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-particle-of-unit-mass-is-moving-along-the-xaxis-under-the-influence-of-a-force-and-its-total-energy-is-conserved-four-possible-forms-of-the-potentia

slug:
a-particle-of-unit-mass-is-moving-along-the-xaxis-under-the-influence-of-a-force-and-its-total-energy-is-conserved-four-possible-forms-of-the-potentia-63223

question:
<div class="question">A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></math> are constants). Match the potential energies in column I to the corresponding statement(s) in column II.<table border="1"> <tbody> <tr> <td> </td> <td><strong>Column I</strong></td> <td> </td> <td><strong>Column II</strong></td> </tr> <tr> <td>A.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>1</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math></td> <td>P.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td>B.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math></td> <td>Q.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mn>0</mn></math> .</td> </tr> <tr> <td>C.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>3</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>⁡</mo><mrow><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></mrow></mrow></math></td> <td>R.</td> <td>The force acting on the particle is zero at <math><mi mathvariant="normal">x</mi><mo>=</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td>D.</td> <td><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>4</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfenced></math></td> <td>S.</td> <td>The particle experiences an attractive force towards <math><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi><mo>=</mo><mi mathvariant="normal"> </mi><mn>0</mn></math> in the region <math><mfenced close="|" open="|" separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>&lt;</mo><mi mathvariant="normal">a</mi></math></td> </tr> <tr> <td> </td> <td> </td> <td>T.</td> <td>The particle with total energy <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac></math> can oscillate about the point <math><mi mathvariant="normal">x</mi><mo>=</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math></td> </tr> </tbody></table></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">a-s,t;b-s;c-s,t;d-t;</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">a-s;b-r,s;c-r,s,t;d-s,t;</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">a-r,s;b-q;c-q,r;d-t;</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;</span> </div>

answer:
<div class="solution"><math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>1</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mn>2</mn><mfenced close="]" open="[" separators="|"><mrow><mn>1</mn><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></math><br/><math><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mi mathvariant="normal">x</mi></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mi mathvariant="normal">x</mi><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mo>-</mo><mi mathvariant="normal">x</mi></mrow></mfenced><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math> , at <math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mo>-</mo><mi mathvariant="normal">a</mi></math><br/><math><mfenced close="}" open="" separators="|"><mrow><mtable><mtr><mtd><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mn>0</mn><mo>,</mo></mtd></mtr><mtr><mtd><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></mtd></mtr></mtable></mrow></mfenced><mo>â</mo></math> Oscillate when total ME is less then <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></math><br/><math><mi mathvariant="normal">A</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">T</mi></math><br/>B- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">B</mi><mo>â</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">S</mi></math><br/>C- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>3</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow><mo>+</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow></mrow></mfenced></math><br/><math><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfrac><mrow><mn>2</mn><mi mathvariant="normal">x</mi></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><mrow><mrow><mi mathvariant="normal">exp</mi></mrow><mo>â¡</mo><mrow><mfenced separators="|"><mrow><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfenced></mrow></mrow><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced></math><br/><math><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow></mfrac><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></msup><mi mathvariant="normal"> </mi><mfenced close="]" open="[" separators="|"><mrow><mi mathvariant="normal">x</mi><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi><mo>+</mo><mi mathvariant="normal">a</mi></mrow></mfenced><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi><mo>-</mo><mi mathvariant="normal">a</mi></mrow></mfenced></mrow></mfenced></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mn>0</mn></math> even function hence minima<br/><math><mi mathvariant="normal">C</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">Q</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">S</mi></math><br/>D- <math><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>4</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">x</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>3</mn><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac></mrow></mfenced></math><br/><math><mi mathvariant="normal">F</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mi mathvariant="normal">a</mi></mrow></mfrac><mo>-</mo><mfrac><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></mfrac><mfenced close="]" open="[" separators="|"><mrow><msup><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mi mathvariant="normal"> </mi></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">F</mi><mo>=</mo><mn>0</mn></math><br/><math><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mo>-</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=</mo><mo>+</mo><mi mathvariant="normal">a</mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">U</mi><mo>=</mo><mo>+</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac></math><br/>Hence oscillate about <math><mi mathvariant="normal">x</mi><mo>=</mo><mo>-</mo><mi mathvariant="normal">a</mi></math> if <math><mi mathvariant="normal">T</mi><mo>.</mo><mi mathvariant="normal">M</mi><mo>.</mo><mi mathvariant="normal">E</mi><mo>&lt;</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">U</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>3</mn></mrow></mfrac></math><br/><math><mi mathvariant="normal">D</mi><mo>â</mo><mi mathvariant="normal">P</mi><mo>,</mo><mi mathvariant="normal">R</mi><mo>,</mo><mi mathvariant="normal">T</mi></math></div>


tags: ['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-particle-p-starts-from-the-point-z-0-1-2-i-where-i-1-it-moves-first-horizontally-away-from-origin-by-5-units-and-then-vertically-away-from-origin-by

slug:
a-particle-p-starts-from-the-point-z-0-1-2-i-where-i-1-it-moves-first-horizontally-away-from-origin-by-5-units-and-then-vertically-away-from-origin-by-14396

question:
<div class="question">A particle $P$ starts from the point $z_0=1+2 i$, where $i=\sqrt{-1}$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z_1$. From $z_1$ the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and then it moves through an angle $\frac{\pi}{2}$ in anti-clockwise direction on a circle with centre at origin, to reach a point $z_2$. The point $z_2$ is given by</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$6+7 i$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-7+6 i$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$7+6 i$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$-6+7 i$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$-6+7 i$</span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7V-wZrcaDesQMMcdHT0vrL9hcmdmdeq6EEypYnkMDJU.original.fullsize.png"/><br/><br/>$$<br/>z_2^{\prime}=\left(6+\sqrt{2} \times \cos 45^{\circ}, 5+\sqrt{2} \sin 45^{\circ}\right)=(7,6)=7+6 i<br/>$$<br/>By rotation about $(0,0)$,<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{z_2}{z_2^{\prime}}=e^{i \frac{\pi}{2}} \\<br/>&amp; \Rightarrow \quad z_2=z_2^{\prime}\left(e^{i \frac{\pi}{2}}\right) \\<br/>&amp; \Rightarrow \quad z_2=(7+6 i)\left(\cos \frac{\pi}{-}+i \sin \frac{\pi}{-}\right)=(7+6 i)(i)=-6+7 i<br/>\end{aligned}<br/>$$</div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-person-blows-into-openend-of-a-long-pipe-as-a-result-a-high-pressure-pulse-of-air-travels-down-the-pipe-when-this-pulse-reaches-the-other-end-of-the

slug:
a-person-blows-into-openend-of-a-long-pipe-as-a-result-a-high-pressure-pulse-of-air-travels-down-the-pipe-when-this-pulse-reaches-the-other-end-of-the-49024

question:
<div class="question">A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe.
<br/> <br/>When this pulse reaches the other end of the pipe,</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value">a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open., a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.</span> </div>

answer:
<div class="solution">When a sound pulse is reflected from open end of a pipe, phase changes by $180^{\circ}$. A high pressure pulse i.e., compression is reflected as a low pressure pulse i.e., rarefaction. When sound pulse is reflected through a rigid boundary (closed end of a pipe), no phase change occurs so a high pressure pulse is reflected as a high pressure pulse.</div>


tags: ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">8</span> </div>
source_db: MarksBatch1_P1.db
title:
a-piece-of-ice-heat-capacity-2100-j-kg-1-c-1-and-latent-heat-336-1-0-5-j-kg-1-of-mass-m-gram-is-at-5-c-at-atmospheric-pressure-it-is-given-420-j-of-he

slug:
a-piece-of-ice-heat-capacity-2100-j-kg-1-c-1-and-latent-heat-336-1-0-5-j-kg-1-of-mass-m-gram-is-at-5-c-at-atmospheric-pressure-it-is-given-420-j-of-he-50454

question:
<div class="question">A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ gram is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \mathrm{~g}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is</div>

answer:
<div class="solution">Language of question is slightly wrong. As heat capacity and specific heat are two different physical quantities. Unit $\mathrm{J}-\mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity given in the question is really the specific heat. Now applying the heat exchange equation.<br/>$$<br/>\begin{array}{r}<br/>420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \\<br/>\left(3.36 \times 10^5\right)<br/>\end{array}<br/>$$<br/>Solving this equation we get,<br/>$$<br/>m=8 \mathrm{~g}<br/>$$<br/>$\therefore$ The correct answer is 8 .</div>


tags: ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-piece-of-wire-is-bent-in-the-shape-of-a-parabola-y-k-x-2-yaxis-vertical-with-a-bead-of-mass-m-on-it-the-bead-can-slide-on-the-wire-without-friction--1

slug:
a-piece-of-wire-is-bent-in-the-shape-of-a-parabola-y-k-x-2-yaxis-vertical-with-a-bead-of-mass-m-on-it-the-bead-can-slide-on-the-wire-without-friction-1-90961

question:
<div class="question">A piece of wire is bent in the shape of a parabola $y=k x^2$ (y-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with the constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the $y$-axis is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{a}{g k}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{a}{2 g k}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{2 a}{g k}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{a}{4 g k}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{a}{2 g k}$<br/></span> </div>

answer:
<div class="solution">$N \sin \theta=m g$<br/>$$<br/>\begin{aligned}<br/>N \cos \theta &amp; =m a \\<br/>\tan \theta &amp; =\frac{g}{a} \\<br/>\cot \theta &amp; =\frac{a}{g}=\tan \left(90^{\circ}-\theta\right)=\frac{d y}{d x}=2 k x<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/kDkPeNyYC5fMi_8lwwdUnIYuwQGNAzZla9ZDheSXGmA.original.fullsize.png"/><br/><br/>$\therefore \quad x=\frac{a}{2 k g}$</div>


tags: ['Mathematics', 'Three Dimensional Geometry', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-plane-passes-through-1-2-1-and-is-perpendicular-to-two-planes-2-x-2-y-z-0-and-x-y-2-z-4-then-the-distance-of-the-plane-from-the-point-1-2-2-is

slug:
a-plane-passes-through-1-2-1-and-is-perpendicular-to-two-planes-2-x-2-y-z-0-and-x-y-2-z-4-then-the-distance-of-the-plane-from-the-point-1-2-2-is-82024

question:
<div class="question">A plane passes through $(1,-2,1)$ and is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, then the distance of the plane from the point $(1,2,2)$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>0<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>1<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{2}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$2 \sqrt{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$2 \sqrt{2}$</span> </div>

answer:
<div class="solution">Let the equation of plane be,<br/>$$<br/>a(x-1)+b(y+2)+c(z-1)=0<br/>$$<br/>which is perpendicular to $2 x-2 y+z=0$ and $x-y+2 z=4$<br/>$$<br/>\begin{array}{llrl}<br/>\Rightarrow &amp; 2 a-2 b+c &amp; =0 \text { and } a-b+2 c=0 \\<br/>\text { or } &amp; \frac{a}{-2} 1 &amp; =\frac{b}{2}=\frac{c}{-2} \\<br/>\Rightarrow &amp; -12 &amp; 1 &amp; -1 \\<br/>\Rightarrow &amp; \frac{a}{-3} &amp; =\frac{b}{-3}=\frac{c}{0} \\<br/>&amp; \frac{a}{1}=\frac{b}{1}=\frac{c}{0}<br/>\end{array}<br/>$$<br/>So, the equation of plane is,<br/>$x-1+y+2=0$ or $x+y+1=0$, its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}$.</div>


tags: ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2020 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">2</span> </div>
source_db: MarksBatch1_P1.db
title:
a-point-charge-q-of-mass-m-is-suspended-vertically-by-a-string-of-length-l-a-point-dipole-of-dipole-moment-p-is-now-brought-towards-q-from-infinity-so

slug:
a-point-charge-q-of-mass-m-is-suspended-vertically-by-a-string-of-length-l-a-point-dipole-of-dipole-moment-p-is-now-brought-towards-q-from-infinity-so-57461

question:
<div class="question"><p>A point chargeÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> of massÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> is suspended vertically by a string of lengthÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mo>.</mo></math> A point dipole of dipole momentÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>p</mi><mo>â</mo></mover></math> is now brought towardsÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position isÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi><mo>Ã</mo><mo>(</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo>)</mo></math>, whereÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi></math> is the acceleration due to gravity, then the value ofÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math> is ____________ (Note that for three coplanar forces keeping a point mass in equilibrium,Â <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi mathvariant="normal">F</mi><mrow><mi>sin</mi><mi>Î¸</mi></mrow></mfrac></math> is the same for all forces, whereÂ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi></math> is any one of the forces and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Î¸</mi></math>Â is the angle between the other two forces)</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/66b1f109-9a81-453a-8162-902331dfe560-image.png"/></p></div>

answer:
<div class="solution"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/d05922c2-b8c0-4772-8864-2f402f721035-image.png"/></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo mathvariant="italic">=</mo><mi>q</mi><mi>V</mi><mo mathvariant="italic">+</mo><mi>m</mi><mi>g</mi><mi>h</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo mathvariant="italic">=</mo><mi>q</mi><mfenced><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn>2</mn></msup></mfrac></mfenced><mo mathvariant="italic">+</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>1</mn></mfenced></math></p><p>For equilibrium</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mi>Î²</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>sin</mi><mfenced><mrow><mn>90</mn><mo>-</mo><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mrow></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mo>(</mo><mn>180</mn><mo>-</mo><mi>Î±</mi><mo>)</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>;</mo><mo>Â Â </mo><mfrac><mrow><mi>q</mi><mi>E</mi></mrow><mrow><mi>sin</mi><mi>Î±</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>9</mn><mi>E</mi></mrow><mrow><mn>2</mn><mi>sin</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced><mi>cos</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mrow><mi>cos</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac><mo>;</mo><mo>Â Â </mo><mi>q</mi><mi>E</mi><mo>=</mo><mn>2</mn><mo mathvariant="italic">mg</mo><mi>sin</mi><mfenced><mfrac><mi>Î±</mi><mn>2</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi><mfenced><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn mathvariant="italic">2</mn></msup></mfrac></mfenced><mo mathvariant="italic">=</mo><mi>m</mi><mi>g</mi><mi>x</mi><mi>sin</mi><mstyle mathvariant="italic"><mrow><mo>(</mo><mfrac><mi>Î±</mi><mn>2</mn></mfrac><mo>)</mo></mrow></mstyle><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mfenced><mfrac><mi>Î±</mi><mn mathvariant="italic">2</mn></mfrac></mfenced><mo mathvariant="italic">Â·</mo><mfrac><mi>h</mi><mrow><mi>sin</mi><mfenced><mfrac><mi mathvariant="normal">Î±</mi><mn>2</mn></mfrac></mfenced></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi><mfrac><mrow><mi>k</mi><mi>p</mi></mrow><msup><mi>x</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>Â </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mn>2</mn></mfenced></math></p><p>From (1) and (2)</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo>=</mo><mn>2</mn><mi>m</mi><mi>g</mi><mi>h</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>2</mn></math></p><p>Â </p></div>


tags: ['Physics', 'Oscillations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-point-mass-is-subjected-to-two-simultaneous-sinusoidal-displacements-in-x-direction-x-1-t-a-sin-t-and-x-2-t-a-sin-t-3-2-adding-a-third-sinusoidal-di

slug:
a-point-mass-is-subjected-to-two-simultaneous-sinusoidal-displacements-in-x-direction-x-1-t-a-sin-t-and-x-2-t-a-sin-t-3-2-adding-a-third-sinusoidal-di-26735

question:
<div class="question">A point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction, $x_1(t)=A \sin \omega t$ and $x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B \sin (\omega t+\phi)$ bring the mass to a complete rest. The values of $B$ and $\phi$ are</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\sqrt{2 A}, \frac{3 \pi}{4}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$A, \frac{4 \pi}{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} A, \frac{5 \pi}{6}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$A, \frac{\pi}{3}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$A, \frac{4 \pi}{3}$<br/></span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aUgqUfC-c6FUMydFnyL-WX3gfqcC22f0LCjoN10xBxo.original.fullsize.png"/><br/><br/>Resultant amplitude of $x_1$ and $x_2$ is $A$ at angle $\left(\frac{\pi}{3}\right)$ from $A_1$. To make resultant of $x_1, x_2$ and $x_3$ to be zero. $A_3$ should be equal to $A$ at angle $\phi=\frac{4 \pi}{3}$ as shown in figure.<br/>$\therefore$ Correct answer is (b).<br/>Alternate Solution<br/>If we substitute,<br/>$$<br/>x_1+x_2+x_3=0<br/>$$<br/>or $A \sin \omega t+A \sin \left(\omega t+\frac{2 \pi}{3}\right)$<br/>$$<br/>+B \sin (\omega t+\phi)=0<br/>$$<br/>Then, by applying simple mathematics, we can prove that<br/>$$<br/>\begin{array}{rlrl}<br/>B &amp; =A \\<br/>\text { and } &amp; \phi &amp; =\frac{4 \pi}{3}<br/>\end{array}<br/>$$<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) Question can be solved by applying mathematics also.<br/>(iii) Amplitudes of two or more sine or cosine functions of same frequency $\omega$ can be added by vector method.</div>


tags: ['Physics', 'Center of Mass Momentum and Collision', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-point-mass-of-1-kg-collides-elastically-with-a-stationary-point-mass-of-5-kg-after-their-collision-the-1-kg-mass-reverses-its-direction-and-moves-wi

slug:
a-point-mass-of-1-kg-collides-elastically-with-a-stationary-point-mass-of-5-kg-after-their-collision-the-1-kg-mass-reverses-its-direction-and-moves-wi-26616

question:
<div class="question">A point mass of $1 \mathrm{~kg}$ collides elastically with a stationary point mass of $5 \mathrm{~kg}$. After their collision, the $1 \mathrm{~kg}$ mass reverses its direction and moves with a speed of $2 \mathrm{~ms}^{-1}$. Which of the following statement(s) is/are correct for the system of these two masses?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Momentum of $5 \mathrm{~kg}$ mass after collision is $4 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Total kinetic energy of the system is $4 \mathrm{~J}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$<br/>, <br/>Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$<br/></span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/z_PhW8UfShvxIqh556cc7MWcXwR9YnrJGM4z7hNa3Po.original.fullsize.png"/><br/><br/>$$<br/>\begin{aligned}<br/>&amp; v_1^{\prime}=\left(\frac{m_1-m_5}{m_1+m_5}\right) v_1+\left(\frac{2 m_5}{m_1+m_5}\right) v_5 \\<br/>&amp; -2=\left(\frac{1-5}{1+5}\right) v_1+0 \quad\left(\text { as } v_5=0\right) \\<br/>&amp; \therefore \quad v_1=3 \mathrm{~ms}^{-1} \\<br/>&amp; v_5^{\prime}=\left(\frac{m_5-m_1}{m_1+m_5}\right) v_5+\left(\frac{2 m_1}{m_1+m_2}\right) v_1 \\<br/>&amp; =0+\left(\frac{2 \times 1}{6}\right)(3)=1 \mathrm{~ms}^{-1} \\<br/>&amp; P_{\mathrm{CM}}=P_i=(1)(3) \\<br/>&amp; =3 \frac{\mathrm{kg}-\mathrm{m}}{\mathrm{s}} \\<br/>&amp; P_5{ }^{\prime}=(5)(1)=5 \frac{\mathrm{kg}-\mathrm{m}}{5} \\<br/>&amp; K_{\mathrm{CM}}=\frac{P_{\mathrm{CM}}^2}{2 M_{\mathrm{CM}}} \\<br/>&amp; =\frac{9}{2 \times 6}=0.75 \mathrm{~J} \\<br/>&amp; K_{\text {total }}=\frac{1}{2} \times 1 \times(3)^2 \\<br/>&amp; =4.5 \mathrm{~J} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct options are (a) and (c).</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-point-object-is-placed-at-distance-of-20-cm-from-a-thin-planoconvex-lens-of-focal-length-15-cm-the-plane-surface-of-the-lens-is-now-silvered-the-ima

slug:
a-point-object-is-placed-at-distance-of-20-cm-from-a-thin-planoconvex-lens-of-focal-length-15-cm-the-plane-surface-of-the-lens-is-now-silvered-the-ima-23586

question:
<div class="question">A point object is placed at distance of $20 \mathrm{~cm}$ from a thin planoconvex lens of focal length $15 \mathrm{~cm}$. The plane surface of the lens is now silvered. The image created by the system is at<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/yc5nxE-emhqaMWcy037V8GiTt-GsnpNEs0_7ETm70lE.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$60 \mathrm{~cm}$ to the left of the system<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$60 \mathrm{~cm}$ to the right of the system<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$12 \mathrm{~cm}$ to the left of the system<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$12 \mathrm{~cm}$ to the right of the system</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$12 \mathrm{~cm}$ to the left of the system<br/></span> </div>

answer:
<div class="solution">Refraction from lens<br/>$$<br/>\begin{aligned}<br/>\frac{1}{v_1}-\frac{1}{-20} &amp; =\frac{1}{15} \\<br/>v &amp; =60 \mathrm{~cm}<br/>\end{aligned}<br/>$$ <img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/rBYnE5k12tvD8EtQ2q85ORZ8rV5TAkDX7qj-t3E8GmM.original.fullsize.png"/><br/><br/><br/>i.e., first image is formed at $60 \mathrm{~cm}$ to the right of lens system.<br/>Reflection from mirror<br/>After reflection from the mirror, the second image will be formed at a distance of $60 \mathrm{~cm}$ to the left of lens system.<br/>Refraction from lens<br/>$$<br/>\begin{aligned}<br/>\frac{1}{v_3}-\frac{1}{60} &amp; =\frac{1}{15} \\<br/>v_3 &amp; =12 \mathrm{~cm}<br/>\end{aligned}<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GTmt2gH5CzNqArz52bKt-jDaojEvoq9Fy1CFfBtJbOY.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-police-car-with-a-siren-of-frequency-8-khz-is-moving-with-uniform-velocity-36-km-h-towards-a-tall-building-which-reflects-the-sound-waves-the-speed-

slug:
a-police-car-with-a-siren-of-frequency-8-khz-is-moving-with-uniform-velocity-36-km-h-towards-a-tall-building-which-reflects-the-sound-waves-the-speed-23247

question:
<div class="question">A police car with a siren of frequency $8 \mathrm{kHz}$ is moving with uniform velocity $36 \mathrm{~km} / \mathrm{h}$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 \mathrm{~m} / \mathrm{s}$. The frequency of the siren heard by the car driver is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$8.50 \mathrm{kHz}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$8.25 \mathrm{kHz}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$7.75 \mathrm{kHz}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$7.50 \mathrm{kHz}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$8.50 \mathrm{kHz}$<br/></span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/N3B2m51q3dVOLcZ5J59iF2tGdvIQA5PNpL1HMwO7NI0.original.fullsize.png"/><br/><br/>Apparent frequency of sound heard by car driver (observer) reflected from the building will be<br/>$$<br/>\begin{aligned}<br/>f^{\prime} &amp; =f\left(\frac{v+v_0}{v-v_s}\right) \\<br/>&amp; =8\left(\frac{320+10}{320-10}\right) \\<br/>&amp; =8.5 \mathrm{kHz}<br/>\end{aligned}<br/>$$<br/>$\therefore$ Correct option is (a).<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) Driver will listen two sounds, direct and reflected. Direct sound will be of $8 \mathrm{kHz}$ as driver has no relative motion with the car. But reflected sound is of increased frequency because driver and image of car both are approaching towards each other.</div>


tags: ['Chemistry', 'Chemical Kinetics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-positron-is-emitted-from-11-23-na-the-ratio-of-the-atomic-mass-and-atomic-number-of-the-resulting-nuclide-is

slug:
a-positron-is-emitted-from-11-23-na-the-ratio-of-the-atomic-mass-and-atomic-number-of-the-resulting-nuclide-is-87482

question:
<div class="question">A positron is emitted from ${ }_{11}^{23} \mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>22/10<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>22/11<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>23/10<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$23 / 12$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>23/10<br/></span> </div>

answer:
<div class="solution">$$<br/>{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e<br/>$$</div>


tags: ['Physics', 'Dual Nature of Matter', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch1_P1.db
title:
a-proton-is-fired-from-very-far-away-towards-a-nucleus-with-charge-q-120-e-where-e-is-the-electronic-charge-it-makes-a-closest-approach-of-10-fm-to-th

slug:
a-proton-is-fired-from-very-far-away-towards-a-nucleus-with-charge-q-120-e-where-e-is-the-electronic-charge-it-makes-a-closest-approach-of-10-fm-to-th-57892

question:
<div class="question">A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 \mathrm{fm}$ to the nucleus. The de Broglie wavelength (in units of $\mathrm{fm}$ ) of the proton at its start is: (take the proton mass, $m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times$ $10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}$;
<br/> <br/>$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}$</div>

answer:
<div class="solution">From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system
<br/> <br/>$\begin{array}{c}
<br/> <br/>\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\
<br/> <br/>\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}
<br/> <br/>\end{array}$
<br/> <br/>Putting the values of $4 \pi \varepsilon_{0}, r, h, q_{1}, q_{2}$ and $m$ we get, deBroglie wavelength of proton, $\lambda=7 \mathrm{fm}$</div>


tags: ['Physics', 'Nuclear Physics', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-radioactive-sample-s-1-having-an-activity-of-5-ci-has-twice-the-number-of-nuclei-as-another-sample-s-2-which-has-an-activity-of-10-ci-the-half-lives-1

slug:
a-radioactive-sample-s-1-having-an-activity-of-5-ci-has-twice-the-number-of-nuclei-as-another-sample-s-2-which-has-an-activity-of-10-ci-the-half-lives-1-93785

question:
<div class="question">A radioactive sample $S_1$ having an activity of $5 \mu \mathrm{Ci}$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu \mathrm{Ci}$. The half lives of $S_1$ and $S_2$ can be</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>20 yr and 5 yr, respectively<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>20 yr and 10 yr, respectively<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>10 yr each<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>5 yr each</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>20 yr and 5 yr, respectively<br/></span> </div>

answer:
<div class="solution">Activity of $S_1=\frac{1}{2}$ (activity of $S_2$ )<br/>or<br/>$$<br/>\lambda_2 N_1=\frac{1}{2}\left(\lambda_2 N_2\right) \text { or } \frac{\lambda_1}{\lambda_2}=\frac{N_2}{2 N_1}<br/>$$<br/>or<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{T_1}{T_2}=\frac{2 N_1}{N_2}\left(T=\text { half-life }=\frac{\ln 2}{\lambda}\right) \\<br/>&amp; N_1=2 N_2 \\<br/>&amp; \frac{T_1}{T_2}=4<br/>\end{aligned}<br/>$$<br/>Given<br/>$\therefore$ correct option is (a).</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-ray-of-light-travelling-in-water-is-incident-on-its-surface-open-to-air-the-angle-of-incidence-is-which-is-less-than-the-critical-angle-then-there-w

slug:
a-ray-of-light-travelling-in-water-is-incident-on-its-surface-open-to-air-the-angle-of-incidence-is-which-is-less-than-the-critical-angle-then-there-w-16051

question:
<div class="question">A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>only a reflected ray and no refracted ray<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>only a refracted ray and no reflected ray<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>a reflected ray and a refracted ray and the angle between them would be greater than $180^{\circ}-2 \theta$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$<br/></span> </div>

answer:
<div class="solution">Since $\theta &lt; \theta_c$, both reflection and refraction will take place. From the figure we can see that angle between reflected and refracted rays $\alpha$ is less than $180^{\circ}-2 \theta$.<br/>$\therefore$ Option (c) is correct.<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8tpGTJwd4YDjGKQxEU46fhTSHh_eezAmuy-VX6ti1gw.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-ray-op-of-monochromatic-light-is-incident-on-the-face-a-b-of-prism-a-bc-d-near-vertex-b-at-an-incident-angle-of-6-0-see-figure-if-the-refractive-ind

slug:
a-ray-op-of-monochromatic-light-is-incident-on-the-face-a-b-of-prism-a-bc-d-near-vertex-b-at-an-incident-angle-of-6-0-see-figure-if-the-refractive-ind-79168

question:
<div class="question">A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/Y0fLkA7kftBwNbLUTcG8moZnPGiATLL8kEguqIkDJW4.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>The ray gets totally internally reflected at face $C D$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>The ray comes out through face $A D$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>The angle between the incident ray and the emergent ray is $90^{\circ}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>The angle between the incident ray and the emergent ray is $120^{\circ}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>The ray gets totally internally reflected at face $C D$<br/>, <br/>The ray comes out through face $A D$<br/>, <br/>The angle between the incident ray and the emergent ray is $90^{\circ}$<br/></span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\<br/>&amp; \therefore r=30^{\circ}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OqCuJa-SSDWftpM_SKULPN9o12Ag40EDBSrs9Q8ivPw.original.fullsize.png"/><br/><br/>$$<br/>\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)<br/>$$<br/>or $\sin \theta_C=\frac{1}{\sqrt{3}}=0.577$<br/>At point $Q$, angle of incidence inside the prism is $i=45^{\circ}$.<br/>Since $\sin i=\frac{1}{\sqrt{2}}$ is greater than $\sin \theta_C=\frac{1}{\sqrt{2}}$, ray gets totally internally reflected at face $C D$. Path of ray of light after point $Q$ is shown in figure.<br/>From the figure, we can see that angle between incident ray $O P$ and emergent ray $R S$ is $90^{\circ}$.<br/>Therefore, correct options are (a), (b) and (c).</div>


title: a-real-gas-behaves-like-an-ideal-gas-if-its
slug: a-real-gas-behaves-like-an-ideal-gas-if-its-37638
question: <div class="question">A real gas behaves like an ideal gas if its</div>
tags: ['Physics', 'Kinetic Theory of Gases', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
source_db: MarksBatch1_P1.db
options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>pressure and temperature are both high<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>pressure and temperature are both low<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>pressure is high and temperature is low<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>pressure is low and temperature is high</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>pressure is low and temperature is high</span> </div>

answer:
<div class="solution">A real gas behaves like an ideal gas at low pressure and high temperature. $\therefore$ correct option is (d).</div>


tags: ['Physics', 'Current Electricity', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected--1

slug:
a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected-1-23356

question:
<div class="question">A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$3 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4 \Omega$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$5 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$6 \Omega$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$3 \Omega$<br/></span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; R &amp; &gt;2 \Omega \\<br/>\therefore &amp; 100-x &amp; &gt;x \\<br/>\text { Applying } &amp; \frac{P}{Q} &amp; =\frac{R}{S}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bj5wXDgKnTfN4Vg1PIwLyg-p-M4Vw5b76zBexKtpjvI.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k0HDe0hHZT2EmygS2hd4fIRONvnb06ZJCeC2L7TIA9I.original.fullsize.png"/><br/><br/>We have<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{2}{R}=\frac{x}{100-x} \\<br/>&amp; \frac{R}{2}=\frac{x+20}{80-x}<br/>\end{aligned}<br/>$$<br/>Solving Eqs. (i) and (ii), we get<br/>$$<br/>R=3 \Omega<br/>$$<br/>$\therefore$ Answer is (a).</div>


tags: ['Physics', 'Current Electricity', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected-

slug:
a-resistance-of-2-is-connected-across-one-gap-of-a-meterbridge-the-length-of-the-wire-is-100-cm-and-an-unknown-resistance-greater-than-2-is-connected-19203

question:
<div class="question">A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$3 \Omega$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4 \Omega$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$5 \Omega$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$6 \Omega$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$3 \Omega$<br/></span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; R &amp; &gt;2 \Omega \\<br/>\therefore &amp; 100-x &amp; &gt;x \\<br/>\text { Applying } &amp; \frac{P}{Q} &amp; =\frac{R}{S}<br/>\end{aligned}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bj5wXDgKnTfN4Vg1PIwLyg-p-M4Vw5b76zBexKtpjvI.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k0HDe0hHZT2EmygS2hd4fIRONvnb06ZJCeC2L7TIA9I.original.fullsize.png"/><br/><br/>We have<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{2}{R}=\frac{x}{100-x} \\<br/>&amp; \frac{R}{2}=\frac{x+20}{80-x}<br/>\end{aligned}<br/>$$<br/>Solving Eqs. (i) and (ii), we get<br/>$$<br/>R=3 \Omega<br/>$$<br/>$\therefore$ Answer is (a).</div>


tags: ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-satellite-is-moving-with-a-constant-speed-v-in-a-circular-orbit-about-the-earth-an-object-of-mass-m-is-ejected-from-the-satellite-such-that-it-just-

slug:
a-satellite-is-moving-with-a-constant-speed-v-in-a-circular-orbit-about-the-earth-an-object-of-mass-m-is-ejected-from-the-satellite-such-that-it-just-78503

question:
<div class="question">A satellite is moving with a constant speed $v$ in a circular orbit about the earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{2} m v^2$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$m v^2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{2} m v^2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$2 m v^2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$m v^2$<br/></span> </div>

answer:
<div class="solution">In circular orbit of a satellite, potential energy<br/>$$<br/>\begin{aligned}<br/>&amp; =-2 \times(\text { kinetic energy) } \\<br/>&amp; =-2 \times \frac{1}{2} m v^2=-m v^2<br/>\end{aligned}<br/>$$<br/>Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be $+m v^2$.<br/>$\therefore$ Correct answer is (b).<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In circular orbit of satellite $U=-\frac{G M m}{r}$ and $K=\frac{G M m}{2 r}$ or $U=-2 K$<br/>Here, $U$ is potential energy and $K$ is kinetic energy.</div>


tags: ['Physics', 'Alternating Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-series-r-c-circuit-is-connected-to-a-c-voltage-source-consider-two-cases-a-when-c-is-without-a-dielectric-medium-and-b-when-c-is-filled-with-dielect

slug:
a-series-r-c-circuit-is-connected-to-a-c-voltage-source-consider-two-cases-a-when-c-is-without-a-dielectric-medium-and-b-when-c-is-filled-with-dielect-12486

question:
<div class="question">A series $R$ - $C$ circuit is connected to $A C$ voltage source. Consider two cases; $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$I_R^A&gt;I_R^B$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$I_R^A &lt; I_R^B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$V_C^A&gt;V_C^B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$V_C^A &lt; V_C^B$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$I_R^A &lt; I_R^B$, <br/>$V_C^A&gt;V_C^B$<br/></span> </div>

answer:
<div class="solution">$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}$<br/>In case (b) capacitance $C$ will be more.<br/>Therefore, impedance $Z$ will be less.<br/>Hence, current will be more.<br/>$\therefore$ Option (b) is correct.<br/>Further, $\quad \begin{aligned} V_C &amp; =\sqrt{V^2-V_R^2} \\ &amp; =\sqrt{V^2-(I R)^2}\end{aligned}$<br/>In case (b), since current $I$ is more.<br/>Therefore, $V_C$ will be less.<br/>$\therefore$ Option (c) is correct.<br/>$\therefore$ Correct options are (b) and (c).<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In my opinion problems of alternating currents are not very difficult.<br/>(iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.</div>


tags: ['Physics', 'Alternating Current', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">4</span> </div>
source_db: MarksBatch1_P1.db
title:
a-series-r-c-combination-is-connected-to-an-ac-voltage-of-angular-frequency-500-rad-s-if-the-impedance-of-the-r-c-circuit-is-r-125-the-time-constant-i

slug:
a-series-r-c-combination-is-connected-to-an-ac-voltage-of-angular-frequency-500-rad-s-if-the-impedance-of-the-r-c-circuit-is-r-125-the-time-constant-i-48889

question:
<div class="question">A series $R-C$ combination is connected to an $\mathrm{AC}$ voltage of angular frequency $\omega=500 \mathrm{rad} / \mathrm{s}$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is</div>

answer:
<div class="solution">$Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}$<br/>$$<br/>\therefore \quad R^2+X_C^2=1.25 R^2<br/>$$<br/>or $\quad X_C=\frac{R}{2}$<br/>or $\quad \frac{1}{\omega C}=\frac{R}{2}$<br/>$\begin{aligned} \therefore \text { Time constant } &amp; =C R=\frac{2}{\omega} \\ &amp; =\frac{2}{500} \mathrm{~s}=4 \mathrm{~ms}\end{aligned}$<br/>$\therefore$ Answer is 4 .<br/>Analysis of Question<br/>(i) Question is very simple.<br/>(ii) I think this is one of the simplest formula based question of this paper.</div>


tags: ['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-ship-is-fitted-with-three-engines-e-1-e-2-and-e-3-the-engines-function-independently-of-each-other-with-respective-probabilities-2-1-4-1-and-4-1-for

slug:
a-ship-is-fitted-with-three-engines-e-1-e-2-and-e-3-the-engines-function-independently-of-each-other-with-respective-probabilities-2-1-4-1-and-4-1-for-12411

question:
<div class="question">A ship is fitted with three engines $E_{1}, E_{2}$ and $E_{3}$. The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{4}$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_{1}, X_{2}$ and $X_{3}$ denote respectively the events that the engines $E_{1}, E_{2}$ and $E_{3}$ are functioning. Which of the following is(are) true?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$P\left[X_{1}^{c} \mid X\right]=\frac{3}{16}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$P\left[X \mid X_{2}\right]=\frac{5}{16}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$P\left[X \mid X_{1}\right]=\frac{7}{16}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value">$P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$, $P\left[X \mid X_{1}\right]=\frac{7}{16}$</span> </div>

answer:
<div class="solution">Given that $P\left(X_{1}\right)=\frac{1}{2}, \mathrm{P}\left(X_{2}\right)=\frac{1}{4}, \mathrm{P}\left(X_{3}\right)=\frac{1}{4}$ $P(X)=P$ (at least 2 engines are functioning)
<br/> <br/>$\begin{aligned}=P\left(X_{1} \cap X_{2}\right.&amp;\left.\cap X_{3}^{C}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right) \\ &amp;+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}\right) \end{aligned}$
<br/> <br/>$=\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{4}$
<br/> <br/>(a) $P\left(X_{1}^{C} / X\right)=\frac{P\left(X_{1}^{C} \cap X\right)}{P(X)}=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)}{P(X)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$
<br/> <br/>$\therefore$ (a) is not true.
<br/> <br/>(b) $P$ [Exactly two engines are functioning $/ X]$
<br/> <br/>$=\frac{P[(\text { Exactly two engines are functioning }) \cap X]}{P(X)}$
<br/> <br/>$=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P(X)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8}$
<br/> <br/>$\therefore$ (b) is true.
<br/> <br/>(c) $P\left(X / X_{2}\right)=\frac{P\left(X \cap X_{2}\right)}{P\left(X_{2}\right)}$
<br/> <br/>$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{2}\right)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8}$
<br/> <br/>$\therefore$ (c) is not true.
<br/> <br/>(d) $P\left(X / X_{1}\right)=\frac{P\left(X \cap X_{1}\right)}{P\left(X_{1}\right)}$
<br/> <br/>$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{1}\right)}$
<br/> <br/>$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$
<br/> <br/>$\therefore$ (d) is true.</div>


tags: ['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-signal-which-can-be-green-or-red-with-probability-5-4-and-5-1-respectively-is-received-by-station-a-and-then-transmitted-to-station-b-the-probabilit

slug:
a-signal-which-can-be-green-or-red-with-probability-5-4-and-5-1-respectively-is-received-by-station-a-and-then-transmitted-to-station-b-the-probabilit-82953

question:
<div class="question">A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $A$ and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac{3}{4}$. If the signal received at station $B$ is green, then the probability that the original signal green is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{3}{5}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{6}{7}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{20}{23}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{9}{20}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{20}{23}$<br/></span> </div>

answer:
<div class="solution">From the tree-diagram it follows that<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/l4-0jmsEW2b3RYr9-D6RJ2U3VsRy7evmHsKptiune_4.original.fullsize.png"/><br/><br/>$$<br/>\begin{gathered}<br/>P\left(B_G \mid G\right)=\frac{10}{16}=\frac{5}{8} \\<br/>\therefore \quad P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2} \\<br/>P\left(G \mid B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}<br/>\end{gathered}<br/>$$</div>


tags: ['Physics', 'Dual Nature of Matter', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch1_P1.db
title:
a-silver-sphere-of-radius-1-cm-and-work-function-47-ev-is-suspended-from-an-insulating-thread-in-freespace-it-is-under-continuous-illumination-of-200--2

slug:
a-silver-sphere-of-radius-1-cm-and-work-function-47-ev-is-suspended-from-an-insulating-thread-in-freespace-it-is-under-continuous-illumination-of-200-2-35541

question:
<div class="question">A silver sphere of radius $1 \mathrm{~cm}$ and work function $4.7 \mathrm{eV}$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 \mathrm{~nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 &lt; A &lt; 10$ ). The value of $Z$ is</div>

answer:
<div class="solution">Photoemission will stop when potential on silver sphere becomes equal to the stopping potential.<br/>$\therefore \quad \frac{h c}{\lambda}-W=e V_0$<br/>Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$<br/>$$<br/>\begin{aligned}<br/>&amp; \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\<br/>&amp;=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\<br/>&amp;(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\<br/>&amp; \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7<br/>\end{aligned}<br/>$$<br/>$\therefore$ Answer is 7 .<br/>Analysis of Question<br/>(i) Question is moderately difficult.<br/>(ii) In $\mathrm{eV}_0$ if we substitute the value of $e=1.6 \times 10^{-19} \mathrm{C}$, then answer is in J. If we leave it then, answer is in $e \mathrm{~V}$.</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-simple-telescope-used-to-view-distant-objects-has-eyepiece-and-objective-lens-of-focal-lengths-f-e-and-f-0-respectively-then

slug:
a-simple-telescope-used-to-view-distant-objects-has-eyepiece-and-objective-lens-of-focal-lengths-f-e-and-f-0-respectively-then-14648

question:
<div class="question">A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths $f_e$ and $f_0$, respectively. Then<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ijYR7qiWVf7-qQhgeKaI74NNU25wHAxcWA_wEDPcLlc.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>(a) p, (b) r, (c) r, (d) p,q,s<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>(a) q, (b) r, (c) s, (d) p,s<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>(a) p, (b) p, r, (c) q, (d) p,q<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>(a) q, (b) s, (c) r, (d) p,q,s</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>(a) p, (b) r, (c) r, (d) p,q,s<br/></span> </div>

answer:
<div class="solution">(a) $\rightarrow p$<br/>(b) $\rightarrow r$<br/>(c) $\rightarrow r$<br/>(d) $\rightarrow$ p, q, s</div>


tags: ['Physics', 'Motion In Two Dimensions', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-1

slug:
a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-1-70589

question:
<div class="question">A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8JfHbEDRw9CzUF3KhntqqOIXAPC3H9itqgYrnpnqqJI.original.fullsize.png"/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\sqrt{51} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\sqrt{52} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\sqrt{53} \mathrm{~m} / \mathrm{s}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> </div>

answer:
<div class="solution">Time of flight of projectile,
<br/> <br/>$T=\frac{2 V \sin \theta}{g}$
<br/> <br/>$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
<br/> <br/>Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$</div>


tags: ['Physics', 'Motion In Two Dimensions', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l

slug:
a-small-block-is-connected-to-one-end-of-a-massless-spring-of-unstretched-length-49-m-the-other-end-of-the-spring-see-the-figure-is-fixed-the-system-l-85437

question:
<div class="question">A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8JfHbEDRw9CzUF3KhntqqOIXAPC3H9itqgYrnpnqqJI.original.fullsize.png"/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\sqrt{51} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\sqrt{52} \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\sqrt{53} \mathrm{~m} / \mathrm{s}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\sqrt{50} \mathrm{~m} / \mathrm{s}$</span> </div>

answer:
<div class="solution">Time of flight of projectile,
<br/> <br/>$T=\frac{2 V \sin \theta}{g}$
<br/> <br/>$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
<br/> <br/>Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$</div>


tags: ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-small-block-of-mass-of-01-kg-lies-on-a-fixed-inclined-plane-pq-which-makes-an-angle-with-the-horizontal-a-horizontal-force-of-1-n-acts-on-the-block-

slug:
a-small-block-of-mass-of-01-kg-lies-on-a-fixed-inclined-plane-pq-which-makes-an-angle-with-the-horizontal-a-horizontal-force-of-1-n-acts-on-the-block-82917

question:
<div class="question">A small block of mass of $0.1 \mathrm{~kg}$ lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of $1 \mathrm{~N}$ acts on the block through its centre of mass as shown in the figure.
<br/> <br/>The block remains stationary if (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eyWmHlpMbKkU5KNCI11wped3TI_abWjiG9pQcOAesZI.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\theta=45^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\theta&gt;45^{\circ}$ and a frictional force acts on the block towards P.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\theta&gt;45^{\circ}$ and a frictional force acts on the block towards Q.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\theta &lt; 45^{\circ}$ and a frictional force acts on the block towards Q.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value">$\theta=45^{\circ}$, $\theta&gt;45^{\circ}$ and a frictional force acts on the block towards Q.</span> </div>

answer:
<div class="solution">The various forces acting on the block are as shown in the figure.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aswDGCZDDEq0RtQdOpKwJtq4gJ_Yh9XUhPtzcG6z2p4.original.fullsize.png"/><br/> <br/> <br/>When $\theta=45^{\circ}, \sin \theta=\cos \theta$
<br/> <br/>The block will remain stationary and the frictional force is zero.
<br/> <br/>When $\theta&gt;45^{\circ}, \sin \theta&gt;\cos \theta$
<br/> <br/>Therefore a frictional force acts towards $Q$.
<br/> <br/>When $\theta &lt; 45^{\circ}, \cos \theta&gt;\sin \theta$
<br/> <br/>Therefore a frictional force acts towards $P$.</div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-1

slug:
a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-1-58555

question:
<div class="question"><p>A small circular loop of area <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math> and resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is fixed on a horizontal <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math>-plane with the center of the loop always on the axis <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> of a long solenoid. The solenoid has <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> turns per unit length and carries current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> counter clockwise as shown in the figure. The magnetic field due to the solenoid is in <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> direction. List-I gives time dependences of <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> in terms of a constant angular frequency <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ω</mi></math>. List-II gives the torques experienced by the circular loop at time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mfrac><mi>π</mi><mrow><mn>6</mn><mi>ω</mi></mrow></mfrac></math>, Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>α</mi><mo>=</mo><mfrac><mrow><msup><mi>A</mi><mn>2</mn></msup><msubsup><mi>μ</mi><mn>0</mn><mn>2</mn></msubsup><msup><mi>m</mi><mn>2</mn></msup><msup><mi>I</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac></math>.</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3dbd710a-2aec-4891-a220-baef9f053b38-image.png" style="width: 220px; height: 150px;"/></p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-I</strong></td> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-II</strong></td> </tr> <tr> <td>(i)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(p)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math></td> </tr> <tr> <td>(ii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(q)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(r)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iv)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(s)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></td> </tr> <tr> <td> </td> <td> </td> <td>(t)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> </tbody></table><p>Which one of the following options is correct?</p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">t</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">p</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </div>

answer:
<div class="solution"><blockquote><p>Note: This question was given bonus by JEE council.</p><p>The magnetic field in a solenoid is given by, <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math>. Therefore, the flux through the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfenced><mrow><mi>A</mi><mover><mi>k</mi><mo>^</mo></mover></mrow></mfenced><mo>·</mo><mfenced><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p>For case I:</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The emf will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>Therefore, the current in the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The magnetic moment of the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mi>i</mi><mi>A</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p>Therefore, the torque</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mrow><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></mrow></mfenced><mo>=</mo><mfrac><mrow><mo>-</mo><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi mathvariant="normal">I</mi></mfenced><mo>→</mo><mi>q</mi></math></p><p>For case II</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mn>0</mn></math></p><p>Therefore, we can directly write <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mn>0</mn></math></p><p>Hence <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>II</mi></mfenced><mo>→</mo><mi>p</mi></math></p><p>Following the same process as in case I for case III</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mo>×</mo><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mi>ω</mi><mi>t</mi><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>III</mi><mo>→</mo><mi>s</mi></math></p><p>Following the same process as in case I for case IV</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mi>ω</mi><mi>t</mi><mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><mo>·</mo><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>α</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>IV</mi></mfenced><mo>→</mo><mi>r</mi></math></p></blockquote></div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2022 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi

slug:
a-small-circular-loop-of-area-a-and-resistance-r-is-fixed-on-a-horizontal-x-y-plane-with-the-center-of-the-loop-always-on-the-axis-n-of-a-long-solenoi-86013

question:
<div class="question"><p>A small circular loop of area <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math> and resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> is fixed on a horizontal <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math>-plane with the center of the loop always on the axis <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> of a long solenoid. The solenoid has <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> turns per unit length and carries current <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi></math> counter clockwise as shown in the figure. The magnetic field due to the solenoid is in <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> direction. List-I gives time dependences of <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math> in terms of a constant angular frequency <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ω</mi></math>. List-II gives the torques experienced by the circular loop at time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mfrac><mi>π</mi><mrow><mn>6</mn><mi>ω</mi></mrow></mfrac></math>, Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>α</mi><mo>=</mo><mfrac><mrow><msup><mi>A</mi><mn>2</mn></msup><msubsup><mi>μ</mi><mn>0</mn><mn>2</mn></msubsup><msup><mi>m</mi><mn>2</mn></msup><msup><mi>I</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac></math>.</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3dbd710a-2aec-4891-a220-baef9f053b38-image.png" style="width: 220px; height: 150px;"/></p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-I</strong></td> <td style="text-align: center;"> </td> <td style="text-align: center;"><strong>List-II</strong></td> </tr> <tr> <td>(i)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(p)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math></td> </tr> <tr> <td>(ii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(q)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iii)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(r)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> <tr> <td>(iv)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>2</mn></msqrt></mfrac><mfenced><mrow><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></math></td> <td>(s)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></td> </tr> <tr> <td> </td> <td> </td> <td>(t)</td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mrow><mn>3</mn><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></td> </tr> </tbody></table><p>Which one of the following options is correct?</p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">t</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">p</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">t</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">i</mi><mo>→</mo><mi mathvariant="normal">q</mi><mo>,</mo><mi>ii</mi><mo>→</mo><mi mathvariant="normal">p</mi><mo>,</mo><mi>iii</mi><mo>→</mo><mi mathvariant="normal">s</mi><mo>,</mo><mi>iv</mi><mo>→</mo><mi mathvariant="normal">r</mi></math></span> </div>

answer:
<div class="solution"><blockquote><p>Note: This question was given bonus by JEE council.</p><p>The magnetic field in a solenoid is given by, <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></math>. Therefore, the flux through the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfenced><mrow><mi>A</mi><mover><mi>k</mi><mo>^</mo></mover></mrow></mfenced><mo>·</mo><mfenced><mrow><msub><mi>μ</mi><mn>0</mn></msub><mi>m</mi><mi>I</mi><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p>For case I:</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The emf will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ε</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>Therefore, the current in the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p>The magnetic moment of the loop will be,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mi>i</mi><mi>A</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p>Therefore, the torque</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">n</mi><mo>^</mo></mover></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced close="]" open="["><mrow><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></mrow></mfenced><mo>=</mo><mfrac><mrow><mo>-</mo><mi>α</mi></mrow><mn>4</mn></mfrac><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi mathvariant="normal">I</mi></mfenced><mo>→</mo><mi>q</mi></math></p><p>For case II</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mn>0</mn></math></p><p>Therefore, we can directly write <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mn>0</mn></math></p><p>Hence <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>II</mi></mfenced><mo>→</mo><mi>p</mi></math></p><p>Following the same process as in case I for case III</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mo>×</mo><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>sin</mi><mi>ω</mi><mi>t</mi><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mfenced><mrow><mi>sin</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mi>cos</mi><mi>ω</mi><mi>t</mi><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><msup><mi>sin</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mfrac><mi>α</mi><mn>4</mn></mfrac><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>III</mi><mo>→</mo><mi>s</mi></math></p><p>Following the same process as in case I for case IV</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo>=</mo><mfrac><mrow><mi>B</mi><mi>A</mi></mrow><msqrt><mn>2</mn></msqrt></mfrac><mi>sin</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><mi>A</mi><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>m</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><mi>B</mi><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><msqrt><mn>2</mn></msqrt><mi>R</mi></mrow></mfrac><mi>cos</mi><mi>ω</mi><mi>t</mi><mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mover><mi>m</mi><mo>→</mo></mover><mo>×</mo><mover><mi>B</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mrow><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>×</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mrow><mi>ω</mi><mi>t</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>A</mi><mn>2</mn></msup><mi>ω</mi></mrow><mrow><mn>2</mn><mi>R</mi></mrow></mfrac><mfenced><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></mfenced><mo>·</mo><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mfrac><mi>π</mi><mn>6</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>τ</mi><mo>→</mo></mover><mo>=</mo><mo>+</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>α</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>IV</mi></mfenced><mo>→</mo><mi>r</mi></math></p></blockquote></div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-small-mass-m-is-attached-to-a-massless-string-whose-other-end-is-fixed-at-p-as-shown-in-the-figure-the-mass-is-undergoing-circular-motion-in-the-x-y

slug:
a-small-mass-m-is-attached-to-a-massless-string-whose-other-end-is-fixed-at-p-as-shown-in-the-figure-the-mass-is-undergoing-circular-motion-in-the-x-y-93373

question:
<div class="question">A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $\vec{L}_{O}$ and $\vec{L}_{P}$ respectively, then
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8I6vSvu0X5rikcmCVJdUb7jfJ20GlgxqNY9J1KxpsgU.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\vec{L}_{O}$ and $\vec{L}_{P}$ do not vary with time</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\vec{L}_{O}$ varies with time while $\vec{L}_{P}$ remains constant</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\vec{L}_{O}$ and $\vec{L}_{P}$ both vary with time</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time</span> </div>

answer:
<div class="solution">For all locations of $\mathrm{m}$ the angular momentum of the mass $m$ about $O$ i.e., $L_{o}$ is $m r^{2} \omega$ and is directed toward $+z$ direction. The angular momentum of mass $m$ about $P$ i.e., $L_{p}$ is $m v l$ and is directed for the given location of $m$ as shown in the figure. For different location of $m$, the direction of $\vec{L}_{P}$ remains changing.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bwsNwamWN-cQ7HKxrMmC3itcaB_EkdPHkLoI6EWocck.original.fullsize.png"/></div>


tags: ['Physics', 'Work Power Energy', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to--2

slug:
a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to-2-54739

question:
<div class="question">A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$<br/>\text { position. The object is }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5hO_SeRZPfuRbMzeqNADHQSXsGxc3QBbS3m0w8QuJ4E.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>ring<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>solid sphere<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>hollow sphere<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>disc</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>disc</span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 &amp; =m g\left(\frac{3 v^2}{4 g}\right) \\<br/>I &amp; =\frac{1}{2} m R^2<br/>\end{aligned}<br/>$$<br/>So, body is disc.<br/>The correct option is (d).</div>


tags: ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to-

slug:
a-small-object-of-uniform-density-rolls-up-a-curved-surface-with-an-initial-velocity-v-it-reaches-up-to-a-maximum-height-of-4-g-3-v-2-with-respect-to-35455

question:
<div class="question">A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$<br/>\text { position. The object is }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5hO_SeRZPfuRbMzeqNADHQSXsGxc3QBbS3m0w8QuJ4E.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>ring<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>solid sphere<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>hollow sphere<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>disc</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>disc</span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 &amp; =m g\left(\frac{3 v^2}{4 g}\right) \\<br/>I &amp; =\frac{1}{2} m R^2<br/>\end{aligned}<br/>$$<br/>So, body is disc.<br/>The correct option is (d).</div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2020 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-small-roller-of-diameter-20-cm-has-an-axle-of-diameter-10-cm-see-figure-below-on-the-left-it-is-on-a-horizontal-floor-and-a-meter-scale-is-positione

slug:
a-small-roller-of-diameter-20-cm-has-an-axle-of-diameter-10-cm-see-figure-below-on-the-left-it-is-on-a-horizontal-floor-and-a-meter-scale-is-positione-20970

question:
<div class="question">A small roller of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo>Â </mo><mi>cm</mi></math> has an axle of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>10</mn><mo>Â </mo><mi>cm</mi></math> (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>50</mn><mo>Â </mo><mi>cm</mi><mo>,</mo></math> the position of the scale will look like (figures are schematic and not drawn to scale)
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/4f447cbf-b054-43c8-93b9-d22dc77ca59c-image.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/aab3bcfa-9e87-46ad-a299-0d0c725c3c1a-image.png"/></p></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e10ff687-0ab8-40fb-8e7b-5fcfbe1e8c0a-image.png"/></p></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/dafb5633-4f65-49be-9a08-b083eee0b470-image.png"/></p></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/12b2502c-ee2b-4354-871e-c3f54559e6fc-image.png"/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e10ff687-0ab8-40fb-8e7b-5fcfbe1e8c0a-image.png"/></p></span> </div>

answer:
<div class="solution"><p>Angular displacement <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Î¸</mi><mo>=</mo><mfrac><msub><mi>S</mi><mn>0</mn></msub><mn>10</mn></mfrac><mo>=</mo><mn>5</mn></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/e339f447-da08-4479-99ca-b33b00d22109-image.png"/></p><p>Displacement of point <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi></math>Â is,<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mi>r</mi><mo>.</mo><mi>Î¸</mi><mo>=</mo><mn>15</mn><mo>Ã</mo><mn>5</mn><mo>=</mo><mn>75</mn><mo>Â </mo><mi>cm</mi></math></p><p>Displacement of roller <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>50</mn><mo>Â </mo><mi>cm</mi></math></p><p>Relative displacement <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>75</mn><mo>-</mo><mn>50</mn><mo>=</mo><mn>25</mn><mo>Â </mo><mi>cm</mi></math></p><p>Â </p></div>


title: a-solid-sphere-is-in-pure-rolling-motion-on-an-inclined-surface-having-inclination
slug: a-solid-sphere-is-in-pure-rolling-motion-on-an-inclined-surface-having-inclination-33574
tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
question:
<div class="question">A solid sphere is in pure rolling motion on an inclined surface having inclination $\theta$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/SqpFTGNOh5r0Id1yrfTHMgHqm61n25jCGlaCyaWC2gQ.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>frictional force acting on sphere is $f=\mu m g \cos \theta$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$f$ is disspative force<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>friction will increase its angular velocity and decreases its linear velocity<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>If $\theta$ decreases, friction will decrease</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>friction will increase its angular velocity and decreases its linear velocity<br/>, <br/>If $\theta$ decreases, friction will decrease</span> </div>

answer:
<div class="solution">In case of pure rolling,<br/>$$<br/>\begin{aligned}<br/>&amp; f=\frac{m g \sin \theta}{1+\frac{m R^2}{I}} \\<br/>\therefore &amp; f \propto \sin \theta<br/>\end{aligned}<br/>$$<br/>(upwards)<br/>Therefore, as $\theta$ decreases force of friction will also decrease.</div>


tags: ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">2</span> </div>
source_db: MarksBatch1_P1.db
title:
a-solid-sphere-of-radius-r-has-a-charge-q-distributed-in-its-volume-with-a-charge-density-k-r-a-where-k-and-a-are-constants-and-r-is-the-distance-from

slug:
a-solid-sphere-of-radius-r-has-a-charge-q-distributed-in-its-volume-with-a-charge-density-k-r-a-where-k-and-a-are-constants-and-r-is-the-distance-from-94247

question:
<div class="question">A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=k r^a$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$.</div>

answer:
<div class="solution">From Gauss theorem,<br/>$$<br/>\begin{aligned}<br/>&amp; E \propto \frac{q}{r^2} \quad(q=\text { charge enclosed }) \\<br/>&amp; \therefore \quad \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \\<br/>&amp; 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>Solving this equation we get, $a=2$</div>


tags: ['Physics', 'Rotational Motion', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess--1

slug:
a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess-1-93161

question:
<div class="question">$$<br/>\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }<br/>$$<br/>melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/lSgwN8rcG18285o-EaZ9esu_dYPU3wCgkbldDfNUel8.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2}{\sqrt{5}} R$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{\sqrt{15}} R$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{\sqrt{3}}{\sqrt{15}} R$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> </div>

answer:
<div class="solution">$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$</div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess-

slug:
a-solid-sphere-of-radius-r-has-moment-of-inertia-j-about-its-qeometrical-axis-if-it-is-m-e-lt-e-d-in-t-o-a-d-i-sco-f-r-a-d-i-u-s-r-an-d-t-hi-c-kn-ess-10772

question:
<div class="question">$$<br/>\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }<br/>$$<br/>melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/lSgwN8rcG18285o-EaZ9esu_dYPU3wCgkbldDfNUel8.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2}{\sqrt{5}} R$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{\sqrt{15}} R$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{\sqrt{3}}{\sqrt{15}} R$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{2}{\sqrt{15}} R$<br/></span> </div>

answer:
<div class="solution">$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$</div>


tags: ['Chemistry', 'Practical Chemistry', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-solution-of-a-metal-ion-when-treated-with-ki-gives-a-red-precipitate-which-dissolved-in-excess-ki-to-give-a-colourless-solution-moreover-the-solutio

slug:
a-solution-of-a-metal-ion-when-treated-with-ki-gives-a-red-precipitate-which-dissolved-in-excess-ki-to-give-a-colourless-solution-moreover-the-solutio-36957

question:
<div class="question">A solution of a metal ion when treated with $\mathrm{KI}$ gives a red precipitate which dissolved in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitqte. The metal ion is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{Pb}^{2+}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{Hg}^{2+}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{Cu}^{2+}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{Co}^2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\mathrm{Hg}^{2+}$<br/></span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; \mathrm{Hg}^{2+}+2 \mathrm{I}^{-} \longrightarrow \mathrm{HgI}_2, \mathrm{HgI}_2+\mathrm{H}^{-} \\<br/>&amp; \text {(Excess) }\left.\longrightarrow \mathrm{HgI}_4\right]^{2-} \\<br/>&amp; \text { Soluble }^{2-} \\<br/>&amp; \mathrm{Hg}^{2+}+\mathrm{Co}(\mathrm{SCN})_2 \longrightarrow \mathrm{Hg}(\mathrm{SCN})_2<br/>\end{aligned}<br/>$$<br/>Blue crystalline ppt.</div>


tags: ['Chemistry', 'd and f Block Elements', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-solution-of-colourless-salt-h-on-boiling-with-excess-naoh-produces-a-nonflammable-gas-the-gas-evolution-ceases-after-sometime-upon-addition-of-zn-du

slug:
a-solution-of-colourless-salt-h-on-boiling-with-excess-naoh-produces-a-nonflammable-gas-the-gas-evolution-ceases-after-sometime-upon-addition-of-zn-du-14813

question:
<div class="question">A solution of colourless salt $\mathrm{H}$ on boiling with excess $\mathrm{NaOH}$ produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of $\mathrm{Zn}$ dust to the same solution, the gas evolution restarts. The colourless salt (s) $\mathrm{H}$ is (are)</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{NO}_3$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{NO}_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{NH}_4 \mathrm{Cl}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\mathrm{NH}_4 \mathrm{NO}_3$<br/>, <br/>$\mathrm{NH}_4 \mathrm{NO}_2$<br/></span> </div>

answer:
<div class="solution">$$<br/>\begin{aligned}<br/>&amp; \mathrm{NH}_4 \mathrm{NO}_3+\mathrm{NaOH} \longrightarrow \mathrm{NaNO}_3+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\<br/>&amp; \mathrm{NaNO}_3+8[\mathrm{H}] \mathrm{NaOH}+\mathrm{NH}_3+2 \mathrm{H}_2 \mathrm{O} \\<br/>&amp; \mathrm{NH}_4 \mathrm{NO}_2+\mathrm{NaOH} \longrightarrow \mathrm{NaHO}_2+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\<br/>&amp; \mathrm{NaNO}_2+6[\mathrm{H}] \longrightarrow \mathrm{NaOH}+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Practical Chemistry', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-solution-when-diluted-with-h-2-o-and-boiled-it-gives-a-white-precipitate-on-addition-of-excess-nh-4-cl-nh-4-oh-the-volume-of-precipitate-decreases-l

slug:
a-solution-when-diluted-with-h-2-o-and-boiled-it-gives-a-white-precipitate-on-addition-of-excess-nh-4-cl-nh-4-oh-the-volume-of-precipitate-decreases-l-42315

question:
<div class="question">A solution when diluted with $\mathrm{H}_2 \mathrm{O}$ and boiled, it gives a white precipitate. On addition of excess $\mathrm{NH}_4 \mathrm{Cl} / \mathrm{NH}_4 \mathrm{OH}$, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in $\mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl}$.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{Zn}(\mathrm{OH})_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{Al}(\mathrm{OH})_3$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{Mg}(\mathrm{OH})_2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{Ca}(\mathrm{OH})_2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\mathrm{Zn}(\mathrm{OH})_2$<br/></span> </div>

answer:
<div class="solution">A solution of $\mathrm{Zn}^{2+}$ ion reacts with water which on boiling give white ppt. of $\mathrm{Zn}(\mathrm{OH})_2$. This precipitates dissolve in excess of $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ because $\mathrm{Zn}(\mathrm{OH})_2$ on reaction with $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ forms tetrammine zinc (II) complex (soluble).<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/uZilZltjS4fVUCxfYQUh2bfU6By7vtSKZ1G1L66vvg4.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/54xVnRJAct4YK6nXe_KVzZIA81hhc1znvIPmpFiT-TQ.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-sphere-is-rolling-without-slipping-on-a-fixed-horizontal-plane-surface-in-the-figure-a-is-the-point-of-contact-b-is-the-centre-of-the-sphere-and-c-i

slug:
a-sphere-is-rolling-without-slipping-on-a-fixed-horizontal-plane-surface-in-the-figure-a-is-the-point-of-contact-b-is-the-centre-of-the-sphere-and-c-i-26004

question:
<div class="question">A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, $A$ is the point of contact. $B$ is the centre of the sphere and $C$ is its topmost point. Then,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7Xsr-fGTyJjgVVwNEX_nZqDunz93zW95YxOhMA7BOEM.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A=2\left(\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right)$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=4\left|\overrightarrow{\mathbf{v}}_B\right|$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$<br/>, <br/>$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$<br/></span> </div>

answer:
<div class="solution">$$<br/>v_A=0, v_B=v \text { and } v_C=2 v<br/>$$<br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/CyOWdRHTYq2MSjWUtaeoFm7lHACpL2peoHF_5bBE4g8.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-spherical-metal-shell-a-of-radius-r-a-and-a-solid-metal-sphere-b-of-radius-r-b-r-a-are-kept-far-apart-and-each-is-given-charge-q-now-they-are-connec

slug:
a-spherical-metal-shell-a-of-radius-r-a-and-a-solid-metal-sphere-b-of-radius-r-b-r-a-are-kept-far-apart-and-each-is-given-charge-q-now-they-are-connec-17365

question:
<div class="question">A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B &lt; \left(R_A\right)$ are kept far apart and each is given charge $+Q$. Now, they are connected by a thin metal wire. Then</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$E_A^{\text {inside }}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$Q_A&gt;Q_B$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$E_A^{\text {on surface }} &lt; E_B^{\text {on surface }}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$E_A^{\text {inside }}=0$<br/>, <br/>$Q_A&gt;Q_B$<br/>, <br/>$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$<br/>, <br/>$E_A^{\text {on surface }} &lt; E_B^{\text {on surface }}$</span> </div>

answer:
<div class="solution">Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.<br/>$$<br/>\begin{array}{ll}<br/>\therefore &amp; V_A=V_B \\<br/>\text { or } &amp; \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)<br/>\end{array}<br/>$$<br/>Since, $R_A&gt;R_B$<br/>$$<br/>\therefore \quad Q_A&gt;Q_B<br/>$$<br/>$\therefore$ Option (b) is correct.<br/>Potential is also equal to,<br/>$$<br/>\begin{array}{rlrl}<br/>V &amp; =\frac{\sigma R}{\varepsilon_0} \\<br/>V_A &amp; =V_B \\<br/>\therefore \quad \sigma_A R_A &amp; =\sigma_B R_B \\<br/>\text { or } \quad \sigma_B &amp; \frac{\sigma_A}{\sigma_B} &amp; =\frac{R_B}{R_A}<br/>\end{array}<br/>$$<br/><br/>$\therefore$ Option (c) is correct.<br/>Electric field on surface,<br/>$$<br/>E=\frac{\sigma}{E_0} \text { or } E \propto \sigma<br/>$$<br/>or $\quad \sigma_A &lt; \sigma_B$<br/>Since, $\quad \sigma_A &lt; \sigma_B$<br/>$\therefore \quad E_A &lt; E_B$<br/>$\therefore$ Option (d) is also correct.<br/>$\therefore$ Correct options are (a), (b), (c) and (d).<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) $E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}$<br/>(On surface)<br/>As, $\sigma=$ surface charge density<br/>$$<br/>=\frac{Q}{4 \pi R^2}<br/>$$<br/>(iii) $V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}$</div>


tags: ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-spherical-portion-has-been-removed-from-a-solid-sphere-having-a-charge-distributed-uniformly-in-its-volume-as-shown-in-the-figure-the-electric-field-1

slug:
a-spherical-portion-has-been-removed-from-a-solid-sphere-having-a-charge-distributed-uniformly-in-its-volume-as-shown-in-the-figure-the-electric-field-1-25523

question:
<div class="question">A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/w32mClG-c0ydnRkgWN7fvwcQKA3ZiypT8WiDkA-9iB4.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>zero everywhere<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>non-zero and uniform<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>non-uniform<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>zero only at its centre</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>non-zero and uniform<br/></span> </div>

answer:
<div class="solution">Inside the cavity, field at any point is uniform and non-zero. Therefore, correct option is (b).</div>


tags: ['Physics', 'Gravitation', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-1

slug:
a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-1-16691

question:
<div class="question">A spherically symmetric gravitational system of particles has a mass density<br/>$$<br/>\rho=\left\{\begin{array}{l}<br/>\rho_0 \text { for } r \leq R \\<br/>0 \text { for } r&gt;R<br/>\end{array}\right.<br/>$$<br/>where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 &lt; r &lt; \infty)$ from the centre of the system is represented by</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LA_vMBB7WhU2Vip7y_eiK1-zkU_SDBCOiBh6fpNcRvE.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OSf9wp55TuxJM9G6WaYwibu2xUu2N2abQyF2IXtTUak.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c7v7l3sB9fKCfhxFSsS39jb0MepdzAJ1UM82vwi3us8.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">For $r \leq R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}<br/>$$<br/>Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$<br/>Substituting in Eq. (i) we get, $v \propto r$<br/>i.e., $v-r$ graph is a straight line passing through origin.<br/>For $r&gt;R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}<br/>$$<br/>or $\quad V \propto \frac{1}{\sqrt{r}}$<br/>The corresponding $v-r$ graph will be as shown in option (c).<br/>$\therefore$ correct option is (c).</div>


tags: ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u

slug:
a-spherically-symmetric-gravitational-system-of-particles-has-a-mass-density-0-for-r-r-0-for-r-r-w-h-ere-rho0-i-s-a-co-n-s-t-an-t-a-t-es-t-ma-ssc-an-u-24122

question:
<div class="question">A spherically symmetric gravitational system of particles has a mass density<br/>$$<br/>\rho=\left\{\begin{array}{l}<br/>\rho_0 \text { for } r \leq R \\<br/>0 \text { for } r&gt;R<br/>\end{array}\right.<br/>$$<br/>where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 &lt; r &lt; \infty)$ from the centre of the system is represented by</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LA_vMBB7WhU2Vip7y_eiK1-zkU_SDBCOiBh6fpNcRvE.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OSf9wp55TuxJM9G6WaYwibu2xUu2N2abQyF2IXtTUak.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c7v7l3sB9fKCfhxFSsS39jb0MepdzAJ1UM82vwi3us8.original.fullsize.png"/><br/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9r4XEvZ2nmCJLJy-w_gGqxYumnEQB0ILHwTYuH9voJY.original.fullsize.png"/><br/></span> </div>

answer:
<div class="solution">For $r \leq R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}<br/>$$<br/>Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$<br/>Substituting in Eq. (i) we get, $v \propto r$<br/>i.e., $v-r$ graph is a straight line passing through origin.<br/>For $r&gt;R$ :<br/>$$<br/>\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}<br/>$$<br/>or $\quad V \propto \frac{1}{\sqrt{r}}$<br/>The corresponding $v-r$ graph will be as shown in option (c).<br/>$\therefore$ correct option is (c).</div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch1_P1.db
title:
a-stationary-source-is-emitting-sound-at-a-fixed-frequency-f-0-which-is-reflected-by-two-cars-approaching-the-source-the-difference-between-the-freque-1

slug:
a-stationary-source-is-emitting-sound-at-a-fixed-frequency-f-0-which-is-reflected-by-two-cars-approaching-the-source-the-difference-between-the-freque-1-50848

question:
<div class="question">A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2 \%$ of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is $330 \mathrm{~ms}^{-1}$.</div>

answer:
<div class="solution">Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/D31ONv5MKojv2wN20KPcx_S2EmT-1I0UavavMug4QkI.original.fullsize.png"/><br/><br/>Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$<br/>$$<br/>\begin{aligned}<br/>f_2 &amp; =f_0\left(\frac{v+v_2}{v-v_2}\right) \\<br/>\therefore \quad f_1-f_2 &amp; =\left(\frac{1.2}{100}\right) f_0 \\<br/>&amp; =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\<br/>\text { or }\left(\frac{1.2}{100}\right) f_0 &amp; =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0<br/>\end{aligned}<br/>$$<br/>As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$<br/>$$<br/>\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0<br/>$$<br/>or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$<br/>$$<br/>\begin{aligned}<br/>&amp; =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\<br/>&amp; =7.128 \mathrm{kmh}^{-1}<br/>\end{aligned}<br/>$$<br/>$\therefore$ The nearest integer is 7 .<br/>Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$<br/>$$<br/>\begin{aligned}<br/>f_2 &amp; =f_0\left(\frac{v+v_2}{v-v_2}\right) \\<br/>\therefore \quad f_1-f_2 &amp; =\left(\frac{1.2}{100}\right) f_0 \\<br/>&amp; =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\<br/>\text { or }\left(\frac{1.2}{100}\right) f_0 &amp; =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0<br/>\end{aligned}<br/>$$<br/>As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$<br/>$$<br/>\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0<br/>$$<br/>or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$<br/>$$<br/>\begin{aligned}<br/>&amp; =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\<br/>&amp; =7.128 \mathrm{kmh}^{-1}<br/>\end{aligned}<br/>$$<br/>$\therefore$ The nearest integer is 7 .</div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Main']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch1_P1.db
title:
a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic--1

slug:
a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic-1-20260

question:
<div class="question">A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.</div>

answer:
<div class="solution">Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.<br/>$$<br/>r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-Tnw_k9oHJNYG2G47lZ6gOLhOzmflmnXKOJwQUhvb5g.original.fullsize.png"/><br/></div>


tags: ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch1_P1.db
title:
a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic-

slug:
a-steady-current-i-goes-through-a-wire-loop-pqr-having-shape-of-a-right-angle-triangle-with-pq-3-x-pr-4-x-and-qr-5-x-if-the-magnitude-of-the-magnetic-95144

question:
<div class="question">A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.</div>

answer:
<div class="solution">Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.<br/>$$<br/>r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-Tnw_k9oHJNYG2G47lZ6gOLhOzmflmnXKOJwQUhvb5g.original.fullsize.png"/><br/></div>


tags: ['Mathematics', 'Straight Lines', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-1

slug:
a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-1-78188

question:
<div class="question">A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$y+\sqrt{3} x+2-3 \sqrt{3}=0$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} y-x+3+2 \sqrt{3}=0$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sqrt{3} y+x-3+2 \sqrt{3}=0$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> </div>

answer:
<div class="solution">A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BbwFhwlBUeWDKh1kwXK76q7lwHtU9fgCoUCgBopRkvA.original.fullsize.png"/><br/><br/>where, $\quad \sqrt{3} x+y=1$<br/>$$<br/>\Rightarrow \quad y=-\sqrt{3} x+1<br/>$$<br/>Then, $\quad \tan \theta=-\sqrt{3}$<br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\<br/>\Rightarrow &amp; \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\<br/>\text { and } &amp; \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; y+2=0 \\<br/>\text { and } &amp; \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\<br/>\Rightarrow &amp; y+2=\sqrt{3} x-3 \sqrt{3}<br/>\end{array}<br/>$$<br/>Neglecting, $y+2=0$ as it does not intersect $Y$-axis.</div>


tags: ['Mathematics', 'Straight Lines', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is

slug:
a-straight-line-l-through-the-point-3-2-is-inclined-at-an-angle-6-0-to-the-line-3-x-y-1-if-l-also-intersects-the-x-axis-then-the-equation-of-l-is-70203

question:
<div class="question">A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$y+\sqrt{3} x+2-3 \sqrt{3}=0$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{3} y-x+3+2 \sqrt{3}=0$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sqrt{3} y+x-3+2 \sqrt{3}=0$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$y-\sqrt{3} x+2+3 \sqrt{3}=0$<br/></span> </div>

answer:
<div class="solution">A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BbwFhwlBUeWDKh1kwXK76q7lwHtU9fgCoUCgBopRkvA.original.fullsize.png"/><br/><br/>where, $\quad \sqrt{3} x+y=1$<br/>$$<br/>\Rightarrow \quad y=-\sqrt{3} x+1<br/>$$<br/>Then, $\quad \tan \theta=-\sqrt{3}$<br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\<br/>\Rightarrow &amp; \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\<br/>\text { and } &amp; \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; y+2=0 \\<br/>\text { and } &amp; \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\<br/>\Rightarrow &amp; y+2=\sqrt{3} x-3 \sqrt{3}<br/>\end{array}<br/>$$<br/>Neglecting, $y+2=0$ as it does not intersect $Y$-axis.</div>


tags: ['Mathematics', 'Straight Lines', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-straight-line-through-the-vertex-p-of-a-pqr-intersects-the-side-qr-at-the-point-s-and-the-circumcircle-of-the-pqr-at-the-point-t-if-s-is-not-the-cen

slug:
a-straight-line-through-the-vertex-p-of-a-pqr-intersects-the-side-qr-at-the-point-s-and-the-circumcircle-of-the-pqr-at-the-point-t-if-s-is-not-the-cen-98308

question:
<div class="question">A straight line through the vertex $P$ of a $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the $\triangle P Q R$ at the point $T$. If $S$ is not the centre of the circumcircle, then</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T} &lt; \frac{2}{\sqrt{Q S \cdot S R}}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T}&gt;\frac{2}{\sqrt{Q S \cdot S R}}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T} &lt; \frac{4}{Q R}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{P S}+\frac{1}{S T}&gt;\frac{4}{Q R}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\frac{1}{P S}+\frac{1}{S T}&gt;\frac{2}{\sqrt{Q S \cdot S R}}$<br/>, <br/>$\frac{1}{P S}+\frac{1}{S T}&gt;\frac{4}{Q R}$</span> </div>

answer:
<div class="solution">Let a straight line through the vertex $P$ of a given $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of $\triangle P Q R$ at the point $T$.<br/>Points $P, Q, R, T$ are concyclic, hence<br/>$$<br/>\begin{array}{rlrl}<br/>P S \cdot S T &amp; =Q S \cdot S R &amp; &amp; \\<br/>\Rightarrow \quad P_1 &amp; \geq 2 \sqrt{P S \cdot S T} &amp; \\<br/>&amp; \text { Now, } \frac{P S+S T}{2} &amp; \geq \sqrt{P S \cdot S T} &amp; {[\because \mathrm{AM} \geq \mathrm{GM}]}<br/>\end{array}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/vrafhqNxAoheGeJLHlvdpbhAM3nv4TAWfk7gv8-lXXs.original.fullsize.png"/><br/><br/>and $\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}$<br/>Also, $\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R}$<br/>$\Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\<br/>\text { Hence, } &amp; \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} .<br/>\end{array}<br/>$$</div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-student-is-performing-the-experiment-of-resonance-column-the-diameter-of-the-column-tube-is-4-cm-the-frequency-of-the-tuning-fork-is-512-hz-the-air-

slug:
a-student-is-performing-the-experiment-of-resonance-column-the-diameter-of-the-column-tube-is-4-cm-the-frequency-of-the-tuning-fork-is-512-hz-the-air-48462

question:
<div class="question">A student is performing the experiment of resonance column. The diameter of the column tube is $4 \mathrm{~cm}$. The frequency of the tuning fork is $512 \mathrm{~Hz}$. The air temperature is $38^{\circ} \mathrm{C}$ in which the speed of sound is $336 \mathrm{~m} / \mathrm{s}$. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$14.0 \mathrm{~cm}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$15.2 \mathrm{~cm}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$16.4 \mathrm{~cm}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$17.6 \mathrm{~cm}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$15.2 \mathrm{~cm}$</span> </div>

answer:
<div class="solution">Considering the end correction $[e=0.3 D$ where $\mathrm{D}=$ diameter $]$
<br/> <br/>$f=n\left[\frac{v}{4(l+e)}\right]$ For first resonance, $n=1$
<br/> <br/>$\therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D$
<br/> <br/>or, $l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}$</div>


tags: ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-student-performed-the-experiment-of-determination-of-focal-length-of-a-concave-mirror-by-u-v-method-using-an-optical-bench-of-length-15-m-the-focal-

slug:
a-student-performed-the-experiment-of-determination-of-focal-length-of-a-concave-mirror-by-u-v-method-using-an-optical-bench-of-length-15-m-the-focal-37259

question:
<div class="question">A student performed the experiment of determination of focal length of a concave mirror by $u-v$ method using an optical bench of length $1.5 \mathrm{~m}$. The focal length of the mirror used is 24 $\mathrm{cm}$. The maximum error in the location of the image can be $0.2 \mathrm{~cm}$.<br/>The 5 sets of $(u, v)$ values recorded by the student (in $\mathrm{cm})$ are : $(42,56),(48$, $48),(60,40),(66,33),(78,39)$. The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is(are)</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$(42,56)$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$(48,48)$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$(66,33)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$(78,39)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$(66,33)$<br/>, <br/>$(78,39)$</span> </div>

answer:
<div class="solution">Values of options (c) and (d) don't match with the mirror formula,<br/>$$<br/>\frac{1}{v}+\frac{1}{u}=\frac{1}{f}<br/>$$</div>


tags: ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-student-performed-the-experiment-to-measure-the-speed-of-sound-in-air-using-resonance-aircolumn-method-two-resonances-in-the-aircolumn-were-obtained

slug:
a-student-performed-the-experiment-to-measure-the-speed-of-sound-in-air-using-resonance-aircolumn-method-two-resonances-in-the-aircolumn-were-obtained-62775

question:
<div class="question">A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air column is the second resonance. Then,</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>the intensity of the sound heard at the first resonance was more than that at the second resonance<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>the prongs of the tuning fork were kept in a horizontal place above the resonance tube<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>the amplitude of vibration of the ends of the prongs is typically around $1 \mathrm{~cm}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>the length of the air-column at the first resonance was somewhat shorter than $1 / 4$ th of the wavelength of the sound in air</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>the intensity of the sound heard at the first resonance was more than that at the second resonance<br/>, <br/>the length of the air-column at the first resonance was somewhat shorter than $1 / 4$ th of the wavelength of the sound in air</span> </div>

answer:
<div class="solution">$1 &lt; \frac{\lambda}{4}$<br/>Further, larger the length of air column, feebler is the intensity.<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0_QzsCyIrBeMEzEd77BCqJOEofh6_Ukn13vWiVWjlOw.original.fullsize.png"/><br/></div>


tags: ['Chemistry', 'Some Basic Concepts of Chemistry', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>
source_db: MarksBatch1_P1.db
title:
a-student-performs-a-titration-with-different-burettes-and-finds-titre-values-of-252-ml-2525-ml-and-250-ml-the-number-of-significant-figures-in-the-av

slug:
a-student-performs-a-titration-with-different-burettes-and-finds-titre-values-of-252-ml-2525-ml-and-250-ml-the-number-of-significant-figures-in-the-av-97272

question:
<div class="question">A student performs a titration with different burettes and finds titre values of $25.2 \mathrm{~mL}, 25.25 \mathrm{~mL}$, and $25.0 \mathrm{~mL}$.<br/>The number of significant figures in the average titre value is</div>

answer:
<div class="solution">Average<br/>$$<br/>\begin{aligned}<br/>&amp; =\frac{25.2+25.25+25.0}{3}=\frac{75.45}{3} \\<br/>&amp; =25.15=25.2 \mathrm{~mL}<br/>\end{aligned}<br/>$$<br/>Number of significant figure is 3 .<br/>Basics in chemistry<br/>Rule application<br/>III (Difficulty level is high because of shear ignorance of such topics)</div>


tags: ['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2006']
source_db: MarksBatch1_P1.db
title:
a-student-performs-an-experiment-for-determination-of-g-t-2-4-2-l-l-1-m-and-he-commits-an-error-of-l-for-t-he-takes-the-time-of-n-oscillations-with-th

slug:
a-student-performs-an-experiment-for-determination-of-g-t-2-4-2-l-l-1-m-and-he-commits-an-error-of-l-for-t-he-takes-the-time-of-n-oscillations-with-th-39361

question:
<div class="question">A student performs an experiment for determination of $g\left(=\frac{4 \pi^2 l}{T^2}\right), l \approx 1 \mathrm{~m}$, and he commits an error of $\Delta l$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 \mathrm{~s}$. For which of the following data, the measurement of $g$ will be most accurate?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\Delta L=0.5, \Delta T=0.1, n=20$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\Delta L=0.5, \Delta T=0.1, n=50$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\Delta L=0.5, \Delta T=0.01, n=20$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\Delta L=0.1, \Delta T=0.05, n=50$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\Delta L=0.1, \Delta T=0.05, n=50$</span> </div>

answer:
<div class="solution">$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{n}$<br/>In option (d) error in $\Delta g$ is minimum and number of observations made are maximum. Hence, in this case error in $g$ will be minimum.</div>


tags: ['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-student-performs-an-experiment-to-determine-the-youngs-modulus-of-a-wire-exactly-2-m-long-by-searles-method-in-a-particular-reading-the-student-meas

slug:
a-student-performs-an-experiment-to-determine-the-youngs-modulus-of-a-wire-exactly-2-m-long-by-searles-method-in-a-particular-reading-the-student-meas-28627

question:
<div class="question">A student performs an experiment to determine the Young's modulus of a wire, exactly $2 \mathrm{~m}$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 \mathrm{~mm}$ with an uncertainty of $\pm 0.05 \mathrm{~mm}$ at a load of exactly $1.0 \mathrm{~kg}$. The student also measures the diameter of the wire to be $0.4 \mathrm{~mm}$ with an uncertainty of $\pm 0.01 \mathrm{~mm}$. Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ (exact). The Young's modulus obtained from the reading is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$(2.0 \pm 0.3) 10^{11} \mathrm{~N} / \mathrm{m}^2$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$(2.0 \pm 0.1) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$(2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$<br/></span> </div>

answer:
<div class="solution">$\begin{aligned} Y &amp; =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \\ &amp; =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\end{aligned}$<br/>Further $\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)$<br/>$$<br/>\begin{aligned}<br/>\therefore \quad \Delta Y &amp; =\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times 2.0 \times 10^{11} \\<br/>&amp; =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2<br/>\end{aligned}<br/>$$<br/>The correct option is (b).</div>


tags: ['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-student-uses-a-simple-pendulum-of-exactly-1-m-length-to-determine-g-the-acceleration-due-to-gravity-he-uses-a-stop-watch-with-the-least-count-of-1-s

slug:
a-student-uses-a-simple-pendulum-of-exactly-1-m-length-to-determine-g-the-acceleration-due-to-gravity-he-uses-a-stop-watch-with-the-least-count-of-1-s-55593

question:
<div class="question">A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{~s}$ for this and records $40 \mathrm{~s}$ for 20 oscillations. For this observation, which of the following statement(s) is/are true?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Error $\Delta T$ in measuring $T$, the time period, is $1 \mathrm{~s}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Percentage error in the determination of $g$ is $5 \%$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Percentage error in the determination of $g$ is $2.5 \%$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$<br/>, <br/>Percentage error in the determination of $g$ is $5 \%$<br/></span> </div>

answer:
<div class="solution">$T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}$.<br/>Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$<br/>$\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}$<br/>or $\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}$<br/>Further, $T=2 \pi \sqrt{\frac{l}{g}}$ or $\quad T \propto g^{-1 / 2}$<br/>$$<br/>\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100<br/>$$<br/>or $\%$ error in determination of $g$ is<br/>$$<br/>\begin{aligned}<br/>\frac{\Delta g}{g} \times 100 &amp; =-200 \times \frac{\Delta T}{T} \\<br/>&amp; =-\frac{200 \times 0.05}{2}=-5 \%<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct options are (a) and (c).</div>


tags: ['Mathematics', 'Circle', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-tangent-pt-is-drawn-to-the-circle-x-2-y-2-4-at-the-point-p-3-1-a-straight-line-l-perpendicular-to-pt-is-a-tangent-to-the-circle-x-3-2-y-2-1-question-1

slug:
a-tangent-pt-is-drawn-to-the-circle-x-2-y-2-4-at-the-point-p-3-1-a-straight-line-l-perpendicular-to-pt-is-a-tangent-to-the-circle-x-3-2-y-2-1-question-1-33934

question:
<div class="question">A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^{2}+y^{2}-1$.
<br/> <br/><strong> Question:</strong> A possible equation of $L$ is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$x-\sqrt{3} y=1$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$x+\sqrt{3} y=1$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$x-\sqrt{3} y=-1$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$x+\sqrt{3} y=5$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$x-\sqrt{3} y=1$</span> </div>

answer:
<div class="solution">Equation of tangent $P T$ to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$ is $x \sqrt{3}+y=4$ Let the line $L$, perpendicular to tangent $P T$ be $x-y \sqrt{3}+\lambda=0$
<br/> <br/>As it is tangent to the circle $(x-3)^{2}+y^{2}=1$ $\therefore \quad$ Length of perpendicular from centre of circle to the Tangent $=$ radius of circle.
<br/> <br/>$\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text { or }-5$
<br/> <br/>$\therefore$ Equation of $L$ can be $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$</div>


