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NCERT Solutions for Class 10 S...
NCERT Solutions for Class 10 Science Chapter 12 - Electricity
Class 10 Science Chapter 12 Electricity is an important chapter for all students. It is a highly detailed chapter with several aspects of electricity discussed in it. NCERT Solutions for Class 10 Science Electricity can be of immense help to the students. Several eminent minds put in effort together in the preparation of this solution set. They have prepared the PDF version of the NCERT Solutions for Class 10 Science Chapter 12, easily downloadable from Vedantu website. This solution set is a simple, concise, yet detailed representation of Class 10 Science Chapter Electricity. The students can also download the Class 10 Maths NCERT Solutions at their convenience and use it to prepare for their examination.
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Intext Exercise 1
1. What Does an Electric Circuit Mean?
Ans: An electric circuit can be defined as a continuous and closed path of an electric current.
The circuit consists of various electric devices.
2. Define the Unit of Current.
Ans: One ampere is constituted by the flow of one coulomb of charge per second. i.e.,
\[1A=\frac{1C}{1s}\]
3. Calculate the Number of Electrons Constituting One Coulomb of Charge.
Ans: Charge of $1$ electron \[=n\times e=1.6\text{ }\times \text{ }{{10}^{-19}}\text{ }C\]
Assume $n$ electrons make one coulomb.
\[n=\frac{1}{e}\]
\[\Rightarrow n=\frac{1}{\left( 1.6\times {{10}^{-19}} \right)}~\]
\[\Rightarrow n=6.25\times {{10}^{18}}\]
Therefore, $\text{1}$ coulomb of charge constitutes \[\text{6}\text{.25}\times \text{1}{{\text{0}}^{\text{18}}}\] electrons.
1. Name a Device That Helps To Maintain a Potential Difference Across a Conductor.
Ans: Any device like a cell or a battery can maintain potential difference in a circuit and across a conductor.
2. What Is Meant by Saying That the Potential Difference Between Two Points is \[\text{1 V}\]?
Ans: It means that\[~\text{1 J}\] of work is done in moving \[\text{1 C}\] charge from one point to the other.
3. How Much Energy Is Given to Each Coulomb of Charge Through a \[\text{6 V}\] battery?
Ans: We can find the energy given to a charge by finding the work it does in the circuit.
Work done in circuit is given by,
Work Done = Potential Difference × Charge
Charge \[=\text{ }1\text{ }C~\]
Potential difference \[=\text{ }6\text{ }V~\]
Work Done \[=\text{ }6\text{ }\times \text{ }1\text{ }=\text{ }6\text{ }J~\]
Therefore, $\text{6J}$ of energy is given to each coulomb of charge passing through a battery of \[\text{6 V}\].
1. On What Factors Does the Resistance of a Conductor Depend?
Ans: The resistance of a conductor depends upon the following factors:
Length of the conductor \[\left( \text{l} \right)\].
Cross-sectional area of the conductor \[\left( \text{A} \right)\text{ }\!\!~\!\!\text{ }\].
Nature of Material of the conductor \[\left( \text{ }\!\!\rho\!\!\text{ } \right)\].
Temperature of the conductor \[\left( T \right)~\].
2. Will Current Flow More Easily Through a Thick Wire or a Thin Wire of the Same Material, When Connected to the Same Source? Why?
Ans: The relation between resistance and the area of cross section can be given as: $\text{R}\propto \frac{\text{1}}{\text{A}}$.
Resistance is inversely proportional to the area of cross-section of the wire. As the resistance decreases, the current increases.
Thicker the wire, less current will pass through it whereas thinner the wire, more current will pass.
3. Let the Resistance of an Electrical Component Remain Constant While the Potential Difference Across the Two Ends of the Component Decreases to Half of Its Former Value. What Change Will Occur in the Current Through It?
Ans: The change in the current flowing through the component is given by Ohm's law as,
\[V=IR~\]
$\Rightarrow I=\frac{V}{R}$
\[\text{R}\] is Resistance of the electrical component.
\[V\] is the Potential Difference.
\[\text{I}\] is the Current
If the potential difference is halved keeping \[\text{R}\]the same,
${{I}_{2}}=\frac{{{V}_{2}}}{R}$
Where ${{V}_{2}}=\frac{V}{2}$
$\Rightarrow {{I}_{2}}=\frac{V}{2R}$
But we know that $I=\frac{V}{R}$
$\Rightarrow {{I}_{2}}=\frac{I}{2}$
That is, the current gets halved.
4. Why are Coils of Electric Toasters and Electric Irons Made of an Alloy Rather Than a Pure Metal?
Ans: The alloys are made such that their resistivity is higher than metals. And they don't melt at high temperatures.
5. Use the Data in Table To Answer the Following:
Table: Electrical resistivity of some substances at \[\text{20 }\!\!{}^\circ\!\!\text{ C}\]
Resistivity $\Omega m$
$1.60\times 10^{-8}$
$10.0\times 10^{-8}$
Constantan (alloy of Cu and Ni)
$44\times 10^{-6}$
Manganin (alloy of Cu, Mn and Ni)
Nichrome (alloy of Ni, Cr, Mn and Fe)
$100\times 10^{-6}$
$10^{10}\sim 10^{14}$
Hard rubber
Paper (Dry)
$10^{12}$
a) Which Among Iron and Mercury is a Better Conductor?
Ans: Resistivity of iron$=10.0\times {{10}^{-8}}\Omega m$
Resistivity of mercury$=94.0\times {{10}^{-8}}\Omega m$
Resistivity of mercury is more than that of iron. So, iron is a better conductor than mercury because resistivity is inversely proportional to conductivity.
b) Which Material is the Best Conductor?
Ans: Metal with least resistivity is the best conductor. In the above
question, according to the table, Silver is the best conductor.
1. Draw a Schematic Diagram of a Circuit Consisting of a Battery of Three Cells of \[\text{2 V}\] each, a \[\text{5 }\!\!\Omega\!\!\text{ }\] resistor, an \[\text{8 }\!\!\Omega\!\!\text{ }\] resistor, and a \[\text{12 }\!\!\Omega\!\!\text{ }\] resistor, and a Plug Key, All Connected in Series.
Ans: The following circuit diagram shows three resistors of resistances $\text{5}\Omega $, \[\text{8 }\!\!\Omega\!\!\text{ }\] and \[\text{12 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] respectively connected in series with a battery of potential \[\text{6 V}\].
(Image will be uploaded soon)
2. Redraw the Circuit of Question $1$, Putting in an Ammeter To Measure the Current Through the Resistors and a Voltmeter To Measure Potential Difference Across the \[12\text{ }\Omega \] resistor. What Would Be the Readings in the Ammeter and the Voltmeter?
Ans: An ammeter should be connected in series with the resistors in the circuit, to measure the current flowing through the resistors.
To measure the potential difference across the\[12\text{ }\Omega \] resistor, a voltmeter should be connected parallel across the\[12\text{ }\Omega \] resistor, as shown in the figure below.
Let's find the total current in the circuit. That should be the same as the current through each resistor and the ammeter.
\[{{R}_{eq}}=\text{ }R1\text{ }+\text{ }R2\text{ }+R3~\]
\[\Rightarrow {{R}_{eq}}=5+8+12~\]
\[\Rightarrow {{R}_{eq}}=25\Omega ~\]
Hence, \[{{I}_{eq}}=\frac{V}{{{R}_{eq}}}~\]
\[\Rightarrow {{I}_{eq}}=\frac{6}{25}=0.24\text{ }A~\]
Thus, the ammeter will read \[0.24A\].
Voltage across \[{{R}_{3}}=\text{ }{{V}_{3}}\]
\[\Rightarrow {{V}_{3}}={{I}_{eq}}\times {{R}_{3}}~~\]
\[\Rightarrow {{V}_{3}}=0.24\times 12\text{ }~\]
\[\Rightarrow {{V}_{3}}=2.88\]
Thus, the voltmeter will read \[2.88V~\].
1. Judge the Equivalent Resistance When the Following Are Connected in Parallel:
a) \[1 \Omega \] and \[{{10}^{6}}\Omega \]
Ans: In parallel connection, we can state that, equivalent resistance is lesser than the least resistance.
So, in both the cases, the resistance should be less than \[1\Omega \]. Verification of these statements can be done as below:
When \[1\text{ }\Omega \]and \[{{10}^{6}}\Omega \]are connected in parallel:
Let $R$ be the equivalent resistance.
$\frac{1}{R}=\frac{1}{1}+\frac{1}{{{10}^{6}}}$
$\Rightarrow R=\frac{{{10}^{6}}}{{{10}^{6}}+1}\approx \frac{{{10}^{6}}}{{{10}^{6}}}$
$\Rightarrow R=1\Omega $
Therefore, equivalent resistance is less than \[1\text{ }\Omega ~\].
b) \[1\text{ }\Omega ,\text{ }{{10}^{3}}\Omega \] and \[{{10}^{6}}\Omega .~\]
Ans: When \[1\text{ }\Omega ,\text{ }{{10}^{3}}\Omega \] and \[{{10}^{6}}\Omega .~\]are connected in parallel:
$\frac{1}{R}=\frac{1}{1}+\frac{1}{{{10}^{3}}}+\frac{1}{{{10}^{6}}}=\frac{{{10}^{6}}+{{10}^{3}}+1}{{{10}^{6}}}$
$\Rightarrow R=\frac{1000000}{1000001}=0.999\Omega $
Therefore, equivalent resistance \[=\text{ }0.999\text{ }\Omega ~\], which is less than\[\text{ }1\Omega ~\].
2. An Electric Lamp of \[\text{100 }\!\!\Omega\!\!\text{ }\], a Toaster of Resistance \[\text{50 }\!\!\Omega\!\!\text{ }\], and a Water Filter of Resistance \[\text{500 }\!\!\Omega\!\!\text{ }\] are Connected in Parallel to a \[\text{220 V}\]Source. What is the Resistance of an Electric Iron Connected to the Same Source That Takes as Much Current as all Three Appliances, and What is the Current Through it?
Ans: Given that,
Resistance of electric lamp, \[{{\text{R}}_{\text{1}}}\text{= 100 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Resistance of toaster, \[{{\text{R}}_{\text{2}}}\text{= 50 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Resistance of water filter, \[{{\text{R}}_{\text{3}}}\text{= 500 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Voltage of the source, \[\text{V = 220 V }\!\!~\!\!\text{ }\]
These are connected in parallel, as shown in the following figure.
Let $R$ be the equivalent resistance of the circuit.
$\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{5+10+1}{500}=\frac{16}{500}$
$\Rightarrow R=\frac{500}{16}\Omega $
According to Ohm's law,
\[\text{V = IR }\!\!~\!\!\text{ }\]
Current flowing through the circuit \[\text{= I }\!\!~\!\!\text{ }\]
$\Rightarrow I=\frac{220} {\frac{500} {16}}=\frac{200\times 16} {500}=6.4A$
Hence, \[\text{6}\text{.4 A}\]of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential\[\text{(220 V)}\] \[\text{=6}\text{.4 A }\!\!~\!\!\text{ }\]
Let $\text{R }\!\!'\!\!\text{ }$ be the resistance of the electric iron. According to Ohm's law,
\[\text{V = I{R}'}\]
$\Rightarrow R'=\frac{V}{I}=\frac{220}{6.4}=34.375\Omega $
Therefore, the resistance of the electric iron is \[\text{34}\text{.375 }\!\!\Omega\!\!\text{ }\] and the current flowing through it is \[\text{6}\text{.4A}\text{. }\!\!~\!\!\text{ }\]
3. What Are the Advantages of Connecting Electrical Devices in Parallel With the Battery Instead of Connecting Them in Series?
Ans: All the devices receive the same voltage in parallel connection.
Failure of one device results does not disrupt the complete circuit.
Overall resistance is reduced.
4. How Can Three Resistors of Resistances \[\text{2 }\!\!\Omega\!\!\text{ , 3 }\!\!\Omega\!\!\text{ }\] and \[\text{6 }\!\!\Omega\!\!\text{ }\] be Connected to Give a Total Resistance of
a) \[\text{4 }\!\!\Omega\!\!\text{ }\]?
Ans: There are three resistors of resistances \[\text{2 }\!\!\Omega\!\!\text{ , 3 }\!\!\Omega\!\!\text{ }\] and \[\text{6 }\!\!\Omega\!\!\text{ }\] respectively.
Consider the circuit below:
Here, \[\text{6 }\!\!\Omega\!\!\text{ }\] and \[\text{3 }\!\!\Omega\!\!\text{ }\]
resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
$\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}$
$\Rightarrow R=\frac{1}{\frac{1}{6}+\frac{1}{3}}=\frac{6\times 3}{6+3}=2\Omega $
This combination is connected to a \[\text{2 }\!\!\Omega\!\!\text{ }\] resistor in series.
Therefore, equivalent resistance of the circuit \[=\text{ }2\text{ }\Omega \text{ }+\text{ }2\text{ }\Omega \text{ }=\text{ }4\text{ }\Omega ~\]
Hence, the total resistance of the circuit is \[\text{4 }\!\!\Omega\!\!\text{ }\].
b) \[\text{ 1 }\!\!\Omega\!\!\text{ }\]
Ans: The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in parallel. Therefore, their equivalent resistance will be given as
$R=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=\frac{1}{\frac{3+2+1}{6}}=\frac{6}{6}=1\Omega $
Therefore, the total resistance of the circuit is \[\text{1 }\!\!\Omega\!\!\text{ }\text{. }\!\!~\!\!\text{ }\]
5. What is
a) The Highest Total Resistance that can be Secured by Combinations of Four Coils of Resistance \[\text{4 }\!\!\Omega\!\!\text{ , 8 }\!\!\Omega\!\!\text{ , 12 }\!\!\Omega\!\!\text{ , 24 }\!\!\Omega\!\!\text{ }\]?
Ans: There are four coils of resistances \[\text{4 }\!\!\Omega\!\!\text{ , 8 }\!\!\Omega\!\!\text{ , 12 }\!\!\Omega\!\!\text{ , 24 }\!\!\Omega\!\!\text{ }\] respectively.
To get highest resistance we must connect these resistances in series. Their equivalent resistance is given by the sum \[4\text{ }+\text{ }8\text{ }+\text{ }12\text{ }+\text{ }24\text{ }=\text{ }48\text{ }\Omega ~\].
b) The Lowest Total Resistance that can be Secured by Combinations of Four Coils of Resistance \[\text{4 }\!\!\Omega\!\!\text{ , 8 }\!\!\Omega\!\!\text{ , 12 }\!\!\Omega\!\!\text{ , 24 }\!\!\Omega\!\!\text{ }\]?
Ans: For lowest resistance connect them in parallel.
Here, \[{{\text{R}}_{eq}}\]is given by
${{R}_{eq}}=\frac{1}{\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}}=\frac{1}{\frac{6+3+2+1}{24}}=\frac{24}{12}=2\Omega $
Therefore,\[~\text{2 }\!\!\Omega\!\!\text{ }\] is the lowest possible equivalent resistance.
1. Why Does the Cord of an Electric Heater Not Glow While the Heating Element Does?
Ans: Heat produced in a system is proportional to its resistance.
The resistance of the heating element is more; so more heat is produced and so it glows.
Also, the resistance of wire is less; so, the heat produced is also less. Hence, the wire does not glow but the heating element does.
2. Compute the Heat Generated While Transferring \[\text{96000}\]coulomb of Charge in One Hour Through a Potential Difference of \[\text{50 V}\].
Ans: The amount of heat \[\left( \text{H} \right)\]produced is given by the Joule's law of heating as
\[H\text{ }=\text{ }VIt~\]
Voltage, \[V=\text{ }50\text{ }V~\]
Time, \[t\text{ }=\text{ }1\text{ }h\text{ }=\text{ }1\text{ }\times \text{ }60\text{ }\times \text{ }60\text{ }s~\]
$H=V.I.t$
$\Rightarrow H=V\cdot \frac{q}{t}\cdot t$
$\Rightarrow H=V\cdot q$
$\Rightarrow H=50\times 96000$
$\Rightarrow H=4.8\times {{10}^{6}}J$
Thus, heat produced while transferring the charge is $H=4.8\times {{10}^{6}}J$.
3. An Electric Iron of Resistance \[\text{20 }\!\!\Omega\!\!\text{ }\] Takes a Current of \[\text{5 A}\]. Calculate the Heat Developed in \[\text{30 s}\].
Ans: The amount of \[\left( \text{H} \right)\]heat produced is given by the Joule's law of heating as \[H=VIt={{I}^{2}}Rt\].
Current, \[\text{I = 5 A }\!\!~\!\!\text{ }\]
Time, \[\text{t = 30 s }\!\!~\!\!\text{ }\]
Resistance \[\text{R = 20}\Omega \text{ }\!\!~\!\!\text{ }\]
$H={{I}^{2}}Rt$
$\Rightarrow H={{5}^{2}}\times 20\times 30$
Therefore, the amount of heat developed in the electric iron is $H=1.5\times {{10}^{4}}J$.
1. What Determines the Rate at Which Energy is Delivered by a Current?
Ans: The rate at which energy is delivered by a current or consumed by the appliance is the power of the appliance.
2. An Electric Motor Takes \[\text{5 A}\] from a \[\text{220V}\] line. Determine the Power of the Motor and the Energy Consumed in $\text{2}$ hrs.
Ans: Power \[\left( \text{P} \right)\]is given by the expression \[\text{P}=\text{VI }\!\!~\!\!\text{ }\].
Voltage, \[\text{V= 220 V }\!\!~\!\!\text{ }\]
\[\text{P = 220 }\!\!\times\!\!\text{ 5 = 1100 W }\!\!~\!\!\text{ }\]
Energy consumed by the motor$=P\times t~$
Time, \[t=2h=2\times 60\times 60=7200s~\]
\[\Rightarrow P=1100\times 7200~\]
Therefore, the power of the motor \[=\text{1100 W }\!\!~\!\!\text{ }\]
Energy consumed by the motor $=\text{7}\text{.92 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{J}$.
NCERT Exercise
1. A Piece of Wire of Resistance $\text{R}$ is Cut into Five Equal Parts. These parts are then Connected in Parallel. If the Equivalent Resistance of this Combination is \[\text{R }\!\!'\!\!\text{ }\], then the Ratio \[~\text{R/R }\!\!'\!\!\text{ }\] is
a. $\frac{\text{1}}{\text{25}}$
b. $\frac{\text{1}}{\text{5}}$
c. $\text{5}$
d. $\text{25}$
Ans: Relation between resistance and current is $R\propto I$.
If the wire is cut into five equal parts, the resistance of each part \[=\frac{R}{5}\].
All the five parts are connected in parallel. Hence, equivalent resistance \[\left( {\text{{R}'}} \right)\]is given as
$\frac{1}{R'}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}=\frac{5+5+5+5+5}{R}=\frac{25}{R}$
$\Rightarrow \frac{R}{R'}=25$
Therefore, the ratio \[\frac{R}{{{R}'}}\] is $25$.
2. Which of the Following Does Not Represent Electric Power in a Circuit?
a. \[{{\text{I}}^{\text{2}}}\text{R}\]
b. \[~\text{I}{{\text{R}}^{\text{2}}}\text{ }\!\!~\!\!\text{ }\]
c. $VI$
d. $\frac{{{\text{V}}^{\text{2}}}}{\text{R}}$
Ans: The correct option is (b).
Electrical power is given by the expression, \[\operatorname{P} = VI\]…(i)
According to Ohm's law, \[~\text{V = IR}\] … (ii)
Clearly, apart from ($\text{b}$), all the other equations can be converted to \[\text{V I }\!\!~\!\!\text{ }\].
3. An Electric Bulb is Rated \[\text{220 V}\] and \[\text{100 W}\]. When it is operated on \[\text{110 V}\],the Power Consumed Will be
a). \[\text{100 W }\!\!~\!\!\text{ }\]
b). \[\text{75 W }\!\!~\!\!\text{ }\]
c). \[\text{50 W }\!\!~\!\!\text{ }\]
d). \[\text{25 W}~\]
Ans: The correct answer is option (d).
Energy consumed by an appliance is given by the expression,
\[\text{P = VI = }\!\!~\!\!\text{ }\frac{{{\text{V}}^{\text{2}}}}{\text{R}}\]
The resistance of the bulb remains constant if the supply voltage is reduced to \[\text{110 V}\]. So, resistance should be found first.
$\Rightarrow R=\frac{{{V}^{2}}}{P}$
Given \[\text{P = 100 W }\!\!~\!\!\text{ }\];
\[\text{V = 220 V }\!\!~\!\!\text{ }\];
\[\Rightarrow R=~\frac{{{\left( 220 \right)}^{2}}}{100}=484\Omega \]
Now if the bulb is operated on \[\text{110 V}\], then the energy consumed by it is given by the expression for power as
$\Rightarrow P'=\frac{{{\left( V' \right)}^{2}}}{R}=\frac{{{\left( 110 \right)}^{2}}}{484}=25W$.
4. Two Conducting Wires of the Same Material and of Equal Lengths and Equal Diameters Are First Connected in Series and Then Parallel in a Circuit Across the Same Potential Difference. The Ratio of Heat Produced in Series and Parallel Combinations Would be
a). \[\text{1:2 }\!\!~\!\!\text{ }\]
b). \[\text{2:1 }\!\!~\!\!\text{ }\]
c). \[\text{1:4 }\!\!~\!\!\text{ }\]
d). \[\text{4:1 }\!\!~\!\!\text{ }\]
Ans: The ratio of heat produced in series and parallel combinations would be \[\text{1:4 }\!\!~\!\!\text{ }\]and the correct answer is option (c).
For the same $\text{V}$ and $\text{t}$; ratio of heat produced in series \[{{\text{H}}_{S}}\] to heat produced in parallel \[{{\text{H}}_{P}}\] is given by,
$\frac{{{H}_{S}}}{{{H}_{P}}}=\frac{\frac{{{V}^{2}}}{{{R}_{S}}}t}{\frac{{{V}^{2}}}{{{R}_{P}}}t}=\frac{{{R}_{P}}}{{{R}_{S}}}$
Now, \[{{R}_{s}}=R+R=2R~\];
$\Rightarrow {{R}_{p}}=\frac{1}{\frac{1}{R}+\frac{1}{R}}=\frac{R}{2}$
Hence, ratio:
$\frac{{{H}_{S}}}{{{H}_{P}}}=\frac{\frac{R}{2}}{2R}=\frac{1}{4}$
Therefore, the ratio of heat produced in series and parallel combinations is \[\text{1:4}\text{. }\!\!~\!\!\text{ }\]
5. How is a Voltmeter Connected in the Circuit To Measure the Potential Difference Between Two Points?
Ans: The voltmeter should be connected in parallel to measure the potential difference between two points.
This is because the voltage across the circuit in parallel can be determined by the voltmeter.
On the other hand, for a series circuit, voltmeters cannot be used. Instead, an ammeter is used in a series circuit.
6. A Copper wire has Diameter \[\text{0}\text{.5 mm}\]and resistivity of \[\text{1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-8}}}\text{ }\!\!\Omega\!\!\text{ m}\text{.}\] What Will be the Length of this Wire to Make its Resistance \[\text{10 }\!\!\Omega\!\!\text{ }\]? How Much Does the Resistance Change if the Diameter is Doubled?
Ans: Resistance \[\left( \text{R} \right)\]of a copper wire of length \[lm\]and cross-section \[A{{m}^{2}}\] is given by the expression,
$R=\rho \frac{l}{A}$
Resistivity of copper, \[\rho =1.6\times {{10}^{-8}}\Omega \text{ }m~\]
Area of cross-section of the wire, \[A=\pi {{\left( \frac{Diameter}{2} \right)}^{2}}\]
Diameter\[\text{= 0}\text{.5 mm = 0}\text{.0005 m }\!\!~\!\!\text{ }\]
Resistance, \[\text{R= 10 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Hence, length of the wire,$l=\frac{RA}{\rho }=\frac{10\times 3.14\times {{\left( \frac{0.005}{2} \right)}^{2}}}{1.6\times {{10}^{-8}}}=\frac{10\times 3.14\times 25}{4\times 1.6}=122.65m$
If the diameter of the wire is doubled, new diameter\[D'=2\times 0.5=1mm=0.001\text{ }m\]
Now, let's find the new resistance with a new diameter (twice the old one). i.e.,
\[D'=2D~\]
\[\Rightarrow A'=p\text{ }\left( D'/2 \right)2=4A~\]
Now new resistance will also be reduced 4 times.
\[\text{{R}' = R/4 }\!\!~\!\!\text{ }\]
\[\Rightarrow \text{{R}'}=\text{10/4}=\text{2}\text{.5 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Therefore, the length of the wire is \[\text{122}\text{.65 m}\] and the new resistance is \[\text{2}\text{.5 }\!\!\Omega\!\!\text{ }\text{. }\!\!~\!\!\text{ }\]
7. The Values of Current $\text{I}$ Flowing in a Given Resistor for the Corresponding Values of Potential Difference $\text{V}$ Across the Resistor are Given Below:
$\text{I}$(ampere)
$0.5$
$\text{V}$(volts)
$10.2$
Plot a Graph Between $\text{V}$ and $\text{I}$ and Calculate the Resistance of That Resistor.
Ans: The plot between voltage and current is called $\mathbf{I}\text{-V}$ characteristic.
The voltage is plotted on the x-axis and current is plotted on the y-axis.
The values of the current for different values of the voltage are shown in the given table.
$\text{V}$ (volts)
$\text{I}$ (ampere)
The VI characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance \[\left( \text{R} \right)\]as,
Slope $=\frac{\text{1}}{\text{R}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{2}}{\text{6}\text{.8}}$
\[\Rightarrow \text{R}=\frac{\text{6}\text{.8}}{\text{2}}=\text{3}\text{.4 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Therefore, the resistance of the resistor is \[\text{3}\text{.4 }\!\!\Omega\!\!\text{ }\].
8. When a \[\text{12 V}\] Battery is Connected Across an Unknown Resistor, There is a Current of \[\text{2}\text{.5mA}\] in the Circuit. Find the value of the Resistance of the Resistor.
Resistance \[\left( \text{R} \right)\]of a resistor is given by Ohm's law as,
$\Rightarrow \text{R=}\frac{\text{V}}{\text{I}}$
Potential difference, \[\text{V = 12 V }\!\!~\!\!\text{ }\]
Current in the circuit, \[\text{I = 2}\text{.5 mA =2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{A }\!\!~\!\!\text{ }\]
\[R=\text{ }\frac{12}{2.5\times {{10}^{-3}}}=4.8\times {{10}^{3}}\Omega =4.8\text{ }k\Omega ~\]
Therefore, the resistance of the resistor is \[\text{4}\text{.8 k }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\].
9. A Battery of \[\mathbf{9V}\] is Connected in Series with Resistors of \[\text{0}\text{.2 }\!\!\Omega\!\!\text{ , 0}\text{.3 }\!\!\Omega\!\!\text{ , 0}\text{.4 }\!\!\Omega\!\!\text{ ,}\]\[\text{0}\text{.5 }\!\!\Omega\!\!\text{ }\] and \[\text{12 }\!\!\Omega\!\!\text{ }\] , respectively. How Much Current Would Flow through the \[\text{12 }\!\!\Omega\!\!\text{ }\] resistor?
Ans: The same current should flow through all the resistances and the circuit as everything is connected in series.
\[\text{V = IR}\]
Where $\text{R}$ is the equivalent resistance.
The sum of the resistances will give the value of $\text{R}$. \[\Rightarrow R=0.2+0.3+0.4+0.5+12=13.4\text{ }\Omega ~\]
Potential difference, \[\text{V = 9 V }\!\!~\!\!\text{ }\];
$\Rightarrow I=\frac{9}{13.4}=0.671A$
Therefore, the current that would flow through the \[\text{12 }\!\!\Omega\!\!\text{ }\] resistor is \[\text{0}\text{.671 A}\text{. }\!\!~\!\!\text{ }\]
10. How Many \[\text{176 }\!\!\Omega\!\!\text{ }\] resistors (in parallel) are Required to carry \[\text{5 A}\] on a \[\text{220 V}\] Line?
Ans: Let's assume the number of resistances to be $x$.
The equivalent resistance of resistors connected in parallel is given by
\[\frac{1}{R}=~x\times \frac{1}{176}\]
$\Rightarrow R=\frac{176}{x}$
Now, Ohm's law is given as
\[V=IR\]
$\Rightarrow R=\frac{V}{I}$
Supply voltage, \[\text{V = 220 V }\!\!~\!\!\text{ }\]
Current, \[I\text{ }=\text{ }5\text{ }A~\]
So, the equation can be written as,
\[\frac{176}{x}=\frac{220}{5}\]
\[\Rightarrow x=\frac{176}{44}\]
\[\Rightarrow x=4\]
Therefore, four resistors of \[\text{176 }\!\!\Omega\!\!\text{ }\] are required to draw the given amount of current.
11. Show How You Would Connect Three Resistors, Each of Resistance
\[\text{6 }\!\!\Omega\!\!\text{ }\], so that the combination has a resistance of
a). \[\text{9 }\!\!\Omega\!\!\text{ }\]
Two Resistors in Parallel: Consider the following figure.
Two \[\text{6 }\!\!\Omega\!\!\text{ }\] resistors are connected in parallel. Their equivalent resistance will be
$R=\frac{1}{\frac{1}{6}+\frac{1}{6}}=\frac{6\times 6}{6+6}=3\Omega $
The third \[6\text{ }\Omega \] resistor is connected in series with \[\text{3 }\!\!\Omega\!\!\text{ }\].
Hence, the equivalent resistance of the circuit is \[\text{6 }\!\!\Omega\!\!\text{ + }\!\!~\!\!\text{ 3 }\!\!\Omega\!\!\text{ = 9 }\!\!\Omega\!\!\text{ }\text{. }\!\!~\!\!\text{ }\]
b). \[\text{4 }\!\!\Omega\!\!\text{ }\]
Two Resistors in Series: Consider the following figure.
Two \[\text{6 }\!\!\Omega\!\!\text{ }\] resistors are connected in series.
Their equivalent resistance will be the sum \[R=6+6=12\Omega ~\]
The third \[\text{6 }\!\!\Omega\!\!\text{ }\] resistor is in parallel with \[\text{12 }\!\!\Omega\!\!\text{ }\].
Hence, equivalent resistance will be
$\Rightarrow{{R}_{eq}}=\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{12\time
s 6}{12+6}=4\Omega $
Therefore, the total resistance is \[\text{4 }\!\!\Omega\!\!\text{ }\].
12. Several Electric Bulbs Designed to be Used on a \[\text{220 V}\] Electric Supply Line, are Rated \[~\text{10W}\]. How Many Lamps can be Connected in Parallel with Each Other Across the Two Wires of \[\text{220 V}\] line if the maximum allowable current is \[\text{5 A}\]?
Ans: Resistance \[{{\text{R}}_{\text{1}}}\] of one bulb is given by the expression,
${{P}_{1}}=\frac{{{V}^{2}}}{{{R}_{1}}}$
$\Rightarrow {{R}_{1}}=\frac{{{V}^{2}}}{{{P}_{1}}}$
Maximum allowable current, \[\text{I = 5 A }\!\!~\!\!\text{ }\].
Rating of an electric bulb, \[{{\text{P}}_{\text{1}}}\text{= 10 W }\!\!~\!\!\text{ }\].
Resistance of one bulb is ${{R}_{1}}=\frac{{{\left( 220 \right)}^{2}}}{10}=4840\Omega $.
Apply Ohm's law to circuit with all the bulbs,
$\text{R}$is the total resistance of the circuit for x number of electric bulbs.
$\Rightarrow R=\frac{V}{I}=\frac{220}{5}=44\Omega $
Resistance of each electric bulb, \[{{\text{R}}_{\text{1}}}\text{= 4840 }\!\!\Omega\!\!\text{ }\].
Clearly, since the resistors are connected in parallel,
$\frac{1}{R}=\frac{1}{{{R}_{1}}}\times x$
$\Rightarrow x=\frac{{{R}_{1}}}{R}=\frac{4840}{44}=110$
Therefore, $\text{110}$ electric bulbs can be connected in parallel.
13. A Hot Plate of an Electric Oven Connected to a \[\text{220 V}\]Line has Two Resistance Coils $\text{A}$ and $\text{B}$ , Each of \[\text{24 }\!\!\Omega\!\!\text{ }\] Resistances, Which May be Used Separately, in Series, or in Parallel. What are the Currents in the Three Cases?
Ans: The given data in the question is as follows:
Supply voltage, \[\text{V=220 V }\!\!~\!\!\text{ }\].
Resistance of one coil, \[\text{R=24 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\].
a) Coils are Used Separately:
\[\text{V = }{{\text{I}}_{\text{1}}}{{\text{R}}_{\text{1}}}\text{ }\!\!~\!\!\text{ }\]
\[{{\text{I}}_{\text{1}}}\]is the current flowing through the coil.
$\Rightarrow {{\text{I}}_{\text{1}}}\text{=}\frac{\text{V}}{{{\text{R}}_{\text{1}}}}\text{=}\frac{\text{220}}{\text{24}}\text{=9}\text{.166A}$
Therefore, \[\text{9}\text{.16 A}\]current will flow through each coil when used separately.
b) Coils are Connected in Series:
Total resistance, \[{{\text{R}}_{\text{2}}}\text{= 24 }\!\!\Omega\!\!\text{ + 24 }\!\!\Omega\!\!\text{ = 48 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
\[\text{V=}{{\text{I}}_{\text{2}}}{{\text{R}}_{\text{2}}}\text{ }\!\!~\!\!\text{ }\]
${{\text{I}}_{\text{2}}}$ is the current flowing through the series circuit.
$\Rightarrow {{\text{I}}_{\text{2}}}\text{=}\frac{\text{V}}{{{\text{R}}_{\text{2}}}}\text{=}\frac{\text{220}}{\text{48}}\text{=4}\text{.58A}$
Therefore, \[\text{4}\text{.58 A}\]current will flow through the circuit when the coils are connected in series.
c) Coils are Connected in Parallel:
Total resistance,${{\text{R}}_{\text{3}}}$ is given as
${{R}_{3}}=\frac{\text{1}}{\frac{\text{1}}{\text{24}}\text{+}\frac{\text{1}}{\text{24}}}\text{=}\frac{\text{24}}{\text{2}}\text{=12 }\!\!\Omega\!\!\text{ }$
${{\text{I}}_{\text{3}}}$is the current flowing through the circuit.
$\Rightarrow {{\text{I}}_{\text{3}}}\text{=}\frac{\text{V}}{{{\text{R}}_{\text{3}}}}\text{=}\frac{\text{220}}{\text{12}}\text{=18}\text{.33A}$
Therefore, \[\text{18}\text{.33 A}\]current will flow through the circuit when coils are connected in parallel.
14. Compare the Power Used in the \[\text{2 }\!\!\Omega\!\!\text{ }\] Resistor in Each of the Following Circuits:
a). \[\text{6 V}\] battery in series with \[\text{1 }\!\!\Omega\!\!\text{ }\] and \[\text{2 }\!\!\Omega\!\!\text{ }\] resistors, and
Ans: In the question it is given that \[\text{V = 6 V }\!\!~\!\!\text{ }\].
\[\text{1 }\!\!\Omega\!\!\text{ }\] and \[\text{2 }\!\!\Omega\!\!\text{ }\] resistors are connected in series.
Therefore, equivalent resistance of the circuit,
\[~\text{R=1+2=3 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
$\text{I}$ is the current through the circuit
\[\Rightarrow \text{I =}\frac{\text{6}}{\text{3}}\text{= 2 A }\!\!~\!\!\text{ }\]
In series the same current flows in the entire circuit.
Hence, current flowing through the \[\text{2 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] resistor is \[\text{2 A}\] .
Power is given by the expression,
\[\text{P=}{{\text{I}}^{\text{2}}}\text{R = }{{\text{2}}^{\text{2}}}\times \text{ 2 = 8 W }\!\!~\!\!\text{ }\]
b). \[\text{4 V}\] battery in parallel with \[\text{12 }\!\!\Omega\!\!\text{ }\] and \[\text{2 }\!\!\Omega\!\!\text{ }\] resistors.
Ans: Potential difference, \[\text{V = 4 V }\!\!~\!\!\text{ }\].
$\text{12 }\!\!\Omega\!\!\text{ }$ and \[\text{2 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] resistors are connected in parallel.
The voltage across each component of a parallel circuit remains the same. Hence, the voltage across \[\text{2 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] resistor will be \[\text{4 V}\].
Power consumed by \[\text{2 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] resistor is given by
\[\text{P=}\frac{{{\text{V}}^{\text{2}}}}{\text{R}}\text{=}\frac{{{\text{4}}^{\text{2}}}}{\text{2}}\text{= 8 W }\!\!~\!\!\text{ }\]
Therefore, the power used by \[\text{2 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\] resistor is $\text{8W}$.
15. Two Lamps, One Rated \[\text{100 W}\] at \[\text{220 V}\], and the Other \[\text{60 W}\]at\[\text{220 V}\], are Connected in Parallel to Electric Mains Supply. What Current is Drawn From the Line if the Supply Voltage is\[\text{220 V}\]?
Ans: Both the bulbs are in parallel. Therefore, potential difference across each of them will be \[\text{220 V}\], because voltage same in a parallel circuit.
Current drawn by the bulb of rating \[\text{100 W}\]is given by,
Power = Voltage × Current
Current$\text{ =}\frac{\text{Power}}{\text{Voltage}}\text{=}\frac{\text{100}}{\text{220}}\text{A}$
Similarly, current drawn by the bulb of rating \[60W\]is given by,
Current $\text{=}\frac{\text{Power}}{\text{Voltage}}\text{=}\frac{\text{60}}{\text{220}}\text{A}$
Hence, total current drawn from the line $\text{=}\frac{\text{100}}{\text{220}}\text{+}\frac{\text{60}}{\text{220}}\text{=0}\text{.727A}$.
16. Which Uses More Energy, a \[\text{250 W}\] TV set in $\text{1}$hr, or a \[\text{1200 W}\] Toaster in $\text{10}$ Minutes?
Ans: Energy consumed by an electrical appliance is given by the expression,
\[\text{H = Pt }\!\!~\!\!\text{ }\]
Power of the appliance\[~\text{= P }\!\!~\!\!\text{ }\]
Time \[\text{= t }\!\!~\!\!\text{ }\]
Energy consumed by a TV set of power \[\text{250 W}\] in $\text{1}$h \[{{\text{E}}_{TV}}\text{= 250 }\!\!\times\!\!\text{ 3600 = 9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{J}\]
Energy consumed by a toaster of power \[\text{1200 W}\]in $\text{10}$ minutes \[{{\text{E}}_{toaster}}\text{= 1200 }\!\!\times\!\!\text{ 600= 7}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{J}\]
Therefore, the energy consumed by a TV set of power \[\text{250 W}\] in $\text{1}$h is more than the energy consumed by a toaster of power \[\text{1200 W}\] in $\text{10}$minutes.
17. An Electric Heater of Resistance \[\text{8 }\!\!\Omega\!\!\text{ }\] draws \[\text{15 A}\] from the service mains $\text{2}$ hours. Calculate the rate at which heat is developed in the heater.
Ans: Rate of heat produced by a device is given by the expression for power as
\[\text{P = }{{\text{I}}^{\text{2}}}\text{R }\!\!~\!\!\text{ }\]
Resistance of the electric heater, \[\text{R = 8 }\!\!\Omega\!\!\text{ }\!\!~\!\!\text{ }\]
Current drawn, \[\text{I = 15 A }\!\!~\!\!\text{ }\]
$\Rightarrow \text{P}=\text{1}{{\text{5}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 8=1800J/s}$
Therefore, heat is produced by the heater at the rate of \[\text{1800 J/s}\text{. }\!\!~\!\!\text{ }\]
18. Explain the following.
a). Why is Tungsten Used Almost Exclusively for Filament of Electric Lamps?
Ans: Tungsten has a very high melting point and resistivity.
Due to these properties', tungsten can produce and maintain high temperatures.
b). Why are the Conductors of Electric Heating Devices, Such As Bread-Toasters and Electric Irons, Made of an Alloy Rather Than a Pure Metal?
Ans: Alloys have higher melting points than a pure metal. They also are preferred in heating devices because of the heat they produce. This is because of their reduced conductivity which also prevents an electric shock since metals are good conductors while alloys are not.
c). Why is the Series Arrangement Not Used for Domestic Circuits?
Ans: In series arrangement, failure of one component results in failure of the complete circuit. And in series, each device will get different voltages. So parallel connection is preferred over series connection.
d). How Does the Resistance of a Wire Vary With its Area of Cross-Section?
Ans: Resistance \[\left( \text{R} \right)\]of a wire is inversely proportional to its area of cross-section \[\left( \text{A} \right)\text{,}\]i.e., $\text{R}\propto \frac{\text{1}}{\text{A}}$.
e). Why are Copper and Aluminium Wires Usually Employed for Electricity Transmission?
Ans: Copper and aluminium wires are good conductors of electricity because of their low resistivity. Thus, they are usually employed for transmission of electricity.
NCERT Solutions for Class 10 Science Free PDF Download
NCERT Solutions Chapter 12 Class 10 Science is an effective solution set prepared by famous teachers worldwide. They have prepared the PDF version of the NCERT Solution of Science Class 10 Electricity and have made it available for download. The students can download the solution set as per their convenience on their computer, tablets, or mobiles and study them in their preparation for their respective examinations.
You can Find the Solutions of All the Science Chapters below.
Chapter 1 - Chemical Reactions and Equations
Chapter 2 - Acids, Bases and Salts
Chapter 3 - Metals and Non-metals
Chapter 4 - Carbon and Its Compounds
Chapter 5 - Periodic Classification of Elements
Chapter 6 - Life Processes
Chapter 7 - Control and Coordination
Chapter 8 - How do Organisms Reproduce?
Chapter 9 - Heredity and Evolution
Chapter 10 - Light Reflection and Refraction
Chapter 11 - Human Eye and Colourful World
Chapter 12 - Electricity
Chapter 13 - Magnetic Effects of Electric Current
Chapter 14 - Sources of Energy
Chapter 15 - Our Environment
Chapter 16 - Management of Natural Resources
Science Class 10 Chapter 12 deals with various concepts revolving around the main topic of electricity. Let us look into some of these concepts.
Electric Current and Charge
Science Chapter 12 Class 10 begins with a detailed description of electric current and charge. The NCERT class 10 science chapter 12 solution defines every keyword belonging to the topic. It also describes the key formulas by which these concepts are linked. The concepts are properly represented with schematics and diagrams so that the students understand the details properly. At the end of this section, Class 10 Electricity Exercise Solutions contain some self-evaluating questions. The students can try answering these questions to check their level of understanding of the concepts.
Potential and Potential Difference
In this segment of NCERT Solutions for Class 10 Science Chapter 12, the students will learn about the flow of charge. They are introduced to the concepts of potential and potential difference. The NCERT Solutions for Class 10 Science Electricity will also provide a thorough description of the related parameters' connection, using formulas to link them. The solution set also shows the application of such formulas. Like the previous segment, the solution set for science NCERT Class 10 Chapter 12 for this segment finishes with several questions for students to evaluate themselves.
Circuit Diagrams
Circuit diagrams are important to represent any electrical circuit. In this segment, the students will learn about the different symbols used to prepare a circuit diagram. This segment includes several diagrams and schematics to elaborate on the details.
Ohm's Law and Resistance
Another important concept in Class 10 Science Chapter Electricity is resistance. The solution set for 10th Science Chapter 12 explains resistance and its related concepts in detail. The different resistances for different metals have been included. The factors affecting the resistance of a material is also discussed with proper mathematical formulas. The concept of resistors and their combinations are also considered in this solution set. The creators of these notes have included several diagrams to explain the ideas. They have also described Ohm's law. The mathematical formula associated with it and the application of such laws are also being described elaborately.
The solution set has been oriented towards exam preparation, and therefore, it also contains several solved examples. These solved mathematical problems are the perfect representation of tough questions that are generally encountered in the examinations. The students can practice these examples and the unsolved questions provided at the end of this segment if they wish to prepare well for their examinations.
Heating Effect of Electric Current and its Application
In the next segment, the heating effect of electric current is discussed in detail. The passage of electrical energy produces heat, and the Joule's effect measures such a process. The details of Joule's effect are covered in this section. There are several mathematical formulas associated with Joule's effect. All these formulas are also included in the solution set for Science Chapter 12 Class 10. The students must practice these solutions hard so that they can get a grasp over the subject.
Along with the details, the solution set for Science Class 10 Chapter 12 also covers the application of such heating effects. The students will learn about real-world applications, which are often referred to during most examinations.
The final concept covered in this solution set is electric power. Before learning about electric power, it is important to understand the individual concepts covered above. As a result, the student will face less difficulty understanding this segment's details and practice the mathematical formulas associated with it. The solution set also includes several figures and solved and unsolved examples to practice the problems. Such problems carry high marks in examinations, and students can score full marks in these questions if they practice and clear their concepts properly.
At the end of the solution set, all the chapter's salient points are summarised and listed in a pointwise manner. The students can learn about the entire chapter by studying these points. Such pointwise listing is helpful for students to prepare on the night before the examination. You can find such representations in top online study forums covering Electricity Class 10 like Vedantu. At the end of the solution set, several questions and answers related to the chapter are provided. These questions will help the students build confidence in answering similar questions they face in the examination.
Exercise Unit 5 total Solutions: 6 Questions (2 short questions 4 Long questions)
Key Features of NCERT Solutions for Class 10 Science Chapter 12
The key features of the NCERT Solutions for Class 10 Science Chapter 12 include:
The entire solution set is written down in a simple language that the students can easily understand.
Several diagrams and schematics are used to explain the concepts to the students.
Exam-oriented questions (both solved and unsolved) are being provided.
Several mathematical problems are being included to help the students in their preparation.
Several direct and conceptual questions are being provided at the end of every segment of the chapter.
The students can easily download the PDF version and study from the solution set at their convenience.
1. Why should you study the concepts of Potential and Potential Difference from NCERT solutions?
Ans: The potential and potential difference are important aspects of the chapter on Electricity. Most students considered this chapter has a high-scoring one since most direct questions come from this chapter. To answer any question from this chapter, a student must learn about potential and potential differences and solve problems associated with the concepts. NCERT solutions concisely describe all the concepts without leaving out any important information. The students will be able to grasp the main ideas within a few words and be equipped to answer all these topics mentioned in this chapter.
2. Why should you learn about resistance, resistors and practice problems from the NCERT solution sets?
Ans: One of the most common questions in all the Board examinations from the chapter on Electricity is the circuits' mathematical problem. To answer any of these mathematical questions, you need to understand the concepts of resistance and resistors. You will also need an idea about current and potential differences. Another important concept that you need to get cleared is how the resistors are connected- in series or parallel. NCERT Solutions are designed to clear these concepts and provide different types of problems associated with the concepts. It also provides a list of solved examples to help students learn about the different kinds of problems that come in the board examinations.
3. List out the topics covered in Chapter 12 Electricity of NCERT Solutions for Class 10 Science.
Ans: Chapter 12, Electricity, of Class 10 Science is a diversified topic consisting of detailed aspects of the concept. This chapter includes topics such as Electric current and charge, Potential difference, Ohm's law, Resistance, Electric Power, Heating effect, Applications, and Circuit diagrams. Vedantu explains the concepts in a simplified manner. It also targets the numerical problems and their step-by-step solutions to enhance the students' preparation.
4. Why should the students download the NCERT Solutions for Class 10 Science Chapter 12 PDF?
Ans: Vedantu offers chapter-wise solutions for all the exercises in the NCERT Textbook on the official website and the Vedantu app. The students should consider downloading the Vedantu NCERT Solutions for Class 10 Science Chapter 12 PDF for the following reasons:
The language is simple and easy to understand. The advanced concepts are explained more thoroughly.
It is easily accessible to the students at any time. They can either opt to study online or download it for free.
Includes numerical problems and their solutions.
5. What is a conductor?
Ans: Any cell or battery that is connected to measure the potential difference between the two ends of the circuit is called a conductor. Resistance of a conductor depends on several factors, such as the material nature, temperature of the conductor, area of cross-section of the conductor, and the length of the conductor. It is said that the metal that has the least reactivity is the best conductor. Vedantu provides the best solution guide for all the students for free. They can download the solutions in PDF format and use them for their online reference.
6. How should a student prepare for the Electricity chapter of Class 10 Science?
Ans: As Vedantu offers simplified answers and revision notes for all the exercises, you can refer to them as they are provided by the subject matter experts. As for Electricity, it is a pretty interesting topic that includes all about resistances, conductors, potential difference, electricity, electric power, heating effect, and application of electricity in our day to day lives. This chapter forms the basis for the chapters that will be taught in the higher standards.
7. How does Electricity find applications in real life?
Ans: Electricity finds applications in our day-to-day lives such as operating laptops, computers, mobile phones, or other electronics. The other applications may include cooling, refrigeration, lighting, etc. The Vedantu study guide covers a wide range of examples and solutions to all the exercises in this chapter. The experienced experts also provide revision notes, important questions, and mock tests to improvise the student's learning.


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Ferrarri 308QV On Dyno At 8000 RPM
Published by Dani on June 25, 2015 June 25, 2015
Song Ferrarri 308QV On Dyno At 8000 RPM •et; Performed by Cirith Ungol
On releases
Servants of Chaos (2001) Compilation appearances, DVD, and Live
Back to Song
Ferrarri 308QV On Dyno At 8000 RPM - Cirith Ungol (Google search or WP search)
Ferrari 308 QV (Quattrovalvole) is a V6/V8 1981-84 model of the car. This track is basically the sound of a Ferrariengine, Rob and Greg's hymn to their favourite car. It is by far not their only reference to Ferrari-cars. See also Ferrari.
Greg and I met in 1971 and we instantly hit it off as we both had a love for sports cars, especially Ferrari's! This is a friend's car (Gary Reed) on a dyno. What other engine can compare to a Ferrari? The answer is none!!!!!
Robert Garven, linernotes
Metal Heart: It's no secret anymore that you are big Ferrari-Fan. And I heard you even got one…Isn't that a quite expensive hobby? Which type and model do you own? (I'm asking that cause I'm as well a big fan of those cars…)
Robert Garven: Yes, I am and have been an enormous fan of Ferrari since the age of 11! Racing and cars were an inspiration and theme for several of our songs over the years. I originally started talking to Greg, as he and I were fellow auto enthusiasts at a young age. I was really hoping that if the band got big one of my dreams was buy many Ferrari's as I feel that they are not only cars but also works of art. I now consider myself very lucky to have just one red 1975 Dino 308 gt4. I am not wealthy yet I managed to get married (no kids) and buy a small house and own a classic Ferrari. (Not bad for a failed rock musician!) I do all the work on it myself as a hobby and this summer I performed a major engine service. There are many things I do not like about our country, but for some reason the car is king here and there are many affordable Ferrari's! Where I live in Southern California it hardly rains or gets cold so it is the perfect environment to own and drive a car as this. Although we do not have the Autobahn, unfortunately! I do not have much money but I do have my priorities!
Diabolical Conquest: I understand that you have always been deeply into Ferraris, and finally satiated a lifetime yen for your very own Ferrari. What is it about the car that has fascinated you for so many years? It would appear that a Ferrari is a strange rock n'roll fetish, when most blue-collar rock n'roll outlaws champion the Harley Davidson.
Well don't get me started! I was attracted to Ferrari as a child, being born on the same day that Dino Ferrari died in Italy 50 years ago. I remember the first time I saw a Ferrari in San Francisco, I was around 11 at the time. I remember getting down on my hands and knees in front of it looking at the prancing horse badge and thinking that this is the most beautiful thing I have ever seen. Many Ferraristi have similar stories. It is funny as we have some popularity in Italy and I have a love for all things Italian. I even have a Roman soldiers outfit that I wear around sometimes! My main love is the sound of the engines. The heart of every Ferrari is the engine and unless you have been in a 12 cylinder Ferrari at 8000 rpm redline there is no way to explain the mechanical symphony produced. My Dino is only an 8 cylinder but it has 4 double throat Weber carburetors and it has its own scream which I an enamored with. At night flames shoot out of the exhaust pipes! Actually many rocker and drummers have a love for the cars, as they are the pinnacle of automotive achievement! Period!
Eternal Flames: Imagine you are winning a free-holiday-journey – where would you prefer to go: To Egypt to watch the pyramids, a week of fishing somewhere up in Canada (very quiet and restful) or a trip to the French seaside at autumn where it's cold and windy?
Robert Garven: I am a Ferrari enthusiast, so I would have to say my dream vacation would be a trip to the Ferrari factory then to see the Italian Grand Prix at Monza, where hopefully Ferrari would come in first and second with Michael Schumacher on top of the podium!
Spaceland Productions, USA

redpajama_set_name: RedPajamaCommonCrawl


text:
Another irritant for Trump as Duterte courts the Russians
Duterte on board the Russian warship.http://news.abs-cbn.com/news/04/21/17/duterte-the-russians-are-with-me
The Philippine president boarded the Russian guided missile warship Varyag in the Philippines and will be visiting Moscow next month
by MG Singh
April 22, 2017 at 3:23 AM April 22, 2017 at 3:23 AM
President Duterte of the Philippines is a law unto himself. He is a strong willed man who practices what he believes. Duterte for long has felt that overdependence on the US has been to the detriment of the Philippines and after coming to power has charted a fresh course in the foreign policy of the Philippines. This includes making overtures to China and Russia and also loosening the strings with the US.
Duterte has been particularly incensed that despite an alliance with the US, the latter did nothing to stop the Chinese encroachments in the South China Sea.
He has a point when he asks as to why the US did not send five aircraft carriers to the South China Sea and nip the Chinese occupation and subsequent build up in the bud. He has now opened out to the Russians who he claims " are with him."
Recently the Russian Guided Missile destroyer visited the Philippines, and the president boarded the ship at the invitation of the Russian commander Captain Alexei Ulyanenko. This is part of the president's policy to try and have an independent foreign policy that is not tied to the US. CNN has reported this news.
Onboard Russian warships
The president boarded the Russian warship and was suitably impressed. Earlier also in January, the president had boarded the Russian destroyer Admiral Tributs.
The Russian warships are now on a goodwill visit to the Philippines. In contrast to Duterte's visits to the Russian ships, the president sent only three officials to visit the US carrier, Carl Vinson, when it docked in the Philippines. He avoided visiting the ship.
Duterte to visit Moscow
President Duterte is also slated to visit Moscow next month when a number of agreements are likely to be signed.
The Philippines will not do not have a formal military alliance, but they are likely to ask for modern military hardware from Russia. On the anvil are drones, Night-vision gear, sniper rifles and even helicopters. Duterte is the first Philippine President to chart an independent foreign policy. America would be alarmed, but the Philippine president won't scrap the defense pact with the US.
He will explore independent options as he has a problem on his hands with China claiming almost the entire South China Sea.
President Donald is aware of the scenario. The US has also toned down its rhetoric on the Chinese military buildup on the Spratly islands, and Duterte has noted this. He has thus decided on an independent policy that is not tied to the US apron strings.
MG Singh
Early retired Air Force Officer who loves to write and has over 10000 articles on the net on over a dozen sites like Hubpages.com, expertscolumn.com, etc He is a prolific writer and has published over 60 short stories.
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Read more on the same topic from MG Singh:
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Q: How to update the following rows after the sum of the previous rows reach a threshold? MySQL I want to update the following rows after the sum of the previous rows reach a defined threshold. I'm using MySQL, and trying to think of a way to solve this using SQL only.
Here's an example. Having the threshold 100. Iterating through the rows, when the sum of the previous rows amount >= 100, set the following rows to checked.

Before the operation:

| id | amount | checked |
| 1  | 50     | false   |
| 2  | 50     | false   |
| 3  | 20     | false   |
| 4  | 30     | false   |

After the operation:

| id | amount | checked |
| 1  | 50     | false   |
| 2  | 50     | false   | <- threshold reached (50 + 50 >= 100)
| 3  | 20     | true*   |
| 4  | 30     | true*   |

Is it possible to do it with just a SQL query? Do I need a stored procedure? How could I implement it using either solution?

A: You can do this by calculating the cumulative amount and using update, and join:
update table t join
       (select t.*, (select sum(amount) from table t2 where t2.id <= t.id) as cum
        from table t
       ) tcum
       on tcum.id = t.id and tcum.cum >= 100
    set checked = true;

EDIT:
For faster performance, you can use variables.  The following should be a correct way to do this:
update table t join
       (select t.*, (@cum := @cum + amount) as cum
        from table t cross join
             (select @cum := 0) vars
        order by t.id
       ) tcum
       on tcum.id = t.id and tcum.cum >= 100
    set checked = true;

A: Something like this? haven't tested so it may need a tweak or two but that should do what you want.
UPDATE table t,
    (SELECT
      @a := @a + amount AS cumulative_sum
    FROM table
    JOIN (SELECT @a := 0) as whatever
) temp
  SET t.checked = true WHERE temp.cumulative_sum >= 100 ;

redpajama_set_name: RedPajamaStackExchange


Top by Calvin Klein, Crew neck, It's classic you, CK chest logo, Regular cut, Fits you just right. With a rich company history spanning back to 1968, Calvin Klein originally started life as a coat store in New York City. The now iconic label, famed for their minimalist approach, offers up covetable diffusion lines CK by Calvin Klein and Calvin Klein Jeans, both of which channel the label's inimitable style.


Emergency Loans - Quick Decision - Bad Credit OK!
Apply for Emergency Loans with our Fast Application!
You never know what is going to happen even in your closest future. It means that you cannot be prepared for anything. An unexpected situation can easily change your plans. Sometimes it may even require a lot of attention and effort from. What are those situations? It can be almost everything what can meet you in your daily life, no matter what job you have, how much money you earn and where you live. Every person experience several emergency situations during their life. The most important is to know how to deal with them and what to do to let them finish as soon as possible. How to deal with those emergency situations? There are, of course several ways. The easiest one is to apply for emergency loans. Only in this way you can receive a certain amount of money, which can be used in the most difficult situations. What are they? Check out the following list.
Medical emergencies – you visited a doctor and it appeared that you have to make some additional examinations. The problem is that you do not have enough money this month. If it is an urgent situation, you should not let yourself wait. Apply for an emergency loan to pay your medical bills. Those kinds of situations happen very often, so banks and institutions responsible for lending money are aware of their clients being sometimes in an urgent need.
House-related issues – it is the most popular issue of applying for a loan by more and more people. What is it caused by? There may be a lot of various reasons. First of all, every house has several devices, which are used every day. It is obvious that sooner or later they will stop fulfilling their duties properly and their owner will be in an urgent need of buying a new fridge, washing machine or cooker. Living without them can be difficult, so it is better to avoid this kind of situation. Emergency loans are just a perfect solution. Every house owner can easily borrow some money, which can be spent for a new device. You just have to prove what you spent your money for. A bill or, even better, an invoice is totally enough.
Emergency car repair – imagine a situation when your car gets broken in the middle of the street. You have to transport it to the mechanic and fix it as soon as possible, but you do not have money. Do not hesitate to check out the positive side of emergency loans. They may be very helpful in those kind of situations. No need to buy a new car.
What do all emergency loans have in common?
- They are known for having high interest rates, what definitely does not have a positive impact on people who decide to use them. Unfortunately, not many people are aware of it. If you are in an emergency situation, you do not have time to analyze anything. You just take the money, which you can borrow from a bank or any other financial institution. In return you have to pay much more.
- It is not difficult to get them. You just need to show your personal documents, such as ID card or passport and you can be sure that you will receive your money even the same day.
- They can be offered by two main types of financial institutions – banks and payday loan institutions. The choice is yours. Just be sure that you know the details of the offer you are the most interested in.


If I change a page (in the section "Muffin builder") then it cannot be saved. Saving does not end.
I didn't change anything in the theme files, except to set the "PHP Max Input Vars" up in php.ini.
I deactivated the plugins, deleted the cache. But it still doesn't work.


Men's 'Mazo' green polo shirt from Dissident. Crafted from soft jersey fabric, the military-inspired polo shirt is detailed with a logo-print on the left chest, contrasting shoulder panels and a three-button placket. Complete with raised tipping on the ribbed polo collar and cuffs.


This is the premiere of "The Wonderful Tapestry of Life." It is a film that looks into the life of Ghanaian Ambassador to Italy, Evelyn Anita Stokes-Hayford, as a way to highlight development in Africa. Throughout her years in the international arena, she has supported many projects regarding women and children.
These efforts inspired director Lia Beltrami to create the film, as she believes the world needs to detach itself from stereotypes of the past to form a more just African society of the future.
Development is not always found in economic spheres, but also rooted in the faith and well being of man. For Ambassador Stokes-Hayford, this is another key factor in Ghana's advancement.
These men are living examples of the optimism that she speaks of. However, the film also wants to be sure it is integral, and spreads to this little one's generation in light of current challenges.
Just this year, the Vatican celebrated the fiftieth anniversary of Pope Paul VI's encyclical, "Populorum Progressio." In it, he wrote "development is the new name for peace." Years later, it appears to still be the case.


text:
Q: Fullscreen ad causing keyboard to not appear My iOS app sometimes displays a Greystripe fullscreen ad on startup. If this ad shows, then when I dismiss it, later I will tap a textfield in a UIWebView but the keyboard won't show up. However, if the Greystripe ad DOESN'T appear, then all is well and the keyboard will show up as expected.
After doing some research I think it has something to do with Greystripe making itself the first responder, but I'm not sure how to fix this.
Edit:
I use adwhirl and so to get the startup ad I do exactly as it says here: http://wiki.greystripe.com/index.php/AdWhirl
The UIWebView is inside another controller that is presented with presentModalViewController:animated: from the main view.
I narrowed the problem down: the problem is caused after initiating Greystripe, which occurs either from the full screen startup ad that I initiate manually or from AdWhirl automatically initiating it to display a 320x50 Greystripe banner.

A: After hours of debugging I finally figured it out. I use SVProgressHUD to show loading status while my UIWebView is loading. SVProgressHUD.m makes itself the key window, then when it is dismissed, it returns the key window status to the "topmost" window. For some reason Greystripe, unlike all the other ad networks I'm using, makes itself the topmost window. So the problem could be Greystripe making itself too important, or SVProgressHUD miscalculating the topmost window!
To solve this, I had to manually make my view controller which contains my UIWebView the key window every time after dismissing SVProgressHUD:
[SVProgressHUD dismiss];
[self.view.window makeKeyWindow];


text:
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A review of machine learning applications in wildfire science and management
Jain, Piyush, Coogan, Sean C P, Subramanian, Sriram Ganapathi, Crowley, Mark, Taylor, Steve, Flannigan, Mike D
arXiv.org Machine Learning Mar-1-2020
Artificial intelligence has been applied in wildfire science and management since the 1990s, with early applications including neural networks and expert systems. Since then the field has rapidly progressed congruently with the wide adoption of machine learning (ML) in the environmental sciences. Here, we present a scoping review of ML in wildfire science and management. Our objective is to improve awareness of ML among wildfire scientists and managers, as well as illustrate the challenging range of problems in wildfire science available to data scientists. We first present an overview of popular ML approaches used in wildfire science to date, and then review their use in wildfire science within six problem domains: 1) fuels characterization, fire detection, and mapping; 2) fire weather and climate change; 3) fire occurrence, susceptibility, and risk; 4) fire behavior prediction; 5) fire effects; and 6) fire management. We also discuss the advantages and limitations of various ML approaches and identify opportunities for future advances in wildfire science and management within a data science context. We identified 298 relevant publications, where the most frequently used ML methods included random forests, MaxEnt, artificial neural networks, decision trees, support vector machines, and genetic algorithms. There exists opportunities to apply more current ML methods (e.g., deep learning and agent based learning) in wildfire science. However, despite the ability of ML models to learn on their own, expertise in wildfire science is necessary to ensure realistic modelling of fire processes across multiple scales, while the complexity of some ML methods requires sophisticated knowledge for their application. Finally, we stress that the wildfire research and management community plays an active role in providing relevant, high quality data for use by practitioners of ML methods.
deep learning, global ecology and biogeography, species distribution modeling, (27 more...)
arXiv.org Machine Learning
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Thirty Years of Machine Learning:The Road to Pareto-Optimal Next-Generation Wireless Networks
Wang, Jingjing, Jiang, Chunxiao, Zhang, Haijun, Ren, Yong, Chen, Kwang-Cheng, Hanzo, Lajos
arXiv.org Machine Learning Jan-23-2019
Next-generation wireless networks (NGWN) have a substantial potential in terms of supporting a broad range of complex compelling applications both in military and civilian fields, where the users are able to enjoy high-rate, low-latency, low-cost and reliable information services. Achieving this ambitious goal requires new radio techniques for adaptive learning and intelligent decision making because of the complex heterogeneous nature of the network structures and wireless services. Machine learning algorithms have great success in supporting big data analytics, efficient parameter estimation and interactive decision making. Hence, in this article, we review the thirty-year history of machine learning by elaborating on supervised learning, unsupervised learning, reinforcement learning and deep learning, respectively. Furthermore, we investigate their employment in the compelling applications of NGWNs, including heterogeneous networks (HetNets), cognitive radios (CR), Internet of things (IoT), machine to machine networks (M2M), and so on. This article aims for assisting the readers in clarifying the motivation and methodology of the various machine learning algorithms, so as to invoke them for hitherto unexplored services as well as scenarios of future wireless networks.
computer based training, ieee transaction, renewable energy, (28 more...)
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Explainable Artificial Intelligence: a Systematic Review
Vilone, Giulia, Longo, Luca
arXiv.org Artificial Intelligence Jun-4-2020
This has led to the development of a plethora of domain-dependent and context-specific methods for dealing with the interpretation of machine learning (ML) models and the formation of explanations for humans. Unfortunately, this trend is far from being over, with an abundance of knowledge in the field which is scattered and needs organisation. The goal of this article is to systematically review research works in the field of XAI and to try to define some boundaries in the field. From several hundreds of research articles focused on the concept of explainability, about 350 have been considered for review by using the following search methodology. In a first phase, Google Scholar was queried to find papers related to "explainable artificial intelligence", "explainable machine learning" and "interpretable machine learning". Subsequently, the bibliographic section of these articles was thoroughly examined to retrieve further relevant scientific studies. The first noticeable thing, as shown in figure 2 (a), is the distribution of the publication dates of selected research articles: sporadic in the 70s and 80s, receiving preliminary attention in the 90s, showing raising interest in 2000 and becoming a recognised body of knowledge after 2010. The first research concerned the development of an explanation-based system and its integration in a computer program designed to help doctors make diagnoses [3]. Some of the more recent papers focus on work devoted to the clustering of methods for explainability, motivating the need for organising the XAI literature [4, 5, 6].
computer based training, deep learning, explanation, (31 more...)
arXiv.org Artificial Intelligence
North America > Canada (1.00)
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Automobiles & Trucks (0.92)
Leisure & Entertainment (0.92)
Information Technology > Artificial Intelligence > Representation & Reasoning > Rule-Based Reasoning (1.00)
Artificial Musical Intelligence: A Survey
Liebman, Elad, Stone, Peter
arXiv.org Artificial Intelligence Jun-17-2020
Computers have been used to analyze and create music since they were first introduced in the 1950s and 1960s. Beginning in the late 1990s, the rise of the Internet and large scale platforms for music recommendation and retrieval have made music an increasingly prevalent domain of machine learning and artificial intelligence research. While still nascent, several different approaches have been employed to tackle what may broadly be referred to as "musical intelligence." This article provides a definition of musical intelligence, introduces a taxonomy of its constituent components, and surveys the wide range of AI methods that can be, and have been, brought to bear in its pursuit, with a particular emphasis on machine learning methods.
deep learning, music, neural network, (22 more...)
Country: North America > United States (1.00)
Genre: Overview (1.00)
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Information Technology > Artificial Intelligence > Representation & Reasoning > Personal Assistant Systems (1.00)
Augmenting Physiological Time Series Data: A Case Study for Sleep Apnea Detection
Nikolaidis, Konstantinos, Kristiansen, Stein, Goebel, Vera, Plagemann, Thomas, Liestøl, Knut, Kankanhalli, Mohan
arXiv.org Machine Learning May-22-2019
The quantity of labelled data is small due to privacy concerns and the cost of data acquisition and labelling by a medical expert. Furthermore, it is quite common that collected data are unbalanced and getting enough data to personalize models for individuals is very expensive or even infeasible. This paper addresses these problems by (1) designing a recurrent Generative Adversarial Network to generate realistic synthetic data and to augment the original dataset, (2) enabling the generation of balanced datasets based on heavily unbalanced dataset, and (3) to control the data generation in such a way that the generated data resembles data from specific individuals. We apply these solutions for sleep apnea detection and study in the evaluation the performance of four well-known techniques, i.e., K-Nearest Neighbour, Random Forest, Multi-Layer Perceptron, and Support Vector Machine.
classifier, deep learning, vascular disease, (23 more...)
Genre: Research Report > Experimental Study (1.00)
Health & Medicine > Therapeutic Area > Cardiology/Vascular Diseases (0.93)
Health & Medicine > Therapeutic Area > Sleep (0.70)
Health & Medicine > Therapeutic Area > Pulmonary/Respiratory Diseases (0.61)
Health & Medicine > Therapeutic Area > Neurology (0.61)
Information Technology > Artificial Intelligence > Machine Learning > Statistical Learning > Support Vector Machines (0.69)
Information Technology > Artificial Intelligence > Machine Learning > Neural Networks > Deep Learning (0.69)
Information Technology > Artificial Intelligence > Machine Learning > Neural Networks > Perceptrons (0.54)
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Regional Health Command Pacific
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Joint Services Spread Suicide Prevention Awareness
Story by Anna Marie G. General on 09/25/2019 The Military and Family Support Center (MFSC) at Joint Base Pearl Harbor-Hickam hosted a Joint Services Suicide Prevention Proclamation ceremony and Human Awareness Ribbon Sept. 10 in support of National Suicide Awareness Month. Service members from every branch of service, Department of read more
Previously Unidentified Pearl Harbor Sailor Returned Home To Kansas After Nearly 78 Years
Story by PO2 Justin Pacheco on 09/25/2019 Nearly 78 years after he died in the Japanese attack on Pearl Harbor by Japanese forces on December 7, 1941, Navy Seaman 2nd Class Wilbur Clayton Barrett was finally laid to rest in his hometown of El Dorado, Kansas, September 14. Three-hundred and read more
Story by PO2 Devin Langer on 09/25/2019 PEARL HARBOR Cheers and motivational words filled the humid air. Sweat dripped down the face of someone who was once a first class petty officer. While carrying a fellow Sailor on top of her shoulders, she hustled down a grassy field. She took read more
Advanced Anti-Submarine Warfare Officer Students Train In The Navy's Newest Combat Trainer
Naval Surface and Mine Warfighting Development Center (SMWDC) and the Center for Surface Combat Systems (CSCS) hosted 10 students in the Navy's newest combat trainer, combined integrated air & missile defense/ anti-submarine warfare trainer (CIAT), to increase operator lethality in the Advanced Anti-Submarine Warfare Officer (ASWO) school September 20, 2019. read more
Shariki Soldiers Defend Homeland, Allies
Story by Jason Cutshaw on 09/25/2019 SHARIKI COMMUNICATIONS SITE, Japan Nestled in the northern edge of the Japanese island of Honshu, one small radar site performs a mighty mission. The Army Navy/Transportable Radar Surveillance and Control Model 2, or AN/TPY-2 forward-based mode radar is operated by U.S. Army Space and read more
"pilot For A Day" At The 179th Airlift Wing
Story by TSgt Joseph Harwood on 09/17/2019 Ayden Houston is named an honorary "Pilot for a Day" as part of a special event, Sept. 15, 2019, at the 179th Airlift Wing, Mansfield, Ohio. The 164th Airlift Squadron collaborated with "A Special Wish Foundation" of Cleveland, Ohio, to provide Ayden with read more
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text:
La rosa nera è un album di Gigliola Cinquetti, pubblicato dalla CGD nel 1967.

Tracce
LP
Lato A

Lato B

Musicisti
 Gigliola Cinquetti – voce
 Franco Monaldi – direzione orchestra, arrangiamenti
 Alberto Baldan – direzione orchestra, arrangiamenti

Classifica
Singoli

Note

Collegamenti esterni

redpajama_set_name: RedPajamaWikipedia


Ibanez has historically offered acoustic instruments that electric guitarists would find inviting and easy to play. Altstar models are the perfect option for the aspiring electric player with a drive to explore the world of acoustic guitars. The Altstar series features a compact dreadnought body, 15.7" fingerboard radius, tighter string spacing and 25.5" scale neck, all designed to offer electric guitarists a seamless transition to acoustic playing. The single-cut Altstar also features a neck joint at the 16th fret, rather than the 12th fret, affording unparalleled upper fret access. Whether you're just looking to delve into the world of acoustic guitars, or you are seeking a quieter practice option while maintaining an electric guitar feel, this is your axe!


text:
One of Disney's lost Oswald cartoons, "Neck 'n' Neck," discovered in Japan
Posted on November 15, 2018 by Cristina Sanza
"Sleigh Bells," discovered in 2015
It may be Mickey's 90th birthday this weekend, but his predecessor Oswald the Lucky Rabbit is also in the spotlight this week. A 1928 Walt Disney cartoon featuring Oswald, thought to be lost forever, has been discovered in Japan.
The 16mm copy of "Neck 'n' Neck" was in the possession of anime historian Yasushi Watanabe, who purchased the cartoon for about $4.40 when he was in high school almost 70 years ago. Watanabe bought the film at a toy wholesaler near his home in Osaka, where it was labeled "Mickey Manga Spide." He didn't realize the significance of the cartoon until recently when he read "Oswald the Lucky Rabbit: The Search for the Lost Disney Cartoons" by David Bossert, a longtime Disney animator.
The original runtime of "Neck 'n' Neck" was five minutes, but it was cut down to two minutes for the 16mm release. In the short, Oswald and Ortensia ("Sadie") are out for a ride in his car when they're pursued by a dog policeman. During the chase, the vehicles comically stretch and contract around steep turns.
Oswald in "Sleigh Bells," discovered in 2015
Watanabe's copy is currently being held at Kobe Planet Film Archive, one of the largest private film collections in Japan. "As I've been a Disney fan for many years, I'm happy that I was able to play a role," Watanabe, 84, told the Asahi Shimbun newspaper.
According to the Telegraph, Bossert called the find "very exciting" and hopes to screen the film in Los Angeles for a group of animation scholars. Becky Cline, director of the Walt Disney Archives, said, "We are absolutely delighted to learn that a copy of the lost film exists."
Another lost film from the Oswald collection, "Sleigh Bells," was discovered in 2015 at the British Film Institute's National Archives by a researcher. The film was restored and its world premiere was hosted by Walt Disney Animation Studios at BFI Southbank in December 2015. Over the years, a few other previously-lost cartoons in the series have been rediscovered.
Walt Disney and Ub Iwerks created the character of Oswald the Lucky Rabbit in 1927, with a series of cartoons distributed by Universal Studios through 1928. Due to a contract dispute with Disney's distributor, Charles Mintz, Disney lost control of the character in 1928. The experience inspired him to create a brand-new character: Mickey Mouse. This time, Disney retained the rights to his creation.
In 2006, Disney CEO Bob Iger made a trade with NBC Universal and regained control of the Oswald character. In exchange for minor assets, including the rights to the lucky rabbit, Disney allowed sportscaster Al Michaels to move from Disney-owned ABC and ESPN to NBC Sports. Seventy-nine years after Oswald was created, The Walt Disney Company now owned the rights to the character and the original 26 short films made by Walt Disney.
Six of Disney's original Oswald cartoons remain lost.
Source: Hollywood Reporter
Tagged:Mickey MouseNeck 'n' NeckOswald the Lucky RabbitUb IwerksWalt DisneyYasushi Watanabe
Holidays at Disneyland one-day itinerary: Your family's guide to the best day ever
Posted on November 15, 2018 by Bailee Abell
New on shopDisney (11/15/18): 5 Mickey Mouse Jewelry Pieces to Celebrate His 90th Birthday
Disney's "Dumbo" live-action film gets new trailer, poster
Posted on November 15, 2018 by Sean Sposato
Older Older post: Holidays at Universal Orlando Resort bring festive Grinch, Harry Potter, Mannheim Steamroller experiences
Newer Newer post: Top 7 food items at the Festival of Holidays at Disney California Adventure


Star Wars has always been a source of inspiration to dreamers and filmmakers the world over, but what if this next film ruins it? What if Luke turns to the Dark Side?
When I was little, I used to pretend to be a Princess Leia, always asking my mom to put my hair into two giant buns and using our flashlight as a light saber. I've always loved Star Wars and it influenced me to become a filmmaker.
But I'm nervous about this next installment.
The film will already be a bit tragic. It'll lack the presence of the best princess a girl could look up to, and we'll also be missing every boy's favourite hero, Han Solo. So when I heard that Mark Hamill announced it being a possibility that Luke turns to the Dark Side, I was shocked. I don't know if my little heart can cope with this.
Watching the trailer, I was enthralled by the amazing cinematography, the beautiful scenery and the little hints of the story to come, but Luke's words cut through me.
What do you think? Will Luke turn to the Dark Side? Will The Last Jedi be the end of the world of Star Wars as we know it? Or is there still hope for The Force to right the chaos and bring unity back to the galaxy?
Watch the trailer and let us know what you think in the comments below. Let's be dreamers and bring some hope to each other, the world of Star Wars isn't over yet!
There was a lot of this kind of speculation prior to The Force Awakens because Luke wasn't really in the trailer but the reason for that turned out to be that he was hardly in the film, so I think it's reading a bit too much into the comment in this trailer to assume that it means he's turning to the dark side.
One of the main reasons I think it's unlikely that Luke will turn to the dark side is that it would be so cliched. Like father, like son. That's a battle he fought in the original trilogy – the lure to turn. Having him go there again would be repeating more old ground, something the creators were rightly criticised for with The Force Awakens. I can't see them being that obvious, especially since Kylo Ren (the grandson) is also there.
But it's clear Luke's going through something unusual in this movie. Mark Hamill has said that he isn't happy with some of the developments in his character. Honestly, rather than this meaning he's turned to the dark side (which I think Mark would like because of the acting challenges), it's more likely that he's become disillusioned and is reluctant to do anything for fear of ruining everything; that kind of thing. I think it more likely that Mark would be annoyed by a sulky Luke determined to do nothing than one who turned to the dark side which would provide some really interesting scenes to play.
Also, he's said in another interview that they've tried to steer clear of everything that's already been written about Luke, including fan fiction. I'd be surprised if there weren't multiple fan fictions dedicated to Luke turning to the dark side, so again, I think it's unlikely.
Of course, Mark has also freely said that he's become really good at lying about all this, so maybe he's deliberately trying to lead people astray.
Exactly what the comment in the trailer means and what Mark means when he says he's disappointed in the direction they've chosen for Luke is something that none of us will know the answer to until we get to December.
Hi Charis – What a chilling question. I tend to agree with Lynne. Besides with Kylo Ren already turned – it would make it a bit unbalanced if Luke also turned. I took Luke's statement 'It's time for the Jedi to end' to mean 'it's time to do things differently' as obviously it hasn't been working. Well, that's my take anyhow. We will have to see. It will be different not have Hans and Leila (or at least not Carrie Fisher) in this next film.


text:
Singer, songwriter, actress. Born on December 29, 1946, in London, England. Starting out as a singer when she was in her teens in 1960s, Marianne Faithfull has remained an inventive and original performer, despite many personal and professional highs and lows during her long career. The daughter of an Austro-Hungarian baroness and a former British intelligence officer, she spent some of her early years at a communal farm established by her father. Her parents later split and Faithfull went with her mother to live in Reading.
Educated at a convent, she met Andrew Loog Oldham, the Rolling Stones manager, at a party in 1964 while she was still in school. He produced her first hit, "As Tears Go By," which he helped to write, along with Mick Jagger and Keith Richards of the Rolling Stones. The song made a strong showing on both the British and American charts.


text:
// Copyright (c) Microsoft Corporation. All rights reserved.
// Licensed under the MIT License.
// Code generated by Microsoft (R) AutoRest Code Generator.

package com.azure.resourcemanager.appservice.models;

import com.azure.core.annotation.Fluent;
import com.fasterxml.jackson.annotation.JsonProperty;

/** The configuration settings of the platform of App Service Authentication/Authorization. */
@Fluent
public final class AuthPlatform {
    /*
     * true if the Authentication / Authorization feature is
     * enabled for the current app; otherwise, false.
     */
    @JsonProperty(value = "enabled")
    private Boolean enabled;

    /*
     * The RuntimeVersion of the Authentication / Authorization feature in use
     * for the current app.
     * The setting in this value can control the behavior of certain features
     * in the Authentication / Authorization module.
     */
    @JsonProperty(value = "runtimeVersion")
    private String runtimeVersion;

    /*
     * The path of the config file containing auth settings if they come from a
     * file.
     * If the path is relative, base will the site's root directory.
     */
    @JsonProperty(value = "configFilePath")
    private String configFilePath;

    /**
     * Get the enabled property: <code>true</code> if the Authentication / Authorization feature is enabled
     * for the current app; otherwise, <code>false</code>.
     *
     * @return the enabled value.
     */
    public Boolean enabled() {
        return this.enabled;
    }

    /**
     * Set the enabled property: <code>true</code> if the Authentication / Authorization feature is enabled
     * for the current app; otherwise, <code>false</code>.
     *
     * @param enabled the enabled value to set.
     * @return the AuthPlatform object itself.
     */
    public AuthPlatform withEnabled(Boolean enabled) {
        this.enabled = enabled;
        return this;
    }

    /**
     * Get the runtimeVersion property: The RuntimeVersion of the Authentication / Authorization feature in use for the
     * current app. The setting in this value can control the behavior of certain features in the Authentication /
     * Authorization module.
     *
     * @return the runtimeVersion value.
     */
    public String runtimeVersion() {
        return this.runtimeVersion;
    }

    /**
     * Set the runtimeVersion property: The RuntimeVersion of the Authentication / Authorization feature in use for the
     * current app. The setting in this value can control the behavior of certain features in the Authentication /
     * Authorization module.
     *
     * @param runtimeVersion the runtimeVersion value to set.
     * @return the AuthPlatform object itself.
     */
    public AuthPlatform withRuntimeVersion(String runtimeVersion) {
        this.runtimeVersion = runtimeVersion;
        return this;
    }

    /**
     * Get the configFilePath property: The path of the config file containing auth settings if they come from a file.
     * If the path is relative, base will the site's root directory.
     *
     * @return the configFilePath value.
     */
    public String configFilePath() {
        return this.configFilePath;
    }

    /**
     * Set the configFilePath property: The path of the config file containing auth settings if they come from a file.
     * If the path is relative, base will the site's root directory.
     *
     * @param configFilePath the configFilePath value to set.
     * @return the AuthPlatform object itself.
     */
    public AuthPlatform withConfigFilePath(String configFilePath) {
        this.configFilePath = configFilePath;
        return this;
    }

    /**
     * Validates the instance.
     *
     * @throws IllegalArgumentException thrown if the instance is not valid.
     */
    public void validate() {
    }
}

redpajama_set_name: RedPajamaGithub


Flesh and fire are subtlety employed to express the symbiosis of physical and metaphysical. A crackling, spitting burning fuse traverses the stage in suspended installation constructed during the intense and delicate movements of the body. All the while an omnipresent voice take you on a journey involving the intangible nature of knowledge and emotional experience, that we absorb, reformulate and impart during our existence.
Presented at VAULT Festival 2014.
Performed by the Artful Badger arts and theater company.


text:
West Germany, Cold War Europe and the Algerian War
By: Bulow, Mathilde Von
Part of the New Studies in European History series
An illuminating and provocative account of Germany's role as sanctuary for Algerian nationalists during their fight for independence from France between 1954 and 1962.
The book explores key issues such as the impact of external sanctuaries on French counterinsurgency efforts; the part played by security and intelligence services in efforts to eliminate these sanctuaries; the Algerian War's influence on West German foreign and security policy; and finally, the emergence of West German civic engagement in support of Algeria's independence struggle, which served to shape the newly independent country's perception of its role and place in international society.
Mathilde Von Bulow sheds new light on the impact of FLN activities, the role of anti-colonial movements and insurgencies in the developing world in shaping the dynamics of the Cold War, as well as the manner in which the Algerian War was fought and won.
1D Europe, 1DFGW West Germany, 1HBA Algeria, HBJD European history, HBJH African history, HBLW3 Postwar 20th century history, from c 1945 to c 2000, HBTW The Cold War
Ebook £108.00
Laminated £67.99


When they first emerged back in 2010, the Vaccines were hailed as the saviours of British rock music. Now, five years later, they're on top of the world with their third album "English Graffiti". Jimmy Volts had a natter with frontman Justin Young ahead of their UK tour to see whether he feels the hype surrounding the band was justified.
Catching up with Justin whilst he played tourist, visiting an old Nazi jail in Cologne, he's quick to dive right in and discuss his thoughts on the hype surrounding his band in the early days.
"I don't think we were the saviours of British rock, but I do think we surpassed our own expectations as well as those of the people commenting on the band at the time. Yes, I suppose in the wake of the first album there was this new wave of British rock bands. But I think it would be foolish to credit us as being responsible. Let's call it coincidence." He says assuredly.
Modesty aside, in the eyes of the music industry, the Vaccines helped prise open the door to a new era for British rock. Still, not many would've predicted the extent of their success. They've been accepted internationally as the poster boys for the now thriving UK rock scene and their latest LP went to number 2 in the UK album chart. So, what's the story behind the album title?
The conversation shifts in the direction of the band's gigging life. By the time this interview is published, the Vaccines will have gigged four times in Australia before jetting back to continue their UK leg. Justin explains how the shows have been going down with the fans.
It's clear the Vaccines are living the dream. But surely, with this much popularity, the cliches of the rock & roll lifestyle are never far away?
Here's to hoping the crowd at their upcoming date at the Brighton Centre display their excitement with a little more dignity. Speaking of which, how are the band feeling about performing here again?


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Home Finance Fortress Investment Buys Fremont Warehouse for $80MM
Fortress Investment Buys Fremont Warehouse for $80MM
Kato Road
(EDITOR'S NOTE: According to sources familiar with the deal, the sellers were private owners and the cap rate on the deal was 3.5 percent. The deal originally closed in October.)
Institutional buyers show strong favor for well-located industrial assets.
Fremont, Calif. – Leading diversified professional services and investment management company Colliers announced the successful sale of an industrial warehouse in Fremont at 47550 Kato Road for $80,000,000.
Colliers' Executive Vice President Ed Hofer and Senior Vice President Casey Ricksen represented the seller, CW Kato LLC, and procured Fortress Investment Group LLC as the buyer. Fortress Investment Group LLC is a leading, highly diversified global investment manager with an expanding portofolio of business in the Bay Area.
At the time of the sale, the 259,737 square-foot warehousing/distribution center was leased to a tenant with less than five years remaining on their lease.
"Fortress Investment Group's acquisition of 47550 Kato Road is just one of many transactions contributing to the industrial sector's dominance above all other asset classes," said Hofer. "On top of the fact the asset's prime location in a core market, the building was sold for a relatively low cap rate given its age and length of the tenant's lease."
As the Silicon Valley's second-largest submarket for industrial activity and supply, Fremont showed 42,407,819 square-feet of total inventory by the end of the third quarter of 2021. The area's desirability for industrial warehousing and distribution due to its centralized location was highlighted by an availability rate of 6.2 percent and vacancy rate of 3.6 percent, 1.8 percentage points less than the 5.4 percent of the previous quarter. Fremont's year-to-date net absorption reached 1,013,584 square-feet with 478,042 square-feet of inventory absorbed in the third quarter. The area's supply of new development continues to grow with 61,098 square-feet under construction.
CW Kato LLC
Kato Industrial Park
MetLife Investment Management Sells 251,996 SQFT California Chase Bank Portfolio for $138MM
All Fab Sheet Metal Puts Manufacturing Building on the Market in San Jose Seeking $19MM
Phelan Development Breaks Ground on 284,850 SQFT Industrial Project in Sacramento
Highbridge Equity Partners Offers Fully Occupied 1515 Webster in Oakland up for Sale Without Pricing Guidance
Stellantis Sells 28,425 SQFT San Leandro Auto Dealership for $11.9MM
60-Unit Victoria Park in Fremont Sold for $27MM


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Home » Rethinking and scaling future business in the Web3 – So much more than blockchain and crypto currencies.
Rethinking and scaling future business in the Web3 – So much more than blockchain and crypto currencies.
March 21, 2023, 10:00 a.m. PDT / 7:00 p.m. CEST
Registration will open soon
Web3 is still nascent that promise nothing less than significant change for the future of business: rebalancing of the digital economy and putting the power into the hands of the users/customers. Web3 refers to the transition of digital services from platform companies to blockchain-based smart contracts and protocols. We will discuss the key trends in these technology areas and explore the opportunities for companies, investors, and an ethical reordering of the digital economy.
This event is directed towards entrepreneurs, investors, government, research institutions and corporations.
Thomas Neubert
GM @ Datacenter and AI Cloud Execution and Strategy Group, Intel US HQ,
Founder/Evangelist Transatlantic AI eXchange
Olaf Groth, Ph.D.
CEO Cambrian Futures, Inc., Faculty at Hult International Business School and Haas School of Business at UC Berkeley
This will be an online event.
January 26, 2022=3
10:00 a.m. PDT / 7:00 p.m. CEST
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