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  # Category Archives: Problems
  
  ## A light year is the distance light travels in one year . How many meters are there in one light year?
  
  The speed of light in vacuum is 3 x 10^8 m/s. That means in one second it travels 300000000 metres.
  
  As we know, in one year, there are 365 days; in each day there are 4 hours; in each hour there are 60 minutes and in each minute there are 60 seconds.
  
  Therefore, the distance traveled by light in one year = 3 x 10^8 x 365 x 24 x 60 x 60 = 1 light year = 9.4605284 × 1015 meters
  
  ## Learn Integral Calculus from Scratch
  
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  ## Factors affecting frequency of sound produced by a stretched string
  
  Study how the frequency of sound produced will change in each case with the following strings of length 15cms when the strings are tied between  2 ends-
  
  • aluminium string
  • copper string
  • cotton string
  • metallic string
  • jute string
  
  Also study how the pitch changes when the strings are made taught and loose. Study how the frequency of sound changes with thickness of the following strings
  
  • cotton strings
  • copper strings
  
  This seems to be a homework question or a project question. Therefore I am not giving a detailed answer so as not to tamper the basic aim of assigning a project.
  
  The frequency of sound produced by a stretched string depends on the following factors:
  
  1. the length of the string
  2. the linear mass density (i.e; the mass per unit length) of the string
  3. the tension in the string
  
  When you are using strings of different materials, the factor which changes is the mass per unit length and the same is true when you are changing the thickness.
  
  When you make the string more taut, the tension increases and vice versa.
  
  The question is given for a constant length. Therefore the case of effect of changing length does not come into picture.
  
  The formula showing the relationship is $\large \fn_jvn f=\frac{1}{2L}\sqrt{\frac{T}{m}}$
  
  it is evident from the formula that the frequency of sound is
  
  • inversely proportional to the  length
  • directly proportional to the square root of tension in the string and
  • inversely proportional to the square root of linear density of the string.
  
  on proper substitution, the formula can be recast as
  
  $\large \fn_jvn f=\frac{1}{Ld}\sqrt{\frac{T}{\pi \rho }}$
  
  and this will be more convenient for you to answer the questions.
  
  I recommend that you try to explore by actually performing the experiments.
  
  ## Change in Resistance due to stretching a wire
  
  “A piece of wire is redrawn without any change in volume so that its radius become half the original. Compare the new resistance with the original value.”
  
  When we redraw the wire, the volume remains constant and the resistivity also remains constant. SO, the variables are
  
  1. area of cross section and
  2. the length
  
  When radius becomes half, the area of cross section increases in such a way that A1l1 = A2l2. This implies that when length is halved, area of cross section is doubled.
  
  We know that $\large \inline \fn_cs R=\frac{\rho l}{a}$
  
  Therefore, the new resistance becomes (1/4) th the original value.
  
  So, R2:R1 = 1:4
  
  ## More Problems from Kinematics
  
  1. “On a 120 km track, a train moves the first 30 km at a uniform speed of 30 km/h. How fast must the train in the next 90 km so as to have average speed 60 km/h for the entire trip?”
  2. “A person goes form one place to another on bike
  with uniform speed 40 km/h and returns back at
  uniform speed of 60 km/h. Find average speed
  during the journey.”
  
  ## A numerical from Centre of mass
  
  Two objects, of mass 1 kg and 2 kg, are moving with velocities equal to +2 m/sec and -3 m/sec. The center of mass of the two objects is moving at velocity
  
  =(1×2+2x(-3)/(1+2)=(2-6)/3=-4/3=-1.33 m/s (-ve sign shows that the centre of mass will be moving in the direction of the second body)
  
  ## A numerical Problem from optics
  
  When an object is kept at a distance of 60cm from a concave mirror, the magnification is 1/2. Where should the object be placed to get magnification of 1/3??
  
  Ans:
  
  In the first part, u=-60, m=(-)1/2 implies v=-30 cm (Since the image must be real as concave lens form virtual image of bigger size only)
  
  Substituting in eqn
  
  or
  
  we get f = –20 cm
  
  In II case
  
  m=-1/3
  
  v=u/3
  
  f=-20cm
  
  substituting in
  
  we get u=4f=4 x (-20)=-80 cm
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