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JEE Physics 2024 Question with Solution

A bob of mass ‘mm’ is suspended by a light string of length ‘LL’. It is imparted a minimum horizontal velocity at the lowest point AA such that it just completes a half-circle reaching the topmost position BB. The ratio of kinetic energies (K.E.)A:(K.E.)B(\text{K.E.})_A : (\text{K.E.})_B is:

  • A

    3:23:2

  • B

    5:15:1

  • C

    2:52:5

  • D

    1:51:5

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A bob of mass mm attached to a light string of length LL is given the minimum horizontal speed at the lowest point AA so that it just reaches the topmost point BB.

Find: The ratio (K.E.)A(K.E.)B\dfrac{(\text{K.E.})_A}{(\text{K.E.})_B}.

For just completing the motion up to the top point, the speed at the top must be the minimum needed to keep the string taut:

VB2=gLV_B^2 = gL

Using conservation of mechanical energy between AA and BB:

12mVL2=12mVB2+mg(2L)\frac{1}{2} m V_L^2 = \frac{1}{2} m V_B^2 + mg(2L)

So,

VL2=VB2+4gLV_L^2 = V_B^2 + 4gL

Substituting VB2=gLV_B^2 = gL,

VL2=gL+4gL=5gLV_L^2 = gL + 4gL = 5gL

Hence,

(K.E.)A=12mVL2=12m(5gL)(\text{K.E.})_A = \frac{1}{2} m V_L^2 = \frac{1}{2} m (5gL)

and

(K.E.)B=12mVB2=12m(gL)(\text{K.E.})_B = \frac{1}{2} m V_B^2 = \frac{1}{2} m (gL)

Therefore,

(K.E.)A(K.E.)B=VL2VB2=5gLgL=5\frac{(\text{K.E.})_A}{(\text{K.E.})_B} = \frac{V_L^2}{V_B^2} = \frac{5gL}{gL} = 5

So the ratio is 5:15:1. The correct option is B.

Common mistakes

  • Using VB=0V_B = 0 at the top is incorrect. At the topmost point the bob must still have minimum non-zero speed to keep the string taut. Use the condition VB2=gLV_B^2 = gL instead.

  • Ignoring the gain in gravitational potential energy from AA to BB gives a wrong energy equation. The bob rises through a height of 2L2L, so the increase in potential energy is mg(2L)mg(2L).

  • Taking the ratio of velocities instead of kinetic energies is incorrect. Since kinetic energy is proportional to the square of speed, use (K.E.)A(K.E.)B=VA2VB2\dfrac{(\text{K.E.})_A}{(\text{K.E.})_B} = \dfrac{V_A^2}{V_B^2}.

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