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Physics · Work, Power & Energy

JEE Main 2024 — 29 January, Shift 2 — Question 41

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 half circle reaching the top most position B. The ratio of kinetic energies ( K.E. )A( K.E. )B\frac{(\text { K.E. })_{A}}{(\text { K.E. })_{B}} is :

figure

  1. Option A:

    3:23: 2

  2. Option B:

    5:15: 1

    Correct
  3. Option C:

    2:52: 5

  4. Option D:

    1:51: 5

Answer: B

Step-by-step solution

Apply energy conservation between A & B 12mVL2=12mVH2+mg⁡(2 L)\frac{1}{2} \mathrm{mV}_{\mathrm{L}}^{2}=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^{2}+\operatorname{mg}(2 \mathrm{~L}) ∵VL=5gL\because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}

So, VH=gL\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}

(K.E)A(K.E)B=12 m(5gL)212 m(gL)2=51\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^{2}}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^{2}}=\frac{5}{1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Physics
Chapter
Work, Power & Energy
Topic
Vertical Circular Motion

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