Physics · Work, Power & Energy

JEE Main 2025 — 22 January, Morning Shift — Question 57

A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$

and left to perform vertical motion (circular) as shown in figure. Initially, by applying

horizontal velocity $v_{0}$ at the point ' A '. the string becomes slack when, the bob

reaches at the point ' $D$ '. The ratio of the kinetic energy of the bob at the

points B and C is $\qquad$ .

Option A:
2
Correct
Option B:
1
Option C:
4
Option D:
3

Answer: A

Step-by-step solution

$\frac{1}{2} \mathrm{mv}_{\mathrm{A}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^{2}+\mathrm{mgh}$

$\Rightarrow \frac{1}{2} \mathrm{~m}(5 \mathrm{~g} \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^{2}+\mathrm{mg} \frac{\ell}{2}$

$\Rightarrow \frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}}$ $\Rightarrow \mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell$

$\frac{1}{2} m v_{\mathrm{C}}^{2}=\frac{1}{2} m v_{\mathrm{D}}^{2}+m g \frac{\ell}{2}$

$\Rightarrow \mathrm{KE}_{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell$

$\Rightarrow \frac{\mathrm{KE}_{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Vertical Circular Motion