Physics · Work, Power & Energy

JEE Main 2025 — 29 January, Morning Shift — Question 35

A body of mass ' m ' connected to a massless and unsearchable string goes in vertical circle of radius ' R ' under gravity g . The other end of the string is fixed at the centre of circle. If velocity at top of circular path is $n \sqrt{g R}$ , where, $n \geq 1$ , then ratio of kinetic energy of the body at bottom to that at top of the circle is

Option A:
$\frac{n}{n+4}$
Option B:
$\frac{n+4}{n}$
Option C:
$\frac{n^{2}}{n^{2}+4}$
Option D:
$\frac{n^{2}+4}{n^{2}}$
Correct

Answer: D

Step-by-step solution

$\mathrm{V}_{\text {Top }}=\sqrt{\mathrm{n}^{2} \mathrm{gR}}$

$\mathrm{V}_{\text {Botom }}=\sqrt{\mathrm{n}^{2} \mathrm{gR}+4 \mathrm{gR}}$

Ratio $=\frac{\mathrm{n}^{2}+4}{\mathrm{n}^{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Vertical Circular Motion