Physics · Work, Power & Energy

JEE Main 2024 — 27 January, Shift 2 — Question 37

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

Option A:
$\tan ^{-1}(\sqrt{2})$
Option B:
$2 \tan ^{-1}\left(\frac{1}{2}\right)$
Correct
Option C:
$\tan ^{-1}\left(\frac{1}{2}\right)$
Option D:
$2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Answer: B

Step-by-step solution

Loss in kinetic energy $=$ Gain in potential energy

$\Rightarrow \frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} \ell(1-\cos \theta)$

$\Rightarrow \frac{\mathrm{v}^{2}}{\ell}=2 \mathrm{~g}(1-\cos \theta)$

Acceleration at lowest point $=\frac{\mathrm{v}^{2}}{\ell}$

Acceleration at extreme point $=\mathrm{g} \sin \theta$

Hence, $\frac{\mathrm{v}^{2}}{\ell}=\mathrm{g} \sin \theta$

$\therefore \sin \theta=2(1-\cos \theta)$

$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$

Answer key and solution verified before publishing.

Practise Work, Power & Energy

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Work, Power & Energy
Topic: Vertical Circular Motion