Physics · Horizontal Circular Motion

NEET (UG) 2025 — Question 6

A bob of heavy mass $m$ is suspended by a light string of length $l$ . The bob is given a horizontal velocity $\mathrm{v}_{0}$ as shown in figure. If the string gets slack at some point P making an angle $\theta$ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed $\mathrm{v}_{0}$ is :

Option A:
$(\sin \theta)^{\frac{1}{2}}$
Option B:
$\left(\frac{1}{2+3 \sin \theta}\right)^{\frac{1}{2}}$
Option C:
$\left(\frac{\cos \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}$
Option D:
$\left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{\frac{1}{2}}$
Correct

Answer: D

Step-by-step solution

C.O.M.E. $\frac{1}{2}mv_0^2 = mg\ell(1 + \sin\theta) + \frac{1}{2}mv_p^2 \quad \text{......(i)}$

At pt P $\text{T}_p + mg \sin \theta = \frac{mv_p^2}{\ell} \quad (\text{as } \text{T}_p = 0)$

$mg \sin \theta = \frac{mv_p^2}{\ell} \implies mv_p^2 = mg \ell \sin\theta \quad \text{......(ii)}$

From (i) & (ii)

$\frac{1}{2}mv_0^2 = mg\ell(1 + \sin\theta) + \frac{1}{2}mg\ell\sin\theta$

$v_0^2 = 2g\ell(1 + \sin\theta) + g\ell\sin\theta$

$v_0 = \sqrt{2g\ell + 3g\ell\sin\theta} \quad \text{......(iii)}$ $v_p = \sqrt{g\ell\sin\theta}$

$\frac{v_p}{v_0} = \sqrt{\frac{g\ell\sin\theta}{2g\ell + 3g\ell\sin\theta}}$

$\frac{v_p}{v_0} = \sqrt{\frac{\sin\theta}{2 + 3\sin\theta}}$

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Horizontal Circular Motion
Topic: Problems involving application of circular motion