In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30 ° shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration)

Physics · Work, Power & Energy

JEE Main 2026 — 24 January, Evening Shift — Question 30

In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle $30^{\circ}$ shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration)

Option A:
$\sqrt{5 \mathrm{gr}}$
Option B:
$\sqrt{\frac{7}{2} \mathrm{gr}}$
Correct
Option C:
$\sqrt{4 \mathrm{gr}}$
Option D:
$\sqrt{\frac{5}{2} \mathrm{gr}}$

Answer: B

Step-by-step solution

$\mathrm{T}+\mathrm{mg} \cos 60^{\circ}=\frac{\mathrm{mV}^{2}}{\ell}$ $\mathrm{T}=0$ $\mathrm{V}^{2}=\frac{\mathrm{g} \ell}{2}$ here V is the speed at point A M.E.C. $\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mg}\left(\ell+\ell \cos 60^{\circ}\right)+\frac{1}{2} \mathrm{mV}^{2}$ $\mathrm{u}^{2}=3 \mathrm{~g} \ell+\frac{\mathrm{g} \ell}{2}$ $\mathrm{u}=\sqrt{\frac{7 \mathrm{~g} \ell}{2}}$

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 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Vertical Circular Motion