Physics · Rotational Dynamics

JEE Main 2026 — 22 January, Evening Shift — Question 43

Two masses m and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$ . The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$ . (Assume string does not slip and $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Using energy conservation $\frac{1}{2} m v^{2}+\frac{1}{2} 2 m v^{2}+\frac{1}{2} \frac{30 m R^{2}}{2} \times \frac{v^{2}}{R^{2}}=m g h$ $9 \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}=\sqrt{\frac{\mathrm{gh}}{9}}=\sqrt{\frac{10 \times 3.6}{9}}$ $\mathrm{v}=\sqrt{4}=2 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

Practise Rotational Dynamics

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: 22 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Work-Energy Theorem in General Motion