Physics · Rotational Dynamics

JEE Main 2025 — 2 April, Evening Shift — Question 64

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of $2 \mathrm{rad} / \mathrm{s}^{2}$ . Moment of inertia of the wheel is $\qquad$ $\mathrm{kgm}^{2}$ . (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

$\tau=F R=/ \alpha$

I=\frac{F R}{\alpha}=\frac{10 \times 0.2}{2}=1

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is

Answer: 1

Step-by-step solution

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