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Physics Question 46 – JEE-MAIN 2025

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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 rad/s2. Moment of inertia of the wheel is _______ kg m2. (Acceleration due to gravity =10 m/s2)

This problem involves the relationship between force, torque, angular acceleration, and moment of inertia for a rotating body.

Step 1: Calculate the Torque✦ Active

The force applied to the rim of the wheel creates a torque about its center. The torque (τ) is calculated as the product of the force (F) and the radius (R) at which it is applied.

F=10 N R=0.2 m τ=F×R=10 N×0.2 m=2 N m
Step 2: Calculate the Moment of Inertia○ Expand

The torque produced causes an angular acceleration (α). The relationship between torque, moment of inertia (I), and angular acceleration is given by Newton's second law for rotation.

τ=Iα I=τα I=2 N m2 rad/s2=1 kg m2
💡 Teacher's Secret Hint

Ensure units are consistent for all calculations. The acceleration due to gravity is extraneous information in this problem.

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