JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- Numerical

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 _____ \(\mathrm{kg} \mathrm{m}^2\).
(Acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\))
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Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation

\(\begin{aligned} & \mathrm{FR}=\mathrm{I} \alpha \\ & \Rightarrow \mathrm{I}=\frac{\mathrm{FR}}{\alpha}=\frac{10 \times 0.2}{2}=1 \mathrm{~kg}-\mathrm{m}^2\end{aligned}\)
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