JEE MAIN - Physics (2025 - 2nd April Evening Shift - No. 23)
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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 $2 \mathrm{rad} / \mathrm{s}^2$. Moment of intertia of the wheel is___________ $\mathrm{kg} \mathrm{}\,\, \mathrm{m}^2$. (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )
Answer
1
Explanation
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$$\begin{aligned} & \mathrm{FR}=\mathrm{I} \alpha \\ & \Rightarrow \mathrm{I}=\frac{\mathrm{FR}}{\alpha}=\frac{10 \times 0.2}{2}=1 \mathrm{~kg}-\mathrm{m}^2 \end{aligned}$$
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