Q 11-06-255JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
A wheel of radius $0.2\ \text{m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $10\ \text{N}$ as shown in the figure. The established torque produces an angular acceleration of $2\ \text{rad/s}^2$. The moment of inertia of the wheel is ______ $\text{kg m}^2$. (Acceleration due to gravity $= 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 1
The string leaves the rim tangentially, so the torque about the centre is $\tau = FR = 10\times0.2 = 2\ \text{N m}$.
$$\tau = I\alpha \Rightarrow I = \frac{2}{2} = 1\ \text{kg m}^2$$
Solution by Sreeraj P, M.Sc Physics