Q 11-06-094JEE MainJEE Main 2024 (9 Apr, Shift 1)Easy
A string is wrapped around the rim of a wheel of moment of inertia $0.40\ \text{kg m}^2$ and radius $10\ \text{cm}$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40\ \text{N}$. The angular velocity of the wheel after $10\ \text{s}$ is $x\ \text{rad/s}$, where $x$ is ______.
Numerical value type. Enter your answer.
Answer: 100
$$\tau = FR = 40\times0.1 = 4\ \text{N m},\qquad \alpha = \frac\tau I = \frac{4}{0.4} = 10\ \text{rad s}^{-2}$$
$$\omega = \alpha t = 10\times10 = 100\ \text{rad/s}$$
Solution by Sreeraj P, M.Sc Physics