JEE Physics 2023 Question with Solution

Given: mass of the hollow cylinder $m = 5 \, \text{kg}$, radius $R = 70 \, \text{cm} = 0.7 \, \text{m}$, and pulling force $F = 52.5 \, \text{N}$.

Find: the angular acceleration $\alpha$ of the cylinder.

The torque acting on the hollow cylinder is

$$\tau = FR$$

For a hollow cylinder about its axis, the moment of inertia is

$$I = mR^2$$

Using rotational dynamics,

$$\alpha = \frac{\tau}{I}$$

Substituting $\tau = FR$ and $I = mR^2$,

$$\alpha = \frac{FR}{mR^2}$$

So,

$$\alpha = \frac{F}{mR}$$

Now substitute the values,

$$\alpha = \frac{52.5}{5 \cdot 0.7} = \frac{52.5}{3.5} = 15 \, \text{rad/s}^2$$

Therefore, the angular acceleration of the cylinder is $15 \, \text{rad/s}^2$. The correct option is D.

The answer key lists option $A$, but the solution explicitly concludes $15 \, \text{rad/s}^2$, so the solution is used as the authority.