JEE Main 2025 — Rotational Motion Question with Solution

Mass of the disk, $m=1\text{kg}$.

Initial angular velocity, ${\omega }_i=1800\text{rpm}$.

Final angular velocity, ${\omega }_f=2100\text{rpm}$.

External torque, ${\tau }_{\text{ext}}=25\pi \text{Nm}$.

Time, $t=40\text{seconds}$.

First, convert the rotational speeds from revolutions per minute (rpm) to radians per second (rad/s):

${\omega }_i=1800\times \frac{2\pi }{60}=60\pi \text{rad/s}$

${\omega }_f=2100\times \frac{2\pi }{60}=70\pi \text{rad/s}$

Next, use the equation of motion for rotation to find the angular acceleration $\alpha$:

${\omega }_f={\omega }_i+\alpha t$

$70\pi =60\pi +\alpha (40)$

Solving for $\alpha$:

$\alpha =\frac{70\pi -60\pi }{40}=\frac{10\pi }{40}=\frac{\pi }{4}{\text{rad/s}}^2$

The torque and moment of inertia relationship is given by:

$\tau =I\alpha$

For a thin solid disk rotating along its diameter, the moment of inertia $I$ is $\frac{m{R}^2}{4}$. Thus:

$25\pi =\frac{1\times R^2}{4}\times \frac{\pi }{4}$

Solving for $R$:

$25\pi =\frac{\pi R^2}{16}$

$R^2=\frac{25\pi \times 16}{\pi }=400$

$R=20\text{m}$

The diameter of the disk is:

$\text{Diameter}=2R=2\times 20=40\text{m}$