JEE Physics 2025 Question with Solution

Given: mass of the disk is $1 \, \text{kg}$, initial angular speed is $1800 \, \text{rpm}$, final angular speed is $2100 \, \text{rpm}$, applied torque is $25\pi \, \text{Nm}$, and time is $40 \, \text{s}$.

Find: the diameter of the disk.

Convert angular speeds from rpm to rad/s:

$$\omega_0 = \frac{1800 \times 2\pi}{60} = 60\pi \; \text{rad/s}$$ $$\omega = \frac{2100 \times 2\pi}{60} = 70\pi \; \text{rad/s}$$

For a thin solid disk rotating about its diameter, the moment of inertia is:

$$I = \frac{1}{4}MR^2 = \frac{R^2}{4}$$

Using torque relation:

$$\tau = I\alpha$$ $$25\pi = \frac{R^2}{4}\alpha$$ $$\alpha = \frac{100\pi}{R^2}$$

Now use rotational kinematics:

$$\omega = \omega_0 + \alpha t$$ $$70\pi = 60\pi + \left(\frac{100\pi}{R^2}\right) \times 40$$ $$10\pi = \frac{4000\pi}{R^2}$$ $$R^2 = 400$$ $$R = 20 \, \text{m}$$

Therefore, the diameter is:

$$D = 2R = 40 \, \text{m}$$

So, the required answer is $40$.

Given: torque $\tau = 25\pi \, \text{Nm}$ acts for $40 \, \text{s}$ and angular speed changes from $1800 \, \text{rpm}$ to $2100 \, \text{rpm}$.

Find: the diameter of the disk.

Use angular impulse:

$$\tau \, dt = I \, \Delta \omega$$

Here,

$$\Delta \omega = (2100 - 1800) \times \frac{2\pi}{60} = 300 \times \frac{2\pi}{60}$$

Substitute values:

$$25\pi \times 40 = I \left(300 \times \frac{2\pi}{60}\right)$$

This gives:

$$I = \frac{25 \times 60 \times 40}{300 \times 2} = 100$$

For a disk about its diameter:

$$I = \frac{MR^2}{4}$$

Since $M = 1 \, \text{kg}$,

$$100 = \frac{R^2}{4}$$ $$R^2 = 400$$ $$R = 20 \, \text{m}$$

Hence,

$$D = 2R = 40 \, \text{m}$$

Therefore, the required diameter is $40 \, \text{m}$.