Q56 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
A solid sphere of radius $4$ cm and mass $5$ kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of $1200$ rpm. It is brought to rest in $10$ s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _______ and _______ respectively.
- A$0.128\pi$ Nm, $100$
- B$0.0128\pi$ Nm, $50$
- C$0.128\pi$ Nm, $50$
- Correct answerD$0.0128\pi$ Nm, $100$
Solution
Initial angular velocity $\omega_0 = 1200 \text{ rpm} = \dfrac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}$
Final angular velocity $\omega = 0$
Time $t = 10 \text{ s}$
Angular acceleration $\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{0 - 40\pi}{10} = -4\pi \text{ rad/s}^2$
Moment of inertia of the solid sphere $I = \dfrac{2}{5}MR^2 = \dfrac{2}{5} \times 5 \times (0.04)^2 = 0.0032 \text{ kg m}^2$
Torque $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi \text{ Nm}$
Angle rotated $\theta = \dfrac{\omega_0 + \omega}{2}t = \dfrac{40\pi + 0}{2} \times 10 = 200\pi \text{ rad}$
Number of rotations $N = \dfrac{\theta}{2\pi} = \dfrac{200\pi}{2\pi} = 100$
Answer: $0.0128\pi$ Nm, $100$
Final angular velocity $\omega = 0$
Time $t = 10 \text{ s}$
Angular acceleration $\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{0 - 40\pi}{10} = -4\pi \text{ rad/s}^2$
Moment of inertia of the solid sphere $I = \dfrac{2}{5}MR^2 = \dfrac{2}{5} \times 5 \times (0.04)^2 = 0.0032 \text{ kg m}^2$
Torque $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi \text{ Nm}$
Angle rotated $\theta = \dfrac{\omega_0 + \omega}{2}t = \dfrac{40\pi + 0}{2} \times 10 = 200\pi \text{ rad}$
Number of rotations $N = \dfrac{\theta}{2\pi} = \dfrac{200\pi}{2\pi} = 100$
Answer: $0.0128\pi$ Nm, $100$
Hence, the correct answer is option: 'D'.