Physics Q31 – JEE-MAIN 2026

Home Physics Rotational MotionQuestion #444

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.

Clue 1 of 3: Rotational Kinematics and Dynamics

This problem involves converting units, calculating moment of inertia, and applying rotational kinematic equations along with Newton's second law for rotation.

🥷

Ninja Strategy•Sequential Calculation and Elimination

First, calculate the torque magnitude ( $0.0128 π Nm$ ) to eliminate options 1 and 3. Then, calculate the number of rotations (100) to distinguish between the remaining options and identify the correct answer.

🧠 Full Solution Path

Step 1: Calculate Moment of Inertia and Angular Acceleration✦ Active

First, convert the given values to SI units: radius $R = 4 cm = 0.04 m$ and initial angular velocity $ω_{0} = 1200 rpm = 1200 \times \frac{2 π}{60} rad/s = 40 π rad/s$ . The final angular velocity is $ω_{f} = 0 rad/s$ . The moment of inertia for a solid sphere about an axis through its center is $I = \frac{2}{5} M R^{2}$ .

I = \frac{2}{5} (5 kg) (0.04 m)^{2} = 2 \times 0.0016 {kg m}^{2} = 0.0032 {kg m}^{2}

Next, calculate the angular acceleration $α$ using the rotational kinematic equation $ω_{f} = ω_{0} + α t$ .

0 = 40 π + α (10 s) ⟹ α = - 4 π {rad/s}^{2}

Step 2: Calculate Torque Applied○ Expand

The magnitude of the constant torque $τ$ applied is given by Newton's second law for rotation: $τ = I α$ .

τ = (0.0032 {kg m}^{2}) \times (4 π {rad/s}^{2}) = 0.0128 π Nm

💡 Teacher's Secret Hint

Remember to use the magnitude of angular acceleration for torque calculation unless direction is explicitly asked.

Step 3: Calculate Number of Rotations○ Expand

Calculate the total angular displacement $θ$ using the rotational kinematic equation $θ = \frac{(ω_{0} + ω_{f})}{2} t$ .

θ = \frac{(40 π + 0)}{2} \times 10 s = 20 π \times 10 = 200 π rad

Finally, convert the angular displacement from radians to the number of rotations $N$ using $N = \frac{θ}{2 π}$ .

N = \frac{200 π rad}{2 π rad/rotation} = 100 rotations

Thus, the torque applied is $0.0128 π Nm$ and the number of rotations is $100$ . This corresponds to option 4.

💡 Teacher's Secret Hint

Ensure consistent units throughout the calculation, especially when converting between rpm and rad/s.

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Physics Question 31 – JEE-MAIN 2026

Clue 1 of 3: Rotational Kinematics and Dynamics

🧠 Full Solution Path