A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ π × 10⁻¹ Nm.

Solution

Physics sir · 4 steps

Convert rpm to rad/s first, then torque is just I times the steady angular deceleration.

1. Moment of inertia. For a solid disc about its central axis, I = 1/2MR² = 1/2 × 10 × (0.2)² = 0.2 kg m². Remember to take R = 20 cm = 0.2 m.
2. Angular speed. One revolution is 2π rad and one minute is 60 s, so ω = 600 × 2π/60 = 20π rad s⁻¹.
3. Angular deceleration. The disc goes from 20π rad s⁻¹ to rest in 10 s, so α = 20π/10 = 2π rad s⁻².

   τ = Iα = 0.2 × 2π = 0.4π N m

4. So. τ = 0.4π = 4π × 10⁻¹ N m, so the blank is 4.

The common slip: 8. Taking I = MR² (the ring formula) instead of 1/2MR² doubles the torque to 0.8π N m and gives 8.

The idea to keep. Rotational problems run just like linear ones: τ = Iα mirrors F = ma. Always convert rpm to rad/s and cm to m before you plug in.

Answer from the NTA final answer key. Solution by PrepWiser.