A thin solid disc of mass 1 kg rotates about a diameter at 1800 rpm. A constant external torque 25π N·m acts for 40 s, increasing its speed to 2100 rpm. Find the diameter of the disc in metres.
Physics · Rotational Motion · torque and rotational equilibrium
JEE Mains 2025 8 April Evening Shift (Shift 2)

Question

A thin solid disc of mass 1 kg rotates about a diameter at 1800 rpm. A constant external torque 25π N·m acts for 40 s, increasing its speed to 2100 rpm. Find the diameter of the disc in metres.

G1 Physics · Accepted Answer

Convert angular speeds to rad/s: ωᵢ = 1800×(2π/60) = 60π rad/s. ω_f = 2100×(2π/60) = 70π rad/s. Using ω_f = ωᵢ + αt: 70π = 60π + α(40) ⇒ α = π/4 rad/s². For a thin solid disc about a diameter, I = MR²/4. Torque: τ = Iα. Thus, 25π = (1×R²/4)(π/4). Cancel π: 25 = R²/16 ⇒ R² = 400 ⇒ R = 20 m. Diameter = 2R = 40 m. Answer: 40.