A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of $25 \pi \mathrm{Nm}$ for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ______ m.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Rotational Mechanics

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of

25 π Nm

for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ______ m.

Answer: 40

Solution

The spin rates give

ω_{i} = 1800 \times \frac{2 π}{60} = 60 π rad s^{- 1}

and

ω_{f} = 2100 \times \frac{2 π}{60} = 70 π rad s^{- 1}

, so the angular acceleration is

α = \frac{70 π - 60 π}{40} = \frac{π}{4} rad s^{- 2}

. A disk about a diameter has

I = \frac{m R ^{2}}{4}

, and

τ = I α

gives

25 π = \frac{R ^{2}}{4} \cdot \frac{π}{4}

, so

R^{2} = 400

and

R = 20 m

. The diameter of the disk is

40 m

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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