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 25π N m for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ___________ m. Numerical answer type. Work it out, then open the answer.

Answer: 40. About a diameter: I = 1/4MR² = R²/4. Δω = 300 rpm = 10π rad/s, α = (10π)/(40) = π/4 rad/s² τ = Iα ⇒ 25π = R²/4·π/4 ⇒ R² = 400, R = 20 m Diameter = 2R = 40 m.

A thin solid disk of 1 kg is rotating along its diameter axis at the… | JEE Main 2025 (8 Apr, Shift 2)

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Rotational Kinematics & Dynamics

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 $25\pi\text{ N m}$ for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ___________ m.

Numerical answer type. Work it out, then open the answer.

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Answer: 40

About a diameter: $I = \frac{1}{4}MR^2 = \frac{R^2}{4}$ .
$\Delta\omega = 300\text{ rpm} = 10\pi\text{ rad/s}$ , $\alpha = \frac{10\pi}{40} = \frac{\pi}{4}\text{ rad/s}^2$
$\tau = I\alpha \Rightarrow 25\pi = \frac{R^2}{4}\cdot\frac{\pi}{4} \Rightarrow R^2 = 400$ , $R = 20$ m
Diameter $= 2R = 40$ m.

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