Physics · Rotational Dynamics

JEE Main 2025 — 8 April, Evening Shift — Question 70

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 40s, the speed increases to 2100 rpm . The diameter of the disk is $\qquad$ m.

Answer: 40

Numerical answer — enter this value.

$\tau d t=I \Delta w$

\begin{aligned} & \Rightarrow 25 \pi \times 40=I(300) \times \frac{2 \pi}{60} \\ & I=\frac{25 \times 60 \times 40}{300 \times 2}=100=\frac{M R^{2}}{4} \\ & R^{2}=400 & R=20 \mathrm{~m} \\ & D=40 \mathrm{~m} \end{aligned}

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