Physics · Rotational Dynamics

JEE Main 2025 — 4 April, Evening Shift — Question 64

A solid sphere with uniform density and radius RR is rotating initially with constant angular velocity (ω1)\left(\omega_{1}\right) about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius become R/2R / 2 is xω1x \omega_{1}. The value of xx is \qquad -.

Answer: 32

Numerical answer — enter this value.

Step-by-step solution

Angular momentum will remain conserve. I1ω1=I2ω2I_{1} \omega_{1}=I_{2} \omega_{2}

25MR2ω1=25(M8)(R2)2ω2\frac{2}{5} M R^{2} \omega_{1}=\frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^{2} \omega_{2}

32ω1=ω232 \omega_{1}=\omega_{2} =32=32

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Rotational Dynamics
Topic
Angular Momentum and its Conservation