JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 21)
A solid sphere with uniform density and radius $R$ is rotating initially with constant angular velocity $\left(\omega_1\right)$ 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 $\mathrm{R} / 2$ is $x \omega_1$. The value of $x$ is _________.
Answer
32
Explanation
When sphere is of radius $R$, its mass is $M$, when radius is reduced to $\frac{R}{2}$, mass will reduced to $\frac{M}{8}$
Now by conservation of angular momentum
$$\begin{aligned} & \left(\tau_{\mathrm{ext}}=0\right) \\ & \mathrm{L}_1=\mathrm{L}_2 \\ & \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\ & \left(\frac{2}{5} \mathrm{MR}^2\right) \omega_1=\left(\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right)\left(\frac{\mathrm{R}}{2}\right)^2\right) \omega_2 \end{aligned}$$
$\omega_2=32 \omega_1 \quad$ value of x is 32
Answer is 32
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