Q64 / 75 · JEE Main 2025 (4 Apr Shift 2) · Physics
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 $R / 2$ is $x \omega_1$. The value of $x$ is ________
Correct answer: 32
Solution
When sphere is of radius $R$, its mass is $M$, when radius is reduced to $\frac{\mathrm{R}}{2}$, mass will reduced to $\frac{\mathrm{M}}{8}$ Now by conservation of angular momentum
$\begin{aligned}<br/>& \left(\tau_{e x t}=0\right) \\ & L_1=L_2 \\ & I_1 \omega_1=I_2 \omega_2 \\ & \left(\frac{2}{5} M R^2\right) \omega_1=\left(\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right)\left(\frac{\mathrm{R}}{2}\right)^2\right) \omega_2<br/>\end{aligned}$
$\omega_2=32 \omega_1$ value of x is 32
Answer is 32
$\begin{aligned}<br/>& \left(\tau_{e x t}=0\right) \\ & L_1=L_2 \\ & I_1 \omega_1=I_2 \omega_2 \\ & \left(\frac{2}{5} M R^2\right) \omega_1=\left(\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right)\left(\frac{\mathrm{R}}{2}\right)^2\right) \omega_2<br/>\end{aligned}$
$\omega_2=32 \omega_1$ value of x is 32
Answer is 32
Hence, the correct answer is 32.