Q 11-06-117JEE MainJEE Main 2023 (11 Apr, Shift 1)Medium
A solid sphere of mass $500\ \text{g}$ and radius $5\ \text{cm}$ is rotated about one of its diameters with angular speed of $10\ \text{rad s}^{-1}$. If the moment of inertia of the sphere about its tangent is $x\times10^{-2}$ times its angular momentum about the diameter, then the value of $x$ will be ______.
Numerical value type. Enter your answer.
Answer: 35
$I_{tangent}=\dfrac75MR^2=1.4\times0.5\times0.0025=1.75\times10^{-3}\ \text{kg m}^2$.
$L=\dfrac25MR^2\omega=0.4\times0.5\times0.0025\times10=5\times10^{-3}\ \text{kg m}^2\,\text{s}^{-1}$.
Ratio $=\dfrac{1.75\times10^{-3}}{5\times10^{-3}}=0.35=35\times10^{-2}$.
Solution by Sreeraj P, M.Sc Physics