Q 11-06-114JEE MainJEE Main 2023 (15 Apr, Shift 1)Easy
A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively $(k_{sph}:k_{cyl})$ is $2:\sqrt x$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 5
$k_{sph}=\sqrt{\tfrac25}R$, $k_{cyl}=\sqrt{\tfrac12}R$ (about their symmetry axes).
$$\frac{k_{sph}}{k_{cyl}}=\sqrt{\frac{2/5}{1/2}}=\sqrt{\frac45}=\frac2{\sqrt5}$$
So $x=5$.
Solution by Sreeraj P, M.Sc Physics