Q 11-06-074JEE MainJEE Main 2025 (28 Jan, Shift 1)Medium
The moment of inertia of a solid disc rotating about its diameter is $2.5$ times the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotates in a similar way is $n$ times the moment of inertia of the given ring. Here, $n$ = ______. Consider all the bodies to have equal masses.
Numerical value type. Enter your answer.
Answer: 4
About a diameter: disc $\tfrac{1}{4}MR_d^2$, ring $\tfrac{1}{2}MR_r^2$, solid sphere $\tfrac{2}{5}MR^2$.
$$\frac{1}{4}MR_d^2 = 2.5\times\frac{1}{2}MR_r^2 \Rightarrow R_d^2 = 5R_r^2$$
Sphere with radius $R_d$:
$$I_s = \frac{2}{5}M(5R_r^2) = 2MR_r^2 = 4\times\frac{1}{2}MR_r^2$$
$n = 4$
Solution by Sreeraj P, M.Sc Physics