Q 11-06-229JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
Let the moment of inertia of a hollow cylinder of length $30$ cm (inner radius $10$ cm and outer radius $20$ cm), about its axis be $I$. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$, is:
Answer: (A) $16$ cm
Thick hollow cylinder: $I = \dfrac M2(R_1^2 + R_2^2) = \dfrac M2(100 + 400) = 250M$ (cm²).
Thin cylinder: $I = MR^2$, so $R^2 = 250$ and $R \approx 15.8 \approx 16$ cm.
Solution by Sreeraj P, M.Sc Physics