Q122 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
The radius of gyration of a solid sphere of radius 'R' and mass 'M' about its diameter is $K_d$ and that about a tangent of a solid sphere is $K_t$. The ratio of $K_d$ to $K_t$ is
- A$\left(\dfrac{7}{5}\right)^{1/2}$
- B$\left(\dfrac{7}{2}\right)^{1/2}$
- Correct answerC$\left(\dfrac{2}{7}\right)^{1/2}$
- D$\left(\dfrac{2}{5}\right)^{1/2}$
Solution
The moment of inertia of a solid sphere about its diameter is $I_d = \dfrac{2}{5}MR^2$.
The radius of gyration about the diameter is $K_d = \sqrt{\dfrac{2}{5}}R$.
Using the parallel axis theorem, the moment of inertia of the solid sphere about a tangent is $I_t = I_d + MR^2 = \dfrac{2}{5}MR^2 + MR^2 = \dfrac{7}{5}MR^2$.
The radius of gyration about the tangent is $K_t = \sqrt{\dfrac{7}{5}}R$.
The ratio of $K_d$ to $K_t$ is $\dfrac{K_d}{K_t} = \dfrac{\sqrt{\dfrac{2}{5}}R}{\sqrt{\dfrac{7}{5}}R} = \left(\dfrac{2}{7}\right)^{1/2}$.
Answer: $\left(\dfrac{2}{7}\right)^{1/2}$
The radius of gyration about the diameter is $K_d = \sqrt{\dfrac{2}{5}}R$.
Using the parallel axis theorem, the moment of inertia of the solid sphere about a tangent is $I_t = I_d + MR^2 = \dfrac{2}{5}MR^2 + MR^2 = \dfrac{7}{5}MR^2$.
The radius of gyration about the tangent is $K_t = \sqrt{\dfrac{7}{5}}R$.
The ratio of $K_d$ to $K_t$ is $\dfrac{K_d}{K_t} = \dfrac{\sqrt{\dfrac{2}{5}}R}{\sqrt{\dfrac{7}{5}}R} = \left(\dfrac{2}{7}\right)^{1/2}$.
Answer: $\left(\dfrac{2}{7}\right)^{1/2}$
Hence, the correct answer is option: 'C'.