NEETPhysicsRotational MotionMCQ+4 / −1
The radius of gyration of a solid sphere of mass about is as shown in figure. The radius of the sphere is , then the value of is:

- A5
- B
- C
- D
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Correct answer: D
$$I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} \times 5 R^2=7 R^2\quad \text{.... (1)}$$
$$\begin{aligned} & I_{X Y}=M K^2=5 \times 5^2 \quad \ldots(2) \\ & \therefore 5 \times 5^2=7 \times R^2 \quad \text { [From (1) and (2)] } \\ & \Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad \text { (Given) } \\ & \therefore x=\sqrt{5} \end{aligned}$$
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