The radius of gyration of a solid sphere of mass 5 kg about $X Y$ is 5 m as shown in figure. The rad

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The radius of gyration of a solid sphere of mass 5 kg about $X Y$ is 5 m as shown in figure. The radius of the sphere is $\frac{5 x}{\sqrt{7}} \mathrm{~m}$, then the value of $x$ is

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(d) : As we know that the moment of inertia of a solid sphere about $X Y$ is, $I_{X Y}=\frac{7}{5} M R^2$ \ Also, $I_{X Y}=M K^2$ So, $M K^2=\frac{7}{5} M R^2$ $\Rightarrow R^2=\frac{5}{7} K^2$ \ or $\quad R=K \sqrt{\frac{5}{7}} \quad$ or $\quad R=\frac{5 \sqrt{5}}{\sqrt{7}}$ $(\because K=5 \mathrm{~m})$ \ On comparing with $R=\frac{5 x}{\sqrt{7}}$, the value of $x$ is $\sqrt{5}$.