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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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The correct answer is:
$\sqrt{5}$
$I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} \times 5 R^2=7 R^2$...(1)
$I_{X Y}=M K^2=5 \times 5^2$...(ii)
$\therefore 5 \times 5^2=7 \times R^2 \quad[$ From (1) and (2)]
$\Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad$ (Given)
$\therefore x=\sqrt{5}$
$I_{X Y}=M K^2=5 \times 5^2$...(ii)
$\therefore 5 \times 5^2=7 \times R^2 \quad[$ From (1) and (2)]
$\Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad$ (Given)
$\therefore x=\sqrt{5}$
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