The radius of gyration of a solid sphere of radius $R$ about a is tangential is:

\sqrt{\frac{7}{5}} R

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\sqrt{\frac{2}{5}} R

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\sqrt{\frac{5}{7}} R

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R

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Solution

The correct option is A $\sqrt{\frac{7}{5}} R$
$I_{a x i s} = M k^{2} (k i s r a d i u s o f g y r a t i o n) I f o r s o l i d s p h e r e a b o u t t a n g e n t = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2} \frac{7}{5} M R^{2} = M k^{2} k = \sqrt{\frac{7}{5}} R$

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