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The moment of inertia of a sphere of mass M and radius R about an axis passing through centre is $\large\frac{2}{5}\;$$MR^2$. The radius of gyration of the sphere about a parallel axis to the sphere is

\[\begin {array} {1 1} (a)\;\frac{7}{5}\;R & \quad (b)\;\frac{3}{5}\;R \\ (c)\;\sqrt {\frac{7}{5}}\;R & \quad  (d)\;\sqrt {\frac{2}{5}}\;R \end {array}\]

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1 Answer

Parallel areas theorem
$I= I_{cm}+Md^2$
$I_{AB}=\large\frac{2}{5} $$ MR^2+MR^2$
$\qquad= \large\frac{7}{5}$$MR^2$
answered Nov 26, 2013 by meena.p
edited Jun 18, 2014 by lmohan717
 

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