(C) The moment of inertia of a sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5}MR^2$.Using the parallel axis theorem,the momen

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(C) The moment of inertia of a sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5}MR^2$.Using the parallel axis theorem,the moment of inertia $I$ about a tangent axis is $I = I_{cm} + MR^2$.Substituting the values,$I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.The radius of gyration $K$ is defined by $I = MK^2$.Therefore,$MK^2 = \frac{7}{5}MR^2$.Solving for $K$,we get $K = \sqrt{\frac{7}{5}}R$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

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

A
$ \frac{7}{5}R $
B
$ \frac{3}{5}R $
C
$ \sqrt{\frac{7}{5}} R $
D
$ \sqrt{\frac{3}{5}} R $

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System of Particles and Rotational Motion

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Moment of Inertia and Radius of gyration

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

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