If the radius of a solid sphere is 35,cm, cal | vedclass

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(B) The moment of inertia of a solid sphere about an axis passing through its center of mass is $I_{cm} = \frac{2}{5}MR^2$.According to the parallel a

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$7\sqrt{35}\,cm$

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(B) The moment of inertia of a solid sphere about an axis passing through its center of mass is $I_{cm} = \frac{2}{5}MR^2$.According to the parallel axis theorem,the moment of inertia about a tangent is $I_{tan} = I_{cm} + MR^2$.Substituting the value of $I_{cm}$,we get $I_{tan} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.The radius of gyration $k$ is defined by the relation $I = Mk^2$.Therefore,$Mk^2 = \frac{7}{5}MR^2$.$k = \sqrt{\frac{7}{5}}R$.Given $R = 35\,cm$,we have $k = \sqrt{\frac{7}{5}} 	imes 35 = \sqrt{\frac{7}{5}} 	imes \sqrt{35} 	imes \sqrt{35} = \sqrt{\frac{7 	imes 35}{5}} 	imes \sqrt{35} = \sqrt{49} 	imes \sqrt{5} = 7\sqrt{5}\,cm$. Wait,let's re-calculate: $k = \sqrt{\frac{7}{5}} 	imes 35 = \sqrt{7 	imes 5 	imes 7} = 7\sqrt{5}$. Looking at the options,let's re-evaluate $k = \sqrt{\frac{7}{5}} 	imes 35 = \sqrt{\frac{7}{5} 	imes 1225} = \sqrt{7 	imes 245} = \sqrt{1715} = \sqrt{49 	imes 35} = 7\sqrt{35}\,cm$.

If the radius of a solid sphere is $35\,cm,$ calculate the radius of gyration when the axis is along a tangent.

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