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(D) For a solid sphere,the moment of inertia is $I_{S} = \frac{2}{5} MR^2$. For a solid cylinder,the moment of inertia is $I_{C} = \frac{1}{2} MR^2$.

Vedclass · Accepted Answer

$\frac{14}{15}$

Vedclass · Answer

(D) For a solid sphere,the moment of inertia is $I_{S} = \frac{2}{5} MR^2$. For a solid cylinder,the moment of inertia is $I_{C} = \frac{1}{2} MR^2$. The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$. Therefore,the ratio of the acceleration of the cylinder '$a_{c}$' to that of the sphere '$a_{s}$' is: $\frac{a_{c}}{a_{s}} = \frac{1 + \frac{I_{S}}{MR^2}}{1 + \frac{I_{C}}{MR^2}} = \frac{1 + \frac{2}{5}}{1 + \frac{1}{2}} = \frac{\frac{7}{5}}{\frac{3}{2}} = \frac{7}{5} 	imes \frac{2}{3} = \frac{14}{15}$.

$A$ solid cylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The ratio of the acceleration of the cylinder '$a_{c}$' to that of sphere '$a_{s}$' is

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