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(D) The acceleration $a$ of a body rolling down an incline without slipping is given by:$a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$For a cylinder,t

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$\frac{15}{14}$

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(D) The acceleration $a$ of a body rolling down an incline without slipping is given by:$a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$For a cylinder,the moment of inertia $I_c = \frac{1}{2} M_c R^2$. Thus,the acceleration $a_c$ is:$a_c = \frac{g \sin 	heta_c}{1 + \frac{1}{2}} = \frac{g \sin 	heta_c}{3/2} = \frac{2}{3} g \sin 	heta_c$For a solid sphere,the moment of inertia $I_s = \frac{2}{5} M_s R^2$. Thus,the acceleration $a_s$ is:$a_s = \frac{g \sin 	heta_s}{1 + \frac{2}{5}} = \frac{g \sin 	heta_s}{7/5} = \frac{5}{7} g \sin 	heta_s$Given that the accelerations are the same $(a_c = a_s)$:$\frac{2}{3} g \sin 	heta_c = \frac{5}{7} g \sin 	heta_s$Rearranging to find the ratio:$\frac{\sin 	heta_c}{\sin 	heta_s} = \frac{5/7}{2/3} = \frac{5}{7} 	imes \frac{3}{2} = \frac{15}{14}$

$A$ cylinder of mass $M_c$ and a sphere of mass $M_s$ are placed at points $A$ and $B$ of two inclines,respectively (See Figure). If they roll on the incline without slipping such that their accelerations are the same,then the ratio $\frac{\sin \theta_c}{\sin \theta_s}$ is

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