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(D) For a body rolling down an inclined plane,the acceleration of the center of mass is given by:$a_{cm} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$Fo

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(D) For a body rolling down an inclined plane,the acceleration of the center of mass is given by:$a_{cm} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$For a solid sphere,the moment of inertia is $I = \frac{2}{5} MR^2$.Substituting this into the acceleration formula:$a_{cm} = \frac{g \sin 	heta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + 2/5} = \frac{5}{7} g \sin 	heta$The time taken to travel a distance $L$ along the incline is given by $L = \frac{1}{2} a_{cm} t^2$,which implies $t = \sqrt{\frac{2L}{a_{cm}}}$.Substituting $a_{cm}$:$t = \sqrt{\frac{2L}{\frac{5}{7} g \sin 	heta}} = \sqrt{\frac{14L}{5g \sin 	heta}}$As seen from the expression,the time $t$ depends only on the length of the incline $L$,the angle of inclination $	heta$,and the acceleration due to gravity $g$. It is independent of the mass $M$,radius $R$,and density $ho$ of the sphere. Thus,the time is independent of all these factors.

Two solid spheres of different mass,radii,and density roll down a rough inclined plane under identical situations. Their time to come down is independent of their ............

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