A disc and a sphere of same radius but differ | vedclass

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(A) The time $t$ taken by a body to roll down an inclined plane of length $l$ and inclination $	heta$ without slipping is given by:$t = \sqrt{\frac{2

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(A) The time $t$ taken by a body to roll down an inclined plane of length $l$ and inclination $	heta$ without slipping is given by:$t = \sqrt{\frac{2l(1 + \frac{k^2}{R^2})}{g \sin 	heta}}$where $k$ is the radius of gyration and $R$ is the radius of the object.Since $l$,$	heta$,and $g$ are the same for both,the time depends on the factor $(1 + \frac{k^2}{R^2})$.For a disc,$I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.5 = 1.5$.For a solid sphere,$I = \frac{2}{5}MR^2$,so $k^2 = \frac{2}{5}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.4 = 1.4$.Since $1.4 Therefore,the sphere reaches the bottom first.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

$A$ disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

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