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(C) The acceleration of an object rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a given height and ang

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Sphere,disc,hollow sphere,ring

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(C) The acceleration of an object rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a given height and angle,the time taken to reach the bottom is $t = \sqrt{\frac{2s}{a}}$,where $s$ is the distance along the incline.Thus,$t \propto \sqrt{1 + \frac{I}{MR^2}}$.The object with the smallest moment of inertia $(I)$ coefficient will have the largest acceleration and reach the bottom first.The coefficients of $I$ for the objects are:$1$. Sphere: $I = \frac{2}{5}MR^2 = 0.4 MR^2$$2$. Disc: $I = \frac{1}{2}MR^2 = 0.5 MR^2$$3$. Hollow sphere: $I = \frac{2}{3}MR^2 \approx 0.67 MR^2$$4$. Ring: $I = MR^2 = 1.0 MR^2$Since the order of coefficients is $0.4 < 0.5 < 0.67 < 1.0$,the order of reaching the bottom is Sphere,Disc,Hollow sphere,Ring.

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

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