The following bodies,(1) a ring(2) a disc(3) | vedclass

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(C) For a body of mass $m$ and radius $R$ rolling down an inclined plane of angle $	heta$,the acceleration $a$ is given by $a = \frac{g \sin 	heta}{

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$4$

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(C) For a body of mass $m$ and radius $R$ rolling down an inclined plane of angle $	heta$,the acceleration $a$ is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$,where $k$ is the radius of gyration.The time taken to reach the bottom of an inclined plane of length $S$ is $t = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2S}{g \sin 	heta} \left(1 + \frac{k^2}{R^2}ight)}$.For the time $t$ to be minimum,the ratio $\frac{k^2}{R^2}$ must be minimum.Comparing the values of $\frac{k^2}{R^2}$ for the given bodies:$(1)$ Ring: $k^2 = R^2 \Rightarrow \frac{k^2}{R^2} = 1$$(2)$ Disc: $k^2 = \frac{R^2}{2} \Rightarrow \frac{k^2}{R^2} = 0.5$$(3)$ Solid cylinder: $k^2 = \frac{R^2}{2} \Rightarrow \frac{k^2}{R^2} = 0.5$$(4)$ Solid sphere: $k^2 = \frac{2R^2}{5} \Rightarrow \frac{k^2}{R^2} = 0.4$Since the solid sphere has the minimum value of $\frac{k^2}{R^2}$,it will have the maximum acceleration and will reach the bottom of the inclined plane first.

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