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(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a ring, $k^2 = R^2$, so

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a ring, $k^2 = R^2$, so $a_R = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.For a disc, $k^2 = \frac{R^2}{2}$, so $a_D = \frac{g \sin 	heta}{1 + 0.5} = \frac{2g \sin 	heta}{3}$.The distance traveled along the incline is $d = \frac{h}{\sin 	heta}$.Using $d = \frac{1}{2} a t^2$, we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin 	heta}}$.For the ring: $t_R = \sqrt{\frac{2h}{(g \sin 	heta / 2) \sin 	heta}} = \sqrt{\frac{4h}{g \sin^2 	heta}}$.For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin 	heta / 3) \sin 	heta}} = \sqrt{\frac{3h}{g \sin^2 	heta}}$.Given $	heta = 60^{\circ}$, $\sin 	heta = \frac{\sqrt{3}}{2}$, so $\sin^2 	heta = \frac{3}{4}$.$t_R = \sqrt{\frac{4h}{g(3/4)}} = \sqrt{\frac{16h}{3g}}$ and $t_D = \sqrt{\frac{3h}{g(3/4)}} = \sqrt{\frac{4h}{g}}$.Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$, we have $\sqrt{\frac{h}{g}} \left( \frac{4}{\sqrt{3}} - 2 ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{\frac{h}{10}} \left( \frac{4-2\sqrt{3}}{\sqrt{3}} ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} ight) = 2-\sqrt{3} \Rightarrow \sqrt{h} = \frac{\sqrt{3}}{2} \Rightarrow h = \frac{3}{4} = 0.75 \,m$.

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