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Question

(B) The acceleration $a$ 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,$k^2 = R^2$,s

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(B) The acceleration $a$ 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,$k^2 = R^2/2$,so $a_D = \frac{g \sin 	heta}{1+0.5} = \frac{2g \sin 	heta}{3}$.The distance to be covered 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: $t_R = \sqrt{\frac{2h}{(g \sin 	heta / 2) \sin 	heta}} = \sqrt{\frac{4h}{g \sin^2 	heta}} = \sqrt{\frac{4h}{g (3/4)}} = \sqrt{\frac{16h}{3g}}$.For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin 	heta / 3) \sin 	heta}} = \sqrt{\frac{3h}{g \sin^2 	heta}} = \sqrt{\frac{3h}{g (3/4)}} = \sqrt{\frac{4h}{g}}$.Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$,we have $\sqrt{\frac{h}{g}} \left( \sqrt{\frac{16}{3}} - 2 ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{\frac{h}{10}} \left( \frac{4 - 2\sqrt{3}}{\sqrt{3}} ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} ight) = 2-\sqrt{3}$.$\sqrt{h} = \frac{\sqrt{3}}{2} \implies h = \frac{3}{4} = 0.75 \ m$.

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