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(A) For a smooth inclined plane,the acceleration is $a_1 = g \sin 	heta$. The distance $s$ is given by $s = \frac{1}{2} a_1 t^2 = \frac{1}{2} (g \sin

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$\frac{3}{4}	an 	heta $

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(A) For a smooth inclined plane,the acceleration is $a_1 = g \sin 	heta$. The distance $s$ is given by $s = \frac{1}{2} a_1 t^2 = \frac{1}{2} (g \sin 	heta) t^2$. For a rough inclined plane,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The distance $s$ is $s = \frac{1}{2} a_2 (2t)^2 = \frac{1}{2} g(\sin 	heta - \mu \cos 	heta) (4t^2)$. Equating the two expressions for $s$: $\frac{1}{2} g \sin 	heta t^2 = 2 g(\sin 	heta - \mu \cos 	heta) t^2$. Dividing by $g t^2$: $\frac{1}{2} \sin 	heta = 2(\sin 	heta - \mu \cos 	heta)$. $\frac{1}{4} \sin 	heta = \sin 	heta - \mu \cos 	heta$. $\mu \cos 	heta = \sin 	heta - \frac{1}{4} \sin 	heta = \frac{3}{4} \sin 	heta$. Therefore,$\mu = \frac{3}{4} 	an 	heta$.

$A$ body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough,the time taken now is $2t$. The coefficient of friction of the rough surface is

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