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(C) Let $t_a$ be the time of ascent and $t_d$ be the time of descent. Given $t_a = \frac{1}{2} t_d$,which implies $t_d = 2 t_a$.Since the distance $s$

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$3$

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(C) Let $t_a$ be the time of ascent and $t_d$ be the time of descent. Given $t_a = \frac{1}{2} t_d$,which implies $t_d = 2 t_a$.Since the distance $s$ covered is the same,$s = \frac{1}{2} a_a t_a^2 = \frac{1}{2} a_d t_d^2$.Substituting $t_d = 2 t_a$,we get $a_a t_a^2 = a_d (2 t_a)^2$,so $a_a = 4 a_d$.The acceleration during ascent is $a_a = g \sin 	heta + \mu g \cos 	heta = g \sin 30^{\circ} + \mu g \cos 30^{\circ} = \frac{g}{2} + \frac{\sqrt{3}}{2} \mu g$.The acceleration during descent is $a_d = g \sin 	heta - \mu g \cos 	heta = g \sin 30^{\circ} - \mu g \cos 30^{\circ} = \frac{g}{2} - \frac{\sqrt{3}}{2} \mu g$.Substituting these into $a_a = 4 a_d$: $\frac{g}{2} + \frac{\sqrt{3}}{2} \mu g = 4 (\frac{g}{2} - \frac{\sqrt{3}}{2} \mu g)$.Dividing by $g$: $\frac{1}{2} + \frac{\sqrt{3}}{2} \mu = 2 - 2\sqrt{3} \mu$.$\frac{5\sqrt{3}}{2} \mu = \frac{3}{2} \implies \mu = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$.Comparing $\mu = \frac{\sqrt{3}}{5}$ with $\frac{\sqrt{x}}{5}$,we get $x = 3$.

$A$ body of mass $m$ is launched up on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$. If the time of ascent is half of the time of descent,the value of $x$ is ..... .

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