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

Let us assume that,

T_A be the time of ascent of body and TD be the time of descent of body.

From second equation of motion, we have

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

During the time of ascent, the distance covered by the body,

$\mathrm{s}=\frac{1}{2}{\mathrm{a}}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}^2$                 ...(i)

and during the time of descent, the distance covered by the body,

$\mathrm{S}=\frac{1}{2}{\mathrm{a}}_{\mathrm{D}}{\mathrm{t}}_{\mathrm{D}}^2$             ...(ii)

$\therefore$ From Eqs. (i) and (ii), we get

$\begin{array}{l}\frac{1}{2}{\mathrm{a}}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}^2=\frac{1}{2}{\mathrm{a}}_{\mathrm{D}}{\mathrm{t}}_{\mathrm{D}}^2\\ \Rightarrow {\mathrm{a}}_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}^2={\mathrm{a}}_{\mathrm{D}}{\mathrm{t}}_{\mathrm{D}}^2\Rightarrow \frac{{\mathrm{t}}_{\mathrm{A}}^2}{{\mathrm{t}}_{\mathrm{D}}^2}=\frac{{\mathrm{a}}_{\mathrm{D}}}{{\mathrm{a}}_{\mathrm{A}}}\\ \Rightarrow \frac{{\mathrm{t}}_{\mathrm{A}}^2}{{\mathrm{t}}_{\mathrm{D}}^2}=\frac{\mathrm{gsin}\mathrm{\theta }-\mathrm{\mu gcos}\mathrm{\theta }}{\mathrm{gsin}\mathrm{\theta }+\mathrm{\mu gcos}\mathrm{\theta }}\\ \Rightarrow \frac{{\mathrm{t}}_{\mathrm{A}}}{{\mathrm{t}}_{\mathrm{D}}}=\sqrt{\frac{\mathrm{gsin}\mathrm{\theta }-\mathrm{\mu gcos}\mathrm{\theta }}{\mathrm{gsin}\mathrm{\theta }+\mathrm{\mu gcos}\mathrm{\theta }}}=\sqrt{\frac{\mathrm{gsin}{30}^{\circ }-\mathrm{\mu gcos}{30}^{\circ }}{\mathrm{gsin}{30}^{\circ }+\mathrm{\mu gcos}{30}^{\circ }}}\\ \Rightarrow \frac{1}{2}=\sqrt{\frac{1-\sqrt{3}\mathrm{\mu }}{1+\sqrt{3}\mathrm{\mu }}}\left\{\because \frac{{\mathrm{t}}_{\mathrm{A}}}{{\mathrm{t}}_{\mathrm{D}}}=\frac{1}{2}\left(\text{ given }\right)\right\}\\ \\ \\ \Rightarrow \mathrm{\mu }=\frac{\sqrt{3}}{5} ....\left(\mathrm{iii}\right)\end{array}$

According to question, the coefficient of friction between the body and plane is $\frac{\sqrt{\mathrm{x}}}{5}$, therefore, on comparing it with Eq. (iii), we can write x = 3.