A block starts moving up an inclined plane of inclination 30 ° with an initial velocity of   v 0 .…

$\mathrm{A} \mathrm{to} \mathrm{B}$

${\mathrm{a}}_1=\mathrm{g} \sin 30°+\mathrm{\mu g} \cos 30°$

$=\frac{\mathrm{g}}{2}+\frac{\mathrm{\mu g}\sqrt{3}}{2} ; \mathrm{g}=10 \mathrm{m} {\mathrm{s}}^2$

${\mathrm{v}}_0^2-2{\mathrm{a}}_1\left(\mathrm{s}\right)=0$

$\mathrm{s}=\frac{{\mathrm{v}}_0^2}{{\mathrm{a}}_1}$    .....(i)

$\mathrm{B} \mathrm{to} \mathrm{A}$

${\mathrm{a}}_2=\frac{\mathrm{g}}{2}-\frac{\mathrm{\mu }\sqrt{3}}{2}\mathrm{g}$

${\left(\frac{{\mathrm{V}}_0}{2}\right)}^2=2{\mathrm{a}}_2\left(\mathrm{s}\right)$

$\mathrm{s}=\frac{{\mathrm{V}}_0^2}{4{\mathrm{a}}_2}$  .....(ii)

From equation (i) and (ii)

$\frac{{\mathrm{V}}_0^2}{{\mathrm{a}}_1}=\frac{{\mathrm{V}}_0^2}{4{\mathrm{a}}_2}$

$\Rightarrow {\mathrm{a}}_1=4{\mathrm{a}}_2$

$\Rightarrow 5+5\sqrt{3}\mathrm{\mu }=4\left\{5-5\sqrt{3}\mathrm{\mu }\right\}$

$\Rightarrow 5+5\sqrt{3}\mathrm{\mu } \Rightarrow 25 \sqrt{3}\mathrm{\mu }=15 \Rightarrow \mathrm{\mu }=\frac{\sqrt{3}}{5}=0.346=\frac{346}{1000}$

So, $\frac{1}{1000}=\frac{346}{1000}$