Now, $\frac{1}{2} a_{1} t^{2}=\frac{1}{2} a_{2}(2 t)^{2} \Rightarrow a_{2}=\frac{a_{1}}{4}$         $...(i)$

Clearly $\quad \mathrm{a}_{1}=\frac{\mathrm{mg} \sin \theta}{\mathrm{m}}=\mathrm{g} \sin \theta$            $...(ii)$

and $a_{2}=\frac{m g \sin \theta-\mu m g \cos \theta}{m}=g \sin \theta-\mu g \cos \theta$      $...(iii)$

From $( i ),$ $(ii)$ and $(iii),$

we get $\mu=0.75$