case is $a_2$

Now $\quad \frac{1}{2} a_{1} t^{2}=\frac{1}{2} a_{2}(2 t)^{2}$

$\Rightarrow \quad a_{2}=\frac{a_{1}}{4}$        $...(i)$

Clearly $a_{1}=\frac{\operatorname{mg} \sin \theta}{m}=g \sin \theta \quad$        $...(ii)$

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

$\mu=0.75$