The time taken by an object to slide down 45 ° rough inclined plane is n times as it takes to slide…

Let angle of inclination is $\theta =45°$.

Then, acceleration along the incline in case of smooth incline will be, $a_1=g \sin \theta =\frac{g}{\sqrt{2}}$.

In case of rough incline, kinetic friction will act against the direction of motion. Normal force acting on the object will be $mg\cos \theta$. Therefore, acceleration along the incline in case of rough incline will be,

$a_2=g \sin \theta -\mu g \cos \theta =\frac{g}{\sqrt{2}}-\frac{\mu g}{\sqrt{2}}$.

Now as per question,

$t_2=n{t}_1$

and as the displacement covered is the same in both case, $0+\frac{1}{2}{a}_1{t}_1^2=0+\frac{1}{2}{a}_2{t}_2^2$

$\Rightarrow \frac{g}{\sqrt{2}}{t}_1^2=\left(\frac{g}{\sqrt{2}}-\frac{\mu g}{\sqrt{2}}\right)n^2{t}_1^2$

$\Rightarrow \mu =1-\frac{1}{n^2}$