The time taken by an object to slide down 45 ° rough inclined plane...

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

Then, acceleration along the incline in case of smooth incline will be, ${a}_{1} = gsin\theta = \dfrac{g}{\sqrt{2}}$.

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

${a}_{2} = gsin\theta - \mu gcos\theta = \dfrac{g}{\sqrt{2}} - \dfrac{\mug}{\sqrt{2}}$.

Now as per question,

${t}_{2} = n{t}_{1}$

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

$⇒ \dfrac{g}{\sqrt{2}}{t}_{1}^{2} = \left(\dfrac{g}{\sqrt{2}} - \dfrac{\mug}{\sqrt{2}}\right){n}^{2}{t}_{1}^{2}$

$⇒ \mu = 1 - \dfrac{1}{{n}^{2}}$