$\begin{array}{1 1}0.25\\0.50\\0.75\\1.0\end{array} $

Answer : 0.75

The accelerations of the block sliding down a smooth and rough $45^{\large\circ}$ inclined planes respectively are

$a_1=g\sin 45^{\large\circ}=\large\frac{g}{\sqrt 2}$

$a_2=g(\sin 45^{\large\circ}-\mu_k\cos 45^{\large\circ})=\large\frac{g}{\sqrt 2}$$(1-\mu_k)$

Where $\mu_k$ is the coefficient of kinetic friction.

Now,we know that the square of the time of slide is inversely proportional to the acceleration.

Therefore $\large\frac{t_2^2}{t_1^2}=\frac{a_1}{a_2}=\frac{1}{1-\mu_k}$

Since $t_2=2t_1$,we have

$4=\large\frac{1}{1-\mu_k}$

$\mu_k=0.75$

Hence the correct answer is 0.75

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