A block takes t time to slide down a plane inclined at 45° to the...

Let the length of the inclined plane be $s$.

For the smooth inclined plane, the acceleration of the block is $a_1 = g \sin \theta$.

The time taken is $t_1 = \dfrac{t}{2}$. Using the second equation of motion $s = ut + \dfrac{1}{2}at^2$ with $u = 0$:

$s = \dfrac{1}{2} (g \sin \theta) \left(\dfrac{t}{2}\right)^2 = \dfrac{1}{8} g \sin \theta t^2$

For the rough inclined plane, the acceleration of the block is $a_2 = g(\sin \theta - \mu \cos \theta)$.

The time taken is $t_2 = t$. Using the equation of motion again:

$s = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

Equating the two expressions for $s$:

$\dfrac{1}{8} g \sin \theta t^2 = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

$\dfrac{1}{4} \sin \theta = \sin \theta - \mu \cos \theta$

Given $\theta = 45^{\circ}$, we have $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$. Substituting these values:

$\dfrac{1}{4} = 1 - \mu$

$\mu = 1 - \dfrac{1}{4} = \dfrac{3}{4} = 0.75$

We are given $\mu = \dfrac{\alpha}{100}$.

$\dfrac{\alpha}{100} = 0.75 \Rightarrow \alpha = 75$

Answer: $75$