A block takes \(t\) time to slide down a plane inclined at \(45°\) to the horizontal. If the surface is made smooth (frictionless), the block takes time \(\dfrac{t}{2}\) to slide down the plane. The coefficient of friction between the block and the inclined plane is \(\left(\dfrac{\alpha}{100}\right)\). The value of \(\alpha\) is __________.

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: \(75\)