Explanation

(b) The various forces acting on the block are as shown
From Newton’s law
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$mg\sin{\theta - f = ma\ \ \ldots.(i)}$
Where $f$ is frictional force and $a$ the acceleration downwards.
Since, there is no motion perpendicular to surface, we have
$R - mg\cos{\theta = 0}$
$\Longrightarrow \ \ R = mg\cos{\theta\ \ \ \ldots.(ii)}$
Also, $f = \mu R = \mu\ mg\cos\theta$
Putting the value in Eq. (i) we get
$mg\sin{\theta - \mu\ mg\cos{\theta = ma}}$
$\Longrightarrow \ \ a = g\sin{\theta - \mu\ g\cos\theta}$
Now, velocity at bottom
$v^{2} = u^{2} - 2as$
Since, $v = 0$
$\therefore u = \sqrt{2as}$
$Given,\ \ s = l,\ \ a = g\sin{\theta - g\mu\cos\theta}$
$\therefore\ \ u = \sqrt{2l(g\sin{\theta - g\mu\cos\theta})}$
$u = \sqrt{2gl(\sin{\theta - \mu\cos\theta}})$