Explanation

(b) [image:5_image236.emf]
Let $L$ be the length and $H$ be height of the inclined plane respectively
Acceleration of the block slide down the smooth incline plane is
$a = g\cos{60{^\circ}}$
$\therefore L = \frac{1}{2}g\cos{60{^\circ}}t_{1}^{2}\ \ \ \ \ \lbrack\because u = 0\rbrack$ …(i)
Acceleration of another block dropped vertically down from the same inclined plane is
$a = g$
$\therefore H = \frac{1}{2}at_{2}^{2} = \frac{1}{2}gt_{2}^{2}$ $\lbrack\because u = 0\rbrack$
From figure,
$\cos{60{^\circ}} = \frac{H}{L} \Rightarrow H = L\cos{60{^\circ}}$
$\therefore L\cos{60{^\circ}} = \frac{1}{2}gt_{2}^{2}$ …(ii)
Divide (i) by (ii), we get
$\frac{t_{1}^{2}\cos{60{^\circ}}}{t_{2}^{2}} = \frac{1}{\cos{60{^\circ}}}$
$\frac{t_{1}^{2}}{t_{2}^{2}} = \frac{1}{\cos^{2}{60{^\circ}}} = \frac{4}{1} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{2}{1}$