Explanation

(a) Volume of water in the vessel of base area $A'$ and height $h$ is$V = A'h$. Average velocity of out flowing water when height of water changes from $h$ to 0 is
$v = \frac{\sqrt{2\ gh} + 0}{2} = \frac{\sqrt{2gh}}{2}$
$\therefore V = A\ v\ t$ …(i)
When vessel is filled to height 4 $h$, then volume in vessel
$= 4V = 4A\ vt = 4A\frac{\sqrt{2gh}}{2} \times t\ $
If $t$ is the time taken for the out flowing liquid and $v_{1}$ is the average velocity of out flowing liquid then
$4V = A\ v_{1}t_{1}$
or $t_{1} = \frac{4\ V}{A{\ v}_{1}} = \frac{4\ A\sqrt{2gh} \times t \times 2}{2 \times A \times \sqrt{2g \times 4h}} = 2t$