Initially at $t = 0\,;\,h = h$
$t = t\,;h = 0$
$Then,\,A\left( { - \frac{{dh}}{{dt}}} \right) = \pi {a^2}.v$
$\left[ {Velocity\,of\,efflux\,of\,liquid\,v = \sqrt {2gh} } \right]$
Integrating both sides
$\int\limits_0^t {dt = - \frac{A}{{\sqrt {2g} \pi {a^2}}}{{\int\limits_h^0 h }^{ - 1/2}}dh} $
$\left[ t \right]_0^t = - \frac{A}{{\sqrt {2g} \pi {a^2}}} \cdot \left[ {\frac{{{h^{1/2}}}}{{1/2}}} \right]_h^0$
$t = \frac{{\sqrt 2 A}}{{\pi {a^2}}}\sqrt {\frac{h}{g}} $

| Column - $\mathrm{I}$ | Column - $\mathrm{II}$ |
| $(a)$Rain drops moves downwards with constant velocity. | $(i)$ Viscous liquids |
| $(b)$ Floating clouds at a height in air. | $(ii)$ Viscosity |
| $(iiii)$ Less density |
(Given density of air $=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$ )