Explanation

(b) Let $p_{0}$ is the atmospheric pressure, $\rho$ the density of liquid and$v$ the velocity at which water is coming out. Applying this Bernoulli’s theorem just inside and outside the hole.
$P_{inside} + \rho gh + 0 = P_{outside} + \frac{{\rho v}^{2}}{2}$
$\Longrightarrow \ \ \ \ \ P_{0} + \ \rho gh = P_{0} + \frac{{\rho v}^{2}}{2}$
$\Longrightarrow \ \ \ \ \ v = \sqrt{2gh}$
$= \sqrt{2 \times 10 \times 20}$
$= 20\ {ms}^{- 1}$