The area of cross-section of a large tank is 0 . 5 m ^ 2 . It has a...

Pressure at point $1$ is ${P}_{1} = {P}_{0}$, Pressure at point $2$ is ${P}_{2} = {P}_{0} + \dfrac{mg}{A}$

Applying Bernoulli's equation at top and bottom

${P}_{1} + \dfrac{1}{2}ρ{{v}_{1}}^{2} + ρg{h}_{1} = {P}_{2} + \dfrac{1}{2}ρ{{v}_{2}}^{2} + ρg{h}_{2} ⇒ {P}_{0} + \dfrac{1}{2}ρ{v}^{2} + 0 = {P}_{0} + \dfrac{mg}{A} + 0 + ρgh$

$⇒ \dfrac{250}{5} + {10}^{3} \times 10 \times 0 . 4 = \dfrac{1}{2} \times {10}^{3} \times {v}^{2}$

$v = 3m{s}^{- 1} = 300cm{s}^{- 1}$