There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm=105 N/m2) , then the velocity of water flowing from hole is:

Pressure at the bottom of tank

$P = D = 3 \times 10^{5} N / m^{2} =$ Atmospheric pressure + Pressure of liquid columns

$= P_{0} + h ρ g$

Pressure due to liquid column = Pressure difference between bottom and top

$P - P_{0} = P_{1} = 3 \times 10^{5} - 1 \times 10^{5} = 2 \times 10^{5} = h ρ g$

And velocity of water, $v = \sqrt{2 g h}$

or $v = \sqrt{\frac{2 P_{1}}{ρ}} = \sqrt{\frac{2 \times 2 \times 10^{5}}{10^{3}}} = \sqrt{400} m / s$

There is a hole in the bottom of tank having water. If total pressure at bottom is $3 a t m$ $(1 a t m = 10^{5} N / m^{2})$ , then the velocity of water flowing from hole is: