There is a hole in the bottom of tank having water. If total pressure at bottom is $3atm$ $\left(1atm={10}^{5}N/m^2\right)$ , 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

$=P_0+h\rho g$

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

$P-P_0=P_1=3\times {10}^5-1\times {10}^5=2\times {10}^5=h\rho g$

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

or $v=\sqrt{\frac{2P_1}{\rho }}=\sqrt{\frac{2\times 2\times {10}^5}{{10}^3}}=\sqrt{400}m/s$