A cylindrical container has water and closed at the top. If $v_1, v_2, v_3$ represent the efflux velocities, then

                 $0+\frac{1}{2}\rho v^2+\rho gh=P_a+\frac{1}{2}\rho v_e^2+0$

For large container $v<<v_e,$ and so can be neglected.

$\therefore v_e=\sqrt{2\left(\rho gh-P_a\right)}$

For $v_e\ge 0,$

or $\left(\rho gh-P_a\right)\ge 0$

or                                   $h\ge \frac{P_a}{\rho g}\simeq \frac{1\times {10}^5}{1000\times 10}=10m$

Thus $v_1=0$ and $v_2=0$