$\mathrm{P}_{0}+\rho \mathrm{gh}+1 / 2 \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{0}+1 / 2 \rho \mathrm{v}_{2}^{2}$

By continuity equation

$A_{1} v_{1}=A_{2} v_{2}$

$v_{1}=\frac{v_{2}}{2}$

$g h+1 / 2\left(\frac{v_{2}}{2}\right)^{2}=\frac{1}{2} v_{2}^{2}$

$g h=\frac{v_{2}^{2}}{2}[1-1 / 4]=3 / 8 v_{2}^{2}$

$v_{2}=\sqrt{\frac{8 g h}{3}} \Rightarrow v_{1}=\sqrt{\frac{2 g h}{3}}=\sqrt{\frac{2}{3}}$

volume flow rate $=\sqrt{\frac{2}{3}} \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}$

$=60 \sqrt{\frac{2}{3}} \times 10^{-4} \mathrm{m}^{3} / \mathrm{min}$

$=60 \sqrt{\frac{2}{3}} \times \frac{1}{10} \mathrm{litre} / \mathrm{min}$

$=4.9$ $litre/min$