\({p_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2\)

\(\therefore v_2^2 - v_1^2 = 2gh\)                        \(...\left( 1 \right)\)

According to the equation of continuty

\({A_1}{v_1} = {A_2}{v_2}\)                              \(...(2)\)

\(\frac{{{A_1}}}{{{A_2}}} = \frac{{6\,m{m^2}}}{{10\,m{m^2}}}\)

\(From\,equation\,\left( 2 \right),\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = \frac{6}{{16}}\)

\(or,\,\,{v_2} = \frac{6}{{10}}{v_1}\)

Putting this value of \(v_2\) in equation \((1)\)

\({\left( {\frac{6}{{10}}{v_1}} \right)^2} - {\left( {{v_1}} \right)^2} = 2 \times {10^3} \times 5\)

                                   \(\left[ \begin{array}{l}
g = 10m/{s^2} = {10^3}cm/{s^2}\\
and\,h = 5\,cm
\end{array} \right]\)

\(Solving\,we\,get\,{v_1} = \frac{{10}}{8}\)

Therefore the rate at which water flows through the tube

\( = {A_1}{v_1} = {A_2}{v_2} = \frac{{6 \times 10}}{8} = 7.5cc/s\)