A pipe is running full of water. At a point A it tapers from $\text{4}00{\text{cm}}^{\text{2}}$  cross-sectional area to $\text{2}00{\text{cm}}^{\text{2}}$  at other point B. If the pressure difference between A and B is 30cm of water column. The rate of volume of water through the pipes is $\left[g=10m/s^2\right]$

$$\begin{gathered} A_1{v}_1=A_2{v}_2 \\ 400v_1=200v_2 \\ {v}_2=2v_1 \end{gathered}$$

 $$\begin{gathered} p_1-p_2=\frac{\rho }{2}(v_2^2-v_1^2) \\ h\rho g=\frac{\rho }{2}(v_2^2-v_1^2) \end{gathered}$$                                         

$30\times {10}^{-2}\times {10}^3\times 10=\frac{1000}{2}\left({\left(2v_1\right)}^2-v_1^2\right)$                                   

$3\times {10}^3=500\left(4v_1^2-v_1^2\right)\Rightarrow 3v_1^2=\frac{3000}{500}$                                                

$v_1^2=\frac{1000}{500}\Rightarrow v_1=\sqrt{2}$                                                                

Rate of flow $=A_1{v}_1=400\times {10}^{-4}\times \sqrt{2}$  $=56.56\times {10}^{-3}=56.5$ liters