Water is flowing through a horizontal pipe of variable cross - section. The rate of volume flow is 0.2 lit/sec. The pressure is 30, 000 Pa at a point where the area of cross - section is $5\times {10}^{-5}{\mathrm{m}}^2$. The pressure at the point where the area of cross - section is 10^-4 Pa is

Given,

$\mathrm{Rate} \mathrm{of} \mathrm{volume} \mathrm{flow} = 0.2\mathrm{Lit}/\sec =0.2\times {10}^{-3} {\mathrm{m}}^3/\sec$

$$\begin{gathered} P_1=3000Pa ; \\ {A}_1=5\times {10}^{-5}{m}^2 ; \\ {A}_2={10}^{-4}{m}^2 \end{gathered}$$

By Bernoulli's theorem,

$\mathrm{P}+\frac{1}{2}{\mathrm{\rho V}}^2+\mathrm{\rho gh}=\mathrm{constant}$

The pipe will now have zero potential energy because it is horizontal. 

$\Rightarrow \mathrm{P}+\frac{1}{2}{\mathrm{\rho V}}^2=\mathrm{constant}$

$\therefore {\mathrm{P}}_1+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_1}^2={\mathrm{P}}_2+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_2}^2\mathrm{\_\_\_\_\_\_\_}\left(1\right)$

We also know that the area times the velocity equals the flow rate.

$Q=A_1{v}_1=A_2{v}_2$

substituting the value in equation (1),

$P_1+\frac{1}{2}\rho {\left(\frac{Q}{A_1}\right)}^2=P_2+\frac{1}{2}\rho {\left(\frac{Q}{A_2}\right)}^2$

$P_2=P_1+\frac{1}{2}{Q}^2\rho \left(\frac{1}{{A_1}^2}-\frac{1}{{A_2}^2}\right)$

${\mathrm{P}}_2=30000+\frac{1}{2}\times {(0.2\times {10}^{-3})}^2\left(1000\right)\left(\frac{1}{{\left(5\times {10}^{-5}\right)}^2}-\frac{1}{{\left(10\times {10}^{-5}\right)}^2}\right)=36000 \mathrm{Pa}$