A piston is pressed with a force $F$ on a hydroulic cylinder containing water $\left(\rho ={10}^3 kg/m^3\right)$. The cross-sectional area of the cylinder is $A=100 c{m}^2$. The water is forced into pipe with a cross-sectional area of $1 c{m}^2$ that reses a height $h=50 cm$ (Atmospheric pressure $P_0={10}^5 N/m^2$ )

                                                                  $P=\frac{500}{100\times {10}^{-4}}=5\times {10}^4 N/m^2$

Using Bernoulli’s equations, we have

                                                      $$\begin{gathered} P+0+0=P_a+\frac{1}{2}\rho v^2+\rho gh \\ =1\times {10}^5+\frac{1}{2}\times {10}^3\times {10}^2+{10}^3\times 10\times 0.50 \\ =1.55\times {10}^5 N/m^2 \end{gathered}$$

If $F$ is the force applied then

$$\begin{gathered} P_a+\frac{F}{A}=1.55\times {10}^5 \\ \text{or} 1\times {10}^5+\frac{F}{A}=1.55\times {10}^5 \\ \therefore F=0.55\times {10}^5\times 100\times {10}^{-4}=550 N \end{gathered}$$