Question

On the opposite sides of a wide vertical vessel filled with water two identical holes are opened, each having the cross-sectional area $S=0.50c{m}^2$. The height difference between them is equal to $\Delta h=51cm$. If the resultant force of reaction of the water flowing out of the vessel.

Answer 1

Let the velocity of water, flowing through $A$ be $v_A$ and that through $B$ be $v_B$, then discharging rate through $A=Q_A=S{v}_A$ and similarly through $B=S{v}_B$.
Now, force of reaction at $A$,
$F_A=\rho Q_A{v}_A=\rho S{v}_B^2$

Hence, the net force, $F=\rho S(v_A^2-v_B^2)$ as $\overrightarrow{F_A}\uparrow \downarrow \overrightarrow{F_B}$ (1)
Applying Bernoulli's theorem to the liquid flowing out of $A$ we get
$P_0+\rho gh=P_0+\frac{1}{2}\rho v_A^2$
and similarly at $B$
$P_0+\rho g(h+\Delta h)=P_0+\frac{1}{2}\rho v_B^2$
Hence $(v_B^2-v_A^2)\frac{\rho }{2}=\Delta h\rho g$
Thus $F=2\rho gS\Delta h=0.50N$