Water (density $\rho$ ) is flowing through the uniform horizontal tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)

 

Given
Density of water $=\rho$
Cross-sectional area = A
Constant speed = v
Along x and y directions along in momentum of water is
$\left|\overline{\Delta }\overline{P}x\right|mv\sin {60}^{\circ }=\frac{\sqrt{3}}{2}mv$
 
$\begin{array}{l}\left|\overline{\Delta }\overline{P}x\right|=\frac{mv}{2}+mv=\frac{\sqrt{3}}{2}mv\\ \Rightarrow \left|\overline{\Delta }\overline{P}net\right|=\sqrt{\Delta P_x^2+\Delta P_y^2}=\sqrt{(\frac{9}{4}+\frac{3}{4})mv}\\ \Rightarrow \left|\overline{\Delta }\overline{P}net\right|=\sqrt{3}mv\end{array}$
Force on bend is rate of change of momentum
$\Rightarrow \left|\overline{\Delta }\overline{P}net\right|=\sqrt{3}\left(\frac{dm}{dt}\right).v=\sqrt{3}\rho A{v}^2(\because \frac{dm}{dt}=\rho Av)$