Question

Liquid of density $\rho$ flows along a horizontal pipe of uniform area of cross section $a$ with a velocity $v$ through a right angled bend. What force should be applied to the bend to hold it in equilibrium?

• A. $2a\rho v^2$
• B. $a\rho v^2\sqrt{2}$
• C. $\sqrt{2}a\rho v^2$
• D. $a\rho v^2$

Answer 1

Answer: (C)

Mass of liquid flowing/sec, $M=a\rho v$
On either side of bend, $p_1=p_2=mv$
As turning is through $90^{\circ }$, therefore
$\Delta p=\sqrt{p_1^2+p_2^2}=\sqrt{2}mv\left(\because p_1{p}_2\right)$
$F=\frac{\Delta p}{\Delta t}=\frac{\sqrt{2}mv}{\Delta t}$
$\sqrt{2}mv=\frac{\sqrt{2}\left(a\rho v\right)v}{1}$
$=\sqrt{2}a\rho v^2$