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)

$\sqrt{2}a\rho v^{2}$

Answer 2

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 $\Delta p = \sqrt{p_{1}^{2} +p_{2}^{2}} = \sqrt{2} mv \left(\because p_{1 } p_{2}\right)$ $F = \frac{\Delta p}{\Delta t} = \frac{\sqrt{2}mv}{\Delta t}$ $\sqrt{2} mv =\frac{ \sqrt{2}\left(a\rho v\right)v}{1}$ $= \sqrt{2} a\rho v^{2}$