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

Answer 1

The correct option is C $\sqrt{2}A\rho v^2$

lets $\rho =$ density of liquid

$A=$ cross sectional area

The magnitude of change in momentum,

$dP=P_f-P_i=\sqrt{2}d\Delta mv$

From the second law of newtons

$F=\frac{dP}{dt}=\sqrt{2}v\frac{dm}{dt}$. . . . .(2)

density $\rho =\frac{dm}{Adl}$

$dm=\rho Adl$

Equation (1)

$F=\sqrt{2}v\rho A\frac{dl}{dt}$

$F=\sqrt{2}\rho A{v}^2$ $(\frac{dl}{dt}=v)$

The correct option is C