Question

A fire hydrant delivers water of density at a volume rate L. The water travels vertically upwards through the hydrant and then does 90 turn to emerge horizontally at speed v. The pipe and nozzle have uniform cross section throughout. The force exerted by water on the corner of the hydrant is

• A. zero
• B. $\rho vL$
• C. $\sqrt{2}\rho vL$
• D. $2\rho vL$

Answer 1

The correct option is C $\sqrt{2}\rho vL$

In time $△t$,momentum of water entering the hydrant
$\overrightarrow{p_1}=\left(pL△t\right)v\hat{j}$
Momentum of water while leaving the hydrant in time
$△t$ is $\overrightarrow{p_2}=\left(\rho L△t\right)v\left(\widehat{-}i\right)$
Change in momentum in time $△t$ is
$△\overrightarrow{p}=\overrightarrow{p_2}-\overrightarrow{p_1}=\rho L△tv(-\hat{i}-\hat{j})$
$\mid △\overrightarrow{p}\mid =\rho L△tv\sqrt{(-1{)}^2+(-1{)}^2}$
$=\sqrt{2}\rho L△tv$
Force exerted by water $F=\frac{\mid △\overrightarrow{p}\mid }{△t}$
$=\sqrt{2}\rho Lv$