Question

A fire hydrant deliver water of density $\rho$ at a volume rate $L$. The water travels vertically upward through the hydrant and then does ${90}^{\circ }$ turn to emerge horizontally at speed $V$. The pipe and nozzle have uniform cross section throughout. The force exerted by the water on the corner of the hydrant is

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

Answer 1

Answer: (D)

$\sqrt{2}\rho VL$

Change in momentum in vertical direction in time t = momentum final - momentum initial in vertical direction = $-\rho LVt$

Change in momentum in horizontal direction in time t = momentum final - momentum initial in horizontal direction =$\rho LVt$

So net change in momentum in time t =

Force = Change in momentum per time = $\sqrt{2}\rho LV$

Answer D