Question

A fire hydrant delivers water of density $\rho$ at a volume rate $L$. The water travels vertically upwards 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

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


Given volume rate is $L$
Mass of the water crossing in time $t$ $m=\rho Lt$

From the diagram above,
Change in momentum in vertical direction $=-mv=\rho Ltv$
Change in momentum in horizontal direction $=mv=\rho Ltv$
$\therefore$ Total change in momentum $=\sqrt{(-mv{)}^2+(mv{)}^2}=\sqrt{2}mv=\sqrt{2}\rho Ltv$

Force exerted by the water on the corner = change in momentum per unit time$=\sqrt{2}\rho vL$