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
ρVL
B
zero
C
2ρVL
D
2ρVL

Solution

The correct option is C $$\sqrt { 2 } \rho VL$$Change in momentum in vertical direction in time t = momentum final - momentum initial in vertical direction = $$- \rho LV t$$Change in momentum in horizontal direction in time t =  momentum final - momentum initial in horizontal direction = $$\rho LV t$$So net change in momentum in time t = $\sqrt{(\rho LVt)^{2} + (\rho LVt)^{2}} = \sqrt{2} \rho LVt$Force = Change in momentum per time = $$\sqrt{2} \rho LV$$Answer DPhysics

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