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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 

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A
ρVL
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B
zero
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C
2ρVL
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D
2ρVL
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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 = 

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

Answer D

Physics

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