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A fire hydrant delivers water of density \mathrm{\rho} at a volume rate \mathrm{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 :

Option: 1

\rho V L


Option: 2

zero


Option: 3

2 p V L


Option: 4

\sqrt{2} \rho \mathrm{VL}


Answers (1)

best_answer

Force exerted by the water on the corner

\mathrm{ \begin{aligned} & =\text { change in momentum in } 1 \mathrm{sec} \\ & =\sqrt{2} \mathrm{mv} \\ & =\sqrt{2} \rho \mathrm{VL} \end{aligned} }

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Rakesh

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