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Question

A water pipe has an internal diameter of 10 cm. Water flows through it at the rate of 20 m/sec. The water jet strikes normally on a wall and falls dead. Find the force on the wall (in newtons). (round off to the nearest integer)
[ρw=1000 kg/m3]

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Solution

Mass of water flowing through the tube per second m=Avρ
where A= area of cross section and v= velocity of water of density ρ.

Momentum change/second for water striking the wall =m(0v)=Avρv=Av2ρ

We know that, force acting on the jet is equal to the rate of change of momentum [Newton's 2nd law].

& Force exerted by the jet on the wall is equal and opposite to this [Newton's 3rd law].

Thus, the force on the wall =Av2ρ
=π×(5100)2×(20)2×1000=3141.593142 N

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