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Mechanical Properties of Fluids
Question
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A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by

  • πrρv

  • πrρv2

  • πr2ρv

  • πr2ρv2

Answer

D.

πr2ρv2

Cross-sectional area A = πr2

Volume of liquid flowing per second = AV = πr2v

Mass of the liquid flowing out per second = πr2vρ

Initial momentum of liquid per second = mass of liquid flowing x speed of liquid

= πr2vρ x v = πr2v2ρ

Since the liquid does not rebound after impact, the momentum after impact is zero.

Therefore, the rate of change of momentum = πr2v2ρ

According to Newton's second law, the force exerted on wall = rate of change of momentum

=  πr2ρv2

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