Fluid Mechanics

JEE Main level, single correct, medium

81 of 192

Question 81 of 192

A pressure-pump has a horizontal tube of cross-sectional area 10cm210{cm}^{2} for the outflow of water at a speed of 20ms−120m{s}^{-1}. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given : density of water =1000kgm−3=1000kg{m}^{-3}]

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Answer: (D) 400N400N

Volume of water coming out of the pipe per sec =Av=10×10−4×20=0.02m3s−1=Av=10\times {10}^{-4}\times 20=0.02{m}^{3}{s}^{-1}
Mass of water hitting the wall per second m=0.02×103=20kgs−1m=0.02\times {10}^{3}=20kg{s}^{-1}
As the water hits the wall normal to the wall and stops, the change of momentum of water hitting the wall is =m(0−(−v))=mv=m\left(0-\left(-v\right)\right)=mv
=20×20=400kgms−2=20\times 20=400kgm{s}^{-2}
Since, the change of momentum obtained is per unit time, it is same as the force.
Thus, the force acting on the wall by water == force experienced by water jet=400N=400N.

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