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Question

A rubber pipe with a diameter of 10cm is connected to a nozzle 2cm in diameter. Water flowing through the pipe at a speed of 0.6ms−1 comes out like a jet through the nozzle. The backward force of the nozzle is about:

A
7.7N
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B
67.9N
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C
Zero
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D
2.8N
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Solution

The correct option is B 67.9N
Given : Diameter of pipe , D=10cm=0.1m ,
diameter of nozzle , d=2cm=0.02m ,
velocity through the pipe ,v1=0.6m/s ,
Let ,velocity through the nozzle is v2 .
We have , equation of continuity ,
A1v1=A2v2 ,
π(D/2)2×0.6=π(d/2)2v2 ,
or v2=(0.05)2×0.6(0.01)2=15m/s ,
hence , backward force exerted by nozzle ,
F=mΔv ,
where , m= mass flow rate ,
Δv= change in velocity ,
now, F=ρA1v1(150.6) ,(ρ=1000kg/m3, density of water) ,
or F=1000×π×(D/2)2×0.6×14.4 ,
or F=1000×3.14×(0.1/2)2×0.6×14.4=67.9N

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