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A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected, 25% loses all of its momentum and 25% of the liquid comes back with the same speed. The resultant pressure on the mesh will be

A
ρv2
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B
12ρv2
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C
14ρv2
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D
34ρv2
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Solution

The correct option is D 34ρv2
Let us suppose m is the total mass of the liquid coming out of the hose pipe in time t.


So, m=ρV
m=ρavt
Now it's momentum, p=mv=ρav2t.

For 25% of liquid which losses all of its momentum
Δp=pfpi=0ρav2t4=ρav2t4 .....(1)

For other 25% of liquid which comes back with the same speed,
Δp=pfpi=ρav2t4ρav2t4=ρav2t2 ....(2)

For remaining 50% of liquid which passes unaffected through the mesh,
Δp=0 ........(3)

From Eqs.(1), (2) and (3)
Total change in momentum =ρav2t4+(ρav2t2)+0
=3ρav2t4

So, Force on mesh F=|Δp|Δt
F=3ρav2t4t=3ρav24
Now, Pressure on the mesh
p=FA
p=3ρav24a
p=3ρv24
Hence, option (d) is the correct answer.

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