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Question

# 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 34ρv2Let 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=pf−pi=0−ρav2t4=−ρav2t4 .....(1) For other 25% of liquid which comes back with the same speed, Δp=pf−pi=−ρav2t4−ρav2t4=−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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