Fluid Mechanics

JEE Main level, single correct, medium

68 of 192

Question 68 of 192

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

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Answer: (B) 34ρv2\frac{3}{4} \rho v^{2}

Mass per unit time of the liquid = \rhoav Momentum per second carried by liquid =ρav×v=\rho a v \times v

Net force due to bounced back liquid, F1=2×[14ρav2]\mathrm{F}_{1}=2 \times\left[\frac{1}{4} \rho \mathrm{av}^{2}\right]

Net force due to stopped liquid, F2=14\mathrm{F}_{2}=\frac{1}{4} pav 2^{2} Total force,

F=F1+F2=12ρav2+14pav2=34pav2\mathrm{F}=\mathrm{F}_{1}+\mathrm{F}_{2}=\frac{1}{2} \rho \mathrm{av}^{2}+\frac{1}{4} \mathrm{pav}^{2}=\frac{3}{4} \mathrm{pav}^{2}

Net pressure =34ρv2=\frac{3}{4} \rho v^{2}

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