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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2019 · 11 January · Morning Shift · Physics · Mechanical Properties of Fluids

A liquid of density is coming out of a hose pipe of radius a with...

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:

Options

  1. A14ρv2\frac{1}{4} \rho v^{2}
  2. B34ρv2\frac{3}{4} \rho v^{2}
  3. C12ρv2\frac{1}{2} \rho v^{2}
  4. Dρv2\rho v^{2}

Correct answer

B. 34ρv2\frac{3}{4} \rho v^{2}

Step-by-step solution

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×\mathrm{F}_{1}=2 \times\left[\frac{1}{4} \rho \mathrm{av}^{2}\right]Netforceduetostoppedliquid,Net force due to stopped liquid,\mathrm{F}_{2}=\frac{1}{4}pavpav^{2}Totalforce,Total force,\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}NetpressureNet pressure=\frac{3}{4} \rho v^{2}
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