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 % looses all of its momentum and 25 % comes back with the same speed. The resultant pressure on the mesh will be:

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 % looses all of its momentum and 25 % comes back with the same speed. The resultant pressure on the mesh will be: (A) (1/2)ρ v2 (B) ρ v2 (C) (3/4)ρ v2 (D) (1/4)ρ v2

Tardigrade Admin · Accepted Answer

Mass per unit time =ρ Av Force due to momentum loss =(1/4)ρ Av× v Force due to bounce back =(1/4)ρ Av× 2v Pressure =((ρ A v2/4) + (ρ A v2/2)/A)=(3/4)ρ v2

Q. 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. $25%$ looses all of its momentum and $25%$ comes back with the same speed. The resultant pressure on the mesh will be:

$\frac{1}{2} ρ v^{2}$

$ρ v^{2}$

$\frac{3}{4} ρ v^{2}$

$\frac{1}{4} ρ v^{2}$

Mass per unit time $= ρ A v$
Force due to momentum loss $= \frac{1}{4} ρ A v \times v$
Force due to bounce back $= \frac{1}{4} ρ A v \times 2 v$
Pressure $= \frac{\frac{ρ A v ^{2}}{4} + \frac{ρ A v ^{2}}{2}}{A} = \frac{3}{4} ρ v^{2}$

Q. 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% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

21​ρv2

ρv2

43​ρv2

41​ρv2