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) pv2 (B) (3/4) pv2 (C) (1/2) pv2 (D) (1/4) pv2

Tardigrade Admin · Accepted Answer

Momentum per second carried by liquid per second is

ρ a v^{2}

net force due to reflected liquid =

2 \times [\frac{1}{4} ρ a v^{2}]

net force due to stopped liquid =

\frac{1}{4} ρ a v^{2}

Total force =

\frac{3}{4} ρ a v^{2}

net pressure =

\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 :

$p v^{2}$

$\frac{3}{4} p v^{2}$

$\frac{1}{2} p v^{2}$

$\frac{1}{4} p v^{2}$

Momentum per second carried by liquid per second is $ρ a v^{2}$
net force due to reflected liquid = $2 \times [\frac{1}{4} ρ a v^{2}]$
net force due to stopped liquid = $\frac{1}{4} ρ a v^{2}$
Total force = $\frac{3}{4} ρ a v^{2}$
net pressure = $\frac{3}{4} ρ v^{2}$

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 :

pv2

43​pv2

21​pv2

41​pv2