Question

A liquid of density $\rho$ 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. $\frac{1}{4}\rho v^2$
• B. $\frac{3}{4}\rho v^2$
• C. $\frac{1}{2}\rho v^2$
• D. $\rho v^2$

Answer 1

The correct option is B $\frac{3}{4}\rho v^2$

$\text{Mass per unit time of the liquid}=\rho \pi a^{2}v$
$\text{Momentum per second carried by liquid}\frac{p}{t}=\rho \pi a^{2}v\times v$

Net force due to $25\%$ bounce back liquid,
$F_1=\frac{1}{4}\frac{2p}{t}=\left[\frac{1}{2}\rho \pi a^2{v}^2\right]$

Net force due to $25\%$ stopped liquid,
$F_2=\frac{1}{4}\frac{p}{t}=\frac{1}{4}\rho \pi a^2{v}^2$

Total force,
$F=F_1+F_2=\frac{1}{2}\rho \pi a^2{v}^2+\frac{1}{4}\rho \pi a^2{v}^2$

$\Rightarrow F=\frac{3}{4}\rho \pi a^2{v}^2$

$\text{Net pressure},P=\frac{F}{A}=\frac{3}{4}\frac{\rho \pi a^2{v}^2}{\pi a^2}$

$\Rightarrow P=\frac{3}{4}\rho v^2$

Hence, option $\left(B\right)$ is correct.