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, $25\%$ loses all of its momentum and $25\%$ of the liquid comes back with the same speed. The resultant pressure on the mesh will be

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

Answer 1

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

Let us suppose $m$ is the total mass of the liquid coming out of the hose pipe in time $t$.


So, $m=\rho V$
$\Rightarrow m=\rho avt$
Now it's momentum, $p=mv=\rho a{v}^{2}t$.

For $25\%$ of liquid which losses all of its momentum
$\Delta p=p_f-p_i=0-\frac{\rho a{v}^{2}t}{4}=-\frac{\rho a{v}^{2}t}{4}.....\left(1\right)$

For other $25\%$ of liquid which comes back with the same speed,
$\Delta p=p_f-p_i=-\frac{\rho a{v}^{2}t}{4}-\frac{\rho a{v}^{2}t}{4}=-\frac{\rho a{v}^{2}t}{2}....\left(2\right)$

For remaining 50% of liquid which passes unaffected through the mesh,
$\Delta p=0........\left(3\right)$

From Eqs.(1), (2) and (3)
Total change in momentum $=-\frac{\rho a{v}^{2}t}{4}+(-\frac{\rho a{v}^{2}t}{2})+0$
$=\frac{-3\rho a{v}^{2}t}{4}$

So, Force on mesh $F=\frac{|\Delta p|}{\Delta t}$
$\Rightarrow F=\frac{\frac{3\rho a{v}^{2}t}{4}}{t}=\frac{3\rho a{v}^2}{4}$
Now, Pressure on the mesh
$p=\frac{F}{A}$
$\Rightarrow p=\frac{\frac{3\rho a{v}^2}{4}}{a}$
$\Rightarrow p=\frac{3\rho v^2}{4}$
Hence, option $\left(d\right)$ is the correct answer.