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. $\rho v^2$
• B. $\left(3/4\right)\rho v^2$
• C. $\left(1/2\right)\rho v^2$
• D. $\left(1/4\right)\rho v^2$

Answer 1

The correct option is B

$\left(3/4\right)\rho v^2$

Step 1: Given data

$\%$ of liquid which passes unaffected $=50\%$

$\%$ of liquid which looses its momentum $=25\%$

$\%$ of liquid which come back with same speed$=25\%$

Step 2: The resultant pressure on the mesh

Momentum per second carried by liquid per second $=\rho a{v}^2\left[Where,v=speed,a=area,\rho =densityofliquid\right]$

net force due to reflected liquid $=2\times \left[\left(1/4\right)\rho a{v}^2\right]$

net force due to stopped liquid $=\left(1/4\right)\rho a{v}^2$

Total force $=2\times \left[\left(1/4\right)\rho a{v}^2\right]+\left[\left(1/4\right)\rho a{v}^2\right]$

$=(3/4)\rho a{v}^2$

Therefore, pressure, $P=\frac{force}{area}$

$$\begin{gathered} =\frac{(3/4)\rho a{v}^2}{a} \\ =(3/4)\rho v^2 \end{gathered}$$

Hence, option (B) is correct.