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

Answer 1

Answer: (C)

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