Question

A stream of a liquid of density $\rho$ flowing horizontally with a speed $v$ gushes out of a tube of radius $r$ and hits at a vertically wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of liquid is given by

• A. $\pi r\rho v$
• B. $\pi r\rho v^2$
• C. $\pi r^2\rho v$
• D. $\pi r^2\rho v^2$

Answer 1

Answer: (D)

$\pi r^2\rho v^2$

area of pipe $A-\pi r^2$ and velocity $vm/s$

let the cross section be length $vm$ crossed by water in $1sec$.

Therefore volume is $V=\pi r^{2}v$

We know density is mass divided by volume

$\rho =\frac{m}{V}$

$m=\rho \pi r^{2}v$

Force is change in momentum $F=ma=m\times \frac{v}{t}$

$F=\rho \pi r^2{v}^2$