Question

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

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

Cross-sectional area $A=\pi r^2$,
Volume of liquid flowing per second$=AV=\pi r^{2}v$
Mass of the liquid flowing out per second $=\pi r^{2}v\rho$
Initial momentum of liquid per second
$=$ mass of liquid flowing $\times$ Speed of liquid
$=\pi r^{2}v\rho \times v=\pi r^2{v}^2\rho$
Since the liquid does not rebound after impact, the momentum after impact is zero.
$\therefore$ Rate of change of momentum $=\pi r^2{v}^2\rho$
According to Newton's second law, the force exerted on wall$=$ rate of change of momentum $=\pi r^2\rho v^2$.