Q99 / 180 · JIPMER 2017 · Physics
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$\pi r^2 \rho v$
- Correct answerD$\pi r^2 \rho v^2$
Solution
Cross-sectional area $A=\pi r^2$,
Volume of liquid flowing per second $=A V=\pi r^2 V$
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 $=\pi r^2 \rho v^2$
Hence, the correct answer is option: 'D'.