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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
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The correct answer is:
$\pi r^2 \rho v^2$
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 $=\pi r^2 \rho v^2$
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 $=\pi r^2 \rho v^2$
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