A liquid of density p is flowing with a speed v through a pipe of cros

1.	A liquid of density p is flowing with a speed v through a pipe of cross-sectional area a. The pipe is bent in the shape of a right angle as shown. What force should be exerted on the pipe at the corner to keep it fixed?
Answer» <P>Answer: The force exterted is $\sqrt{2}pav^2$ Explanation: Density of the liquid = p Speed = v Cross-sectional area of the PIPE = a The liquid will exert force on the pipe at the bend, this is the required force we need to calculate Mass of the liquid flowing per second through the pipe = density × area × velocity = p × a × v = pav This is in the HORIZONTAL direction Therefore, the momentum will be = $(pav)v\hat i$ = $pav^2\hat i$ After the bend, the same AMOUNT of liquid will be flowing in the downwards direction Therefore the momentum will be = $-pav^2\hat j$ Force EXERTED = Change in momentum or, $\vec F= pav^2\hat i-(-pav^2\hat j)$ $\implies \vec F= pav^2\hat i+pav^2\hat j$ Thus, the magnitude of the force = $\sqrt{2}pav^2$