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A horizontal tube of uniform cross-secti...

A horizontal tube of uniform cross-sectional area `A` is bent in the form of `U` as shown in figure. If the liquid of density `rho` enters and leaves the tube with velcity v, then the extermal force `F` requried to hold the bend stationary is

A

`F=0`

B

`rhoAv^(2)`

C

`2rhoAv^(2)`

D

`(1)/(2) rhoAv^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

`F=(Delta p)/(Delta t)=((Delta m)/(Delta t))(Delta v)`
`=rho((Delta V)/(Delta t))(2v)`
`=(rho)(Av)(2v)=2rhoAv^(2)`.
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