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A horizontally oriented bent thin tube A...

A horizontally oriented bent thin tube `ABC` of length `1.4 m` rotates with a constant angular velocity `omega` about a stationary vertical axis `OO'` passing through the end `A` as shown in figure. The portion `BC` of tube is parallel to `OO'` and its length is equal `0.4 m`. At `t=0`, `AB` part of tube is filled with an ideal fluid. The end `A` of tube is open, the closed end `C` has a very small orifice. Find the maximum height reached by fluid relative to end `C`. (take `g=10//s^(2)` and `omega=10 rad//s`)

Text Solution

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Let any time `t` fluid just fills the portion `BC` and comes out with velocity `v`.
`l=1m`, `x=0.4m`
In rotating frame, centrifugal force on portion `A'B`,
`F=rhoAint_(x)^(l)omega^(2)r.dr=(rhoAomega^(2)(l^(2)-x^(2)))/(2)`
Due to pressure difference difference between `C` and `A'`
`DeltaP=(1)/(2)rho omega^(2)(l^(2)-x^(2)-rhogx)`
Now `(1)/(2)rhovV^(2)=DeltaP=(rhoomega^(2))/(2)(l^(2)-x^(2))-rhogx`....`(1)`
From Bermoulli theorem,
Let it reaches upto a height `h`, then,
`rhogh=(1)/(2)rhov^(2)=(rhoomega^(2)(l^(2)-x^(2)))/(2)-rhogx`
`rhoxx10xxh=rhoxx((10)^(2))/(2)(1^(2)-(.4)^(2))-.4xx10xxrhorArrh=3.8m`.
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Knowledge Check

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