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Figure shows as thin tube open at one en...

Figure shows as thin tube open at one end rotating with constant angular speed `omega`. The other end of the tube has a small opening. Tube contains a fluid of density `rho` which effuses with speed `u` relative to the tube. Let `rho_(A)` be the pressure at the point in the tube just behind the opening. Assume the atmospheric pressure to be `p_(0)` and pick correct option(s) for the instant when `h=l/2`

A

`p_(A)=p_(0)+(rho omega^(2)l^(2))/4`

B

`p_(A)=p_(0)+3/8rho omega^(2) l^(2)`

C

`u=(omega l)/2`

D

`u=(sqrt(3))/2 omegal`

Text Solution

Verified by Experts

The correct Answer is:
B, D

By Newton's second law,
`(p_(A)-p_(0))s=rhoshomega^(2)(l-h/2)`
`impliesp_(A)=p_(0)+rho omega^(2)h(l- h/2)`
By Bernoulli's equation
`u=omegasqrt(2h(l-h/2))`
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