A cylindrical vessel of radius r containing a liqiud is rotating about a vertical axis through the centre of circular base, If the vessel is rotating with angular velocity `omega` then what is the difference of the heights of liquid at centre of vessel and edge?

Options

A. `(romega)/(2g)`
B. `(r^(2)omega^(2))/(2g)`
C. `sqrt(2gromega)`
D. `(omega^(2))/(2gr^(2))`

Explanation

(b) From Bernoulli's theorem,
`p_(A)+(1)/(2)dv_(A)^(2)+dgh_(A)=oe+(1)/(2)dv_(B)^(2)+dgh_(B)`
Here. `h_(A)=h_(B)`
`p_(A)=(1)/(2)dv_(A)^(2)=p_(B)+(1)/(2)dv_(B)^(2)`
`p_(A)-p_(B)=(1)/(2)d(v_(B)^(2)-v_(A)^(2))`

Now `v_(A)=o,v_(B)=romega`
`p_(A)-p_(B)=hdg`
`hdg=(1)/(2)dr^(2)omega^(2)`
or `h=(r^(2)omega^(2))/(2g)`