A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base .If the redius of the vessel is r and angular velocity of rotation is `omega`, then the difference in the heights of the liquid at the centre of the vessel and the edge is .....

Options

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

Answer

b

Explanation

According to Bernoulli.s equation,
`P_(A)+(1)/(2)rhov_(A)^(2)+rhogh_(A)`
`=P_(B)+(1)/(2)rhov_(B)^(2)+rhogh_(B)`
Here , `h_(A)=h_(B)`
`thereforeP_(A)+(1)/(2)rhov_(A)^(2)`
`=P_(B)+(1)/(2)rhov_(B)^(2)`
`thereforeP_(A)-P_(B)=(1)/(2)rho[v_(B)^(2)-v_(A)^(2)]`
Now ,`v_(A)=0,v_(B)-romegaandP_(A)-P_(B)=rhog`
`thereforehrhog=(1)/(2)rhor^(2)omega^(2)orh=(r^(2)omega^(2))/(2g)`