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

Answer

B

Explanation

From Bernoulli's theorem, `P_(A)+1/2dv_(A)^(2)+dgh_A=P_(B)+1/2dv_(H)^(2)=P_(H)+1/2dv^2H+dgh_(H)`
Here `h_(A)=H_(B)`
`therefore 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)=0, v_(B)=romega` and `P_(A)-P_(B)=hdg`
`therefore hdg=1/2 dr(2)omega^(2)or h=(r^(2)omega^(2))/(2g)`