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 r omega)`
D. `(omega^2)/(2 g r^2)`

Answer

B

Explanation

`P_A + (1)/(2) rho v_A^2 + rho gh_A = P_B + (1)/(2) rho v_B^2 + rho gh_B`
`h_A = h_B`
`P_A - P_B = (1)/(2) rho (v_B^2 - v_A^2)`
`v_A = 0, v_B = r omega, P_A - P_B = rho gh`
`rho gh = (1)/(2) rho r^2 omega^2`
`h = (r^2 omega^2)/(2 g)`.
.