A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides, if the radius of vessel is ` 0.05 m` and the speed of rotation is `2 rev//s`, find difference in the height of the liquid at the centre of the vessel and its sides.

Explanation

Consider a particle at position (x,y) as shown in Fig. 7.71 . For its equilibrium along y-axis
`R "cos" theta = "mg" ……. (i)`
And for its circular motion in horizontal plane
R sin `theta = mx omega^(2)` ……… (ii)
So , dividing Eqn . (ii) by (i)
tan `theta = (omega^(2) //g) x ` ....... (iii)
But from differential calculus we know that
slope = `(dy)/(dx) = "tan" theta` ....... (iv)
So , from Eqns . (iii) and (iv)
`(dy)/(dx) = (omega^(2))/(g) x,` i.e., `" " dy = (omega^(2))/(g) ` x dx
which on integration yields
`y = (omega^(2))/(g) - (x^(2))/(2) + C`
But as at x = 0 , y = 0 , C = 0
`therefore " " y = (omega^(2))/(2g) x^(2) `....... (v)
Which is the equation of a parabola . So , the surface of revolution is a paraboloid and for a point on the rim y = h and x = r . So ,
`h = (omega r)^(2)//2g ` ......... (vi)
Here , `omega = (2 pi xx 2//1) = 4 pi` rad/s
and `r = 0.05 `m
So , `h = ((4pi)^(2) xx (0.5)^(2))/(2 xx 9.8) = 2 xx 10^(-2) m = 2 cm `