Rotating Vessel

Let, consider a point on the curved surface of water.

Let , the mass of the point is m.

The force acted on the mass m is F which makes an angle ¢ with the vertical

So that Fcos¢=mg

$Fsin¢=mw^2x$

$tan¢=w^2x/g$

$dy/dx=w^2x/g$

$Y=w^2×x^2/2g$ [integrating both side]

$Y=5m{w=5,x=2,g=10}$

The height of the highest point from the base is =5+1=6m

Formula for the question is

Y= $\frac{R^{2}w^{2}}{2g}$ + $\y_{min}$ Substitute the given values in formula you get y minimum is 1 metre in the question

dy/dx=w^2x/2g.THEN integrate it.

Applying Bernoulli's equation considering two points : one at the vertex of parabola and one at the edge of the tube where water is at highest position.

You will get $h' = 5 m, h' \rightarrow$ height with respect to vertex of parabola.

Means from base it is $\boxed{5 + 1 = 6}$

H(max)=h(min)+ (w^2).)R^2)/2g : g is acceleration due to gravity= 10m/s -->H = 1+ (5x5)(2x2)/2x10 ---> = 6

At first find the pressure on both the ends of the tube. Substrac two equations to get this formula H-h=w^2/2g(R^2). put the values and get the answer!