Consider a small element AB of the liquid of length dx at a distance x from the axis ofrotation. Due to centripetal force, the liquid rises to height h1 in the left arm and to a height h2 in the right arm.
Pressure at A is p1 = h1 ρg Pressure at B is p2 = h2 ρg Pressure difference Δp = p1−p2 = (h1−h2)ρg If a is the cross-section area of the tube, the net force of element AB is F = aΔp = (h1−h2)aρg Mass of element dm = ρadx ∴ Centripetal force is Fe =
x1
∫
x2
dmxω2 = ω2ρa
x1
∫
x2
xdx =
1
2
ω2 αa (x12−x22) (ii) Equating (i) and (ii) we get h1−h2 =