Consider a small element AB of the liquid of length dx at a distance x fromthe axis of rotationDue to centripetal force, the liquid rises to height h1 in the left ann and to aheight h2 in the right arm
Pressure at A is p1 = h1ρg Pressure at B is p2 = h2ρg Pressure difference Δp = p1−p2 = (h1−h2)ρg If a is the cross - sectional area of the tube, the net force on element AB is F = a Δp = (h1−h2) aρg (i) Mass of element dm = ρadx ∴ Centipetal force is Fe =
x1
∫
x2
dmxω2 = ω2ρa
x1
∫
x2
xdx =
1
2
ω2ρa(x12−x22) ... (ii) Equating (i) and (ii), we get h1−h2 =