JEE Advanced Model Paper 2 with solutions for online practice

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 =

ω2

2g

(x12−x22) ⇒ 0.25 =

ω2

2×10

[(0.6)2 - (0.4)2]
⇒ ω = 5 rad s−1