JEE Mains Model Paper 2

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 =

ω2

2g

(x12−x22)
⇒ 0.25 =

ω2

2×10

|(0.6)2−(0.4)2|
⇒ ω = 5 rads−1