JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Elasticity & SHM

A tank contains two immiscible liquids of densities

6 ρ

and

2 ρ

. The higher density liquid is filled up to a height

L /2

from the bottom. A thin rod of density

ρ

and length

L

is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium, the time period of small oscillations is

\frac{2 π}{n} \frac{L}{g}

, where

g

is the acceleration due to gravity. The value of

n

is:

Answer: 1.73

Solution

For small angular displacement

θ

, the buoyant forces are:

From lower liquid (density $6 ρ$ ): $F_{B 1} \approx 6 ρ \frac{L}{2} Ag$ acting at $\frac{L}{4}$ from hinge
From upper liquid (density $2 ρ$ ): $F_{B 2} \approx 2 ρ \frac{L}{2} Ag$ acting at $\frac{3 L}{4}$ from hinge

Net restoring torque:

τ = ρ AgL^{2} [\frac{3}{4} + \frac{3}{4} - \frac{1}{2}] θ = ρ AgL^{2} θ

Moment of inertia:

I = \frac{ρ LAL ^{2}}{3}

Angular acceleration:

α = - \frac{3 ρ AgL ^{2}}{ρ AL ^{3}} θ = \frac{- 3 g}{L} θ

Time period:

T = 2 π \frac{L}{3 g} = \frac{2 π}{3} \frac{L}{g}

Therefore,

n = 3 = 1.732 \approx 1.73

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Elasticity & SHM

Solution

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