Fluids · Small Oscillations

Turn buoyancy into restoring torque and find n

The hinged rod is immersed in two layers: the lower half in density 6ρ and the upper half in density 2ρ. For a small angular displacement, the buoyant forces and weight create a net restoring torque proportional to θ.

2026 Paper 1 Q9 · hinged rod in two liquids
Torque view
Small angle θ 8°
Two buoyant halves
The lower half displaces liquid of density 6ρ and the upper half displaces liquid of density 2ρ.
RelationF₁ = 3ρALg, F₂ = ρALg
Resultn = √3
Dense liquiddensity 6ρ, lower half
Light liquiddensity 2ρ, upper half
Restoring torqueopposes displacement

Derivation path

01 / BUOYANT FORCES

Each half displaces a different liquid

The lower half is in the 6ρ liquid and the upper half is in the 2ρ liquid.

F₁ = 6ρ(A L/2)g = 3ρALg, F₂ = 2ρ(A L/2)g = ρALg
02 / TORQUE ABOUT HINGE

Use lever arms L/4, 3L/4 and L/2

For small θ, each vertical force has perpendicular lever arm rθ.

τ = [3ρALg(L/4) + ρALg(3L/4) − ρALg(L/2)]θ
03 / SIMPLE HARMONIC MOTION

Compare τ with Iα

The rod has I = mL²/3 = ρAL³/3 about the hinge.

ω² = 3g/L, so T = 2π√(L/3g)
Final value
Given T = (2π/n)√(L/g), and from torque balance T = 2π√(L/3g).
Therefore n = √3.
Answer: √3