Question

Two non-viscous, incompressible and immiscible liquids of densities ρ and 1.5 ρ are poured into two limbs of a circular tube of radius R and small cross-section kept fixed in a vertical plane as shown in Fig. 1.24. Each liquid occupies one-fourth the circumference of the tube. (a) Find the angle θ that the radius vector to the interface makes with the vertical in equilibrium position. (b) If the whole liquid is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations.

Answer 1

(a) Since each liquid occupies one-fourth the circumference of the tube, 

The pressure P₁ at D due to liquid on the left limb is

P₁ = (R – R sinθ) 1.5 ρg

The pressure P₂ at D due to liquid on the right limb is

P₂ = (R – R cos θ) 1.5 ρg + (R sin θ + R cos θ)ρg

At equilibrium P₁ = P₂. Thus, we have 

(1 – sin θ) 1.5 = (1 – cos θ) 1.5 + sin θ + cos θ 

Solving this equation, we get 2.5 sin θ = 0.5 cos θ, 

(b) When the liquid is given a small upward displacement y = BB′ in the right limb [Fig. 1.25 (b)], then y = Rα where α = ∠B′OB, and A goes to A′ and C goes to C′. The pressure difference at D is

Thus, Restoring force = – 2.55 ρgy × A, 

where A is the area of cross-section of the tube. Mass of the liquid in the tube is

The acceleration of the liquid column is

which shows that the motion is simple harmonic. The time period of oscillations is given by