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 ** 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 ** of oscillations is given by