A container has two immiscible liquids of densities ρ1 and ρ2 (ρ2 > ρ1). A capillary of radius r is inserted in the liquid so that its bottom reaches upto the denser liquid. The denser liquid rises in the capillary and attains a height equal to h, which is equal to the column length of the lighter liquid. Assuming zero contact angle, find the surface tension of the heavier liquid?

Engineering

Physics

Buoyancy

Surface Tension and Surface Energy

Question

A container has two immiscible liquids of densities ρ₁ and ρ₂ (ρ₂ > ρ₁). A capillary of radius r is inserted in the liquid so that its bottom reaches upto the denser liquid. The denser liquid rises in the capillary and attains a height equal to h, which is equal to the column length of the lighter liquid. Assuming zero contact angle, find the surface tension of the heavier liquid?

2πr(ρ₂ – ρ₁)gh

$\frac{r}{2} (ρ_{2} - ρ_{1}) gh$

2πrρ₂gh

$\frac{{rρ}_{2} gh}{2}$

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When a capillary tube is inserted through a lighter liquid (density ρ₁) into a denser liquid (density ρ₂), the denser liquid rises. The capillary rise h is supported by the surface tension T of the denser liquid. The net upward force due to surface tension is $2 π r T$ . This force balances the weight of the column of fluid it is lifting. The effective weight of the fluid column is the weight of the denser liquid of height h minus the buoyant force from the displaced lighter liquid, which is $π r^{2} h g (ρ_{2} - ρ_{1})$ . Setting the forces equal gives the surface tension.

Force balance: $2 π r T = π r^{2} h g (ρ_{2} - ρ_{1})$

Solving for T: $T = \frac{r h g (ρ_{2} - ρ_{1})}{2}$

Final Answer: $\frac{r}{2} (ρ_{2} - ρ_{1}) g h$