Surface tension of two liquids (having same densities), T₁ and T₂, are measured using capillary rise method utilizing two tubes with inner radii of r₁ and r₂ where r₁ > r₂. The measured liquid heights in these tubes are h₁ and h₂ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h₁ and h₂ and surface tensions T₁ and T₂ satisfy the relation :
Options

Question

Surface tension of two liquids (having same densities), T₁ and T₂, are measured using capillary rise method utilizing two tubes with inner radii of r₁ and r₂ where r₁ > r₂. The measured liquid heights in these tubes are h₁ and h₂ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h₁ and h₂ and surface tensions T₁ and T₂ satisfy the relation :

Options

G1 Physics · Accepted Answer

For capillary rise, h = 2T/(ρgr). For the comparison used in the given paper, the surface tension is taken to be the same in the two measurements: T₁ = T₂. The densities are also the same. Therefore, h ∝ 1/r. Given, r₁ > r₂, we obtain h₁ < h₂. Hence, h₁ < h₂ and T₁ = T₂. Correct option: A.