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 above the lowest point of miniscus]. The heights h₁ and h₂ and surface tensions T₁ and T₂ satisfy the relation :

Solution

Physics sir · 2 steps

Use Jurin's law: hr is fixed by T. Test each option and keep the only one consistent with r₁ > r₂.

1. Capillary rise. Ignoring the liquid above the bottom of the meniscus, the rise is h = (2T cos θ)/(rρ g). The densities are the same, and we take the same contact angle for both, so (2 cos θ)/(ρ g) is common.

   T₁/T₂ = (h₁ r₁)/(h₂ r₂)

   Option	With r₁ > r₂, the heights give	Option claims	Consistent?
   (a) h₁ > h₂	h₁ r₁ > h₂ r₂, so T₁ > T₂	T₁ = T₂	No
   (b) h₁ < h₂	h₁ r₁ can equal h₂ r₂	T₁ = T₂	Yes
   (c) h₁ > h₂	h₁ r₁ > h₂ r₂, so T₁ > T₂	T₁ < T₂	No
   (d) h₁ = h₂	h₁ r₁ > h₂ r₂, so T₁ > T₂	T₁ = T₂	No

2. So. With equal surface tensions, the product hr is the same in both tubes, so the wider tube has the lower column: h₁ < h₂ and T₁ = T₂. The answer is (b).

The idea to keep. In capillary rise, hr = (2T cos θ)/(ρ g). For the same liquid properties, a narrower tube always lifts the liquid higher. Height and radius are inversely related.

Answer from the NTA final answer key. Solution by PrepWiser.