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: (a) h₁ > h₂ and T₁ = T₂ (b) h₁ h₂ and T₁ < T₂ (d) h₁ = h₂ and T₁ = T₂

Answer: (b). Capillary rise h = (2Tcos θ)/(rρ g), so (h₁)/(h₂) = (T₁)/(T₂)·(r₂)/(r₁) (same ρ). If T₁ = T₂: (h₁)/(h₂) = (r₂)/(r₁) r₂, so h₁ h₂ is impossible. This rules out (c). So the answer is (b) h₁ < h₂ and T₁ = T₂.

Surface tension of two liquids (having same densities), T₁ and T₂… | JEE Main 2026 (21 Jan, Shift 2)

Q13JEE Main 2026 · 21 JAN · SHIFT 2Medium

Angle of Contact & Capillarity

Surface tension of two liquids (having same densities), $T_1$ and $T_2$ , are measured using capillary rise method utilizing two tubes with inner radii of $r_1$ and $r_2$ where $r_1 > r_2$ . The measured liquid heights in these tubes are $h_1$ and $h_2$ respectively. [Ignore the weight of the liquid above the lowest point of miniscus]. The heights $h_1$ and $h_2$ and surface tensions $T_1$ and $T_2$ satisfy the relation:

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Answer: (b)

Capillary rise $h = \dfrac{2T\cos\theta}{r\rho g}$ , so $\dfrac{h_1}{h_2} = \dfrac{T_1}{T_2}\cdot\dfrac{r_2}{r_1}$ (same $\rho$ ).
If $T_1 = T_2$ : $\dfrac{h_1}{h_2} = \dfrac{r_2}{r_1} < 1$ since $r_1 > r_2$ , so $h_1 < h_2$ . This rules out (a) and (d).
If $T_1 < T_2$ : $\dfrac{h_1}{h_2} < 1$ again, so $h_1 > h_2$ is impossible. This rules out (c).
So the answer is (b) $h_1 < h_2$ and $T_1 = T_2$ .

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Angle of Contact & Capillarity question from JEE Main 2026 (21 Jan, Shift 2)

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