Surface tension of two liquids (having same densities), $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$, 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 $\mathrm{h}_{1}$ and $\mathrm{h}_{2}$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights $\mathrm{h}_{1}$ and $\mathrm{h}_{2}$ and surface tensions $T_{1}$ and $T_{2}$ satisfy the relation :

(A) $h_{1}<h_{2}$ and $T_{1}=T_{2}$

(B) $h_{1}=h_{2}$ and $T_{1}=T_{2}$

(C) $h_{1}>h_{2}$ and $T_{1}=T_{2}$

(D) $h_{1}>h_{2}$ and $T_{1}<T_{2}$

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Fluid Mechanics

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 about the lowest point of miniscus]. The heights

h_{1}

and

h_{2}

and surface tensions

T_{1}

and

T_{2}

satisfy the relation :

Answer: (A)

Neglecting the weight of the liquid above the lowest point of the meniscus, the capillary rise is

h = \frac{2 T}{ρ g r}

, so for a given liquid

h \propto \frac{1}{r}

. With equal densities and equal surface tension

T_{1} = T_{2}

, and

r_{1} > r_{2}

, the narrower tube gives the greater rise, so

h_{1} < h_{2}

. The only internally consistent pair is

h_{1} < h_{2}

with

T_{1} = T_{2}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option