Physics · Mechanical Properties of Matter

JEE Main 2026 — 21 January, Evening Shift — Question 37

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 $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ satisfy the relation :

Option A:
$\mathrm{h}_{1}<\mathrm{h}_{2}$ and $\mathrm{T}_{1}=\mathrm{T}_{2}$
Correct
Option B:
$\mathrm{h}_{1}=\mathrm{h}_{2}$ and $\mathrm{T}_{1}=\mathrm{T}_{2}$
Option C:
$\mathrm{h}_{1}>\mathrm{h}_{2}$ and $\mathrm{T}_{1}=\mathrm{T}_{2}$
Option D:
$h_{1}>h_{2}$ and $T_{1}<T_{2}$

Answer: A

Step-by-step solution

$h=\frac{2 T}{\rho g r}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ If $\mathrm{r}_{1}>\mathrm{r}_{2} \Rightarrow \mathrm{~h}_{2}>\mathrm{h}_{1}$

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy

Answer: A

Step-by-step solution

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