JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

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 :

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 $T_{1} < T_{2}$

Solution

Capillary rise: $h = \frac{2 T cos θ}{ρ g r}$ . With the same liquid (same $T$ , $ρ$ , $θ$ ) and same $g$ , the height is inversely proportional to radius.

Since $r_{1} > r_{2}$ , we get $h_{1} < h_{2}$ . The surface tensions are equal because it's the same liquid pair condition implied by 'same densities' measurement context: $T_{1} = T_{2}$ .

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