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Fluid Mechanics

JEE Main level, single correct, medium

94 of 192

Question 94 of 192

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 $r_{1}\gt r_{2}$ . 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 surfaces tensions $T_{1}$ and $T_{2}$ satisfy the relation:

Show answer and solution

Answer: (D) $h_{1} \lt h_{2} \text { and } T_{1}=T_{2}$

For capillary rise, the height of liquid in a tube is given by:

h = \frac{2T\cos\theta}{\rho g r}

, where

T

is surface tension and

r

is the tube radius.

For two liquids with equal density and contact angles:

h_1 = \frac{2T_1\cos\theta}{\rho g r_1}

and

h_2 = \frac{2T_2\cos\theta}{\rho g r_2}

Taking the ratio:

\frac{h_1}{h_2} = \frac{T_1}{T_2} \times \frac{r_2}{r_1}

Since

r_1 \gt r_2

, we have

\frac{r_2}{r_1} \lt 1

. The capillary rise is inversely proportional to tube radius, so the smaller radius tube will have greater height. Since the liquids are measured to have heights

h_1

and

h_2

with

r_1 \gt r_2

, and they have the same density, the empirical relation shows that both liquids rise to different heights inversely proportional to their radii. For liquids with the same density measured in different tubes, if the heights and radii satisfy the capillary rise equation, the surface tensions must be equal:

T_1 = T_2

, and

h_1 \lt h_2

since

r_1 \gt r_2

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