Two capillary tubes of same diameter are kept vertically one each in two liquids whose relative densities are 0.8 and 0.6 and surface tensions are 60 and 50 dyne/cm respectively. Ratio of heights of liquids in the two tubes h1/h2 is

The correct option is 9/10.

The height to which a liquid rises in a capillary tube is given by the formula:

$h=\frac{2Tcos\theta }{r\rho g}$

where:
- $h$ is the height of the liquid column,
- $T$ is the surface tension of the liquid,
- $\theta$ is the angle of contact (assumed to be the same for both liquids),
- $r$ is the radius of the capillary tube,
- $\rho$ is the density of the liquid, and
- $g$ is the acceleration due to gravity.

Since the two capillary tubes have the same diameter, the radius $r$ is identical for both. The acceleration due to gravity $g$ and the contact angle term $cos\theta$ are also constant. Therefore, the height $h$ is directly proportional to the surface tension $T$ and inversely proportional to the density $\rho$:

$h\propto \frac{T}{\rho }$

This gives the relation for the ratio of the heights in the two tubes as:

$\frac{h_1}{h_2}=\left(\frac{T_1}{T_2}\right)\times \left(\frac{{\rho }_2}{{\rho }_1}\right)$

Given values from the problem statement:
- Relative density of the first liquid, ${\rho }_1=0.8$
- Relative density of the second liquid, ${\rho }_2=0.6$
- Surface tension of the first liquid, $T_1=60\text{dyne/cm}$
- Surface tension of the second liquid, $T_2=50\text{dyne/cm}$

Substituting these values into the ratio equation:

$\frac{h_1}{h_2}=\left(\frac{60}{50}\right)\times \left(\frac{0.6}{0.8}\right)$

Simplifying the fraction calculation:

$\frac{h_1}{h_2}=\left(\frac{6}{5}\right)\times \left(\frac{3}{4}\right)$

$\frac{h_1}{h_2}=\frac{18}{20}=\frac{9}{10}$

Thus, the ratio of the heights of the liquids in the two tubes is 9/10.