Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.5cm in the same capillary tube. If the density of mercury is 13.6 gm/cc and its angle of contact is 135° and density of water is 1 gm/cc and its angle of contact is 0° , then the ratio of surface tensions of the two liquids is (cos135° = 0.7)

To find the ratio of the surface tensions of water and mercury, we can use the capillary rise formula. The height $h$ to which a liquid rises (or falls) in a capillary tube of radius $r$ is given by:
$h=\frac{2T\cos \theta }{r\rho g}$
where:
- $T$ is the surface tension of the liquid,
- $\theta$ is the angle of contact,
- $\rho$ is the density of the liquid,
- $g$ is the acceleration due to gravity, and
- $r$ is the radius of the capillary tube.

Since both liquids are in the same capillary tube, the radius $r$ and the acceleration due to gravity $g$ are constant. Therefore, we can express the surface tension $T$ as proportional to:
$T\propto \frac{h\rho }{\cos \theta }$

Let us write down the given values for water (denoted with subscript 1) and mercury (denoted with subscript 2):
For water:
- Height of rise, $h_1=10\text{cm}$
- Density of water, ${\rho }_1=1\text{g/cm}{3}^{}$
- Angle of contact, ${\theta }_1=0°$, so $\cos {\theta }_1=\cos 0°=1$

For mercury:
- Depth of fall, $h_2=3.5\text{cm}$ (we consider the magnitude of depression)
- Density of mercury, ${\rho }_2=13.6\text{g/cm}{3}^{}$
- Angle of contact, ${\theta }_2=135°$, so $\left|\cos {\theta }_2\right|=|\cos 135°|=0.7$ (using the magnitude)

Now, we find the ratio of their surface tensions $\frac{T_1}{T_2}$:
$\frac{T_1}{T_2}=\frac{h_1}{h_2}\times \frac{{\rho }_1}{{\rho }_2}\times \frac{\cos {\theta }_2}{\cos {\theta }_1}$

Substitute the values into the equation:
$\frac{T_1}{T_2}=\frac{10}{3.5}\times \frac{1}{13.6}\times \frac{0.7}{1}$

Let's simplify this expression:
$\frac{T_1}{T_2}=\frac{10}{3.5}\times 0.7\times \frac{1}{13.6}$
Since $\frac{0.7}{3.5}=\frac{1}{5}$, we get:
$\frac{T_1}{T_2}=\frac{10}{5}\times \frac{1}{13.6}$
$\frac{T_1}{T_2}=2\times \frac{1}{13.6}$
(multiplying numerator and denominator by 10)
$\frac{T_1}{T_2}=\frac{20}{136}=\frac{5}{34}$ (dividing both by 4)

Thus, the ratio of the surface tensions of the two liquids is 5:34.