Water rises to a height of 9cm in a certain capillary tube. In the same tube the level of mercury surface is depressed by 3 cm . Compute the ratio of surface tension of water and mercury . Density of mercury `13.6 xx 10^(3) kg m^(-3)`. Angle of contact of water is zero and that for mercury is `135^(@)`.

Explanation

To solve the problem, we need to compute the ratio of the surface tension of water (SW) to the surface tension of mercury (SM) using the given heights and angles of contact.

### Step-by-step Solution:

1. **Identify the given values:**
- Height of water (HW) = 9 cm = 0.09 m
- Height of mercury (HM) = 3 cm = 0.03 m (depressed, so we will consider it as negative)
- Density of mercury (ρM) = 13.6 × 10^3 kg/m³
- Density of water (ρW) = 1 × 10^3 kg/m³ (approximation)
- Angle of contact for water (θ1) = 0°
- Angle of contact for mercury (θ2) = 135°

2. **Use the formula for surface tension:**
The formula for surface tension (ST) in a capillary tube is given by:
\[ ST = \frac{h \cdot \rho \cdot g}{\cos(\theta)} \]
where:
- h = height of the liquid column
- ρ = density of the liquid
- g = acceleration due to gravity (approximately 9.81 m/s²)
- θ = angle of contact

3. **Calculate the surface tension of water (SW):**
\[ SW = \frac{HW \cdot \rhoW \cdot g}{\cos(θ1)} \]
Substituting the values:
\[ SW = \frac{0.09 \cdot (1 \times 10^3) \cdot g}{\cos(0°)} = \frac{0.09 \cdot 1000 \cdot g}{1} = 0.09 \cdot 1000 \cdot g \]

4. **Calculate the surface tension of mercury (SM):**
\[ SM = \frac{HM \cdot \rhoM \cdot g}{\cos(θ2)} \]
Substituting the values (remembering HM is negative):
\[ SM = \frac{-0.03 \cdot (13.6 \times 10^3) \cdot g}{\cos(135°)} \]
Since \(\cos(135°) = -\frac{1}{\sqrt{2}}\):
\[ SM = \frac{-0.03 \cdot (13.6 \times 10^3) \cdot g}{-\frac{1}{\sqrt{2}}} = 0.03 \cdot (13.6 \times 10^3) \cdot g \cdot \sqrt{2} \]

5. **Calculate the ratio of surface tension of water to surface tension of mercury:**
\[ \frac{SW}{SM} = \frac{0.09 \cdot 1000 \cdot g}{0.03 \cdot (13.6 \times 10^3) \cdot g \cdot \sqrt{2}} \]
The \(g\) cancels out:
\[ \frac{SW}{SM} = \frac{0.09 \cdot 1000}{0.03 \cdot (13.6 \times 10^3) \cdot \sqrt{2}} \]
Simplifying:
\[ = \frac{0.09 \cdot 1000}{0.03 \cdot 13.6 \cdot 1000 \cdot \sqrt{2}} = \frac{0.09}{0.03 \cdot 13.6 \cdot \sqrt{2}} = \frac{3}{13.6 \cdot \sqrt{2}} \]

6. **Calculate the numerical value:**
\[ \sqrt{2} \approx 1.414 \Rightarrow 13.6 \cdot \sqrt{2} \approx 19.2 \]
Thus:
\[ \frac{SW}{SM} \approx \frac{3}{19.2} \approx 0.156 \]
This can be approximated to:
\[ \frac{SW}{SM} \approx \frac{2}{13} \]

### Final Answer:
The ratio of surface tension of water to surface tension of mercury is approximately \( \frac{2}{13} \).