Water rises up in a glass capillary upto a height of `9.0cm`, while mercury falls down by `3.4 cm` in the same capillary. Assume angles of contact for water glass and mercury glass `0^(@)` and `135^(@)` respectively. Determine the ratio of surface tension of mercury and water `(cos 135^(@)=-0.71)`.

Explanation

To solve the problem of determining the ratio of surface tension of mercury and water, we will use the formula for capillary rise and fall in a capillary tube.

### Step-by-Step Solution:

1. **Understand the Problem**:
- Water rises in a glass capillary tube to a height of \( h_1 = 9.0 \, \text{cm} \).
- Mercury falls in the same capillary tube to a height of \( h_2 = -3.4 \, \text{cm} \) (negative because it falls).
- The angle of contact for water with glass is \( \theta_1 = 0^\circ \) and for mercury with glass is \( \theta_2 = 135^\circ \).

2. **Use the Capillary Rise and Fall Formula**:
The height \( h \) in a capillary tube is given by:
\[ h = \frac{2S \cos \theta}{\rho g R} \]
where:
- \( S \) is the surface tension,
- \( \theta \) is the angle of contact,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( R \) is the radius of the capillary tube.

3. **Set Up the Equations**:
For water (rising):
\[ h_1 = \frac{2S_1 \cos \theta_1}{\rho_1 g R} \]
For mercury (falling):
\[ h_2 = \frac{2S_2 \cos \theta_2}{\rho_2 g R} \]

4. **Divide the Two Equations**:
\[ \frac{h_1}{h_2} = \frac{S_1 \cos \theta_1}{S_2 \cos \theta_2} \cdot \frac{\rho_2}{\rho_1} \]

5. **Rearranging for Surface Tension Ratio**:
\[ \frac{S_1}{S_2} = \frac{h_1}{h_2} \cdot \frac{\cos \theta_2}{\cos \theta_1} \cdot \frac{\rho_1}{\rho_2} \]

6. **Substituting Known Values**:
- \( h_1 = 9.0 \, \text{cm} \)
- \( h_2 = -3.4 \, \text{cm} \) (take absolute value for calculation)
- \( \theta_1 = 0^\circ \) (thus \( \cos 0^\circ = 1 \))
- \( \theta_2 = 135^\circ \) (given \( \cos 135^\circ = -0.71 \))
- \( \rho_1 = 1 \, \text{g/cm}^3 \) (density of water)
- \( \rho_2 = 13.6 \, \text{g/cm}^3 \) (density of mercury)

Now substituting these values:
\[ \frac{S_1}{S_2} = \frac{9.0}{-3.4} \cdot \frac{-0.71}{1} \cdot \frac{1}{13.6} \]

7. **Calculate the Ratio**:
- First calculate \( \frac{9.0}{-3.4} \approx -2.647 \)
- Then calculate \( \frac{-0.71}{1} = -0.71 \)
- Finally, calculate \( \frac{1}{13.6} \approx 0.07353 \)

Thus:
\[ \frac{S_1}{S_2} = -2.647 \cdot -0.71 \cdot 0.07353 \approx 0.135 \]

8. **Final Ratio**:
Therefore, the ratio of surface tension of mercury to water is:
\[ \frac{S_2}{S_1} \approx 7.23 \]

### Conclusion:
The ratio of surface tension of mercury to that of water is approximately \( 7.23 \).