Calculate the ratio of surface tension of mercury and water if in a capillary tube, mercury is depressed by 3.45 cm and water rises by 9.8 cm in the same tube. The angle of contact for mercury is `140^@` and that for water is `0^@.`

Answer

To calculate the ratio of the surface tension of mercury (\( \sigma_M \)) to that of water (\( \sigma_W \)), we can use the capillary rise and depression formulas for both liquids.

### Step-by-Step Solution:

1. **Understand the Capillary Action Formula**:
The height of liquid in a capillary tube is given by the formula:
\[ h = \frac{2 \sigma \cos \theta}{\rho g R} \]
where:
- \( h \) = height (or depression) of the liquid column,
- \( \sigma \) = surface tension of the liquid,
- \( \theta \) = contact angle,
- \( \rho \) = density of the liquid,
- \( g \) = acceleration due to gravity,
- \( R \) = radius of the capillary tube.

2. **Set Up the Equations for Mercury and Water**:
- For mercury (depressed by 3.45 cm):
\[ -3.45 = \frac{2 \sigma_M \cos(140^\circ)}{\rho_M g R} \]
- For water (raised by 9.8 cm):
\[ 9.8 = \frac{2 \sigma_W \cos(0^\circ)}{\rho_W g R} \]

3. **Substituting Known Values**:
- The density of mercury (\( \rho_M \)) is \( 13.6 \, \text{g/cm}^3 \).
- The density of water (\( \rho_W \)) is \( 1 \, \text{g/cm}^3 \).
- The cosine values are:
- \( \cos(140^\circ) = -\cos(40^\circ) \) (since \( \cos(140^\circ) \) is negative),
- \( \cos(0^\circ) = 1 \).

4. **Rearranging the Equations**:
- From the mercury equation:
\[ \sigma_M = \frac{-3.45 \cdot \rho_M g R}{2 \cos(140^\circ)} \]
- From the water equation:
\[ \sigma_W = \frac{9.8 \cdot \rho_W g R}{2 \cos(0^\circ)} \]

5. **Taking the Ratio of Surface Tensions**:
\[ \frac{\sigma_M}{\sigma_W} = \frac{\frac{-3.45 \cdot \rho_M g R}{2 \cos(140^\circ)}}{\frac{9.8 \cdot \rho_W g R}{2 \cos(0^\circ)}} \]
- The \( g \) and \( R \) cancel out, and we get:
\[ \frac{\sigma_M}{\sigma_W} = \frac{-3.45 \cdot \rho_M \cdot \cos(0^\circ)}{9.8 \cdot \rho_W \cdot \cos(140^\circ)} \]
- Substituting the values:
\[ \frac{\sigma_M}{\sigma_W} = \frac{-3.45 \cdot 13.6 \cdot 1}{9.8 \cdot 1 \cdot (-\cos(40^\circ))} \]

6. **Calculating the Final Ratio**:
- Since \( \cos(140^\circ) = -\cos(40^\circ) \), we can simplify:
\[ \frac{\sigma_M}{\sigma_W} = \frac{3.45 \cdot 13.6}{9.8 \cdot \cos(40^\circ)} \]
- After calculating, you will find:
\[ \frac{\sigma_M}{\sigma_W} \approx 6.25 \]

### Final Answer:
The ratio of the surface tension of mercury to that of water is approximately \( 6.25 \).