Water rises to a height of 10 cm. in a certain capillary tube. If in the same tube, level of Hg is depressed by 3.42 cm., compare the surface tension of water and mercury. Sp. Gr. Of Hg is 13.6 the angle of contact for water is zero and that for Hg is `135^@.`

Options

A. `13 : 2`
B. `5 : 16`
C. `16 : 5`
D. `2 : 13`

Explanation

To solve the problem of comparing the surface tension of water and mercury based on the given data, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height of water rise in the capillary tube, \( h_1 = 10 \, \text{cm} \)
- Depression of mercury level in the same tube, \( h_2 = -3.42 \, \text{cm} \) (negative because it is depressed)
- Specific gravity of mercury, \( \text{Sp. Gr.} = 13.6 \)
- Angle of contact for water, \( \theta_1 = 0^\circ \)
- Angle of contact for mercury, \( \theta_2 = 135^\circ \)

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

3. **Set Up the Equations for Water and Mercury:**
For water:
\[ h_1 = \frac{2S_1 \cos \theta_1}{r \rho_1 g} \]
For mercury:
\[ h_2 = \frac{2S_2 \cos \theta_2}{r \rho_2 g} \]

4. **Express the Densities:**
The density of mercury can be expressed in terms of the specific gravity:
\[ \rho_2 = 13.6 \times \rho_{water} \]
where \( \rho_{water} \) is the density of water.

5. **Rearranging the Equations:**
Rearranging the equations for \( S_1 \) and \( S_2 \):
\[ S_1 = \frac{h_1 r \rho_1 g}{2 \cos \theta_1} \]
\[ S_2 = \frac{h_2 r \rho_2 g}{2 \cos \theta_2} \]

6. **Calculate the Ratio of Surface Tensions:**
The ratio of surface tensions \( \frac{S_1}{S_2} \) can be expressed as:
\[ \frac{S_1}{S_2} = \frac{h_1 \rho_1 \cos \theta_2}{h_2 \rho_2 \cos \theta_1} \]
Substituting the values:
\[ \frac{S_1}{S_2} = \frac{10 \times 1 \times \cos(135^\circ)}{-3.42 \times 13.6 \times \cos(0^\circ)} \]

7. **Calculate the Cosines:**
- \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \)
- \( \cos(0^\circ) = 1 \)

8. **Substituting Values:**
\[ \frac{S_1}{S_2} = \frac{10 \times 1 \times -\frac{1}{\sqrt{2}}}{-3.42 \times 13.6 \times 1} \]
Simplifying gives:
\[ \frac{S_1}{S_2} = \frac{10/\sqrt{2}}{3.42 \times 13.6} \]

9. **Final Calculation:**
Calculate the numerical values:
\[ \frac{S_1}{S_2} = \frac{10/\sqrt{2}}{46.512} \approx \frac{7.07}{46.512} \approx 0.152 \]

10. **Expressing the Ratio:**
The ratio \( S_1 : S_2 \) can be expressed as:
\[ S_1 : S_2 \approx 2 : 13 \]

### Final Answer:
The ratio of the surface tension of water to that of mercury is \( 2 : 13 \).