Water rises to a height of `10 cm` in a capillary tube and mercury falls to a depth of `3.42 cm` in the same capillary tube. If the density of mercury is `13.6 g//c.c.` and the angles of contact for mercury and for water are `135^@` and `0^@`, respectively, the ratio of surface tension for water and mercury is

Options

A. `1:3`
B. `1:4`
C. `1:5.5`
D. `1:6.5`

Explanation

To find the ratio of surface tension for water (S1) to that of mercury (S2), we can use the formula derived from the capillary rise and fall concepts.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height of water rise (h1) = 10 cm
- Depth of mercury fall (h2) = 3.42 cm
- Density of mercury (ρ2) = 13.6 g/cm³
- Angle of contact for water (θ1) = 0°
- Angle of contact for mercury (θ2) = 135°

2. **Use the Capillary Rise and Fall Formula:**
The height of liquid in a capillary tube is given by the formula:
\[ h = \frac{2S \cos \theta}{\rho g r} \]
Where:
- S = surface tension
- θ = angle of contact
- ρ = density of the liquid
- g = acceleration due to gravity (we can cancel this out later)
- r = radius of the capillary tube (we can also cancel this out)

3. **Set Up the Ratios:**
From the capillary rise and fall for water and mercury, we can write:
\[ \frac{h1}{h2} = \frac{2S1 \cos \theta1}{\rho1 g r} \cdot \frac{\rho2 g r}{2S2 \cos \theta2} \]
Simplifying this gives:
\[ \frac{h1}{h2} = \frac{S1 \cos \theta1 \cdot \rho2}{S2 \cos \theta2 \cdot \rho1} \]

4. **Rearranging for the Surface Tension Ratio:**
Rearranging the equation to find the ratio of surface tensions:
\[ \frac{S1}{S2} = \frac{h1}{h2} \cdot \frac{\cos \theta2}{\cos \theta1} \cdot \frac{\rho1}{\rho2} \]

5. **Substituting the Values:**
- For water, we can assume its density (ρ1) is approximately 1 g/cm³.
- Substitute the known values:
\[ \frac{S1}{S2} = \frac{10}{3.42} \cdot \frac{\cos(135°)}{\cos(0°)} \cdot \frac{1}{13.6} \]

6. **Calculate the Cosine Values:**
- \(\cos(135°) = -\frac{1}{\sqrt{2}} \approx -0.7071\)
- \(\cos(0°) = 1\)

7. **Final Calculation:**
\[ \frac{S1}{S2} = \frac{10}{3.42} \cdot \left(-0.7071\right) \cdot \frac{1}{13.6} \]
\[ = \frac{10 \cdot -0.7071}{3.42 \cdot 13.6} \]
\[ = \frac{-7.071}{46.512} \approx -0.152 \]

8. **Taking the Absolute Value:**
Since we are interested in the ratio of magnitudes:
\[ \frac{S1}{S2} \approx 0.152 \implies S1 : S2 \approx 1 : 6.5 \]

### Conclusion:
The ratio of surface tension for water to that of mercury is approximately **1 : 6.5**.