The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius `r_(1)` while water rises the same amount h in a capillary tube of radius `r_(2)`. The ratio, `(r_(1)//r_(2))` , is then close to

Options

A. `2/5`
B. `2/3`
C. `4/5`
D. `3/5`

Explanation

To solve the problem, we need to analyze the height of liquid rise or depression in a capillary tube for both mercury and water. We will use the capillary rise formula:

### Step-by-Step Solution:

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

2. **Setting Up the Equations for Mercury and Water**:
For water (which rises):
\[ h = \frac{2T_{\text{water}} \cos(0)}{\rho_{\text{water}} g r_2} \]
For mercury (which is depressed):
\[ h = \frac{2T_{\text{Hg}} \cos(135^\circ)}{\rho_{\text{Hg}} g r_1} \]

3. **Equating the Heights**:
Since both heights are equal (both are \( h \)):
\[ \frac{2T_{\text{water}} \cdot 1}{\rho_{\text{water}} g r_2} = \frac{2T_{\text{Hg}} \cdot \cos(135^\circ)}{\rho_{\text{Hg}} g r_1} \]

4. **Simplifying the Equation**:
Cancel out common terms (like \( 2 \) and \( g \)):
\[ \frac{T_{\text{water}}}{\rho_{\text{water}} r_2} = \frac{T_{\text{Hg}} \cdot \cos(135^\circ)}{\rho_{\text{Hg}} r_1} \]

5. **Substituting Known Values**:
We know:
- \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \) (but we take the modulus, so it becomes \( \frac{1}{\sqrt{2}} \))
- Given ratios:
- \( \frac{T_{\text{Hg}}}{T_{\text{water}}} = 7.5 \)
- \( \frac{\rho_{\text{Hg}}}{\rho_{\text{water}}} = 13.6 \)

Substitute these into the equation:
\[ \frac{T_{\text{water}}}{\rho_{\text{water}} r_2} = \frac{7.5 \cdot \frac{1}{\sqrt{2}}}{\rho_{\text{Hg}} r_1} \]

6. **Rearranging to Find the Ratio**:
Rearranging gives:
\[ \frac{r_1}{r_2} = \frac{T_{\text{Hg}} \cdot \cos(135^\circ) \cdot \rho_{\text{water}}}{T_{\text{water}} \cdot \rho_{\text{Hg}}} \]
Substituting the ratios:
\[ \frac{r_1}{r_2} = \frac{7.5 \cdot \frac{1}{\sqrt{2}} \cdot \rho_{\text{water}}}{\rho_{\text{Hg}} \cdot 1} \]

7. **Calculating the Final Ratio**:
Now substituting the density ratio:
\[ \frac{r_1}{r_2} = \frac{7.5 \cdot \frac{1}{\sqrt{2}}}{13.6} \]

8. **Final Calculation**:
\[ \frac{r_1}{r_2} = \frac{7.5}{13.6 \sqrt{2}} \approx \frac{7.5 \cdot 1.4}{2 \cdot 13.6} \approx \frac{10.5}{27.2} \approx 0.386 \approx 0.4 \]

### Conclusion:
Thus, the ratio \( \frac{r_1}{r_2} \) is approximately \( 0.4 \).