The depression of mercury in a capillary tube of radius `R_(1)` is observed to be equal to the rise of water in another capillary tube of radius `R_(2)`. If the ratio of surface tension of mercury and water is `7.5`, ratio of their density `(p_(Hg))/(p_("water"))=13.6` and their angle of contact are `theta_(Hg)=135^(@)` and `theta_("water")=0^(@)` in the respective tubes then `R_(1)//R_(2)` is :

Options

A. `0.2`
B. `0.4`
C. `0.7`
D. `0.8`

Answer

To solve the problem, we need to find the ratio \( \frac{R_1}{R_2} \) given the conditions of mercury and water in capillary tubes.

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
For mercury (depression):
\[ h_{Hg} = \frac{2T_{Hg} \cos \theta_{Hg}}{\rho_{Hg} g R_1} \]
For water (rise):
\[ h_{water} = \frac{2T_{water} \cos \theta_{water}}{\rho_{water} g R_2} \]
Given that \( h_{Hg} = h_{water} \), we can set the two equations equal to each other:
\[ \frac{2T_{Hg} \cos \theta_{Hg}}{\rho_{Hg} g R_1} = \frac{2T_{water} \cos \theta_{water}}{\rho_{water} g R_2} \]

3. **Canceling Common Terms**:
We can cancel \( 2 \) and \( g \) from both sides:
\[ \frac{T_{Hg} \cos \theta_{Hg}}{\rho_{Hg} R_1} = \frac{T_{water} \cos \theta_{water}}{\rho_{water} R_2} \]

4. **Cross Multiplying**:
Rearranging gives:
\[ T_{Hg} \cos \theta_{Hg} \cdot R_2 = T_{water} \cos \theta_{water} \cdot R_1 \cdot \frac{\rho_{Hg}}{\rho_{water}} \]
Rearranging gives:
\[ \frac{R_1}{R_2} = \frac{T_{Hg} \cos \theta_{Hg}}{T_{water} \cos \theta_{water}} \cdot \frac{\rho_{water}}{\rho_{Hg}} \]

5. **Substituting Given Values**:
- Given:
- \( \frac{T_{Hg}}{T_{water}} = 7.5 \)
- \( \frac{\rho_{Hg}}{\rho_{water}} = 13.6 \)
- \( \theta_{Hg} = 135^\circ \) (thus, \( \cos 135^\circ = -\frac{1}{\sqrt{2}} \))
- \( \theta_{water} = 0^\circ \) (thus, \( \cos 0^\circ = 1 \))

Plugging these values into the equation:
\[ \frac{R_1}{R_2} = \frac{7.5 \cdot \left(-\frac{1}{\sqrt{2}}\right)}{1} \cdot \frac{1}{13.6} \]
Simplifying further:
\[ \frac{R_1}{R_2} = \frac{-7.5}{\sqrt{2} \cdot 13.6} \]

6. **Calculating the Final Value**:
- Calculate \( \frac{7.5}{13.6} \approx 0.55 \)
- Therefore, \( \frac{R_1}{R_2} \approx -\frac{0.55}{\sqrt{2}} \)
- Since we are interested in the absolute ratio, we take the positive value:
\[ \frac{R_1}{R_2} \approx 0.4 \]

### Final Answer:
\[ \frac{R_1}{R_2} \approx 0.4 \]