Water rises to a height 0f 10 cm in a capillary tube, and mercuryfalls to a depth of 3.42 cm in the same capillary tube. IF the density of mercury is 13.6 and the angle of contact is `135^(@)`, the ratio of surface tension for water and mercury is

Options

A. `1 : 0.5`
B. `1 : 3`
C. `1 : 6.5`
D. `1.5 : 1`

Explanation

To find the ratio of surface tension for water and mercury, we can use the capillary rise and fall formulas. Here’s the step-by-step solution:

### Step 1: Write the formula for capillary rise and fall
The height \( h \) through which a liquid rises or falls in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{R \rho g} \]
where:
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the angle of contact,
- \( R \) is the radius of the capillary tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step 2: Apply the formula for water
For water, the height raised is \( h_1 = 10 \, \text{cm} \). The equation becomes:
\[ h_1 = \frac{2T_1 \cos \theta_1}{R \rho_1 g} \]
where:
- \( T_1 \) is the surface tension of water,
- \( \theta_1 \) is the angle of contact for water (assumed to be \( 0^\circ \)),
- \( \rho_1 \) is the density of water (approximately \( 1 \, \text{g/cm}^3 \)).

### Step 3: Apply the formula for mercury
For mercury, the height fallen is \( h_2 = -3.42 \, \text{cm} \) (negative because it falls). The equation becomes:
\[ h_2 = \frac{2T_2 \cos \theta_2}{R \rho_2 g} \]
where:
- \( T_2 \) is the surface tension of mercury,
- \( \theta_2 = 135^\circ \),
- \( \rho_2 = 13.6 \, \text{g/cm}^3 \).

### Step 4: Set up the ratio of the two equations
Dividing the equation for water by the equation for mercury gives:
\[ \frac{h_1}{h_2} = \frac{T_1 \cos \theta_1}{T_2 \cos \theta_2} \cdot \frac{\rho_2}{\rho_1} \]

### Step 5: Substitute known values
Substituting the known values:
- \( h_1 = 10 \, \text{cm} \)
- \( h_2 = -3.42 \, \text{cm} \)
- \( \theta_1 = 0^\circ \) (thus \( \cos \theta_1 = 1 \))
- \( \theta_2 = 135^\circ \) (thus \( \cos \theta_2 = -\frac{1}{\sqrt{2}} \))
- \( \rho_1 = 1 \, \text{g/cm}^3 \)
- \( \rho_2 = 13.6 \, \text{g/cm}^3 \)

The equation becomes:
\[ \frac{10}{-3.42} = \frac{T_1 \cdot 1}{T_2 \cdot -\frac{1}{\sqrt{2}}} \cdot \frac{13.6}{1} \]

### Step 6: Simplify the equation
This simplifies to:
\[ \frac{10}{3.42} = \frac{T_1 \cdot 13.6 \sqrt{2}}{T_2} \]

### Step 7: Solve for the ratio \( \frac{T_1}{T_2} \)
Cross-multiplying gives:
\[ T_1 = \frac{10 \cdot T_2}{3.42} \cdot \frac{1}{13.6 \sqrt{2}} \]
Thus,
\[ \frac{T_1}{T_2} = \frac{10}{3.42 \cdot 13.6 \sqrt{2}} \]

### Step 8: Calculate the numerical value
Calculating the numerical value:
1. Calculate \( 3.42 \cdot 13.6 \approx 46.512 \).
2. Calculate \( \sqrt{2} \approx 1.414 \).
3. Thus, \( 3.42 \cdot 13.6 \cdot \sqrt{2} \approx 46.512 \cdot 1.414 \approx 65.8 \).
4. Finally, \( \frac{T_1}{T_2} \approx \frac{10}{65.8} \approx 0.152 \).

### Final Result
The ratio of surface tension for water and mercury is:
\[ \frac{T_1}{T_2} \approx 0.152 \quad \text{or} \quad \frac{1}{6.58} \]