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 water n for water and are `135^(2)` and `0^@`, respectively, the ratio of surface, tension for water and mercury is

Options

A. `1:0.15`
B. `1:3`
C. `1:6.5`
D. `1.5:1`

Explanation

To find the ratio of surface tension for water and mercury using the capillary rise and fall, we can follow these steps:

### Step 1: Write the Capillary Rise and Fall Equations
For a liquid in a capillary tube, the height of the liquid column (h) is given by the formula:

\[ h = \frac{2T \cos \theta}{r \rho g} \]

Where:
- \( h \) = height of the liquid column (capillary rise for water and capillary fall for mercury)
- \( T \) = surface tension of the liquid
- \( \theta \) = angle of contact
- \( r \) = radius of the capillary tube
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity

### Step 2: Set Up the Equations for Water and Mercury
For water:
\[ h_w = \frac{2T_w \cos(0^\circ)}{r \rho_w g} \]
Given that \( h_w = 10 \, \text{cm} \), \( \rho_w \) (density of water) is approximately \( 1 \, \text{g/cm}^3 \), and \( \cos(0^\circ) = 1 \):

\[ 10 = \frac{2T_w}{r \cdot 1 \cdot g} \]

For mercury:
\[ h_m = \frac{2T_m \cos(135^\circ)}{r \rho_m g} \]
Given that \( h_m = -3.42 \, \text{cm} \) (negative because it falls), \( \rho_m = 13.6 \, \text{g/cm}^3 \), and \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \):

\[ -3.42 = \frac{2T_m \left(-\frac{1}{\sqrt{2}}\right)}{r \cdot 13.6 \cdot g} \]

### Step 3: Rearranging the Equations
From the water equation:
\[ T_w = \frac{10rg}{2} \]

From the mercury equation:
\[ -3.42 = \frac{-2T_m}{r \cdot 13.6 \cdot g \cdot \sqrt{2}} \]
This simplifies to:
\[ T_m = \frac{3.42 \cdot r \cdot 13.6 \cdot g \cdot \sqrt{2}}{2} \]

### Step 4: Finding the Ratio of Surface Tensions
Now, we can find the ratio of surface tensions \( \frac{T_w}{T_m} \):

\[ \frac{T_w}{T_m} = \frac{\frac{10rg}{2}}{\frac{3.42 \cdot r \cdot 13.6 \cdot g \cdot \sqrt{2}}{2}} \]

Canceling out common terms:
\[ \frac{T_w}{T_m} = \frac{10}{3.42 \cdot 13.6 \cdot \sqrt{2}} \]

### Step 5: Calculate the Numerical Value
Now, substituting the values:
\[ \sqrt{2} \approx 1.414 \]
Calculating:
\[ \frac{T_w}{T_m} = \frac{10}{3.42 \cdot 13.6 \cdot 1.414} \]

Calculating the denominator:
\[ 3.42 \cdot 13.6 \cdot 1.414 \approx 67.5 \]
Thus,
\[ \frac{T_w}{T_m} \approx \frac{10}{67.5} \approx 0.148 \]

### Final Answer
The ratio of surface tension for water to mercury is approximately \( 0.148 \).

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