The angle of contact between glass and water is `0^@` and water (surface tension `70 dyn//cm`) rises in a glass capillary up to `6 cm`. Another liquid of surface tension `140 dyn//cm`, angle of contact `60^@` and relative density `2` will rise in the same capillary up to

Options

A. `12cm`
B. `24cm`
C. `3cm`
D. `6cm`

Explanation

To solve the problem, we will use the formula for capillary rise, which is given by:

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

where:
- \( h \) = height of capillary rise
- \( T \) = surface tension of the liquid
- \( \theta \) = angle of contact
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity

### Step 1: Understand the given data for the first liquid (water)
- Surface tension \( T_1 = 70 \, \text{dyn/cm} \)
- Angle of contact \( \theta_1 = 0^\circ \)
- Capillary rise \( h_1 = 6 \, \text{cm} \)

### Step 2: Calculate the density of water
The density of water \( \rho_1 \) is approximately \( 1 \, \text{g/cm}^3 \).

### Step 3: Use the capillary rise formula for the first liquid
Using the formula for capillary rise for water:

\[ h_1 = \frac{2T_1 \cos \theta_1}{\rho_1 g} \]

Substituting the known values:

\[ 6 = \frac{2 \cdot 70 \cdot \cos 0^\circ}{1 \cdot g} \]

Since \( \cos 0^\circ = 1 \), we can simplify this to:

\[ 6 = \frac{140}{g} \]

### Step 4: Rearranging to find \( g \)
From the equation, we can express \( g \):

\[ g = \frac{140}{6} \approx 23.33 \, \text{cm/s}^2 \]

### Step 5: Understand the given data for the second liquid
- Surface tension \( T_2 = 140 \, \text{dyn/cm} \)
- Angle of contact \( \theta_2 = 60^\circ \)
- Relative density \( \text{RD} = 2 \) implies \( \rho_2 = 2 \, \text{g/cm}^3 \)

### Step 6: Use the capillary rise formula for the second liquid
Using the formula for the second liquid:

\[ h_2 = \frac{2T_2 \cos \theta_2}{\rho_2 g} \]

Substituting the known values:

\[ h_2 = \frac{2 \cdot 140 \cdot \cos 60^\circ}{2 \cdot g} \]

### Step 7: Calculate \( \cos 60^\circ \)
Since \( \cos 60^\circ = \frac{1}{2} \):

\[ h_2 = \frac{2 \cdot 140 \cdot \frac{1}{2}}{2 \cdot g} \]

This simplifies to:

\[ h_2 = \frac{140}{2g} \]

### Step 8: Substitute \( g \) from Step 4
Now substituting \( g \):

\[ h_2 = \frac{140}{2 \cdot \frac{140}{6}} = \frac{140 \cdot 6}{280} = \frac{6}{2} = 3 \, \text{cm} \]

### Final Answer
The height to which the second liquid will rise in the capillary is:

\[ \boxed{3 \, \text{cm}} \]