The angle of contact between glass and water is `0^@` and surface tension is `70 dyn//cm`. Water 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. (a) 12 cm
B. (b) 24 cm
C. (c) 3 cm
D. (d) 6 cm

Explanation

To solve the problem, we will use the formula for the height of liquid rise in a capillary tube, which is given by:

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

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

### Step 1: Write down the known values for water
For water, we have:
- Surface tension \( T_1 = 70 \, \text{dyn/cm} \)
- Angle of contact \( \theta_1 = 0^\circ \)
- Height of rise \( h_1 = 6 \, \text{cm} \)
- Density of water \( \rho_1 = 1 \, \text{g/cm}^3 \) (since relative density is 1)
- Radius \( r_1 = r \)

### Step 2: Write down the known values for the second liquid
For the second liquid, we have:
- Surface tension \( T_2 = 140 \, \text{dyn/cm} \)
- Angle of contact \( \theta_2 = 60^\circ \)
- Relative density \( = 2 \) implies \( \rho_2 = 2 \, \text{g/cm}^3 \)
- Radius \( r_2 = r \)

### Step 3: Set up the ratio of heights
Using the formula for height in a capillary tube for both liquids, we can set up the ratio:

\[ \frac{h_2}{h_1} = \frac{T_2 \cos \theta_2}{T_1 \cos \theta_1} \cdot \frac{\rho_1}{\rho_2} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:

\[ \frac{h_2}{6} = \frac{140 \cdot \cos 60^\circ}{70 \cdot \cos 0^\circ} \cdot \frac{1}{2} \]

### Step 5: Calculate the cosines
We know:
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \cos 0^\circ = 1 \)

Substituting these values gives:

\[ \frac{h_2}{6} = \frac{140 \cdot \frac{1}{2}}{70 \cdot 1} \cdot \frac{1}{2} \]

### Step 6: Simplify the equation
This simplifies to:

\[ \frac{h_2}{6} = \frac{140 \cdot \frac{1}{2}}{70} \cdot \frac{1}{2} \]

Calculating the right side:

\[ \frac{h_2}{6} = \frac{70}{70} \cdot \frac{1}{2} = \frac{1}{2} \]

### Step 7: Solve for \( h_2 \)
Now, multiply both sides by 6:

\[ h_2 = 6 \cdot \frac{1}{2} = 3 \, \text{cm} \]

### Final Answer
The height to which the second liquid will rise in the capillary is \( 3 \, \text{cm} \).

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