A liquid of density ` rho` and surface tension ` sigma ` rises in a capillary tube of diameter d. Angle of contact between the tube and liquid is zero. The weight of the liquid in the capillary tube is:

Options

A. (a)` 2pi sigma rho d `
B. (b)` (pi sigma d )/( 2 ) `
C. (c)` pi sigma d `
D. (d) ` ( pi sigma d )/(rho ) `

Explanation

To find the weight of the liquid in a capillary tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a liquid with density \( \rho \) and surface tension \( \sigma \) rising in a capillary tube of diameter \( d \).
- The angle of contact between the liquid and the tube is zero, meaning the liquid wets the tube completely.

2. **Identify the Forces Acting on the Liquid**:
- The upward force acting on the liquid due to surface tension is given by \( F_t \).
- The downward force acting on the liquid is its weight \( W \), which can be expressed as \( W = mg \), where \( m \) is the mass of the liquid.

3. **Relate Surface Tension to Weight**:
- The force due to surface tension can be calculated using the formula:
\[ F_t = \sigma \cdot L \]
where \( L \) is the length of the contact line. For a circular tube, \( L \) is the circumference of the tube, which is given by:
\[ L = \pi d \]
- Therefore, the surface tension force becomes:
\[ F_t = \sigma \cdot \pi d \]

4. **Equate the Forces**:
- At equilibrium, the upward force due to surface tension equals the downward gravitational force (weight) acting on the liquid:
\[ F_t = W \]
- Substituting the expressions we have:
\[ \sigma \cdot \pi d = mg \]

5. **Express Mass in Terms of Density**:
- The mass \( m \) of the liquid can be expressed in terms of its volume and density:
\[ m = \rho V \]
- The volume \( V \) of the liquid in the tube can be calculated as:
\[ V = A \cdot h \]
where \( A \) is the cross-sectional area of the tube and \( h \) is the height of the liquid column. The cross-sectional area \( A \) is given by:
\[ A = \frac{\pi d^2}{4} \]
- Thus, the volume becomes:
\[ V = \frac{\pi d^2}{4} \cdot h \]
- Therefore, the mass is:
\[ m = \rho \cdot \frac{\pi d^2}{4} \cdot h \]

6. **Substitute Mass Back into Weight Equation**:
- Now substituting \( m \) back into the weight equation:
\[ \sigma \cdot \pi d = \rho \cdot \frac{\pi d^2}{4} \cdot h \cdot g \]

7. **Solve for Weight**:
- Rearranging gives us the weight of the liquid in the capillary tube:
\[ W = \sigma \cdot \pi d \]

### Final Answer:
The weight of the liquid in the capillary tube is:
\[ W = \sigma \cdot \pi d \]