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. ` 2pi sigma rho d `
B. ` (pi sigma d )/( 2 ) `
C. ` pi sigma d `
D. ` ( pi sigma d )/(rho ) `

Explanation

To find the weight of the liquid in a capillary tube of diameter \(d\), we can follow these steps:

### Step 1: Understand the situation
The liquid rises in a capillary tube due to surface tension. Given that the angle of contact is zero, the liquid will rise uniformly in the tube.

### Step 2: Identify the forces acting on the liquid
The upward force due to surface tension must balance the weight of the liquid column in the tube. The force due to surface tension can be calculated using the formula:

\[ F_t = \sigma \cdot L \]

where \(F_t\) is the force due to surface tension, \(\sigma\) is the surface tension of the liquid, and \(L\) is the length of the contact line around the tube.

### Step 3: Calculate the length of the contact line
The contact line length \(L\) for a circular tube is given by the circumference of the tube:

\[ L = \pi d \]

where \(d\) is the diameter of the tube.

### Step 4: Substitute the contact line length into the force equation
Now, substituting \(L\) into the force equation gives us:

\[ F_t = \sigma \cdot \pi d \]

### Step 5: Relate the force to the weight of the liquid
The weight of the liquid column can be expressed as:

\[ W = mg \]

where \(m\) is the mass of the liquid and \(g\) is the acceleration due to gravity. The mass \(m\) can be expressed in terms of the volume \(V\) and density \(\rho\):

\[ m = \rho V \]

### Step 6: Calculate the volume of the liquid in the capillary tube
The volume \(V\) of the liquid column in the tube can be expressed 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 \(A\) for a circular tube is:

\[ A = \frac{\pi d^2}{4} \]

### Step 7: Substitute the volume into the weight equation
Thus, the weight of the liquid can be expressed as:

\[ W = \rho \cdot \left(\frac{\pi d^2}{4}\right) \cdot h \cdot g \]

### Step 8: Set the forces equal
Setting the upward force due to surface tension equal to the weight of the liquid gives us:

\[ \sigma \cdot \pi d = \rho \cdot \left(\frac{\pi d^2}{4}\right) \cdot h \cdot g \]

### Step 9: Solve for the weight of the liquid
From the above equation, we can isolate \(W\):

\[ W = \sigma \cdot \pi d \]

This shows that the weight of the liquid in the capillary tube is directly proportional to the surface tension and the diameter of the tube.

### Final Answer
Thus, the weight of the liquid in the capillary tube is:

\[ W = \sigma \cdot \pi d \]

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