A vertical capillary is brought in contact with the water surface (surface tension = T). The radius of the capillary is r and the contact angle `theta = 0^(@)`. The increase in potential energy of the water (density `=rho`) is

Options

A. independent of `rho`
B. independent of r
C. independent of T
D. zero

Answer

To solve the problem step by step, we will analyze the situation involving a vertical capillary tube in contact with water and calculate the increase in potential energy of the water.

### Step 1: Understand the Given Parameters
- Surface tension of water, \( T \)
- Radius of the capillary tube, \( r \)
- Contact angle, \( \theta = 0^\circ \)
- Density of water, \( \rho \)

### Step 2: Determine the Height of Water Rise in the Capillary
The height \( h \) to which water rises in a capillary tube can be calculated using the formula:
\[ h = \frac{2T \cos \theta}{r \rho g} \]
Since the contact angle \( \theta = 0^\circ \), we have:
\[ \cos(0^\circ) = 1 \]
Thus, the formula simplifies to:
\[ h = \frac{2T}{r \rho g} \]

### Step 3: Calculate the Volume of Water in the Capillary
The volume \( V \) of water that has risen in the capillary tube can be expressed as:
\[ V = A \cdot h \]
where \( A \) is the cross-sectional area of the capillary tube. The area \( A \) for a circular cross-section is given by:
\[ A = \pi r^2 \]
Therefore, the volume becomes:
\[ V = \pi r^2 \cdot h = \pi r^2 \cdot \frac{2T}{r \rho g} = \frac{2\pi r T}{\rho g} \]

### Step 4: Calculate the Mass of Water
The mass \( m \) of the water that has risen in the capillary can be calculated using the density:
\[ m = \rho V = \rho \cdot \frac{2\pi r T}{\rho g} = \frac{2\pi r T}{g} \]

### Step 5: Calculate the Increase in Potential Energy
The increase in potential energy \( PE \) of the water is given by:
\[ PE = mgh \]
Substituting the values of \( m \) and \( h \):
\[ PE = \left(\frac{2\pi r T}{g}\right) \cdot g \cdot \left(\frac{2T}{r \rho g}\right) \]
This simplifies to:
\[ PE = \frac{2\pi r T}{g} \cdot g \cdot \frac{2T}{r \rho g} = \frac{4\pi T^2}{\rho g} \]

### Conclusion
The increase in potential energy of the water in the capillary tube is:
\[ PE = \frac{4\pi T^2}{\rho g} \]