A capillary tube of radius r is lowered into a liquid of surface tension T and density `rho`. Given angle of contact `=0^(@)`.
What is the potential energy acquired by the liquid in the capillary?

Options

A. `(pi T^(2))/(2 rho g)`
B. `T^(2)/(2 rho g)`
C. `T^(2)/(rho g)`
D. `(2 pi T^(2))/(rho g)`

Answer

To find the potential energy acquired by the liquid in a capillary tube, we can follow these steps:

### Step 1: Determine the height of the liquid column
The height \( h \) to which the liquid rises in the capillary tube can be calculated using the formula:
\[ h = \frac{2T \cos(\theta)}{\rho g r} \]
where:
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the contact angle (which is \( 0^\circ \) in this case, so \( \cos(0) = 1 \)),
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

Since the contact angle is \( 0^\circ \), the formula simplifies to:
\[ h = \frac{2T}{\rho g r} \]

### Step 2: Calculate the mass of the liquid in the capillary
The volume \( V \) of the liquid that rises in the capillary tube can be calculated as:
\[ V = \pi r^2 h \]
The mass \( m \) of the liquid is given by:
\[ m = \rho V = \rho \pi r^2 h \]

### Step 3: Find the center of mass of the liquid column
The center of mass of the liquid column is at a height of \( \frac{h}{2} \) from the bottom of the capillary tube.

### Step 4: Calculate the potential energy
The potential energy \( PE \) acquired by the liquid can be calculated using the formula:
\[ PE = mgh_{cm} \]
where \( h_{cm} = \frac{h}{2} \) is the height of the center of mass. Substituting the expression for mass \( m \) and height \( h_{cm} \):
\[ PE = \left(\rho \pi r^2 h\right) g \left(\frac{h}{2}\right) \]
Substituting \( h = \frac{2T}{\rho g r} \):
\[ PE = \left(\rho \pi r^2 \cdot \frac{2T}{\rho g r}\right) g \left(\frac{\frac{2T}{\rho g r}}{2}\right) \]

### Step 5: Simplify the expression
Substituting and simplifying:
\[ PE = \left(\frac{2\pi r T}{g}\right) g \left(\frac{T}{\rho g r}\right) \]
\[ PE = \frac{2\pi T^2}{\rho g} \]

### Final Result
Thus, the potential energy acquired by the liquid in the capillary tube is:
\[ PE = \frac{2\pi T^2}{\rho g} \]