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.
In the above problem, the increase in gravitational potential energy of the liquid due to rise in the capillary is :

Options

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

Explanation

To find the increase in gravitational potential energy of a liquid that rises in a capillary tube, we can follow these steps:

### Step 1: Determine the height of the liquid rise (H)
The height to which the liquid rises in a capillary tube can be calculated using the formula for capillary rise:

\[ H = \frac{2\sigma \cos \theta}{\rho g r} \]

Given that the angle of contact \(\theta = 0\), we have \(\cos \theta = 1\). The radius \(r\) is half of the diameter \(d\), so \(r = \frac{d}{2}\). Substituting these values into the formula gives:

\[ H = \frac{2\sigma \cdot 1}{\rho g \cdot \frac{d}{2}} = \frac{4\sigma}{\rho g d} \]

### Step 2: Calculate the mass of the liquid that has risen
The volume \(V\) of the liquid that has risen in the capillary tube can be expressed as:

\[ V = \text{Area} \times \text{Height} = \left(\frac{\pi d^2}{4}\right) H \]

Substituting for \(H\):

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

The mass \(m\) of the liquid can be calculated using the density \(\rho\):

\[ m = \rho V = \rho \left(\frac{\pi d \sigma}{\rho g}\right) = \frac{\pi d \sigma}{g} \]

### Step 3: Calculate the increase in potential energy (U)
The increase in gravitational potential energy \(U\) of the liquid can be calculated using the formula:

\[ U = mgh_{cm} \]

Where \(h_{cm}\) is the height of the center of mass of the liquid. Since the liquid rises to height \(H\), the center of mass will be at \(\frac{H}{2}\):

\[ h_{cm} = \frac{H}{2} = \frac{2\sigma}{\rho g d} \]

Substituting \(m\) and \(h_{cm}\) into the potential energy formula:

\[ U = \left(\frac{\pi d \sigma}{g}\right) g \left(\frac{2\sigma}{\rho g d}\right) \]

Simplifying this expression:

\[ U = \frac{\pi d \sigma}{g} \cdot g \cdot \frac{2\sigma}{\rho g d} = \frac{2\pi \sigma^2}{\rho g} \]

### Final Result
Thus, the increase in gravitational potential energy of the liquid due to rise in the capillary tube is:

\[ U = \frac{2\pi \sigma^2}{\rho g} \]