A tube of radius r and sufficient height is dipped in a liquid of surface tension S and density `rho`. Heat developed upto steady state will be
Note: `theta` is the contact angle

Options

A. `(pi S^(2) cos^(2) theta)/(rho g)`
B. `(pi S^(2) cos^(2) theta)/(2rho g)`
C. `(2pi S^(2) cos^(2) theta)/(rho g)`
D. `(4pi S^(2) cos^(2) theta)/(rho g)`

Answer

To solve the problem of determining the heat developed up to steady state when a tube of radius \( r \) is dipped in a liquid with surface tension \( S \), density \( \rho \), and contact angle \( \theta \), we will follow these steps:

### Step 1: Determine the height of capillary rise
The height \( h \) to which the liquid rises in the tube due to capillary action is given by the formula:
\[ h = \frac{2S \cos \theta}{\rho g r} \]
where:
- \( S \) is the surface tension,
- \( \theta \) is the contact angle,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the tube.

### Step 2: Calculate the work done by surface tension
The work done by the surface tension when the liquid rises to height \( h \) can be calculated as:
\[ W_{\text{surface tension}} = \text{Force} \cdot \text{Displacement} \]
The force due to surface tension acting around the circumference of the tube is \( 2\pi r S \), and the displacement is \( h \). Therefore,
\[ W_{\text{surface tension}} = 2\pi r S \cdot h \cdot \cos \theta \]

Substituting the expression for \( h \):
\[ W_{\text{surface tension}} = 2\pi r S \cdot \left(\frac{2S \cos \theta}{\rho g r}\right) \cdot \cos \theta \]
This simplifies to:
\[ W_{\text{surface tension}} = \frac{4\pi S^2 \cos^2 \theta}{\rho g} \]

### Step 3: Calculate the work done by gravity
The work done against gravity when the liquid rises can be calculated using the change in potential energy. The volume of the liquid raised is \( \pi r^2 h \), and the mass is given by \( \rho \cdot \text{Volume} \):
\[ \text{Mass} = \rho \cdot \pi r^2 h \]
The gravitational potential energy change when the center of mass rises to height \( h/2 \) is:
\[ W_{\text{gravity}} = -\left(\rho \cdot \pi r^2 h\right) \cdot g \cdot \left(\frac{h}{2}\right) \]
Substituting for \( h \):
\[ W_{\text{gravity}} = -\left(\rho \cdot \pi r^2 \cdot \frac{2S \cos \theta}{\rho g r}\right) \cdot g \cdot \left(\frac{1}{2} \cdot \frac{2S \cos \theta}{\rho g r}\right) \]
This simplifies to:
\[ W_{\text{gravity}} = -\frac{\rho \cdot \pi r^2 \cdot 2S \cos \theta}{\rho g r} \cdot g \cdot \frac{S \cos \theta}{\rho g r} \]
After simplification, we find:
\[ W_{\text{gravity}} = -\frac{2\pi S^2 \cos^2 \theta}{\rho g} \]

### Step 4: Calculate total heat developed
The total heat developed \( Q \) is the sum of the work done by surface tension and the work done against gravity:
\[ Q = W_{\text{surface tension}} + W_{\text{gravity}} \]
Substituting the expressions we derived:
\[ Q = \frac{4\pi S^2 \cos^2 \theta}{\rho g} - \frac{2\pi S^2 \cos^2 \theta}{\rho g} \]
This simplifies to:
\[ Q = \frac{2\pi S^2 \cos^2 \theta}{\rho g} \]

### Final Answer
Thus, the heat developed up to steady state is:
\[ Q = \frac{2\pi S^2 \cos^2 \theta}{\rho g} \]