A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `sigma` is the surface tension the diameter of the drop of the liquid is

Options

A. `sqrt((sigma)/(g(2rho-d)))`
B. `sqrt((2sigma)/(g(2rho-d)))`
C. `sqrt((6sigma)/(g(2rho-d)))`
D. `sqrt((12sigma)/(g(2rho-d)))`

Explanation

To find the diameter of a drop of liquid of density \( \rho \) that is floating half-immersed in another liquid of density \( d \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces Acting on the Drop**:
- The drop is floating half-immersed, which means that the weight of the drop is balanced by the buoyant force and the surface tension force acting on the drop.
- The forces acting on the drop are:
- Weight of the drop (\( mg \)) acting downwards.
- Buoyant force (\( F_b \)) acting upwards.
- Surface tension force (\( F_s \)) acting upwards.

2. **Weight of the Drop**:
- The weight of the drop can be expressed as:
\[ mg = \rho V g \]
- Here, \( V \) is the volume of the drop.

3. **Volume of the Drop**:
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Since the drop is half-immersed, the volume submerged is half of the total volume:
\[ V_{\text{submerged}} = \frac{1}{2} V = \frac{2}{3} \pi r^3 \]

4. **Buoyant Force**:
- The buoyant force can be calculated using Archimedes' principle:
\[ F_b = d \cdot V_{\text{submerged}} \cdot g = d \cdot \left(\frac{2}{3} \pi r^3\right) g \]

5. **Surface Tension Force**:
- The surface tension force is given by:
\[ F_s = \sigma \cdot L \]
- The length \( L \) in contact with the liquid is the circumference of the drop at the water surface, which is \( 2\pi r \):
\[ F_s = \sigma \cdot (2\pi r) \]

6. **Setting Up the Equation**:
- Since the drop is floating, the weight is equal to the sum of the buoyant force and the surface tension force:
\[ mg = F_b + F_s \]
- Substituting the expressions we derived:
\[ \rho \cdot \left(\frac{4}{3} \pi r^3\right) g = d \cdot \left(\frac{2}{3} \pi r^3\right) g + \sigma \cdot (2\pi r) \]

7. **Simplifying the Equation**:
- Cancel \( g \) and \( \frac{2}{3} \pi r^3 \) from both sides:
\[ \rho \cdot \frac{4}{3} = d + \frac{3\sigma}{r} \]
- Rearranging gives:
\[ \frac{3\sigma}{r} = \rho \cdot \frac{4}{3} - d \]

8. **Solving for \( r \)**:
- Rearranging for \( r \):
\[ r = \frac{3\sigma}{\rho \cdot \frac{4}{3} - d} \]

9. **Finding the Diameter**:
- The diameter \( D \) of the drop is given by:
\[ D = 2r = 2 \cdot \frac{3\sigma}{\rho \cdot \frac{4}{3} - d} = \frac{6\sigma}{\rho \cdot \frac{4}{3} - d} \]

### Final Expression for Diameter:
\[ D = \frac{6\sigma}{\rho \cdot \frac{4}{3} - d} \]