A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `rho` 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 solve the problem of finding the diameter of a drop of liquid of density \( \rho \) that is floating half-immersed in a liquid of density \( d \), we will follow these steps:

### Step 1: Understand the Forces Acting on the Drop
The drop is half-immersed in the liquid, which means that:
- The weight of the drop (\( Mg \)) acts downwards.
- The buoyant force (\( F_b \)) acts upwards.
- The surface tension force (\( F_s \)) also acts upwards.

### Step 2: Write the Expressions for the Forces
1. **Weight of the drop** (\( Mg \)):
\[ M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \rho \]
Therefore, the weight is:
\[ Mg = \frac{4}{3} \pi r^3 \rho g \]

2. **Buoyant Force** (\( F_b \)):
The volume of the drop that is submerged is half of the total volume, so:
\[ F_b = \text{Volume submerged} \times \text{Density of liquid} \times g = \left(\frac{1}{2} \times \frac{4}{3} \pi r^3\right) d g = \frac{2}{3} \pi r^3 d g \]

3. **Surface Tension Force** (\( F_s \)):
The force due to surface tension acts along the circumference of the drop:
\[ F_s = \sigma \times \text{Length in contact} = \sigma \times 2 \pi r \]

### Step 3: Set Up the Equilibrium Condition
At equilibrium, the total upward forces (buoyant force + surface tension force) equal the downward force (weight of the drop):
\[ F_b + F_s = Mg \]
Substituting the expressions we derived:
\[ \frac{2}{3} \pi r^3 d g + \sigma \times 2 \pi r = \frac{4}{3} \pi r^3 \rho g \]

### Step 4: Simplify the Equation
Rearranging the equation gives:
\[ \sigma \times 2 \pi r = \frac{4}{3} \pi r^3 \rho g - \frac{2}{3} \pi r^3 d g \]
Factoring out \( \frac{2}{3} \pi r^3 g \):
\[ \sigma \times 2 \pi r = \frac{2}{3} \pi r^3 g (2\rho - d) \]

### Step 5: Solve for \( r \)
Dividing both sides by \( 2 \pi \):
\[ \sigma = \frac{r^2 g (2\rho - d)}{3} \]
Rearranging gives:
\[ r^2 = \frac{3\sigma}{g(2\rho - d)} \]
Taking the square root:
\[ r = \sqrt{\frac{3\sigma}{g(2\rho - d)}} \]

### Step 6: Find the Diameter
The diameter \( D \) of the drop is:
\[ D = 2r = 2 \sqrt{\frac{3\sigma}{g(2\rho - d)}} \]

### Final Answer
\[ D = 2 \sqrt{\frac{3\sigma}{g(2\rho - d)}} \]