Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if `T` is the surface tension is

Options

A. `2^(5//3)pir^(2)`
B. `4pir^(2)T`
C. `2pir^(2)T`
D. `2^(8//3)pir^(2)`

Explanation

To find the surface energy of the bigger drop formed by merging two smaller mercury drops, we can follow these steps:

### Step 1: Understand the Volume Conservation
When two mercury drops of radius \( r \) merge to form a bigger drop, the volume of the smaller drops must equal the volume of the larger drop. The volume \( V \) of a sphere is given by the formula:

\[ V = \frac{4}{3} \pi r^3 \]

For two smaller drops, the total volume is:

\[ V_{small} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

Let the radius of the bigger drop be \( R \). The volume of the bigger drop is:

\[ V_{big} = \frac{4}{3} \pi R^3 \]

Setting these volumes equal gives us:

\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 2: Solve for the Radius of the Bigger Drop
We can simplify the equation by canceling \( \frac{4}{3} \pi \) from both sides:

\[ 8r^3 = 4R^3 \]

Dividing both sides by 4:

\[ 2r^3 = R^3 \]

Taking the cube root of both sides gives us:

\[ R = 2^{1/3} r \]

### Step 3: Calculate the Surface Energy of the Bigger Drop
The surface energy \( SE \) of a sphere is given by the formula:

\[ SE = T \times A \]

where \( A \) is the surface area. The surface area \( A \) of a sphere is given by:

\[ A = 4 \pi R^2 \]

Substituting \( R = 2^{1/3} r \):

\[ A = 4 \pi (2^{1/3} r)^2 = 4 \pi (2^{2/3} r^2) = 4 \cdot 2^{2/3} \pi r^2 \]

Now substituting this into the surface energy formula:

\[ SE = T \times 4 \cdot 2^{2/3} \pi r^2 \]

### Step 4: Final Expression for Surface Energy
Thus, the surface energy of the bigger drop is:

\[ SE = 4 \cdot 2^{2/3} \pi r^2 T \]