A certain number of spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R` and volume `V`. If `T` is the surface tension of the liquid, then

Options

A. energy `= 4VT((1)/(r)-(1)/(R))` is released
B. energy `= 3VT((1)/(r)+(1)/(R))` is absorbed
C. energy `= 3VT((1)/(r)-(1)/(R))` is released
D. energy is neither released nor absorbed

Explanation

To solve the problem, we need to find the change in surface energy when a certain number of smaller spherical drops of radius \( r \) coalesce to form a larger spherical drop of radius \( R \). The surface tension of the liquid is given as \( T \).

### Step-by-Step Solution:

1. **Calculate the Surface Area of the Smaller Drops:**
Each smaller drop has a radius \( r \). The surface area \( A \) of a single spherical drop is given by:
\[ A = 4\pi r^2 \]
If there are \( n \) smaller drops, the total surface area of these drops is:
\[ A_{small} = n \times 4\pi r^2 = 4\pi n r^2 \]

2. **Calculate the Surface Area of the Larger Drop:**
The larger drop formed has a radius \( R \). The surface area of the larger drop is:
\[ A_{large} = 4\pi R^2 \]

3. **Calculate the Change in Surface Energy:**
The surface energy \( E \) is given by the product of surface area and surface tension \( T \). Therefore, the surface energy of the smaller drops is:
\[ E_{small} = T \times A_{small} = T \times 4\pi n r^2 \]
The surface energy of the larger drop is:
\[ E_{large} = T \times A_{large} = T \times 4\pi R^2 \]
The change in surface energy \( \Delta E \) when the drops coalesce is:
\[ \Delta E = E_{large} - E_{small} = T \times 4\pi R^2 - T \times 4\pi n r^2 \]
Simplifying this gives:
\[ \Delta E = T \times 4\pi (R^2 - n r^2) \]

4. **Relate the Volumes of the Drops:**
Since the smaller drops coalesce to form the larger drop, the volume before and after must remain the same. The volume \( V \) of a single smaller drop is:
\[ V_{small} = \frac{4}{3}\pi r^3 \]
Therefore, the total volume of \( n \) smaller drops is:
\[ V_{total} = n \times \frac{4}{3}\pi r^3 \]
The volume of the larger drop is:
\[ V_{large} = \frac{4}{3}\pi R^3 \]
Setting the volumes equal gives:
\[ n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]
Simplifying this, we find:
\[ n r^3 = R^3 \quad \Rightarrow \quad n = \frac{R^3}{r^3} \]

5. **Substituting \( n \) in the Change of Surface Energy:**
Substitute \( n \) back into the change in surface energy:
\[ \Delta E = T \times 4\pi \left( R^2 - \frac{R^3}{r^3} r^2 \right) \]
Simplifying further:
\[ \Delta E = T \times 4\pi \left( R^2 - \frac{R^2}{r^3} R^3 \right) = T \times 4\pi R^2 \left( 1 - \frac{R}{r^3} \right) \]

### Final Result:
The change in surface energy when the drops coalesce is:
\[ \Delta E = 4\pi T \left( R^2 - \frac{R^3}{r^3} \right) \]