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 of energy released or absorbed when a certain number of spherical drops of a liquid coalesce to form a single larger drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial and Final States**:
- We have `n` small drops, each with a radius `r`.
- These drops coalesce to form one larger drop with radius `R`.

2. **Calculate the Volume**:
- The volume of one small drop is given by:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume of `n` small drops is:
\[ V_{\text{total}} = n \cdot V_{\text{small}} = n \cdot \frac{4}{3} \pi r^3 \]
- The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
- Since volume is conserved, we set these equal:
\[ n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
- Simplifying gives:
\[ n r^3 = R^3 \quad \Rightarrow \quad n = \frac{R^3}{r^3} \]

3. **Calculate Surface Energy**:
- The surface energy of a drop is given by:
\[ \text{Surface Energy} = \text{Surface Area} \times \text{Surface Tension} \]
- For one small drop:
\[ \text{Surface Energy}_{\text{small}} = 4 \pi r^2 T \]
- For `n` small drops:
\[ \text{Total Surface Energy}_{\text{small}} = n \cdot 4 \pi r^2 T = \frac{R^3}{r^3} \cdot 4 \pi r^2 T = 4 \pi R^3 \frac{T}{r} \]
- For the larger drop:
\[ \text{Surface Energy}_{\text{large}} = 4 \pi R^2 T \]

4. **Calculate Change in Surface Energy**:
- The change in surface energy (ΔE) when the drops coalesce is:
\[ \Delta E = \text{Surface Energy}_{\text{large}} - \text{Total Surface Energy}_{\text{small}} \]
- Substituting the expressions we found:
\[ \Delta E = 4 \pi R^2 T - 4 \pi R^3 \frac{T}{r} \]
- Factoring out \(4 \pi T\):
\[ \Delta E = 4 \pi T \left( R^2 - \frac{R^3}{r} \right) \]

5. **Final Expression**:
- This can be simplified to:
\[ \Delta E = 4 \pi T \left( R^2 - R^3 \frac{1}{r} \right) \]
- This expression represents the energy absorbed or released during the coalescence process.