n drops of a liquid, each with surface energy E. joining to form a single drop
(a). Some energy will be released in the process
(b). Some energy will be absorbed in the process
(c). The energy released or absorbed will be `E(n-n^(2//3))`
(d). the energy released or absorbed will be `nE(2^(2//3)-1)`

Explanation

To solve the problem of n drops of a liquid each with surface energy E joining to form a single drop, we can follow these steps:

### Step 1: Understand the Problem
We have n small drops, each with a surface energy E. When these drops coalesce to form a single larger drop, we need to analyze the change in surface energy.

### Step 2: Calculate the Volume
The volume of each small drop is given by:
\[ V_{small} = \frac{4}{3} \pi r^3 \]
Thus, for n small drops, the total volume is:
\[ V_{total} = n \cdot V_{small} = n \cdot \frac{4}{3} \pi r^3 \]

When these drops combine into a single larger drop, the volume of the larger drop (with radius R) is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]

Setting the two volumes equal gives:
\[ n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
From this, we can simplify to find:
\[ R^3 = n \cdot r^3 \]
Thus:
\[ R = n^{1/3} r \]

### Step 3: Calculate Initial Surface Energy
The surface energy of a single small drop is:
\[ E = 4 \pi r^2 T \]
Therefore, the total initial surface energy of n small drops is:
\[ E_{initial} = n \cdot E = n \cdot (4 \pi r^2 T) = 4 \pi n r^2 T \]

### Step 4: Calculate Final Surface Energy
The surface energy of the larger drop is:
\[ E' = 4 \pi R^2 T \]
Substituting R from Step 2:
\[ E' = 4 \pi (n^{1/3} r)^2 T = 4 \pi n^{2/3} r^2 T \]

### Step 5: Compare Initial and Final Surface Energies
Now, we compare the initial and final surface energies:
- Initial energy: \( E_{initial} = 4 \pi n r^2 T \)
- Final energy: \( E' = 4 \pi n^{2/3} r^2 T \)

### Step 6: Calculate Energy Released
The energy released during the coalescence process is given by:
\[ Q = E_{initial} - E' \]
Substituting the expressions we found:
\[ Q = 4 \pi n r^2 T - 4 \pi n^{2/3} r^2 T \]
Factoring out common terms:
\[ Q = 4 \pi r^2 T (n - n^{2/3}) \]

### Step 7: Relate to Given Options
The energy released can be expressed in terms of E:
Since \( E = 4 \pi r^2 T \), we can rewrite the released energy as:
\[ Q = E(n - n^{2/3}) \]

### Conclusion
The correct answer is that some energy will be released in the process, and the energy released is given by:
\[ Q = E(n - n^{2/3}) \]

Thus, the correct option is (c).

---