Calculate the energy released when 1000 small water drops each of same radius `10^(-7)m` coalesce to form one large drop. The surface tension of water is `7.0xx10^(-2)N//m`.

Explanation

To calculate the energy released when 1000 small water drops coalesce to form one large drop, we can follow these steps:

### Step 1: Calculate the radius of the large drop
When 1000 small drops of radius \( r \) coalesce to form one large drop, the volume of the large drop is equal to the total volume of the small drops.

The volume \( V \) of a single small drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For 1000 small drops, the total volume \( V_{total} \) is:
\[ V_{total} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]

Let \( R \) be the radius of the large drop. The volume of the large drop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]

Setting the two volumes equal gives:
\[ \frac{4000}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 1000 r^3 = R^3 \]
Thus, we can find \( R \):
\[ R = (1000)^{1/3} r = 10 r \]

### Step 2: Calculate the surface area of the small and large drops
The surface area \( A \) of a single small drop is:
\[ A_{small} = 4 \pi r^2 \]
For 1000 small drops, the total surface area \( A_{total} \) is:
\[ A_{total} = 1000 \times 4 \pi r^2 = 4000 \pi r^2 \]

The surface area of the large drop is:
\[ A_{large} = 4 \pi R^2 = 4 \pi (10r)^2 = 400 \pi r^2 \]

### Step 3: Calculate the change in surface area
The change in surface area \( \Delta A \) when the drops coalesce is:
\[ \Delta A = A_{total} - A_{large} = 4000 \pi r^2 - 400 \pi r^2 = 3600 \pi r^2 \]

### Step 4: Calculate the energy released
The energy \( E \) released due to the change in surface area is given by:
\[ E = \text{Surface Tension} \times \Delta A \]
Substituting the values:
\[ E = \gamma \Delta A = (7.0 \times 10^{-2} \, \text{N/m}) \times (3600 \pi r^2) \]

### Step 5: Substitute the radius of the small drop
Given \( r = 10^{-7} \, \text{m} \):
\[ E = (7.0 \times 10^{-2}) \times (3600 \pi (10^{-7})^2) \]
Calculating \( (10^{-7})^2 = 10^{-14} \):
\[ E = (7.0 \times 10^{-2}) \times (3600 \pi \times 10^{-14}) \]
Calculating \( 3600 \times 7.0 \):
\[ E = 25200 \pi \times 10^{-14} \, \text{J} \]

### Step 6: Final calculation
Using \( \pi \approx 3.14 \):
\[ E \approx 25200 \times 3.14 \times 10^{-14} \approx 79008 \times 10^{-14} \approx 7.9 \times 10^{-10} \, \text{J} \]

### Final Answer
The energy released when 1000 small water drops coalesce to form one large drop is approximately:
\[ E \approx 7.9 \times 10^{-10} \, \text{J} \]

---