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 into one large drop, we can follow these steps:

### Step 1: Understand the Problem
We have 1000 small water drops, each with a radius of \( r = 10^{-7} \, \text{m} \). We need to find the energy released when these drops combine to form one large drop.

### Step 2: Calculate the Volume of Small Drops
The volume \( V \) of one small drop can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r = 10^{-7} \, \text{m} \):
\[ V_{\text{small}} = \frac{4}{3} \pi (10^{-7})^3 = \frac{4}{3} \pi (10^{-21}) \, \text{m}^3 \]

### Step 3: Calculate the Total Volume of 1000 Small Drops
The total volume of 1000 small drops is:
\[ V_{\text{total}} = 1000 \times V_{\text{small}} = 1000 \times \frac{4}{3} \pi (10^{-21}) = \frac{4000}{3} \pi (10^{-21}) \, \text{m}^3 \]

### Step 4: Calculate the Radius of the Large Drop
The volume of the large drop \( V_{\text{large}} \) is equal to the total volume of the small drops:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting the two volumes equal:
\[ \frac{4000}{3} \pi (10^{-21}) = \frac{4}{3} \pi R^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 4000 \times 10^{-21} = R^3 \]
Taking the cube root:
\[ R = (4000 \times 10^{-21})^{1/3} = 10^{-7} \times (4000)^{1/3} \]

Calculating \( (4000)^{1/3} \):
\[ (4000)^{1/3} \approx 15.874 \implies R \approx 15.874 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the Change in Surface Area
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Calculating the surface area of the large drop:
\[ A_{\text{large}} = 4 \pi R^2 = 4 \pi (15.874 \times 10^{-7})^2 \]
Calculating the surface area of one small drop:
\[ A_{\text{small}} = 4 \pi (10^{-7})^2 \]
The total surface area of 1000 small drops:
\[ A_{\text{total small}} = 1000 \times A_{\text{small}} = 1000 \times 4 \pi (10^{-7})^2 \]

### Step 6: Calculate the Change in Area
The change in area \( \Delta A \):
\[ \Delta A = A_{\text{large}} - A_{\text{total small}} \]

### Step 7: Calculate the Energy Released
The energy released \( U \) due to the change in surface area is given by:
\[ U = \text{Surface Tension} \times \Delta A \]
Substituting the values:
\[ U = 7.0 \times 10^{-2} \, \text{N/m} \times |\Delta A| \]

### Final Calculation
After calculating \( \Delta A \) and substituting it into the energy formula, we find:
\[ U \approx 7.9 \times 10^{-12} \, \text{J} \]

### Final Answer
The energy released when 1000 small water drops coalesce to form one large drop is approximately \( 7.9 \times 10^{-12} \, \text{J} \).

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