.If `1000` droplets of water of surface tension `0.07N/m`,having same radius `1mm` each,combine to from a single drop.In the process the released surface energy is : (Take `pi=(22)/(7)`

Options

A. `8.8 xx 10^-5 J`
B. `7.92 xx 10^-4 J`
C. `7.92 xx10^-6 J`
D. `9.68 xx 10^-4 J`

Answer

To solve the problem of calculating the released surface energy when 1000 droplets of water combine into a single drop, we will follow these steps:

### Step 1: Understand the Given Data
- Number of droplets, \( n = 1000 \)
- Surface tension, \( \sigma = 0.07 \, \text{N/m} \)
- Radius of each droplet, \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the Initial Volume of the 1000 Droplets
The volume \( V \) of a single droplet is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
For 1000 droplets, the total initial volume \( V_i \) is:
\[ V_i = n \cdot V = 1000 \cdot \frac{4}{3} \pi r^3 \]

### Step 3: Calculate the Final Volume of the Single Large Drop
Let the radius of the final large drop be \( R \). The volume of the single large drop \( V_f \) is:
\[ V_f = \frac{4}{3} \pi R^3 \]
By the principle of volume conservation, we have:
\[ V_i = V_f \]
Thus,
\[ 1000 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ 1000 \cdot r^3 = R^3 \]

### Step 4: Solve for the Radius of the Final Drop
Substituting \( r = 1 \times 10^{-3} \, \text{m} \):
\[ R^3 = 1000 \cdot (1 \times 10^{-3})^3 = 1000 \cdot 10^{-9} = 10^{-6} \]
Taking the cube root:
\[ R = (10^{-6})^{1/3} = 10^{-2} \, \text{m} = 10 \, \text{mm} \]

### Step 5: Calculate the Surface Energy of the Initial and Final Drops
The surface energy \( U \) is given by:
\[ U = \sigma \cdot A \]
where \( A \) is the surface area. The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]

#### Initial Surface Energy \( U_1 \)
For the 1000 droplets:
\[ U_1 = n \cdot \sigma \cdot (4 \pi r^2) = 1000 \cdot 0.07 \cdot (4 \pi (1 \times 10^{-3})^2) \]
Calculating \( U_1 \):
\[ U_1 = 1000 \cdot 0.07 \cdot (4 \pi \cdot 10^{-6}) = 0.07 \cdot 4 \cdot \frac{22}{7} \cdot 10^{-3} \]
\[ U_1 = 0.07 \cdot 4 \cdot 3.14 \cdot 10^{-3} \approx 0.07 \cdot 12.56 \cdot 10^{-3} \approx 0.000878 \, \text{J} \]

#### Final Surface Energy \( U_2 \)
For the single large drop:
\[ U_2 = \sigma \cdot (4 \pi R^2) = 0.07 \cdot (4 \pi (10 \times 10^{-3})^2) \]
Calculating \( U_2 \):
\[ U_2 = 0.07 \cdot (4 \pi \cdot 10^{-4}) = 0.07 \cdot 4 \cdot \frac{22}{7} \cdot 10^{-4} \]
\[ U_2 = 0.07 \cdot 12.56 \cdot 10^{-4} \approx 0.000878 \cdot 10^{-1} \approx 0.0000878 \, \text{J} \]

### Step 6: Calculate the Released Surface Energy
The released surface energy \( U_{\text{released}} \) is given by:
\[ U_{\text{released}} = U_1 - U_2 \]
Calculating:
\[ U_{\text{released}} = 0.000878 - 0.0000878 \approx 0.0007902 \, \text{J} \]

### Final Answer
The released surface energy when the droplets combine to form a single drop is approximately:
\[ \boxed{0.0007902 \, \text{J}} \]