A water drop of radius 1 cm is broken into 1000 equal droplets. If the surface tension of water is 0.075 N/m , then the gain in surface enrgy will be

Options

A. 0
B. `8.5xx10^(-4)` J
C. `7.5xx10^(-4)` J
D. infinite

Answer

To find the gain in surface energy when a water drop of radius 1 cm is broken into 1000 equal droplets, we can follow these steps:

### Step 1: Calculate the initial surface energy of the original droplet.
The formula for the surface energy (SE) of a sphere is given by:
\[ \text{SE} = \text{T} \times \text{A} \]
where \( T \) is the surface tension and \( A \) is the surface area. For a sphere, the surface area \( A \) is given by:
\[ A = 4 \pi r^2 \]
For the original droplet:
- Radius \( r = 1 \text{ cm} = 0.01 \text{ m} \)
- Surface tension \( T = 0.075 \text{ N/m} \)

Thus, the initial surface energy \( \text{SE}_{\text{initial}} \) is:
\[ \text{SE}_{\text{initial}} = T \times 4 \pi r^2 = 0.075 \times 4 \pi (0.01)^2 \]
Calculating this:
\[ \text{SE}_{\text{initial}} = 0.075 \times 4 \pi \times 0.0001 = 0.075 \times 4 \times 3.14 \times 0.0001 \]
\[ \text{SE}_{\text{initial}} = 0.075 \times 0.001256 = 9.42 \times 10^{-5} \text{ J} \]

### Step 2: Calculate the radius of each smaller droplet.
When the original droplet is broken into 1000 equal smaller droplets, the volume remains the same. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Let the radius of each smaller droplet be \( r' \). The volume of the original droplet is equal to the total volume of the 1000 smaller droplets:
\[ \frac{4}{3} \pi (0.01)^3 = 1000 \times \frac{4}{3} \pi (r')^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ (0.01)^3 = 1000 \times (r')^3 \]
\[ (0.01)^3 = 10^{-6} = 1000 \times (r')^3 \]
\[ (r')^3 = \frac{10^{-6}}{1000} = 10^{-9} \]
Taking the cube root:
\[ r' = 10^{-3} \text{ m} = 0.001 \text{ m} = 0.1 \text{ cm} \]

### Step 3: Calculate the final surface energy of the 1000 smaller droplets.
Now, we calculate the surface energy of the 1000 smaller droplets:
\[ \text{SE}_{\text{final}} = 1000 \times T \times 4 \pi (r')^2 = 1000 \times 0.075 \times 4 \pi (0.001)^2 \]
Calculating this:
\[ \text{SE}_{\text{final}} = 1000 \times 0.075 \times 4 \pi \times 10^{-6} \]
\[ = 1000 \times 0.075 \times 0.00001256 = 0.000942 \text{ J} \]

### Step 4: Calculate the gain in surface energy.
The gain in surface energy is given by:
\[ \Delta \text{SE} = \text{SE}_{\text{final}} - \text{SE}_{\text{initial}} \]
Substituting the values:
\[ \Delta \text{SE} = 0.000942 - 9.42 \times 10^{-5} \]
\[ \Delta \text{SE} = 0.000942 - 0.0000942 = 0.0008478 \text{ J} \approx 8.478 \times 10^{-4} \text{ J} \]

### Conclusion
The gain in surface energy when the water drop is broken into 1000 droplets is approximately \( 8.478 \times 10^{-4} \text{ J} \).

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