A big drop of water whose diameter is 0.2 cm, is broken into 27000 small drops of equal volume. Work done in this process will be (surface tension of water is `7xx10^(-2)N//m`)

Options

A. `5xx10^(5)` joule
B. `2.9xx10^(-5)` joule
C. `2.55xx10^(-5)` joule
D. zero

Answer

To solve the problem, we need to calculate the work done when a big drop of water is broken into smaller drops. Here’s a step-by-step solution:

### Step 1: Find the radius of the big drop
The diameter of the big drop is given as 0.2 cm. Therefore, the radius \( R \) is:
\[ R = \frac{0.2 \, \text{cm}}{2} = 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 1 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the volume of the big drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting \( R = 1 \times 10^{-3} \, \text{m} \):
\[ V = \frac{4}{3} \pi (1 \times 10^{-3})^3 = \frac{4}{3} \pi (1 \times 10^{-9}) \, \text{m}^3 \]

### Step 3: Find the radius of the small drops
The big drop is broken into \( N = 27000 \) small drops of equal volume. The volume of one small drop \( v \) is:
\[ v = \frac{V}{N} = \frac{\frac{4}{3} \pi (1 \times 10^{-9})}{27000} \]
The volume of a small drop can also be expressed as:
\[ v = \frac{4}{3} \pi r^3 \]
Equating the two volumes:
\[ \frac{4}{3} \pi r^3 = \frac{\frac{4}{3} \pi (1 \times 10^{-9})}{27000} \]
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ r^3 = \frac{1 \times 10^{-9}}{27000} \]
Calculating \( r \):
\[ r^3 = \frac{1 \times 10^{-9}}{27 \times 10^3} = \frac{1 \times 10^{-9}}{2.7 \times 10^4} \approx 3.7 \times 10^{-14} \]
Taking the cube root:
\[ r \approx (3.7 \times 10^{-14})^{1/3} \approx 3.3 \times 10^{-5} \, \text{m} \]

### Step 4: Calculate the initial and final surface energies
The initial surface energy \( E_i \) of the big drop is:
\[ E_i = T \cdot A_i = T \cdot (4 \pi R^2) \]
Substituting \( R = 1 \times 10^{-3} \, \text{m} \):
\[ E_i = (7 \times 10^{-2}) \cdot (4 \pi (1 \times 10^{-3})^2) = (7 \times 10^{-2}) \cdot (4 \pi \cdot 1 \times 10^{-6}) \]

The final surface energy \( E_f \) of the small drops is:
\[ E_f = T \cdot A_f = T \cdot (N \cdot 4 \pi r^2) \]
Substituting \( N = 27000 \) and \( r \):
\[ E_f = (7 \times 10^{-2}) \cdot (27000 \cdot 4 \pi (3.3 \times 10^{-5})^2) \]

### Step 5: Calculate the work done
The work done \( W \) in breaking the drop is the change in surface energy:
\[ W = E_f - E_i \]

### Final Calculation
1. Calculate \( E_i \) and \( E_f \) using the values obtained.
2. Substitute these values into the equation for \( W \).