A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)

Options

A. `4.35xx10^-2J`
B. `4.35xx10^-3J`
C. `4.35xx10^-6J`
D. `4.35xx10^-8J`

Explanation

To find the work done when a mercury drop of radius 1 cm is broken into \(10^6\) smaller droplets of equal size, we will calculate the change in surface energy. The work done is equal to the change in surface energy, which is given by:

\[ W = S_f - S_i \]

where \(S_f\) is the final surface energy and \(S_i\) is the initial surface energy.

### Step 1: Calculate the initial volume of the mercury drop
The volume \(V\) of a sphere is given by the formula:

\[ V = \frac{4}{3} \pi r^3 \]

For the initial drop with radius \(r = 1 \text{ cm}\):

\[ V_i = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \text{ cm}^3 \]

### Step 2: Calculate the radius of the smaller droplets
Since the volume is conserved when the drop is broken into \(10^6\) smaller droplets, the total volume of the smaller droplets must equal the volume of the original drop.

Let \(r\) be the radius of each smaller droplet. The volume of one smaller droplet is:

\[ V_d = \frac{4}{3} \pi r^3 \]

The total volume of \(10^6\) droplets is:

\[ V_f = 10^6 \cdot \frac{4}{3} \pi r^3 \]

Setting the initial volume equal to the final volume:

\[ \frac{4}{3} \pi = 10^6 \cdot \frac{4}{3} \pi r^3 \]

Cancelling \(\frac{4}{3} \pi\) from both sides:

\[ 1 = 10^6 r^3 \]

Solving for \(r\):

\[ r^3 = 10^{-6} \implies r = 10^{-2} \text{ cm} = 0.01 \text{ cm} \]

### Step 3: Calculate the final surface energy \(S_f\)
The surface area \(A\) of a sphere is given by:

\[ A = 4 \pi r^2 \]

The final surface area for \(10^6\) droplets is:

\[ S_f = 10^6 \cdot 4 \pi (0.01)^2 \cdot T \]

Substituting \(T = 35 \times 10^{-2} \text{ N/m}\):

\[ S_f = 10^6 \cdot 4 \pi (0.01)^2 \cdot (35 \times 10^{-2}) \]

Calculating:

\[ S_f = 10^6 \cdot 4 \pi (0.0001) \cdot (0.35) = 10^6 \cdot 4 \pi \cdot 0.000035 \]

\[ S_f = 4 \pi \cdot 35 \times 10^{-6} \text{ J} \]

### Step 4: Calculate the initial surface energy \(S_i\)
Using the initial radius of 1 cm:

\[ S_i = 4 \pi (1)^2 \cdot T \]

Substituting \(T\):

\[ S_i = 4 \pi (1) \cdot (35 \times 10^{-2}) = 4 \pi \cdot 0.35 \]

### Step 5: Calculate the work done \(W\)
Now, we can calculate the work done:

\[ W = S_f - S_i \]

Substituting the values we found:

\[ W = (4 \pi \cdot 35 \times 10^{-6}) - (4 \pi \cdot 0.35) \]

Factoring out \(4 \pi\):

\[ W = 4 \pi (35 \times 10^{-6} - 0.35) \]

Calculating the difference:

\[ W = 4 \pi (35 \times 10^{-6} - 350 \times 10^{-6}) = 4 \pi (-315 \times 10^{-6}) \]

\[ W = -1260 \pi \times 10^{-6} \text{ J} \]

### Final Result
The work done is:

\[ W \approx -3.96 \times 10^{-3} \text{ J} \]