A mercury drop of radius 1.0 cm is sprayed into `10^(6)` droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is `32 xx 10^(-2) Nm^(-1)`)

Options

A. `3.98 xx 10^(-4) J`
B. `8.46 xx 10^(-4) J`
C. `3.98 xx 10^(-2) J`
D. `8.46 xx 10^(-2) J`

Explanation

To solve the problem of energy expended when a mercury drop of radius 1.0 cm is sprayed into \(10^6\) droplets of equal sizes, we can follow these steps:

### Step 1: Understand the Problem
We have a large mercury drop with a radius \(R = 1.0 \, \text{cm} = 0.01 \, \text{m}\) that is converted into \(10^6\) smaller droplets of equal radius \(r\). We need to calculate the energy expended in this process, given the surface tension of mercury \(S = 32 \times 10^{-2} \, \text{N/m}\).

### Step 2: Calculate the Volume of the Large Drop
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting \(R = 0.01 \, \text{m}\):
\[ V = \frac{4}{3} \pi (0.01)^3 = \frac{4}{3} \pi \times 10^{-6} \, \text{m}^3 \]

### Step 3: Calculate the Volume of the Small Droplets
Let the radius of each small droplet be \(r\). The total volume of \(10^6\) droplets is:
\[ V_{\text{small}} = 10^6 \times \frac{4}{3} \pi r^3 \]
Since the volume is conserved, we have:
\[ \frac{4}{3} \pi R^3 = 10^6 \times \frac{4}{3} \pi r^3 \]
Cancelling \(\frac{4}{3} \pi\) from both sides:
\[ R^3 = 10^6 r^3 \]

### Step 4: Solve for \(r\)
From the equation \(R^3 = 10^6 r^3\):
\[ r^3 = \frac{R^3}{10^6} \]
Taking the cube root:
\[ r = \frac{R}{10^2} = \frac{0.01}{100} = 0.0001 \, \text{m} = 0.1 \, \text{mm} \]

### Step 5: Calculate the Change in Surface Area
The surface area \(A\) of a sphere is given by:
\[ A = 4 \pi r^2 \]
The initial surface area \(A_i\) of the large drop is:
\[ A_i = 4 \pi R^2 = 4 \pi (0.01)^2 = 4 \pi \times 10^{-4} \, \text{m}^2 \]
The final surface area \(A_f\) of \(10^6\) small droplets is:
\[ A_f = 10^6 \times 4 \pi r^2 = 10^6 \times 4 \pi (0.0001)^2 = 10^6 \times 4 \pi \times 10^{-8} \, \text{m}^2 \]

### Step 6: Calculate the Change in Surface Area
The change in surface area \(\Delta A\) is:
\[ \Delta A = A_f - A_i = 10^6 \times 4 \pi \times 10^{-8} - 4 \pi \times 10^{-4} \]
Factoring out \(4 \pi\):
\[ \Delta A = 4 \pi \left(10^6 \times 10^{-8} - 10^{-4}\right) = 4 \pi \left(10^{-2} - 10^{-4}\right) = 4 \pi \left(0.01 - 0.0001\right) = 4 \pi \times 0.0099 \]

### Step 7: Calculate the Energy Expended
The energy expended \(E\) is given by:
\[ E = S \times \Delta A \]
Substituting the values:
\[ E = 32 \times 10^{-2} \times 4 \pi \times 0.0099 \]
Calculating \(E\):
\[ E \approx 32 \times 10^{-2} \times 4 \times 3.14 \times 0.0099 \approx 3.98 \times 10^{-2} \, \text{J} \]

### Final Answer
The energy expended in this process is approximately:
\[ E \approx 3.98 \times 10^{-2} \, \text{J} \]

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