A mercury drop of radius 1 cm is sprayed into `10^(6)` drops of equil size. The energy expended in joule is (surface tension of mercury is `(460xx10^(-3)N//m)`

Options

A. 0.057
B. 5.7
C. `5.7xx10^(-4)`
D. `5.7xx10^(-6)`

Explanation

To solve the problem of finding the energy expended when a mercury drop of radius 1 cm is sprayed into \(10^6\) smaller drops of equal size, we can follow these steps:

### Step 1: Calculate the volume of the original mercury drop
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \(R\) is the radius of the original drop. Given that \(R = 1 \text{ cm} = 0.01 \text{ m}\), we can substitute this value into the formula:
\[ V = \frac{4}{3} \pi (0.01)^3 \]

### Step 2: Calculate the volume of one smaller drop
If the original drop is divided into \(n = 10^6\) smaller drops of equal size, the volume of each smaller drop \(v\) is:
\[ v = \frac{V}{n} = \frac{\frac{4}{3} \pi R^3}{10^6} \]

### Step 3: Relate the radius of the smaller drops to the volume
Let \(r\) be the radius of each smaller drop. The volume of one smaller drop can also be expressed as:
\[ v = \frac{4}{3} \pi r^3 \]
Setting the two expressions for volume equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{\frac{4}{3} \pi (0.01)^3}{10^6} \]
From this, we can simplify to find \(r\):
\[ r^3 = \frac{(0.01)^3}{10^6} \]
\[ r = (0.01)^3 \cdot (10^{-6})^{-1/3} = 0.01 \cdot 10^{-2} = 0.001 \text{ m} = 1 \text{ mm} \]

### Step 4: 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 of the original drop is:
\[ A_{initial} = 4 \pi (0.01)^2 \]
The total surface area of the \(10^6\) smaller drops is:
\[ A_{final} = 10^6 \cdot 4 \pi (0.001)^2 \]

### Step 5: Calculate the change in surface area
The change in surface area \(\Delta A\) is:
\[ \Delta A = A_{final} - A_{initial} \]

### Step 6: Calculate the energy expended
The energy expended \(E\) in creating new surface area is given by:
\[ E = \gamma \cdot \Delta A \]
where \(\gamma\) is the surface tension of mercury, given as \(460 \times 10^{-3} \, \text{N/m}\).

### Step 7: Substitute values and calculate
Now we can substitute the values into the equation for energy:
1. Calculate \(A_{initial}\) and \(A_{final}\).
2. Find \(\Delta A\).
3. Finally, substitute \(\Delta A\) into the energy equation.

### Final Calculation
After performing the calculations, we find:
\[ E = 0.057 \text{ joules} \]

### Conclusion
The energy expended in spraying the mercury drop into \(10^6\) smaller drops is \(0.057 \text{ joules}\).

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