A spherical drop of mercury of radius 3mm falls to the ground and breaks into 27 smaller drops of equal size. Calculate the amount of work done. Surfae tension of mercury `= 0.472 Nm^(-1)`

Answer

To solve the problem, we need to calculate the work done when a spherical drop of mercury breaks into smaller drops. Here’s a step-by-step breakdown:

### Step 1: Understand the Problem
A spherical drop of mercury with a radius of 3 mm falls and breaks into 27 smaller drops of equal size. We need to find the work done in this process, given the surface tension of mercury.

### Step 2: Convert Units
Convert the radius from millimeters to meters:
- Radius \( R = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)

### Step 3: Calculate the Volume of the Original Drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the value of \( R \):
\[ V = \frac{4}{3} \pi (3 \times 10^{-3})^3 \]

### Step 4: Calculate the Volume of the Smaller Drops
Since the original drop breaks into 27 smaller drops, the volume of each smaller drop \( v \) is:
\[ v = \frac{V}{27} \]
Using the volume formula for the smaller drops:
\[ v = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the smaller drops.

### Step 5: Relate the Volumes
Since the total volume remains constant:
\[ \frac{4}{3} \pi R^3 = 27 \left( \frac{4}{3} \pi r^3 \right) \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 27 r^3 \]
Thus, we can express \( r \) in terms of \( R \):
\[ r = \frac{R}{3} \]

### Step 6: Calculate the Increase in Surface Area
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
The surface area of the original drop:
\[ A_{\text{original}} = 4 \pi R^2 \]
The surface area of the 27 smaller drops:
\[ A_{\text{smaller}} = 27 \times 4 \pi r^2 = 27 \times 4 \pi \left(\frac{R}{3}\right)^2 = 27 \times 4 \pi \frac{R^2}{9} = 12 \pi R^2 \]

### Step 7: Calculate the Increase in Surface Area
The increase in surface area \( \Delta A \) is:
\[ \Delta A = A_{\text{smaller}} - A_{\text{original}} = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2 \]

### Step 8: Calculate the Work Done
The work done \( W \) is given by the formula:
\[ W = \text{Surface Tension} \times \Delta A \]
Substituting the values:
\[ W = 0.472 \, \text{N/m} \times 8 \pi R^2 \]
Substituting \( R = 3 \times 10^{-3} \, \text{m} \):
\[ W = 0.472 \times 8 \times 3.142 \times (3 \times 10^{-3})^2 \]

### Step 9: Calculate the Final Value
Calculating the above expression:
\[ W = 0.472 \times 8 \times 3.142 \times 9 \times 10^{-6} \]
\[ W \approx 1.067 \times 10^{-4} \, \text{J} \]

### Final Answer
The amount of work done is approximately \( 1.067 \times 10^{-4} \, \text{J} \).

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