Calculate the work done in spraying a spherical drop of mercury of 1 mm radius into a million droplets of equal size. The surface tension of mercury `=550 xx10^(-3) Nm^(-1)`

Answer

To solve the problem of calculating the work done in spraying a spherical drop of mercury of 1 mm radius into a million droplets of equal size, we can follow these steps:

### Step 1: Understand the Concept of Surface Energy
The work done in creating new droplets from a larger droplet is equal to the increase in surface energy. The surface energy (E) is given by the formula:
\[ E = \text{Surface Area} \times \text{Surface Tension} \]
where the surface tension (T) of mercury is given as \( 550 \times 10^{-3} \, \text{N/m} \).

### Step 2: Calculate the Initial Surface Area (Si)
The initial drop is a sphere with a radius \( R_i = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \). The surface area \( S_i \) of a sphere is calculated using the formula:
\[ S_i = 4 \pi R_i^2 \]
Substituting the value of \( R_i \):
\[ S_i = 4 \pi (1 \times 10^{-3})^2 = 4 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Final Surface Area (Sf)
When the original droplet is split into a million smaller droplets, the volume remains constant. The volume of the original droplet is:
\[ V_i = \frac{4}{3} \pi R_i^3 \]
The volume of one of the smaller droplets (with radius \( R_f \)) is:
\[ V_f = \frac{4}{3} \pi R_f^3 \]
Since there are a million droplets, the total volume of the smaller droplets is:
\[ V_f \text{ (total)} = 10^6 \times \frac{4}{3} \pi R_f^3 \]
Setting the initial volume equal to the final volume:
\[ \frac{4}{3} \pi R_i^3 = 10^6 \times \frac{4}{3} \pi R_f^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R_i^3 = 10^6 R_f^3 \]
Taking the cube root:
\[ R_f = \frac{R_i}{10^2} = \frac{1 \times 10^{-3}}{10^2} = 1 \times 10^{-5} \, \text{m} \]

Now, calculate the final surface area \( S_f \):
\[ S_f = 10^6 \times 4 \pi R_f^2 = 10^6 \times 4 \pi (1 \times 10^{-5})^2 = 10^6 \times 4 \pi \times 10^{-10} = 4 \pi \times 10^{-4} \, \text{m}^2 \]

### Step 4: Calculate the Work Done (W)
The work done is equal to the increase in surface energy, which can be expressed as:
\[ W = S_f \cdot T - S_i \cdot T \]
Substituting the values:
\[ W = (4 \pi \times 10^{-4} - 4 \pi \times 10^{-6}) \times (550 \times 10^{-3}) \]
\[ W = 4 \pi \times (10^{-4} - 10^{-6}) \times 550 \times 10^{-3} \]
\[ W = 4 \pi \times 0.000399 \times 550 \times 10^{-3} \]
Calculating this gives:
\[ W \approx 6.844 \times 10^{-4} \, \text{J} \]

### Final Answer
The work done in spraying the spherical drop of mercury into a million droplets is approximately:
\[ W \approx 6.844 \times 10^{-4} \, \text{J} \]

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