A spherical mercury drop of `10^(-3)m` radius is sprayed into million drops of the same size. Calculate the energy used in doing so. Given surface tension of mercury `= 0.55 Nm^(-1)`

Answer

To solve the problem of calculating the energy used in spraying a spherical mercury drop of radius \(10^{-3} \, m\) into a million drops of the same size, we will follow these steps:

### Step 1: Calculate the Initial Surface Area
The initial surface area \(S_i\) of the mercury drop can be calculated using the formula for the surface area of a sphere:

\[ S_i = 4 \pi r_i^2 \]

Where \(r_i = 10^{-3} \, m\).

\[ S_i = 4 \pi (10^{-3})^2 = 4 \pi (10^{-6}) \, m^2 \]

### Step 2: Calculate the Final Surface Area
When the original drop is split into a million smaller drops, the radius of each smaller drop \(r_f\) can be determined using the volume conservation principle. Since the volume of a sphere is given by:

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

The volume of the original drop is:

\[ V_i = \frac{4}{3} \pi r_i^3 \]

And the volume of the million smaller drops is:

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

Setting these two volumes equal gives:

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

This simplifies to:

\[ r_i^3 = 10^6 r_f^3 \]

Thus, we find:

\[ r_f = r_i \cdot 10^{-2} = 10^{-3} \cdot 10^{-2} = 10^{-5} \, m \]

Now, we can calculate the final surface area \(S_f\):

\[ S_f = 10^6 \cdot 4 \pi r_f^2 \]

Substituting \(r_f\):

\[ S_f = 10^6 \cdot 4 \pi (10^{-5})^2 = 10^6 \cdot 4 \pi (10^{-10}) = 4 \pi (10^{-4}) \, m^2 \]

### Step 3: Calculate the Change in Surface Area
The change in surface area \(\Delta S\) is given by:

\[ \Delta S = S_f - S_i \]

Substituting the values we calculated:

\[ \Delta S = 4 \pi (10^{-4}) - 4 \pi (10^{-6}) = 4 \pi (10^{-4} - 10^{-6}) = 4 \pi (0.0001 - 0.000001) = 4 \pi (0.000099) \, m^2 \]

### Step 4: Calculate the Work Done (Surface Energy)
The work done (or surface energy) \(E\) is given by:

\[ E = \text{Surface Tension} \times \Delta S \]

Given the surface tension \(\sigma = 0.55 \, N/m\):

\[ E = 0.55 \times 4 \pi (0.000099) \, J \]

Calculating this value:

\[ E \approx 0.55 \times 4 \times 3.14 \times 0.000099 \approx 6.844 \times 10^{-4} \, J \]

### Final Result
Thus, the energy used in spraying the mercury drop into a million smaller drops is approximately:

\[ E \approx 6.85 \times 10^{-4} \, J \]