Calculating Surface Energy Change of Breaking Droplet

The problem asks for the value of α representing the change in surface energy when a large liquid drop breaks into smaller droplets.

Initial Conditions

• Initial drop diameter, $d = 2\text{ mm}$.
• Initial drop radius, $R = d/2 = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$.
• Number of smaller droplets, $N = 512$.
• Surface tension of the liquid, $T = 0.08\text{ N/m}$.

Step 1: Relate Radii using Volume Conservation

Assume the total volume of the liquid remains constant.

Volume of the large drop, $V_1 = \frac{4}{3}\pi R^3$.

Let the radius of each small droplet be $r$. The total volume of $N$ small droplets is $V_2 = N \times \frac{4}{3}\pi r^3$.

Equating volumes: $V_1 = V_2 \implies \frac{4}{3}\pi R^3 = N \times \frac{4}{3}\pi r^3$.

This simplifies to $R^3 = N r^3$, so $r = \frac{R}{N^{1/3}}$.

Since $N = 512 = 8^3$, we have $N^{1/3} = 8$. Therefore, $r = \frac{R}{8}$.

Step 2: Calculate Change in Surface Area

Initial surface area, $A_1 = 4\pi R^2$.

Final total surface area, $A_2 = N \times 4\pi r^2$. Substitute $r = R/N^{1/3}$:

$A_2 = N \times 4\pi \left(\frac{R}{N^{1/3}}\right)^2 = N \times 4\pi \frac{R^2}{N^{2/3}} = 4\pi R^2 N^{1/3}$.

The change in surface area is $\Delta A = A_2 - A_1$.

$\Delta A = 4\pi R^2 N^{1/3} - 4\pi R^2 = 4\pi R^2 (N^{1/3} - 1)$.

Substitute known values: $R = 10^{-3}\text{ m}$ and $N^{1/3} = 8$.

$\Delta A = 4\pi (10^{-3}\text{ m})^2 (8 - 1) = 4\pi \times 10^{-6} \times 7 = 28\pi \times 10^{-6}\text{ m}^2$.

Step 3: Calculate Change in Surface Energy

The change in surface energy is given by $\Delta E = T \times \Delta A$.

$\Delta E = (0.08\text{ N/m}) \times (28\pi \times 10^{-6}\text{ m}^2)$.

$\Delta E = (8 \times 10^{-2}) \times (28\pi \times 10^{-6}) = 224\pi \times 10^{-8}\text{ J}$.

$\Delta E = 2.24\pi \times 10^{-6}\text{ J}$.

Using the approximation $\pi \approx 3.14$:

$\Delta E \approx 2.24 \times 3.14 \times 10^{-6} \approx 7.0336 \times 10^{-6}\text{ J}$.

Step 4: Determine the Value of α

The problem states that the change in surface energy is $\alpha \times 10^{-6}\text{ J}$.

Comparing our calculated $\Delta E$ with the given expression: $\alpha \times 10^{-6}\text{ J} \approx 7.0336 \times 10^{-6}\text{ J}$.

Therefore, $\alpha \approx 7.0336$.

Rounding to the nearest integer, the value of $\alpha$ is 7.