Calculate Change in Surface Energy When Drop Splitting Into Droplets

To calculate the change in surface energy when a single liquid drop splits into smaller droplets, we use the principle of volume conservation and the relationship between surface energy and surface area.

Step 1: Conservation of Volume

Let $R$ be the radius of the initial big drop and $r$ be the radius of each of the $N = 512$ smaller droplets.

Since the total volume remains constant: $\frac{4}{3}\pi R^3 = N \times \frac{4}{3}\pi r^3$

Taking the cube root on both sides: $R = N^{1/3} r \implies r = \frac{R}{N^{1/3}}$

Given $N = 512$: $N^{1/3} = (512)^{1/3} = 8$

Thus, the radius of each smaller droplet is: $r = \frac{R}{8}$

Step 2: Calculate the Change in Surface Area

The initial surface area of the single drop is: $A_i = 4\pi R^2$

The total final surface area of the $N$ droplets is: $A_f = N \times 4\pi r^2 = 512 \times 4\pi \left(\frac{R}{8}\right)^2 = 512 \times 4\pi \times \frac{R^2}{64} = 8 \times 4\pi R^2$

The change in total surface area ($\Delta A$) is: $\Delta A = A_f - A_i = 8(4\pi R^2) - 4\pi R^2 = 7 \times 4\pi R^2 = 28\pi R^2$

Step 3: Calculate the Change in Surface Energy

The change in surface energy ($\Delta U$) is equal to the surface tension ($T$) multiplied by the change in surface area ($\Delta A$): $\Delta U = T \times \Delta A = T \times 28\pi R^2$

Given values:

• Diameter of initial drop, $D = 2\text{ mm} \implies R = 1\text{ mm} = 10^{-3}\text{ m}$
• Surface tension, $T = 0.08\text{ N/m}$

Substitute these values into the expression: $\Delta U = 0.08 \times 28\pi \times (10^{-3})^2\text{ J}$ $\Delta U = 2.24\pi \times 10^{-6}\text{ J}$

Using $\pi \approx 3.1416$: $\Delta U \approx 2.24 \times 3.1416 \times 10^{-6}\text{ J} \approx 7.037 \times 10^{-6}\text{ J} \approx 7 \times 10^{-6}\text{ J}$

Comparing with $\Delta U = \alpha \times 10^{-6}\text{ J}$, we get: $\alpha \approx 7$

Correct Answer: B ($7$)