A water drop of radius $$1 \text{ cm}$$ is broken into $$729$$ equal droplets. If surface tension of water is $$75 \text{ dyne cm}^{-1}$$, then the gain in surface energy upto first decimal place will be (Given $$\pi = 3.14$$)

A water drop of radius $$R = 1 \text{ cm}$$ is broken into $$729$$ equal droplets. The surface tension of water is $$S = 75 \text{ dyne/cm}$$. We need to find the gain in surface energy.

By conservation of volume:

$$\frac{4}{3}\pi R^3 = 729 \times \frac{4}{3}\pi r^3$$

$$R^3 = 729 \cdot r^3$$

$$r = \frac{R}{9} = \frac{1}{9} \text{ cm}$$

Initial surface area (one big drop):

$$A_i = 4\pi R^2 = 4\pi (1)^2 = 4\pi \text{ cm}^2$$

Final surface area (729 small drops):

$$A_f = 729 \times 4\pi r^2 = 729 \times 4\pi \times \frac{1}{81} = 36\pi \text{ cm}^2$$

$$\Delta A = A_f - A_i = 36\pi - 4\pi = 32\pi \text{ cm}^2$$

$$\Delta E = S \times \Delta A = 75 \times 32\pi \text{ dyne/cm} \times \text{cm}^2 = 2400\pi \text{ erg}$$

Using $$\pi = 3.14$$:

$$\Delta E = 2400 \times 3.14 = 7536 \text{ erg}$$

Since $$1 \text{ erg} = 10^{-7} \text{ J}$$:

$$\Delta E = 7536 \times 10^{-7} \text{ J} = 7.536 \times 10^{-4} \text{ J}$$

Rounding to the first decimal place: $$\Delta E \approx 7.5 \times 10^{-4} \text{ J}$$.

The correct answer is Option C: $$7.5 \times 10^{-4} \text{ J}$$.