A spherical water drop of diameter 2 mm splits in to 8 small drops of equal size in air at temperature of 25°C. The surface tension coefficient of water in air is 0.073 N/m and dynamic viscosity is 0.01 poise. The approximate work done in splitting up the drop is

Correct option is B

​$$\textbf{Given:} \\\text{Diameter of original drop, } D = 2 \ \text{mm} = 2 \times 10^{-3} \ \text{m} \\\text{Radius of original drop, } R = 1 \times 10^{-3} \ \text{m} \\\text{Number of smaller drops formed } = 8 \\\text{Surface tension, } \sigma = 0.073 \ \text{N/m} \\\text{Viscosity is not required for this problem.} \\\\\textbf{Volume Conservation} \\\text{Let radius of each smaller drop be } r. \\\text{Since volume is conserved:} \\\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 \Rightarrow R^3 = 8r^3 \Rightarrow r = \left(\frac{R^3}{8}\right)^{1/3} = \frac{R}{2} = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \ \text{m} \\\\\textbf{Surface Area Calculation} \\\text{Surface area of original drop:} \\A_1 = 4 \pi R^2 = 4 \pi (1 \times 10^{-3})^2 = 4 \pi \times 10^{-6} \ \text{m}^2 \\\text{Surface area of 8 small drops:} \\A_2 = 8 \times 4 \pi r^2 = 8 \times 4 \pi (0.5 \times 10^{-3})^2 = 8 \times 4 \pi \times 0.25 \times 10^{-6} = 8 \pi \times 10^{-6} \ \text{m}^2 \\\\\textbf{Work Done = Increase in Surface Energy} \\W = \sigma (A_2 - A_1) = 0.073 \times (8 \pi \times 10^{-6} - 4 \pi \times 10^{-6}) = 0.073 \times 4 \pi \times 10^{-6} \\W = 0.073 \times 4 \times 3.1416 \times 10^{-6} = 0.073 \times 12.5664 \times 10^{-6} = 0.918 \times 10^{-6} \ \text{J} \\\boxed{W \approx 0.92 \ \mu \text{J}}$$​​