AIIMS 2019 Physics Energy and Surface Tension MCQ Question

When a mercury drop of radius 1 cm is divided into $10^6$ smaller drops, the radius of each smaller drop becomes r = \frac{1 \, \text{cm}}{10^6^{1/3}} \approx 0.001 \, \text{cm}. The energy associated with the surface tension can be calculated using the formula $E = \Delta A \cdot \gamma$, where $\Delta A$ is the change in surface area and $\gamma$ is the surface tension. The initial surface area of the larger drop is $A_{\text{initial}} = 4\pi R^2$ and the total surface area of the smaller drops is $A_{\text{final}} = 10^6 \cdot 4\pi r^2$. Calculating the energy change yields approximately $0.057 \, \text{J}$, making option A correct. The other options are incorrect as they either underestimate or overestimate the energy change based on incorrect calculations of the surface area or the surface tension.