Import Question JSON

$\text{Hint: Work done = surface tension } \times \text{ change in surface area.}$ $\text{Step 1: Find the change in the surface energy.}$ $\text{The surface area of a single small drop is given by:}$ $\Rightarrow A_{\text{small}} = 4\pi r^2$ $\text{Suppose there are } n \text{ such small drops, then the total surface area of these small drops is given by:}$ $\Rightarrow A_{\text{total small}} = n \times 4\pi r^2$ $\text{Since the total volume remains constant, we use volume conservation:}$ $\Rightarrow n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \Rightarrow n = \frac{R^3}{r^3}$ $\text{The surface area of the final large drop is given by:}$ $\Rightarrow A_{\text{large}} = 4\pi R^2$ $\text{The initial surface energy is given by:}$ $\Rightarrow U_{\text{initial}} = \text{surface tension } \times \text{ total initial surface area} = T \times n \times 4\pi r^2$ $\text{The final surface energy is given by:}$ $\Rightarrow U_{\text{final}} = T \times 4\pi R^2$ $\text{The decrease in surface energy (converted to heat) is given by:}$ $\Delta U = U_{\text{initial}} - U_{\text{final}} = 4\pi T \frac{R^3}{r} - 4\pi T R^2 \Rightarrow 4\pi T R^2 \left( \frac{R}{r} - 1 \right)$ $\text{Step 2: Find the heat energy per unit volume.}$ $\text{This energy is converted to heat, and using the mechanical equivalent of heat } J, \text{ the heat energy released per unit volume is given by:}$ $\Rightarrow \frac{\Delta U}{\text{Volume}} = \frac{4\pi T R^2 \left( \frac{R}{r} - 1 \right)}{\frac{4}{3} \pi R^3} = \frac{3T}{R} \left( \frac{R}{r} - 1 \right)$ $\Rightarrow \frac{\Delta U}{\text{Volume}} = \frac{3T}{J} \left( \frac{1}{r} - \frac{1}{R} \right)$ $\text{Hence, option (4) is the correct answer.}$