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(B) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the total volume remains constant, we have $n 	imes (4/3) \

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Surface area decreases and heat is released

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(B) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the total volume remains constant, we have $n 	imes (4/3) \pi r^3 = (4/3) \pi R^3$, which gives $R = n^{1/3} r$.The initial surface area of $n$ drops is $A_i = n 	imes 4 \pi r^2$.The final surface area of the big drop is $A_f = 4 \pi R^2 = 4 \pi (n^{1/3} r)^2 = 4 \pi n^{2/3} r^2$.Since $n^{2/3}  1$, the final surface area $A_f$ is less than the initial surface area $A_i$. Thus, the surface area decreases.Surface energy is given by $U = T 	imes A$, where $T$ is the surface tension. Since the surface area decreases, the surface energy of the system decreases.According to the law of conservation of energy, the decrease in surface energy is released as heat. Therefore, surface area decreases and heat is released.

When $n$ identical mercury drops combine to form a single big drop,

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