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(C) When $n$ small drops of radius $r$ coalesce to form a larger drop of radius $R$,the volume remains conserved: $n 	imes (4/3) \pi r^3 = (4/3) \pi

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Temperature increases

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(C) When $n$ small drops of radius $r$ coalesce to form a larger drop of radius $R$,the volume remains conserved: $n 	imes (4/3) \pi r^3 = (4/3) \pi R^3$,which implies $R = n^{1/3} r$.Since $n = 10^6$,$R = (10^6)^{1/3} r = 100r$.The surface area of the small drops is $A_i = n 	imes 4 \pi r^2$ and the surface area of the large drop is $A_f = 4 \pi R^2 = 4 \pi (n^{1/3} r)^2 = n^{2/3} 4 \pi r^2$.The change in surface area is $\Delta A = A_f - A_i = 4 \pi r^2 (n^{2/3} - n)$.Since $n^{2/3} Because surface energy $U = T 	imes A$ (where $T$ is surface tension),a decrease in surface area leads to a release of surface energy.This released energy is converted into internal energy,causing the temperature of the drop to increase.

When $10^6$ small drops coalesce to make a new larger drop,then the drop:

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