A) \[{{2}^{1/2}}:1\]
B) \[1:1\]
C) \[{{2}^{2/3}}:1\]
D) None of these
Correct Answer: C
Solution :
Volume remains constant after coalescing. Thus, \[\frac{4}{3}\pi {{R}^{3}}=2\times \frac{4}{3}\pi {{r}^{3}}\] where\[R\]is radius of bigger drop and\[r\]is radius of each smaller drop. \[\therefore \] \[R={{2}^{1/3}}r\] Now, surface energy per unit surface area is the surface tension. So, surface energy,\[W=T\Delta \,\,A\] or \[W=4\pi {{R}^{2}}T\] Therefore, surface energy of bigger drop \[{{W}_{1}}=4\pi {{({{2}^{1/3}}r)}^{2}}T\] \[=({{2}^{2/3}})4\pi {{r}^{2}}T\] Surface energy of smaller drop \[{{W}_{2}}=4\pi {{r}^{2}}T\] Hence, required ratio \[\frac{{{W}_{1}}}{{{W}_{2}}}={{2}^{2/3}}:1\]
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