A) \[20\times {{10}^{-6}}J\]
B) \[2.7\times {{10}^{-6}}J\]
C) \[4\times {{10}^{-6}}J\]
D) \[5.4\times {{10}^{-6}}J\]
Correct Answer: B
Solution :
As, \[\Delta A=4\pi {{r}^{2}}\,({{n}^{1/3}}-1)\] \[\Rightarrow \] \[\Delta \omega =S\Delta A\] (Here, work is required to change the area) \[=72\times {{10}^{-3}}\times 4\pi \times {{({{10}^{-3}})}^{2}}\,({{64}^{1/3}}-1)\] \[=2.7\times {{10}^{-6}}J\] (Just consider about surface energy7).
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