A) \[23.4\mu J\]
B) \[18.5\mu J\]
C) \[26.8\mu J\]
D) \[16.8\mu J\]
Correct Answer: A
Solution :
Increase in surface energy or work done in splitting a big drop \[=4\pi {{R}^{2}}T({{n}^{1/3}}-1)\] \[\Rightarrow W=4\pi \times {{(2\times {{10}^{-3}})}^{2}}\times 0.465({{8}^{1/3}}-1)=23.4\ \mu J\]
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