A) \[8\pi r\Delta S\]
B) \[4\pi r\Delta rS\]
C) \[16\pi r\Delta rS\]
D) \[2\pi r\Delta rS\]
Correct Answer: A
Solution :
[a] Change in surface energy \[\Delta \,E=\left[ 4\,\pi \,{{(r+\Delta r)}^{2}}-4\pi {{r}^{2}} \right]S\] \[=\left[ 4\,\pi \,{{(\Delta r)}^{2}}-8\Delta \pi r \right]S\] Ar is too small, \[{{(\Delta r)}^{2}}\]is still smaller and can be neglected increase in surface energy = \[8\pi r\Delta rS\]
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