A) 1 : 10
B) 1 : 15
C) 1 : 20
D) 1 : 25
Correct Answer: C
Solution :
Surface energy \[W=AT\] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{n\times 4\pi {{r}^{2}}}{4\pi {{R}^{2}}}=\frac{n{{r}^{2}}}{{{R}^{2}}}\] But, \[R=r\times {{n}^{1/3}}\] \[\therefore \] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{n{{r}^{2}}}{{{r}^{2}}{{n}^{2/3}}}={{n}^{1/3}}={{(8000)}^{1/3}}\] \[\frac{{{W}_{2}}}{{{W}_{1}}}=\frac{1}{20}\]
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