A) \[8\pi {{R}^{2}}T\]
B) \[12\pi {{R}^{2}}T\]
C) \[8{{\pi }^{2}}{{R}^{2}}T\]
D) \[12{{\pi }^{2}}{{R}^{2}}T\]
Correct Answer: A
Solution :
: Change in surface energy\[=T\times \]change in area. Volume of one big drop = Volume of 27 small drops \[\therefore \] \[\frac{4}{3}\pi {{R}^{3}}=27\times \frac{4}{3}\pi {{r}^{3}}\] or \[R=3r\] .... (i) Surface area of 27 drops\[=27\times 4\pi {{r}^{2}}\] Surface area of one bigger drop\[=4\pi {{R}^{2}}\] \[\therefore \]change in area\[=27\times 4\pi {{r}^{2}}-4\pi {{R}^{2}}\] \[=4\pi [27{{r}^{2}}-{{R}^{2}}]=4\pi \left[ 27\times \frac{{{R}^{2}}}{9}-{{R}^{2}} \right]\] \[=4\pi \times 2{{R}^{2}}=8\pi {{R}^{2}}\]. \[\therefore \]Change in surface energy\[=8\pi {{R}^{2}}T\].
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