A) \[3\pi T{{r}^{2}}\]
B) \[6\pi T{{r}^{2}}\]
C) \[12\pi T{{r}^{2}}\]
D) \[24\pi T{{r}^{2}}\]
Correct Answer: D
Solution :
Initially area of soap bubble \[{{A}_{1}}=4\pi {{r}^{2}}\] Under isothermal condition radius becomes 2r, Then, area \[{{A}_{2}}=4\pi {{(2r)}^{2}}\] \[=4\pi .4{{r}^{2}}\] \[=16\pi {{r}^{2}}\] Increase in surface area \[\Delta A=2({{A}_{2}}-{{A}_{1}})\] \[=2(16\pi {{r}^{2}}-4\pi {{r}^{2}})\] \[=24\pi {{r}^{2}}\] Energy spent \[W=T\times \Delta A\] \[=T.24\pi {{r}^{2}}\] or \[W=24\pi T{{r}^{2}}\,J\]
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