A) \[\frac{1}{r}\]
B) \[\frac{1}{{{r}^{2}}}\]
C) \[\because \]Surface area \[s=4\pi {{r}^{2}}\] and \[\frac{dr}{dt}=2\] \ \[\frac{ds}{dt}=4\pi \times 2r\frac{dr}{dt}\] = \[8\pi r\times 2=16\pi r\]Þ \[\frac{ds}{dt}\propto r\].
D) \[{{r}^{2}}\]
Correct Answer: B
Solution :
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