A) Square of radius
B) Square root of radius
C) Inversely proportional to radius
D) Cube of the radius
Correct Answer: C
Solution :
[c] Let volume \[=V=\frac{4}{3}\pi {{r}^{3}}\] ?(1) and surface area \[=S=4\pi {{r}^{2}}\] ?(2) Now, \[(1)\Rightarrow \frac{dv}{dt}=\frac{4}{3}\times 3\pi {{r}^{2}}\times \frac{dr}{dt}\] \[=4\pi {{r}^{2}}\frac{dr}{dt}...(3)\] \[(2)\Rightarrow \frac{ds}{dt}=4\pi \times 2\times r\frac{dr}{dt}=\frac{8\pi {{r}^{2}}}{r}\frac{dr}{dt}\] \[=\frac{2}{r}\left[ 4\pi {{r}^{2}}\frac{dr}{dt} \right]=\frac{2}{r}\frac{dv}{dt}\] (from 3)
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