A) constant
B) proportional to \[\sqrt{r}\]
C) proportional to \[{{r}^{2}}\]
D) proportional to \[r\]
Correct Answer: D
Solution :
Surface area, \[S=4\pi {{r}^{2}}\] \[\frac{ds}{dt}=8\,\pi \,r\,\frac{dr}{dt}\,\,\,\Rightarrow \,\,8\pi r\frac{dr}{dt}=8\] \[\left\{ \frac{ds}{dt}=8\,c{{m}^{2}}/s \right\}\] \[\Rightarrow \] \[\frac{dr}{dt}=\frac{1}{\pi r}\] Now, \[v=\frac{4\pi {{r}^{3}}}{3}\Rightarrow \,\frac{dv}{dt}=4\pi {{r}^{2}}\frac{dr}{dt}=4\pi {{r}^{2}}.\frac{1}{\pi r}=4r\] \[\therefore \] \[\frac{dv}{dt}\propto r\]
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