A) \[2\pi \,sq\,cm/s\]
B) \[\pi \,sq\,cm/s\]
C) \[\frac{3\pi }{2}sq\,cm/s\]
D) \[\frac{\pi }{2}sq\,cm/s\]
Correct Answer: A
Solution :
Given rate of increase volume of sphere is \[\frac{dV}{dt}=\pi \] ?...(i) We know that, Volume of sphere; \[V=\frac{4}{3}\,\pi {{r}^{3}}\] \[\Rightarrow \] \[\frac{dV}{dt}=\frac{4}{3}\pi .3{{r}^{2}}.\frac{dr}{dt}=4\pi {{r}^{2}}\frac{dr}{dt}\] \[\Rightarrow \] \[\pi =4\,\pi {{r}^{2}}\frac{dr}{dt};\] [from Eq.(i)] \[\Rightarrow \] \[\frac{dr}{dt}=\frac{1}{4{{r}^{2}}}\] ??(ii) Also, we know that, surface of sphere, \[S=4\pi {{r}^{2}}\] \[\Rightarrow \] \[\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi r.\frac{1}{4{{r}^{2}}}\] [from Eq. (ii)] \[\Rightarrow \] \[\pi =4\,\pi {{r}^{2}}\frac{dr}{dt};\] \[[\because \,\,\,given\,\,r=1]\] \[\Rightarrow \] \[\frac{dS}{dt}=2\pi \] So, rate of increase in surface \[=2\pi \,sq\,cm/s\]
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