A) \[\frac{1}{54\pi }\]cm/min
B) \[\frac{5}{6\pi }\] cm/min
C) \[\frac{1}{36\pi }\] cm/min
D) \[\frac{1}{18\pi }\] cm/min
Correct Answer: D
Solution :
\[V=\frac{4}{3}\pi \,{{(x+10)}^{3}}\] where x is thickness of ice. \[\therefore \] \[\frac{dV}{dt}=4\pi {{(10+x)}^{2}}\frac{dx}{dt}\] \[={{a}^{x}}{{b}^{2x-1}}{{(\log a{{b}^{2}})}^{2}}\] At \[x=5\],\[\left( \frac{dx}{dt} \right)=\frac{1}{18\pi }cm/\min .\]
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