A) 1 mm/s
B) 0.001 cm/s
C) 2 mm/s
D) 0.002 cm/s
Correct Answer: D
Solution :
Let A be the height, i be the radius of the base and V be the volume of the cone at time t. Then, \[V=\frac{1}{3}\pi {{r}^{2}}h\]\[\Rightarrow \]\[V=\frac{2}{3}\pi {{r}^{3}}\]\[[\because h=2r,given]\] On differentiating both sides w.r.t. t, we get \[\frac{dV}{dt}=2\pi {{r}^{2}}\frac{dr}{dt}\]\[\Rightarrow \]\[40=2{{(10)}^{4}}\frac{dr}{dt}\] \[[\because \pi {{r}^{2}}=1{{m}^{2}}={{10}^{4}}c{{m}^{2}}and\frac{dV}{dt}=40c{{m}^{3}}/s,given]\]\[\Rightarrow \]\[\frac{dr}{dt}=\frac{2}{1000}cm/s=0.002cm/s\]
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