question_answer

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is\[{{\tan }^{-1}}\left( \frac{1}{2} \right).\]Water is poured into it at a constant rage of 5 cubic meter per minute. The the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is :-                                                 [JEE Main 9-4-2019 Afternoon]

A) \[2/\pi \]                                   

B) \[1/5\pi \]

C) \[1/10\pi \]                   

D) \[1/15\pi \]

Correct Answer: B

Solution :

\[\tan \theta =\frac{1}{2}=\frac{r}{h}\]           \[r=\frac{h}{2}\]           \[V=\frac{1}{3}\pi {{r}^{2}}h\]           \[V=\frac{1}{3}\pi .\frac{{{h}^{3}}}{4}\]           \[\frac{dV}{dt}=\frac{\pi }{12}{{(3h)}^{2}}\left( \frac{dh}{dt} \right)\]           \[5=\frac{\pi }{4}.(100)\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{5\pi }\]