question_answer

Water is dropped at the rate of \[2\,{{m}^{3}}\] per second into a cone with semi-vertical angle \[45{}^\circ \]. The rate at which circumference of water (in contact with air) changes when height of the water in the cone is 2 m, is

A)  1 m/s                   

B)  2 m/s

C)  3 m/s                   

D)  4m/s

Correct Answer: A

Solution :

\[\tan {{45}^{o}}=\frac{r}{h}\Rightarrow \,h=r\] \[V=\frac{1}{3}\pi {{r}^{2}}\cdot \,r=\frac{1}{3}\pi {{r}^{3}}\] \[\Rightarrow \]               \[\frac{dV}{dt}=\pi {{r}^{2}}\frac{dr}{dt}\] \[\Rightarrow \]               \[\frac{dr}{dt}=\frac{2}{\pi \cdot 4}=\frac{1}{2\pi }\] \[\Rightarrow \]               \[2\pi \frac{dr}{dt}=1\,m/\sec \]