Question

A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm³/sec. At what rate is the radius increasing when its circular base area is 1 m²?

पर्याय

• 1 mm/sec

• 0.001 cm/sec

• 2 mm/sec

•  0.002 cm/sec

Answer 1

0.002 cm/sec

\[\text { Let r be the radius,h be the height and V be the volume of the cone at any time t. Then },\]
\[V = \frac{1}{3}\pi r^2 h\]
\[ \Rightarrow V = \frac{2}{3}\pi r^3 \left[ \because h = 2r \right]\]
\[ \Rightarrow \frac{dV}{dt} = 2 \times {10}^4 \frac{dr}{dt} \left[ \because \pi r^2 = 1 m^2 \text { or } {10}^4 {cm}^2 \right] \]
\[ \Rightarrow \frac{dr}{dt} = \frac{1}{2 \times {10}^4}\frac{dV}{dt}\]
\[ \Rightarrow \frac{dr}{dt} = \frac{40}{2 \times {10}^4}\]
\[ \Rightarrow \frac{dr}{dt} = 0 . 002 \ cm/\sec\]