Question

A cone whose height is always equal to the diameter, is increasing in volume at the rate of 40 cm${}^3$/s. At what rate is the radius increasing when its circular base are is $1m^2$?

• A. 1 mm/s
• B. 0.001 cm/s
• C. 2 mm/s
• D. 0.002 cm/s

Answer 1

The correct option is D 0.002 cm/s

Let the radius of the cone be $r$

Let base area $=A$

Volume of cone $=V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^{2}d=\frac{1}{3}\pi r^2\left(2r\right)=\frac{2}{3}\pi r^3$

$\frac{dV}{dt}=2\pi r^2\frac{dr}{dt}$

Given, $\frac{dV}{dt}=40{cm}^3/s,A=\pi r^2=1m^2=10000{cm}^2$

$\Rightarrow 40=2\left(10000\right)\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=0.002cm/s$

So, option D is correct.