Sand is pouring from a pipe at the rate of 12 $cm^3/s$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

The correct answer is $\dfrac{1}{48 \pi} cm/s$

Let r be the radius, h be the height and V be the volume of sand cone at time t.

Then, $V = \dfrac{1}{3} \pi r^2h$ ----(1)

Volume of sand pouring per second = $\dfrac{dV}{dt}$ = 12 $cm^3/s$

Height of cone is always $\dfrac{1}{6}$ of base radius ⇒ $h = \dfrac{1}{6}r$


Hence equation (1) can be re-writtten as follows

$V = \dfrac{1}{3} \pi (6h)^2h$ = $12 \pi h^3$

∴ Differentiating above equation

$\dfrac{dV}{dt}$ = $12 \pi$ $\left( 3h^2 \dfrac{dh}{dt} \right) $

$12$ = $12 \pi$ $\left( 3h^2 \dfrac{dh}{dt} \right) $

⇒ $\dfrac{dh}{dt}$ = $\dfrac{1}{3 \pi h^2}$

$ \left( \dfrac{dh}{dt} \right)_{h=4}$ = $\dfrac{1}{3 \pi 4^2}$ = $\dfrac{1}{48 \pi} cm/s$