Sand is pouring from a pipe at the rate of $15 \text{ cm}^3$ per minute. The falling sand forms a cone on the ground such that the height of the cone is always one-third of the radius of the base. How fast is the height of the sand cone increasing at the instant when the height is $4 \text{ cm}$?

$\frac{5}{48\pi}$

The correct answer is Option (1) → $\frac{5}{48\pi}$ ##

Given, $\frac{dV}{dt} = 15 \text{ cm}^3/\text{min}$

$h = \frac{1}{3}r, h = 4$

Since, $V = \frac{1}{3}\pi r^2 h$

$= \frac{1}{3}\pi(3h)^2 \times h$

$V = \frac{9}{3}h^3\pi = 3h^3\pi$

$\frac{dV}{dt} = 9h^2\pi \frac{dh}{dt}$

$15 = 9h^2\pi \frac{dh}{dt}$

$\frac{15}{9 \times (4)^2\pi} = \frac{dh}{dt} ⇒\frac{dh}{dt} = \frac{5}{48\pi}$