Question

Sand is pouring from a pipe at the rate of $12$ cc/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always ${\frac{1}{6}}^{th}$ of the radius of the base How fast is the height of the sand cone increasing when the height is $4$ cm.

• A. $\frac{1}{24\pi }cm/s$
• B. $\frac{1}{48\pi }cm/s$
• C. $\frac{1}{42\pi }cm/s$
• D. $\frac{48}{\pi }cm/s$

Answer 1

The correct option is B $\frac{1}{48\pi }cm/s$

$v=\frac{1}{3}\pi r^{2}h$ but $h=\frac{r}{6}$
$\Rightarrow v=\frac{1}{3}\pi {\left(6h\right)}^{2}h\Rightarrow v=12\pi h^2$
$\frac{dv}{dt}=36\pi h^2\frac{dh}{dt}$
When $\frac{dv}{dt}=12{cm}^2/r$ and $h=4$ cm
$\frac{dh}{dt}=\frac{12}{36\pi {\left(4\right)}^2}=\frac{1}{48\pi }cm/sec$