Sand is poured from a pipe at the rate of \(24\pi\) cm3/min. The falling sand forms a cone on the ground whose base radius is always twice its height. The rate at which the height of the sand cone is increasing when the height is 3 cm is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{3}\) cm/min
Let h be the height and r = 2h the base radius at time t. Writing the volume in terms of h alone:
\( \displaystyle V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (2h)^{2}h \)
\( \displaystyle = \frac{4}{3}\pi h^{3} \)
Differentiating with respect to t by the chain rule:
\( \displaystyle \frac{dV}{dt} = 4\pi h^{2}\frac{dh}{dt} \)
With \(\frac{dV}{dt} = 24\pi\) and h = 3:
\( 24\pi = 4\pi \times 9 \times \frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt} = \frac{24}{36} = \frac{2}{3} \) cm/min