Sand is poured at the rate of \(16\pi\) cm3/s and forms a cone on the ground whose height is always half the radius of its base. The rate at which the height of the cone is increasing when the height is 2 cm is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 cm/s
Let the height be h and the radius r. Since \(h = \frac{r}{2}\), we have r = 2h, and the volume can be written in h alone:
\( \displaystyle V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2h)^2 h \)
\( \displaystyle V = \frac{4}{3}\pi h^3 \)
Differentiating with respect to time (chain rule):
\( \displaystyle \frac{dV}{dt} = 4\pi h^2\frac{dh}{dt} \)
With \(\frac{dV}{dt} = 16\pi\) and h = 2:
\( \displaystyle 16\pi = 4\pi(4)\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt} = 1 \)
So the height is increasing at 1 cm/s.
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{4}\) cm/s — Incorrect. This comes from taking r = 4h (height one-fourth of the radius) instead of r = 2h.
Option (2) 4 cm/s — Incorrect. This comes from taking r = h, so that \(V = \frac{1}{3}\pi h^3\) and \(16\pi = 4\pi\frac{dh}{dt}\).
Option (4) \(\displaystyle\frac{1}{2}\) cm/s — Incorrect. With the correct volume \(\frac{4}{3}\pi h^3\), the rate at h = 2 is exactly 1 cm/s, not half of it.