Sand is pouring from a pipe at the rate of 12 cm3/sec. 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?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{48\pi} \) cm/sec
The height is one-sixth of the radius: \( h=\frac{r}{6} \), so r = 6h. In terms of h only, the volume is
\( \displaystyle V=\frac{1}{3}\pi r^2h \)
\( \displaystyle =\frac{1}{3}\pi(6h)^2h \)
\( \displaystyle =12\pi h^3 \)
Differentiating with respect to time t (rate of change):
\( \displaystyle \frac{dV}{dt}=36\pi h^2\frac{dh}{dt} \)
With \( \frac{dV}{dt}=12 \) and h = 4:
\( \displaystyle 12=36\pi(16)\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt}=\frac{12}{576\pi}=\frac{1}{48\pi}\ \text{cm/sec} \)
NCERT (2026-27 edition):
Rate of Change of Quantities
