QPaperGen

Sand is pouring from a pipe at the rate of \(12,\text{cm}^3/\text{s}\).
The falling sand forms a cone on the ground such that the height is always one-sixth of the radius of the base.
We are to find how fast the height of the cone is increasing when the height is \(4,\text{cm}\).\[\]

The volume of a cone is given by:
\[V = \dfrac{1}{3} \pi r^2 h\]

Since the height is one-sixth of the radius,
\[h = \dfrac{1}{6}r \Rightarrow r = 6h\]

Substitute into the volume formula:

\[V = \dfrac{1}{3} \pi (6h)^2 h = \dfrac{1}{3} \pi \cdot 36h^2 \cdot h = 12\pi h^3\]

Differentiate both sides with respect to time \(t\):

\[\dfrac{dV}{dt} = \dfrac{d}{dt}(12\pi h^3) = 12\pi \cdot 3h^2 \cdot \dfrac{dh}{dt} = 36\pi h^2 \dfrac{dh}{dt}\]

We are given: At \(h = 4,\text{cm}\): \(\dfrac{dV}{dt} = 12,\text{cm}^3/\text{s}\), So


\[12 = 36\pi (4)^2 \dfrac{dh}{dt} = 36\pi \cdot 16 \cdot \dfrac{dh}{dt}\]
\[\Rightarrow \dfrac{dh}{dt} = \dfrac{12}{576\pi} = \dfrac{1}{48\pi},\text{cm/s}\]

Final Answer:
The height of the cone is increasing at a rate of \(\dfrac{1}{48\pi},\text{cm/s}\) when the height is \(4,\text{cm}\).