$A$ cylindrical bucket of height $32 \,cm$ and base radius $18 \,cm$ is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $24 \,cm$,find the radius and slant height of the heap.

(N/A) Given,radius of the base of the bucket $(R) = 18 \,cm$.
Height of the bucket $(H) = 32 \,cm$.
Volume of the sand in the cylindrical bucket $= \pi R^2 H = \pi \times (18)^2 \times 32 = 10368 \pi \,cm^3$.
Let the radius of the conical heap be $r \,cm$ and its height be $h = 24 \,cm$.
Volume of the conical heap $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 \times 24 = 8 \pi r^2 \,cm^3$.
Since the volume of sand remains the same,$8 \pi r^2 = 10368 \pi$.
$r^2 = \frac{10368}{8} = 1296$.
$r = \sqrt{1296} = 36 \,cm$.
The slant height $l$ of the cone is given by $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}$.
$l = 12\sqrt{13} \approx 43.27 \,cm$.
Thus,the radius of the heap is $36 \,cm$ and the slant height is approximately $43.27 \,cm$.