$A$ cylindrical bucket,$32 \, cm$ high and with a base radius of $18 \, cm$,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. [Take $\pi = \frac{22}{7}$]

(N/A) Height $(h_1)$ of the cylindrical bucket $= 32 \, cm$.
Radius $(r_1)$ of the base of the bucket $= 18 \, cm$.
Height $(h_2)$ of the conical heap $= 24 \, cm$.
Let the radius of the base of the conical heap be $r_2$.
Since the volume of sand remains the same:
Volume of sand in the cylindrical bucket $=$ Volume of sand in the conical heap.
$\pi \times r_1^2 \times h_1 = \frac{1}{3} \pi \times r_2^2 \times h_2$
$\pi \times 18^2 \times 32 = \frac{1}{3} \pi \times r_2^2 \times 24$
$18^2 \times 32 = r_2^2 \times 8$
$r_2^2 = \frac{324 \times 32}{8} = 324 \times 4 = 1296$
$r_2 = \sqrt{1296} = 36 \, cm$.
Now,the slant height $(l)$ of the cone is given by $l = \sqrt{r_2^2 + h_2^2}$.
$l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}$.
$l = \sqrt{144 \times 13} = 12\sqrt{13} \, cm$.
Thus,the radius is $36 \, cm$ and the slant height is $12\sqrt{13} \, cm$.