A cylindrical bucket of height 32 cm

Given, radius of the base of the bucket = 18 cm

Height of the bucket = 32 cm                                         ‘

So, volume of the sand in cylindrical bucket = πr2h= π (18)2 x 32 = 10368 π

Also, given height of the conical heap (h) = 24 cm

Let radius of heap be r cm.

Then, volume of the sand in the heap $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi r^{2} \times 24=8 \pi r^{2}$

According to the question,

Volume of the sand in cylindrical bucket = Volume of the sand in conical heap

$\Rightarrow$ $10368 \pi=8 \pi r^{2}$

$\Rightarrow$ $10368=8 r^{2}$

$\Rightarrow$ $r^{2}=\frac{10368}{8}=1296$

$\Rightarrow$ $r=36 \mathrm{~cm}$

Again, let the slant height of the conical heap $=l$

Now, $l^{2}=h^{2}+r^{2}=(24)^{2}+(36)^{2}$

$=576+1296=1872$

$\therefore$ $l=43.267 \mathrm{~cm}$

Hence, radius of conical heap of sand = 36 cm

and slant height of conical heap = 43.267 cm