Given , radius of the base of the bucket = 18 cm
Height of the bucket = 32 cm
So, that sand in cylindrical bucket `= pir^(2)h = pi (18)^(2) xx 32 = 10368pi`
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 `= (1)/(3) pir^(2)h`
`= (1)/(3)pir^(2)xx 24 = 8 pir^(2)`
According to the question,
Volume of the sand in cylindrical bucket = Volume of the sand in conical heap
`rArr" "10368 pi = 8 pi r^(2)`
`rArr" " 10368 = 8 r^(2)`
`rArr" "r^(2) = (10368)/(8) = 1296`
`rArr" " = r = 36 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 cm`
Hence, radius of conical heap of sand = 36 cm
and slant height of conical heap = 43.267 cm