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Answer :
Volume of the sand = Volume of the cylindrical bucket \(= \ \pi r^2 h \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32 \) cm3
Volume of the conical heap \(= \ \frac{1}{3} \pi r^2 h \) where, h = 24 cm
\(= \ \frac{1}{3} \pi r^2 \ × \ 24 \ = \ 8\pi r^2 \)
The volume of the conical heap will be equal to that of sand.
\( therefore 8\pi r^2 \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32 \)
\(\Rightarrow \ r^2 \ = \ 18 \ × \ 18 \ × \ 4 \ = \ 18^2 \ × \ 2^2 \)
\(\Rightarrow \ r \ = \ 18 \ × \ 2 \ = \ 36 \)
Here, slant height, \(l \ = \ \sqrt{r^2 \ + \ h^2} \)
\(\Rightarrow \ l \ = \ \sqrt{36^2 \ + \ 24^2} \ \)
\( = \ 12 \sqrt{9 \ + \ 4} \ \)
\( = \ 12 \sqrt{13} \)
\(\therefore \) The radius of the conical heap is 36 cm and its slant height is \( 12 \sqrt{13} \) cm