Answer

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**Hint:**In order to solve this problem we need to compare the volume of sand in a cylindrical bucket with Volume of sand in a conical heap. Doing this will solve your problem and will give you the right answer.

**Complete step-by-step answer:**

It is given to us that,

Height $\left( {{h_1}} \right)$ of cylindrical bucket= 32cm

Radius $\left( {{r_1}} \right)$ of circular end of bucket= 18cm

Height $\left( {{h_2}} \right)$ of conical heap= 24cm

Let the radius of the circular end of the conical heap be ${r_2}$ .

Therefore, the Volume of sand in the cylindrical bucket that will be equal to the volume of sand in the conical heap.

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

And hence on putting the values we have,

$

\Rightarrow \pi \times {r_1}^2 \times {h_1} = \dfrac{1}{3}\pi \times {r_2}^2 \times {h_2} \\

\\

$

Now on putting the given values in the above equation, we have

$ \Rightarrow \pi \times {\left( {18} \right)^2} \times 32 = \dfrac{1}{3}\pi \times {\left( {{r_2}} \right)^2} \times 24$

And hence on doing the simplification, we have

$

\Rightarrow {\left( {{r_2}} \right)^2} = \dfrac{{3 \times {{18}^2} \times 32}}{{24}} = {18^2} \times 4 \\

{\text{And hence ,}} \\

\Rightarrow {{\text{r}}_2} = 18 \times 2 = 36cm \\

$

Now, Slant height=$\sqrt {{r_2}^2 + {h_2}^2} = \sqrt {{{12}^2}({3^2} + {2^2})} = 12\sqrt {13} \;cm$

Therefore the radius and the slant height of the conical heap is 36cm and $12\sqrt {13} \;cm$ respectively.

**Note:**This question is based on mensuration and hence first of all we have to compare the Volume of sand in the cylindrical bucket with Volume of sand in conical heap and with the help of that we can find the value of radius of the circular end of the conical heap and later on slant height. Proceeding like this will take you towards the right answer.

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