Question

A cylindrical bucket, 32cm high with the radius of the base 18cm, 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 24cm, find the radius and slant height of the conical heap of sand.

Answer 1

Hint: Use the fact that the volume of the sand will remain constant. Use the fact that the volume of a cylinder of radius r and height h is given by $\pi {{r}^{2}}h$ , and the volume of a cone of radius r and height h is given by $\dfrac{1}{3}\pi {{r}^{2}}h$. Assume that the radius of the cone be r. Hence find the volume of the cone in terms of r. Compare this volume to the volume of the cylinder and hence find the value of r. Use the fact that the slant height of a right circular cone of radius r and height h is given by $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$. Hence find the slant height of the cone. Verify your answer.

Complete step-by-step answer:

The radius of the cylinder r = 18cm and the height of the cylinder = 32cm.
Hence the volume of the sand = volume of the cylinder.
We know that the volume of a cylinder with radius r and height h is given by $\pi {{r}^{2}}h$.
Hence the volume of the cylinder $=\pi {{r}^{2}}h=\pi {{\left( 18 \right)}^{2}}\left( 32 \right)=10368\pi $
Hence the volume of sand $=10368\pi $ cubic centimetres
Let the radius of the conical heap be r.

Here radius = r and height h = 24cm
We know that the volume of a cone of radius r and height h is given by $\dfrac{1}{3}\pi {{r}^{2}}h$
Hence the volume of the cone $=\dfrac{1}{3}\pi {{r}^{2}}\left( 24 \right)=8{{r}^{2}}\pi $
Hence the volume of sand $=8\pi {{r}^{2}}$
But the volume of sand $10368\pi $
Hence we have $8\pi {{r}^{2}}=10368\pi $
Dividing by $\pi $ on both sides, we get
$8{{r}^{2}}=10368$
Dividing by 8 on both sides, we get
${{r}^{2}}=1296$
Hence we have $r=\sqrt{1296}=36$
Hence the radius of the conical heap = 36cm.
We know that the slant height of a right circular cone of radius r and height h is given by $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$.
Hence we have
$l=\sqrt{{{\left( 24 \right)}^{2}}+{{\left( 36 \right)}^{2}}}=\sqrt{576+1296}=12\sqrt{ 13}$
Hence the slant height of the cone is 60cm.
Note: [1] Do not substitute the value of $\pi $ in these types of questions as it will lead to incorrect results if you approximate the net value of volume.
[2] Verification:
Volume of cone $=\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi {{\left( 36 \right)}^{2}}24=10368\pi $
Volume of cylinder $=\pi {{r}^{2}}h=\pi {{\left( 18 \right)}^{2}}32=10368\pi $
Hence the volumes of the cylinder and the cone are equal. Hence the result is verified.