Question

A metallic sphere of radius $4.2cm$ is melted and recast into the shape of a cylinder of radius $6cm$. Find the height of the cylinder.

Answer 1

Hint: The volume of the solid will remain unchanged before and after recasting. The volume of the sphere is $\dfrac{4}{3}\pi {r^3}$ and that of the cylinder is $\pi {r^2}h$. Use these formulas and compare the volume in both the scenarios to get the height of the cylinder.

Complete Step-by-Step solution:
The radius of the sphere is given in the question which is $4.2cm$.
$ \Rightarrow {\left( r \right)_{sphere}} = 4.2cm$
We know the volume of the sphere is $\dfrac{4}{3}\pi {r^3}$. So, we have:
$
   \Rightarrow {\left( V \right)_{sphere}} = \dfrac{4}{3}\pi {\left( {{r_{sphere}}} \right)^3}, \\
   \Rightarrow {\left( V \right)_{sphere}} = \dfrac{4}{3}\pi \times {\left( {4.2} \right)^3} .....(i) \\
$
The radius of the cylinder formed after recasting is $6cm$.
$ \Rightarrow {\left( r \right)_{cylinder}} = 6cm$
 And we know that the volume of the cylinder is $\pi {r^2}h$. So, we have:
$
   \Rightarrow {\left( V \right)_{cylinder}} = \pi {r^2}h, \\
   \Rightarrow {\left( V \right)_{cylinder}} = \pi \times {\left( 6 \right)^2} \times h .....(ii) \\
$
But the volume of the solid will remain unchanged before and after recasting. Then we have:
$ \Rightarrow {\left( V \right)_{sphere}} = {\left( V \right)_{cylinder}}$
Putting values from equations $(i)$ and $(ii)$, we’ll get:
\[
   \Rightarrow \dfrac{4}{3}\pi \times {\left( {4.2} \right)^3} = \pi \times {\left( 6 \right)^2} \times h \\
   \Rightarrow 36h = \dfrac{4}{3} \times 74.088, \\
   \Rightarrow h = \dfrac{{296.352}}{{108}}, \\
   \Rightarrow h = 2.744 \\
\]
Therefore, the height of the cylinder formed is $2.744cm$.

Note: Whenever a solid is melted and converted into some other shape, the volume of the solid remains constant because recasting is done using the same amount of material. In the above problem if the solid is converted into another shape like cone or hemisphere then also we have to use the same approach.