Question

A spherical shell of lead, whose external diameter is$18cm$, is melted and recast into a right circular cylinder, whose height is$8cm$ and diameter is$12cm$. Determine the internal diameter of the shell.

Answer 1

Hint: Consider a symbols form internal radii of shell and since it is melted and recast
volume of spherical shell=volume of right circular cylinder and solve it.
So first of all we have to find the internal diameter of the shell.
So for spherical shell,
So in question it is given that the spherical shell of lead was melted and then recast into a right circular cylinder.
So now we have given the external diameter of the spherical shell of lead as$18cm$.
So we know that radius is half of diameter.
Let us assume ${{R}_{1}}$as an external radius of a spherical shell.
So external radius of spherical shell can be calculated as,
${{R}_{1}}=\dfrac{18}{2}=9cm$
So we get external radius${{R}_{1}}$as$9cm$,
So let us assume that${{R}_{2}}$is the internal radius of the spherical shell.
So for right circular cylinder,
Let us assume$H$as the height of the right circular cylinder .
It is given that the height of the right circular cylinder $H$is$8cm$.
So the diameter of the right circular cylinder is given as$12cm$.
So the radius is half of diameter.
Let us assume $R$ as the radius of the right circular cylinder.
So the radius is,
$R=\dfrac{12}{2}=6cm$
As the spherical shell is melted and recast to form a right circular cylinder.
So their volumes will be equal.
So volume of spherical shell=volume of right circular cylinder……..(1)
We know,
Volume of spherical shell$=\dfrac{4}{3}\pi (R_{1}^{3}-R_{2}^{3})$……….(2)
Volume of right circular cylinder$=\pi {{R}^{2}}H$………….(3)
So from (1), (2) and (3), we get,
$\dfrac{4}{3}\pi (R_{1}^{3}-R_{2}^{3})=\pi {{R}^{2}}H$
$\dfrac{4}{3}(R_{1}^{3}-R_{2}^{3})={{R}^{2}}H$
Now substituting the given values for ${{R}_{1}},RandH$We get,
$\begin{align}
  & \dfrac{4}{3}({{9}^{3}}-R_{2}^{3})=({{6}^{2}})8 \\
 & \dfrac{4}{3}({{9}^{3}}-R_{2}^{3})=36\times 8 \\
 & \dfrac{4}{3}({{9}^{3}}-R_{2}^{3})=288 \\
 & 4({{9}^{3}}-R_{2}^{3})=288\times 3 \\
 & ({{9}^{3}}-R_{2}^{3})=\dfrac{288\times 3}{4} \\
 & ({{9}^{3}}-R_{2}^{3})=216 \\
\end{align}$
Now rearranging the equation we get,
$\begin{align}
  & {{9}^{3}}-216=R_{2}^{3} \\
 & 729-216=R_{2}^{3} \\
 & 513=R_{2}^{3} \\
\end{align}$
$\begin{align}
  & R_{2}^{3}=513 \\
 & {{R}_{2}}={{(513)}^{\dfrac{1}{3}}} \\
\end{align}$
So we can write ${{(513)}^{\dfrac{1}{3}}}={{(27\times 19)}^{\dfrac{1}{3}}}$we get,
$\begin{align}
  & {{R}_{2}}={{(27\times 19)}^{\dfrac{1}{3}}} \\
 & {{R}_{2}}={{({{3}^{3}}\times 19)}^{\dfrac{1}{3}}} \\
\end{align}$
Now taking ${{3}^{3}}$outside the bracket we get,
${{R}_{2}}=3{{(19)}^{\dfrac{1}{3}}}$
So we have got the internal radius of spherical shell${{R}_{2}}$as $3{{(19)}^{\dfrac{1}{3}}}$.
So we know that diameter is twice of radius, we get,
Let internal diameter be ${{D}_{2}}$,
So ${{D}_{2}}=2{{R}_{2}}=2\times 3{{(19)}^{\dfrac{1}{3}}}=6{{(19)}^{\dfrac{1}{3}}}$
So ${{D}_{2}}=6{{(19)}^{\dfrac{1}{3}}}cm$
So we get that the internal diameter of the spherical shell of lead is $6{{(19)}^{\dfrac{1}{3}}}cm$.
Note: Read the question in a thorough manner. It is given that the spherical shell is converted to a right
 circular cylinder. You must know the formula of volume of spherical shell$=\dfrac{4}{3}\pi (R_{1}^{3}-R_{2}^{3})$and volume of right circular cylinder$=\pi {{R}^{2}}H$. Give a proper notation for internal and external radius and diameter so that you do not get confused.