Question

A cylinder whose height is two-third of its diameter has the same volume as a sphere of radius $4cm$ . Calculate the radius of the base of the cylinder.

Answer 1

Hint: At first, we assume the height of the cylinder is “h” and its diameter is “d”. We can write, $h=\dfrac{2}{3}d$ . The volume of the cylinder becomes ${{V}_{c}}=\dfrac{1}{4}\pi {{d}^{2}}\left( \dfrac{2}{3}d \right)=\dfrac{1}{6}\pi {{d}^{3}}$ . The volume of a sphere of radius r is given as \[{{V}_{s}}=\dfrac{4}{3}\pi {{\left( 4 \right)}^{3}}=268.08cc\]. Equating ${{V}_{c}},{{V}_{s}}$ , we get the value of d and then we take $\dfrac{d}{2}$ to get the radius.

Complete step by step answer:
In this problem we are given the details of the dimensions of a cylinder and the details of the dimension of a sphere. It is said that the height of the cylinder is two-third of its diameter. The volume of the cylinder is the same as that of the sphere whose radius is $4cm$ .
Let the height of the cylinder is “h” and its diameter is “d”. Since, the height of the cylinder is two-third of its diameter, we can write,
$h=\dfrac{2}{3}d$
The volume of a cylinder of diameter d and height h is ${{V}_{c}}=\dfrac{1}{4}\pi {{d}^{2}}h$ . Substituting the value of h in the volume formula, we get,
$\Rightarrow {{V}_{c}}=\dfrac{1}{4}\pi {{d}^{2}}\left( \dfrac{2}{3}d \right)=\dfrac{1}{6}\pi {{d}^{3}}$
The volume of a sphere of radius r is given as ${{V}_{s}}=\dfrac{4}{3}\pi {{r}^{3}}$ . Putting the value of r as $4cm$ , we get,
\[{{V}_{s}}=\dfrac{4}{3}\pi {{\left( 4 \right)}^{3}}=268.08cc\]
It is given that the volume of the cylinder and that of the sphere are equal. So, we can write,
$\begin{align}
  & {{V}_{c}}={{V}_{s}} \\
 & \Rightarrow \dfrac{1}{6}\pi {{d}^{3}}=268.08cc \\
 & \Rightarrow d=\sqrt[3]{\dfrac{268.08\times 6}{\pi }}=8cm \\
\end{align}$
Thus, we can conclude that the radius of the base of the cylinder is $\dfrac{d}{2}=4cm$ .

Note: The one basic thing of these types of problems is that we need to have a clear idea of which quantity to take as the variable. It makes the whole problem look easy. We should take care of the cube root and should not mistakenly do the square root. Also, a special care of the units must be taken.