Question

Find the ratio of the volume of a cube to that of a sphere which will fit inside it.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{shere}\mathrm{be}R\mathrm{and}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cube}\mathrm{be}a. \\ \mathrm{As},\mathrm{the}\mathrm{sphere}\mathrm{is}\mathrm{fit}\mathrm{inside}\mathrm{the}\mathrm{cube}. \\ \mathrm{So},\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{sphere}=\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cube} \\ \Rightarrow 2R=a.....\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cube}\mathrm{to}\mathrm{that}\mathrm{of}\mathrm{the}\mathrm{sphere}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cube}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{sphere}} \\ =\frac{a^3}{\left({\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3\right)} \\ =\frac{{\left(2R\right)}^3}{\left({\displaystyle \frac{4}{3}\mathrm{\pi }{R}^3}\right)}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =\frac{3\times 8R^3}{{\displaystyle 4\mathrm{\pi }{R}^3}} \\ =\frac{6}{\mathrm{\pi }} \\ =6:\mathrm{\pi } \end{gathered}$$

So, the ratio of the volume of the cube to that of the sphere is 6 : $\mathrm{\pi }$.