Question

The volumes of two spheres are in the ratio $64:27$. Find the difference of their surface area if the sum of their radii is $7cm$.

Answer 1

Step 1: Find the radius of the two spheres

Given that, the ratio of the volumes of two spheres is $64:27$

The volume of a sphere of radius $r$ $=\frac{4}{3}{\mathrm{\pi r}}^3$

Let us consider $r_1$ and $r_2$ be the radius of the two spheres.

The ratio of the volume of the two spheres$=\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\left({\mathrm{r}}_1\right)}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^3}=\frac{64}{27}$

$\Rightarrow \frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3}=\frac{64}{27}$

$\Rightarrow \frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3}=\frac{4^3}{3^3}$

$\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{4}{3}$

and $r_1+r_2=7$

Let $r_1=4x$ and $r_2=3x$

$$\begin{gathered} \Rightarrow 3x+4x=7 \\ \Rightarrow 7x=7 \\ \Rightarrow x=1 \end{gathered}$$

So, $r_1=4cm$ and $r_2=3cm$

Therefore, the radius of first sphere $r_1=4cm$ and radius of second sphere $r_2=3cm$

Step 2: Find the difference of surface areas of the two spheres

The surface area of sphere of radius $r$ $=4{\mathrm{\pi r}}^2$

The difference of the surface area of the two spheres$=4\mathrm{\pi }{\left({\mathrm{r}}_1\right)}^2-4\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^2$

$$\begin{gathered} =4\mathrm{\pi }\left({\mathrm{r}}_1^2-{\mathrm{r}}_2^2\right) \\ =4\mathrm{\pi }\left(4^2-3^2\right) \\ =4\mathrm{\pi }\left(16-9\right) \\ =4\times \frac{22}{7}\times 7\left(\because \mathrm{\pi }=\frac{22}{7}\right) \\ =4\times 22 \\ =88{\mathrm{cm}}^2 \end{gathered}$$

Hence, the difference of surface area of the two spheres is $88c{m}^2$.