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 7.

Answer 1

Given that, the ratio of the volumes of two spheres is $64:27$ and sum of their radii is $7$.

To find out,

Difference of the surface areas of two spheres

Let $r_1$ and $r_2$ be the radii of two spheres.

$\therefore r_1+r_2=7........\left(i\right)$

We know that, $\text{Volume of a sphere}=\frac{4}{3}\pi r^3$

$\therefore \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}=\frac{64}{27}$

$\Rightarrow {\left(\frac{r_1}{r_2}\right)}^3=\frac{64}{27}\Rightarrow \frac{r_1}{r_2}={\left(\frac{64}{27}\right)}^{1/3}=\frac{4}{3}$

$\therefore \frac{r_1}{r_2}=\frac{4}{3}$

$\Rightarrow r_1=\frac{4}{3}{r}_2.........\left(ii\right)$

Using $\left(i\right)$ and $\left(ii\right)$, we get:

$\frac{4}{3}{r}_2+r_2=7$

$\Rightarrow \frac{7}{3}{r}_2$

$\therefore r_2=3$

Using $\left(ii\right)$, we get:

$r_1=\frac{4}{3}\times 3$

$\therefore r_1=4$

We know that, $\text{Surface area of a sphere}=4\pi r^2$

$\therefore$ difference between the surface areas will be:

$4\pi {r_1}^2-4\pi {r_2}^2=4\pi ({r_1}^2-{r_2}^2)$

$\Rightarrow 4\times \frac{22}{7}(16-9)$ [Using $\pi =\frac{22}{7},r_1=4\text{and}{r}_2=3$]

$\Rightarrow \frac{88}{7}\times 7$

$=88\text{square units}$

Hence, the difference of the surface areas of the two given spheres is $88$.