Question

The volumes of two spheres are in the ratio 64:27 and the sum of their radii is 7 cm. What is the difference in their total surface areas?

• A. $88c{m}^2$
• B. $44c{m}^2$
• C. $84c{m}^2$
• D. $100c{m}^2$

Answer 1

The correct option is A $88c{m}^2$

Let $r_1$ and $r_2$ be the radii of two spheres, the ratio of volumes of which is 64 : 27.
i.e., $\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}=\frac{64}{27}$
$⟹\frac{r_1^3}{r_2^3}=\frac{4^3}{3^3}$
$⟹(\frac{r_1}{r_2}{)}^3=(\frac{4}{3})³$
$⟹\frac{r_1}{r_2}=\frac{4}{3}$

Since the sum of the radii is 7 cm, we should have $r_1=4cm$ and $r_2=3cm$
Then, the difference between the surface areas of the two spheres is
= 4πr₁² - 4πr₂²
= 4π (r₁² - r₂²)
= 4π (4² - 3²)
= 4π (16-9)
= 4π (7)
= 4$\times \frac{22}{7}\times$ 7
= $88c{m}^2$