Question

The volume of the two spheres are in the ratio $64:27$. The difference of their surface areas, if the sum of their radii is $7$ , is

• A. $28{\mathrm{cm}}^2$
• B. $88{\mathrm{cm}}^2$
• C. $64\pi {\mathrm{cm}}^2$
• D. $36\pi {\mathrm{cm}}^2$

Answer 1

The correct option is B

$88{\mathrm{cm}}^2$

Step 1: Given data

The ratio of the volumes of the two spheres, $V_1:V_2=64:27$

Sum of the radii of the two spheres, $r_1+r_2=7cm$

Differences in the surface area of the two spheres, $S_1-S_2=?$

Step 2: Calculating the difference in the surface area of the two spheres

Volume of a sphere$=\frac{4}{3}{\mathrm{\pi r}}^3$

Surface area of a sphere$=4{\mathrm{\pi r}}^2$

$$\begin{gathered} \frac{V_1}{V_2}=\frac{64}{27} \\ \Rightarrow \frac{\frac{4}{3}\pi {\left(r_1\right)}^3}{\frac{4}{3}\pi {\left(r_2\right)}^3}=\frac{64}{27}={\left(\frac{4}{3}\right)}^3 \\ \Rightarrow \frac{r_1}{r_2}=\frac{4}{3} \end{gathered}$$

Since, $r_1+r_2=7cm$

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

Required difference, $S_1-S_2=4\mathrm{\pi }{\left({\mathrm{r}}_1\right)}^2-4\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^2=4\mathrm{\pi }[{\left(4\right)}^2-{\left(3\right)}^2]$

$\Rightarrow {\mathrm{S}}_1-{\mathrm{S}}_2=4\mathrm{\pi }[16-9]=4\mathrm{\pi }\left[7\right]=28{\mathrm{\pi cm}}^2=28\times \frac{22}{7}=88{\mathrm{cm}}^2$

Hence, the difference of their surface areas will be $88{\mathrm{cm}}^2$

Hence, option B is correct.