The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm.

(d) 88 cm2

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:

$\frac{\frac{4}{3} \pi(7-r)^{3}}{\frac{4}{3} \pi r^{3}}=\frac{64}{27}$

$\Rightarrow \frac{(7-\tau)}{r}=\frac{4}{3}$

$\Rightarrow 21-3 r=4 r$

$\Rightarrow 21=7 r$

$\Rightarrow r=3 \mathrm{~cm}$

Now, the radii of the two spheres are 3 cm and 4 cm.

$\therefore$ Required difference $=4 \times \frac{22}{7} \times 4^{2}-4 \times \frac{22}{7} \times 3^{2}$

$\therefore$ Required difference $=4 \times \frac{22}{7} \times 4^{2}-4 \times \frac{22}{7} \times 3^{2}$

$=4 \times \frac{22}{7} \times(16-9)$

$=4 \times \frac{22}{7} \times 7$

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