The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm. The difference of their total surface areas is
(a) 38 cm2
(b) 58 cm2
(c) 78 cm2
(d) 88 cm2
(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 $=4 \times \frac{22}{7} \times 4^{2}-4 \times \frac{22}{7} \times 3^{2}$
$=4 \times \frac{22}{7} \times(16-9)$
$=4 \times \frac{22}{7} \times 7$
$=88 \mathrm{~cm}^{2}$
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