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Question 91
If the radius of a sphereis increased by 4 cm, its surface areais increased by $$464 \pi cm^2$$ . What is the volume(in $$cm^3$$) of the original sphere?
Solution
Difference in the surface area = $$464 \pi$$
$$4\pi(r + 4)^2 - 4\pi r^2 = 464 \pi$$
$$4\pi[r^2 + 16 + 8r -Â r^2] =Â 464 \pi$$
16 + 8r = 116
r = 100/8 = 25/2Â cmÂ
Volume of the the sphere = $$\frac{4}{3} \pi r^3$$
=Â $$\frac{4}{3} \pi (25/2)^3$$ = $$\frac{15625}{6} \pi$$
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