A hollow iron sphere with an outer radius of 14 cm is melted and recast into 64 identical solid spheres, each of radius 3 cm. Find the inner radius (nearest to the integer) of the original sphere.

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Accepted Answer

1. Volume equality: (4/3)π(outer^3 - inner^3) = 64 × (4/3)π × (3^3).

2. Cancel (4/3)π and compute: $14^3 - r^3 = 64	imes27 = 1728$. So $r^3 = 2744 - 1728 = 1016$.

3. Take cube root: $r=\sqrt[3]{1016}\approx10.05$. Nearest integer is **10 cm**.

FINAL_ANSWER: a

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