A solid metallic sphere of radius 8 cm is melted and recast into 64 identical solid spheres. The diameter of each smaller sphere formed is :

1. 4 cm

2. 2 cm

3. 8 cm

4. 1 cm

Given,

Radius of larger metallic sphere (R) = 8 cm

Let radius of each smaller sphere be r cm.

Given,

A solid metallic sphere of radius 8 cm is melted and recast into 64 identical solid spheres.

∴ Volume of larger metallic sphere = 64 × Volume of smaller metallic sphere

$\Rightarrow \dfrac{4}{3}πR^3 = 64 \times \dfrac{4}{3}πr^3 \\[1em] \Rightarrow 64 = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} \\[1em] \Rightarrow 64 = \dfrac{R^3}{r^3} \\[1em] \Rightarrow 64 = \dfrac{8^3}{r^3} \\[1em] \Rightarrow r^3 = \dfrac{8^3}{64} \\[1em] \Rightarrow r^3 = \dfrac{512}{64} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} = 2 \text{ cm}.$

Diameter of smaller sphere = 2 × 2 = 4 cm.

Hence, Option 1 is the correct option.