The surface area of a solid metallic sphere is 616 cm² . It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?

Surface area of a metallic sphere = 616 cm²

Let the radius of this sphere be R cm.

∴ 4πR² = 616

$\Rightarrow 4 \times \dfrac{22}{7} \times \text{R}^2 = 616 \\[1em] \Rightarrow \dfrac{88}{7} \times \text{R}^2 = 616 \\[1em] \Rightarrow \text{R}^2 = \dfrac{616 \times 7}{88} \\[1em] \Rightarrow \text{R}^2 = \dfrac{4312}{88} \\[1em] \Rightarrow \text{R}^2 = 49 \\[1em] \Rightarrow \text{R} = \sqrt{49} \\[1em] \Rightarrow \text{R} = 7 \text{ cm.}$

Given,

Big sphere is melted and recast into smaller spheres of diameter 3.5 cm.

Radius, r = $\dfrac{\text{diameter}}{2} = \dfrac{3.5}{2}$ = 1.75 cm

Let the number of smaller spheres formed be n.

Volume of big sphere = n × Volume of each small sphere

$\Rightarrow \dfrac{4}{3}π\text{R}^3 = \text{n} \times \dfrac{4}{3}π\text{r}^3 \\[1em] \text{Dividing both sides by 4π and multiplying by 3, we get :} \\[1em] \Rightarrow \text{R}^3 = \text{n} \times \text{r}^3 \\[1em] \Rightarrow 7^3 = \text{n} \times 1.75^3 \\[1em] \Rightarrow 343 = \text{n} \times 5.359375 \\[1em] \Rightarrow \text{n} = \dfrac{343}{5.359375} \\[1em] \Rightarrow \text{n} = 64.$

Hence, 64 small spheres can be formed.