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.

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

Given, big spheres are converted into smaller spheres of diameter = 3.5 cm or radius = $\dfrac{3.5}{2}$.

Let the number of smaller spheres formed be n.

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

$\therefore \dfrac{4}{3}πR^3 = n \times \dfrac{4}{3}πr^3$

Dividing both sides by 4π and multiplying by 3 we get,

$\Rightarrow R^3 = nr^3 \\[1em] \Rightarrow 7^3 = n \times \Big(\dfrac{3.5}{2}\Big)^3 \\[1em] \Rightarrow n = \dfrac{7^3 \times 2^3}{3.5^3} \\[1em] \Rightarrow n = 2^3 \times 2^3 \\[1em] \Rightarrow n = 64.$

Hence, 64 small spheres can be formed.