A large solid iron sphere of diameter 10m is melted and 2/5 of the melted amount is recast into several small spheres of diameter 2m. What is the ratio of the total surface area of the smaller spheres and the surface area of the large sphere?

Volume of large sphere$$=\frac{4\pi\ }{3}\times\ \left(\ \frac{\ 10}{2}\right)^3=\frac{4\pi\times\ 5^3\ }{3}$$

2/5 of the volume is recast into several spheres of 2m diameter.

Let the number of small spheres be '$$n$$'

$$n\times\ \frac{4\pi}{3}\times\ \left(\frac{2}{2}\right)^3\ =\frac{2}{5}\times\ \frac{4\pi\times\ 5^3\ }{3}$$

On solving, we get $$n=50$$.

The ratio of the total surface area of the smaller spheres and the surface area of the large sphere= $$\left(50\times\ 4\pi r^2\right)\ :\ \left(4\pi R^2\right)$$

= $$\left(50\times\ 4\pi\left(1\right)^2\right)\ :\ \left(4\pi\left(5\right)^2\right)$$

$$=2\ :\ 1$$