The total surface area of a cube of edge length 2 cm is $12\times {10}^{-16}$ and the total surface area of the same material if it were subdivided into colloidal size cubes, each having an edge length of l0^-8 cm is $12\times {10}^x$. The value of x is___

Surface area of cube of edge length 1 cm - 6 × 2 cm²

Surface area of cube of edge length 2 × l0^-8 cm

$=12\times {10}^{-16}{\mathrm{cm}}^2$

Let n cubes of 12 × 10^-16 cm² are present on breaking the cube of 6 cm²

$\therefore \mathrm{n}\times \text{ volume of cube of edge length }{10}^{-8}\mathrm{cm}=\text{ volume of cube }$of edge length 2 cm.

$\begin{array}{l}\mathrm{n}\times \frac{4}{3}\times \pi \times {\left[{10}^{-8}\right]}^3=\frac{4}{3}\times \pi \times (2{)}^3\\ \therefore \mathrm{n}=8\times {10}^{+24}\end{array}$

New total surface area

$\therefore ={10}^{24}\times 12\times {10}^{-16}=12\times {10}^8$