3 Tutor System
Starting just at 265/hour

# Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find thei) radius r' of the new sphere,ii) ratio of S and S'.

i) Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere = $$\frac{4}{3}{\pi}{r}^3$$
$$\therefore$$ Volume of 27 solid iron spheres
= 27 × $$\frac{4}{3}{\pi}{r}^3$$

27 solid iron spheres are melted to form 1 iron sphere.

Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres.

Now, Let the radius of this new sphere be r'.

$$\therefore$$ Volume of new solid iron sphere = $$\frac{4}{3}{\pi}{r'}^3$$
$$\Rightarrow$$ $$\frac{4}{3}{\pi}{r'}^3$$ = 27 × $$\frac{4}{3}{\pi}{r}^3$$
$$\Rightarrow$$ $${r'}^3 = 27 × {r}^3$$
$$\Rightarrow$$ r' = 3r.

ii) Now, we have,
surface area of 1 solid iron sphere of radius r = $$4{\pi}{r}^2$$
So, Surface area of iron sphere of radius r'
= $$4{\pi}{r'}^2$$
= $$4{\pi}{3r}^2$$
= $$36{\pi}{r}^2$$

Therefore, ratio of S and S'
= $$\frac{S}{S'}$$
= $$\frac{4{\pi}{r}^2}{36{\pi}{r}^2}$$
= $$\frac{1}{9}$$
= 1 : 9