Question

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is $\frac{1}{4}$ of the radius of the original ball, how many such balls are made ? Compare the surface area, of all the smaller balls combined together with that of the original ball.

Answer 1

Let R and r be the radius of the original ball and small ball respectively.

So, $r=\frac{1}{4}R=>R=4r----\left(1\right)$

$Noofballsmade=\frac{Volumeoforiginalball}{Volumeofeachsmallball}$

$=\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}$

$=\frac{R^3}{r^3}$

Using (1),

$=\frac{(4r{)}^3}{r^3}=64$

$\frac{TotalSurfaceareaof64smallball}{Surfaceareaoforiginalball}=\frac{\left(64\right)4\pi r^2}{4\pi R^2}=\frac{64r^2}{(4r{)}^2}=\frac{64(r{)}^2}{16r^2}=\frac{4}{1}$

∴ Required ratio = 4: 1