An iron s$rac{S_{2}}{S}=rac{64 imes 4 pi imes(r)^{2}}{4 pi imes(4 r)^{2}}=4$pherical ball has been melted and recast into smaller balls of equal size.
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 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.
Let the radius of the big metallic ball is 4r. Therefore, the volume of the big metallic ball is
$V=\frac{4}{3} \pi \times(4 r)^{3}$
The metallic sphere is melted to produce small balls of radius $\frac{4 r}{4}=r$. Then, the volume of each of the small balls is
$V_{1}=\frac{4}{3} \pi \times(r)^{3}$
Since, the volume of the big metallic ball is equal to the sum of the volumes of the small balls, we have the number of produced small balls is
$\frac{V}{V_{1}}=\frac{\frac{4}{3} \pi \times(4 r)^{3}}{\frac{4}{3} \pi \times(r)^{3}}$
$=(4)^{3}$
$=64$
Hence, the number of small balls is 64
The surface area of the big ball is
$S=4 \pi \times(4 r)^{2}$
The surface area of each of the small ball is
$S_{1}=4 \pi \times(r)^{2}$
Therefore, the total surface area of the 64 small balls is
$S_{2}=64 \times 4 \pi \times(r)^{2}$
Now, we compute the following ratio