Twenty seven solid iron spheres, each of radius $ r $ and surface area $ \mathrm{S} $ are melted to form a sphere with surface area $ S^{\prime} $. Find the
(i) radius $ r^{\prime} $ of the new sphere,
(ii) ratio of $ \mathrm{S} $ and $ \mathrm{S}^{\prime} $.

Given:

Twenty seven solid iron spheres, each of radius \( r \) and surface area \( \mathrm{S} \) are melted to form a sphere with surface area \( S^{\prime} \).

To do:

We have to find the

(i) radius \( r^{\prime} \) of the new sphere,
(ii) ratio of \( \mathrm{S} \) and \( \mathrm{S}^{\prime} \).

Solution:

(i) Radius of each solid iron sphere $=r$

This implies,

Volume of each solid iron sphere $=\frac{4}{3} \pi r^{3}$

Volume of each 27 solid iron sphere $=27\times\frac{4}{3} \pi r^{3}$

$=36 \pi r^{3}$

Radius of the new sphere $=r^{\prime}$

Therefore,

Volume of the new sphere $=\frac{4}{3} \pi r^{\prime 3}$

$\frac{4}{3} \pi r^{\prime 3}=36 \pi r^{3}$

$r^{\prime 3}=\frac{36 \times 3 \times r^{3}}{4}$

$=27 r^{3}$

$=(3 r)^{3}$

$r^{\prime}=3 r$

The radius \( r^{\prime} \) of the new sphere is $3r$.

(ii) Surface area of each iron sphere of radius $r$ is $S =4\pi r^2$

Surface area of the iron sphere of radius $r^{\prime}= 4 \pi r^{\prime 2}$

Ratio of $S$ and $S^{\prime}=4\pi r^2:4 \pi r^{\prime 2}$

$=r^2:(3r)^2$

$=r^2:9r^2$

$=1:9$

The ratio of $S$ and $S^{\prime}$ is $1: 9$.