Twenty seven solid iron spheres, each of radius $r$ and surface area $S$, are melted to form a sphere with surface area $S'$. Find the
(a) Radius $r'$ of the new sphere and
(b) Ratio of $S$ and $S'$

Hint: In this question, we know that 27 solid spheres are melted to form a larger sphere. So, the volume of the larger sphere must be equal to the sum of volumes of all the 27 smaller spheres. So, if we equalize the volume of the larger sphere with the sum of the volumes of the 27 smaller spheres, we can find a relation between $r$ and $r'$. Then, using this value we can get the surface area of the larger sphere in terms of $r$ and thus find the ratio of $S$ and $S'$.


Complete step-by-step answer:
In this question, we are given that twenty seven solid iron spheres, each of radius $r$ and surface area $S$, are melted to form a sphere with surface area $S'$.
We are asked to find out the values of radius $r'$ of the new sphere and ratio of $S$ and $S'$.
We know that the formula for surface area of a sphere is equal to $4\pi {{r}^{2}}$……….(1.1)
and that of its volume is $\dfrac{4}{3}\pi {{r}^{3}}$…………(1.2).
So, the surface area of the smaller spheres with radius $r$ and surface area $S$ is equal to
$S=4\pi {{r}^{2}}.............(1.3)$ (using equation 1.1)
Now, volume of the new larger sphere is equal to the sum of the volumes of the 27 smaller spheres, that is:
$\begin{align}
& 27\times \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}\text{ }(\text{using equation 1}\text{.2}) \\
& \Rightarrow 27{{r}^{3}}={{\left( r' \right)}^{3}} \\
& \Rightarrow {{\left( 3r \right)}^{3}}={{\left( r' \right)}^{3}} \\
& \Rightarrow 3r=r'.....................(1.4) \\
\end{align}$So now,
$\begin{align}
& S'=4\pi {{\left( r' \right)}^{2}}...............(\text{using equation 1}\text{.1}) \\
& =4\pi {{\left( 3r \right)}^{2}}.................(\text{from equation 1}\text{.3}) \\
& =4\pi \times 9{{r}^{2}} \\
& =9\left( 4\pi {{r}^{2}} \right) \\
& =9S...............(1.5) \\
\end{align}$Now, the ratio of $S$ and $S'$ is equal to
$\dfrac{S}{S'}=\dfrac{S}{9S}=\dfrac{1}{9}$
So, the ratio of $S$ and $S'$ is equal to $1:9$.
.


Note:Note that in this case, we should use the fact that the total volume of the iron remains the same but the surface area is decreased by 9 times. However, we do not need to calculate the actual surface area as only the ratios are required in this question.