Question

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 radius r’ of the new sphere.

Answer 1

Hint: To find the radius r’, we find the volume of the new sphere and then we find the volume of 27 old spheres. Then we equate them to find r’.

Complete step-by-step answer:

We know that
Surface area of sphere =$4\pi {{\text{r}}^2}{\text{ sq units}}$ and
Volume of the sphere = $\dfrac{4}{3}\pi {{\text{r}}^3}{\text{ cubic units}}$

Given Data, r be the radius of each solid iron sphere and r' the radius of new solid iron sphere, then

Volume of new sphere = $\dfrac{4}{3}\pi {\text{r}}{{\text{'}}^3}{\text{ cubic units}}$

Volume of old sphere = $\dfrac{4}{3}\pi {{\text{r}}^3}{\text{ cubic units}}$

Now, volume of 27 solid sphere of radius r = 27 × $\dfrac{4}{3}\pi {{\text{r}}^3}{\text{ cubic units}}$ = $36\pi {{\text{r}}^3}{\text{ cubic units}}$

 Given, 27 solid iron spheres are melted to form a new sphere with radius r'.

Hence, volume of new sphere with radius r’ = Volume of 27 old spheres of radius r

⟹$\dfrac{4}{3}\pi {\text{r}}{{\text{'}}^3}{\text{ cubic units}}$ = $36\pi {{\text{r}}^3}{\text{ cubic units}}$
$
   \Rightarrow \dfrac{4}{3}\pi {\text{r}}{{\text{'}}^3} = {\text{ 36}}\pi {{\text{r}}^3} \\
   \Rightarrow {\text{r}}{{\text{'}}^3} = {\text{ 27}}{{\text{r}}^3} \\
   \Rightarrow {\text{r' = 3r}} \\
$
Hence, radius r' of the new sphere = 3r

Note: The key in solving such problems is to know that the volume of a new sphere with radius r’ is equal to the volume of 27 solid spheres of radius r. whenever one shape is melted to the other the volume remains constant. Surface area occupies two dimensional space hence it is measured in square units, whereas volume occupies a three dimensional space so it is measured in cubic units.