Question

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

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

Answer 1

(i)

Radius of $1$ solid iron sphere $=r$

Volume of $1$ solid iron sphere $=\frac{4}{3}\pi r^3$

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

$27$ solid iron spheres are melted to form $1$ iron sphere.

Therefore, the volume of this iron sphere will be equal to the volume of $27$ solid iron spheres.

Let the radius of this new sphere be $r^{\prime }$.

Volume of new solid iron sphere $=\frac{4}{3}\pi (r^{\prime }{)}^3$

$\frac{4}{3}\pi (r^{\prime }{)}^3=27\times \frac{4}{3}\pi r^3$

$\Rightarrow (r^{\prime }{)}^3=27r^3$

$\Rightarrow (r^{\prime }{)}^3=(3r{)}^3$

$\Rightarrow r^{\prime }=3r$

(ii)

Surface area of $1$ solid iron sphere of radius $\left(S\right)=4\pi r^2$

Surface area of iron sphere of radius $\left(S^{\prime }\right)=4\pi (r^{\prime }{)}^2$

$\Rightarrow S^{\prime }=4\pi (3r{)}^2$

$\Rightarrow S^{\prime }=36\pi r^2$

Now,

$\frac{S}{S^{\prime }}=\frac{4\pi r^2}{36\pi r^2}$

$\Rightarrow \frac{S}{S^{\prime }}=\frac{1}{9}$

$\therefore S:S^{\prime }=1:9$