Question

The radius of a solid iron sphere is $8cm$. Eight rings of iron plate of external radius $6\frac{2}{3}cm$ and thickness $3cm$ are made by melting this sphere. Find the internal diameter of each ring.

Answer 1

We have
Volume of solid iron sphere $\frac{4}{3}\pi \times 8^3{cm}^3=\frac{2048}{2}\pi {cm}^3$

External radius of each iron ring $=6\frac{2}{3}cm=\frac{20}{3}cm$

Let the internal radius of each ring be $r$ cm

Since each ring forms a hollow cylindrical shell of external and internal radii $\frac{20}{3}cm$ and $r$ cm respectively and height $3cm$

Volume of each ring $=\pi \{{\left(\frac{20}{3}\right)}^2-r^2\}\times 3{cm}^3$

volume of 8 such rings
$=8\pi (\frac{400}{9}-r^2)\times 3{cm}^3=24\pi (\frac{400}{9}-r^2){cm}^3$

Clearly volume of 8 rings= volume of the sphere

$\Rightarrow 24\pi (\frac{400}{9}-r^2)=\frac{2048}{2}\pi$

$\Rightarrow \frac{400}{9}-r^2=\frac{2048}{2}\pi \times \frac{1}{24\pi }$

$\Rightarrow r^2=16$

$\Rightarrow r=4cm$