Answer:
Volume of iron used is 58982cm
Explanation:
For hollow cylinder,
\bold{Height=9cm}Height=9cm
\bold{Internal \ radius=21cm}Internal radius=21cm
\bold{Externa \ radius=28cm}Externa radius=28cm
volume \: of \: cylinder = \pi \times {r}^{2} \times h \:volumeofcylinder=π×r
2
×h
\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part
\bold{=\frac{22}{7}×28^2×9}=
7
22
×28
2
×9
\bold{=22×4×28×9}=22×4×28×9
\bold{=71456cm^3}=71456cm
3
\bold{Volume \ of \ hollow \ part}Volume of hollow part
\bold{=\frac{22}{7}×21^2×9}=
7
22
×21
2
×9
\bold{=22×3×21×9}=22×3×21×9
\bold{=12474cm^3}=12474cm
3
\bold{Volume \ of \ iron \ used=}Volume of iron used=
\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part
\bold{- Volume \ of \ hollow \ part}−Volume of hollow part
{=71456-12474}=71456−12474
{=58982cm^3}=58982cm
3
\bold\red{Volume \ of \ iron \ used \ is \ 58982cm^3}Volume of iron used is 58982cm
3