A solid iron pole consists of a cylinder of height $220 \mathrm{~cm}$ and base diameter $24 \mathrm{~cm}$, which is surmounted by another cylinder of height $60 \mathrm{~cm}$ and radius $8 \mathrm{~cm}$. Find the mass of the pole, given that $1 \mathrm{~cm}^{3}$ of iron has approximately $8 \mathrm{~g}$ mass.
(Use $\pi=3.14$ )
First cylindrical part has height 220 cm and radius 12 cm.
Its volume $=\pi \times(12)^{2} \times 220 \mathrm{~cm}^{3}$.
Second cylindrical part has height $60 \mathrm{~cm}$ and radius $8 \mathrm{~cm}$.
Its volume $=\pi \times(8)^{2} \times 60 \mathrm{~cm}^{3}$
Total volume $=\{144 \times 220+64 \times 60\} \pi \mathrm{cm}^{3}$
$=35520 \pi \mathrm{cm}^{3}=35520 \times 3.14 \mathrm{~cm}^{3}$
$=111532.8 \mathrm{~cm}^{3}$
Total weight (at the rate of $8 \mathrm{gm}$ per $1 \mathrm{~cm}^{3}$ )
$=\frac{\mathbf{1 1 1 5 3 2 . 8} \times \mathbf{8}}{\mathbf{1 0 0 0}} \mathbf{k g}=111.5328 \times 8 \mathrm{~kg}=892.2624 \mathrm{~kg}$