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A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that $1\ cm^3$ of iron has approximately 8 g mass. (Use $\pi = 3.14$)
A solid iron pole consists of a cylinder with a height of 220 cm and a base diameter of 24 cm, which is surmounted by another cylinder with a height of 60 cm and a radius of 8 cm.
We have to find the mass of the pole, given that $1\ cm^3$ of iron has approximately 8 g mass.
The radius of the 1st cylinder $= 12\ cm$
Height of the 1st cylinder $= 220\ cm$
Volume of 1st cylinder $= \pi r^2h$
$= \pi (12)^2 (220)$
$= 144 \times 220 \pi$
$= 144 \times 220 \times 3.14$
$= 99475.2\ cm^3$.......(i)
The radius of the 2nd cylinder $= 8\ cm$
Height of the 2nd cylinder $= 60\ cm$
Volume of 2nd cylinder $= \pi r^2h$
$= \pi (8)^2 (60)$
$= 64 \times 60 \pi $
$= 64 \times 60 \times 3.14$
$= 12057.6\ cm^3$........(ii)
The total volume of the solid $=$ Volume of 1st cylinder $+$ Volume of 2nd cylinder
$= 99475.2 + 12057.6$
$= 111532.8\ cm^3$
Mass of $1\ cm^3$ of iron $= 8\ g$
Mass of $111532.8\ cm^3$ of iron $= 111532.8 \times 8$
$= 892262.4\ g$
$= 892.262\ kg$.
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