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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 $\pi = 3.14$)

Given:

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.

To do:

We have to find the mass of the pole, given that $1\ cm^3$ of iron has approximately 8 g mass.

Solution:

The radius of the 1st cylinder $= 12\ cm$

Height of the 1st cylinder $= 220\ cm$

Therefore,

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 $= 60\ cm$

This implies,

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)

Therefore,

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 $= 8\ g$

This implies,

Mass of $111532.8\ cm^3$ of iron $= 111532.8 \times 8$

$= 892262.4\ g$

$= 892.262\ kg$.

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