Question

A solid iron pole consists of a cylinder of height $220cm$ and base diameter $24cm$ which is surmounted by another cylinder of height $60cm$ and radius $8cm$. Find the mass of the pole, given that $1{cm}^3$ of iron has approximately $8g$ mass (Use $\pi =3.14$)

Answer 1

Given:

Height of the lower cylinder $=220cm$

Height of the upper cylinder $=60cm$

Diameter of the lower cylinder $=24cm$

Diameter of the upper cylinder $=16cm$

Mass of iron in $1c{m}^3$ of iron $=8g$

We know that the volume of a cylinder is,

$V=\pi r^{2}h$

Therefore,

Total volume of iron in two cylinders

$=\{\pi \times {12}^2\times 220+\pi \times 8^2\times 60\}{cm}^3=35520{cm}^3=35520\times 3.14{cm}^3=111532.8{cm}^3$

Total mass of the iron pole is

$=111532.8\times 8g=\frac{11532.8\times 8}{1000}kg=892.2624kg$

Hence $892.2624kg$ is the required mass of the iron in the pole.