Question

A solid iron pole having a cylindrical portion $110\text{cm}$ high and of base diameter $12\text{cm}$ is surmounted by a cone $9\text{cm}$ high. Find the mass of the pole, given that the mass of $1{\text{cm}}^3$ of iron is $8\text{gm}$.

Answer 1

Given: Diameter of cone $=12\text{cm}$

So, Radius of cone $=6\text{cm}$

and height of cone $=9\text{cm}$

Volume of cone $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times 6\times 6\times 9$

$=339.428{\text{cm}}^3$

Radius of the cylinder, $R=$ Radius of cone $=6\text{cm}$

Height of the cylinder, $H=110\text{cm}$

Volume of cylinder $=\pi R^{2}H$

$=\frac{22}{7}\times 6\times 6\times 110$

$=12445.714{\text{cm}}^3$

Hence,

Volume of pole $=$ Volume of the conical part $+$ Volume of the cylindrical part

$=339.43+12445.714$

$=12785.142{\text{cm}}^3$

Required mass of pole $=8\times 12785.142$

$=102281.13\text{gm}$

$=102.281\text{kg}$