Question

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 ${1cm}^3$ of iron has approximately 8 g mass. (Use $\pi =3.14$) [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks

Radius of big cylinder
$=r_1=\frac{24}{2}=12cm$

Radius of small cylinder
$r=8cm$

Height of big cylinder
$=h_1=220cm$

Height of small cylinder
$=h=60cm$

Volume of big cylinder
$=\pi .(r_1{)}^2.h_1=3.14\times (12{)}^2\times 220=99475.2c{m}^3$

Volume of small cylinder
$=\pi .r^2.h=3.14\times (8{)}^2\times 60=12057.6c{m}^3$

Volume of Pole = Volume of Big Cylinder + Volume of small Cylinder

$=99475.2+12057.6=111532.8c{m}^3$

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

$111532.8c{m}^{3}ofiron=111532.8\times 8$

$=892262.4g=892.262kg$ (1000gm= 1kg)