Question

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm.The length of the pipe is 35 cm. Find the mass of the pipe, if $1c{m}^3$ of wood has a mass of 0.6 g.
$[Assume\pi =\frac{22}{7}]$

Answer 1

Inner radius,
$\left(r_1\right)$ of the cylindrical pipe $=\left(\frac{24}{2}\right)cm=12cm$

Outer radius,
$\left(r_2\right)$ of the cylindrical pipe $=\left(\frac{28}{2}\right)cm=14cm$

Height (h) of the pipe = length of the pipe = 35 cm

Volume of the pipe
$=\pi (r_2^2-r_1^2)hc{m}^3=110\times 52c{m}^3=5720c{m}^3$

Mass of $1c{m}^{3}wood=0.6g$

Mass of 5720 $c{m}^{3}wood=(5720\times 0.6)g=3432g$

= 3.432 kg (1000 g = 1 kg)