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 $1$ cm${}^3$ of wood has a mass of $0.6$ g.

Answer 1

Inner radius $=\frac{24}{2}=12$cm
Outer radius $=\frac{28}{2}=14$cm
Height $=$ Length of pipe $=35$cm
Volume of pipe $=\pi (r_2^2-r_1^2)h$
$=\frac{22}{7}\times ({14}^2-{12}^2)\times 35$cm${}^3$
$=110\times 52$cm${}^3$
$=5720$cm${}^3$
Mass of $1$cm${}^3$ wood $=0.6$kg
Mass of $5720$cm${}^3$ wood $=(5720\times 0.6)g$
$=3432$g
$=3.432$kg