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 gm.

Answer 1

Inner diameter of cylindrical wooden pipe = 24 cm

$\therefore$ Inner radius (r) = $\frac{24}{2}$ = 12 cm

and outer radius (R) = $\frac{28}{2}$ = 14 cm

Lenght of pipe (h) = 35 cm

$\therefore$ Volume of the pipe = $\pi h(R^2-r^2)$

=$\frac{22}{7}\times 35({14}^2-{12}^2)c{m}^3$

= $22\times 5(196-144)c{m}^3$

= $110\times 52=5720c{m}^3$

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

$\therefore$ Total mass = 0.6 $\times$ 5720 g

= 3432 g = 3.432 kg