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 cm3 of wood has a mass of 0.6 gm.
Solution:
Given,
Inner radius (r1) of a cylindrical pipe = 24/2 = 12 cm
Outer radius (r2) of a cylindrical pipe = 24/2 = 14 cm
Height of pipe (h) = length of pipe = 35 cm
Mass of pipe $=$ volume $\times$ density $=\pi\left(\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}\right) \mathrm{h}$
$=22 / 7\left(14^{2}-12^{2}\right) 35=5720 \mathrm{~cm}^{3}$
Mass of $1 \mathrm{~cm}^{3}$ wood $=0.6 \mathrm{gm}$
Therefore, mass of $5720 \mathrm{~cm}^{3}$ wood $=5720 \times 0.6=3432 \mathrm{gm}=3.432 \mathrm{~kg}$