The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm.

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 $=5720 \times 0.6=3432 \mathrm{gm}=3.432 \mathrm{~kg}$

 

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