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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 $0.6\ gm$.
Given:
The inner diameter of a cylindrical wooden pipe is $24\ cm$ and its outer diameter is $28\ cm$. The length of the pipe is $35\ cm$.
$1\ cm^3$ of wood has a mass of $0.6\ g$.
To do:
We have to find the mass of the pipe.
Solution:
The inner diameter of the cylindrical wooden pipe $= 24\ cm$
This implies,
Inner radius $(r)=\frac{24}{2}$
$=12 \mathrm{~cm}$
The outer diameter of the cylindrical wooden pipe $= 28\ cm$
Outer radius $(R)=\frac{28}{2}$
$=14 \mathrm{~cm}$
Length of the pipe $(h)=35 \mathrm{~cm}$
Therefore,
Mass of the pipe used $=\pi h(\mathrm{R}^{2}-r^{2})$
$=\frac{22}{7} \times 35 (14^{2}-12^{2})$
$=22 \times 5(196-144)$
$=110 \times 52$
$=5720 \mathrm{~cm}^{3}$
Therefore,
Total mass of the pipe $=0.6 \times 5720 \mathrm{~g}$
$=3432 \mathrm{~g}$
$=3.432 \mathrm{~kg}$
Hence, the mass of the pipe is $3.432 \mathrm{~kg}$.