# The inner diameter of a cylindrical wooden pipe is $24 \mathrm{~cm}$ and its outer diameter is $28 \mathrm{~cm}$. The length of the pipe is $35 \mathrm{~cm}$. Find the mass of the pipe, if $1 \mathrm{~cm}^{3}$ of wood has a mass of $0.6 \mathrm{~g}$ .

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 $3.432 \mathrm{~kg}$.

Updated on: 10-Oct-2022

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