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A cylindrical road roller made of iron is $ 1 \mathrm{~m} $ long. Its internal diameter is $ 54 \mathrm{~cm} $ and the thickness of the iron sheet used in making the roller is $ 9 \mathrm{~cm} $. Find the mass of the roller, if $ 1 \mathrm{~cm}^{3} $ of iron has $ 7.8 \mathrm{gm} $ mass. (Use $ \left.\pi=3.14\right) $
Given:
A cylindrical road roller made of iron is \( 1 \mathrm{~m} \) long.
Its internal diameter is \( 54 \mathrm{~cm} \) and the thickness of the iron sheet used in making the roller is \( 9 \mathrm{~cm} \).
\( 1 \mathrm{~cm}^{3} \) of iron has \( 7.8 \mathrm{gm} \) mass.
To do:
We have to find the mass of the roller.
Solution:
Length of cylindrical road roller $h = 1\ m$
$= 100\ cm$
Inner diameter of the roller $= 54\ cm$
Thickness of the iron sheet $= 9\ cm$
This implies,
Inner radius of the roller $r = \frac{54}{2}$
$= 27\ cm$
Outer radus of the roller $R = 27 + 9$
$= 36\ cm$
Therefore,
Volume of the roller $= \pi R^2 h - \pi r^2 h$
$=\pi h (R^2- r^2)$
$= 3.14 \times 100 (36^2 - 27^2)$
$= 314 \times (1296- 729)$
$= 314 \times 567$
$= 178038\ cm^3$
Mass of $1\ cm^3$ of iron $= 7.8\ gm$
Therefore,
Total mass of iron $= 178038 \times 7.8\ gm$
$= 1388696.4\ gm$
$=1388.6964\ kg$
$= 1388.7\ kg$
The mass of the roller is $1388.7\ kg$.