# 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 $1388.7\ kg$.

Updated on: 10-Oct-2022

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