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 $\pi=3.14$ )
We have to find the mass of the roller.
Radius of inner cylinder $\left(r_{1}\right)=27 \mathrm{~cm}$
Radius of outer cylinder
$\left(r_{2}\right)=(27+9) \mathrm{cm}$
$=36 \mathrm{~cm}$
Length of the cylinder $(h)=100 \mathrm{~cm}$
So, volume of iron,
$=\pi h\left(r_{2}^{2}-r_{1}^{2}\right)$
$=(3.14)(100)(1296-729)$
$=178038 \mathrm{~cm}^{3}$
It is given that, $1 \mathrm{~cm}^{3}$ of iron has a mass of $7.8 \mathrm{gm}$.
So the mass of iron used,
$=(178038)(7.8) \mathrm{gm}$
$=1388696.4 \mathrm{gm}$
$=1388.7 \mathrm{~kg}$