A cylindrical road roller made of iron is 1 m long,

Solution:

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,

$=(178038)(7.8) \mathrm{gm}$

 

$=1388696.4 \mathrm{gm}$

$=1388.7 \mathrm{~kg}$