A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm, and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller if 1 cm³ of iron has a mass of 7.8 g. (Use π = 22/7)

Given:
Cylindrical road roller height (h) = 1 m = 100 cm
Cylindrical road roller internal diameter = 54 cm
Cylindrical road roller internal radius, $r_1$= 54/2 = 27 cm
The thickness of the road roller (t) = 9 cm

Suppose the outer radii of the cylindrical road roller be $r_2$.

So,
$r_2=r_1+\mathrm{t}$
$r_2=36\mathrm{ cm}$

Therefore, the volume of the roller is:

$\mathrm{V}=\pi \left({\mathrm{r}}_2^2-{\mathrm{r}}_1^2\right)\times \mathrm{h}$

$\Rightarrow \mathrm{V}=\frac{22}{7}\left({36}^2-{27}^2\right)\times 100$

$\mathrm{V}=1,78,200{\mathrm{ cm}}^3$

As, 1 cm³ of iron has 7.8 g mass. So the density,

$\mathrm{\rho }=7.8 \frac{\mathrm{g}}{{\mathrm{cm}}^3}$

Therefore, the mass of the roller,

$\mathrm{m}=\mathrm{V}\times \mathrm{\rho }$

$\Rightarrow \mathrm{m}=1,78,200\times 7.8$

$\Rightarrow \mathrm{m}=13,89,960 \mathrm{g}$

$\Rightarrow \mathrm{m}=1389.960 \mathrm{kg}$

Therefore, the mass of the roller is 1390 kg (Approx.).