Question

A cylindrical road roller made of iron is 1m 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 $c{m}^3$ of iron has 7.8 gm mass.$(Use\pi =3.14)$

Answer 1

Solution:

As per the parameters given in the question, we have]

Height of the cylindrical road roller = h = 1 m = 100 cm

Internal Diameter of the cylindrical road roller = 54 cm

Internal radius of the cylindrical road roller = 27 cm = r (as we know that the radius is half of the diameter)

Given the thickness of the road roller (T) = 9 cm

Let us assume that the outer radii of the cylindrical road roller be R.

T = R – r

9 = R – 27

R = 27 + 9

R = 36 cm

Now,

The volume of the iron sheet ,

$V=\pi \times (R^2-r^2)\times h$

$=\pi \times ({36}^2-{27}^2)\times 100=1780.38c{m}^3$

So, the volume of the iron sheet = V = 1780.38 cm³

Mass of 1 cm³ of the iron sheet = 7.8 gm

So, the mass of 1780.38 cm³ of the iron sheet = 1388696.4gm = 1388.7 kg

Hence, the mass of the road roller (m) = 1388.7 kg