Question

A 890kg culvert is made of a hollow cylindrical material with outer radius of 76cm and an inner radius of 62cm. It crosses a road of width 3m. Determine the density of the material used in its construction in Kg/m3 correct to one decimal place.

Answer 1

Given :

Mass of culvert = 890 kg

Hollow cylinder

Outer radius = 76 cm

Inner radius = 62 cm

Width of road = 3 m

To find :

Density of material used.

Solution :

Density of a material is equal to mass of that material per unit volume of that material.

So, The formula of Density :

$\\ \bf \rho \: = \frac{mass}{volume}$

(equation 1)

It is given that,

Outer radius = 76 cm = 0.76 m

Inner radius = 62 cm = 0.62 m

The formula of volume of Hollow cylinder :

$\bf \: V = \pi({r_1}^{2} - {r_2}^{2}) \times h$

(equation 2)

where, in this case,

r1 = outer radius of hollow cylinder = 0.76 m

r2 = inner radius of hollow cylinder = 0.62 m

h = Height of hollow cylinder = 3m

V = Volume of hollow cylinder

Putting all the values in equation 2,

The volume of hollow cylinder:

$\bf \: V = \pi({0.76}^{2} - {0.62}^{2}) \times 3 \\ \bf \: V = \pi(0.1932) \times 3 \\ \bf \: V = \pi \times 0.5796 \\ \bf \: V = 1.8208 \: {m}^{3}$

so,

Volume of hollow cylinder = 1.8208 cubic meters.

We, know that mass of hollow cylinder = 890 kg

So,

putting values of mass and volume in equation 1,

The density of material :

$\rho = \frac{890}{1.8208} = 488.796 \: \frac{kg}{ {m}^{3} }$

Converting it upto only one decimal place,

ρ = 488.8 kg per cubic meters.