The figure shows a vertical section through the axis of a solid paper weight made in the shape of a right circular cone. Its height is 8 cm and the diameter of its base is 4 cm. The shaded portion is made of lead 2 cm thick while the unshaded portion is made of wood. The density of lead and wood are 11.3 g/cm and 0.9 g/cm respectively.

Find,
(i) the total volume of the paper weight,
(ii) the volume of the wooden portion,
(iii) the total mass of the paper weight.​

i. 33.51 cm³

ii. 14.14 cm³

iii. 231.61 g

Step-by-step explanation:

(i) the total volume of the paper weight,

Since the paper weight is in the form of a cone, its volume is the volume of a cone V = πD²h/12 where D = diameter of cone = 4 cm and h = height of cone = 8 cm

So, V = πD²h/12

V = π(4 cm)² × 8 cm/12

V = 16π cm² × 8 cm/12

V = 128π cm³/12

V = 402.124 cm³/12

V = 33.51 cm³

(ii) the volume of the wooden portion,

The volume of the wooden portion, V' = πr'²h'/3 where r' = radius of wooden portion and h' = height of wooden portion

From similar triangles in the figure, we have

height of wooden portion, h'/radius of wooden portion, r' = height of paper weight, h/radius of paper weight, r

h'/r' = h/r where r = D/2 where D =diameter of paper weight = 4 cm. So, r = 4 cm/2 = 2 cm

and h' = height of paper weight - height of lead portion = 8 cm - 2 cm = 6 cm

So,

r' = rh'/h = 2 cm × 6 cm/8 cm = 12 cm²/8 cm = 1.5 cm

So, V' = πr'²h'/3

V' = π(1.5 cm)² × 6 cm/3

V' = π2.25 cm² × 2 cm

V' = 4.5π cm³

V' = 14.14 cm³

(iii) the total mass of the paper weight.

The total mass of paper weight, m = mass of wooden portion + mass of lead portion = ρ'V' + ρ"V" where ρ' = density of wood = 0.9 g/cm³, V' = volume of wooden portion = , ρ" = density of lead = 11.3 g/cm³ and V" = volume of lead portion = V - V' = 33.51 cm³ - 14.14 cm³ = 19.37 cm³

So, m = 0.9 g/cm³ × 14.14 cm³ + 11.3 g/cm³ × 19.37 cm³

m = 12.726 g + 218.881 g

m = 231.607

m ≅ 231.61 g

it has to have an angle of 90 degrees