Given,

Radius of lower end, r₂ = 10 cm

Radius of upper end, r₁ = 30 cm

Height of bucket, h = 24 cm

(i) As we know

volume of frustum of a cone =

Where, h = height, r₁ and r₂ are radii of two ends (r₁ > r₂)

Capacity of bucket =

=3.14 × 8 × (100 + 900 + 300)

= 3.14 × 8 × 1300 = 32656 cm³

(ii) Area of metal used to make bucket = CSA of frustum + base area

We know that,

Curved surface area of frustum = πl(r₁ + r₂)

Where, r₁ and r₂ are the radii of two ends (r₁ > r₂)

And l = slant height and

l = √(h²+ (r₁ – r₂ )²)

So, we have

Slant height, l=√(24² + (30 – 10)²)

⇒ l =√(576 + 100)

⇒ l =√676

⇒ l =26 cm

And as the base has lower end,

Base area = πr₂², where r₂ is the radius of lower end

Therefore,

Area of metal sheet used = πl(r₁ + r₂) + πr₂²

= π(26)[10 + 30] + π(10)²

= 1040π + 100π = 1140π

= 1140(3.14) = 3579.6 cm²