a—(q), b—(s), c—(p), d—(r)

a) Given: The radii of the circular ends of a bucket are 20 cm and 10 cm respectively.

Height of the bucket is 30cm

Bucket is in the shape of frustum.

Let V be the Volume of the Bucket(Frustum)

Volume of the frustum is given by: × h × (R² + r² + Rr) (here r and R are the radii of smaller and larger circular ends respectively)

∴ V = × h × (R² + r² + R × r)

⇒ V = × 30 × (20² + 10² + 20 × 10)

⇒ V = × 30 × (400 + 100 + 200) = × 30 × (700)

⇒ V = × 30 × (700) = 22000 cm³

∴ The capacity of the bucket is: 22000 cm³

b) Given: Height of the frustum of a cone: 15 cm

radii of the Circular ends: 28cm and 20 cm.

Here slant height h can be found by using Pythagoras theorem.

∴ s² = h² + (R-r)² (here R is 28cm and r is 20cm)

⇒ s² = 15² + (28—20)²

⇒ s² = 15² + (8)²

⇒ s² = 225 + 64

⇒ s² = 289

⇒ s = √289 = 17

∴ Slant height 0f the Frustum is 17cm

c) Given: The radii of the end of a bucket are 33 cm and 27 cm and its slant height is 10 cm

TSA of a frustum of a cone = πl(r₁ + r₂) + πr₁² + πr₂² (here l , r₁, r₂ are the slant height, radii of the frustum)

Let S be the TSA of the Frustum.

∴ S = πl(r₁ + r₂) + π(r₂)² + π(r₁)²

⇒ S = π × 10 × (33 + 27) + π(33)² + π(27)²

⇒ S = π(600 + 1089 + 729) = 2418π

∴ TSA of frustum is 2418π.

d) Given: Three solid metallic sphere of radii 3 cm, 4 cm and 5 cm.

Volume of the Solid sphere is: πr³ (here r is the radius of the sphere).

Let V₁ be the volume of the sphere with radius 3cm.

Let V₂ be the volume of the sphere with radius 4cm.

Let V₃ be the volume of the sphere with radius 5cm.

∴ V₁ = πr³ = π(3)³

∴ V₂ = πr³ = π(4)³

∴ V₃ = πr³ = π(5)³

Here,

Let V be the Volume of the sphere that is formed by melting the spheres with volumes V₁ , V₂ , V₃ .

∴ V = V₁ + V₂ + V₃ = π(3)³ + π(3)³ + π(3)³

⇒ V = π(3³ + 4³ + 5³) = π(27 + 64+ 125) = π(216)

⇒ V = π(216) = π(6)³

Here, we can see that radius of the Sphere is 6cm, ∴ diameter = 2 × 6 = 12cm

∴ Diameter of sphere formed by melting spheres with volumes V₁, V₂ , V₃ is 12cm