Match the following columns:

Column I	Column II
(a) The radii of the circular ends of a bucket in the form of frustum of a cone of height 30 cm are 20 cm and 10 cm respectively. The capacity of the bucket is ….. cm3. [Take π = 22/7]	(p) 2418π
(b) The radii of the circular ends of a conical bucket of height 15 cm are 28 and 20 cm respectively. The slant height of the bucket is …. cm.	(q) 22000
(c) The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is …. Cm2.	(r) 12
(d) Three solid metallic sphere of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is … cm.	(s) 17

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