Radius of upper part of container (r₁) = 20 cm
Radius of lower part of container (r₂) = 8 cm
Height of container (h) = 16 cm
Slant height of container = √((r₁ – r₂)² + h²)
= √((20 – 8)² + 16²) = √(144+256) = √400 = 20 cm
Capacity of container = Volume of frustum = 1/3πh(r²₁ + r²₂ + r₁r₂)
= 1/3 × 3.14 × 16 × (20² + 8² + 20 × 8 ) = 1/3 × 3.14 × 16 × (400 + 64 + 160)
= 1/3 × 3.14 × 16 × 624 = 104449.92 cm³ = 10.45 litres.
Cost of 1 litre of milk = Rs 20
Therefore, the cost of 10.45 litres of milk = 10.45 × Rs 20 = Rs 209
The area of metal sheet used to make the container
= π(r₁+ r₂)l + π(r₂)²
= π(20 + 8) 20 + π(8)²
= 560π + 64 π = 624 π cm²
Cost of 100 cm² metal sheet = Rs 8
Cost of cm² metal sheet = Rs 8/100
Therefore, the cost of 624 π cm² metal sheet = Rs 8/100 × 624π = Rs 8/100 × 624 × 3.14 = Rs 156.75
Hence, the cost of the milk which can completely fill the container is Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.

Radius of pipe (r₁) = 20 / 200 = 0.1 m
Area of cross- section of pipe = πr₁² = π(0.1)² = 0.01π m³
Speed of water = 3 km/h = 3000/60 = 50 m/min
Volume of water flows through in 1 minute = 50 × 0.01π = 0.5πm³
Volume of water flows through in t = t × 0.5π m³
Radius of cylindrical tank (r₂) = 5 m
Height of cylindrical tank (h₂) = 2 m
Let the tank will fill completely in t minutes.
Therefore,
The volume of water flows out in t minutes = volume of cylindrical tank
⇒ t × 0.5π = π ×(r₂)² ×h₂
⇒ t × 0.5 = 5² × 2
⇒ t = 100
Hence, the tank will completely fill in 100 minutes.

Radius of bucket, (r₁) = 18/ 2 = 9 cm, Height of bucket (h₁) = 32 cm
Height of conical heap (h₂) = 24 cm,
Let the radius of conical heap = (r₂)
Therefore, volume of sand in the bucket = volume of conical heap
⇒ πr₁²h₁ = 1/3πr₂²h₂ ⇒ π × 18² × 32 = 1/3π × r₂²× 24
⇒ r₂² = (3 × 18 × 18 × 32/24 = 18 × 18 × 4 ⇒ r₂ = 36 cm
slant height = √(36² + 24²) = √(1296 + 576) = √1872 = 12√13 cm.
Hence, the radius of conical heap is 36 cm and its slant height is 12√13 cm.

Radius of first sphere (r₁) = 6 cm, Radius of second sphere (r₂) = 8 cm
Radius of third sphere (r₃) = 10 cm,
Let the radius of the new sphere = r
According to question, volumes of three spheres = Volume of new sphere
⇒ 4/3π(r₁³ + r₂³ + r₃³) 4/3πr ³ ⇒ 4/3π (6³+ 8³ + 10³) = 4/3πr³
⇒ 216 + 512 + 1000 = r³ ⇒ r³ = 1728 ⇒ r = 12 cm
Hence, the radius of resulting sphere is 12 cm.

Radius of well (r) = 3/2 m, Depth of well h₁ = 14 m, Width of embankment = 14 m
The embankment is in the form of hollow cylinder with internal radius r₂ = 3/2 m and Outer radius r₁ = 3/2 + 4 = 11/2 m
Let the height of embankment = H
According to question, the earth dug out from the well = volume of embankment
⇒ πr₁² = π (r₁² – r₂² )h
⇒ π(3/2)² 14 = π [(11/2)² – (3/2)²] × h ⇒ 9/4 14 = 112/4h ⇒ h = 9/8 = 1.125m
Hence, the height of embankment is 1.125 m.