Question

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20 cm respectively. Find

1. the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm²
2. the cost of milk at the rate of ₹ 35 per litre which can fill it completely.

Answer 1

Given: A frustum of a cone, open at the top, with height h = 16 cm, lower radius r = 8 cm and upper radius R = 20 cm; cost of metal = ₹ 10 per 100 cm² and milk = ₹ 35 per litre.

Step-wise calculation:

1. Slant height (l) `l = sqrt((R - r)^2 + h^2)`

= `sqrt((20 - 8)^2 + 16^2)`

= `sqrt(12^2 + 16^2)`

= `sqrt(144 + 256)`

= `sqrt(400)`

= 20 cm

2. Metal sheet area (open top: curved surface + bottom)

Curved surface area = π (R + r) l 

= π × (20 + 8) × 20

= π × 28 × 20

= 560π cm² 

Bottom area = πr²

= π × 8²

= 64π cm² 

Total area = π(560 + 64)

= 624π cm²   ...(Take π = 3.14)

⇒ Total area = 3.14 × 624 

= 1959.36 cm²

3. Cost of metal sheet rate = ₹ 10 per 100 cm² 

= ₹ 0.10 per cm² 

Cost = 1959.36 × 0.10

= ₹ 195.936 ≈ ₹ 195.94

4. Volume of the frustum (capacity)

`V = (1/3) πh(R^2 + r^2 + Rr)` 

= `(1/3) xx π xx 16 xx (400 + 64 + 160)` 

= `(π/3) xx 16 xx 624` 

= 3328π cm³ 

With π = 3.14

⇒ V = 3328 × 3.14 

= 10449.92 cm³ 

= 10.44992 litres ≈ 10.45 L

5. Cost of milk to fill it:

Rate = ₹ 35 per litre. 

Cost = 10.44992 × 35 

= ₹ 365.7472 ≈ ₹ 365.75

1. Cost of metal sheet ≈ ₹ 195.94.
2. Cost of milk to fill the container ≈ ₹ 365.75.