Question

A container made of a metal sheet open at the top is of the form of a 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

(i) The cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm².

(ii) The cost of milk at the rate of Rs. 35 per liter which can fill it completely.

Answer 1

Radius of lower end = r = 8 cm

Radius of upper end = R = 20 cm

Height of container frustum = h = 16 cm

Cost of 100 cm² metal sheet = Rs 10

So, Cost of 1 cm² metal sheet = 10/100 = Rs 0.1

Let l be the slant height.

l² = (R-r)² + h²

= (20-8)² + 16²

= 256 + 144

= 400

or l = 20 cm

Surface area of frustum of the cone = πr² + π(R + r)l cm²

= π[ (20 + 8 )20 + 8²]

= π[560 + 64]

= 624 X ( 22/7 )

= 1961.14 cm²

(i) Find cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm²

Cost of metal sheet per 100 cm² = Rs. 10

Cost of metal for Rs. 1961.14 cm² = (1961.14 x10)/100 = Rs. 196.114

(ii) Find the volume of frustum:

Volume of frustum = 1/3 πh(r² + R² + rR)

= 1/3 x 22/7 x 16(8² + 20² + 8×20)

= 1/3 x 22/7 x 16(64 + 400 + 160)

= 10345.4208 cm³

= 10.345 liters

Using

1000 cm³ = 1 litre

1 cm³ = 1/1000 litre

Cost of 1 liter milk = Rs. 35

Cost of 10.345 liter milk = Rs. 35 x 10.345 = Rs. 363 (approx)