Question

container, opened from the top and made up of a metal sheet, is in the form of a
rustun of a cone of height 16cm with radii of its lower and upper ends as 8 cm and
20 cm, respectively. Find the cost of the milk which can completely fill the container,
at the rate of 40 per litre. Also find the cost of metal sheet used to make the
container, if it costs 8 per 100 sq cm (take = 3.14)
duct an activity as done in the previous section. Take a​

Answer 1

For frustum of a cone,

h= 16 cm

R= 20 cm

r= 8 cm

l= √h2 + (R-r)2

= √16^2 + (20-8)^2

= √256 + 12^2

= √256 + 144

l= √400

l= 20 cm

Area of metal sheet = CSA of frustum + Area of base

= π×l(R+r)

=3.14 × 20 (20+8)

=3.14 × 20 × 28

=3.14 × 616

=1934.24 cm2

Cost of metal sheet = 8/100 × 1934.24

= Rs. 154.7392

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

= 1/3 ×3.14×16(20^2 + 8^2 + 20×8)

= 1/3 × 50.24(400 + 64 + 160)

= 1/3 × 50.24 × 624

= 1/3 × 31349.76

= 10449.92 cm3

= 10449.92/1000

= 10.44992 litre

Cost of milk = 40 × 10.44992

= Rs. 417996.8

Please mark it as Brainliest.