A container, opened from the top and made up of a metal sheet, is in the form of a
frustum of a cone of height 16 cm 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 Rs 20 per litre. Also find the curved surface area of metal sheet (Take t =3.14)
Answers
Answer:
Radius of the upper end of the container = r1 = 20 cm
Radius of the lower end of the container = r2 = 8 cm
Height of the container = 16 cm
Slant height, l = √{h2 + (r1 – r2)2}
= √{(16)2 +(20 – 8)2
= √256 + (12)2
= √256 + 144
= √400
= 20 cm
Now,
= 10445.76 cm3
= 10.44576 Litre
A litre of milk cost Rs 40
So, total cost of filling the container with milk = Rs 40 × 10.44576
= Rs 417.83
Now, we need to find cost of metal, for that we need to find the area of container
Since container is closed from bottom,
Surface Area of the container
= CSA of the frustum + Area of circular base
= πl(r1+r2) + πr22
= 3.14 × [{20 × (20 + 8)} + (8)2]
= 3.14 × [400 + 160 + 64]
= 3.14 × 624
= 1959.36 cm2
Cost of 100cm2 of metal sheet = Rs 10
= Rs 195.93
Hence,
Cost of milk = Rs 417.83
Cost of metal sheet = Rs 195.93
happy to help
Answer:
please make it brainliest
