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 and the radii of its lower amd upper circular edges are 8 cm and 20 cm respectively. Find the cost of metal sheet used to make rhe container if the cost rupees 10 per 100 cm square. And the cost of milky at the rate of rupees 35 per litre which can fill it completely.

Answer 1

CSA of frustum = πl(r1+r2)

l = √ (r2-r1)^2 + h^2

l = √ (20-8)^2 + 16^2

l = √ (12)^2 + 256

l = √ 144 + 256

l = √ 400

l = 20cm

CSA of frustum = πl(r1+r2)

= π ×20 (20+8)

= π ×20 ×28

=22/7 × 20 × 28

= 22 × 20 × 4

= 22 × 80

= 1760cm^2 + πr2

= 1760 + 22/7 × 20

= 1760 + 440/7

= 12760/7

= 1822.85cm^2

cost of per 100cm^2 = ₹10

for = 1822.85 × ₹10

= ₹18228.5

Volume of frustum = 1/3πh[(r1)^2+(r2)^2+(r1r2)]

= 1/3π× 16(64+400+160)

= 1/3π × 16 × 624

= 1/3 × 22/7 × 16 × 624

= 10459.42cm^3

1cm^3 = 0.001liter

so 10.45942 liter

cost of per liter = ₹35

for = 10.45942 × ₹35

= ₹366.07970

Answer 2

Answer:

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