Question

an open container made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper end as 4 cm and 10 cm. Find the cost of oil which can completely fill the container at the rate of Rs 50 per litre
Also find the cost of Rs 5 per 100 cm square
Please anybody answer this
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Answer 1

Given:-

• r1= 4cm
• r2 = 10cm
• h = 8cm
• Cost for 1 litre = Rs 50

To Find:-

• Cost of metal used

Solution :-

Volume of frustrum container = 1/3 π h[ r1² + r2² +(r1×r2)]

= 1/3 × 22/7 × 8cm [(4)² +(10)² + (4×10)]

= 2614.2cm³

Volume = 2614.2 ml = 2.6 litres

Cost = 2.6 × 50 = Rs 120

Use of metal sheet = CSA of container

= π(r1 + r2)l

= 22/7 (4+10)10

= 440cm²

Cost of per 100cm² = Rs 5

Cost for per cm² = 5/100 = 1/20

Cost of 440cm² = 1/20×440 = Rs 22

Answer 2

Given :-

• Height of the container (h) ➜ 8 cm.
• Radius of the bases (R )➜ 10 cm and r ➜ 4 cm

Let the slant height be 'l'

$\sf\implies{l = \sqrt{(R - r) {}^{2} + h {}^{2} }}$

$\sf\implies{ \sqrt{(10 - 4) {}^{2} + (8) {}^{2} }}$

$\sf\implies{10cm}$

Now,

volume of the container.

$\sf\implies \frac{\pi}{3} (R{}^{2} + r {}^{2} R.r)h$

$\sf\implies \frac{\pi}{3} ((10) {}^{2} + (4) {}^{2} + 10 \times 4 )\times 8$

$\sf\implies \frac{3.14}{3} (100 + 16 + 40) \times 8$

$\sf\implies(3.14 \times 52) \times 8 \: cm {}^{3}$

\

$\sf\implies \: 1.30624$

Therefore,

Cost of oil at the rate of Rs. 50 per litre

= 1.30624×50

= Rs. 65.312

Surface Area of the container

$\sf\implies \: \pi (l \: (R \: + r \: ) + r {}^{2} )$

$\sf\implies \: 3.14 \times (10(10 + 4) + 16)$

$\sf\implies3.14 \times 156$

$\sf\implies489.84 \: cm {}^{2}$