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 ends as 4 cm and 10 cm respectively. Find the cost of oil which can completely fill the container at the rate of Rs. 50 per litre. Also, find the cost of metal used, if it costs Rs. 5 per 100 cm2.
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Answers
Answer:
Let l be the slant height of the frustum
We have
r
1
=20cm,r
2
=8cm;h=16cm
∴l=
(r
1
−r
2
)
2
+h
2
⇒l=
(20−8)
2
+16
2
=20cm
Now, volume of the container
=
3
1
π(r
1
2
+r
2
2
+r
1
r
2
)h
=
3
1
π(20
2
+8
2
+20×8)×16cm
3
=
3
3.14×624×16
cm
3
=10449.92cm
3
=10.45litres
Cost of milk at the rate of 15 Rs. per litre =Rs.(10.45×15)=Rs.156.75
Total surface area of the frustum =π(r
1
+r
2
)l+πr
2
2
={3.14(20+8)×20+3.14×8
2
}cm
2
=3.15×(560+64)cm
2
=3.14×624cm
2
=1959.36cm
2
Cost of metal used =Rs.(
100
1959.36×5
)=Rs.97.96
Step-by-step explanation:
nice question
Answer:
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 ends as 4 cm and 10 cm respectively.
The solution is given in the attached file.
