A Container Opened at 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. - Mathematics

Sum

A container opened at 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 milk which can completely fill the container, at the rate of ₹ 50 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 10 per 100 cm2. (Take π = 3⋅14)

Solution

We have to find the cost of milk which can completely fill the container
The volume of container = Volume of a frustum
= 1/3πh(r12 + r22 + r1r2 )

Here,
height = 16 cm
radius of upper end = 20 cm
And radius of lower end = 8 cm
Plugging the values in the formula we get

Volume of container = 1/3 x 3.14 x 16 ((20)2 + (8)2 + 20 x 8)
= 1/3 x 50.24 (400 + 64 + 160 )
= 1/3 x 50.24 (624)
= 10449.92 cm3
=10.449 litre                 ....(∵ 1 litre=1000 cm3)

Cost of 1-litre milk is Rs 50
Cost of 10.449 litre milk = 50 x 10.449 = Rs 522.45
We will find the cost of metal sheet to make the container
Firstly, we will find the area of container

Area of container = Curved surface area of the frustum + area of bottom circle    ...(∵ container is closed from bottom)

Area of container = π(r1 + r2)l + πr2
Now, we will find l
l = sqrt(h^2 + (r_1 − r_2)^2)

sqrtl = sqrt((16)^2+(20−8)^2)

sqrtl = sqrt((16)^2+(12)^2)

sqrtl = sqrt( 256 + 144)

sqrtl = sqrt(400)
l = 20 cm

Area of frustum = 3.14 x 20(20 + 8)
= 1758.4 cm2

Area of bottom circle = 3.14 x 82 = 200.96 cm2

Area of container = 1758.4 + 200.96  =1959.36 cm2

Cost of making 100 cm2 = Rs 10
Cost of making 1 cm2 = 10/100 = Rs. 110
Cost of making 1959.36 cm2 = 1/10 x 1959.36 = 195.936
Hence, cost of milk is Rs 522.45
And cost of metal sheet is Rs 195.936.

Is there an error in this question or solution?
2018-2019 (March) 30/4/3

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