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Home » 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 cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.
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 cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.
CBSE | Class 10 | Exercise 13.4 | Maths | NCERT | Surface Areas And Volumes
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 cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.

Given,

r1 = 20 cm,

r2 = 8 cm and

h = 16 cm

Ncert solutions class 10 chapter 13-24

    \[\therefore Volume\text{ }of\text{ }the\text{ }frustum\text{ }=\text{ }\left( \right)\times \pi \times h\left( r12+r22+r1r2 \right)\]

Ncert solutions class 10 chapter 13-25

It is given that the pace of milk

    \[=\text{ }Rs.\text{ }20/liter\]

Thus, Cost of milk = 20×volume of the frustum

    \[=\text{ }Rs.\text{ }209\]

Presently, incline tallness will be

Ncert solutions class 10 chapter 13-26

        Thus, CSA of the holder

    \[=\text{ }\pi \left( r1+r2 \right)\times l\]

Ncert solutions class 10 chapter 13-27

Henceforth, the all out metal that would be needed to make holder will be

    \[=\text{ }1758.4\text{ }+\text{ }\left( Area\text{ }of\text{ }base\text{ }circle \right)\]

    \[=\text{ }1758.4+201\text{ }=\text{ }1959.4\text{ }cm2\]

∴ Total expense of metal

    \[=\text{ }Rs.\text{ }\left( 8/100 \right)\text{ }\times \text{ }1959.4\text{ }=\text{ }Rs.\text{ }157\]

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