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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. - Mathematics

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प्रश्न

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. [Take π = 3.14]

उत्तर

Radius (r1) of upper end of container = 20 cm

Radius (r2) of lower end of container = 8 cm

Height (h) of container = 16 cm

Slant height (l) of frustum = `sqrt((r_1-r_2)^2+h^2)`

`=sqrt((20-8)^2+(16)^2)`

`=sqrt((12)^2+(16)^2) = sqrt(144+256)`

= 20 cm

Capacity of container = Volume of frustum

`=1/3pih(r_1^2+r_2^2+r_1r_2)`

`=1/3xx3.14xx16xx[(20)^2+(8)^2+(20)(8)]`

`=1/3xx3.14xx16(400+64+160)`

`=1/3xx3.14xx16xx624`

= 10449.92 cm3

= 10.45 litres.

Cost of 1 litre milk = Rs 20

Cost of 10.45 litre milk = 10.45 × 20

= Rs 209

Area of metal sheet used to make the container

`=pi(r_1+r_2)l + pir_2^2`

= π (20 + 8) 20 + π (8)2

= 560 π + 64 π = 624 π cm2

Cost of 100 cm2 metal sheet = Rs 8

Cost of 624 π cmmetal sheet  = `(624 xx 3.14 xx 8)/100`

= 156.75

Therefore, the cost of the milk which can completely fill the container is

Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.

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अध्याय 13: Surface Areas and Volumes - Exercise 13.4 [पृष्ठ २५७]

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एनसीईआरटी Mathematics [English] Class 10
अध्याय 13 Surface Areas and Volumes
Exercise 13.4 | Q 4 | पृष्ठ २५७

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