Name: Conversion of Solids
Uploaded: 2019-07-19
Duration: 5 min 33 s
Description: What happens when a goldsmith heats a gold bar to create something? The gold bar's shape gets changed. But what about its volume? Does it change too? Find ou...

Question

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 c m

with radii of its lower and upper ends as

8 c m

and

20 c m

, respectively. Find the cost of the milk which can completely fill the container, at the rate of

₹ 20 p e r l i t r e

. Also find the cost of metal sheet used to make the container, if it costs

₹ 8 p e r 100 {cm}^{2}

. (Take

π = 3.14

)

Accepted Answer

It is given that,

The radius $(r_{1})$ of upper end of container $= 20 cm$

The radius $(r_{2})$ of lower end of container $= 8 cm$

The 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

$= \frac{1}{3} π h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$

$= \frac{1}{3} \times 3.14 \times 16 \times [{(20)}^{2} + {(8)}^{2} + (20) (8)]$

$= \frac{1}{3} \times 3 .14 \times 16 \times 624 = 16 .74 \times 624 = 10449 .92 {cm}^{3}$

$= 10 .45 litre$ (Since $1 litre = 1000 c m^{3}$ )

Cost of $1 litre$ milk $= ₹ 20$

Cost of $10.45 litre$ milk $= 10.45 \times 20$ $= ₹ 209$

Area of metal sheet used to make the container $= C S A$ of frustum $+$ area of lower circular end

$= π (r_{1} + r_{2}) l + π r_{2}^{2}$

$= π (20 + 8) 20 + π {(8)}^{2}$

$= 560 π + 64 π = 624 π {cm}^{2}$

Cost of $100 c m^{2}$ metal sheet $= ₹ 8$

Cost of $624 π c m^{2}$ metal sheet $= \frac{624 \times 3 .14 \times 8}{100} = ₹ 156 .75$

Therefore, the cost of the milk which can completely fill the container is $₹ 209$ and the cost of metal sheet used to make the container is $₹ 156 .75$ .

Important Questions on Surface Areas and Volumes

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