Question

a metal container open from the top is in the shape of a frustum of cone of height 21cm with radii of it's lower and upper circular ends are 8cm and 20 cm respectively,find the cost of milk which can completely fill the container at the rate of 35/litre​

Answer 1

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$\underline{\red{\frak{QuestioN}}}$

A metal container open from the top is in the shape of a frustum of cone of height 21 cm with radii of its lower and upper circular ends are 8 cm and 20 cm respectively, Find the cost of milk which can completely fill the container at the rate of 35 per litre.

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$\underline{ \red{ \frak{SolutioN}}}$

$\tt \: height \: of \: frustum \: = h = 21 \: cm$

$\tt \: radius \: of \: upper \: surface = r_1 = 20 \: cm \\ \\ \tt \: radius \: of \: lower \: surface \: = r_2 = 8 \: cm$

$\tt volume \: of \\ \tt \: frustum = \frac{1}{3}\pi \: h( {r_1}^{2} + {r_2}^{2} + r_1r_2)$

$\tt \: putting \: values$

$\tt\large{ \: volume \: of \: \: frustum }\\ \tt \: = \frac{1}{3} \times \frac{22}{7} \times 21( {(20)}^{2} + {(8)}^{2} + (20\times 8 \: )) \\ \\ \tt \: = \frac{22}{21} \times 21(400 + 64 + 160) \\ \\\tt = 22(624) = 13728 \: {cm}^{3}$

SO,

$\large \: \tt \: volume \: of \: frustum = 13728 \: {cm}^{3}$

Now,

we know that

$\tt \: 1 \: litre = 1000 \: {cm}^{3} \\ \\ \tt \: then \\ \\ 1 \: \tt {cm}^{3} = \frac{1}{1000} \: litre = 0.001 \: litre$

So,

$\tt \: 13728 \: {cm}^{3} = 13728 \times 0.001 \: litres \\ \\ \tt 13728 \: {cm}^{3} = 13.728 \: litres$

Rate for filling the container per litre is 35 Rs

So,

Cost for filling the container will be

$\tt \: = 13.728 \times 35 \: Rs \\ \\ = \tt \: 480.48 \: Rs$

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