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A container, open at the top, made up of metal sheet is in the form of the frustum of a cone of height 16 cm with radii of lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs 40 per litre and also find the cost of metal sheet used if it costs Rs 10 per $100\; cm^2$ . (Use $\pi=3.14$ )

Answers (1)

$h=16 \; cm$

$r_1=8 \; cm, r_2=20 \; cm$

$l=\sqrt{h^2+(r_1-r_2)^2}$

$l=\sqrt{16^2+(20-8)^2}$

$l=\sqrt{256+144}$

$l=20 \; cm$

$\text{Volume of frustum}=\frac{1}{3}\pi h (r_1^2+r_2^2+r_1r_2)$ $=\frac{1}{3}\times 3.14 \times 16 (64+400+160)$

$=10449.92 \; cm^3$

$=10.449 \; \text{litre}$

$=10.45 \; \text{litre}$

$\text{Cost of milk}=40\times 10.45 = Rs \; 418$

Surface area of the container

$=\pi l(r_1+r_2)+\pi r_1^2$

$=3.14 \times 20(8+20)+3.14\times 8^2$

$=3.14 \times 20\times 28+3.14\times 64$

$= 1758.63+200.96$

$= 1959.36\; cm^2$

$\text{Cost of metal sheet}=Rs \; 10/100 \; cm^2$

$\text{Cost of metal sheet required}= 1959.36 \times \frac{10}{100}$

$=Rs \; 195.9$

Cost of metal sheet used to make the container is Rs 195.9

Posted by

Ravindra Pindel

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