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 $100\ cm^2$. (Take $\pi = 3.14$)

Given:

A container opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of the height of 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. 

To do:

We have to find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre and the cost of the metal sheet used to make the container, if it costs Rs. 8 per $100\ cm^2$.

Solution:

Radius of the lower end $(r_1) = 8\ cm$

Radius of the upper end $(r_2) = 20\ cm$

Height of the frustum $(h) = 16\ cm$

Volume of the container $=\frac{\pi h}{3}[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}]$

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

$=3.14 \times \frac{16}{3}[64+400+160]$

$=3.14 \times \frac{16}{3} \times 624$

$=10449.92 \mathrm{~cm}^{3}$

$=10.449$ litres            (Since $1 \mathrm{~cm}^{3}=\frac{1}{1000} 1$)

Cost of 1 litre of milk $=Rs.\ 20$

Total cost of milk $=Rs,\ 20 \times 10.449$

$=Rs.\ 208.98$

Slant height of the container $(l)=\sqrt{(r_{2}-r_{1})^{2}+h^{2}}$

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

$=\sqrt{12^{2}+16^{2}}$

$=\sqrt{144+256}$

$=\sqrt{400}$

$=20\ cm$

Curved surface area of the bucket $=\pi l(r_{1}+r_{2})$

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

$=3.14 \times 20 \times 28$

$=1758.4 \mathrm{~cm}^{2}$

The base area of the bucket $=\pi r_{1}^{2}$

$=3.14 \times 8^2$

$=200.96 \mathrm{~cm}^{2}$

The metal sheet used $=$ Curved surface area of the frustum $+$ base area of the bucket

$=1758.4+200.96$

$=1959.36 \mathrm{~cm}^{2}$

The total cost of the metal sheet used $=Rs.\ \frac{8 \times 1959.36}{100}$

$=Rs.\ 156.75$

Tutorialspoint

Updated on: 10-Oct-2022

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