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A bucket is in the form of a frustum of a cone of height $30\ cm$ with radii of its lower and upper ends as $10\ cm$ and $20\ cm$ respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of Rs. 25 per litre. (Use $\pi = 3.14$)
Given:
A bucket is in the form of a frustum of a cone of height $30\ cm$ with radii of its lower and upper ends as $10\ cm$ and $20\ cm$ respectively.
To do:
We have to find the capacity and surface area of the bucket and the cost of milk which can completely fill the container, at the rate of Rs. 25 per litre.
Solution:
Height of the bucket $h=30\ cm$
Upper radius of the bucket $r_1=20\ cm$
Lower radius of the bucket $r_2=10\ cm$
Therefore,
Capacity (volume) of the bucket $=\frac{\pi}{3}[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}]h$
$=\frac{3.14 \times 30}{3}[20^{2}+10^{2}+20 \times 10] \mathrm{cm}^{3}$
$=21980\ cm^3$
$=21.980$ litres
Slant height $l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}$
$=\sqrt{900+100}$
$=31.62 \mathrm{~cm}$
Surface area of the bucket $=$ Curved surface area of the bucket $+$ Surface area of the bottom
$=\pi l(r_{1}+r_{2})+\pi r_{2}^{2}$
$=3.14 \times 31.62(20+10)+\frac{22}{7}(10)^{2}$
$=3.14[948.6+100]$
$=3.14[1048.6]$
$=3292.6 \mathrm{~cm}^{2}$
Cost of 1 litre of milk $=Rs.\ 25$
This implies,
Cost of $21.980$ litres of milk $=Rs.\ 21.980 \times 25$
$=Rs.\ 594.50$