A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm,
respectively. Find the cost of milk at the rate of ? 22 per L which the container can hold.
Given that,height of milk container (h) = 16 cm,
Radius of lower end of milk container (r) = 8 cm
and radius of upper end of milk container (R) = 20 cm
$\therefore$ Volume of the milk container made of metal sheet in the form of a frustum of a cone
$=\frac{\pi h}{3}\left(R^{2}+r^{2}+R r\right)$
$=\frac{22}{7} \times \frac{16}{3}\left[(20)^{2}+(8)^{2}+20 \times 8\right]$
$=\frac{22 \times 16}{21}(400+64+160)$
$=\frac{22 \times 16 \times 624}{21}=\frac{219648}{21} \quad\left[\because 1 \mathrm{~L}=1000 \mathrm{~cm}^{3}\right]$
$=10459.42 \mathrm{~cm}^{3}=10.45942 \mathrm{~L}$
So, volume of the milk container is $10459.42 \mathrm{~cm}^{3}$.
$\because$ Cost of $1 \mathrm{~L}$ milk $=₹ 22$
$\therefore$ Cost of $10.45942 \mathrm{~L}$ milk $=22 \times 10.45942=₹ 230.12$
Hence, the required cost of milk is ₹ $230.12$
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