A container, open at the top, is in the form of a frustum of a cone of height 24 cm

Solution:

We have,

Height, $h=24 \mathrm{~cm}$,

Upper radius, $R=20 \mathrm{~cm}$ and

Lower radius, $r=8 \mathrm{~cm}$

Now,

Volume of the container $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 24 \times\left(20^{2}+8^{2}+20 \times 8\right)$

$=\frac{176}{7} \times(400+64+160)$

$=\frac{176}{7} \times 624$

$=\frac{109824}{7} \mathrm{~cm}^{3}$

$=\frac{109.824}{7} \mathrm{~L} \quad\left(\mathrm{As}, 1000 \mathrm{~cm}^{3}=1 \mathrm{~L}\right)$

So, the cost of the milk in the container $=\frac{109.824}{7} \times 21$

$=329.472$

$\approx ₹ 329.47$

Hence, the cost of milk which can completely fill the container is ₹329.47.