A bucket has top and bottom diameter of 40 cm and 20 cm respectively.

Solution:

The radii of the top and bottom circles are r1 = 20 cm and r2 = 10 cm respectively. The height of the bucket is h = 12 cm. Therefore, the volume of the bucket is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(20^{2}+20 \times 10+10^{2}\right) \times 12$

$=\frac{1}{3} \times \frac{22}{7} \times 700 \times 12$

$=8800 \mathrm{~cm}^{3}$

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(20-10)^{2}+12^{2}}$

$=2 \sqrt{61} \mathrm{~cm}$

The total surface area of the bucket is

$=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^{2}$

$=\frac{22}{7} \times(20+10) \times 2 \sqrt{61}+\frac{22}{7} \times 10^{2}$

$=\frac{1320 \sqrt{61}+2200}{7} \mathrm{~cm}^{2}$

$=\frac{1320 \sqrt{61}+2200}{7 \times 100} \mathrm{dm}^{2}$

The total cost of tin sheet used for making the bucket is

$=1.20 \times\left(\frac{1320 \sqrt{61}+2200}{7 \times 100}\right)$

$=21.40$