A bucket made up of a metal sheet is in the form of frustum of a cone.

Solution:

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l

$=\sqrt{h^{2}+(R-r)^{2}}$

$=\sqrt{24^{2}+(15-5)^{2}}$

$=\sqrt{576+(10)^{2}}$

$=\sqrt{576+100}$

$=\sqrt{676}=26 \mathrm{~cm}$

Capacity of the frustum

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

$=\frac{1}{3} \times 3.14 \times 24\left(15^{2}+5^{2}+15 \times 5\right)$

$=3.14 \times 8 \times 325$

$=8164 \mathrm{~cm}^{3}=8.164$ litres

A litre of milk cost Rs 20.

So, total cost of filling the bucket with milk $=8.164 \times 20=$ Rs $163.28$

Surface area of the bucket

$=\pi r^{2}+\pi l(R+r)$

$=\pi\left[5^{2}+26(15+5)\right]$

$=3.14[25+26(20)]$

$=3.14[25+520]$

$=1711.3 \mathrm{~cm}^{2}$

Cost of 100 cm2 of metal sheet is Rs 10.

So, cost of metal used for making the bucket $=\frac{1711.3}{100} \times 10=$ Rs $171.13$