Question:
A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of Rs. 10 per 100 cm2 .
(Use π = 3.14).
Solution:
The slant height of the bucket is given by
$l=\sqrt{h^{2}+(R-r)^{2}}$
$=\sqrt{(24)^{2}+(15-5)^{2}}$
$=\sqrt{576+100}$
$=\sqrt{676}$
$=26 \mathrm{~cm}$
Surface area of bucket
= Curved surface area of bucket + Area of the smaller circlular base
$=\pi l(R+r)+\pi r^{2}$
$=3.14 \times 26 \times(15+5)+3.14 \times 5 \times 5$
$=1632.8+78.5$
$=1711.3 \mathrm{~cm}^{2}$
Cost of metal sheet used $=\frac{10}{100} \times 1711.3=\mathrm{Rs} 171.13$