 # A bucket made up of a metal sheet is in the form Question:

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used in Rs 20 per 100 cm2 .
(Use π = 3.14)

Solution:

The height of the bucket is h=16cm. The radii of the upper and lower circles of the bucket are r1 =20 cm and r2 = 8 cm respectively.

The slant height of the bucket is

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

$=\sqrt{(20-8)^{2}+16^{2}}$

$=\sqrt{400}$

$=20 \mathrm{~cm}$

The volume of the bucket is

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

$=\frac{1}{3} \pi\left(20^{2}+20 \times 8+8^{2}\right) \times 16$

$=\frac{1}{3} \times 3.14 \times 624 \times 16$

$=3.14 \times 208 \times 16$

$=10449.92 \mathrm{~cm}^{3}$

Hence the volume of the bucket is $10449.92 \mathrm{~cm}^{3}$

The surface area of the used metal sheet to make the bucket is

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

$=\pi \times(20+8) \times 20+\pi \times 8^{2}$

$=\pi \times 28 \times 20+64 \pi$

$=624 \pi \mathrm{cm}^{2}$

Therefore, the total cost of making the bucket is

$=\frac{624 \pi}{100} \times 20$

$=\operatorname{Rs} .391 .9$