A bucket is in the form of a frustum of a cone with a c

SOLUTION

Soln:

Radii of top circular ends $r_1=20cm$

Radii of bottom circular end of bucket $r_2=12cm$

Let height of bucket be ‘h’

Volume of frustum cone $=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h$
$=\frac{1}{3}\pi ({20}^2+{12}^2+20\times 12)h=\frac{784\pi h}{3}c{m}^3$ —-(a)

Given capacity/ volume of the bucket = 12308.8 $c{m}^3$—–(b)

Equating (a) and (b)

$\frac{784\pi h}{3}=1208.8$

$h=\frac{1208.8\times 3}{784\pi }=15cm$

∴∴ height of the bucket (h) = 15 cm

Let ‘L’ be slant height of bucket

$L^2=(r_1-r_2{)}^2+h^2$

$L=\sqrt{(r_1-r_2{)}^2+h^2}=17cm$

∴ length of the bucket/ slant height of the bucket (L) =17 cm

Curved surface area of bucket =

$\pi (r_1+r_2)L+\pi r_2^2$

$\pi (20+12)17+\pi \times {12}^2=\pi (9248+144)=2160.32c{m}^3$

∴ curved surface area = 2160.32 cm²