Question

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig.). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. (Take $\pi =\frac{22}{7}$) [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

The total height of the bucket = 40 cm, which includes the height of the base.

So, the height of the frustum of the cone = (40-6)cm = 34cm

Therefore, the slant height of the frustum $l=\sqrt{h^2+(r_1-r_2{)}^2}$

where $r_1=22.5cm,r_2=12.5cm,h=34cm$

$l=\sqrt{{34}^2+(22.5-12.5{)}^2}cm$

$=\sqrt{{34}^2+{10}^2}=35.44cm$

The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder

$=[\pi \times 35.44(22.5+12.5)+\pi \times (12.5{)}^2+2\pi \times 12.5\times 6]c{m}^2$

$=\frac{22}{7}(1240.4+156.25+150)c{m}^2$

$=4860.9c{m}^2$

Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket )

$=\frac{\pi \times h}{3}\times ({r_1}^2+{r_2}^2+r_1{r}_2)$ (Volume of frustum)

$=\frac{22}{7}\times \frac{34}{3}\times \left[\right(22.5{)}^2+(12.5{)}^2+22.5\times 12.5]c{m}^3$

$=\frac{22}{7}\times \frac{34}{3}\times 943.75=33615.48c{m}^3$

$=33.62litres(approx.)$ $(1000c{m}^3=1litre)$