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 the metal sheet used is Rs 20 per 100 ${\text{c}}{{\text{m}}^2}$. (Use $\pi = 3.14$).

Answer 1

Hint: Here, we will proceed by directly using the formulas for frustum of a cone i.e., Volume of the frustum cone = $\dfrac{\pi }{3}\left( {{R^2} + {r^2} + R \times r} \right)h$ and Curved surface area of the frustum of cone = $\pi \left( {R + r} \right)\sqrt {{{\left( {R - r} \right)}^2} + {h^2}} + \pi {r^2}$.

Complete step-by-step answer:
Given, Height of the frustum of cone h= 16 cm
Diameter of lower circular end, d = 16 cm
Diameter of upper circular end, D = 40 cm
Since, Radius = $\dfrac{{{\text{Diameter}}}}{2}$
Radius of lower circular end, r = $\dfrac{{{\text{Diameter of lower circular end}}}}{2} = \dfrac{{16}}{2} = 8{\text{ cm}}$
Radius of upper circular end, R = $\dfrac{{{\text{Diameter of upper circular end}}}}{2} = \dfrac{{40}}{2} = 20{\text{ cm}}$
As we know that the volume of the frustum of cone is given by
Volume of the frustum cone = $\dfrac{\pi }{3}\left( {{R^2} + {r^2} + R \times r} \right)h{\text{ }} \to {\text{(1)}}$ where R is the radius of the upper circular end, r is the radius of the lower circular end and h is the height of the frustum of cone
By putting the values of r, R, h and $\pi = 3.14$ in the formula given by equation (1), we get
Volume of the bucket in the form of a frustum of cone = $\dfrac{{3.14}}{3}\left( {{{20}^2} + {8^2} + 20 \times 8} \right)16 = \dfrac{{3.14}}{3}\left( {400 + 64 + 160} \right)16$
$ \Rightarrow $ Volume of the bucket in the form of a frustum of cone = $\dfrac{{3.14}}{3}\left( {624} \right)16 = 10449.92{\text{ c}}{{\text{m}}^3}$
We know that the curved surface area of the frustum of cone is given by
Curved surface area of the frustum of cone = $\pi \left( {R + r} \right)\sqrt {{{\left( {R - r} \right)}^2} + {h^2}} {\text{ }} \to {\text{(2)}}$
Also, we know that the area of any circular end of radius r is given by
Area of any circular end with radius r = $\pi {r^2}{\text{ }} \to {\text{(3)}}$
Surface area of bucket = Curved surface area of the frustum of cone + Area of the lower circular end
 Using the formulas given by equations (2) and (3) in the above equation and then substituting the values r, R, h and $\pi = 3.14$, we get
Surface area of bucket = $\pi \left( {R + r} \right)\sqrt {{{\left( {R - r} \right)}^2} + {h^2}} + \pi {r^2} = 3.14\left( {20 + 8} \right)\sqrt {{{\left( {20 - 8} \right)}^2} + {{16}^2}} + 3.14{\left( 8 \right)^2}$
$ \Rightarrow $ Surface area of bucket = $\left[ {3.14\left( {28} \right)\sqrt {400} } \right] + 3.14\left( {64} \right) = 1758.4 + 200.96 = 1959.36{\text{ c}}{{\text{m}}^2}$
It is also given, Cost of metal sheet used for the surface area of 100 ${\text{c}}{{\text{m}}^2}$ = Rs 20
$ \Rightarrow $ Cost of metal sheet used for the surface area of 1 ${\text{c}}{{\text{m}}^2}$ = Rs $\dfrac{{20}}{{100}}$
$ \Rightarrow $ Cost of metal sheet used for the surface area of 1959.36 ${\text{c}}{{\text{m}}^2}$ = Rs $\left( {\dfrac{{20}}{{100}}} \right) \times 1959.36 = {\text{Rs }}391.872$.

Note- In this particular problem, the metal sheet used to make the bucket consists of the curved surface of the frustum of the cone and the area of the lower circular end because the buckets are usually closed from the bottom and in the bottom, there is a circular end having radius as 8 cm and diameter as 16 cm.