Question

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20cm respectively. Find
(i) the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm² [CBSE 2013]
(ii) the cost of milk at the rate of ₹ 35 per litre which can fill it completely. [CBSE 2017]

Answer 1

Let $r=8\mathrm{cm},R=20\mathrm{cm},h=16\mathrm{cm}.$
$$\begin{gathered} \Rightarrow l=\sqrt{{\left(R-r\right)}^2+h^2}=\sqrt{{\left(20-8\right)}^2+{16}^2}=\sqrt{144+256}=\sqrt{400} \\ \Rightarrow l=20\mathrm{cm}. \end{gathered}$$
(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base
$$\begin{gathered} =\mathrm{\pi }\left(r\mathit{+}R\right)l+\mathrm{\pi }{r}^{\mathit{2}} \\ =\mathrm{\pi }\left(8+20\right)20+\mathrm{\pi }{\left(8\right)}^2=560\mathrm{\pi }+64\mathrm{\pi }=624\mathrm{\pi }=1961.14{\mathrm{cm}}^2. \end{gathered}$$
The cost for 100 cm² of sheet = Rs. 10.
The cost of 1 cm² of sheet = $\frac{10}{100}=\mathrm{Rs}.0.1$.
The cost of 1961.14 cm² of sheet = $1961.14\times 0.1=\mathrm{Rs}.196.11$
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =
$$\begin{gathered} \frac{\mathrm{\pi }h}{3}\left[r^2+R^2+rR\right]=\frac{22}{7}\times \frac{16}{3}\left[8^2+{20}^2+160\right]=\frac{{\displaystyle 22}}{{\displaystyle 7}}\times \frac{{\displaystyle 16}}{{\displaystyle 3}}\left[64+400+160\right] \\ =\frac{{\displaystyle 22}}{{\displaystyle 7}}\times \frac{{\displaystyle 16}}{{\displaystyle 3}}\times 624=\frac{73216}{7}{\mathrm{cm}}^3. \end{gathered}$$
we know that 1 l=1000 cm³ or 1 cm³ = 0.001 l.
$\Rightarrow$Volume=$\frac{73216}{7}\times 0.001=\frac{{\displaystyle 73216}}{{\displaystyle 7}}\times \frac{1}{1000}=\frac{73.216}{7}l.$
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled = $\frac{73.216}{7}\times 35=\mathrm{Rs}366.08.$