A container, opened from the top and made up of a metal sheet,

Question:

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height $16 \mathrm{~cm}$ with radii of its lower and upper ends as $8 \mathrm{~cm}$ and $20 \mathrm{~cm}$, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per $100 \mathrm{~cm}^{2}$.

(Take $\pi=3.14$ ).

Solution:

We have $: r_{1}=20 \mathrm{~cm}$,

$\mathrm{r}_{2}=8 \mathrm{~cm}$ and $\mathrm{h}=16 \mathrm{~cm}$

$\therefore$ Volume of the frustum

$=\frac{\mathbf{1}}{\mathbf{3}} \pi h\left[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right]$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16\left[20^{2}+8^{2}+20 \times 8\right] \mathrm{cm}^{3}$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16 \times[400+64+160] \mathrm{cm}^{3}$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16 \times 624 \mathrm{~cm}^{3}=\left[\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times \mathbf{1 6} \times \mathbf{2 0 3}\right] \mathrm{cm}^{3}$

$=\left[\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times \mathbf{1 6} \times \mathbf{2 0 3}\right] \div 1000$ litres

$=\frac{\mathbf{3 1 4} \times \mathbf{1 6} \times \mathbf{2 0 8}}{\mathbf{1 0 0 0 0 0}}$ litres

$\therefore$ Cost of milk $=` 20 \times \frac{\mathbf{3 1 4} \times \mathbf{1 6} \times \mathbf{2 0 8}}{\mathbf{1 0 0 0 0 0}}$ litres

$=` 208.998 \approx 209$

Now, slant height of the given frustum

$\ell=\sqrt{\mathbf{h}^{2}+\left(\mathbf{r}-\mathbf{r}_{2}\right)^{2}}=\sqrt{\mathbf{1 6}^{2}+(\mathbf{2 0}-\mathbf{8})^{2}}$

$=\sqrt{\mathbf{1 6}^{2}+12^{2}}=\sqrt{\mathbf{2 5 6}+144}=\sqrt{\mathbf{4 0 0}}=20 \mathrm{~cm}$

$\therefore$ Curved surface area $=\pi\left(r_{1}+r_{2}\right) \ell$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}}(20+8) \times 20 \mathrm{~cm}^{2}$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 28 \times 20 \mathrm{~cm}^{2}=1758.4 \mathrm{~cm}^{2}$

Area of the bottom $=\pi r^{2}$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 8 \times 8 \mathrm{~cm}^{2}=200.96 \mathrm{~cm}^{2}$

$\therefore$ Total area of metal required

$=1758.4 \mathrm{~cm}^{2}+200.96 \mathrm{~cm}^{2}=1959.36 \mathrm{~cm}^{2}$

Cost of metal required $=\frac{\mathbf{8}}{\mathbf{1 0 0}} \times 1959.36 \mathrm{~cm}^{2}$

$=156.75$

 

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