A milk container is made of metal sheet in the shape of frustum of a cone whose volume is $ 10459 \frac{3}{7} \mathrm{~cm}^{3} $ The radii of its lower and upper circular ends are $ 8 \mathrm{~cm} $ and $ 20 \mathrm{~cm} $ respectively. Find the cost of metal sheet used in making the container at the rate of $ ₹ 1.40 $ per $ \mathrm{cm}^{2} $. (Use $ \pi=22.7 $ )

Given:

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is \( 10459 \frac{3}{7} \mathrm{~cm}^{3} \).

The radii of its lower and upper circular ends are \( 8 \mathrm{~cm} \) and \( 20 \mathrm{~cm} \) respectively.

To do:

We have to find the cost of metal sheet used in making the container at the rate of \( ₹ 1.40 \) per \( \mathrm{cm}^{2} \).

Solution:

Volume of the frustum $= 10459\frac{3}{7}\ cm^3$

$=\frac{73216}{7}\ cm^3$
Lower radius of the frustum $r_2 = 8\ cm$

Upper radius of the frustum $r_1 = 20\ cm$

Let $h$ be the height of the frustum.

Therefore,

Volume $=\frac{\pi}{3}[r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}] \times h$

$\frac{73216}{7}=\frac{22}{7 \times 3}[20^{2}+20 \times 8+8^{2}] \times h$

$\frac{73216}{7}=\frac{22}{21}[400+160+64] h$

$\frac{73216}{7}=\frac{22}{21} \times 624 h$

$h=\frac{73216}{7} \times \frac{21}{22 \times 624}$

$h=16$

Slant height of the frustum $l=\sqrt{h^{2}+(r_{1}-r_{2})^{2}}$

$=\sqrt{(16)^{2}-(20-8)^{2}}$

$=\sqrt{16^{2}+12^{2}}$

$=\sqrt{256+144}$

$=\sqrt{400}$

$=20 \mathrm{~cm}$

This implies,

Metal sheet used $=\pi (R+r)+\pi r^{2}+\pi R^{2}$

$=\frac{22}{7} \times 20(20+8)+\frac{22}{7} \times 8 \times 8+\frac{22}{7} \times 20 \times 20$

$=\frac{22}{7} \times 560+\frac{22}{7} \times 64+\frac{22}{7} \times 400$

$=\frac{22}{7} \times 1024$

$=3218.28 \mathrm{~cm}^2$

Rate of the sheet $=Rs.\ 1.40$ per $\mathrm{cm}^{2}$

Total cost of the sheet $=Rs.\ 3218.28 \times 1.40$

$=Rs.\ 4505.59$

The cost of metal sheet used in making the container is Rs. 4505.59.

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Updated on: 10-Oct-2022

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