A bucket made up of a metal sheet is in the form of a frustum of a cone of height $ 16 \mathrm{~cm} $ with diameters of its lower and upper ends as $ 16 \mathrm{~cm} $ and $ 40 \mathrm{~cm} $ respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is $ ₹ 20 $ per $ 100 \mathrm{~cm}^{2} $. (Use $ \pi=3.14 $ )

Given:

A bucket made up of a metal sheet is in the form of a frustum of a cone of height \( 16 \mathrm{~cm} \) with diameters of its lower and upper ends as \( 16 \mathrm{~cm} \) and \( 40 \mathrm{~cm} \) respectively. 

To do:

We have to find the volume of the bucket and the cost of the bucket if the cost of metal sheet used is \( ₹ 20 \) per \( 100 \mathrm{~cm}^{2} \).

Solution:

Lower radius of the bucket $r_1 =\frac{16}{2}$

$= 8\ cm$
Upper radius of the bucket $r_2 =\frac{40}{2}$

$= 20\ cm$
Height of the bucket $h = 16\ cm$

Therefore,

Slant height of the bucket $l=\sqrt{h^{2}+(r_2-r_1)^{2}}$

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

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

$=\sqrt{256+144}$

$=\sqrt{400}$

$=20 \mathrm{~cm}$

Volume of the bucket $=\frac{1}{3} \pi (r_2^{2}+r_1^{2}+r_2 r_1) h$

$=\frac{1}{3} \times 3.14[20^{2}+8^{2}+20 \times 8] \times 16$

$=\frac{1}{3} \times 3.14[400+64+160] \times 16$

$=\frac{3.14}{3} \times 624 \times 16$

$=3.14 \times 208 \times 16$

$=10449.92 \mathrm{~cm}^{3}$

Total surface area of the bucket $=\pi(r_2+r_1) l+\pi r_1^{2}$

$=\pi[(r_2+r_1) l+r_1^{2}]$

$=3.14[(20+8) 20+(8)^{2}]$

$=1959.36 \mathrm{~cm}^{2}$

Cost of the metal sheet used per $100 \mathrm{~cm}^{2}=Rs.\ 20$

Total cost $=1959.36 \times \frac{20}{100}$

$=Rs.\ 391 .87$

The volume of the bucket is $10449.92\ cm^3$ and the cost of the bucket is Rs. 391.87.

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Updated on: 10-Oct-2022

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