A bucket has top and bottom diameters of \( 40 \mathrm{~cm} \) and \( 20 \mathrm{~cm} \) respectively. Find the volume of the bucket if its depth is \( 12 \mathrm{~cm} \). Also, find the cost of tin sheet used for making the bucket at the rate of \( ₹ 1.20 \) per \( \mathrm{dm}^{2} \). (Use \( \pi=3.14 \) )

Given:

A bucket has top and bottom diameters of \( 40 \mathrm{~cm} \) and \( 20 \mathrm{~cm} \) respectively.

Depth of the bucket is \( 12 \mathrm{~cm} \).

To do:

We have to find the volume of the bucket and the cost of tin sheet used for making the bucket at the rate of \( ₹ 1.20 \) per \( \mathrm{dm}^{2} \). 

Solution:

Upper diameter of the bucket $= 40\ cm$

Lower diameter of the bucket $= 20\ cm$

This impies,

Upper radius $r_1 = \frac{40}{2}$

$=20\ cm$

Lower radius $r_2 =\frac{20}{2}$ 

$= 10\ cm$

Depth of the bucket $h = 12\ cm$

Therefore,

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

$=\frac{22}{7 \times 3}(20^{2}+20 \times 10+10^{2}) \times 12$

$=\frac{22}{21}(400+200+100) \times 12$

$=\frac{22}{21} \times 700 \times 12$

$=8800 \mathrm{~cm}^{3}$

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

$=\sqrt{12^{2}+(20-10)^{2}}$

$=\sqrt{144+100}$

$=\sqrt{244} \mathrm{~cm}$

Surface area of the bucket $=\pi(r_{1}+r_{2}) l+\pi r_{2}^{2}$

$=3.14(20+10) \sqrt{244}+3.14 \times(10)^{2}$

$=3.14 \times 30 \times 15.62+3.14 \times 100$

$=1471.404+314$

$=1785.404 \mathrm{~cm}^{2}$

$=17.85404 \mathrm{dm}^{2}$

Cost of $1 \mathrm{dm}^{2}$ tin sheet $=Rs.\ 1.20$

Total cost of tin sheet used for making the bucket $=Rs.\ 17.85404 \times 1.20$

$=Rs.\ 21.42$

$=Rs.\ 21.40$

The cost of tin sheet used for making the bucket is Rs. 21.40.

Tutorialspoint

Updated on: 10-Oct-2022

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