If the radii of the circular ends of a conical bucket which is $ 45 \mathrm{~cm} $ high be $ 28 \mathrm{~cm} $ and $ 7 \mathrm{~cm} $, find the capacity of the bucket. ( Use $ \pi=22 / 7 $ ).

Given:

The radii of the circular ends of a conical bucket which is \( 45 \mathrm{~cm} \) high are \( 28 \mathrm{~cm} \) and \( 7 \mathrm{~cm} \).

To do:

We have to find the capacity of the bucket.

Solution:

Let the upper radius of the conical bucket $r_{1}=28 \mathrm{~cm}$ and the lower radius $r_{2}=7 \mathrm{~cm}$

Height of the bucket $h=45 \mathrm{~cm}$

Therefore,

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

$=\frac{22}{7 \times 3}[(28)^{2}+28 \times 7+(7)^{2}] \times 45$

$=\frac{22}{21}[784+196+49] \times 45$

$=\frac{22}{21} \times 1029 \times 45$

$=22 \times 147 \times 15$

$=48510 \mathrm{~cm}^{3}$

The capacity of the bucket is $48510\ cm^3$.

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Updated on: 10-Oct-2022

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