A backet is in the form of a frustum of a cone

Question:

A backet is in the form of a frustum of a cone and holds 28.490 L of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the

height of the bucket.

Solution:

Given, volume of the frustum = 28.49 L = 28.49 x 1000 cm3                                    [∴ 1 L = 1000 cm3]

= 28490 cm3

and radius of the top (r1) = 28 cm

radius of the bottom (r2) = 21 cm

Let height of the bucket = h cm

Now, volume of the bucket $=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)=28490$ [given]

$\Rightarrow \quad \frac{1}{3} \times \frac{22}{7} \times h\left(28^{2}+21^{2}+28 \times 21\right)=28490$

$\Rightarrow \quad h(784+441+588)=\frac{28490 \times 3 \times 7}{22}$

$\Rightarrow \quad 1813 h=1295 \times 21$

$h=\frac{1295 \times 21}{1813}=\frac{27195}{1813}=15 \mathrm{~cm}$

 

 

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