πh3 R2 + Rr + r2

$\Rightarrow \frac{22 h}{7 x 3} (28^{2} + 28 x 21 + 21^{2}) = 28490 \Rightarrow \frac{22}{21} h x 1813 = 28490 \Rightarrow h = \frac{28490 x 21}{22 x 1813} = 15 cm$

Hence the height of bucket is 15 cm.

" />πh3 R2 + Rr + r2

$\Rightarrow \frac{22 h}{7 x 3} (28^{2} + 28 x 21 + 21^{2}) = 28490 \Rightarrow \frac{22}{21} h x 1813 = 28490 \Rightarrow h = \frac{28490 x 21}{22 x 1813} = 15 cm$

Hence the height of bucket is 15 cm.

" />πh3 R2 + Rr + r2

$\Rightarrow \frac{22 h}{7 x 3} (28^{2} + 28 x 21 + 21^{2}) = 28490 \Rightarrow \frac{22}{21} h x 1813 = 28490 \Rightarrow h = \frac{28490 x 21}{22 x 1813} = 15 cm$

Hence the height of bucket is 15 cm.

" />

Surface Areas and Volumes

Question

Wired Faculty App

A bucket is in the form of a frustum of a cone and its can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket. $[use π = \frac{22}{7}]$

Answer

Here, R = 28 cm and r = 21 cm, we need to find h

Volume of frustum = 28.49 L = 28.49 x 1000 cm³ = 28490 cm³

Now, volume of frustum = $\frac{πh}{3} (R^{2} + Rr + r^{2})$

$\Rightarrow \frac{22 h}{7 x 3} (28^{2} + 28 x 21 + 21^{2}) = 28490 \Rightarrow \frac{22}{21} h x 1813 = 28490 \Rightarrow h = \frac{28490 x 21}{22 x 1813} = 15 cm$