Question

A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, then find the height of the bucket. [CBSE 2012]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Radius}\mathrm{of}\mathrm{upper}\mathrm{end},R=28\mathrm{cm}\mathrm{and} \\ \mathrm{Radius}\mathrm{of}\mathrm{lower}\mathrm{end},r=21\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{bucket}\mathrm{be}h. \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{the}\mathrm{bucket}\mathrm{can}\mathrm{hold}=28.49\mathrm{L} \\ \Rightarrow \mathrm{Volume}\mathrm{of}\mathrm{bucket}=28490{\mathrm{cm}}^3\left(\mathrm{As},1\mathrm{L}=1000{\mathrm{cm}}^3\right) \\ \Rightarrow \frac{1}{3}\mathrm{\pi }h\left(R^2+r^2+Rr\right)=28490 \\ \Rightarrow \frac{1}{3}\times \frac{22}{7}\times h\times \left({28}^2+{21}^2+28\times 21\right)=28490 \\ \Rightarrow \frac{22h}{21}\times 49\times \left(16+9+12\right)=28490 \\ \Rightarrow \frac{22h}{3}\times 7\times 37=28490 \\ \Rightarrow h=\frac{28490\times 3}{22\times 7\times 37} \\ \therefore h=15\mathrm{cm} \end{gathered}$$

So, the height of the bucket is 15 cm.