Question

A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5 m. The earth dug out is spread over the remaining portion of the field. How much is the level of the field raised?

Answer 1

Length of the tank = 5 m
Breadth of the tank = 4.5 m
Height of the tank = 2.1 m
Volume of the earth dug out = $5\times 4.5\times 2.1=47.25{\mathrm{m}}^3$

Now, area over which the earth is spread = area of the field − area of the earth dug out
$$\begin{gathered} =13.5\times 2.5-5\times 4.5 \\ =33.75-22.5 \\ =11.25{\mathrm{m}}^2 \end{gathered}$$

∴ Height of the raised field = $\frac{\mathrm{volume}\mathrm{of}\mathrm{earth}\mathrm{dug}\mathrm{o}\mathrm{ut}}{\mathrm{area}\mathrm{over}\mathrm{which}\mathrm{the}\mathrm{earth}\mathrm{is}\mathrm{spread}}=\frac{47.25}{11.25}=4.2\mathrm{m}$