Question

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around the well in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Hint:

To find the volume of a cylindrical ring, subtract the volume of the smaller cylinder from the volume of the larger cylinder.

Answer 1

The volume of the earth dug out of the well is the volume of a cylinder with radius \( r = \frac{3}{2} = 1.5 \, \text{m} \) and height \( h = 14 \, \text{m} \). The volume \( V_{\text{well}} \) is given by: \[ V_{\text{well}} = \pi r^2 h = 3.14 \times (1.5)^2 \times 14 = 3.14 \times 2.25 \times 14 = 99.78 \, \text{m}^3. \]
Step 1: The earth is spread in the form of a circular ring, where the inner radius is \( r = 1.5 \, \text{m} \), and the outer radius is \( r + 4 = 1.5 + 4 = 5.5 \, \text{m} \). The volume \( V_{\text{ring}} \) of the embankment is the volume of the cylindrical ring, given by: \[ V_{\text{ring}} = \pi \left( (r + 4)^2 - r^2 \right) h_{\text{embankment}}. \] Substitute the values: \[ V_{\text{ring}} = 3.14 \left( (5.5)^2 - (1.5)^2 \right) h_{\text{embankment}} = 3.14 \left( 30.25 - 2.25 \right) h_{\text{embankment}} = 3.14 \times 28 h_{\text{embankment}}. \]
Step 2: We know that the volume of the earth dug out is the same as the volume of the embankment, so: \[ 99.78 = 3.14 \times 28 \times h_{\text{embankment}}. \] Solve for \( h_{\text{embankment}} \): \[ h_{\text{embankment}} = \frac{99.78}{3.14 \times 28} = \frac{99.78}{87.92} \approx 1.13 \, \text{m}. \]
Conclusion: The height of the embankment is approximately \( 1.13 \, \text{m} \).