Questions: For the summer cottage will drilled a water well through the rock. The diameter of the circular drilling hole is 118 millimetres and the depth of the circular drilling hole is 214 metres. The mineral aggregate drilled from the hole will be applied evenly to the paths of the summer cottage which widths are 45 centimetres and total length is 42 metres. What is the thickness of the layer? Assuming that the volume of the circular drilling hole is equal with the volume of the mineral aggregate. Answer tolerancy limits have been set ± 1 mm.

Solution Steps

To find the thickness of the layer of mineral aggregate, we need to calculate the volume of the drilled hole and then distribute this volume evenly over the paths. First, calculate the volume of the cylindrical hole using the formula for the volume of a cylinder: \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height (or depth). Convert the diameter from millimetres to metres to find the radius. Then, calculate the area of the paths and use the volume to find the thickness by dividing the volume by the area of the paths.

The diameter of the drilling hole is given as 118 mm. To find the radius in meters, we first convert the diameter to meters:

\[ \text{diameter in meters} = \frac{118}{1000} = 0.118 \, \text{m} \]

The radius is half of the diameter:

\[ r = \frac{0.118}{2} = 0.059 \, \text{m} \]

The volume \( V \) of a cylinder is given by the formula:

\[ V = \pi r^2 h \]

Substituting the known values:

\[ V = \pi (0.059)^2 \times 214 \]

\[ V \approx 2.340 \, \text{m}^3 \]

The width of the path is given as 45 cm. Convert this to meters:

\[ \text{width in meters} = \frac{45}{100} = 0.45 \, \text{m} \]

The area \( A \) of the paths is:

\[ A = \text{width} \times \text{length} = 0.45 \times 42 \]

\[ A = 18.9 \, \text{m}^2 \]

The thickness \( t \) of the layer is found by dividing the volume of the mineral aggregate by the area of the paths:

\[ t = \frac{V}{A} = \frac{2.340}{18.9} \]

\[ t \approx 0.1238 \, \text{m} \]

Convert the thickness to millimeters:

\[ t = 0.1238 \times 1000 = 123.8 \, \text{mm} \]

Final Answer