A cylindrical well of height 80 metres and radius 7 metres is dug in a field 28 metres long and 22 metres wide. The earth taken out is spread evenly on the field. What is the increase (in metres) in the level of the field?

Options

A. 13.33
B. 26.66
C. 18.17
D. 28.17

Answer

To solve the problem step by step, we need to calculate the volume of the cylindrical well and then determine how much the earth taken out will raise the level of the field.

### Step 1: Calculate the Volume of the Cylindrical Well
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where:
- \( r \) is the radius of the cylinder,
- \( h \) is the height of the cylinder.

Given:
- Radius \( r = 7 \) meters,
- Height \( h = 80 \) meters.

Substituting the values into the formula:
\[ V = \pi (7^2)(80) = \pi (49)(80) = 3920\pi \text{ cubic meters} \]

### Step 2: Calculate the Area of the Field
The area \( A \) of the rectangular field is calculated using the formula:
\[ A = \text{length} \times \text{width} \]
Given:
- Length = 28 meters,
- Width = 22 meters.

Calculating the area:
\[ A = 28 \times 22 = 616 \text{ square meters} \]

### Step 3: Determine the Increase in Level of the Field
The volume of earth taken out from the well will be spread evenly over the area of the field. Let \( h \) be the increase in the level of the field. The volume of earth spread over the field can also be expressed as:
\[ \text{Volume} = \text{Area} \times \text{Height increase} \]
This gives us:
\[ 3920\pi = 616h \]

### Step 4: Solve for \( h \)
Rearranging the equation to find \( h \):
\[ h = \frac{3920\pi}{616} \]

Calculating \( h \):
\[ h = \frac{3920 \times 3.14}{616} \approx \frac{12310.8}{616} \approx 20.00 \text{ meters} \]

### Step 5: Calculate the Final Value
Now we can compute the value of \( h \) more accurately:
Using \( \pi \approx 3.14 \):
\[ h \approx \frac{3920 \times 3.14}{616} \approx \frac{12310.8}{616} \approx 20.00 \text{ meters} \]

### Final Answer
The increase in the level of the field is approximately \( 20.00 \) meters.