A well of internal diameter 14 m and depth 14 m is dug and the earth taken out is spread all around it upto 5 m width and form an embankment. Find the height of the embankment.

Answer

To solve the problem, we need to find the height of the embankment formed by the earth taken out from the well. Let's break down the solution step by step.

### Step 1: Calculate the volume of the well
The well is in the shape of a cylinder. The formula for the volume \( V \) of a cylinder is given by:

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

Where:
- \( r \) is the radius of the cylinder (well),
- \( h \) is the height (depth) of the cylinder.

Given:
- Internal diameter of the well = 14 m, so the radius \( r = \frac{14}{2} = 7 \) m.
- Depth of the well \( h = 14 \) m.

Now, substituting the values into the formula:

\[ V = \pi (7)^2 (14) = \pi (49)(14) = 686\pi \, \text{m}^3 \]

### Step 2: Calculate the volume of the embankment
The embankment is also in the shape of a cylinder, and its volume can be found by subtracting the volume of the inner cylinder (the well) from the volume of the outer cylinder (the well plus the embankment).

The outer radius \( R \) is the radius of the well plus the width of the embankment:

\[ R = 7 + 5 = 12 \, \text{m} \]

Now, the volume of the outer cylinder is:

\[ V_{\text{outer}} = \pi R^2 h' = \pi (12)^2 h' = \pi (144) h' \]

The volume of the embankment is then:

\[ V_{\text{embankment}} = V_{\text{outer}} - V_{\text{well}} = \pi (144) h' - 686\pi \]

Factoring out \( \pi \):

\[ V_{\text{embankment}} = \pi (144 h' - 686) \]

### Step 3: Set the volumes equal
Since the volume of the earth taken out from the well is equal to the volume of the embankment, we have:

\[ 686\pi = \pi (144 h' - 686) \]

Dividing both sides by \( \pi \):

\[ 686 = 144 h' - 686 \]

### Step 4: Solve for \( h' \)
Rearranging the equation gives:

\[ 144 h' = 686 + 686 \]
\[ 144 h' = 1372 \]
\[ h' = \frac{1372}{144} \approx 9.5278 \, \text{m} \]

### Step 5: Final result
Thus, the height of the embankment is approximately:

\[ h' \approx 9.53 \, \text{m} \]