There is a square field of area 'N' square metres. A cylindrical ditch of radius 7 metres and depth 4 metres is dug, and the earth is taken out and spread over the remaining part of the square field, the height of square field. goes up by I .54 metres. What is the value of 'N'?

Options

A. `548 m^2`
B. `524 m^2`
C. `518 m^2`
D. `554 m^2`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the Volume of the Cylindrical Ditch
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
- \( h \) is the height (or depth in this case)

Given:
- Radius \( r = 7 \) meters
- Depth \( h = 4 \) meters

Substituting the values into the formula:

\[ V = \pi \times (7)^2 \times 4 \]
\[ V = \pi \times 49 \times 4 \]
\[ V = 196\pi \text{ cubic meters} \]

Using \( \pi \approx \frac{22}{7} \):

\[ V = 196 \times \frac{22}{7} = \frac{4312}{7} \approx 616 \text{ cubic meters} \]

### Step 2: Determine the Area of the Square Field
Let the area of the square field be \( N \) square meters. The side length \( s \) of the square can be expressed as:

\[ s = \sqrt{N} \]

### Step 3: Calculate the Area of the Circular Ditch
The area \( A \) of the circular ditch is given by:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi \times (7)^2 = 49\pi \]

Using \( \pi \approx \frac{22}{7} \):

\[ A = 49 \times \frac{22}{7} = \frac{1078}{7} \approx 154 \text{ square meters} \]

### Step 4: Calculate the Remaining Area of the Square Field
The remaining area of the square field after the ditch is dug is:

\[ \text{Remaining Area} = N - 49\pi \approx N - 154 \]

### Step 5: Calculate the Volume of Earth Spread Over the Remaining Area
The volume of earth taken out (which is equal to the volume of the ditch) is spread over the remaining area, causing the height to increase by 1.54 meters. Thus, we can express the volume of the earth as:

\[ \text{Volume} = \text{Remaining Area} \times \text{Height Increase} \]

Substituting the values:

\[ 616 = (N - 154) \times 1.54 \]

### Step 6: Solve for \( N \)
Now, we can solve for \( N \):

\[ 616 = 1.54(N - 154) \]

Expanding the equation:

\[ 616 = 1.54N - 1.54 \times 154 \]

Calculating \( 1.54 \times 154 \):

\[ 1.54 \times 154 = 237.16 \]

Now substituting back:

\[ 616 = 1.54N - 237.16 \]

Adding \( 237.16 \) to both sides:

\[ 616 + 237.16 = 1.54N \]
\[ 853.16 = 1.54N \]

Now, divide both sides by \( 1.54 \):

\[ N = \frac{853.16}{1.54} \approx 554 \text{ square meters} \]

### Final Answer
Thus, the value of \( N \) is approximately **554 square meters**.

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