What part of a ditch, 48 metres long, 16.5 metres broad and 4 metres deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres ? (Use `pi = (22)/(7)`)

Options

A. `(1)/(9)`
B. `(2)/(9)`
C. `(7)/(9)`
D. `(8)/(9)`

Answer

To solve the problem, we need to find out how much of the ditch can be filled with the earth obtained from digging a cylindrical tunnel. We will follow these steps:

### Step 1: Calculate the volume of the cylindrical tunnel.
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height (or length) of the cylinder.

Given:
- Diameter of the tunnel = 4 meters, so the radius \( r = \frac{4}{2} = 2 \) meters.
- Length of the tunnel \( h = 56 \) meters.

Substituting the values into the formula:
\[ V = \frac{22}{7} \times (2^2) \times 56 \]
Calculating:
\[ V = \frac{22}{7} \times 4 \times 56 \]
\[ V = \frac{22 \times 224}{7} \]
\[ V = \frac{4928}{7} = 704 \text{ cubic meters} \]

### Step 2: Calculate the volume of the ditch.
The volume \( V \) of a rectangular prism (the ditch) is given by:
\[ V = \text{length} \times \text{breadth} \times \text{depth} \]

Given:
- Length of the ditch = 48 meters
- Breadth of the ditch = 16.5 meters
- Depth of the ditch = 4 meters

Substituting the values into the formula:
\[ V = 48 \times 16.5 \times 4 \]
Calculating:
\[ V = 48 \times 66 = 3168 \text{ cubic meters} \]

### Step 3: Determine the part of the ditch that can be filled.
To find out what part of the ditch can be filled with the earth from the tunnel, we need to divide the volume of earth obtained from the tunnel by the volume of the ditch:
\[ \text{Part filled} = \frac{\text{Volume of tunnel}}{\text{Volume of ditch}} = \frac{704}{3168} \]

### Step 4: Simplify the fraction.
To simplify \( \frac{704}{3168} \), we can divide both the numerator and the denominator by their greatest common divisor (GCD).

Calculating:
\[ \frac{704 \div 704}{3168 \div 704} = \frac{1}{4.5} = \frac{2}{9} \]

Thus, the part of the ditch that can be filled is \( \frac{2}{9} \).

### Final Answer:
The part of the ditch that can be filled by the earth obtained from digging the tunnel is \( \frac{2}{9} \).

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