The number of telephone conversations by a fibre of band width 40MHz, with much less intensity loss, are:

Options

A. 1800
B. 2000
C. 2200
D. 2400

Answer

To solve the problem, we need to determine the number of telephone conversations that can be supported by a fiber optic cable with a bandwidth of 40 MHz, assuming there is no intensity loss.

### Step-by-Step Solution:

1. **Identify the bandwidth of the fiber**:
The given bandwidth of the fiber is 40 MHz.
\[ \text{Bandwidth of fiber} = 40 \text{ MHz} = 40 \times 10^6 \text{ Hz} \]

2. **Identify the bandwidth required for a single telephone conversation**:
The typical bandwidth required for a single telephone conversation is approximately 20 kHz (or 2 x 10^4 Hz).
\[ \text{Bandwidth for one conversation} = 20 \text{ kHz} = 2 \times 10^4 \text{ Hz} \]

3. **Calculate the number of simultaneous conversations**:
To find the number of telephone conversations that can be supported, we divide the bandwidth of the fiber by the bandwidth required for one conversation.
\[ \text{Number of conversations} = \frac{\text{Bandwidth of fiber}}{\text{Bandwidth for one conversation}} = \frac{40 \times 10^6 \text{ Hz}}{2 \times 10^4 \text{ Hz}} \]

4. **Perform the calculation**:
\[ \text{Number of conversations} = \frac{40 \times 10^6}{2 \times 10^4} = \frac{40}{2} \times \frac{10^6}{10^4} = 20 \times 10^2 = 2000 \]

5. **Conclusion**:
Therefore, the number of simultaneous telephone conversations that can be supported by the fiber is 2000.

### Final Answer:
The number of telephone conversations by a fiber of bandwidth 40 MHz is **2000**.

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