Consider a noiseless channel with a bandwidth of 5000 Hz transmitting a signal…

Key formula: maximum bit rate = 2 × bandwidth × log2(number of signal levels)

Given: bandwidth = 5000 Hz, number of signal levels = 2

• Compute log2(2) = 1.

• Maximum bit rate = 2 × 5000 × 1 = 10000 bps.

Therefore, the maximum bit rate is 10000 bps.

Why the other numeric choices are incorrect:

• 2500 bps: Too low and inconsistent with the Nyquist formula for binary signaling.

• 5000 bps: Equal to the bandwidth in Hz, but the maximum bit rate for two signal levels is twice the bandwidth, so this understates the maximum.

• 20000 bps: Would require two additional bits per symbol (log2(number of signal levels) = 2), i.e., four signal levels. With only two levels, this rate is not achievable.