To determine the behavior of the output \( Y \) when the select line \( S \) is set to \( 1 \), we must analyze the feedback path of the circuit.

1. Circuit Logic Analysis

• The circuit uses a \( 2 \times 1 \) Multiplexer (MUX). The output equation of a \( 2 \times 1 \) MUX is: \( Y = \overline{S} \cdot I_0 + S \cdot I_1 \)
• When \( S = 1 \), the MUX selects input \( I_1 \).
• Input \( I_1 \) is connected to the output \( Y \) via an inverter (NOT gate).
• Therefore, the logical operation of the circuit becomes: $$Y_{next} = \overline{Y_{current}}$$

This configuration, where the output is fed back to the input with an odd number of inversions, forms a ring oscillator. The output will continuously toggle between '0' and '1', creating a square wave.

2. Determining the Time Period (\( T \))

The time taken for the output to complete one full cycle (going from \( 0 \rightarrow 1 \rightarrow 0 \)) is twice the total propagation delay of the feedback loop.

• Propagation delay of MUX (\( t_{MUX} \)) = \( 10 \text{ ns} \)
• Propagation delay of Inverter (\( t_{Inv} \)) = \( 0 \text{ ns} \)
• Total delay for a single transition (\( \Delta t \)) = \( t_{MUX} + t_{Inv} = 10 \text{ ns} + 0 \text{ ns} = 10 \text{ ns} \)

A full cycle consists of two transitions (low-to-high and high-to-low):

$$T = 2 \times \Delta t = 2 \times 10 \text{ ns} = 20 \text{ ns}$$

3. Calculating the Frequency (\( f \))

The frequency is the reciprocal of the time period:

$$f = \frac{1}{T}$$ $$f = \frac{1}{20 \times 10^{-9} \text{ s}}$$ $$f = \frac{10^9}{20} \text{ Hz} = 50 \times 10^6 \text{ Hz}$$ $$f = 50 \text{ MHz}$$

Final Conclusion

Since the output toggles with a calculated frequency of \( 50 \text{ MHz} \), the correct choice is:

A square wave of frequency 50 MHz