The propagation delay of the $2 \times 1$ MUX shown in the circuit is $10 \mathrm{~ns}$. Consider the propagation delay of the inverter as $0 \mathrm{~ns}$.
If $\text{S}$ is set to $1$ then the output $\text{Y}$ is $\_\_\_\_\_\_$.
- a square wave of frequency $100 \mathrm{MHz}$
- a square wave of frequency $50 \mathrm{MHz}$
- constant at $0$
- constant at $1$