1. Op-amp Analysis

• The op-amp is ideal and has negative feedback. Therefore, the voltages at its two input terminals are equal (virtual short principle).
• The non-inverting input ($V_+$) is connected to a $1\text{ V}$ source: $$V_+ = 1\text{ V}$$
• Due to the virtual short, the inverting input ($V_-$) must also be at $1\text{ V}$: $$V_- = 1\text{ V}$$

2. Transistor ($Q_1$) Analysis

• The emitter of transistor $Q_1$ is connected directly to the inverting input node ($V_-$). Thus, the emitter voltage $V_E$ is: $$V_E = V_- = 1\text{ V}$$
• We are given the base-emitter voltage $V_{BE} = 0.7\text{ V}$. We can find the base voltage $V_B$: $$V_B = V_E + V_{BE} = 1\text{ V} + 0.7\text{ V} = 1.7\text{ V}$$
• The base is connected to a $5\text{ V}$ source through a $33\text{ k}\Omega$ resistor. We calculate the base current $I_B$ using Ohm's Law: $$I_B = \frac{5\text{ V} - V_B}{33\text{ k}\Omega} = \frac{5\text{ V} - 1.7\text{ V}}{33\text{ k}\Omega} = \frac{3.3\text{ V}}{33,000\text{ }\Omega} = 0.0001\text{ A} = 0.1\text{ mA}$$
• The current gain is given as $\beta = 99$. The emitter current $I_E$ is: $$I_E = (\beta + 1) \cdot I_B = (99 + 1) \cdot 0.1\text{ mA} = 100 \cdot 0.1\text{ mA} = 10\text{ mA}$$

3. Calculating the Output Voltage ($V_O$)

• Since the op-amp is ideal, no current enters its inverting terminal ($I_- = 0$). Therefore, the entire emitter current $I_E$ from $Q_1$ must flow through the $100\text{ }\Omega$ feedback resistor ($R_f$) towards the output $V_O$.
• Using Ohm's Law across the feedback resistor: $$V_- - V_O = I_E \cdot R_f$$ $$1\text{ V} - V_O = (10\text{ mA}) \cdot (100\text{ }\Omega)$$ $$1\text{ V} - V_O = (0.01\text{ A}) \cdot (100\text{ }\Omega)$$ $$1\text{ V} - V_O = 1\text{ V}$$
• Solving for $V_O$: $$V_O = 1\text{ V} - 1\text{ V} = 0\text{ V}$$

Conclusion

The output voltage $V_O$ is $0\text{ V}$. The correct option is (C).