1. Use the constant voltage drop model for diodes

   • Germanium diode (Ge): forward drop $=0.3\,\text{V}$
   • Silicon diode (Si): forward drop $=0.7\,\text{V}$

2. Interpret the standard circuit behavior

   In this type of question, the output $V_0$ is determined by the conducting diodes. Reversing the Ge diode changes its role from forward-biased conduction to reverse-biased non-conduction (since breakdown is assumed not to occur).

3. Original case

   In the original circuit, Ge and Si diodes are arranged so that the effective output level involves their forward drops.

   Since the Ge diode is initially forward conducting, its drop contributes $0.3\,\text{V}$.

4. After reversing the Ge diode

   When the Ge diode is reversed, it will no longer conduct in the previous direction. Because its breakdown voltage is large, it behaves like an open circuit in reverse bias.

   Therefore, the output level shifts by the difference between the Si and Ge forward threshold values:

   $\Delta V_0 = 0.7 - 0.3 = 0.4\,\text{V}$

5. Hence

   The value of $V_0$ changes by

   $\boxed{0.4\,\text{V}}$

6. Option check

   • A: $0.8\,\text{V}$ ❌
   • B: $0.6\,\text{V}$ ❌
   • C: $0.2\,\text{V}$ ❌
   • D: $0.4\,\text{V}$ ✅

Therefore, the correct option is D.