Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of `V_(o)` changes by :
(assume that the Ge diode has large breakdown voltage)

Explanation

The effective forward voltage across Ge diode is `12V-0.3=11.7`
This will appear as the output voltage across the load, that is
`V_(0)=11.7V`
The current in the load is
`i=(V_(0))/(R_(L))=(11.7)/(5KOmega)=2.34mA`.
On reversing the connnections of Ge diode, it will be reverse-biased and conduct no current. Only Si diode will conduct. The effective forward voltage across Si diode is `12V-0.7V=11.3`
`V_(0)=11.3V`
The current in the load is
`i=(V_(0))/(R_(L))=(11.3)/(5kOmega)=2.26mA`.