In this circuit, diode $ D_1 $ is reverse biased, meaning it does not conduct current. On the other hand, diode $ D_2 $ is forward biased, so it allows current to flow through it.

Due to this configuration, there is a current flowing through $ D_2 $, and a voltage drop of 5 V occurs across the resistor connected in the circuit. As a result, the output voltage ($ V_{\text{out}} $) is 0 V.