The output voltage in the following circuit is (Consider ideal diode case) :

Both diodes have their p-sides connected to the central junction node ($$V_{\text{out}}$$), which is pulled up toward $$+5\text{ V}$$ via a resistor.

If $$V_{\text{out}}$$ rises above $$0\text{ V}$$, the anode of $$D_2$$ becomes more positive than its cathode ($$0\text{ V}$$). This makes diode $$D_2$$ forward-biased.

For $$D_1$$ to be forward-biased, $$V_{\text{out}}$$ would have to exceed $$+5\text{ V}$$, which is impossible since the maximum supply voltage is $$+5\text{ V}$$. Thus, diode $$D_1$$ remains reverse-biased (open circuit).

Since ideal diode $$D_2$$ is forward-biased, it acts as a perfect closed switch (short circuit) directly to the ground node:

$$V_{\text{out}} = V_{\text{ground}} = 0\text{ V}$$

The entire voltage drop of $$5\text{ V}$$ occurs across the pull-up resistor, clamping the output node securely to $$0\text{ V}$$.