Special Purpose Diodes question from JEE Main 2020 (5 Sep, Shift 2)

Answer: (d)

1. For $V_{in} < 4$ V neither Zener breaks down, so $V_0$ rises with $V_{in}$ at the full rate.
2. For $4 < V_{in} < 6$ V, Zener B conducts with its $100\,\Omega$ in series; this extra branch draws current, so $V_0$ still rises but more slowly.
3. When $V_0$ reaches 6 V, Zener A breaks down and holds $V_0 = 6$ V constant.
4. So the answer is (d): steep rise to 4 V, slower rise to 6 V, then flat at 6 V.