Special Purpose Diodes question from JEE Main 2020 (4 Sep, Shift 1)

Answer: (c)

1. The two Zeners are connected back to back across the output, through series resistor $R_1$.
2. For $|V_{in}| < 6$ V one Zener is reverse biased below breakdown, so no current flows and $V_0 = V_{in}$.
3. For $|V_{in}| > 6$ V the reverse-biased Zener breaks down and the output is clipped at $+6$ V or $-6$ V (forward drop of the other diode neglected).
4. So the output is a sine wave clipped at both $+6$ V and $-6$ V. The answer is (c).