Combination of Resistors question from JEE Main 2021 (31 Aug, Shift 2)

Answer: (d)

1. The left vertical wire shorts the $5\ \Omega$, so the top-left corner is at B's potential.
2. The middle node joins B through $2\ \Omega$ (vertical) and $2\ \Omega$ (top-left), giving $1\ \Omega$; with the next $2\ \Omega$ in series: $3\ \Omega$.
3. This $3\ \Omega$ is in parallel with $3 \parallel 3 = 1.5\ \Omega$: $R_{AB} = \dfrac{3 \times 1.5}{4.5} = 1\ \Omega$.
4. So the answer is (d) $1\ \Omega$.