P-N Junction Diode question from MHT CET 2024 (10 May, Shift 1)

Answer: (b)

1. The right plate of the battery is positive, so current goes up the right side and flows right to left through the branches.
2. $D_1$ points right: reverse biased for this current, no current in the $20\ \Omega$ branch.
3. $D_2$ points left: forward biased, so the $40\ \Omega$ branch conducts.
4. $I = \dfrac{10}{40} = 0.25$ A
5. So the answer is (b) $0.25$ A.