Wheatstone bridge balance condition Steps:

• The voltmeter reads zero across the bridge, indicating balance where potential difference is zero.
• In the bridge, arms are 1Ω and 2Ω on one side, 4Ω and R on the other.
• Balance requires (1Ω × R) = (2Ω × 4Ω).
• Solve: R = (2 × 4) / 1 = 8Ω. Why D is correct:
• Balance condition from Kirchhoff's laws states opposite arm products equal for zero voltage across galvanometer (voltmeter here). Why the others are wrong:
• A: 1Ω gives (1×1)=1 ≠ 8, unbalanced.
• B: 2Ω gives (1×2)=2 ≠ 8, unbalanced.
• C: 4Ω gives (1×4)=4 ≠ 8, unbalanced. Final answer: D