Meter Bridge Measurement

A meter bridge is a practical form of the Wheatstone bridge in which a uniform wire replaces two arms, so the balance condition becomes $\frac{X}{R}=\frac{l}{100-l}$, where $l$ is the balancing length, as demonstrated in NCERT Class 12, Chapter 3 (Current Electricity). Here the unknown $X$ is in the left gap with balancing length $l=40$ cm, and the known $R=6$ ohm. Substituting, $X=R\times \frac{l}{100-l}=6\times \frac{40}{60}=6\times \frac{2}{3}=4$ ohm. The option 9 ohm inverts the length ratio, using $\frac{60}{40}$ instead of $\frac{40}{60}$, which would apply if the unknown sat in the opposite gap. The option 6 ohm wrongly assumes that a balance point at 40 cm implies equal resistances, which is only true at the exact midpoint of 50 cm. The option 2.4 ohm multiplies 6 by 0.4, mistakenly treating 40 cm as a simple fraction of the whole wire rather than as a ratio of the two segments on either side of the balance point. A plausibility check confirms the answer: since the balance point lies closer to the unknown's end of the wire, the unknown resistance should be smaller than the known 6 ohm, and 4 ohm is indeed smaller, fully consistent with the bridge geometry.