Metre Bridge

The metre bridge, a practical form of the Wheatstone bridge described in NCERT Class 12, Chapter 3, is balanced when the ratio of the two unknown-side and standard-side resistances equals the ratio of the two lengths of bridge wire on either side of the jockey. At balance, $\frac{X}{R}=\frac{l}{100-l}$, where $l$ is the balancing length on the unknown side. With the unknown in the left gap and balance at $l=40$ cm, and standard $R=6$ ohm, we get $X=R\times \frac{l}{100-l}=6\times \frac{40}{60}=6\times \frac{2}{3}=4$ ohm. The value 9 ohm is wrong because it inverts the length ratio, using $\frac{60}{40}$ instead. The value 6 ohm is wrong because it assumes the balance sits at the midpoint 50 cm, which would make the resistances equal, but the balance is at 40 cm. The value 2.4 ohm is wrong because it multiplies 6 by the fraction $0.4$ rather than the correct ratio $\frac{40}{60}$. To minimise error, the balance point is best kept near the middle of the wire, because the fractional error in the length ratio is smallest there and the wire's non-uniformity and end resistances have the least influence. The jockey should be pressed only briefly to avoid heating the wire, which would change its resistance. A consistency check confirms that a balance shifted toward the unknown side, below 50 cm, correctly indicates the unknown is smaller than the standard, matching the computed 4 ohm.