• $summary_{type}$: application
• $word_{count}$: 160

The metre bridge implements a Wheatstone bridge using a uniform resistance wire. Balance condition: R/X = $\frac{l}{100-l}$, giving X = R$\frac{100-l}{l}$. The galvanometer shows zero deflection at the null point. For best sensitivity, choose R so that the null point is near l = 50 cm (when X ≈ R). This minimizes percentage error: $Delta_X$/X = $\frac{Delta_R}{R}$ + $Delta_l$(1/l + $\frac{1}{100-l}$). At l = 50: the error term is 2$Delta_l$/50 = $Delta_l$/25, which is minimized. End corrections (alpha, beta) account for resistance of copper strips: corrected formula R/X = $\frac{l+alpha}{(100-l+beta)}$. To find end corrections, use two known resistances. Interchanging R and X (swapping gaps) and averaging results cancels some systematic errors. JEE problems: find X, determine the sensitivity at a given null point, or analyze how errors propagate.