Step 1: In a meter bridge, a known resistance R = 10 Ω in the left gap balances at l = 40.0 cm from the left end with an unknown wire in the right gap. Find the wire's resistance S. Step 2: At balance, R/S = l/(100 − l). Step 3: l/(100 − l) = 40.0/60.0 = 2/3. Step 4: S = R × (100 − l)/l = 10 × 60.0/40.0 = 15.0 Ω. Step 5: The wire is 50.0 cm long with radius 0.25 mm. Resistivity ρ = S·A/L = 15.0 × π(0.25×10⁻³)²/0.500. Step 6: A = 1.963×10⁻⁷ m²; ρ = 15.0 × 1.963×10⁻⁷/0.500 = 5.89×10⁻⁶ Ω·m. Result: S = 15.0 Ω; ρ ≈ 5.9×10⁻⁶ Ω·m (typical of nichrome/constantan-type alloys). Interchanging R and S and averaging removes end-resistance errors.