JEE Main 2024 Physics question
A wire of length 10 cm and radius m is connected across the right gap of a meter bridge. When resistance of 4.5 Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R × 10⁻⁷ Ω m, then value of R is
Answer: (D)
Solution
4.5/60 = S/40 gives S = 3 Ω. ρ = SA/ℓ = 3 × π(√7 × 10⁻⁴)²/0.1 = 66 × 10⁻⁷ Ω m (with π = 22/7), so R = 66.