A wire of length 10 cm and radius √7 × 10⁻⁴ m connected across the right gap of a meter bridge. When a 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 :

Solution

Physics sir · 4 steps

Balance the bridge to get the wire's resistance, then use ρ = RA/L with π = 22/7.

1. Balance condition. At balance, the ratio of the two gap resistances equals the ratio of the bridge-wire lengths on either side. The left gap matches 60 cm, so the right gap matches 100 - 60 = 40 cm.

   4.5/X = 60/40 ⇒ X = 4.5 × 40/60 = 3 Ω

2. Cross-section area. A = π r² = 22/7 × (√7 × 10⁻⁴)² = 22/7 × 7 × 10⁻⁸ = 22 × 10⁻⁸ m².
3. Resistivity. With L = 10 cm = 0.1 m, ρ = XA/L = (3 × 22 × 10⁻⁸)/0.1 = 66 × 10⁻⁷ Ω m.
4. So. R = 66, which is option (c).

The idea to keep. In a metre bridge, each gap pairs with the wire length on its own side. Find the unknown resistance first, then turn it into resistivity with ρ = RA/L, keeping every length in metres.

Answer from the NTA final answer key. Solution by PrepWiser.