Question

In the shown arrangement of the experiment of the meter bridge, if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled? 

• A. \(x\)
• B. \(x/4\)
• C. \(4x\)
• D. 2x

Hint:

In meter bridge experiments: • Balance point depends on ratio of resistances • Uniform change in wire resistance does not shift null point

Answer 1

The Correct Option is A

Step 1: In a meter bridge, the balance (null) condition is: (R₁)/(R₂) = (l)/(100-l) where l is the balancing length from end A.
Step 2: The balancing point depends only on the ratio of resistances, not on the absolute resistance of the bridge wire.
Step 3: Resistance of the wire per unit length: R ∝ (1)/(A) ∝ (1)/(r²) Doubling the radius makes resistance per unit length one–fourth everywhere.
Step 4: Since resistance changes uniformly along the entire wire, the ratio of lengths remains unchanged. ⟹ Null point remains at x