In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be ______________ cm.

Solution

Physics sir · 3 steps

Thicken the whole wire and both segments change by the same factor, so the null point stays put.

1. Balance condition. At null deflection the bridge is balanced, so R₁/R₂ equals the ratio of the resistances of wire segments AC and CB. For a uniform wire, each segment's resistance is ρ l / A.

   R₁/R₂ = (ρ l / A)/(ρ (100 - l) / A) = l/(100 - l)

2. Radius doubled. Doubling the radius makes the area 4A along the whole wire. Both segment resistances drop to one quarter, so ρ and A still cancel and the ratio depends only on l.
3. So. R₁ and R₂ have not changed, so the same l balances the bridge. The balancing length stays 40 cm.

The common slip: 10. Dividing 40 cm by 4 because the wire's resistance becomes one quarter gives 10, but both segments shrink by that same factor, so the ratio is unchanged.

The idea to keep. In a metre bridge only the ratio of the two wire lengths matters. Changing the wire's material, radius or the cell voltage uniformly does not move the null point.

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