In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l = 25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance l' (in cm) will now be________.

Solution

Physics sir · 4 steps

Find how many times bigger the new wire's resistance is, then rebalance the bridge with that new ratio.

1. First balance. At balance the ratio of the gaps equals the ratio of the wire lengths on each side of the jockey: R/S = 25/(100 - 25) = 1/3, so S = 3R.
2. New resistance. With R = (ρ L)/A and A = (π d²)/4, halving the diameter makes the area 1/4 of the old area. Halving the length as well gives R' = (1/2)/(1/4)R = 2R.

   R'/S = 2R/3R = l'/(100 - l')

3. Solve. 3l' = 2(100 - l'), so 5l' = 200 and l' = 40 cm.
4. So. The new balancing length is 40 cm.

The common slip: 25. Treating half the diameter as half the area makes R' = R, so the balance stays at 25 cm, not 40. Area depends on d², so it falls to one quarter.

The idea to keep. Resistance goes as L/d², so the diameter counts twice. In a metre bridge, work out the new resistance ratio first, then solve R'/S = l'/(100 - l').

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