Two resistors 2Ω and 3Ω are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with 3Ω resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ____ Ω.

Solution

Physics sir · 4 steps

Find the first null point, add 22.5 cm, and the new ratio gives the parallel combination.

1. First null point. For a metre bridge of length 100 cm, balance means 2/3 = l/(100 - l). So 3l = 200 - 2l, which gives l = 40 cm from X.
2. New null point. A resistor in parallel with 3 Ω lowers the right-gap resistance, so the null point moves toward Y. It is now at l' = 40 + 22.5 = 62.5 cm, leaving 37.5 cm on the other side.

   2/R' = 62.5/37.5 = 5/3 ⇒ R' = 6/5 = 1.2 Ω

3. Undo the parallel. R' is 3 Ω in parallel with R, so 3R/(3 + R) = 1.2. That gives 3R = 3.6 + 1.2R, so 1.8R = 3.6.
4. So. R = 2 Ω, which is option (d).

The idea to keep. In a metre bridge, the ratio of the gap resistances equals the ratio of the balancing lengths. The new balance gives you the combined resistance first; only then solve for the unknown inside it.

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