In a Wheatstone bridge with 2Ω and 3Ω, the null point is found when R2R1=L−ll.
Initial balance: 32=L−l1l1 gives l1=52L.
For standard wire length L=100 cm, l1=40 cm.
When unknown resistance R is connected in parallel with 3Ω, equivalent resistance becomes 3+R3R.
New null point: 3+R3R2=L−l2l2, which simplifies to 3R2(3+R)=L−l2l2.
Null point shifts 22.5 cm toward Y: l2=40+22.5=62.5 cm.
Substituting: 3R2(3+R)=37.562.5=35
6(3+R)=15R⇒18=9R⇒R=2 Ω
