Meter bridge null point condition: R2R1=100−ll
Initial: R1=2Ω, R2=3Ω, null at l cm
32=100−ll gives 2(100−l)=3l, so l=40 cm
When xΩ connected in parallel with 3Ω: R2′=3+x3xΩ
New null point at (40+10)=50 cm from left:
R2′2=5050=1
3x2(3+x)=1 gives 6+2x=3x
x=6Ω