Initially at null deflection R2R1=32…(i) Finally at null deflection, when null point is shifted R2R1+10=1⇒R1+10=R2...(ii) Solving equations (i) and (ii) we get 32R2+10=R2 10=3R2⇒R2=30Ω &R1=20Ω Now if required resistance is R then 3030+R30×R=32 R=60Ω
