A ring is made of a wire having a resistance 12Ω. The points P and Q as shown in figure, at which a current carrying conductor should be connected, so that the resistance of the sub circuit between these points is equal to 83Ω, divide the ring in lengthsl1and l2.Then

Here, resistance of wire=R=12Ω Equivalent resistance of lengths l1 and l2=83Ω Let, resistance of length l1and l2are R1R2 respectively. ∴Req=R1R2R1+R2 We know, R1+R2=R=12Ω                  ?(i) So, 83=R1R212 or,        R1R2=32                              ?(ii) From (i) and (ii)             R1=4Ωand R2=8Ω             We know, R=ρlA             or         R∝l So, R1R2=l1l2or l1l2=12