In a meter bridge a $30\Omega$ resistance is connected in the left gap and a pair of resistances P and Q in the right gap. Measured from the left, the balance point is$37.5cm$ , when P and Q are in series and $71.4cm$ when they the parallel. The values of P and Q $\left(in\Omega \right)$ are :

detailed solution

Correct option is C

Resistance R=30 Ω Resistance S=P and Q Balancing length l=37.5 cm Balancing length l1=71.4 cm 1st case :when P and Q are in seriesRS=l(100−l) 30P+Q=37.5(100−37.5) 30P+Q=37.562.5 P+Q=30×62.537.5 P+Q=50                         …(i)IInd case :when P and Q are in parallelRS=l1(100−l1) 30PQP+Q=71.4(100−71.4) 30(P+Q)PQ=71.428.6 30×50PQ=71.428.6 PQ=30×50×28.671.4 PQ=600                           …(ii)So, from Eqs. (i) and (ii)P+Q=50 PQ=600 Q=50−P P(50−P)=600 50P−P2=600 P2−50P+600=0 P2−30P−20P+600=0 P(P−30)−20(P−30)=0 P=20,  P=30 ∴    P=30Ω ∴   Q=20Ω