Question

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 $71.4cm$ they are in parallel. The values of $P$ and $Q($ in $\Omega )$ are

• A. 4010
• B. 3515
• C. 3020
• D. 2525

Answer 1

Answer: (C)

Ist case
$\frac{30}{P+Q}=\frac{l}{(100-l)}$
$\frac{30}{P+Q}=\frac{37.5}{(100-37.5)}$
$\frac{30}{P+Q}=\frac{37.5}{62.5}$
$P+Q=\frac{30\times 62.5}{37.5}$
$P+Q=50...(i)$
IInd case
$\frac{30}{\frac{PQ}{P+Q}}=\frac{l}{(100-l)}$
$\frac{30(P+Q)}{PQ}=\frac{71.4}{(100-71.4)}$
$\frac{30\times 50}{PQ}=\frac{71.4}{28.6}$
$PQ=\frac{30\times 50\times 28.6}{71.4}$
$PQ\approx 600...(ii)$
So, from Eqs. (i) and (ii), we get
$P=30\Omega ,Q=20\Omega$