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 $37.5\, cm$, when $P$ and $Q$ are in series and $71.4\, cm$ 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{P Q}{P+Q}} =\frac{l}{(100-l)} $
$\frac{30(P+Q)}{P Q} =\frac{71.4}{(100-71.4)} $
$ \frac{30 \times 50}{P Q} =\frac{71.4}{28.6} $
$ P Q =\frac{30 \times 50 \times 28.6}{71.4} $
$P Q \approx 600\,\,\,...(ii) $
So, from Eqs. (i) and (ii), we get
$P=30\, \Omega, Q=20\, \Omega$