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 $37.5 \mathrm{~cm}$, when $P$ and $Q$ are in series and $71.4 \mathrm{~cm}$ they are in parallel. The values of $P$ and $Q$ (in $\Omega$ ) are

• A

  40 10
• B

  35 15
• Correct answer
  C

  30 20
• D

  25 25

Solution

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

Hence, the correct answer is option: 'C'.