Q. 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 when they are parallel. The values of P and Q (in ohm) are
A) 40, 10
B) 35, 15
C) 30, 20
D) 25, 25
Explanation
(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
$\frac{30}{\frac{PQ}{P + Q}} = \frac{l}{(100 - l)}$
$\frac{30(P + Q)}{PQ} = \frac{71.4}{(100 - 71.4)}$
$\frac{30 \times 50}{PQ} = \frac{71.4}{28.6}$
$PQ = \frac{30 \times 50 \times 28.6}{71.4}$
$P \approx 600\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
So, from Eqs. (i) and (ii)
$P = 30\Omega\ and\ Q = 20\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.