Q. A resistance of 2$\ \Omega$ is connected across one gap of a meter-bridge(the length of the wire is 100cm) and an unknown resistance, greater than 2$\ \Omega$ is connected across the other gap. When these resistances are interchanged, the unknown resistance is
A) 3$\ \Omega$
B) 2$\ \Omega$
C) 4$\ \Omega$
D) 6$\ \Omega$
Explanation
(a) $R > 2\Omega$
$\therefore\ \ 100 - \ x > \ x$
Applying, $\frac{P}{Q} = \frac{R}{S}$
We have
$\frac{2}{R} = \frac{x}{100 - x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$\frac{R}{2} = \frac{x + 20}{80 - x}\ \ \ \ \ \ \ \ \ \ \ldots\ldots..(ii)$
Solving Eqs. (i) and (ii), we get
$R = 3\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.