Q.
In a metre bridge experiment, resistances are connected as shown in figure. The balancing length $l_{1}$ is $55\ cm.$ Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $75\ cm.$ The value of $x$ is
A) $\frac{54}{12}\Omega$
B) $\frac{20}{11}\Omega$
C) $\frac{48}{11}\Omega$
D) $\frac{11}{48}\Omega$
Explanation
(c) For the given meter bridge
$\frac{P}{Q} = \frac{l_{1}}{100 - l_{1}}$
$l_{1} = 55cm \Rightarrow 100 - l_{1} = 45cm$
$\because P = 3Q$
$\Rightarrow Q = 3 \times \frac{45}{55} = 3 \times \frac{9}{11} = \frac{27}{11}\Omega$
When $x$ is connected in series with $P,\ l_{1} = 75cm$
$\Rightarrow \frac{P + x}{Q} = \frac{75\ cm}{25\ cm} \Rightarrow 3 + x = 3 \times \frac{27}{11}$
$\Rightarrow x = \frac{81}{11} - 3 \Rightarrow x = \frac{48}{11}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.