Question

In a metre bridge experiment, resistances are connected as shown in figure. The balancing length $I _{1}$ 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. $\frac{11}{48}\Omega$

Answer 1

Answer: (C)

For the given meter bridge,
$\frac{P}{Q}=\frac{l_{1}}{100-l_{1}}$
$l_{1}=55\, cm \Rightarrow 100-l_{1}=45 \, cm$
$\because P=3\Omega$
$\Rightarrow Q=3\times\frac{45}{55}=3\times\frac{9}{11}=\frac{27}{11}\Omega$
when x is connected in series with $P, l_{1} = 75\, cm$
$\Rightarrow \frac{P+X}{Q}=\frac{75\,cm}{25\,cm}\Rightarrow 3+X=3\times\frac{27}{11}$
$\Rightarrow X=\frac{81}{11}-3\Rightarrow X=\frac{48}{11} \Omega.$