A resistance of $20 \Omega$ is connected in the left gap of a metre bridge and an unknown resistance greater…
A resistance of $20 \Omega$ is connected in the left gap of a metre bridge and an unknown resistance greater than $20 \Omega$ is connected in the right gap. When these resistances are interchanged, the balance point shift by $20 \mathrm{~cm}$. The unknown resistance is
$25 \Omega$
$40 \Omega$
$35 \Omega$
$30 \Omega$
Solution
Let the unknown resistance be $R$ and the balance point be at $l \mathrm{~cm}$ initially.
using, $\frac{R_2}{R_1}=\frac{l}{100-l}$
Case 1: $R_1=20 \Omega$ and $R_2=R$
\(\therefore \frac{20}{R}=\frac{l}{100-l} \quad---(1)\) Case 2: \(R_1=R\) and \(R_2=20 \Omega\) and \(l^{\prime}=l+20 \mathrm{~cm}\) \(\therefore \frac{R}{20}=\frac{l+20}{80-l} \quad---(2)\)
Using equation (1) and (2) we get,
$\frac{l}{100-l}=\frac{80-l}{l+20}$
$\Rightarrow l=40 \mathrm{~cm}$
From equation (1),
$\frac{20}{R}=\frac{40}{100-40}$
$\Rightarrow R=30 \Omega$