The resistances in the left and right gap of a balanced metre bridge are $12 \Omega$ and $36 \Omega$…
The resistances in the left and right gap of a balanced metre bridge are $12 \Omega$ and $36 \Omega$ respectively If the resistances are interchanged the balance point shifts by
$25 \mathrm{~cm}$ towards right,
$50 \mathrm{~cm}$ towards right,
$25 \mathrm{~cm}$ towards left,
$50 \mathrm{~cm}$ towards left,
Solution
In the first case: $R=12 \Omega$ and $S=36 \Omega$
In balanced condition, $\frac{R}{S}=\frac{1}{100-l}$ or $\frac{12}{36}=\frac{l}{100-l}$ or $3 l=100-l$ or $4 l=100 \Rightarrow l=25 \mathrm{~cm}$ Therefore, first balance point is $l_1=25 \mathrm{~cm}$
In the second case: $R=36 \Omega$ and $S=12 \Omega$
In balanced condition, $\frac{R}{S}=\frac{l}{100-l}$ or $\frac{36}{12}=\frac{1}{100-1}$ or $l=300-3 l$ or $4 l=300 \Rightarrow l=l_2=75 \mathrm{~cm}$
Therefore, second balance point is $l_2=75 \mathrm{~cm}$
The shift $=l_2-l_1=(75-25) \mathrm{cm}=50 \mathrm{~cm}$