The given circuit shows a uniform straight wire $AB$ of $100 \mathrm{~cm}$ length fixed at both ends

Question

The given circuit shows a uniform straight wire $AB$ of $100 \mathrm{~cm}$ length fixed at both ends. If the resistors in the left and right gaps are $10 \Omega$ and $15 \Omega$ respectively, in order to get zero reading in the galvanometer $G$, the free end of jockey $J$ must be placed at what distance from end $B$?

Professor PC Thomas Classes · Answer

(c) : The galvanometer shows zero reading in a balanced bridge condition: $\frac{R_1}{R_2} = \frac{x}{l-x}$, where $x$ is the balancing length from $A$. Given $l = 100 \mathrm{~cm}$, $R_1 = 10 \Omega$, and $R_2 = 15 \Omega$. So, $\frac{10}{15} = \frac{x}{100-x} \Rightarrow \frac{2}{3} = \frac{x}{100-x}$. This gives $200 - 2x = 3x \Rightarrow 5x = 200 \Rightarrow x = 40 \mathrm{~cm}$. The distance from $B$ is $100 - 40 = 60 \mathrm{~cm}$.