Please solve this que.

Dear Student,
A metre bridge works on the principle of a balanced wheatstone bridge.
Let $\sigma$ be the resistance per unit length of the metre bridge.
As per the circuit diagram,
$$\begin{gathered} \frac{R}{X}=\frac{40\sigma }{\left(100-40\right)\sigma }=\frac{40}{60}=\frac{2}{3} \\ X=\frac{3R}{2} \end{gathered}$$
Now the values of R and X are doubled and interchanged. Let the balancing point in this case is at a distance y from point A. Then,
$$\begin{gathered} \frac{2X}{2R}=\frac{y}{100-y} \\ \frac{X}{R}=\frac{y}{100-y} \\ Putting the value of X, we get \\ \frac{3R}{2R}=\frac{y}{100-y} \\ 3\left(100-y\right)=2y \\ 300-3y=2y \\ y=60 cm \end{gathered}$$

As the bridge is balanced, the potential at A and B is equal. Hence no current flows through the galvanometer if the galvanometer and the battery are interchanged in the balanced condition.

Regards
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