The resistances in the two gaps of a balanced meter bridge are $X \ \Omega$ and $3X \ \Omega$ respectively. If the resistances are interchanged, the balance point shifts by: (in $\text{ cm}$)

In a metre bridge experiment (see figure), the positions of the cell, $E$, and galvanometer, $G$, are interchanged. We shall observe in the galvanometer:

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire is

In the experimental setup of a meter bridge shown in the figure,the null point is obtained at a distance of $40\,cm$ from $A$. If a $10\,\Omega$ resistor is connected in series with $R_1$,the null point shifts by $10\,cm$. The resistance that should be connected in parallel with $(R_1 + 10)\,\Omega$ such that the null point shifts back to its initial position is .............. $\Omega$.

In a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?

In a meter bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)