In a meter bridge experiment,when resistance $X$ is in the left gap and $Y$ is in the right gap,the null point is found at $20 \ cm$ from the left end. If $X < Y$,what will be the new position of the null point from the same end if a resistance of $4X$ is placed in the left gap instead of $X$?

Two wires $A$ and $B$ of equal lengths are connected in the left and right gaps of a metre bridge. The null point is obtained at $40 \ cm$ from the left end. If the diameters of the wires $A$ and $B$ are in the ratio $3:1$,what is the ratio of the specific resistance (resistivity) of $A$ to that of $B$ (in $:$)?

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

The resistances in the two arms of a meter bridge are $5\,\Omega$ and $R\,\Omega$,respectively. When the resistance $R$ is shunted with an equal resistance,the new balance point is at $1.6\,l_1$. The resistance $R$ is ................. $\Omega$.

$A$ meter bridge is set up as shown to determine the end corrections at $A$ and $B$. When a resistance of $15\,\Omega$ is used in the left gap and $20\,\Omega$ in the right gap,the null point is obtained at a distance of $42\,cm$ from $A$. When these resistances are interchanged,the null point is obtained at a distance of $57\,cm$ from $A$. The values of the end corrections are:

The figure shows a rough sketch of a meter bridge. The galvanometer $(G)$ shows zero deflection at a length $l \, cm$. Now,$R_1$ and $R_2$ are interchanged,and the balancing length increases by $25 \, cm$. Find the ratio $R_1/R_2$.