The figure shows a meter-bridge circuit,with $AB = 100\, cm$,$X = 12\,\Omega$ and $R = 18\,\Omega$,and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$,through what distance will $J$ have to be moved to obtain balance? .............. $cm$

The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The smaller resistance is (in $\Omega$)

In a meter bridge experiment,resistances are connected as shown in the figure. Initially,resistance $P = 4\,\Omega$ and the null point $N$ is at $60\,cm$ from $A$. Now,an unknown resistance $R$ is connected in series to $P$,and the new position of the null point is at $80\,cm$ from $A$. The value of unknown resistance $R$ is

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 $\,cm$)

Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)

In a meter bridge,the balancing length from the left end is found to be $25 \ cm$. The value of the unknown resistance is (assume,standard resistance of $1 \ \Omega$ is in the right gap). (in $\Omega$)