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? .............. $cm$

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 metre bridge experiment, the resistance in the left gap is $20 \text{ } \Omega$ and in the right gap is $60 \text{ } \Omega$. The bridge is balanced. The distance of the null point from the centre of the wire is (in $\text{ cm}$)

Resistance in the two gaps of a meter bridge are $10 \, \Omega$ and $30 \, \Omega$ respectively. If the resistances are interchanged, the balance point shifts by .............. $cm$.

An unknown resistance $R_1$ is connected in series with a resistance of $10 \,\Omega$. This combination is connected to one gap of a meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50 \, cm$. Now,when the $10 \,\Omega$ resistance is removed,the balance point shifts to $40 \, cm$. The value of $R_1$ is (in $\Omega$):

In a meter bridge experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?