In the potentiometer experiment shown in the figure,for the position $X$ of the jockey $J$,there occurs a null deflection in the galvanometer. Then the potential difference between points $A$ and $X$ is ................ $V$.

In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $Omega$)

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

The current in the primary circuit of a potentiometer is $0.2 \, A$. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7} \, \Omega \cdot m$ and $8 \times 10^{-7} \, m^2$ respectively. The potential gradient will be equal to .............. $V/m$.

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is