$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

$A$ potentiometer has a uniform potential gradient. The specific resistance (resistivity) of the material of the potentiometer wire is $10^{-7} \, \Omega \cdot m$, the current passing through it is $0.1 \, A$, and the cross-sectional area of the wire is $10^{-6} \, m^2$. The potential gradient along the potentiometer wire is:

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

In a potentiometer experiment,when key $K_1$ is closed,the balance length is $100 \, cm$. What will be the balancing length when key $K_2$ is closed? ................ $cm$

$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is: