$A$ $6 \, V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $1 \, m$. The positive terminal of another battery of emf $4 \, V$ and internal resistance $1 \, \Omega$ is joined to point $A$ as shown in the figure. The ammeter shows zero deflection when the jockey touches the wire at point $C$. The length $AC$ is equal to:

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

Two cells of emf $E_1$ and $E_2$ $(E_1 > E_2)$ are connected individually to a potentiometer and their corresponding balancing lengths are $625 \, cm$ and $500 \, cm$. Then the ratio $\frac{E_1}{E_2}$ is ...........

The length of a potentiometer wire is $l$. $A$ cell of $emf$ $E$ is balanced at a length $l/3$ from the positive end of the wire. If the length of the wire is increased by $l/2$,at what distance will the same cell give a balance point?

In a potentiometer, the null point is received at the $7^{th}$ wire. If we now want to shift the null point to the $9^{th}$ wire, what should we do?

In the given circuit of a potentiometer,the potential difference $E$ across $AB$ ($10\, m$ length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed),the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now,the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$. The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$,where $a = \dots$ (Refer to the image for balancing lengths $l_{1}$ and $l_{2}$ from point $A$).