If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then:

In an experiment to find the $emf$ of a cell using a potentiometer,the length of the null point for a cell of $emf$ $1.5 \ V$ is found to be $60 \ cm$. If this cell is replaced by another cell of $emf$ $E$,the length of the null point increases by $40 \ cm$. The value of $E$ is $\frac{x}{10} \ V$. The value of $x$ is $............$

The figure shows a potentiometer wire $AB$ having a resistance of $5 \text{ } \Omega$ and a length of $10 \text{ m}$. The e.m.f. of the battery in the primary circuit is $5 \text{ V}$ and the external resistance is $45 \text{ } \Omega$. If the e.m.f. of the cell in the secondary circuit is $0.4 \text{ V}$, find the balancing length $AP$. (Internal resistance is negligible) (in $\text{ m}$)

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

In the figure,the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. The cell $C$'s $emf$ is $\frac{\varepsilon}{2}$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is

$A$ potentiometer is connected between $A$ and $B$ and the balance point is obtained at $203.6 \, cm$. When the end of the potentiometer connected to $B$ is shifted to $C$,then the balance point is obtained at $24.6 \, cm$. If now the potentiometer be connected between $B$ and $C$,the balance point will be at ................. $cm$.