The length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

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 circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

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

In the circuit shown, a four-wire potentiometer is made of a $400\, cm$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r = 0.01\, \Omega /cm$. If an ideal voltmeter is connected as shown with jockey $J$ at $50\, cm$ from end $A$, the expected reading of the voltmeter will be: ............... $V$