The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

In the given potentiometer circuit,the resistance of the potentiometer wire $AB$ is $R_0$. $C$ is a cell of internal resistance $r$. The galvanometer $G$ does not give zero deflection for any position of the jockey $J$. Which of the following cannot be a reason for this?

The length of a wire of a potentiometer is $100 \, cm$,and the $emf$ of its standard cell is $E \, volt$. It is employed to measure the $emf$ of a battery whose internal resistance is $0.5 \, \Omega$. If the balance point is obtained at $l = 30 \, cm$ from the positive end,the $emf$ of the battery is (where $i$ is the current in the potentiometer wire).

$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $3 \text{ V}$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)

$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-