To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

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).

$A$ $10 \, m$ 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 the given arrangement,$E_1 = 5 \, V$ and $E_2 = 7 \, V$. The balancing length is $6 \, m$. If the terminals of $E_2$ are reversed,then the new balancing length will be:

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is

Sensitivity of a potentiometer can be increased by: