Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In an experiment to measure the internal resistance of a cell by a potentiometer,it is found that the balance point is at a length of $2\,m$ when the cell is shunted by a $5\,\Omega$ resistance; and is at a length of $3\,m$ when the cell is shunted by a $10\,\Omega$ resistance. The internal resistance of the cell is,then ................ $\Omega$.

$A$ potentiometer wire has a length of $4 \ m$ and a resistance of $10 \ \Omega$. It is connected to a cell of $2 \ V$ emf. The potential gradient (potential difference per unit length) of the wire is: (in $V/m$)

The material of the wire of a potentiometer is

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively $360 \ cm$ and $420 \ cm$. If the emf of $A$ is $2.4 \ V$,the emf of the second cell $B$ is (in $V$)

$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $\Omega$)