In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

In a potentiometer,a balance point is obtained when:

Define potential gradient and write its $SI$ unit.

The length of a potentiometer wire is $2.5 \ m$ and its resistance is $8 \ \Omega$. $A$ cell of negligible internal resistance and emf of $2.5 \ V$ is connected in series with a resistance of $242 \ \Omega$ in the primary circuit. The potential difference between two points separated by a distance of $20 \ cm$ on the potentiometer wire is (in $mV$)

In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)