$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

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 $mV$)

Let $A$ be the cross-sectional area and $\rho$ be the specific resistance (resistivity) of a potentiometer wire. If $I$ is the current passing through the wire,then the potential gradient along the length of the wire is

$A$ potentiometer is an accurate and versatile device to make electrical measurements of $EMF$ because the method involves:

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.