The potentiometer wire is $5 \ m$ long and a potential difference of $4 \ V$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $200 \ cm$ of the potentiometer wire is: (in $V$)

In the experiment of calibration of a voltmeter,a standard cell of $e.m.f. = 1.1 \text{ V}$ is balanced against $440 \text{ cm}$ of a potentiometer wire. The potential difference across a resistance is found to balance against $220 \text{ cm}$ of the wire. The corresponding reading of the voltmeter is $0.5 \text{ V}$. The error in the reading of the voltmeter will be ................. $V$.

$A$ battery of internal resistance $1 \, \Omega$ and $emf$ $3 \, V$ sends a current through $1 \, m$ of uniform wire of resistance $5 \, \Omega$. The poles of a cell of $emf$ $1.4 \, V$ are connected to two points on the wire such that no current passes through this cell. The potential gradient of the wire is:

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

In the experiment to determine the internal resistance $(r)$ of a cell $(E_1)$ using a potentiometer,the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of the internal resistance $(r)$ of the cell is:

$A$ potentiometer wire of length $1 \ m$ is connected in series with $495 \ \Omega$ resistance and $2 \ V$ battery. If $0.2 \ mV/cm$ is the potential gradient,then the resistance of the potentiometer wire is: (in $\Omega$)