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 $V$)

$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.

In a potentiometer experiment,a wire of length $10 \ m$ and resistance $5 \ \Omega$ is connected to a cell of emf $2.2 \ V$. If the potential difference between two points separated by a distance of $660 \ cm$ on the potentiometer wire is $1.1 \ V$,then the internal resistance of the cell is: (in $Omega$)

The area of cross-section of a potentiometer wire is $6 \times 10^{-7} \ m^2$. The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is $0.15 \ Vm^{-1}$. If the current through the potentiometer wire is $0.3 \ A$,then the resistivity of the material of the potentiometer wire is:

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because

In a potentiometer experiment,two cells of $e.m.f.$ $E_1$ and $E_2$ are used in series and in conjunction,and the balancing length is found to be $58 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $29 \ cm$. The ratio $\frac{E_1}{E_2}$ of the $e.m.f.$ of the two cells is: (in $:1$)