In a potentiometer,a wire of length $10 \ m$ having resistance $50 \ \Omega$ is used. $A$ battery of $5 \ V$ and a resistor of $450 \ \Omega$ are connected in series to the wire. If an unknown battery of emf $E$ balances the potentiometer at $450 \ cm$,then the value of $E$ is (in $V$)

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased by a few centimeters without changing the cell or the driving source, the balancing length will:

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:

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

$A$ $10 \, m$ long wire of resistance $20 \, \Omega$ is connected in series with a battery of e.m.f. $3 \, V$ (negligible internal resistance) and a resistance of $10 \, \Omega$. The potential gradient along the wire is (in $ \, V/m$)