Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625 \, cm$. If the terminals of $E_1$ are reversed,the balancing length obtained is $125 \, cm$. Given $E_2 > E_1$,the ratio $E_1 : E_2$ will be

As shown in the figure,a potentiometer wire of resistance $20\,\Omega$ and length $300\,cm$ is connected with a resistance box ($R$.$B$.) and a standard cell of emf $4\,V$. For a resistance '$R$' of the resistance box introduced into the circuit,the null point for a cell of $20\,mV$ is found to be $60\,cm$. The value of '$R$' is $.....\Omega$

In the given figure,there is a circuit of a potentiometer of length $AB = 10 \, m$. The resistance per unit length is $0.1 \, \Omega/cm$. $A$ battery of $6 \, V$ and an internal resistance of $20 \, \Omega$ is connected across $AB$. The maximum value of emf that can be measured by this potentiometer is (in $V$):

$A$ potentiometer wire has length $10\, m$ and resistance $20\,\Omega$. $A$ $2.5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega$ series resistance. The potential gradient on the wire will be

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 ................. $V$.

$A$ potentiometer wire has length $L$. For a given cell of emf $E$,the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of the potentiometer wire is increased by $50 \%$,then for the same cell,the balance point is obtained at length