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 $\Omega$)

$A$ wire,$10 \ m$ long,has a resistance of $40 \ \Omega$. It is connected in series with a resistance box of resistance $R$ and a $2 \ V$ storage cell. If the potential gradient along the wire is $0.1 \ mV/cm$,then the value of $R$ is (in $\Omega$)

$A$ 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 potentiometer experiments,two 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 $E_1 / E_2$ is

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 $mV$)

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)