$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

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 $:1$)

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

$A$ battery of emf $10 \, V$ is connected to a uniform wire $AB$ of $1 \, m$ length and having a resistance of $10 \, \Omega$ in series with a $10 \, \Omega$ resistor as shown in the figure. Two cells of emf $2 \, V$ each, having internal resistance $2 \, \Omega$ each, are connected in parallel as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is. (in $ \, cm$)

$A$ potentiometer wire of length $1\,m$ and resistance $10\,\Omega$ is connected in series with a cell of $emf$ $2\,V$ with internal resistance $1\,\Omega$ and a resistance box including a resistance $R$. If the potential difference between the ends of the wire is $1\,mV$,the value of $R$ is ............. $\Omega$.

$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.