In the arrangement shown in the figure,when the switch $S_2$ is open,the galvanometer shows no deflection for $l = L/2$. When the switch $S_2$ is closed,the galvanometer shows no deflection for $l = 5L/12$. The internal resistance $(r)$ of the $6\, V$ cell and the $emf$ $E$ of the other battery are respectively:

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

$A$ 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$)

In the given arrangement,$E_1 = 5 \, V$ and $E_2 = 7 \, V$. The balancing length is $6 \, m$. If the terminals of $E_2$ are reversed,then the new balancing length will be:

$A$ battery of $emf$ $E_0 = 12\, V$ is connected across a $4\,m$ long uniform wire having resistance $4\,\Omega /m$. The cells of small $emfs$ $\varepsilon_1 = 2\,V$ and $\varepsilon_2 = 4\,V$ having internal resistance $2\,\Omega$ and $6\,\Omega$ respectively,are connected in parallel as shown in the figure. If the galvanometer shows no deflection at point $N$,the distance of point $N$ from point $A$ is equal to: