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:

$A$ potentiometer has a wire of length $4 \ m$ and resistance $10 \ \Omega$. The potentiometer is connected to a cell of $2 \ V$. The potential drop per unit length is ........... $V/m$.

$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.

In the figure,the potentiometer wire of length $l = 100 \, cm$ and resistance $R = 9 \, \Omega$ is joined to a cell of emf $E_1 = 10 \, V$ and internal resistance $r_1 = 1 \, \Omega$. Another cell of emf $E_2 = 5 \, V$ and internal resistance $r_2 = 2 \, \Omega$ is connected as shown. The galvanometer $G$ will show no deflection when the length $AC$ is ............... $cm$.

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

In an experiment to find the emf of a cell using a potentiometer, the length of the null point for a cell of emf $1.5 \text{ V}$ is found to be $60 \text{ cm}$. If this cell is replaced by another cell of emf $E$, the length of the null point increases by $40 \text{ cm}$. The value of $E$ is $x/10 \text{ V}$. The value of $x$ is