In a potentiometer,a wire of length $10 \ m$ having resistance $50 \ \Omega$ is used. $A$ battery of $5 \ V$ and a resistor of $450 \ \Omega$ are connected in series to the wire. If an unknown battery of emf $E$ balances the potentiometer at $450 \ cm$,then the value of $E$ is (in $V$)

In a potentiometer circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

$A$ 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:

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

$A$ 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 ............... $\Omega$.

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 $\Omega$)