In a potentiometer circuit, there is a cell of $e.m.f.$ $2\, V$, a resistance of $5\, \Omega$ and a wire of uniform thickness of length $1000\, cm$ and resistance $15\, \Omega$. The potential gradient in the wire is:

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?

The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to:

$A$ potentiometer wire of length $100\, cm$ has a resistance of $10\, \Omega$. It is connected in series with a resistance $R$ and an accumulator of emf $2\, V$ and of negligible internal resistance. $A$ source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance $R$?

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 $V$)