For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because:

$A$ 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:

In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is

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 ............... $cm$.

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