The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire,then

$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $Omega$)

In a potentiometer experiment,when three cells $A, B$ and $C$ are connected in series,the balancing length is found to be $420 \ cm$. If cells $A$ and $B$ are connected in series,the balancing length is $220 \ cm$ and for cells $B$ and $C$ connected in series,the balancing length is $320 \ cm$. The emf of cells $A, B$ and $C$ are respectively in the ratio of:

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is

The area of cross-section of a potentiometer wire is $6 \times 10^{-7} \ m^2$. The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is $0.15 \ Vm^{-1}$. If the current through the potentiometer wire is $0.3 \ A$,then the resistivity of the material of the potentiometer wire is:

$A$ $2\,V$ battery,a $15\,\Omega$ resistor,and a potentiometer of $100\,cm$ length are all connected in series. If the resistance of the potentiometer wire is $5\,\Omega$,then the potential gradient of the potentiometer wire is ............... $V/cm$.