$A$ $2 \, V$ battery, a $990 \, \Omega$ resistor, and a potentiometer of $2 \, m$ length are connected in series. If the resistance of the potentiometer wire is $10 \, \Omega$, then the potential gradient of the potentiometer wire is: (in $V m^{-1}$)

What is a potentiometer? Explain the principle of a potentiometer.

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

$A$ potentiometer wire of length $10 \,m$ and resistance $20 \,\Omega$ is connected in series with a $25 \,V$ battery and an external resistance $30 \,\Omega$. $A$ cell of emf $E$ in the secondary circuit is balanced by a $250 \,cm$ long potentiometer wire. The value of $E$ (in volt) is $\frac{x}{10}$. The value of $x$ is.......

In the given circuit of a potentiometer,the potential difference $E$ across $AB$ ($10\, m$ length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed),the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now,the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$. The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$,where $a = \dots$ (Refer to the image for balancing lengths $l_{1}$ and $l_{2}$ from point $A$).

In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.