The length of a potentiometer wire is $1200 \; cm$ and it carries a current of $60 \; mA$. For a cell of $emf \; 5 \; V$ and internal resistance of $20 \; \Omega$,the null point on it is found to be at $1000 \; cm$. The resistance of the whole wire is .............. $\Omega$.

$A$ potentiometer has a uniform potential gradient. The specific resistance (resistivity) of the material of the potentiometer wire is $10^{-7} \, \Omega \cdot m$, the current passing through it is $0.1 \, A$, and the cross-sectional area of the wire is $10^{-6} \, m^2$. The potential gradient along the potentiometer wire is:

$A$ potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

$A$ wire,$10 \ m$ long,has a resistance of $40 \ \Omega$. It is connected in series with a resistance box of resistance $R$ and a $2 \ V$ storage cell. If the potential gradient along the wire is $0.1 \ mV/cm$,then the value of $R$ is (in $\Omega$)

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

$A$ 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: