The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (Figure). $A$ voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

$A$ potentiometer wire of length $4 \text{ m}$ and resistance $5 \Omega$ is connected in series with a resistance of $992 \Omega$ and a cell of e.m.f. $4 \text{ V}$ with internal resistance $3 \Omega$. The length of $0.75 \text{ m}$ on the potentiometer wire balances the e.m.f. of: (in $\text{ mV}$)

In the experiment of calibration of a voltmeter,a standard cell of $e.m.f. = 1.1 \text{ V}$ is balanced against $440 \text{ cm}$ of a potentiometer wire. The potential difference across a resistance is found to balance against $220 \text{ cm}$ of the wire. The corresponding reading of the voltmeter is $0.5 \text{ V}$. The error in the reading of the voltmeter will be ................. $V$.