$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

Resistance of a potentiometer wire is $2 \ \Omega/m$. $A$ cell of e.m.f. $1.5 \ V$ balances at $300 \ cm$. The current through the wire is (in $mA$)

$A$ $6\,V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $100\,cm$. The positive terminal of another battery of $emf$ $4\,V$ and internal resistance $1\,\Omega$ is joined to the point $A$ as shown in the figure. Take the potential at $B$ to be zero. At which point $D$ of the wire $AB$,measured from $A$,is the potential equal to the potential at $C$? ...................... $cm$ (approximately)

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

$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

$A$ $6 \, V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $1 \, m$. The positive terminal of another battery of emf $4 \, V$ and internal resistance $1 \, \Omega$ is joined to point $A$ as shown in the figure. The ammeter shows zero deflection when the jockey touches the wire at point $C$. The length $AC$ is equal to: