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

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is

$A$ 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 a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)