$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

$A$ potentiometer is used to measure the potential difference between $A$ and $B$,and the null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. Find the ratio $\frac{E_{2}}{E_{1}}$,given that $E_{1} > E_{2}$.

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

$A$ $6\,V$ 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)

$A$ $10 \,m$ long potentiometer wire is connected to a battery having a steady voltage. $A$ Leclanche cell is balanced at $4 \,m$ length of the wire. If the length is kept the same,but its cross-section is doubled,the null point will be obtained at ........... $m$.

Explain the method to measure the internal resistance of a cell using a potentiometer.