$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?

Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625 \, cm$. If the terminals of $E_1$ are reversed,the balancing length obtained is $125 \, cm$. Given $E_2 > E_1$,the ratio $E_1 : E_2$ will be

The length of a potentiometer wire is $l$. $A$ cell of emf $E$ is balanced at a length $\left(\frac{l}{3}\right)$ from the positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$,the distance at which the same cell gives the balancing point is (The cell in the primary circuit is ideal and no series resistance is present in the primary circuit.)

$A$ potentiometer wire has length $10\, m$ and resistance $10\, \Omega$. It is connected to a battery of $EMF$ $11\, V$ and internal resistance $1\, \Omega$. The potential gradient in the wire is ............... $V/m$.

$A$ potentiometer wire of length $100\, cm$ has a resistance of $10\, \Omega$. It is connected in series with a resistance $R$ and an accumulator of emf $2\, V$ and of negligible internal resistance. $A$ source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance $R$?

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ and the balancing length is $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is: