In the given figure,a battery $E$ is balanced on $55 \, cm$ length of a potentiometer wire. When a resistance of $10 \, \Omega$ is connected in parallel with the battery,it balances on $50 \, cm$ length of the potentiometer wire. The internal resistance $r$ of the battery is ............. $\Omega$.

$A$ battery of $emf$ $E_0 = 12\, V$ is connected across a $4\,m$ long uniform wire having resistance $4\,\Omega /m$. The cells of small $emfs$ $\varepsilon_1 = 2\,V$ and $\varepsilon_2 = 4\,V$ having internal resistance $2\,\Omega$ and $6\,\Omega$ respectively,are connected in parallel as shown in the figure. If the galvanometer shows no deflection at point $N$,the distance of point $N$ from point $A$ is equal to:

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

$A$ 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 $R$?

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

In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?