$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is

$A$ 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:

In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

$A$ null point is found at $200\,cm$ in a potentiometer when the cell in the secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting,the null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.

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?

In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is