$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs 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:

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

In an experiment to find the $emf$ of a cell using a potentiometer,the length of the null point for a cell of $emf$ $1.5 \ V$ is found to be $60 \ cm$. If this cell is replaced by another cell of $emf$ $E$,the length of the null point increases by $40 \ cm$. The value of $E$ is $\frac{x}{10} \ V$. The value of $x$ is $............$

In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.