When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

$A$ battery of emf $10 \, V$ is connected to a uniform wire $AB$ of $1 \, m$ length and having a resistance of $10 \, \Omega$ in series with a $10 \, \Omega$ resistor as shown in the figure. Two cells of emf $2 \, V$ each, having internal resistance $2 \, \Omega$ each, are connected in parallel as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is. (in $ \, cm$)

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

Balancing point of a potentiometer shifts from a length of $60 \ cm$ to $40 \ cm$ by shunting the cell with a $4 \ \Omega$ resistance. What is the internal resistance of the cell (in $\Omega$)?

For a cell,the balancing length is $0.60 \, m$. For another cell having an electromotive force $(emf)$ $0.1 \, V$ less than the first one,the balancing length is $0.55 \, m$. What are the $emf$ values of the two cells?

In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)