In the given arrangement,$E_1 = 5 \, V$ and $E_2 = 7 \, V$. The balancing length is $6 \, m$. If the terminals of $E_2$ are reversed,then the new balancing length will be:

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

$A$ potentiometer has a wire of length $4 \ m$ and resistance $10 \ \Omega$. The potentiometer is connected to a cell of $2 \ V$. The potential drop per unit length is ........... $V/m$.

In the arrangement shown in the figure,when the switch $S_2$ is open,the galvanometer shows no deflection for $l = L/2$. When the switch $S_2$ is closed,the galvanometer shows no deflection for $l = 5L/12$. The internal resistance $(r)$ of the $6\, V$ cell and the $emf$ $E$ of the other battery are respectively:

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)

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 $............$