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

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

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 ........... $\Omega$.

$A$ potentiometer is connected between $A$ and $B$ and the balance point is obtained at $203.6 \, cm$. When the end of the potentiometer connected to $B$ is shifted to $C$,then the balance point is obtained at $24.6 \, cm$. If now the potentiometer be connected between $B$ and $C$,the balance point will be at ................. $cm$.

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$,then using the same cell,the null point is obtained at:

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is