$A$ potentiometer wire has length $10\, m$ and resistance $20\,\Omega$. $A$ $2.5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega$ series resistance. The potential gradient on the wire will be

$A$ circuit for the measurement of resistance by a potentiometer is shown. The galvanometer is first connected at point $A$ and zero deflection is observed at a length $PJ = 10 \ cm$. In the second case,it is connected at point $C$ and zero deflection is observed at a length of $30 \ cm$ from $P$. Then the unknown resistance $X$ is:

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 . . . . . . $\Omega$.

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 $\Omega$)?

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

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 $60 \; cm$. If the internal resistance of the cell is $\frac{N}{10} \; \Omega$,where $N$ is an integer,then the value of $N$ is: