$A$ $2 \, V$ battery, a $990 \, \Omega$ resistor, and a potentiometer of $2 \, m$ length are connected in series. If the resistance of the potentiometer wire is $10 \, \Omega$, then the potential gradient of the potentiometer wire is: (in $V m^{-1}$)

$A$ $6\,V$ battery is connected to the terminals of a $3\,m$ long uniform wire having resistance $100\,\Omega$. The difference in potential between two points on the wire separated by a distance of $50\,cm$ will be ............. $V$.

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:

The length of a potentiometer wire is $l$. $A$ cell of $emf$ $E$ is balanced at a length $l/3$ from the positive end of the wire. If the length of the wire is increased by $l/2$,at what distance will the same cell give a balance point?

In a potentiometer arrangement,a cell of $emf$ $1.25\; V$ gives a balance point at $35.0\; cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\; cm ,$ what is the $emf$ of the second cell in $V$?

If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then: