The length of a potentiometer wire is $1200 \; cm$ and it carries a current of $60 \; mA$. For a cell of $emf \; 5 \; V$ and internal resistance of $20 \; \Omega$,the null point on it is found to be at $1000 \; cm$. The resistance of the whole wire is .............. $\Omega$.

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 $V$?

$A$ cell in a secondary circuit gives a null deflection for $2.5 \ m$ length of a potentiometer wire having a total length of $10 \ m$. If the length of the potentiometer wire is increased by $1 \ m$ without changing the cell in the primary circuit,the new position of the null point is: (in $m$)

With the help of a potentiometer,we can determine the value of the emf of a given cell. The sensitivity of the potentiometer is:
$(A)$ directly proportional to the length of the potentiometer wire
$(B)$ directly proportional to the potential gradient of the wire
$(C)$ inversely proportional to the potential gradient of the wire
$(D)$ inversely proportional to the length of the potentiometer wire
Choose the correct option for the above statements:

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

Let $A$ be the cross-sectional area and $\rho$ be the specific resistance (resistivity) of a potentiometer wire. If $I$ is the current passing through the wire,then the potential gradient along the length of the wire is