The potentiometer wire is $5 \ m$ long and a potential difference of $4 \ V$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $200 \ cm$ of the potentiometer wire is: (in $V$)

In the given figure,there is a circuit of a potentiometer of length $AB = 10 \, m$. The resistance per unit length is $0.1 \, \Omega/cm$. $A$ battery of $6 \, V$ and an internal resistance of $20 \, \Omega$ is connected across $AB$. The maximum value of emf that can be measured by this potentiometer is (in $V$):

The accurate measurement of $emf$ can be obtained using

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?