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

$A$ potentiometer wire has a length of $100 \ cm$ and is connected to a cell of $emf \ E$. It is used to measure the $emf \ E_0$ of a battery with an internal resistance of $0.5 \ \Omega$. If the balance point is obtained at a distance of $\ell = 30 \ cm$ from the positive terminal,the $emf \ E_0$ of the battery is:

$A$ thermocouple of negligible resistance produces an $e.m.f.$ of $40\,\mu V/^{\circ}C$ in the linear range of temperature. $A$ galvanometer of resistance $10\,\Omega$ whose sensitivity is $1\,\mu A/\text{div}$ is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be (in $^{\circ}C$)

$A$ $10\,m$ long potentiometer wire has a potential gradient of $0.0025\,V/cm$. Calculate the distance of the null point when the wire is connected to a $1.025\,V$ standard cell.

Explain the method to measure the internal resistance of a cell using a potentiometer.

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 $\Omega$)?