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:

$A$ potentiometer wire of length $100\, cm$ has a resistance of $10\, \Omega$. It is connected in series with a resistance $R$ and an accumulator of emf $2\, V$ and of negligible internal resistance. $A$ source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance $R$?

$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $3 \text{ V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \, m$ when the cell is shunted by $3 \, \Omega$ resistance and at a length $1.5 \, m$ when the cell is shunted by $6 \, \Omega$ resistance. The internal resistance of the cell is:

The accurate measurement of $emf$ can be obtained using

The length of a potentiometer wire is $2.5 \ m$ and its resistance is $8 \ \Omega$. $A$ cell of negligible internal resistance and emf of $2.5 \ V$ is connected in series with a resistance of $242 \ \Omega$ in the primary circuit. The potential difference between two points separated by a distance of $20 \ cm$ on the potentiometer wire is (in $mV$)