$A$ potentiometer is an accurate and versatile device to make electrical measurements of $EMF$ because the method involves:

In the given figure,the potentiometer wire $AB$ has a resistance of $5 \, \Omega$ and a length of $10 \, m$. The balancing length $AM$ for the $emf$ of $0.4 \, V$ is ............... $m$.

Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625 \, cm$. If the terminals of $E_1$ are reversed,the balancing length obtained is $125 \, cm$. Given $E_2 > E_1$,the ratio $E_1 : E_2$ will be

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:

Resistance of $100 \, cm$ long potentiometer wire is $10 \, \Omega$. It is connected to a battery of $2 \, V$ and a resistance $R$ in series. $A$ source of $10 \, mV$ gives a null point at $40 \, cm$ length. The value of external resistance $R$ 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 $V$?