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

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

$A$ 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 $m$)

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ 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 ........... $\Omega$.

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.