$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

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 $\Omega$)

In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

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 $60 \, cm$. The internal resistance of the cell in ohms, is

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

$A$ wire of length $100\, cm$ is connected to a cell of $emf$ $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential drop of $1\, mV/cm$ is ............... $\Omega$.