In a potentiometer experiment,a cell is balanced at a length of $240 \ cm$. When the cell is shunted with a resistance of $2 \ \Omega$,it is balanced at a length of $120 \ cm$. The internal resistance of the cell is .......... $\Omega$.

$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

$A$ potentiometer wire,$10 \, m$ long,has a resistance of $40 \, \Omega$. It is connected in series with a resistance box and a $2 \, V$ storage cell. If the potential gradient along the wire is $0.1 \, mV/cm$,the resistance unplugged in the box is .............. $\Omega$.

Two cells having unknown e.m.f.s $E_{1}$ and $E_{2}$ $(E_{1} > E_{2})$ are connected in a potentiometer circuit so as to assist each other. The null point is obtained at $490 \ cm$ from the higher potential end. When cell $E_{2}$ is connected so as to oppose cell $E_{1}$,the null point is obtained at $90 \ cm$ from the same end. The ratio of the e.m.f.s of the two cells $(\frac{E_{1}}{E_{2}})$ is:

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?