Two cells,when connected in series,are balanced on $8 \; m$ on a potentiometer. If the cells are connected such that the polarity of one of the cells is reversed,they balance on $2 \; m$. The ratio of the $e.m.f.$'s of the two cells is:

The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

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 $V$)?

$A$ 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 $\text{ V}$)

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)