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 $\Omega$)

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 -

In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

$A$ $10 \, m$ long wire of resistance $20 \, \Omega$ is connected in series with a battery of e.m.f. $3 \, V$ (negligible internal resistance) and a resistance of $10 \, \Omega$. The potential gradient along the wire is (in $ \, V/m$)

Define potential gradient and write its $SI$ unit.

$A$ potentiometer wire $PQ$ of $1\,m$ length is connected to a standard cell $E_{1}$. Another cell $E_{2}$ of emf $1.02\,V$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open,the null position is obtained at a distance of $49\,cm$ from $Q$. The potential gradient in the potentiometer wire is.......$V/cm$.