$A$ battery of $6\,V$ is connected to the terminals of a $3\,m$ long wire of uniform thickness and resistance of the order of $100\,\Omega$. The difference of potential between two points separated by $50\,cm$ on the wire will be .......... $V$.

Two cells of e.m.f.s $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When the potentiometer is connected between $A$ and $B$,the balancing length of the potentiometer wire is $3.60 \ m$. On connecting the potentiometer between $A$ and $C$,the balancing length is $0.90 \ m$. The ratio $E_1 / E_2$ is

When a cell of $E$.$M$.$F$. '$E_1$' is connected to a potentiometer wire, the balancing length is '$l_1$'. Another cell of $E$.$M$.$F$. '$E_2$' $(E_1 > E_2)$ is connected along with $E_1$ such that the two cells oppose each other, and the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is

Two cells of e.m.f. $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When a potentiometer is connected between points $A$ and $B$, the balancing length of the potentiometer wire is $412 \text{ cm}$. When the same potentiometer is connected between points $A$ and $C$, the balancing length is $103 \text{ cm}$. The ratio $E_1 : E_2$ is:

$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire 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 -