For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

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 such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

$A$ potentiometer circuit shown in the figure is set up to measure the $e.m.f.$ of a cell $E$. As the point $P$ moves from $X$ to $Y$, the galvanometer $G$ shows deflection always in one direction, but the deflection decreases continuously until $Y$ is reached. In order to obtain a balance point between $X$ and $Y$, it is necessary to

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

In the arrangement shown in the figure,when the switch $S_2$ is open,the galvanometer shows no deflection for $l = L/2$. When the switch $S_2$ is closed,the galvanometer shows no deflection for $l = 5L/12$. The internal resistance $(r)$ of the $6\, V$ cell and the $emf$ $E$ of the other battery are respectively: