In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)

$A$ 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

In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)

In a potentiometer experiment,a wire of length $10 \ m$ and resistance $5 \ \Omega$ is connected to a cell of emf $2.2 \ V$. If the potential difference between two points separated by a distance of $660 \ cm$ on the potentiometer wire is $1.1 \ V$,then the internal resistance of the cell is: (in $\Omega$)

$A$ 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 $V/m$)

In the potentiometer experiment shown in the figure,for the position $X$ of the jockey $J$,there occurs a null deflection in the galvanometer. Then the potential difference between points $A$ and $X$ is ................ $V$.