$A$ potentiometer wire of length $10 \,m$ and resistance $20 \,\Omega$ is connected in series with a $25 \,V$ battery and an external resistance $30 \,\Omega$. $A$ cell of emf $E$ in the secondary circuit is balanced by a $250 \,cm$ long potentiometer wire. The value of $E$ (in volt) is $\frac{x}{10}$. The value of $x$ is.......

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased by a few centimeters without changing the cell or the driving source, the balancing length will:

$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (Figure). $A$ voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

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:

In the figure,the potentiometer wire of length $l = 100 \, cm$ and resistance $R = 9 \, \Omega$ is joined to a cell of emf $E_1 = 10 \, V$ and internal resistance $r_1 = 1 \, \Omega$. Another cell of emf $E_2 = 5 \, V$ and internal resistance $r_2 = 2 \, \Omega$ is connected as shown. The galvanometer $G$ will show no deflection when the length $AC$ is ............... $cm$.

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is: