Balancing point of a potentiometer shifts from a length of $60 \ cm$ to $40 \ cm$ by shunting the cell with a $4 \ \Omega$ resistance. What is the internal resistance of the cell (in $\Omega$)?

$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.

$A$ circuit for the measurement of resistance by a potentiometer is shown. The galvanometer is first connected at point $A$ and zero deflection is observed at a length $PJ = 10 \ cm$. In the second case,it is connected at point $C$ and zero deflection is observed at a length of $30 \ cm$ from $P$. Then the unknown resistance $X$ is:

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

$AB$ is a potentiometer wire as shown in the figure. If the value of $R$ is increased,in which direction will the balance point $J$ shift?

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