The potential difference across the ends of a potentiometer wire is constant. Two cells are connected in such a way that they first assist each other and then oppose each other. They are balanced on the potentiometer wire at lengths of $120 \ cm$ and $60 \ cm$ respectively. The ratio of the electromotive forces $(EMF)$ of the cells is:

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$,then using the same cell,the null point is obtained at:

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.