In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

In an experiment to measure the internal resistance of a cell by a potentiometer,it is found that the balance point is at a length of $2\,m$ when the cell is shunted by a $5\,\Omega$ resistance; and is at a length of $3\,m$ when the cell is shunted by a $10\,\Omega$ resistance. The internal resistance of the cell is,then ................ $\Omega$.

If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then:

$A$ potentiometer is used to measure the potential difference between $A$ and $B$,and the null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. Find the ratio $\frac{E_{2}}{E_{1}}$,given that $E_{1} > E_{2}$.

The sliding contact of a potentiometer is in the middle of the potentiometer wire having a total resistance $R_p = 1 \Omega$,as shown in the figure. An external resistance of $R_e = 2 \Omega$ is connected via the sliding contact. Find the total current drawn from the $0.9 \text{ V}$ battery. (in $\text{ A}$)

In the given figure,a battery $E$ is balanced on $55 \, cm$ length of a potentiometer wire. When a resistance of $10 \, \Omega$ is connected in parallel with the battery,it balances on $50 \, cm$ length of the potentiometer wire. The internal resistance $r$ of the battery is ............. $\Omega$.