In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (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 ................ $\Omega$.

$A$ 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 $E_{1} > E_{2}$.

$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

In an experiment of a potentiometer for measuring the internal resistance of a primary cell,a balancing length $\ell$ is obtained on the potentiometer wire when the cell is in an open circuit. Now,the cell is short-circuited by a resistance $R$. If $R$ is equal to the internal resistance of the cell,the balancing length on the potentiometer wire will be:

The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be