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

$A$ galvanometer can be used as a voltmeter by connecting a

$A$ potentiometer wire of length $4 \, m$ and resistance $5 \, \Omega$ is connected in series with a resistance of $992 \, \Omega$ and a cell of e.m.f. $4 \, V$ with internal resistance $3 \, \Omega$. The length of $0.75 \, m$ on the potentiometer wire balances the e.m.f. of (in $ \, mV$)

$A$ potentiometer has a potential gradient of $2 \, mV/cm$. It is used to measure the potential difference across a $10 \, \Omega$ resistor. If the length of the potentiometer wire required to obtain the null point is $50 \, cm$,the current flowing through the $10 \, \Omega$ resistor is ............. $mA$.

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

Two cells of emf $E_1$ and $E_2$ $(E_1 > E_2)$ are connected individually to a potentiometer and their corresponding balancing lengths are $625 \, cm$ and $500 \, cm$. Then the ratio $\frac{E_1}{E_2}$ is ...........