To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $Omega$)

$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (Figure). $A$ voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

$A$ potentiometer has a wire of length $4 \ m$ and resistance $10 \ \Omega$. The potentiometer is connected to a cell of $2 \ V$. The potential drop per unit length is ........... $V/m$.

$A$ battery of emf $10 \, V$ is connected to a uniform wire $AB$ of $1 \, m$ length and having a resistance of $10 \, \Omega$ in series with a $10 \, \Omega$ resistor as shown in the figure. Two cells of emf $2 \, V$ each, having internal resistance $2 \, \Omega$ each, are connected in parallel as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is. (in $ \, cm$)

$A$ wire of length $100\, cm$ is connected to a cell of $emf$ $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential drop of $1\, mV/cm$ is ............... $\Omega$.

Potentiometer measures the potential difference more accurately than a voltmeter because