Potentiometer measures the potential difference more accurately than a voltmeter because

In the given arrangement,$E_1 = 5 \, V$ and $E_2 = 7 \, V$. The balancing length is $6 \, m$. If the terminals of $E_2$ are reversed,then the new balancing length will be:

$A$ wire of length $10 \ m$ and resistance $30 \ \Omega$ is connected to a battery of $emf$ $2.5 \ V$ and internal resistance $5 \ \Omega$ through an external resistance $R$. If the potential gradient along the wire is $50 \ \mu V/mm$,then $R = $ ................. $\Omega$.

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

$A$ potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. $A$ constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is ... $\Omega$.

In the given potentiometer circuit,the balancing length for points $B$ and $C$ is $40 \, cm$. What is the balancing length for points $C$ and $D$ in $cm$?