The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

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:

Potentiometer measures the potential difference more accurately than a voltmeter because

Resistance of $100 \, cm$ long potentiometer wire is $10 \, \Omega$. It is connected to a battery of $2 \, V$ and a resistance $R$ in series. $A$ source of $10 \, mV$ gives a null point at $40 \, cm$ length. The value of external resistance $R$ is ........... $\Omega$.

$A$ potentiometer wire $PQ$ of $1\,m$ length is connected to a standard cell $E_{1}$. Another cell $E_{2}$ of emf $1.02\,V$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open,the null position is obtained at a distance of $49\,cm$ from $Q$. The potential gradient in the potentiometer wire is.......$V/cm$.

In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)