In a potentiometer experiment,when key $K_1$ is closed,the balance length is $100 \, cm$. What will be the balancing length when key $K_2$ is closed? ................ $cm$

$A$ potentiometer wire has a length of $4 \ m$ and a resistance of $10 \ \Omega$. It is connected to a cell of $2 \ V$ emf. The potential gradient (potential difference per unit length) of the wire is: (in $V/m$)

In an experiment with a potentiometer,$V_B = 10 \, V$ and the variable resistance is adjusted to $R = 50 \, \Omega$ (figure). $A$ student wanting to measure the voltage $E_1$ of a battery (approx. $8 \, V$) finds no null point. He then diminishes $R$ to $10 \, \Omega$ and is able to locate the null point on the last $(4^{th})$ segment of the potentiometer wire. Find the resistance of the potentiometer wire and the potential drop per unit length.

In a potentiometer experiment, the null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased without changing the cell or the driving source, the balancing length will:

In the figure,the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. The cell $C$'s $emf$ is $\frac{\varepsilon}{2}$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)