The given figure represents an arrangement of a potentiometer for the calculation of the internal resistance $(r)$ of an unknown battery $(E)$. The balance length is $70.0 \, cm$ with the key open and $60.0 \, cm$ with the key closed. $R$ is $132.40 \, \Omega$. The internal resistance $(r)$ of the unknown cell will be ....... $\Omega$ (Given $E_o > E$):-

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

$A$ 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$)

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

$A$ potentiometer wire has length $10\, m$ and resistance $10\, \Omega$. It is connected to a battery of $EMF$ $11\, V$ and internal resistance $1\, \Omega$. The potential gradient in the wire is ............... $V/m$.