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 of e.m.f.s $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When the potentiometer is connected between $A$ and $B$,the balancing length of the potentiometer wire is $3.60 \ m$. On connecting the potentiometer between $A$ and $C$,the balancing length is $0.90 \ m$. The ratio $E_1 / E_2$ is

$A$ $10\,m$ long potentiometer wire has a potential gradient of $0.0025\,V/cm$. Calculate the distance of the null point when the wire is connected to a $1.025\,V$ standard cell.

In the given circuit of a potentiometer,the potential difference $E$ across $AB$ ($10\, m$ length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed),the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now,the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$. The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$,where $a = \dots$ (Refer to the image for balancing lengths $l_{1}$ and $l_{2}$ from point $A$).

Sensitivity of a given potentiometer can be decreased by

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because