In a potentiometer arrangement,a cell gives a balancing point at $75 \, cm$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3:2$,the difference in the balancing length of the potentiometer wire in above two cases will be ......... $cm$.

In an experiment of a potentiometer for measuring the internal resistance of a primary cell,a balancing length $\ell$ is obtained on the potentiometer wire when the cell is in an open circuit. Now,the cell is short-circuited by a resistance $R$. If $R$ is equal to the internal resistance of the cell,the balancing length on the potentiometer wire will be:

What is a potentiometer? Explain the principle of a potentiometer.

Explain the method to measure the internal resistance of a cell using a potentiometer.

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)

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 $A$).