$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

If the resistivity of a potentiometer wire is $\rho$ and the area of cross-section is $A$,then what will be the potential gradient along the wire,given that a current $I$ flows through it?

It is preferable to measure the $e.m.f.$ of a cell by a potentiometer rather than by a voltmeter because of the following possible reasons.
$(i)$ In the case of a potentiometer,no current flows through the cell.
$(ii)$ The length of the potentiometer wire allows for greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer,when using a potentiometer,is not relevant.
Which of these reasons are correct?

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is: