$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $\Omega$)

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 $A$).

In a potentiometer wire experiment, the $emf$ of a battery in the primary circuit is $20\,V$ and its internal resistance is $5\,\Omega$. There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from $120\,\Omega$ to $170\,\Omega$. The resistance of the potentiometer wire is $75\,\Omega$. Which of the following potential differences can be measured using this potentiometer?

$A$ potentiometer wire has length $L$. For a given cell of emf $E$,the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of the potentiometer wire is increased by $50 \%$,then for the same cell,the balance point is obtained at length

$A$ wire,$10 \ m$ long,has a resistance of $40 \ \Omega$. It is connected in series with a resistance box of resistance $R$ and a $2 \ V$ storage cell. If the potential gradient along the wire is $0.1 \ mV/cm$,then the value of $R$ is (in $\Omega$)

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?