In the following circuit, a $10 \, m$ long potentiometer wire with resistance $1.2 \, \Omega/m$, a resistance $R_1$, and an accumulator of $emf$ $2 \, V$ are connected in series. When the $emf$ of the thermocouple is $2.4 \, mV$, the deflection in the galvanometer is zero at a balancing length of $5 \, m$. The current supplied by the accumulator is:

$A$ galvanometer can be used as a voltmeter by connecting a

$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

In the given figure,there is a circuit of a potentiometer of length $AB = 10 \, m$. The resistance per unit length is $0.1 \, \Omega/cm$. $A$ battery of $6 \, V$ and an internal resistance of $20 \, \Omega$ is connected across $AB$. The maximum value of emf that can be measured by this potentiometer is (in $V$):

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?