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

In the circuit shown, a four-wire potentiometer is made of a $400\, cm$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r = 0.01\, \Omega /cm$. If an ideal voltmeter is connected as shown with jockey $J$ at $50\, cm$ from end $A$, the expected reading of the voltmeter will be: ............... $V$

The figure shows a simple potentiometer circuit for measuring a small $e.m.f.$ produced by a thermocouple. The potentiometer wire $PQ$ has a resistance of $5 \,\Omega$ and the driver cell has an $e.m.f.$ of $2 \, V$. If a balance point is obtained at $0.600 \, m$ along $PQ$ when measuring an $e.m.f.$ of $6.00 \, mV$,what is the value of resistance $R$ in $\Omega$?

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,the same cell will give a balance point at distance $x$. The value of $x$ is

The figure shows a potentiometer with a cell of $2.0 \; V$ and internal resistance $0.40 \; \Omega$ maintaining a potential drop across the resistor wire $AB$. $A$ standard cell which maintains a constant $emf$ of $1.02 \; V$ (for very moderate currents up to a few $mA$) gives a balance point at $67.3 \; cm$ length of the wire. To ensure very low currents are drawn from the standard cell,a very high resistance of $600 \; k \Omega$ is put in series with it,which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown $emf$ $\varepsilon$ and the balance point found similarly,turns out to be at $82.3 \; cm$ length of the wire.
$(a)$ What is the value of $\varepsilon ?$
$(b)$ What purpose does the high resistance of $600 \; k \Omega$ have?
$(c)$ Is the balance point affected by this high resistance?
$(d)$ Would the method work in the above situation if the driver cell of the potentiometer had an $emf$ of $1.0 \; V$ instead of $2.0 \; V ?$
$(e)$ Would the circuit work well for determining an extremely small $emf$,say of the order of a few $mV$ (such as the typical $emf$ of a thermocouple)? If not,how will you modify the circuit?

$A$ potentiometer has a uniform potential gradient. The specific resistance (resistivity) of the material of the potentiometer wire is $10^{-7} \, \Omega \cdot m$, the current passing through it is $0.1 \, A$, and the cross-sectional area of the wire is $10^{-6} \, m^2$. The potential gradient along the potentiometer wire is: