$AB$ is a potentiometer wire of length $100\, cm$ and its resistance is $10\,\Omega$. It is connected in series with a resistance $R = 40\,\Omega$ and a battery of $e.m.f.$ $2\,V$ and negligible internal resistance. If a source of unknown $e.m.f.$ $E$ is balanced by $40\, cm$ length of the potentiometer wire,the value of $E$ is ................. $V$. (in $,V$)

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?

If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then:

In the potentiometer experiment shown in the figure,for the position $X$ of the jockey $J$,there occurs a null deflection in the galvanometer. Then the potential difference between points $A$ and $X$ is ................ $V$.

$A$ potentiometer wire of length $100 \ cm$ and resistance $3 \ \Omega$ is connected in series with a resistance of $8 \ \Omega$ and an accumulator of $4 \ V$ whose internal resistance is $1 \ \Omega$. $A$ cell of e.m.f. $E$ is balanced by $50 \ cm$ length of the wire. The e.m.f. of the cell is: (in $V$)

The length of a potentiometer wire is $1200 \; cm$ and it carries a current of $60 \; mA$. For a cell of $emf \; 5 \; V$ and internal resistance of $20 \; \Omega$,the null point on it is found to be at $1000 \; cm$. The resistance of the whole wire is .............. $\Omega$.