In a potentiometer circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

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?

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

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

In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)