In a potentiometer experiment, when three cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are connected in series, the balancing length is found to be $740 \mathrm{cm}$. If $\mathrm{A}$ and $\mathrm{B}$ are connected in series, balancing length is $440 \mathrm{cm}$ and for $\mathrm{B}$ and $\mathrm{C}$ connected in series, it is $540 \mathrm{cm}$ The emf of $E_{\mathrm{A}}$ , $E_{\mathrm{B}}$ and $E_{\mathrm{C}}$ are respectively. (in volts)

Potentiometer wire of length $1 \mathrm{m}$ is connected in series with $490 \mathrm{\Omega }$ resistance and $2 \mathrm{V}$ battery. If $0.2 \mathrm{mV} {\mathrm{cm}}^{-1}$ is the potential gradient, then resistance of the potentiometer wire is

A potentiometer consists of a wire of length $4 \mathrm{m}$ and resistance $10 \mathrm{\Omega }$. It is connected to a cell of e.m.f. $2 \mathrm{V}$. The potential difference per unit length of the wire will be 

A $2 \mathrm{V}$ battery, a $15 \mathrm{\Omega }$ resistor and a potentiometer of $100 \mathrm{cm}$ length, all are connected in series. If the resistance of the potentiometer wire is $5 \mathrm{\Omega }$, then the potential gradient of the potentiometer wire is 

A potentiometer wire has length $4 \mathrm{m}$ and resistance $8 \mathrm{\Omega }$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2 \mathrm{V}$, to get a potential gradient $1 \mathrm{mV} {\mathrm{cm}}^{-1}$ on the wire is

A wire of length $100 \mathrm{cm}$ is connected to a cell of e.m.f. $2 \mathrm{V}$ and negligible internal resistance. The resistance of the wire is $3 \mathrm{\Omega }$. The additional resistance required to produce a potential drop of $1 \mathrm{millivolt} {\mathrm{cm}}^{-1}$ is 

A potentiometer wire of length $1 \mathrm{m}$ and resistance $10 \mathrm{\Omega }$ is connected in series with a cell of e.m.f. $2 \mathrm{V}$ with internal resistance $1 \mathrm{\Omega }$ and a resistance box including a resistance $R$. If potential difference between the ends of the wire is $1 \mathrm{mV}$, the value of $R$ is 

Figure shows a $10 \mathrm{V}$ potentiometer used for determination of internal resistance of a $1.5 \mathrm{V}$ cell. The balance point of the cell in the open circuit is $76.3 \mathrm{cm}$. When a resistor of $9.5 \mathrm{\Omega }$ is used in the external circuit of the cell, the balance point shifts to $64.8 \mathrm{cm}$ length of the potentiometer wire. The internal resistance of the cell is

Assertion: In the potentiometer circuit shown in figure $E_1$ and $E_2$ are the e.m.f. of batteries $B_1$ and $B_2$ respectively with $E_1>E_2$. Battery $B_1$ has negligible internal resistance. For a given value of resistor $R$, the balance length is $x$. If the diameter of the potentiometer wire $AB$ is increased, the balance length $x$ will decrease.

Reason: At the balance point, the potential difference between $AD$ due to battery $B_1=E_2$ the e.m.f. of battery $B_2$.