Potentiometer

Balancing point of a potentiometer shifts from a length of $60 \mathrm{cm}$ to $40 \mathrm{cm}$ by shunting the cell with a $4 \Omega$ resistance. What is the internal resistance of the cell?

A cell of $2\mathrm{V},\mathrm{ }1\mathrm{\Omega }$ is balanced at $1.9\mathrm{ }\mathrm{m}$. Then what is balanced length for ideal cell of $2\mathrm{V}$ ?

 In the following figure, the $p.d.$ between the points $M$ and $N$ is balanced against $50 \mathrm{cm}$ length of potentiometer wire of total length $100 \text{cm}$. In order to balance the potential difference between the points $N$ and $C$ the jockey to be pressed on potentiometer wire at a distance of $\left(\text{in cm}\right)$:

The potential gradient of potentiometer is $0.2 \mathrm{volt} {\mathrm{m}}^{-1}$. $A$ current of $0.1 \mathrm{amp}$ is flowing through $a$ coil of $2 \mathrm{Ohm}$ resistance. The balancing length in meters for the$\mathrm{p}.\mathrm{d}.$ at the ends of this coil will be

The length of a wire of a potentiometer is $100 \mathrm{cm},$ and the emf of its standard cell is $E \mathrm{volt}$. It is employed to measure the emf of a battery whose internal resistance is $0.5 \mathrm{Ohm}$. If the balance point is

Resistivity of potentiometer wire is ${10}^{-7}\text{Ωm}$ and its area of cross-section is ${10}^{-6}{ \mathrm{m}}^2.$ When a current $i=0.1 \mathrm{A}$ flows through the wire, its potential gradient is :

When a cell is balanced on potentiometer wire, then balancing length is $125 \mathrm{cm}.$ If resistance of $2 \mathrm{ohm}$ is connected across the ends of cell, then balancing length is $100 \mathrm{cm},$ then internal resistance of cell is

A cell is balanced at $100 \mathrm{cm}$ of a potentiometer wire when the total length of the wire is $400 \mathrm{cm}.$ If the length of the potentiometer wire is increased by $100 \mathrm{cm},$ then the new balancing length for the cell will be (Assume pd across potentiometer wire is constant)

Figure shows a $2.0 \mathrm{V}$ potentiometer used for the determination of internal resistance of a $1.5 \mathrm{V}$ cell. The balance point of the cell in 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. Fine the internal resistance of the cell (in $\mathrm{\Omega }$) correct to the nearest integer.

One of the circuits for the measurement of resistance by potentiometer is shown. The galvanometer is connected at point A and zero deflection is observed at length $PJ=30 \mathrm{cm}.$ In second case the secondary cell is changed and zero deflection is observed at length $PJ=10 \mathrm{cm}.$ What is the resistance $R$ (in ohm) ? Take $E_{\mathit{S}}=10 \mathrm{V}$ and $r=1 \mathrm{\Omega }$ in $1^{\mathrm{st}}$ reading and $E_{\mathit{S}}=5 \mathrm{V}$ and $r=2 \mathrm{\Omega }$ in $2^{\mathrm{nd}}$ reading :

A cell of internal resistance of $1 \mathrm{\Omega }$ and emf $5 \mathrm{V}$ balances on a potentiometer wire at length of$2 \mathrm{m}$. The driving battery has emf of $50 \mathrm{V}$. If the cell is shunted by $1 \mathrm{\Omega }$ resistance wire then find by what amount balance point will shift?

The figure shows a potentiometer arrangement used to measure the internal resistance of the cell B. The null point $Q$ is located at $20 \mathrm{cm}$ from point $P$ on a $100 \mathrm{cm}$ wire PR. Find the internal resistance of the cell $B.$

In a potentiometer experiment when terminals of the cell are connected at distance of $52 \mathrm{cm}$ on the wire, then no current flows through it. When $5 \mathrm{\Omega }$ shunt resistance is connected across the cell the balancing length is $40 \mathrm{cm}.$ The internal resistance of the cell (in $\mathrm{\Omega }$) is $\frac{n}{2}$ what is the value of $n$?

A wire $AB$ (of length $1 \mathrm{m}$, area of cross-section $\pi {\mathrm{m}}^2$) is used in the potentiometer experiment to calculate the emf and internal resistance $\left(r\right)$ of the battery. The emf and internal resistance of the driving battery are $15 \mathrm{V}$ and $3 \mathrm{\Omega }$ respectively. The resistivity of wire $AB$ varies as $\rho ={\rho }_{0}x,$ where $x$ is distance from $A$ in meters and ${\rho }_0=24\pi \mathrm{\Omega }$. The distance of null point from $A$ is obtained at $\sqrt{\frac{2}{3}} \mathrm{m}$ when switch $S$ is open. Find the value of $E$.

A $10 \mathrm{m}$ long potentiometer wire has a resistance of $20 \mathrm{ohm}$. It is connected in series with a battery of emf $3 \mathrm{V}$ and a resistance of $10 \mathrm{\Omega }$. The internal resistance of cell is negligible. If the length can be read accurately up to $1 \mathrm{mm}$, the potentiometer can read voltage with an accuracy of $\frac{k}{10}$ (in millivolt). Then the value of $k$ is

A wire $AB$ (of length $1 \mathrm{m},$ area of cross-section $\pi {\mathrm{m}}^2$) is used in potentiometer experiment to calculate emf and internal resistance $\left(r\right)$ of battery. The emf and internal resistance of driving battery are $15 \mathrm{V}$ and $3 \mathrm{\Omega }$, respectively. The resistivity of wire $AB$ varies as $\rho ={\rho }_{0}x,$ where $x$ is distance from $A$ in meters and ${\rho }_0=24\pi \mathrm{\Omega }-\mathrm{m}.$ The distance of null point from $A$ is obtained at $\sqrt{\frac{2}{3}} \mathrm{m}$ when switch $S$ is open. Find the value of $E.$

In a potentiometer, find the value of EMF of a cell in $\mathrm{V}$ if the balancing length for this cell is $65 \mathrm{cm}$. Given that a cell of emf $1.5\mathrm{ }\mathrm{V}$ gave a balanced point at $32\mathrm{ }\mathrm{c}\mathrm{m}$ length of the wire.

If the specific resistance of a potentiometer wire is ${10}^{-7} \mathrm{\Omega } \mathrm{m}$, the current flowing through it is $0.1 \mathrm{A}$ and the cross-sectional area of the wire is ${10}^{-6} {\mathrm{m}}^2$ then, the potential gradient will be

A standard cell of $1.08 \mathrm{V}$ is balanced by the PD across $90\mathrm{ }\mathrm{c}\mathrm{m}$ of a meter long wire supplied by a cell of emf $2 \mathrm{V}$ through a series resistor of resistance $2\mathrm{ }\mathrm{\Omega }$. If the internal resistance of the cell in the primary circuit is zero then the resistance per unit length of potentiometer wire is

The balancing length for a cell is $560 \mathrm{cm}$ in a potentiometer experiment. When an external resistance of $10 \mathrm{\Omega }$ is connected in parallel to the cell, the balancing length changes by $60 \mathrm{cm}.$ If the internal resistance of the ceil is $\frac{\mathrm{N}}{10}\Omega ,$ where $N$ is an integer then value of $N$ is ............