Potentiometer

The potential gradient along the length of 1 metre wire of uniform cross section is $6$ V/m. Let '$B$' and '$C$' be the two points at $40$ cm and $60$ cm marks respectively from one end. The potential difference between points B and C is

A current of $0.01$ A passes through the potentiometer wire of resistivity $10^{6}\ \Omega$ cm and area of cross- section $10^{-2}$ cm. The potential gradient is (in V/m)

A potentiometer wire is 4m long and potential difference of 3V is maintained between the ends. The e.m.f. of the cell which balances against a length of 100 cm of the potentiometer wire is

A potentiometer wire is 4m long and potential difference of 2V is maintained between the ends. The e.m.f. of the cell which balances against a length of 80 cm of the potentiometer wire is

A potentiometer wire is 4 m long and potential difference of 3V is maintained between the ends. The e.m.f. of the cell which balances against a length of 100 cm of the poteniometer wire is

In a potentiometer circuit when two cells of emf $1.6$ V and $1.3$ V are connected to assist each other, the balancing length is $290$ cm. What will be the balancing length in cm when these two cells are connected in opposition?

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

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5$ V is found to be $60$ cm. If this cell is replaced by another cell of emf $E$, the length of null point increases by $40$ cm. The value of $E$ is $x/10$ V.

The value of $x$ is

In a potentiometer, the null point is obtained on $7^{th}$ wire. What should we do to change the null point on $9^{th}$ wire by keeping all other quantities same?

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

The total length of potentiometer wire AB is 50 cm in the arrangement as shown in the figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.

Two cells of e.m.f.s approximately $5$ V and $10$ V are to be compared using a potentiometer (sum and difference method) of length $400$ cm, then,

Two cells of e.m.f. $E_1$ and $E_2$ ($E_1 > E_2$) are connected as shown in figure.

When a potentiometer is connected between points A and B the balancing length of potentiometer wire is $412$ cm. When same potentiometer wire is connected between points A and C the balancing length is $103$ cm. The ratio $E_1 : E_2$ is

When cell of E.M.F. '$E_1$' is connected to potentiometer wire the balancing length is '$l_1$ '. Another cell of E.M.F. '$E_2$' ($E_1 > E_2$) is connected along with $E_1$ so as two cells oppose each other, the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ______

To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at $1$m when cell is shunted by $3\ \Omega$ resistance and at $1.5$m length when cell is shunted by $6\ \Omega$ resistance. The internal resistance of the cell is

Figure shows a potentiometer wire AB having resistance of $5\ \Omega$ and length 10 m. An e.m.f. is $0.4$ V of battery, the balancing length AP is (internal resistance is negligible)

A potentiometer wire of length $4$ m and resistance $5\ \Omega$ is connected in series with a resistance of $992\ \Omega$ and a cell of e.m.f. $4$ V with internal resistance $3\ \Omega$. The length of $0.75$ m on potentiometer wire balances the e.m.f. of

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

A potentiometer wire has length $L$. For a given cell of e.m.f. $E$, the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of potentiometer wire is increased by $50\%$ then for the same cell the balance point is obtained at length (from positive end)

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire 'x' cm long. If length of potentiometer wire is increased by few cm without changing the cell, the balancing length will (Driving source is not changed)

A 2.5 V battery is connected to a potentiometer wire. A cell of e.m.f. 1.08 V is balanced by the voltage drop across 2.16 m of wire. The length of the potentiometer wire is

The potentiometer wire is 5 m long and potential difference of 4 V is maintained between the ends. The e.m.f. of the cell which balances against a length of 200 cm of the potentiometer wire is

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

Two cells of e.m.f.s $E_1$ and $E_2\ (E_1 > E_2)$ are connected as shown in figure.

When the potentiometer is connected between A and B, the balancing length of the potentiometer wire is 3.60 m. On connecting the potentiometer between A and C, the balancing length is 0.90 m. The ratio $E_1/E_2$ is

When cell of e.m.f. ' $E_1$ ' is connected to potentiometer wire, the balancing length is ' $l_1$ '. Another cell of e.m.f. ' $E_2$ ' $(E_1 > E_2)$ is connected so that two cells oppose each other, the balancing length is ' $l_2$ '. The ratio $E_1 : E_2$ is

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by $5\ \Omega$. When a resistance of $15\ \Omega$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of $5\ \Omega$ the balancing length is 250 cm. When the cell is shunted by $20\ \Omega$, the balancing length of potentiometer wire is 400 cm. The internal resistance, of the cell is

The length of a potentiometer wire is ' $L$ '. A cell of e.m.f. ' $E$ ' is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{\mathrm{L}}{3}$, then using the same cell null point is obtained at

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 balance point at distance ' $x$ '. The value of ' $x$ ' is

A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance $100\ \Omega$. The difference of potential between two points 50 cm apart on the wire is

A potentiometer wire of length 1 m is connected in series with $495\ \Omega$ resistance and 2 V battery. If $0.2\text{ mV/cm}$ is the potential gradient, then the resistance of the potentiometer wire is

In potentiometer experiment, cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ the balancing length is 80 cm of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes 20 cm. The ratio $E_1/E_2$ is

When cell of E.M.F. ' $E_1$ ' is connected to potentiometer wire, the balancing length is $l_1$. Another cell of E.M.F. ' $E_2$ ' $(E_1 > E_2)$ is connected so that two cells oppose each other, then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of $5\ \Omega$. But when it is shunted by a resistance of $10\ \Omega$, then balancing length increases by 25 cm. The balancing length when the cell is in an open circuit is

To measure the internal resistance of a battery, potentiometer is used. For $R = 10\ \Omega$, the balance point is observed at $l = 500\text{ cm}$ and for $\mathrm{R} = 1\ \Omega$ the balance point is observed at $l = 400\text{ cm}$. The internal resistance of the battery is approximately:

Resistance of a potentiometer wire is $2\ \Omega/\text{m}$. A cell of e.m.f. $1.5\text{ V}$ balances at $300\text{ cm}$. The current through the wire is

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5\text{ V}$ is found to be $60\text{ cm}$. If this cell is replaced by another cell of emf E, the length-of null point increases by $40\text{ cm}$. The value of $E$ is $\frac{x}{10}\text{ V}$. The value of $x$ is ____________.

With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is

(A) directly proportional to the length of the potentiometer wire

(B) directly proportional to the potential gradient of the wire

(C) inversely proportional to the potential gradient of the wire

(D) inversely proportional to the length of the potentiometer wire

Choose the correct option for the above statements:

In potentiometer experiment, the balancing length is $8\text{ m}$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition the balancing length is $4\text{ m}$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

In potentiometer experiments, two cells of e. m. f. '$E_1$' and '$E_2$' are connected in series $(E_1 > E_2)$, the balancing length is $64\text{ cm}$ of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes $32\text{ cm}$. The ratio $\mathrm{E}_1/\mathrm{E}_2$ is

A $10\text{ m}$ long wire of resistance $20\ \Omega$ is connected in series with a battery of emf $3\text{ V}$ (negligible internal resistance) and a resistance of $10\ \Omega$. The potential gradient along the wire is

A potentiometer wire of length $4\text{ m}$ and resistance $5\ \Omega$ is connected in series with a resistance of $992\ \Omega$ and a cell of e.m.f. $4\text{ V}$ with internal resistance $3\ \Omega$. The length of $0.75\text{ m}$ on potentiometer wire balances the e.m.f. of

The emf of a cell having internal resistance $1\ \Omega$ is balanced against a length of $330\text{ cm}$ on a potentiometer wire. When an external resistance of $2\ \Omega$ is connected across the cell, the balancing length will be:

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$\Omega$. When a resistance of 15$\Omega$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$\Omega$.

A potentiometer wire has length of $5\text{ m}$ and resistance of $16\ \Omega$. The driving cell has an e.m.f. of $5\text{ V}$ and an internal resistance of $4\ \Omega$. When the two cells of e.m.f.s $1.3\text{ V}$ and $1.1\text{ V}$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.

In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3: 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.

As shown in the figure, a potentiometer wire of resistance $20\ \Omega$ and length $300\text{ cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4\text{ V}$. For a resistance '$R$' of resistance box introduced into the circuit, the null point for a cell of $20\text{ mV}$ is found to be $60\text{ cm}$. The value of '$R$' is ___________ $\Omega.$

A potentiometer wire of length $300\text{ cm}$ is connected in series with a resistance 780 $\Omega$ and a standard cell of emf $4\text{ V}$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\text{ mV}$ is found to be $60\text{ cm}$. The resistance of the potentiometer wire is ____________ $\Omega$.

A potentiometer wire of length 10 m and resistance 20 $\Omega$ is connected in series with a 25 V battery and an external resistance 30 $\Omega$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10}$. The value of x is __________.

The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K1' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K1' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ $\Omega$.

A cell, shunted by a 8 $\Omega$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 $\Omega$ resistance. The value of internal resistance of the cell will be ____________ $\Omega$.

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

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 $420\text{ cm}$. If cells $\mathrm{A}$ and $\mathrm{B}$ are connected in series the balancing length is $220\text{ cm}$ and for cells $\mathrm{B}$ and $\mathrm{C}$ connected in series the balancing length is $320\text{ cm}$. The emf of cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively in the ratio of

In potentiometer experiment, cells of e.m.f. '$E_1$ ' and '$E_2$' are connected in series $(E_1 > E_2)$ the balancing length is $64\text{ cm}$ of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes $32\text{ cm}$. The ratio $\frac{E_1}{E_2}$ is

In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where a = _________.

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:

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1\text{ m}$ when shunted by $3\ \Omega$ resistance and at a length $1.5\text{ m}$, when cell is shunted by $6\ \Omega$ resistance The internal resistance of the cell is

In a potentiometer experiment, the balancing length for a cell is $240\text{ cm}$. On shunting the cell with a resistance of $2\ \Omega$, the balancing length becomes half the initial balancing length. The internal resistance of the cell is

In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 $\Omega$ per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is:

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire '$\mathrm{x}$' cm long. If length of potentiometer wire is increased by few centimeter without changing the cell, the balancing length will [Driving source is not changed]

A 10 m long wire of resistance $20\ \Omega$ is connected in series with a battery of emf 3 V and a resistance of $10\ \Omega$. The potential gradient along the wire in $\text{V/m}$ is

A potentiometer wire is $4\text{ m}$ long and potential difference of $3\text{ V}$ is maintained between the ends. The emf of the cell, which balances against a length of $100\text{ cm}$ of the potentiometer wire is

Which of the following instruments is not a direct reading instrument?

Two cells having unknown emfs $E_1$ and $E_2\ (E_1 > E_2)$ are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell $E_2$ is connected, so as to oppose cell $E_1$, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells $\left(\frac{E_1}{E_2}\right)$ is

A potentiometer wire PQ of 1 m length is connected to a standard cell E1. Another cell E2 of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is:

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

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

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 $\Omega$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is:

A potentiometer wire has length $L$ For given cell of emf $E$, the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of potentiometer wire is increased by $50\%$, then for the same cell, the balance point is obtained at length

The resistance of the meter bridge AB in given figure is 4 $\Omega$. With a cell of emf $\varepsilon$ = 0.5 V and rheostat resistance Rh = 2 $\Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon = \varepsilon$2 the same null point J is found for Rh = 6 $\Omega$. The emf $\varepsilon$2 is:

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $\varepsilon$ and internal resistance r. A cell C having emf $\varepsilon$/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –

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 :-

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 $\Omega$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length $\ell$1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length $\ell$2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances $\frac{R_1}{R_2}$, is found to be:

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length $l$ of the potentiometer wire. The e.m.f. E will be given by

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

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity, (ii) 9.5 $\Omega$

the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is

In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potentiometer wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of voltmeter will be:

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at length $l$1 cm and $l$2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10 $\Omega$. Its internal resistance is

In a potentiometer experiment the balancing with a cell is at length $240\text{ cm}.$ On shunting the cell with a resistance of $2\ \Omega,$ the balancing length becomes $120\text{ cm}$. The internal resistance of the cell is

A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be

The length of a wire of a potentiometer is $100\text{ cm}$, and the $\text{e.m.f.}$ of its standard cell is $E$ volt. It is employed to measure the $\text{e.m.f.}$ of a battery whose internal resistance is $0.5\ \Omega.$ If the balance point is obtained at $l = 30\text{ cm}$ from the positive end, the $\text{e.m.f.}$ of the battery is

where $i$ is the current in the potentiometer wire.

If specific resistance of a potentiometer wire is 10-7$\Omega$ m and current flow through it is 0.1 amp., cross-sectional area of wire is 10$-$6 m2 then potential gradient will be

The potentiometer is best for measuring voltage, as

A potentiometer consists of a wire of length $4\text{ m}$ and resistance $10\ \Omega$. It is connected to a cell of emf $2\text{ V}$. The potential gradient of wire is