Cyclotron Working Principle, Applications & Frequency || Class 12 Physics Notes

The cyclotron frequency (or, equivalently, gyro-frequency) is the number of cycles a particle completes around its circular circuit every second.

## Cyclotron Frequency

Under the action of the given magnetic field (B fixed), the given positive ion (e/m fixed) will cover the semi-circular path in a fixed time only, if t in equation is equal to $\frac{T}{2}$

Where T is the time period of the electric field.

Maximum energy of the positive ions

Maximum energy acquired by positive ions depends on the radius R of the dees. After acquire this energy positive ion emerge out the dees and use for hitting the target.

Let $\mathrm{V}_{\mathrm{max}}$ be the velocity acquired by the positive ions, when it moves along the largest circular path i.e. path of radius equal to the radius R of the dees.

Then $\mathrm{B} \mathrm{q} \mathrm{v}_{\max }=\frac{\mathrm{m} \mathrm{v}_{\max }^{2}}{\mathrm{R}}$

OR

$v_{\max }=\frac{B q R}{m}$

Therefore maximum kinetic energy gained by the positive ion,

$$E_{\max }=\frac{1}{2} m v_{\max }^{2}=\frac{1}{2} m\left(\frac{B q R}{m}\right)^{2}$$

OR

$E_{\max }=\frac{1}{2} \frac{B^{2} q^{2} R^{2}}{m}$

The maximum energy acquired by the positive ion can be expressed in another form as given below :

If V is the potential difference applied between the dees and N is the number of times, the positive ion cross the gap between the dees before leaving the dees, then

$E_{\max }=N q V$

$N=2 n$

$n=$ the number of rotations completede.

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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Mind Map of Vectors (Part 2) Physics Class 11, JEE & NEET-Download from here

## Learn all the formulas and important topics of Vectors through this mind map.

Class 11 Rotational Motion [Video Lectures], Complete Revision with Mind Map

Here are the Rotational Motion Video Lectures for quick revision of complete chapter. Revise all the important formulae and key points. Know the tricks and tips to solve questions for JEE, NEET Exams.

The complete chapter comprises 4 Parts:

PART 1

PART 2

PART 3

PART 4

### Introduction to System of Particles and rotational Motion

A rigid body is a solid body of finite size in which deformation is negligible under the effect of deforming forces.

A rigid body is one in which the distances between different particles of the body do not change.

The centre of mass (COM) of a rigid body is the point in or near an object at which the whole mass of the object may be considered to be concentrated.

Collision or impact:
Collision is an isolated event in which a strong force acts between two or more bodies for a short time which results in a change in their velocities. A rigid object can be substituted with a single particle with mass equal to the total mass of the system located at the COM of the rigid object.
On the basis of energy, there are two types of collision-elastic collision and inelastic collision.
On the basis of line of impact, there are two types of collision-head-on collision and oblique collision.

Conservation of linear momentum: The total momentum of an isolated system remains constant. Momentum is conserved for an isolated system regardless of the nature of internal forces. Law of conservation of linear momentum is independent of the frame of reference.

### PART 1

Saransh Sir will be covering the following topics in this video in just 40 minutes:

• Translational Motion of a Rigid Body
• Rotational Motion about Fixed Axis
• Translational Motion
• Pure Rotational Motion
• Combined Rotation & Translational Motion
• Kinetic Energy of Rigid Body performing Rotation
• Moment of Inertia – Continuous Mass Distribution
• Perpendicular Axis Theorem
• Parallel Axis Theorem

### PART 2

Saransh Sir will be covering the following topics in this video in just 40 minutes:

• Cross Product (Vector Product)
• Properties of Cross Product
• Torque Explanation
• Torque Formulae & Numericals
• Condition for Equilibrium
• Force Couple
• Inertial Pulley
• Work energy theorem Applications
• Calculation of Hinge Force
• Work Energy Theorem
• Work Done in Terms of Torque

### PART 3

Saransh Sir will be covering the following topics in this video in just 30 minutes:

• Angular Momentum of Particle
• Conical Pendulum
• Angular Momentum of System of Particle
• Angular Momentum of Rigid Body Performing Pure Rotation About Fixed Axis
• Relation Between Torque and Angular Momentum
• Angular Momentum Conservation Principle
• Toppling
• Condition of Toppling
• Angular Impulse

### PART 4

Rotational Motion – Part 3 | Angular Momentum Mind Map for Class 11, JEE & NEET

Rotational Motion Class 11 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the key formulae and important concepts on finger tips.

Post Office Box Physics || Current Electricity Class 12, JEE & NEET

A Post Office Box can also be used to measure an unknown resistance. It is a Wheatstone Bridge with three arms P, Q and R; while the fourth arm(s) is the unknown resistance. P and Q are known as the ratio arms while R is known as the rheostat arm. Post Office Box, Physics named it so because it has a shape of box and was designed to find resistance of electric cables and telegraph wires. It was used in post offices to determine resistance of transmission lines.

At balance, the unknown resistance

S = (P/Q) R                                                           …… (1)

It is based on principle of Wheatstone bridge.

Unknown resistance is $S =\frac{ Q }{ P } R$ and specific resistance is $\rho=\frac{\pi r^{2} S}{L}$, where r is radius and L is length of wire.

In Post Office box we first press cell key and then press galvanometer key to eliminate induced effects.

It is used to find unknown resistance, specific resistance of a wire, internal resistance of cell, resistance of galvanometer etc.

A typical post office box is in a wooden box with a hinged lid and a metal or Bakelite panel showing circuit connections. Coils of wire are wound non-inductively, mounted in the body of the box, and have a negligible temperature coefficient. Pairs of ratio arms are each 10, 100, 1000 ohms. A resistance arm contains a number of coils from 1 to 5000 ohms with a plug for infinite resistance.

Moving Coil Galvanometer Class 12

Ever wondered how the utility company detects how much power is used each month? The galvanometer is an instrument used to determine the presence, direction, and strength of an electric current in a conductor. Here we will study about the Galvanometer, moving Coil Galvanometer class 12 and its types. These are instruments used for detection and measurement of small currents.

The moving coil galvanometer work on the principle that when a current carrying coil is placed in a magnetic field it experiences a torque.

The different type of moving coil galvanometers are

(a) Pivoted Galvanometer : It consists of a coil of fine insulated wire wound on a metallic frame. The coil is mounted on two jewelled pivots and is symmetrically placed between cylindrical pole pieces of a strong permanent horse-shoe magnet.

(b) Dead beat Galvanometer : Here coil is wound over the metallic frame to make it dead beat. On passing current the galvanometer shows a steady deflection without any oscillation. The damping is produced by eddy currents.

(c) Ballistic Galvanometer : This is used for measurement of charge. Here coil is wound on an insulating frame and oscillates on passing current.

In moving coil galvanometer the deflection produced is proportional to current flowing through galvanometer i.e. they have linear scale of measurement.

In equilibrium deflecting torque = restoring torque

i.e. NIAB $=\mathrm{C} \theta \quad$ or $\quad I=\left(\frac{C}{N A B}\right) \theta \quad$ or $I=K \theta$ where $\mathrm{K}$ is galvanometer constant.

Current Sensitivity

This is defined as the deflection produced in galvanometer when a unit current flows through it.

Current sensitivity $CS =\frac{\theta}{ I }=\frac{ NAB }{ C }$ radian/ampere or division/ampere

(a) The sensitivity can be increased by increasing number of turns in coil (N), area of cross-section of coil (A), magnetic field B and decreasing torsional constant (C).

(b) The reciprocal of current sensitivity is called figure of merit.

Figure of merit $FM =\frac{ I }{\theta}=\frac{ C }{ NAB }$

Voltage Sensitivity

This is defined as the deflection produced in galvanometer when a unit voltage is applied across its terminals.

Voltage sensitivity $VS =\frac{\theta}{ V }=\frac{\theta}{ IR }=\frac{ NAB }{ CR }$ division/volt, where R is resistance of coil.

• Shunting a galvanometer reduces its current sensitivity.

Wheatstone Meter Bridge – Circuit Diagram || Current Electricity Class 12, JEE NEET

A Meter Bridge also called a slide wire bridge is an instrument that works on the principle of a Wheatstone bridge. A Wheatstone meter bridge is used in finding the unknown resistance of a conductor as that of in a Wheatstone bridge.

The wire is made of manganin  because they have low temperature coefficient and large resistivity. The wires are soldered to two thick L shaped copper strips with binding terminals. Between these there is a third copper strip. There are two fixed gaps ab and cd.

(1) It is used to measure unknown resistances $(1 \Omega \text { to } 1000 \Omega)$, for comparision of two unknown resistances and specific resistance of material of wire.

(2) For better accuracy R must be adjusted to obtain L between 40 to 60 cm. It has maximum accuracy when $\mathrm{L} \approx 50 \mathrm{cm}$.

(3) It is more sensitive than PO box but suffers from end resistances.

### SOLVED EXAMPLES

Ex. An unknown resistance S is placed on the left gap and known resistance of $60 \Omega$ in right gap of meter bridge. The null point is obtained at 40 cm from left end of bridge. Find the unknown resistance?

Sol. $S=R \frac{(100-L)}{L}=60 \frac{(100-60)}{60}=40 \Omega$

Ex. In a meterbridge the length of wire is 100 cm. At what point is balance point obtained if two resistances are in ratio 2 : 3?

Sol. $\frac{P}{Q}=\frac{L}{100-L}=\frac{2}{3} \quad$ so $\quad 3 \mathrm{L}=200-2 \mathrm{L} \quad$ or $\quad L=\frac{200}{5}=40 \mathrm{cm}$

Kelvin Bridge – Definition and Diagram || Current Electricity Class 12, JEE & NEET

The Kelvin bridge or Thompson bridge is used for measuring the unknown resistances having a value less than 1Ω. It is the modified form of the Wheatstone Bridge. Here we will study about Kelvin Bridge, its Diagram and the equation of Kelvin Bridge.

### Why there is a requirement of Kelvin Bridge?

Wheatstone bridge use for measuring the resistance from a few ohms to several kilo-ohms. But error occurs in the result when it is used for measuring the low resistance. This is the reason because of which the Wheatstone Bridge is modified, and the Kelvin bridge obtains. The Kelvin bridge is suitable for measuring the low resistance.

### Modification of Wheatstone Bridge

In Wheatstone Bridge, while measuring the low-value resistance, the resistance of their lead and contacts increases the resistance of their total measured value. This can easily be understood with the help of the circuit diagram.

The r is the resistance of the contacts that connect the unknown resistance R to the standard resistance S. The ‘m’ and ‘n’ show the range between which the galvanometer is connected for obtaining a null point.

When the galvanometer is connected to point ‘m’, the lead resistance r is added to the standard resistance S. Thereby the very low indication obtains for unknown resistance R. And if the galvanometer is connected to point n then the r adds to the R, and hence the high value of unknown resistance is obtained. Thus, at point n and m either very high or very low value of unknown resistance is obtained.

So, instead of connecting the galvanometer from point, m and n we chose any intermediate point say d where the resistance of lead r is divided into two equal parts, i.e., r1 and r2

$\frac{r_{1}}{r_{2}}=\frac{P}{Q} \ldots \ldots$ equ ( 1)

The presence of $\mathrm{r}_{1}$ causes no error in the measurement of unknown resistance.
$$R+r_{1}=\frac{P}{Q} \cdot\left(S+r_{2}\right)$$

From equation $(1),$ we get
\begin{aligned} \frac{r_{1}}{r_{1}+r_{2}} &=\frac{P}{P+Q} \\ r_{1}=& \frac{P}{P+Q} \cdot r \end{aligned}

$a S$
\begin{aligned} r_{1}+r_{2} &=r \\ r_{2}=& \frac{Q}{P+Q} \cdot r \\ R+\frac{P}{P+Q} \cdot r &=\frac{P}{Q}\left(S+\frac{Q}{P+Q} r\right) \\ R &=\frac{P}{Q} \cdot S \end{aligned}

The above equation shows that if the galvanometer connects at point d then the resistance of lead will not affect their results.

The above mention process is practically not possible to implement. For obtaining the desired result, the actual resistance of exact ratio connects between the point m and n and the galvanometer connects at the junction of the resistor.

### Kelvin Double Circuit Bridge

The ratio of the arms p and q are used to connect the galvanometer at the right place between the point j and k. The j and k reduce the effect of connecting lead. The P and Q is the first ratio of the arm and p and q is the second arm ratio.

The galvanometer is connected between the arms p and q at a point d. The point d places at the centre of the resistance r between the point m and n for removing the effect of the connecting lead resistance which is placed between the unknown resistance R and standard resistance S.

The ratio of p/q is made equal to the P/Q. Under balance condition zero current flows through the galvanometer. The potential difference between Point $a$  and $b$ is equivalent to the voltage drop between the points $E_{\mathrm{amd}}$.

Now,
$$\begin{array}{c} {E_{a b}=\frac{P}{P+Q} E_{a c}} \\ {E_{a c}=I\left[R+S+\frac{(p+q) r}{p+q+r}\right] \ldots \ldots \text { equ }(1)} \\ {E_{a m d}=I\left[R+\frac{p}{(p+q)}\left\{\frac{(p+q) r}{p+q+r}\right\}\right]} \\ {E_{a c}=I\left[\frac{p r}{p+q+r}\right] \ldots \ldots \ldots e q u(2)} \end{array}$$

For zero galvanometer deflection,
$$\begin{array}{c} {E_{a c}=E_{a m d}} \\ {\frac{P}{P+Q} \cdot I\left[R+\frac{p}{(p+q)}\left\{\frac{(p+q) r}{p+q+r}\right\}\right]=I\left[\frac{p r}{p+q+r}\right]} \\ {R=\frac{P}{Q} \cdot S+\frac{p r}{p+q+r}\left[\frac{P}{Q}-\frac{p}{q}\right]} \end{array}$$

As we known, $\mathbf{P} / \mathbf{Q}=\mathbf{p} / \mathbf{q}$ then above equation becomes
$$R=\frac{P}{Q} \cdot S$$

The above equation is the working equations of the Kelvins bridge. The equation shows that the result obtains from the Kelvin double bridge is free from the impact of the connecting lead resistance.

For obtaining the appropriate result, it is very essentials that the ratio of their arms is equal. The unequal arm ratio causes the error in the result. Also, the value of resistance r should be kept minimum for obtaining the exact result.

The thermo-electric EMF induces in the bridge during the reading. This effect can be reduced by measuring the resistance with the reverse battery connection. The real value of the resistance obtains by takings the means of the two.

Combination of Resistances

Uses of Potentiometer | Comparison of emf of two cells using Potentiometer || Class 12, JEE & NEET

The measuring instrument called a potentiometer is essentially a voltage divider used for measuring electric potential (voltage); the component is an implementation of the same principle, hence its name. Potentiometers are commonly used to control electrical devices such as volume controls on audio equipment. We will study here about Comparison of emf of two cells using Potentiometer, Determination of internal resistance of cell, Calibration of voltmeter, Calibration of ammeter, Measurement of small thermo emf.

The Potentiometer is mainly used for,

PRECAUTIONS IN THE USE OF POTENTIOMETER

### Determination of unknown EMF or Potential Difference

[3] If unknown potential difference V is balanced for length $\ell$ than $V=x \ell=\left(\frac{E_{0}}{\ell_{0}}\right) l$

[4] If the length of potentiometer wire is changed from L to L’ then the new balancing length is $\ell^{\prime}=\left(\frac{L^{\prime}}{L}\right) \ell$ If length is increased L’ > L so $l^{\prime}>l$ and if length is decreased L’ < L so $l^{\prime}<l$ So change in balancing length $\Delta \ell=\left(\ell^{\prime}-\ell\right)$

[ 5] If the current flowing through resistance $\mathrm{R}$ is I then

$$\mathrm{V}=\mathrm{IR}=\mathrm{x} \ell=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \ell \quad \text { so } \quad \mathrm{I}=\frac{\mathrm{x} \ell}{\mathrm{R}}=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \frac{\ell}{\mathrm{R}}$$

For determination of current we use a coil of standard resistance.

Ex. The current flowing through the primary circuit is 2A and resistance per unit length is

0.2 $\Omega / m$. If the potential difference across 10 ohm coil is balanced at 2.5 m then find current flowing through the coil.

Sol. The potential gradient $x=\frac{I R}{L}=2 \times 0.2=0.4 V / m$

Unknown potential $V=x \ell=I^{\prime} R$ so $I^{\prime}=\frac{x \ell}{R}=\frac{0.4 \times 2.5}{10}=0.1 A$

### Comparison of emf of two cells using Potentiometer

[1] No standardisation is necessary.

[2] Let $E_{1}$ emf be balanced at length $\ell_{1}$ and $E_{2}$ emf be balanced at length $\ell_{2}$ then

$E_{1}=x \ell_{1}$ and $E_{2}=x \ell_{2}$ so $\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

[3] If two cells joined in series support each other then the balancing length is $\ell_{1}$ so $E_{1}+E_{2}=x \ell_{1}$

If two cells joined in series oppose each other then the balancing length is $\ell_{2}$ so $E_{1}-E_{2}=x \ell_{2}$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{\ell_{1}}{\ell_{2}}$ or $\frac{E_{1}}{E_{2}}=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}}$

Ex. A uniform potential gradient is established across a potentiometer wire. Two cells of emf $E_{1}$ and $E_{2}$ connected to support and oppose each other are balanced over $\ell_{1}$ = 6m and $\ell_{2}$ = 2m. Find E1/E2.

Sol. $E _{1}+ E _{2}= x \ell_{1}=6 x$ and $E _{1}- E _{2}=2 x$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{6}{2}$ or $\frac{E_{1}}{E_{2}}=\frac{2}{1}$

### Determination of Internal Resistance of Cell

[1] No standardization is necessary.

[2] Keeping $\mathrm{K}_{1}$ open the balancing length $\ell_{1}$ gives emf of cell so E = $x \ell_{1}$

[3] Keeping $K_{1}$ closed the balancing length $\ell_{2}$ for some resistance R gives potential difference so V = $x \ell_{2}$

Internal resistance $r=\frac{E-V}{V} \quad R=\frac{x \ell_{1}-x \ell_{2}}{x \ell_{2}}$

$R=\left(\frac{\ell_{1}-\ell_{2}}{\ell_{2}}\right) R$

### Calibration of Voltmeter

[1] The voltmeters do not given accurate measurements because they do not have infinite resistance.

[2] The error in measurement is found by comparision with readings of potentiometer.

[3] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[4] The unknown potential difference V’ is balanced at length $\ell_{1}$ then $V^{\prime}=x \ell_{1}=\frac{E_{0}}{l_{0}} \ell_{1}$

[5] If reading of voltmeter is V then error is V – V’ which can be positive, negative or zero.

[6] The calibration curve is obtained plotting voltmeter reading V on x axis and error on y axis.

### Calibration of Ammeter

[1] V = IR and if R = 1$\Omega$ then V = I so potential difference across 1$\Omega$ resistance is equal to current through the resistance.

[2] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[3] The current through R or 1$\Omega$ coil is measured by

ammeter () and calculated by potentiometer as $V^{\prime}=I^{\prime} \times 1 \Omega=x \ell_{1}=\left(\frac{E_{0}}{\ell_{0}}\right) \ell_{1}$

[4] The error $I-I^{\prime}$can be found and calibration curve is obtained by plotting ammeter reading  on x axis and error on y axis.

### Measurement of Small Thermo EMF

[1] A high resistance box is used in primary circuit to reduce primary current.

[2] If galvanometer shows no deflection for some $R_{1}$ then potential difference across $R_{1}$ is $V=E_{0}=I R_{1}$ or $|=E_{0} / R_{1}$

[3] If balancing length for small thermo emf E is $\ell$ then $E=x \ell=\frac{E_{0}}{R_{1}} \cdot \frac{R}{L} \ell$

### Precautions in use of Potentiometer

[1] All high potential terminals of primary and secondary circuit should be connected at same point of Potentiometer.

[2] The known emf of primary circuit must be greater than the unknown potential difference to be measured in secondary circuit.

[3] The balancing length is always measured from the point of higher potential.

[4] In state of zero deflection no current flows in galvanometer circuit but it flows in primary circuit.

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

Sensitivity of Potentiometer – Current Electricity || Class 12, JEE & NEET

Sensitivity of Potentiometer means the smallest potential difference it can measure. It can be increased by reducing the potential gradient.

A potentiometer is sensitive if it is capable of measuring the small potential differences and it shows significant change in balancing length for a small change in the potential difference being measured.
Sensitivity of Potentiometer means the smallest potential difference that can be measured by using it and this can be achieved by decreasing the potential gradient by increasing the length of the wire or reducing the current in the potentiometer using rheostat.

[1] Smaller the potential difference that can be measured with a potentiometer more is the sensitivity of the potentiometer.

[2] The sensitivity of potentiometer is inversely proportional to potential gradient $(S \propto 1 / x)$

[3] The sensitivity can be increased by (a) Increasing length of potentiometer wire (b) For a potentiometer wire of fixed length potential gradient is decreased by reducing the current in circuit.

### How Sensitivity of a potentiometer can be increased?

Sensitivity of a potentiometer can be increased by decreasing its potential gradient. This can be done by two ways:

– by increasing the length of potentiometer wire.

– if the potentiometer wire is of fixed length, the potentiometer gradient can be decreased by reducing the current in the potentiometer wire circuit with the help of rheostat and using a single cell.

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

Class 12 Potentiometer Questions for NEET | JEE

Here are the Potentiometer Questions for NEET, IIT JEE and Board Exams.

### Potentiometer Questions for NEET, IIT JEE and Board Exams

Ex. In the primary circuit of a potentiometer a battery of $2 \mathrm{V}$ and a rheostat of $22 \Omega$ are connected. If the length of potentiometer wire is $10 \mathrm{m}$ and its resistance is $18 \Omega$ then find the potential gradient in wire.

Sol. The potential gradient $x=\left(\frac{E}{r+R+R_{h}+R_{e}}\right) \frac{R}{L} \quad r=R_{e}=0 \Omega$

So $x=\left(\frac{2}{0+18+22+0}\right) \frac{18}{10}=0.09 V / m$

Ex. No current flows if the terminals of a cell are joined with 125 cm length of potentiometer wire. If a 20 resistance is connected in parallel to cell balancing length is reduced by 25 cm. Find internal resistance of cell.

Sol. $r=\frac{\ell_{1}-\ell_{2}}{\ell_{2}} \times R =\frac{125-100}{100} \times 20=5 \Omega$

Ex. Figure shows use of potentiometer for comparision of two resistances. The balance point with standard resistance R = 10 is at 58.3 cm, while that with unknown resistance X is 68.5 cm. Find X.

Sol. Let $E_{1}$ and $E_{2}$be potential drops across R and X

so $\frac{E_{2}}{E_{1}}=\frac{1 \times}{1 R}=\frac{x}{R}$ or $X=\frac{E_{2}}{E_{1}} R$

But $\frac{E_{2}}{E_{1}}=\frac{l_{2}}{l_{1}}$ so $X=\frac{\ell_{2}}{\ell_{1}} R=\frac{68.5}{58.3} \times 10=11.75 \Omega$

Ex. The length of potentiometer wire is 40cm. Zero deflection in galvanometer is obtained at point F. Find the balancing length AF.

Sol. Let x be potential gradient and $V_{1}$ and $V_{2}$ be potential difference across 8$\Omega$ and 12$\Omega$ respectively.

$\frac{V_{1}}{V_{2}}=\frac{8}{12}=\frac{x \ell}{x(40-\ell)}$ or $\frac{l}{40-l}=\frac{2}{3}$ or $\ell=16 cm$

Ex. If the current in the primary circuit of a potentiometer wire is 0.2 A, specific resistance of material of wire is $40 \times 10^{-8} \Omega m$ and area of cross-section is $0.8 \times 10^{-6} m ^{2}$ Calculate potential gradient?

Sol. Potential gradient $x=\frac{V}{L}=\frac{I R}{L}=\frac{I \rho}{A}=\frac{0.2 \times 40 \times 10^{-8}}{0.8 \times 10^{-6}}=0.1$ volt $/ m$

Ex. A battery of emf 2V is connected in series with a resistance box and a 10m long potentiometer with resistance $1 \Omega / m$. A 10 mV potential difference is balanced across the entire length of wire. Calculate the current flowing in wire and resistance in resistance box.

Sol. Potential gradient $x=\frac{E}{r+R+R_{e}}\left(\frac{R}{L}\right)=\frac{2}{0+10+R_{e}} \times 1=\frac{2}{10+R_{e}}$

Potential difference $V=x l$ or $10 \times 10^{-3}=\frac{2}{10+ R _{ e }} \times 10$ or $R_{e}=1990 \Omega$

Current flowing $I=\frac{E}{R+R_{e}}=\frac{2}{10+1990}=1 m A$

Ex. While measuring the potential difference between the terminals of a resistance wire the balance point is obtained at 78.4 cm. The same potential difference is measured as 1.2 V with voltmeter. If standard cell of emf 1.018 V is balanced at 63.2 cm then find error in reading of voltmeter.

Sol. $E_{0}=x l_{0}$ and $V=x \ell$ or $V=\frac{E_{0}}{\ell_{0}} \ell$

$V=\frac{1.018 \times 78.4}{63.2}=1.26 volt$

$V=\frac{1.018 \times 78.4}{63.2}=1.26 volt$ so error = 1.2 – 1.26 = –0.06 V

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

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