Ampere’s Circuital Law and it Applications || Magnetic Effects of Current Class 12
Do you know about the Ampere’s Circuital Law, well it is a current distribution that helps us to calculate the magnetic field, And yes, Biot-Savart’s law does the same but Ampere’s law uses the case high symmetry. We will first understand the ampere’s circuital law, its definition, formulae, & Applications of Ampere’s Law in detail,

## Ampere’s Circuital Law

Ampere’s circuital law states that the line integral of magnetic field induction $\overrightarrow{\mathrm{B}}$ around any closed path in a vacuum is equal to $\mu_{0}$ times the total current threading the closed path, i.e.,

This result is independent of the size and shape of the closed curve enclosing a current.

This is known as Ampere’s circuital law.

Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.

Ampere’s law may be derived from the Biot-Savart law and Biot-Savart law may be derived from the Ampere’s law.

Ampere’s law is more useful under certain symmetrical conditions.

Biot-Savart law based on the experimental results whereas Ampere’s law based on mathematical.

## Applications of Ampere’s Law

### (a) Magnetic induction due to a long current-carrying wire.

Consider a long straight conductor Z-Z’ is along the z-axis. Let I will be the current flowing in the direction as shown in Fig. The magnetic field is produced around the conductor. The magnetic lines of force are concentric circles in the XY plane as shown by dotted lines. Let the magnitude of the magnetic field induction produced at a point P at distance r from the conductor is

Consider a close circular loop as shown in the figure.

According to Ampere’s law $\oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d}} \ell=\mu_{0} \sum \mathrm{I}$

The direction of $\overrightarrow{\mathrm{B}}$ at every point is along the tangent to the circle.

Consider a small element $\overrightarrow{\mathrm{d} \ell}$ of the circle of radius r at P. The direction of $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{d} \ell}$ the same. Therefore, angle between them is zero.

Line integral of $\overrightarrow{\mathrm{B}}$ around the complete circular path of radius $\mathrm{r}$ is given by

$\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} \ell=\oint \mathrm{B} \mathrm{d} \ell \cos 0^{\circ}$ $=\quad \mathrm{B} \oint \mathrm{d} \ell=\mathrm{B} \times 2 \pi \mathrm{r}$ $(\oint \mathrm{d} \ell=2 \pi \mathrm{r}=$ cicumference of the circle.) and $\quad \sum I=I$

So we get $\mathrm{B} \times 2 \pi \mathrm{r}=\mu_{0} \mathrm{I}$

### (b) Magnetic field created by a long current carrying conducting cylinder

A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross-section of the wire.

For finding the behavior of magnetic field due to this wire, let us divide the whole region into two parts.

(a) $\mathrm{r} \geq \mathrm{R}$ and

(b) $\mathrm{r}<\mathrm{R}$

$r=$ distance from the centre of the wire.

For $\mathrm{r} \geq \mathrm{R}:$ For closed circular path denoted by ( 1) from symmetry $\overrightarrow{\mathrm{B}}$ must be constant in magnitude and parallel to $\overrightarrow{\mathrm{d} \ell}$ at every point on this circle. Because the total current passing through the plane of the circle is I.

For $\mathrm{r}<\mathrm{R}:$ The current $\mathrm{I}$ passing through the plane of circle 2 is less than the total current I. Because the current is uniform over the cross-section of the wire.

Current through unit area $=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}}$

So current through area enclosed by circle 2 is $\mathrm{I}^{\prime}=\frac{\mathrm{I} \pi \mathrm{r}^{2}}{\pi \mathrm{R}^{2}}$

Now we apply Ampere’s law for circle $2 .$

The magnitude of the magnetic field versus $r$ for this configuration is plotted in figure. Note that inside the wire $\mathrm{B} \rightarrow 0$ as $\mathrm{r} \rightarrow 0 .$ Note also that eqn. (a) and eqn (b) give the same value of the magnetic field at $r=R,$ demonstrating that the magnetic field is continuous at the surface of the wire.

### (c) Magnetic field due to a conducting current carrying hollow cylinder

Consider a conducting hollow cylinder with inner radius $r_{1}$ and outer radius $r_{2} .$ And current $\mathrm{I}$ is flowing through it.

(I) $\quad$ For $r<r_{1}$

$\sum \mathrm{I}=0$ and hence $\quad B=0$

(II) $\quad$ For $r_{1}<r<r_{2}$

Now current I is flowing through area $\left[\pi r_{2}^{2}-\pi r_{1}^{2}\right]$

So, current per unit area $=\frac{I}{\pi\left(r_{2}^{2}-r_{1}^{2}\right)}$

$\therefore$ current flowing through area in bet” $\mathrm{r}_{1}<\mathrm{r}<\mathrm{r}_{2}$ is $\mathrm{I}=\frac{\mathrm{I}}{\pi\left(\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}\right)} \times\left(\pi \mathrm{r}^{2}-\pi \mathrm{r}_{1}^{2}\right)$

by using ampere’s law for circle of radius $\mathrm{r} \oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d}} \vec{\ell}=\mu_{0} \sum \mathrm{I}$

or $\quad \oint B d \ell \cos 0^{\circ}=\mu_{0}\left[\frac{I\left(r^{2}-r_{1}^{2}\right)}{r_{2}^{2}-r_{1}^{2}}\right]$

or $\quad \mathrm{B} \oint \mathrm{d} \ell=\mu_{0} \mathrm{I}\left[\frac{\mathrm{r}^{2}-\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}}\right]$

or $\quad B=\frac{\mu_{0} I}{2 \pi r}\left[\frac{r^{2}-r_{1}^{2}}{r_{1}^{2}-r_{1}^{2}}\right]$

$[\because \oint \mathrm{d} \ell=2 \pi \mathrm{r}]$

(a) For $r=r_{2}$

$B=\frac{\mu_{0} I}{2 \pi r_{2}}$

(b) For $r>r_{2}$

$B=\frac{\mu_{0} I}{2 \pi r}$

Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12

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Magnetic Field at the Centre of a Circular Coil | Circular Current loop as a Magnetic Dipole
Here we will study about the Magnetic Field at the Centre of a Circular Coil. Here we will study about the number of cases

## Magnetic Field at the Center of a Circular Current-Carrying Coil

Consider a circular coil of radius r through which current I is flowing. Let AB be an infinitesimally small element of length d\ell. According to Biot-Savart’s law, the magnetic field dB at the center P of the loop due to this small element d $\ell$ is

$\mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{r^{2}}$ where $\theta$ is the angle between $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$.

$\left.\therefore \quad \mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d \ell \sin 90^{\circ}}{r^{2}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d \ell}{\mathrm{r}^{2}} \quad \text { (for circular loop, } \theta=90^{\circ}\right)$

The loop can be supposed to consists of a number of small elements placed side by side. The magnetic field due to all the elements will be in the same direction. So, the net magnetic field at P is given by

$\mathrm{B}=\sum \mathrm{dB}=\sum \frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} \mathrm{d} \ell}{\mathrm{r}^{2}}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}^{2}} \sum \mathrm{d} \ell$

$\therefore \quad \mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi r^{2}} \times 2 \pi r$
$| \Sigma \mathrm{d} \ell=\text { circumference of the circle }=2 \pi \mathrm{r})$

### Magnetic Field due to part of the current-carrying circular conductor (Arc) :

$B=\frac{\mu_{0} I}{4 \pi r^{2}} \sum d \ell\left(\because \frac{\sum d \ell}{r}=\alpha\right)$

$B=\frac{\mu_{0} I}{4 \pi r} \alpha$

### Magnetic Field on the Axis of a Circular Coil

Consider a circular loop of radius a through which current I is flowing as shown in fig. The point $P$ lies on the axis of the circular current loop i.e., along the line perpendicular to the plane of the loop and passing through its center.

Let $x$ be the distance of the observation point $P$ from the centre 0 of the loop. Let us consider an infinitesimally small element $\mathrm{AB}$ of length d\ell. Radius of the loop $=\mathrm{a}$ According to Biot-Savart’s law, the magnetic field at P due to this small element

\begin{aligned} \overrightarrow{\mathrm{dB}} &=\frac{\mu_{0} \mathrm{I}}{4 \pi r^{3}}[\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{r}}] \\ \mathrm{dB} &=\frac{\mu_{0} \mathrm{I} \mathrm{d} \ell \sin \theta}{4 \pi \mathrm{r}^{2}} \end{aligned}

or $\quad \mathrm{dB}=\frac{\mu_{0} \mathrm{I} \mathrm{d} \ell}{4 \pi \mathrm{r}^{2}}\left(\theta=90^{\circ}\right)$

The direction of $\overrightarrow{\mathrm{dB}}$ is perpendicular to the plane of the current element $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}(\mathrm{CP})$ as shown in fig. by $\overrightarrow{\mathrm{PM}}$

Similarly if we consider another small element just diametrically opposite to this element then

magnetic field due to this at point $P$ is $\overrightarrow{\mathrm{dB}^{\prime}},$ denoted by PN and of the same magnitude. $\mathrm{d} \mathrm{B}=\mathrm{dB}^{\prime}$

Both $\overrightarrow{\mathrm{dB}}$ and $\overrightarrow{\mathrm{dB}^{\prime}}$ can be resolved into two mutually perpendicular components along $\mathrm{PX}$ and $\mathrm{zz}$ :

The components along ZZ’ [dB $\cos \alpha$ and $\left.d B^{\prime} \cos \alpha\right]$ cancel each other as they are equal and opposite in direction.

The same will hold for such other pairs of current elements. over the entire circumference of the loop.

Therefore, due to the various current elements, the components of the magnetic field is only along PX will contribute to the magnetic field due to the whole loop at point P.

### The magnetic dipole moment of the current loop

The current loop can be regarded as a magnetic dipole that produces its magnetic field and the magnetic dipole moment of the current loop is equal to the product of ampere-turns and area of the current loop. we can write

Case II : If the observation point is far far away from the coil, then $a<<x$. So, $a^{2}$ can be neglected in comparison to $x^{2}$.

$\therefore \quad B=\frac{\mu_{0} N I a^{2}}{2 x^{3}}$

terms of magnetic dipole moment, $\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{x}^{3}} \quad\left[\mathrm{B}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{NIA}}{\mathrm{x}^{3}}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{NIA}}{\mathrm{x}^{3}}\right]$
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Helmholtz Coils | Magnetic Field between two Coils | eSaral

Helmholtz coil is named after the German physicist Hermann von Helmholtz. It is comprised of two identical magnetic coils positioned in parallel to each other, and their centers are aligned in the same x-axis. The two coils are separated by a distance equal to the radius like a mirror image as shown in Figure 1. When current is passing through the two coils in the same direction, it generates a uniform magnetic field in a three-dimension region of space within the coils. Helmholtz coils are normally used for scientific experiments, magnetic calibration, to cancel background (earth’s) magnetic field, and for electronic equipment magnetic field susceptibility testing.

## Helmholtz Coils

The two coaxial coils of equal radii placed at a distance equal to the radius of either of the coils and in which the same current in the same direction is flowing are known as Helmholtz coils.

For these coils $x=\frac{a}{2}, I_{1}=I_{2}=I, a_{1}=a_{2}=a$

The two coils are placed mutually parallel to each other these coils are used to produce a uniform magnetic field. In between two coils along the axis at the middle point rate of change of magnetic field is constant, so if distance increases from a coil magnetic field decrease, but the distance from another coil decreases, so magnetic field due to second coil increases and hence the resultant magnetic field produced in the region between two coils remains uniform.

or $\mathrm{B}=0.76 \frac{\mu_{0} n \mathrm{I}}{a} \quad$ or $\mathrm{B}=1.423 \mathrm{B}_{\mathrm{C}}$

(Bc is a magnetic field at the center of a single coil.)\

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Right Hand Palm Rule – Magnetic Effect of Current Class12, JEE & NEET
So far we have described the magnitude of the magnetic force on a moving electric charge, but not the direction. The magnetic field is
a vector field, thus the force applied will be oriented in a particular direction. There is a clever way to determine this direction using nothing more than your right hand. The direction of the magnetic force $F$ is perpendicular to the plane formed by $v$ and $B$, as determined by the right-hand palm rule, which is illustrated in the figure. The right-hand rule states that to determine the direction of the magnetic force on a positive moving charge, $f$, point the thumb of the right hand in the direction of $v,$ the fingers in the direction of $B,$ and a perpendicular to the palm points in the direction of $F$

## Right Hand Palm Rule

If we hold the thumb of the right hand mutually perpendicular to the grip of the fingers such that the curvature of the finger represents the direction of current in the wire loop, then the thumb of the right hand will point in the direction of the magnetic field near the center of the current loop

### Graph of B v/s X

As soon as x increases magnetic field B decreases, dependence of B on x is shown in figure

Rate of change of B with respect to x is different at different values of x

for $x<\pm \frac{a}{2}$ curve is convex and $\quad$ for $x>\pm \frac{a}{2}$ curve is concave

At $x=\pm \frac{a}{2} \quad$ we get $\frac{d B}{d x}=$ const $,$ and $\frac{d^{2} B}{d x^{2}}=0$

So at $x=+\frac{a}{2} \&-\frac{a}{2}$ B varies linearly with $x$

These points are called points of inflexion.

Distance in between these two points is equal to radius of the coil

$B=\frac{B_{C}}{\left(1+\frac{x^{2}}{a^{2}}\right)^{3 / 2}}$

$\because$ Magnetic field at the centre of coil $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{NI}}{2 \mathrm{a}}$

Biot Savart’s Law
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Magnetic Field due to Infinite Straight Conductor | Magnetic Effects of Current Class 12

Hey, do you want to learn about the Magnetic Field due to Infinite Straight Conductor? If yes. Then keep reading.

## Magnetic field due to long straight conductor

Here we will discuss all the cases involved in the magnetic field due to Conductor such as Magnetic Field due to Infinite Straight Conductor, and many more discussed below:
Consider a long straight conductor XY through which current I is flowing from $X$ to Y. Let $P$ be the observation point at a distance ‘r’ ‘from the conductor XY. Let us consider an infinitesimally small current element $\mathrm{CD}$ of length d\ell. Let s be the distance of $\mathrm{P}$ from the mid-point $\odot$ of the current element. Let $\theta$ be the angle that OP makes with the direction of the current. The magnetic field at $P$ due to the current element $C D$ is

$\mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{\mathrm{s}^{2}}[\text { Biot-Savart’s law }]$ The magnetic field at P due to the whole of the conductor XY

## Case I :

If the conductor is infinitely long, then $\theta_{1}=90^{\circ}$ and $\theta_{2}=90^{\circ}$
$\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi}\left[\sin \frac{\pi}{2}+\sin \left(\frac{\pi}{2}\right)\right]=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}[1+1]=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}$ Or

### Case II :

If a conductor is of infinite length but one end is in front of point $P$ i.e. one end of conductor starts from point $N$ then $\theta_{1}=0^{\circ}$ and $\theta_{2}=90^{\circ}$

### Case III :

Conductor is finite length and point P is just in front of the middle of the conductor

### Case IV :

Biot Savart’s Law

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Right hand Thumb Rule in Physics Class 12 | Magnetic Effects of Current
The right-hand thumb rule in Physics states that: to determine the direction of the magnetic force on a positive moving charge,ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

## Right-Hand Thumb Rule:

If we grasp the conductor in the palm of the right hand so that the thumb points in the direction of the flow of current, then the direction in which the fingers curl, gives the direction of magnetic field lines. For the current flowing through the conductor in the direction shown in fig. (a) or (b), both the rules predict that magnetic field lines will be in an anticlockwise direction when seen from above.

The magnetic field produced by a current-carrying straight conductor is of circular symmetry. The magnetic lines of force are concentric circles with the current-carrying conductor passing through their common center. The plane of the magnetic lines of force is perpendicular to the length of the conductor.

Biot Savart’s Law

##### Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Magnetic Effect of Electric Current Class 12 Notes | Introduction
The electricity and magnetism are linked to each other and it is proved when the electric current passes through the copper wire, it produces a magnetic effect. The electromagnetic effects first time noticed by Hans Christian Oersted. Oersted discovered a magnetic field around a conductor carrying an electric current. The magnetic field is a quantity, which has both magnitude and direction. The direction of a magnetic field is usually taken to be the direction in which, a north pole of the compass needle moves inside it. So here you will get Magnetic Effect of Electric Current Class 12 complete Notes to prepare for Boards as well as for JEE & NEET Exams. Oersted discovered a magnetic field around a conductor carrying an electric current. Other related facts are as follows:
(a) A magnet at rest produces a magnetic field around it while an electric charge at rest produces an electric field around it.
(b) A current-carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving with a uniform velocity has an electric as well as a magnetic field around it.
(c) An electric field cannot be produced without a charge whereas a magnetic field can be produced without a magnet.
(d) No poles are produced in a coil carrying current but such a coil shows north and south polarities.
(e) All oscillating or an accelerated charge produces E.M. waves also in addition to electric and magnetic fields.

### Unit of Magnetic field

UNIT OF $\overrightarrow{\mathrm{B}}:$ MKS weber/metre $^{2},$ SI tesla, CGS maxwell cm’ or gauss.

One Tesla $=$ one (weber/m’) $=10^{4}$ (maxwell/cm’) $=10^{4}$ gauss

## Biot-Savart’s Law

With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field $\overrightarrow{\mathrm{d} B}$ at a point $P$ associated with a length element $\overrightarrow{\mathrm{d} \ell}$ of a wire carrying a steady current I.

$\mu_{0}$ is called permeability of free space $\frac{\mu_{0}}{4 \pi}=10^{-7}$ henry/meter.

$1(\mathrm{H} / \mathrm{m})=1 \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}=1 \frac{\mathrm{Wb}}{\mathrm{Am}}=1 \frac{\mathrm{N}}{\mathrm{A}^{2}}=1 \frac{\mathrm{Ns}^{2}}{\mathrm{c}^{2}}$

DIMENSIONS of $\mu_{0}=\left[\mathrm{M}^{\prime} \mathrm{L}^{\prime} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$

For vaccum $: \sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}=\mathrm{c}=3 \times 10^{8} \mathrm{m} / \mathrm{s}$

### Biot-Savart law in Vector form

[Note: Static charge is a source of electric field but not of magnetic field, whereas the moving charge is a source of electric field as well as magnetic field.]

the direction of $\mathrm{d} \mathrm{B}$ is perpendicular to the plane determined by $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ (i.e. if $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ lie in the plane of the paper then $\overrightarrow{\mathrm{dB}}$ is $\perp$ to plane of the paper). In the figure, direction of

$\overrightarrow{\mathrm{dB}}$ is into the page. (Use right hand screw rule).

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Difference Between Potentiometer and Voltmeter – Current Electricity || Class 12, JEE & NEET

The Potentiometer and the Voltmeter both are the voltage measuring devices. The main Difference Between Potentiometer and Voltmeteris that the potentiometer measures the emf of the circuit whereas voltmeter measures the end terminal voltage of the circuit. The other differences between the Potentiometer and the voltmeter are explained below in the comparison chart.

## Standardization of Potentiometer

The process of determination of potential gradient on wire of potentiometer is known as standardisation of potentiometer. A standard cell is one whose emf remains constant. Cadmium cell with emf 1.0186 V at $20^{\circ}$C is used as a standard cell. In laboratory a Daniel cell with emf 1.08 V is usually used as a standard cell.
If $\ell_{0}$ is the balancing length for standard emf $E_{0}$ then potential gradient $x=\frac{E_{0}}{l_{0}}$

### Key Differences:

The following are the key differences between Potentiometer and voltmeter.
1. The Potentiometer is an instrument used for measuring the emf, whereas the voltmeter is a type of meter which measures the terminal voltage of the circuit.
2. The Potentiometer accurately measures the potential difference because of zero internal resistance. Whereas, the voltmeter has a high internal resistance which causes the error in measurement. Thus the voltmeter approximately measures the voltage.
3. The sensitivity of the Potentiometer is very high, i.e. it can measure small potential differences between the two points. The voltmeter has low sensitivity.
4. The Potentiometer uses the null deflection type instrument whereas the voltmeter uses the deflection type instrument.
5. The Potentiometer has infinite internal resistance, whereas the Potentiometer has high measurable resistance.
Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

### Conclusion

The Potentiometer and voltmeter both measures the emf in volts. The Potentiometer is used in a circuit where the accurate value of voltage is required. For approximate calculation, the voltmeter is used.   Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

JEE Main 2020: NTA reopens Application Process | Apply before 24th May

JEE Main 2020 Application Form – NTA has again released JEE Main application form 2020. The forms are available online on JEE Main Official Website. Fresh registered are also open for eligible 10+2 appeared / passed candidates, until May 24, 2020. The process to submit application form of JEE Main 2020 is the same as before.

Candidate need to click on the apply link below and register by entering name, e-mail id, etc. information. Apart from the details, candidates also have to upload images and pay application fee. The amount of JEE Main 2020 application fee varies as per the category and number of papers.

JEE Main application form correction reopens between May 25 to 31. Know more details on JEE Main 2020 application form below.

#### DIRECT LINK TO REGISTER IN JEE MAIN 2020

The application process of JEE Main 2020 includes registration, filling the application form, uploading scanned images, payment of application fee and taking the printout of the filled-in form. For JEE Main 2020 registration, candidates must check the eligibility criteria as prescribed by the NTA. Candidates who have missed the JEE Main 2020 January exam can appear for the JEE Main 2020 2nd session.

#### About JEE Main 2020 Exam:

Joint Entrance Examination is organized by National Testing Agency to offer admission into courses like B.Tech / B.E., B.Arch, B.Plan. JEE Main is a computer based test that takes place at national level. The duration of the exam is 3 hours. It consists of multiple-choice questions and numerical value type questions. There is negative marking in the case of multiple choice questions. After qualifying this exam, the candidates can take admission in IIITs / CFTIs / NITs or can sit for JEE Advance exam.

#### IMPORTANT DATES AND EVENTS TO REMEMBER

• The JEE Main Exam 2020 will be held across the country from July 18 to July 23 this year.
• The application form process for JEE-Mains 2020 will be available on or before May 24, 2020.
• The forms can be completed and submitted only up to 5 pm on May 24, whereas the submission of the fees can be done until 11.50 pm on May 24, 2020.
• Candidates have been asked to pay the application fee online via debit card, net banking, UPI, or Paytm.
• Students qualifying in the JEE-Mains exams will appear for the JEE Advanced exam which will be held later

#### List of Documents Required for Filling Application Form

Candidate needs some documents to fill and submit the application form. Check the list of required documents here:

• scanned image of photograph
• scanned image of signature
• a valid e-mail id
• valid and active mobile number
• past academic mark sheets and certificates

The candidates have to upload Photo and Signatue in the application form of JEE Main 2020 and as per the given specifications:

### JEE Main 2020 Exam Pattern

As per the eligibility criteria for, a few changes in the pattern of the question paper(s) and number of question(s) for B.E./B.Tech has been approved by the JEE Apex Board (JAB) for the conduct of JEE (Main)-2020 Examination.

 Paper Subject with Number of Questions Type of Questions TIMING OF THE EXAMINATION(IST) B.tech Mathematics – 25(20+5) Physics – 25(20+5) Chemistry – 25(20+5) 20 Objective Type – Multiple Choice Questions (MCQs) & 5 Questions with answer as numerical value, with equal weightage to Mathematics, Physics & Chemistry 1st Shift: 09:30 a.m. to 12:30 p.m. 2nd Shift: 02:30 p.m. to 05:30 p.m.

### JEE Main 2020 Dates

Application form availability February 7, 2020
Last date to apply May 24, 2020 (reopens)
Last date to upload images and pay application fee May 24, 2020
Correction in Particulars (last date)
May 31, 2020
Exam date 18 July – 23 July ’20
Release date of answer key and recorded responses To be notified later
Release of answer key To be notified later
Declaration of result To be notified later
Announcement of NTA score To be notified later

### Eligibility Criteria

There is no age limit for the candidates to appear in JEE Main 2020 Examination. The candidates who have passed their 12th Examination in 2018, 2019 and appearing in 2020 are eligible to JEE Main Examination 2020.

Those candidates who cleared the class 12 exam in 2017 or before 2017 are not eligible to appear for JEE Main 2020 exam.

#### Educational Qualification

Candidates must have at least 5 subjects in class 12 exam or equivalent exam. Where Math, Physics and Chemistry are the essential subjects.

### JEE Main 2020 Detailed Syllabus

FREE Revision Series For JEE 2020 | Quick Revision Videos

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Mind Map For Hydrocarbons: Alkenes | Class 11, JEE & NEET – Download from here
Get to learn all important points and reactions of Hydrocarbons – Alkene through these mind maps. Download and share with your friends also.
Mind Maps for Rotational Motion: Torque Revision Class XI, 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.

JEE Main 2020 New Dates Announced | Check Here for Details

The exam dates for NTA JEE Main 2020 announced by the Union Minister of Human Resource Development Ramesh Pokhriyal Nishank on tuesday (5th May’20). The JEE (Main) 2020 exam was earlier re-scheduled by the ministry, due to the nationwide lockdown in the wake of coronavirus outbreak. The new date for the JEE (Main) 2020 exams was since then awaited.

According to the announcement, Union Minister of Human Resource Development Ramesh Pokhriyal Nishank said the examination process for the JEE (Main) 2020 will begin from July this year, while the session will begin by August. The JEE Main exam will be conducted from July 18 to July 23, 2020, while the JEE Advanced in August, Ramesh Pokhriyal Nishank said. Through JEE Main entrance exam, students will be able to get admission to BE/B.tech, B.Plan, and B.Arch degree courses at the various IITs (Indian Institute of Information Technology), NITs (National Institute of Technology) and other Centrally Funded Technical Institutions (CFTIs) across India.

### JEE Main 2020 Exam Pattern

As per the eligibility criteria for, a few changes in the pattern of the question paper(s) and number of question(s) for B.E./B.Tech has been approved by the JEE Apex Board (JAB) for the conduct of JEE (Main)-2020 Examination.
 Paper Subject with Number of Questions Type of Questions TIMING OF THE EXAMINATION(IST) B.tech Mathematics – 25(20+5) Physics – 25(20+5) Chemistry – 25(20+5) 20 Objective Type – Multiple Choice Questions (MCQs) & 5 Questions with answer as numerical value, with equal weightage to Mathematics, Physics & Chemistry 1st Shift: 09:30 a.m. to 12:30 p.m. 2nd Shift: 02:30 p.m. to 05:30 p.m.
There is no age limit for the candidates to appear in JEE Main 2020 Examination. The candidates who have passed their 12th Examination in 2018, 2019 and appearing in 2020 are eligible to JEE Main Examination 2020. Those candidates who cleared the class 12 exam in 2017 or before 2017 are not eligible to appear for JEE Main 2020 exam.

### Quick Revision Videos

#### 👉Physics Revision Series by Saransh Gupta Sir (AIR-41)

JEE MAIN 2020 Question Paper PDF Download | All Shifts (7th, 8th & 9th January)
NTA has successfully conducted JEE Main Jan 2020 exam in all the test centers and we understand that students are waiting for JEE MAIN 2020 Question Paper PDF format. Based on the reviews by the students, it is concluded that the exam is comparatively easy than the previous year. Furthermore, the questions asked in the exam were more conceptual based. So Question Papers PDF of all shifts are available here to download. JEE Main 2020 Question Paper – Candidates can download JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 from this page.

### JEE Main 2020 (January) – All Shifts Question Papers

The candidates can challenge the answer keys online. The window to raise objections will be available for a week. In case, the NTA finds the objections raised are incorrect, they will publish the result of JEE Main. The tentative result declaration date is by January 31.

### Candidates may check their JEE Main question papers from Here!!!

JEE Main 2020 Exam | January Attempt | Students’ Reactions and Reviews

## JEE MAIN 2020 Students Reactions | 7th January (SHIFT-1)

JEE MAIN, one of the biggest examination in the country was scheduled today. The difficulty level and selection ratio are the well-known aspects of this exam but the number of candidates that appear every year, is also the factor that makes it so special. Every year more than ten lakhs of aspirants apply and compete for just only 30-35 thousand seats (approximately) of National Institute of Technologies and IIITs. Today on 7th January 2020, it was the day, when JEE Main 2020 Exam was held in the morning shift from 9:30-12:30 across India. Many of you are also appeared or going to appear in near future and want to know the overall level of today’s exam. So, dedicated to students and in order to help them, eSaral is again here, providing you the details about today’s exam. In this video, our team members have gathered the information about the questions, paper pattern and the difficulty level of the exam. We went to some of the centers in KOTA city, yes!! The coaching city. Get to know about the students reviews and reactions regarding JEE Main 2020 Exam:
So, watch the video to know how they feel about the exam. We asked them regarding the easy to tough subjects based on the questions and topics, also the marking scheme of the numerical type questions and the ratio of 11th and 12th standards topics in the exam. So if you have these kind of doubts or want to know about the detail watch the video and based on the conversation with the aspirants we developed a PDF too. You can download the same to know in detail. Here is the complete analysis of JEE Main 2020, 7th Jan SHIFT-1 Exam by Saransh Gupta Sir with Students who have appeared in Exam.

## JEE MAIN 2020 Students Reactions | 7th January (SHIFT-2)

So, here are the students reviews on the Second shift of 7th January JEE Main Exam. As per the students reviews it can be concluded that both shifts of the day comprised of easy to moderate level questions. Watch out the video and comment your review of JEE Main Exam

## JEE MAIN 2020 Students Reactions | 8th January (SHIFT-2)

JEE Main 2020 Paper Analysis | Discussion with Students

## JEE MAIN 2020 Paper Analysis for 7th January SHIFT-1

Today on 7th January, the national level competition exam JEE MAIN was held across the nation and with the completion of the exam, there are many queries in the mind of aspirants. It is natural to feel a certain kind of anxiety after the exam like how others feel, the difficulty level of the exam and the expected cut-offs. Here, at eSaral we are providing you the first detailed analysis that is available to all the students. Watch the video to now about the questions’ level (according to eSaral students) and the topics wise questions from 11th and 12th standards. The physics faculty at eSaral, Saransh Gupta sir and chemistry faculty Prateek Gupta sir are doing the detailed analysis of the JEE MAIN QUESTION PAPER as per the reviews by the students of eSaral. In the video, the questions from each subject and the topics from where they were asked are explained using prepometer, the tool designed by the eSaral. In the discussion session the students of eSaral told about the questions and difficulty level and the question that they faced in the examination. JEE Main 2020 Paper Analysis for January 07th exam is updated here. Students are sharing their reviews of JEE Main Exam here. Watch out the complete video till the end to know in detail!

#### Watch the Reactions and Reviews of Students outside JEE Main Exam Center

In physics, it is found that many of the questions were direct and formula based related to the memory of the candidate. There were approximately 10 questions form the class 11th syllabus out of 25 questions. Watch out the video if you want to know about the same kind of detailed analysis for chemistry and mathematics. You will also get some idea about specific questions and their solutions that were discussed in the analysis video. So from our analysis we found the overall exam as easy to moderate level. Only a few questions were placed in the difficult to very difficult category and it was found that the question were not much confusing. We have developed a PDF regarding the detailed analysis. Download the file and know about the exam. If you think that this time you could not attempt your best attempt then don’t worry we are going to start our BOUNCE BACK CRASH COURSE for the April month JEE MAIN exam. Enroll and ace the exam. The above PDF contains weightage of number of questions per chapter.

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## JEE MAIN 2020 Paper Analysis for 9th January SHIFT-2

(Shift-2) 9th Jan Download PDF   JEE Main 2020 Question Paper JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 are given below. Download or View from here!!!

### 👉 Click to Join Free Physics Revision Series by Saransh Gupta Sir (AIR-41, IIT-Bombay)

Mind Maps for Modern Physics: Nuclei Revision – Class XII, JEE, NEET
Nuclear Physics in Class 12 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

Mind Maps for Atomic Structure Revision – Class XII, JEE, NEET
Atomic Structure in Class 12 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

Wave Optics – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.     Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions    Download eSaral app  for free study material and video tutorials. Simulator   Previous Years JEE Advanced Questions
Q. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$. In each of these cases $S_{1} P_{0}$ = $S_{2} P_{0}$, $S_{1} P_{1}$$S_{2} P_{1} = \lambda / 4 and S_{1} P_{2}$$S_{2} P_{2}=\lambda / 3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta$ (P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation. [IIT-JEE-2009]

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Sol. ((A) $p, s ;(B) q ;(C) t ;(D) r, s, t$) (A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$ $\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$ $\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$ $\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$ $\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$ $\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$ $\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$ $\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$ $\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$ $(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$ $\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$ $\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$ $8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$ $8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$ $\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$ $\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$ $=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$ $8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$ $\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$

Q. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_{G}, \beta_{R}$ and $\beta_{B},$ respectively. Then (A) $\beta_{G}>\beta_{B}>\beta_{R}$ (B) $\beta_{B}>\beta_{G}>\beta_{R}$ (C) $\beta_{R}>\beta_{B}>\beta_{G}$ (D) $\beta_{R}>\beta_{G}>\beta_{B}$ [IIT-JEE-2012]

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Sol. (D) $\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$ $\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$

Q. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is :- (A) $(2 n+1) \frac{\lambda}{2}$ (B) $(2 n+1) \frac{\lambda}{4}$ (C) $(2 n+1) \frac{\lambda}{8}$ $(D)(2 n+1) \frac{\lambda}{16}$ [JEE Advanced 2013]

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Sol. (B) $\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$ $\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$ $\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$ $\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$ $\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$

Q. A light source, which emits two wavelengths $\lambda_{1}=400 \mathrm{nm}$ and $\lambda_{2}=600 \mathrm{nm},$ is used in a Young’s double slit experiment. If recorded fringe widths for $\lambda_{1}$ and $\lambda_{2}$ are $\beta_{1}$ and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ respectively, then :- (A) $\beta_{2}>\beta_{1}$ (B) $\mathrm{m}_{1}>\mathrm{m}_{2}$ (C) From the central maximum, $3^{\mathrm{rd}}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$ (D) The angular separation of fringes of $\lambda_{1}$ is greater than $\lambda_{2}$ [JEE Advanced 2014]

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Sol. (A,B,C) $\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$ $\mathrm{B}_{2}>\beta_{1}$ $\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$ $\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$

Q. A Young’s double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $=4 / 3$ ) as shown in the figure. The positions of maxima on the surface of water are given by $x^{2}=p^{2} m^{2} \lambda^{2}-d^{2},$ where $\lambda$ is the wavelength of light in air (refractive index $=1$, $2 d$ is the separation between the slits and $m$ is an integer. The value of p is. [JEE Advanced 2015]

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Sol. 3

Q. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of $\mathrm{S}_{1} \mathrm{S}_{2}$, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining $\mathrm{S}_{1} \mathrm{S}_{2}$. Which of the following is (are) true of the intensity pattern on the screen ? (A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction (B) Semi circular bright and dark bands centered at point O (C) The region very close to the point O will be dark (D) Straight bright and dark bands parallel to the x-axis [JEE-Mains 2016]

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Sol. (B,C) Path difference at point O = d = .6003 mm = 600300 nm $=\frac{2001}{2}(600 \mathrm{nm})=1000 \lambda+\frac{\lambda}{2}$ $\Rightarrow$ minima form at point $\mathrm{O}$ Line $S_{1} S_{2}$ and screen are $\perp$ to each other so fringe pattern is circular (semi-circular because only half of screen is available)

Q. Two coherent monochromatic  point sources $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of wavelength $\lambda$ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct ? (A) A dark spot will be formed at the point $\mathrm{P}_{2}$ (B) The angular separation between two consecutive bright spots decreases as we move from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ along the first quadrant (C) At $\mathrm{P}_{2}$ the order of the fringe will be maximum (D) The total number of fringes produced between $P_{1}$ and $\mathrm{P}_{2}$ in the first quadrant is close to 3000 [JEE Advanced 2017]

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Sol. (C,D)

Liquid Solution – JEE Main Previous Year Question of with Solutions
JEE Main Previous Year Question of Chemistry with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Chemistry will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. Get detailed Class 11th &12th Chemistry Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the folloowing statements is correct regarding the behaviour of the solution ? (1) The solution is non-ideal, showing –ve deviation from Raoult’s law (2) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law (3) The solution formed is an ideal solution. (4) The solutionis non-ideal, showing +ve deviation from Raoult’s law [AIEEE-2009]

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Sol. (4) (A) n–heptone : Non Polar (B) Ethanol : Polar $\mathrm{F}_{\mathrm{A}-\mathrm{B}}<\mathrm{F}_{\mathrm{A}-\mathrm{A}}, \mathrm{F}_{\mathrm{B}-\mathrm{B}} \Rightarrow+$ deviation

Q. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively :- (1) 400 and 600 (2) 500 and 600 (3) 200 and 300 (4) 300 and 400 [AIEEE-2009]

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Sol. (1) $550=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{3}{4}$ $560=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{5}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{4}{5}$ $\mathrm{P}_{\mathrm{A}}^{\circ}=400 \quad \mathrm{P}_{\mathrm{B}}^{\circ}=600$ torr

Q. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 of heptane and 35 g of octane will be (molar mass of heptane = 100 g $\mathrm{mol}^{-1}$ and of octane = 114 g $\mathrm{mol}^{-1}$) :- (1) 144.5 kPa           (2) 72.0 kPa             (3) 36.1 kPa            (4) 96.2 kPa [AIEEE-2010]

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Sol. (2) $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}=105 \times \frac{1 / 4}{1 / 4+0.307}$ $=105 \times 0.449=47.13 \mathrm{K} \mathrm{Pa}$ $\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}=45 \times 0.551=24.795$ $\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=71.925 \mathrm{atm}$

Q. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta \mathrm{T}_{\mathrm{f}}\right)$, when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right):$ :- (1) 0.0186 K            (2) 0.0372 K              (3) 0.0558 K              (4) 0.0744 K [AIEEE-2010]

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Sol. (3) $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}$ $=3 \times 1.86 \times 0.01 / 1$ $=0.0558 \mathrm{K}$

Q. The molality of a urea solution in which 0.0100g of urea, $\left.\left[\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]$ is added to 0.3000 $\mathrm{dm}^{3}$ of water at STP is :- (1) 0.555 m (2) $5.55 \times 10^{-4} \mathrm{m}$ (3) 33.3 m (4) $3.33 \times 10^{-2} \mathrm{m}$ [AIEEE-2011]

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Sol. (2) $\mathrm{m}=\frac{\mathrm{n}}{\mathrm{W}(\mathrm{kg})}=\frac{0.01 / 60}{0.3 \mathrm{kg}}=5.55 \times 10^{-4} \mathrm{mol} / \mathrm{kg}$

Q. A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is :- (1) 136.2           (2) 171.2           (3) 68.4            (4) 34.2 [AIEEE-2011]

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Sol. (3) $\pi_{\mathrm{c.s}}=\pi_{\mathrm{Unk}}$ $\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{c.s.}} \mathrm{RT}=\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{unk} .} \mathrm{RT}$ $\frac{5 \times 10}{342}=\frac{1 \times 10}{\mathrm{M}}$ M = 68.4 gm/mol

Q. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – $6^{\circ} \mathrm{C}$ will be : $\left(\mathrm{K}_{\mathrm{f}} \text { for water }=1.86 \mathrm{K} \mathrm{kgmol}^{-1}, \text { and molar mass of ethylene glycol }=62 \mathrm{gmol}^{-1}\right)$ (1) 400.00 g             (2) 304.60 g            (3) 804.32 g            (4) 204.30 g [AIEEE-2011]

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Sol. (3) $6=1.86 \times \frac{\mathrm{w} / 62}{4} \Rightarrow \mathrm{w}=800 \mathrm{gm}$

Q. The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :- $(1) \alpha=\frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{i}-1}$ (2) $\alpha=\frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{i}-1}$ (3) $\alpha=\frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$ (4) $\alpha=\frac{\mathrm{i}-1}{\mathrm{x}+\mathrm{y}+1}$ [AIEEE-2011]

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Sol. (3)

Q. $\mathrm{K}_{\mathrm{f}}$ for water is 1.86 K kg $\mathrm{mol}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to –$-2.8^{\circ} \mathrm{C} ?$ (1) 27 g            (2) 72 g            (3) 93 g             (4) 39 g [AIEEE-2012]

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Sol. (3) $2.8=1.86 \times \frac{\mathrm{w} / 62}{1} \Rightarrow \mathrm{w}=93.33 \mathrm{gm}$

Q. A solution containing 0.85 g of $\mathrm{ZnCl}_{2}$ in 125.0 g of water freezes at $-0.23^{\circ} \mathrm{C}$ . The apparent degree of dissociation of the salt is : ($\mathbf{k}_{f}$ for water = 1.86 K kg $\mathrm{mol}^{-1}$, atomic mass ; Zn = 65.3 and Cl = 35.5) (1) 1.36%            (2) 2.47%              (3) 73.5%             (4) 7.35% [Jee (Main)-2012 online]

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Sol. (3) $0.23=(1+2 \alpha) \times 1.86 \times \frac{0.85 / 134.5}{0.125}$ $\alpha=0.735=73.5 \%$

Q. Liquids A and B form an ideal solution. At $30^{\circ}$C, the total vapour pressure of a solution containing 1 mol of A and 2 moles of B is 250 mm Hg. The total vapour pressure becomes 300 mm Hg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are (1) 450, 150 mm Hg (2) 250, 300 mm Hg (3) 125, 150 mm Hg (4) 150, 450 mm Hg [Jee (Main)-2012 online]

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Sol. (1) $250=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{3}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{2}{3}$ $300=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{1}{2}$ $\mathrm{P}_{\mathrm{A}}^{0}=450 \mathrm{mm}$ $\mathrm{P}_{\mathrm{B}}^{0}=150 \mathrm{mm}$

Q. The freezing point of a 1.00 m aqueous solution of HF is found to be $-1.91^{\circ} \mathrm{C}$. The freezing point constant of water, $\mathrm{K}_{\mathrm{f}}$, is 1.86 K kg $\mathrm{mol}^{-1}$. The percentage dissociation of HF at this concentration is (1) 2.7%             (2) 30%            (3) 10%             (4) 5.2% [Jee (Main)-2012 online]

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Sol. (1) $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$ $1.91=(1+\alpha) \times 1.86 \times 1$ $\alpha=0.027$

Q. How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at 268 K ? $\left(\mathrm{K}_{f} \text { for water is } 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$ (1) 899.04 g              (2) 886.02 g            (3) 868.06 g                 (4) 880.07 g [Jee (Main)-2013 online]

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Sol. (2) $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^{0}-\mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$ $273.15-268=1.86 \times \frac{\mathrm{w} / 32}{10}$ $\mathrm{w}=886.02 \mathrm{g}$

Q. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be : (1) 0.137           (2) 0.205            (3) 0.237            (4) 0.435 [Jee (Main)-2013 online]

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Sol. (3) Benzen $\rightarrow 4$ Toluene $\rightarrow B$ y $_{B}=\frac{P_{B}^{0} \times X_{B}}{P_{B}^{0} X_{B}+P_{A}^{0} X_{A}}=\frac{37 \times 0.5}{37 \times 0.5+119 \times 0.5}=0.237$

Q. A molecule M associates in a given solvent according to the equation M  $(\mathrm{M})_{\mathrm{n}}$. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is : (1) 2              (2) 4               (3) 5               (4) 3 [Jee (Main)-2013 online]

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Sol. (1) $\mathrm{M}=\mathrm{M}_{\mathrm{n}}$ $1-0.2 \quad 0.2 / \mathrm{n}$ $0.9=\frac{1-0.2+0.2 / \mathrm{n}}{1}$ $0.9=0.8+\frac{0.2}{\mathrm{n}}$ $0.1=\frac{0.2}{\mathrm{n}}$ $\mathrm{n}=2$

Q. 12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is : (1) 60           (2) 80             (3) 40            (4) 20 [Jee (Main)-2013 online]

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Sol. (4) $\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=0.1=\frac{12 / \mathrm{m}}{108 / 18} \Rightarrow \mathrm{m}=20$

Q. The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be :- (1) 0.875 M           (2) 1.00 M            (3) 1.75 M            (4) 0.975 M [Jee (Main)-2013]

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Sol. (1) $\mathrm{M}_{\mathrm{f}}=\frac{\mathrm{M}_{1} \mathrm{V}_{1}+\mathrm{M}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}+\mathrm{V}_{2}}=0.875 \mathrm{M}$

Q. The observed osmotic pressure for a 0.10 M solution of Fe$\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2}$ at $25^{\circ} \mathrm{C}$ is 10.8 atm. The expected and experimental (observed) values of Van’t Hoff factor (i) will be respectively : $\left(\mathrm{R}=0.082 \mathrm{L} \mathrm{atm} \mathrm{k}^{-} \mathrm{mol}^{-1}\right)$ (1) 3 and 5.42          (2) 5 and 3.42           (3) 4 and 4.00           (4) 5 and 4.42 [Jee (Main)-2014 online]

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Sol. (4) $\pi_{\mathrm{ob}}=\mathrm{i} \frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$ $10.8=\mathrm{i} \times 0.1 \times 0.082 \times 298$ $\mathrm{i}=4.42$

Q. For an ideal Solution of two components A and B, which of the following is true ? (1) $\Delta \mathrm{H}_{\text {mixing }}<0$ (zero) (2) $\mathrm{A}-\mathrm{A}, \mathrm{B}-\mathrm{B}$ and $\mathrm{A}-\mathrm{B}$ interactions are identical (3) $\mathrm{A}-\mathrm{B}$ interaction is stronger than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ interactions (4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero) [Jee(Main)-2014 online]

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Sol. (2) $\Delta \mathrm{H}_{\operatorname{mix}}=0$

Q. Consider separate solution of $0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}), 0.100 \mathrm{MM} \mathrm{g}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{aq}), 0.250 \mathrm{M} \mathrm{KBr}(\mathrm{aq})$ and 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})$ at $25^{\circ} \mathrm{C}$. Which statement is true about these solutions, assuming all salts to be strong electrolytes ? (1) 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}$ (aq) has the highest osmotic pressure. (2) 0.500 M $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (aq) has the highest osmotic pressure. (3) They all have the same osmotic pressure. (4) 0.100 M $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (aq) has the highest osmotic pressure. [Jee (Main)-2014]

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Sol. (3)

Q. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by : (1) association (2) dissociation (3) complex formation (4) partial ionization [Jee (Main)-2015 online]

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Sol. (1) Acetic acid in non polar solvent (benzene) associates.

Q. A solution at $20^{\circ} \mathrm{C}$ is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively : (1) 38.0 torr and 0.589 (2) 30.5 torr and 0.389 (3) 35.8 torr and 0.280 (4) 35.0 torr and 0.480 [Jee (Main)-2015 online]

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Sol. (1) \begin{aligned} \mathrm{P}_{\mathrm{T}} &=\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0} \mathrm{X}_{\mathrm{B}} \\ &=747 \times \frac{1.5}{5}+22.3 \times \frac{3.5}{5} \\ &=38 \mathrm{torr} \end{aligned}

Q. The vapour pressure of acetone at $20^{\circ}$C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100 g of acetone at $20^{\circ}$$20^{\circ}$C, its vapour pressure was 183 torr. The molar mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$ of the substance is : (1) 128 (2) 488 (3) 32 (4) 64 [Jee (Main)-2015]

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Sol. (4) $\begin{array}{rl}{\frac{185-183}{185}} & {=\frac{1.2 / \mathrm{m}}{100 / 58}} \\ {\mathrm{m}=64} & {\mathrm{gm} / \mathrm{mol}}\end{array}$

Q. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ? (1) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$ (2) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}$ (3) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$ (4) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ [Jee (Main)-2018]

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Sol. (3)

JEE MAIN 2020 Admit Card for January Attempt | Download Hall Ticket for JEE Main 2020 Exam

• Application number.
• Password or date of birth.
JEE Main 2020 Admit Card will look like the image given below:

### What to Carry in Examination Hall

On the exam day, candidate have to carry following documents to the exam hall:

1. JEE Main 2020 hall ticket – Print it on an A4 sheet and make sure all the information should be clear and correct.

2. One passport size photograph – Along with the JEE Main admit card, candidates also have to carry one passport size photograph. This should be the same photo as the one uploaded in the form.

3. Valid Original ID Proof – As per information brochure, candidates have to carry one id proof. However, it has to be carried in original and should contain photograph of the candidate.

It is better that candidates carry the same id proof, the details of which were entered in the application form. List of valid id proofs for JEE Main 2020 is as follows:

• PAN card.
• Voter id card.
• Ration card.
• Passport.
• 12th Class admit card with photograph.
• Bank passbook with photograph.

4. PwD Certificate – Candidates who applied for scribe facility have to carry PwD Certificate with the admit card.

### What Not to Carry in the Examination Hall

Following items are not allowed inside the exam hall:

• Text material.
• Calculator.
• Docu pen.
• Log tables.
• Silde rules.
• Electronic watches.
• Mobile phone.
• Paper,
• Metallic objects etc.

Important Instructions – Candidates should note that if they are planning to carry metallic objects such as Kara and Kirpan, etc. should report to the center at least 1 hour 30 minutes before closing of the gate. NTA might ask the candidates to not take it inside the exam hall.

### Exam Time Schedule(Tentative)

 Particulars Shift 1 Shift 2 Entry in the exam hall 7.30 am – 9.00 am 1.00 pm – 2.00 pm Instruction by invigilators 9.00 am – 9.20 am 2.00 pm – 2.20 pm Login and read the instructions 9.20 am 2.20 pm Exam starts at 9.30 am 2.30 pm B.E.. / B.Tech Exam 9.30 am – 12.30 pm 2.30 pm – 5.30 pm B.Arch Exam 9.30 am – 12.30 pm 2.30 pm – 5.30 pm B.Planning Exam – 2.30 pm – 5.30 pm B.Arch & B.Planning Exam (Both) – 2.30 pm – 6.00 pm

JEE Main 2020 Sample Questions released by NTA are available here:

### MATHEMATICS Sample Questions based on Numerical value by NTA

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