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JEE Main Important Formulas 2025 for Physics, Chemistry, Maths

JEE Main Important Formulas 2025 for Physics, Chemistry, Maths

Edited By Team Careers360 | Updated on Dec 14, 2024 09:46 AM IST | #JEE Main

JEE Main Formulas 2025 - JEE Mains is the competitive entrance exams for engineering programs. To do well on this exam, you must understand the fundamental concepts and formulas of Mathematics, Physics, and Chemistry. Aspirants preparing for the JEE Mains can check the JEE Main formulas 2025 available on this page. JEE Mains is Conducted by the National Testing Agency (NTA), this exam evaluates a candidate's proficiency in Physics, Chemistry, and Mathematics. Since there are many formulas in these three subjects, candidates need a way to recall them for revision purposes. JEE Main important formulas in Math, Physics, and Chemistry are integral to calculating answers for numerical questions.

This Story also Contains
  1. JEE Main Formulas 2025 for Physics, Chemistry, Maths
  2. New Exam Pattern
  3. JEE Main Formulas for Physics 2025
  4. JEE Main formulas for Chemistry 2025
  5. JEE Main formulas for Maths 2025
JEE Main Important Formulas 2025 for Physics, Chemistry, Maths
JEE Main Important Formulas 2025 for Physics, Chemistry, Maths

JEE Main Formulas 2025 for Physics, Chemistry, Maths

Candidates must make a handy note of all important formulas to revise frequently. Candidates must have a good command of each topic and the formulas to crack the JEE Main 2025 exam to ace the test. Through this article, students can find the provided JEE Main formulas for Maths, Physics, and Chemistry. Knowing important formulas in depth can help you solve problems fast and accurately, which is important for scoring well in JEE Main.

Background wave

New Exam Pattern

The exam is divided into two main sections:

Section A: The quizzes contained in this section are 20 MCQs for each course. What the MCQ requires is four options and only one of them is the right answer.

Section B: This section has 5 numerical value questions for each of the subject areas, and the candidate only has to answer all five of these. These numerical value questions have to be answered accurately, sometimes to the second decimal place.

JEE Main Formulas for Physics 2025

Aspirants preparing for JEE Mains must remember that along with concepts one needs to revise and remember the formulas, which are very important while solving any problems. As JEE Main Physics formulas are given below, these formulas need to be memorized daily as direct questions and formulas are asked in exams.

Below are a few important formulas for JEE Main Physics.

Newton’s first law of motion

If Fnet =Oanet=O forces in all directions are zero,i.e, Fx=0,Fy=0,Fz=0

Work, Energy, and Power for Rotating Body

Work-
For translation motion W=Fds
So for rotational motion W=τdθ
2. Rotational kinetic energy-

The energy of a body has by virtue of Its rotational motion is called its rotational kinetic energy.


3. Power =Rate of change of kinetic energy

For translation motion P=F,V
So for rotational motion

P=d(KR)dt=d(12Jι2)dt=Iωdωdt=Iαω=τω


Or P=τω

Newton's Law of Gravitation

Fm1m2r2


Or, F=Gm1m2r2

Where
F Force
G Gravitational constant
m1,m2 Masses
r Distance between masses

Gravitational Potential energy at a point


Then gravitational force on test mass m at a distance r from M is given by F=GMmr2
And the amount of work done in bringing a body from to r

=W=rGMmx2dx=GMmr


And this is equal to gravitational potential energy
SoU =GMmr
U gravitational potential energy
M Mass of source-body
m mass of test body
r distance between two
Note- U is always negative in the gravitational field because Force is attractive in nature.
Means As the distance r increases U becomes less negative
l.e U will increase as r increases

And for r=,U=0 which is maximum

Gravitational Potential energy of discrete distribution of masses

U=G[m1m2r12+m2m3r23+]

U Net Gravitational Potential Energy
r12,r23 The distance of masses from each other

Change of potential energy

If a body of mass m is moved 些, r1 to r2

Then Change of potential energy is given as

ΔU=GMm[1r11r2]

ΔU change of energy
r1,r2 distances
If r1>r2 then the change in potential energy of the body will be negative.
Le To decrease potential energy of a body we have to bring that body closer to the earth.

The relation between Potential and Potential energy

 As U=GMmr=m[GMr] But V=GMr So U=mV


Where V Potential
U Potential energy
r distance

The gravitational potential energy at height 'h' from the earth's surface

Uh=GMmR+h


Using GM=gR2

Uh=gR2mR+hUh=mgR1+hR

Uh The potential energy at the height h
R Radius of earth

Power in AM waves

If Vrms is root mean square value
and R= Resistance
then Power dissipated in any circuit.

P=Vrms2R


So Carrier Power will be given as

Pc=(Ec2)2R=Ec22R

Ec= The amplitude of the carrier wave
R= Resistance
Similarly, Total Power of sidebands will be given as

Pst=(m0Ec22)2R+(mnEc22)2R=ma2Ec24R


Where

ma= modulation index 

Ec= the amplitude of carrier waves

R= resistance 


And this gives Total power of AM wave as

Ptatal =Pc+Psb=Ec22R(1+ma22)

where
ma= modulation index
Ec= the amplitude of carrier waves
R= Resistance
Note-maximum power in the AM wave without distortion Occurs when ma=1
Le Pt=1.5P=3Ps

Frequency modulation

Frequency modulation
- Frequency modulation deviation-The The amount by which carrier frequency is varied from its unmodulated value.

The deviation is made proportional to the instantaneous value of the modulating voitage.
- Value of frequency deviation =δ^=fc

dWax =fmax fc=±KEm

Ex0= modulating amplitude
- The modulation index of frequency modulation-

It is defined as the ratio of maximum frequency deviation to the modulating frequency.

mf=δmuxxfm=±KEmfm

JEE Main formulas for Chemistry 2025

Candidates while studying the chemistry, they need to revise and practice the chemical equations and symbols, to some chemistry is tough subject but when candidates practices chemical equations, revises the properties, formulas and symbols they will have command over the subject Candidates can check the JEE Main Chemistry formulas below

Physical Properties of Alkali Metals

1701851766948

Molar Conductance

1701851768392

Equivalent Conductance

 Equivalent conductance = Molar conductance x where x= Molecular mass  Equivalent mass =n factor 

Method of Preparation of Carboxylic Acid

1701851769429

Kinetic Energy

If at a given temperature, n3 molecules have speed u1,n2, malecules have speed u2,n3 molecules have speed u3 and so an Then, the total kinetic energy ( EK ) of the gas at this temperature is given ly:

EK=12m( m1vS2+n2v22+nsvj2+)

where m is the mass of the molecule. The corresponding average kinetic energy Ek of the gas will be:

EK=12m1(n1v12+n2v22+n3v22+..)(n1+n2+n3+.)


If the verm{n1vi2+n2v23+njvj2+.)(n1+n2+n3+.)=v2
then the average kinetic energy is given by :

EK=12 mv2

where v is ghen by

v=(n1v12+n2v22+n1v32+)(n1+n2+n5+)


This 'v' is known as root-mean-square speed umins 

Molarity(M): No. of moles of solute/ volume of solution in liter

Molality (m): No. of moles of solute/weight of solvent in kg

JEE Main formulas for Maths 2025

Candidates must go through all the formulas and practice the mathematical problems. Without formulas you cannot solve any problem though you know how to solve it. Revising the formulas daily is very important. Here we have provided Mathematics formulas for JEE Mains.

Equation of Circle

  • Centre-Radius Form

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Let P(x,y) be any point on the circle. Then, by definition, |CP|=r.
Using the distance formula, we have

(xh)2+(yk)2=r

i.e. (xh)2+(yk)2=r2

If the centre of the circle is the origin or (0,0) then equation of the circle becomes (x0)2+(y0)2=r2
i.e. x2+y2=r2

- General equation of a circle

The equation of a circle with centre at ( h,k ) and radius r is

(xh)2+(yk)2=r2x2+y22hx2ky+h2+k2r2=0


Which is of the form :

x2+y2+2gx+2fy+c=0


This is known as the general equation of the circle.

Line and Circle
S is a circle with center O and radius r , and L is a straight line in the plane of the circle.
Equation of circle S:x2+y2=a2
Equation of line L: y=mx+c
To find their point(s) of intersection, we can solve these equations simultaneously x2+(mx+c)2=a2
(1+m2)x2+2mcx+c2a2=0


Exponential Limits

(i) limx0ax1x=logea

Proof:

limx0ax1x=limx0(1+x(loga)1!+x2(loga)22!+)1x

[using Taylor series expansion of ax ]

=limx0(loga1!+x(loga)22!+)=logea

(ii) limx0ex1x=1

In General, if limxaf(x)=0, then we have
(a) limxaaf(x)1f(x)=logea
(b) limxaef(x)1f(x)=logee=1

Logarithmic Limits

To evaluate the Logarithmic limit we use the following results:

limx0loge(1+x)x=1


Proof:

limx0loge(1+x)x=limx0xx22+x33x

[using Taylor series expansion of loge(1+x) ]

=limx0(1x2+x23)=1


In General, if limxaf(x)=0, then we have limxaloge(1+f(x))f(x)=1

Frequently Asked Questions (FAQs)

1. What is the formula of molecular mass in terms of vapor density?

The formula of molecular mass in terms of vapor density is 

Molecular mass = 2 * vapor density 

2. How do I remember all the formulas for JEE Main?

Revision is the best way to remember all the formulas. Practice more questions based on formulas and revise the formulas on a daily basis.

3. Can I derive the formula during the exam?

Yes, you can derive the formula during the exam but it is very time-consuming so candidates must learn all the formulas to save time during the exam.

4. What is the general formula for alkanes, alkenes, and alkynes respectively?

General  formula for alkanes is CnH2n+2 , alkenes is CnH2n and  for alkynes is C

nH2n-2 respectively.

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Questions related to JEE Main

Have a question related to JEE Main ?

HEY THERE!!

With 967 IPE marks and 90 percentile in JEE Mains, you have a very high probability of getting a CSE core seat in SASTRA University through the usual merit-based admission process.

Admission Process at SASTRA

  1. Stream 1 (70% seats): It is conducted on a combined score: 75% weightage to your Class 12 (IPE) marks and 25% weightage to your JEE Main score.
  2. Stream 2 (30% seats): Only on Class 12 (IPE) marks.

Your Chances

  • Recent times student reports and expert answers indicate that above 900 IPE marks and a good JEE Main percentile (generally above 85–90) are generally sufficient to obtain a CSE seat in SASTRA.
  • SASTRA's CSE cutoff is likely to be tough, but your 967/1000 in IPE and 90 percentile in JEE Main are strong for the CSE core, especially in Stream 1.
  • Last year's cutoff every year is based on the set of candidates, but your profile is on or above the average admitted student profile for CSE.

Conclusion:

With your marks, you have a strong possibility of securing a CSE core seat at SASTRA University, though allotment will finally be made depending on the competition that year and the availability of seats. Attempt to apply through both streams to maximize chances.

Dear candidate,

You have 93 percentile in AEEE and 92.9 percentile in JEE Main, so there is a strong chance of getting admission in several branches such as :

  1. ECE
  2. EEE
  3. CSE in Amrita Vishwa Vidyapeetham, especially under Group B (higher fee).

Admission for CSE at coimbatore is competitive but there is a chance if you are from home state.

All the best!


HEY THERE!!!

On your marks-56% in JEE Main and 88% in CBSE Board (Tamil Nadu)-you are eligible to apply for the B.Tech. in Electronics Engineering (VLSI Design and Technology) at SASTRA University, Thanjavur, but a seat for the course will be allotted based on the combined merit and competition.

VLSI Course Prospects

  1. Admission is merit-based; there is no cut-off marks-it depends on the pool of candidates each year.
  2. For competitive streams like VLSI, the combined marks in boards are typically higher.
  3. With 88% boards, you have a good opportunity, especially through Stream 2 (on the basis of marks in boards only).
  4. In case you have a poor JEE percentile, you have a better chance through Stream 2.

Summary

You are eligible and stand a good chance for VLSI course in SASTRA Thanjavur, especially under Stream 2 with your good board marks. Admission is not guaranteed but depends on the merit list and competition annually. Apply early and monitor the official releases for counselling and rank lists.

HEY THERE!!!

With 978/1000 in Class 12th (98%) and 80.9 percentile in JEE Mains, you have a very high chance of getting Computer Science Engineering (CSE) with specialization at SASTRA University, Thanjavur.

How SASTRA Admission Works

  1. Stream 1: 70% of seats are filled based on a combined merit score (75% weightage to Class 12 marks + 25% weightage to JEE Main percentile).
  2. Stream 2: 30% of seats are filled based solely on Class 12 marks

Your Chances

  • Your Class 12 mark (978/1000) is good and will place you high in the merit list.
  • Your JEE Main percentile (80.9) is average and combined with your excellent board marks puts your overall merit score in competition.
  • Recent trends suggest that 95%+ board students and 75–85 percentile students in JEE Main have achieved CSE and its specializations at SASTRA, especially during the later rounds or through Stream 2

Conclusion

You are likely to receive CSE with specialization from SASTRA University through Stream 1 or Stream 2 because of your good board marks and average JEE Main percentile. Final allotment depends on competition and availability of seats, but your profile is very competitive to receive CSE and its specializations in SASTRA.

Hii,

A few key documents are required for JEE. Your admit card, which grants you entry to the exam room, comes first. To prove who you are, you also need identification, such as your school ID or Aadhar card. To prove your age, your birth certificate is required.

Your class 12 grade report is required as documentation of your education. A category certificate is necessary if you fall under one of the reserved categories, such as SC, ST, or OBC. Keep these with you on test day.

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