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BITSAT 2025 Formulas for Physics, Chemistry, Maths
  • BITSAT Exam
  • BITSAT 2025 Formulas for Physics, Chemistry, Maths

BITSAT 2025 Formulas for Physics, Chemistry, Maths

#BITSAT
Team Careers360Updated on 11 Aug 2025, 01:01 AM IST

BITSAT Formulas for PCM: BITSAT is a crucial exam for every student who wishes to join the top private engineering college in India - BITS Pilani. It is considered a moderate to difficult exam, inclined more on the difficult side. To properly prepare for the BITS Admission Test (BITSAT), one needs to understand concepts thoroughly along with having a good command of formulas. The exam encompasses three significant subjects - Physics, Chemistry, and Mathematics (PCM) - each one overflowing with key formulas you must remember and use promptly. Within this article, we’ll simplify a few of the BITSAT formulas for Physics, Chemistry, and Maths to help you optimize your study efforts for the BITSAT 2026 exam. Let's dive in!

This Story also Contains

  1. Benefits of Knowing BITSAT Formulas for PCM
  2. BITSAT 2025 Formulas for Physics
  3. BITSAT 2025 Formulas for Chemistry
  4. BITSAT 2025 Formulas for Maths
BITSAT 2025 Formulas for Physics, Chemistry, Maths
BITSAT 2025 Formulas for Physics, Chemistry, Maths

Benefits of Knowing BITSAT Formulas for PCM

The BITSAT examination is famed for its fast and rapid format. During the 180-minute exam, you have to answer 150 multiple-choice questions, giving you very little time to calculate formulas during the test. Remembering major formulas can cut down on precious time, raise accuracy, and help you compete effectively. There exists a unique collection of equations for each subject that allows for fast problem resolution, allowing you to concentrate more on understanding their application. We should analyse the important formulas for each area.

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BITSAT 2025 Formulas for Physics

Physics in BITSAT is heavy on numerical problems, requiring a good grasp of formulas. Understanding these formulas will help you solve problems on topics like Mechanics, Electricity, Magnetism, and Optics. Here are some important BITSAT Physics formulas:

Newton’s Second Law:
Newton’s second law of motion:-

  • It states that the acceleration of the particle measured from an inertial frame is given by the (vector) sum of all the forces acting on the particle divided by its mass (only when mass is constant), i.e.,

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\vec{a}=\frac{\vec{F}}{m}\Rightarrow \vec{F}=m\vec{a}

2.To cross river in the shortest path

condition ( velocity of boat along river flow must be zero)

t= \frac{d}{\sqrt{v^{2}-u^{2}}}

d= width of river

v = Speed of Boat w.r.t. river

u = speed of river

3. W=FScos Theta


Ohm’s Law:

In a conductor, if all external physical conditions like temperature and pressure are kept constant the Current flowing through a conductor is directly proportional to the Potential difference across two ends.

V\propto I

V=IR

R- Electric Resistance


  • The graph between V and I

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1730807916918

The slope gives the resistance


  • The graph between V and I at different temperatures

1730807917133

Here T1>T2. The resistance of a conductor increases with increase in temperature


Kinematic Equations: For uniform acceleration:

Equation of motions

There are three equations of motion

  1. 1st Equation of motion or velocity time equation

Formula

V=u+at

V = Final velocity

u = Initial velocity

A = acceleration

T = time


  1. 2nd Equation of Position time equation

Formula

s= ut +\frac{1}{2}at^{2}

s\rightarrow Displacement

u\rightarrowInitial velocity

a\rightarrowacceleration

t\rightarrow time


  1. 3rd Equation of Velocity –displacement equation

Formula

V^{2}-u^{2}=2as

V\rightarrow Final Velocity

s\rightarrow Displacement

u\rightarrowInitial velocity

a\rightarrowacceleration

  • Displacement in nth second

  • Formula

S_{n}= u+\frac{a}{2}(2n-1)

Where u= Initial velocity

a= uniform acceleration

n= 1730807914586 second


the Net electric field is -


E_{net} =\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{x Q}{\left(x^{2}+R^{2}\right)^{3 / 2}}


Lens Formula:
frac{1}{v}-frac{1}{u}= frac{1}{f}

(f is the focal length, v is the image distance, and u is the object distance)

BITSAT 2025 Formulas for Chemistry

Chemical formulas are of great significance in BITSAT; most of the numerical problems in Physical Chemistry will either make or mar your performance. Chemistry and both Inorganic and Organic Chemistry in particular will also test you on some principles and reaction mechanisms. Here are some BITSAT Chemistry formulas that will come in handy:

Ideal Gas Equation:

PV=nRT


Pressure × Volume equals to number of moles × Gas constant × Temperature

Graham’s Law of Diffusion
According to it "At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or molecular weight". It is applicable only at low pressure.

\mathrm{r} \propto \frac{1}{\sqrt{\mathrm{M}}} \text { or } \frac{1}{\sqrt{\mathrm{d}}}Here r = rate of diffusion or effusion of a gas or liquid. M and d are the molecular weight and density respectively.

For any two gases, the ratio of the rate of diffusion at constant pressure and temperature can be shown as

\mathrm{r}_{1} / \mathrm{r}_{2}=\sqrt{\mathrm{M}_{2} / \mathrm{M}}_{1} \text { or } \sqrt{\mathrm{d}_{2} / \mathrm{d}_{1}}Hence diffusion or effusion of a gas or gaseous mixture is directly proportional to the pressure difference of the two sides and is inversely proportional to the square root of the gas or mixture effusing or diffusing out.

Some Other Relation Based on Graham’s law
As r = V/t = Volume/time, thus:

\\\mathrm{\frac{V_{1} t_{2}}{V_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\text { As } r=\frac{n}{t}=\frac{d}{t}=\frac{w}{t}}\\\\\mathrm{Thus,\: \frac{n_{1} t_{2}}{n_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\frac{w_{1} t_{2}}{w_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\frac{\mathrm{d}_{1} \mathrm{t}_{2}}{\mathrm{d}_{2} \mathrm{t}_{1}}=\sqrt{\mathrm{M}_{2} / \mathrm{M}_{1}}}


Arrhenius Equation:

We know that the first-order equation is given as follows:

\mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}

But there are other forms of rate law also available that we use for different purposes. These forms are mentioned below:

  • Use to solve numericals:

    \mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}\mathrm{\Rightarrow log_{10}\left [ \frac{A_{o}}{A} \right]\: =\: \frac{kt}{2.303}}\mathrm{Thus, t\: =\: \frac{2.303}{k}\, log_{10}\left [ \frac{A_{o}}{A} \right]}

  • Exponential form:

    \mathrm{log_{e}A\: -\: log_{e}A_{o}\: -kt}\\\mathrm{\Rightarrow log\frac{A}{A_{o}}\: =\: -kt}\\\\\mathrm{\Rightarrow \frac{A}{A_{o}}\: =\: e^{-kt}}\\\\\mathrm{Thus,\: A\: =\: A_{o}e^{-kt}}This equation is also known as exponential form.

  • The rate constant is shown by this formula; Rate constant = Frequency factor × Exponential factor related to activation energy

Boyle’s Law:

1730807916666

k1 is the proportionality constant whose value depends upon the following factors.

1730807915415

Pressure and volume are direct and opposite ratio with temperature remaining constant, that is pressure is inversely proportional to volume of the gas.

Molarity Formula:

Molarity (M) is defined as the number of moles of solute dissolved in one litre (or one cubic decimetre) of solution,

\text { Molarity }=\frac{\text { Moles of solute }}{\text { Volume of solution in litre }}

First Law of Thermodynamics:

Heat supplied = Work done by the system + Increase in internal energy

So increase in internal energy = Heat supplied - work done by the system

\textrm{ie. } \Delta =q+w \ \ \ \ \ \ \ [ \because \textup{ work done by the system is -w}]

Nernst Equation:

At T = 298K, the Nernst equation is given as follows:

\mathrm{E_{cell}\: =\: E^{o}_{cell}\: -\: \frac{0.059}{n}log_{10}Q}where n is the number of electrons exchanged.

Thus the Nernst equation for the full cell is given as follows:

\mathrm{E_{cell}\: =\:(P-Q)-\: \frac{0.059}{x}log_{10}\frac{c_{1}}{c_{2}}}

It goes without saying that it is possible to solve any Chemistry question with the help of these basic number schemas rather quickly.

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BITSAT 2025 Formulas for Maths

Mathematics in BITSAT is more of calculation than application where time plays a key role. There are several mathematical formulas that can be very helpful in solving maths problems and mastering the following BITSAT Maths formulas will go along way in helping you solve problems much simpler.

Quadratic equation:

A polynomial equation in which the highest degree of a variable term is 2 is called quadratic equation.

Standard form of quadratic equation is ax2 + bx + c = 0

Where a, b and c are constants (they may be real or imaginary) and called the coefficients of the equation and a\neq0 (a is also called the leading coefficient).

Eg, -5x2 - 3x + 2 = 0, x2 = 0, (1 + i)x2 - 3x + 2i = 0

As degree of quadratic polynomial is 2, so it always has 2 roots (number of real roots + number of imaginary roots = 2)

The root of the quadratic equation is given by the formula:

\\\mathrm{x = \frac{-b \pm \sqrt{D}}{2a}}\\\\\mathrm{or} \\\mathrm{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}

Where D is called the discriminant of the quadratic equation, given by D = b^2 - 4ac ,

Sine Rule

In any \DeltaABC, the side are proportional to sines of the opposite angle

\\\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\a,b,c\;\;are\;sides\;of\;triangle\;ABC


Differential equation:

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable

\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

Integration Formulas:

It is inverse process of differentation.

\frac{d}{dx}\left \{ F(x) \right \}= f(x)


Probability Formula:

P(A)=Number of favorable outcomesTotal number of outcomesP(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}P(A)=Total number of outcomesNumber of favorable outcomes

Group frequency distribution

\dpi{100} Interval\: width= \frac{x_{i}-x_{s+1}}{n}

Permutations and Combinations:

Permutation basically means the arrangement of things. And when we talk about arrangement then the order becomes important if the things to be arranged are different from each other (when things to be arranged are the same then order doesn’t have any role to play). So in permutations order of objects becomes important.

Arranging n objects in r places (Same as arranging n objects taken r at a time) is equivalent to filling r places from n things.

1730807916477

So the number of ways of arranging n objects taken r at a time = n(n - 1) (n - 2) ... (n - r + 1)

\\\mathrm{\frac{n(n-1)(n-2)...(n-r+1)(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!} = ^{n}P_{\;r}}

Where r \leq n\text{ and }r \in W

So, the number of ways arranging n different objects taken all at a time = ^{n}P_{\;n} = n!.

Area under a Curve:

A=∫abf(x)dxA = \int_a^b f(x) dxA=∫abf(x)dx

Tips to All Students:

In This Guide, You Will Come Across Various Formulas That Will Help You Pass in BITSAT

Daily Practice: It should become a discipline, however, to revise these formulas daily for at least 15-20 minutes, anyway. If you are going to utilize these questions in the test, then the more you drill, the more facile it will be for you to conjure it in your mind.

Create Formula Sheets: Pre Almanac Handout Also need to make the Rubik’s cube template a habanero: Give out a formula sheet according to the person and topic. You may set it aside and keep referring to it for final moment reflection.

Mock Tests: These formulas should be used any time one is handling mock tests. This will help you employ them as effectively in a time-bound manner as it is in a bodily exam such as the BITSAT. For students preparing for BITSAT, Careers360 conducts mocks tests.

Frequently Asked Questions (FAQs)

Q: Formulae related to Chemistry in which is important for BITSAT exam.
A:

For BITSAT, it is particularly useful to review Physical Chemistry equations from the Ideal Gas Law, through the application of the Arrhenius equation, the Nernst equation, up to the Faraday’s laws of electrolysis. Classes also help in getting the knowledge of some of the major principles concepts of Thermodynamics and Electrochemistry.

Q: Which formulas are important for solutions in BITSAT Physics?
A:

For the BITSAT examination, the important formulas in Physics are Newton’s laws of motion, kinematics equations Ohm’s law, Coulomb’s law, work energy formula, magnetic field and lenses formula etc. Closely related important topics repeat during the BITSAT examinations, such as Mechanics, Electrodynamics, and Optics.

Q: What can I do about recording the basic equations required for BITSAT?
A:

Prepare the formula sheet all different for Physics fields, Chemistry fields, and Maths fields. Go through these sheets every day and then try and solve tests and past years question papers appear to be solved using these formulas. Returning to the act of writing them down and solving multiple problems also increases the storage of these items in memory.

Q: Why are the BITSAT formulas important to the exam?
A:

In BITSAT which is typically a speed-based exam, you are likely to be faced with 150 questions within 180 minutes only. Maintaining the essential PCM formulas enables to solve concerns, and so, erases the necessity for considering things anew for starters. Understanding these formulas in detail will give you a convenient opportunity to manage time effectively and also improve precision.

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Student is in akash institute for JEE preparation.. how assess her prospect to get rank in JEE mains, BITSAT and Telangana EAMCET

Since the student is in Aakash Institute for JEE prep, her prospects in JEE Mains, BITSAT, and Telangana EAMCET depend on factors like her consistency in mock tests, accuracy, syllabus coverage, and exam temperament. Aakash provides regular All India Test Series (AITS), which is the best way to benchmark her performance nationally scores above 200 in AITS usually indicate a strong JEE Mains rank (<50k), while 300+ marks in BITSAT mocks show good chances of Pilani/Hyderabad/Goa admission. For Telangana EAMCET, securing 80%+ accuracy in state-level practice tests typically ensures a government seat in CSE/ECE. Regular mock analysis, improving weak topics, and comparing her percentile trends will give the most reliable measure of her admission chances.

Read Complete Answer
R
Rithwik Kumar Sahu
23 Aug'25
With 210 score in Bitsat what are the chances in getting M. Sc integrates courses Hyderabad campus

Hello Aspirant,

From your BITSAT score of 210, BITS Hyderabad entry supports a very focused choice of integrated M.Sc. (like M.Sc. Physics, Chemistry, Mathematics, Economics or Biology). Generally with this score, admission into the M.Sc. programs at BITS Hyderabad is likely competitive. Based on past trends, the cutoff's for the M.Sc. programs at Hyderabad have ranged 220 to 250. While 210 will typically be below the cutoff marks for most M.Sc. courses, you might still have a small chance / opportunity during the last / spot rounds, provided the cutoff trend is slightly lower that year.

Read Complete Answer
G
gopi
30 Jun'25
I scored 172 in bitsat and 14110 rank in srmjee phase 2 can i expect something also i have option of vit bhopal and Chandigarh university what should i go for?

Hello Aspirant,

With a BITSAT score of 172 and SRMJEEE Phase 2 rank of 14,110, securing top branches like CSE at BITS Pilani, Goa, or Hyderabad is unlikely, as their cutoffs are much higher (around 240–270+ for lesser-known branches). Similarly, SRM KTR campus for CSE closes within ~2,000–5,000 rank, so core branches at SRM KTR may not be possible, but Ramapuram or Vadapalani campuses might offer CSE or allied branches.

Now, comparing your other two options:

VIT Bhopal vs. Chandigarh University (CU):

  1. VIT Bhopal

    • Part of the reputed VIT group.

    • Newer campus but decent infrastructure.

    • Better academic reputation than CU.

    • Good placements mainly for CSE/IT streams.

  2. Chandigarh University (CU):

    • Known for aggressive placement marketing.

    • Good exposure, foreign tie-ups, and more affordable for North Indian students.

    • Decent placement packages for CSE/AI/ML, but mass placements are often on the lower side.

Which Should You Choose?

If you get CSE or CSE-related branches at VIT Bhopal then you take admission at VIT Bhopal.
If not, Chandigarh University is a decent alternative, especially if you’re from North India and want affordability and exposure.

All the best for your admission process.

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S
Sonakshi Kumari
24 Jun'25
I have to join in B. Tech cyber security. So I need to write BITSAT exam. So can you please tell me how to enter into the exam and get selected

Hello,

To get admission into B.Tech Cyber Security at BITS , you need to follow these steps:

  • Fill out the BITSAT application form on the official website.

  • Prepare for subjects like Physics, Chemistry, Maths, English, and Logical Reasoning.

  • Appear for the BITSAT exam as per the given schedule.

  • Try to score well (around 270+ is usually safe for Cyber Security).

  • After the result, apply for counselling and choose B.Tech Cyber Security as your first preference.

  • If your score meets the cut-off, you will get a seat.

  • Keep checking the official website for updates and important dates.

Hope it helps !

Read Complete Answer
P
Prachi Kumari
24 Jun'25
BITSAT score is 263. Will i get EEE or Electronics and Instrumentation in Hyderbad or Goa this year, given that papers were tough in 2025 as compared to 2024.

Hello,

With a BITSAT score of 263, getting EEE or Electronics and Instrumentation at BITS Hyderabad or Goa may be tough. These branches usually close at a higher score.

Even if the paper was tougher this year, the cut-offs may not drop much. You might have a better chance in later rounds, but it’s still a bit difficult. Keep backup options ready.

Hope it helps !

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P
Prachi Kumari
26 Jun'25
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Popular BITSAT Questions

By mistake i wrote my birthdate wrong in the form and i alreay confromed it. It use birthdate as a password 

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A photon of wavelength 6000 Å has an energy E. The wavelength of the photon of light which corresponds to an energy equal to 2E is

  • Option 1)

    300 Å

  • Option 2)

    3000 Å

  • Option 3)

    30 Å

  • Option 4)

    3 Å

 
Read Complete Answer

A reduction in atomic size with increase in atomic number is a characteristic of elements of:

  • Option 1)

    f-block

  • Option 2)

    radioactive series

  • Option 3)

    high atomic mass

  • Option 4)

    d-block

 
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An element has the configuration 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{2} : To which block in the long form of the periodic table, does this belong?

  • Option 1)

    s-block

  • Option 2)

    p-block

  • Option 3)

    d-block

  • Option 4)

    f-block

 
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An element R forms the highest oxide R_{2}O_{5} . R belong to

  • Option 1)

    4 group

  • Option 2)

    6 group

  • Option 3)

    5 group

  • Option 4)

    8th group

 
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An element X occurs in short period having configuration ns2 np1 . The formula and nature of its oxide are:

  • Option 1)

    X03 , basic

  • Option 2)

    XO3 ,acidic

  • Option 3)

    X2O3 , amphoteric

  • Option 4)

    X2O3, Basic

 
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Elements of group IB and IIB are called:

  • Option 1)

    Normal Elements

  • Option 2)

    Transition elements

  • Option 3)

    Alkaline earth metals 

  • Option 4)

    Alkali metals

 
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Gases begin to conduct electricity at low pressure because

  • Option 1)

    At low pressure gases turn to plasma

  • Option 2)

    Atoms break up into electrons and protons

  • Option 3)

    Electrons in atoms can move freely at low pressure

  • Option 4)

    Colliding electrons can acquire higher kinetic energy due to the increased mean free path leading to ionisation of atoms.

 
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Halogens are placed in same group because they:

  • Option 1)

    Are electronegative

  • Option 2)

    Are most reactive

  • Option 3)

    Are not metals

  • Option 4)

    Have 7 electrons in outermost orbit.

 
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Identify the pairs which are not of isotopes

(i)  _{6}^{12}\textrm{X}, \:_{6}^{13}\textrm{Y}

(ii) _{17}^{35}\textrm{X}, \:_{17}^{37}\textrm{Y}

(iii) _{6}^{14}\textrm{X}, \:_{7}^{14}\textrm{Y}

(iv) _{4}^{8}\textrm{X}, \:_{5}^{8}\textrm{Y}

  • Option 1)

    i and ii

  • Option 2)

    ii and iii

  • Option 3)

    iii and iv

  • Option 4)

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