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.

Also Read:

• Important Formulas for Maths
• Chemistry Important Formulas
• Important Formulas for Physics

BITSAT 2026 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.,

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

2.To cross river in the shortest path

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

$\begin{aligned} & t=\frac{d}{\sqrt{v^2-u^2}} \\ & d={ }_{\text {width of fiver }} \\ & \mathrm{v} \text { = Speed of Baat w.r.t river } \\ & \mathrm{u}=\text { speed of river } \\ & \text { 3. } W=F S \cos \Theta\end{aligned}$

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 the two ends.

$\begin{aligned} & V \propto I \\ & V=I R \\ & R-\text { Electric Resistance }\end{aligned}$

• The graph between V and I

The slope gives the resistance

• The graph between V and I at different temperatures

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

Kinematic Equations: For uniform acceleration:

the Net electric field is -

$$
E_{n e t}=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{x Q}{\left(x^2+R^2\right)^{3 / 2}}
$$

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 2026 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:

$P V=n R T$

Pressure × Volume = 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.

$\mathrm{r} \propto \frac{1}{\sqrt{\mathrm{M}}}$ 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
$\mathrm{r}_1 / \mathrm{r}_2=\sqrt{\mathrm{M}_2 / \mathrm{M}_1}$ 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:

$\begin{aligned} & \frac{\mathrm{V}_1 \mathrm{t}_2}{\mathrm{~V}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \text { As } \mathrm{r}=\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{\mathrm{w}}{\mathrm{t}} \\ & \text { Thus, } \frac{\mathrm{n}_1 \mathrm{t}_2}{\mathrm{n}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \frac{\mathrm{w}_1 \mathrm{t}_2}{\mathrm{w}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \frac{\mathrm{~d}_1 \mathrm{t}_2}{\mathrm{~d}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1}\end{aligned}$

Arrhenius Equation:

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

$\log _{10} \mathrm{~A}=\log _{10} \mathrm{~A}_{\circ}-\frac{\mathrm{kt}}{2.303}$

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:
  
  $\begin{aligned} & \log _{10} \mathrm{~A}=\log _{10} \mathrm{~A}_{\mathrm{o}}-\frac{\mathrm{kt}}{2.303} \\ & \Rightarrow \log _{10}\left[\frac{\mathrm{~A}_{\mathrm{o}}}{\mathrm{A}}\right]=\frac{\mathrm{kt}}{2.303} \\ & \text { Thus, } \mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10}\left[\frac{\mathrm{~A}_{\mathrm{o}}}{\mathrm{A}}\right]\end{aligned}$

• Exponential form:
  
  $\begin{aligned} & \log _{\mathrm{e}} \mathrm{A}-\log _{\mathrm{e}} \mathrm{A}_{\mathrm{o}}-\mathrm{kt} \\ & \Rightarrow \log \frac{\mathrm{A}}{\mathrm{A}_{\mathrm{o}}}=-\mathrm{kt} \\ & \Rightarrow \frac{\mathrm{A}}{\mathrm{A}_{\mathrm{o}}}=\mathrm{e}^{-\mathrm{kt}} \\ & \text { Thus, } \mathrm{A}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-\mathrm{kt}}\end{aligned}$

• This equation is also known as the exponential form.

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

Boyle’s Law:

Boyle's law may be expressed mathematically as
$\mathrm{P} \propto \frac{1}{\mathrm{~V}},($ at constant T and n$)$
or $\mathrm{V} \propto \frac{1}{\mathrm{P}}$, (at constant T and n )
Where,
$\mathrm{T}=$ temperature, $\mathrm{P}=$ pressure of the gas
$\mathrm{n}=$ number of moles of a gas and $\mathrm{V}=$ volume of the gas
$\Rightarrow \mathrm{V}=\mathrm{k}_1 \frac{1}{\mathrm{P}}$

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

$\begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2=\text { constant } \\ & \text { or } \quad \frac{P_1}{P_2}=\frac{V_2}{V_1}\end{aligned}$

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,

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

$\Delta=q+w$ [Since, work done by the system is -w]

Nernst Equation:

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

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.059}{\mathrm{n}} \log _{10} \mathrm{Q}$

where n is the number of electrons exchanged.

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

$\mathrm{E}_{\text {cell }}=(P-Q)-\frac{0.059}{x} \log _{10} \frac{c_1}{c_2}$

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

BITSAT 2026 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 $a \neq 0$ a is also called the leading coefficient.

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

As the degree of the quadratic polynomial is 2, 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:

$\begin{aligned} & \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{D}}}{2 \mathrm{a}} \\ & \text { or } \\ & \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}\end{aligned}$

Where D is called the discriminant of the quadratic equation, given by

$D=b^2-4 a c$

Sine Rule

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 the inverse process of differentation.

$\frac{d}{d x}\{F(x)\}=f(x)$

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

$$
\text { Interval width }=\frac{x_i-x_{s+1}}{n}
$$

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.

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

$$
\frac{n(n-1)(n-2) \ldots(n-r+1)(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}={ }^n P_r
$$

Where $r \leq n$ 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, you can go with the mocks tests.