BITSAT 2025 Formulas for Physics, Chemistry, Maths

Edited By Team Careers360 | Updated on Nov 05, 2024 05:38 PM IST | #BITSAT

BITSAT Formulas for PCM: 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 2025 exam.

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Benefits of Knowing BITSAT Formulas for PCM
BITSAT 2025 Formulas for Physics
BITSAT 2025 Formulas for Chemistry
BITSAT 2025 Formulas for Maths

BITSAT 2025 Formulas for Physics, Chemistry, Maths

Since the exam dates for BITSAT 2025 have already been disclosed, likely set for May 2025, now is the right time to start grasping these important formulas for BITSAT.

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 analyze 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)

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

width of river

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

u = speed of river

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.

$V\propto I$

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

1st Equation of motion or velocity time equation

Formula

V = Final velocity

u = Initial velocity

A = acceleration

T = time

2nd Equation of Position time equation

Formula

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

$s\rightarrow$ Displacement

$u\rightarrow$ Initial velocity

$a\rightarrow$ acceleration

$t\rightarrow$ time

3rd Equation of Velocity –displacement equation

Formula

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

$V\rightarrow$ Final Velocity

$s\rightarrow$ Displacement

$u\rightarrow$ Initial velocity

$a\rightarrow$ acceleration

Displacement in nth second
Formula

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

Where Initial velocity

uniform acceleration

n= 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:

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.

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 ,

Sine Rule

In any $\Delta$ ABC, 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

JEE Main Formulas PDF For Physics, Chemistry, And Mathematics

Important Formulas for Maths

Chemistry Important Formulas

Important Formulas for Physics

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)

1. Why are the BITSAT formulas important to the exam?

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.

2. What can I do about recording the basic equations required for BITSAT?

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.

3. Which formulas are important for solutions in BITSAT Physics?

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.

4. Formulae related to Chemistry in which is important for BITSAT exam.

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.

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Questions related to BITSAT

Have a question related to BITSAT ?

I have completed my 12th grade in the academic year 2023-24 can I give an attempt for bitsat 2025

Hello Greetings

According to the BITSAT eligibility criteria:

Eligibility Criteria

1. *Qualifying Examination*: Candidates should have passed the 12th class examination of 10+2 system from a recognized Central or State board or its equivalent.

2. *Age Limit*: There is no age limit for appearing in the BITSAT examination.

3. *Year of Passing*: Candidates who have passed their 12th examination in 2023 or 2024 are eligible to appear in BITSAT 2025.

Since you completed your 12th grade in the academic year 2023-24, you are eligible to give an attempt for BITSAT 2025.

Additional Tips

1. Ensure you meet the subject-wise eligibility criteria, which typically includes Physics, Chemistry, and Mathematics (PCM) or Physics, Chemistry, and Biology (PCB) in your 12th standard.

2. Check the official BITSAT website for the latest eligibility criteria, exam pattern, and application process.

3. Prepare well for the exam by referring to the prescribed syllabus and practicing with sample papers or mock tests.

Have a great day

Read Complete Answer

Sakshi Badoni 06 January,2025

Manipal and bitsat exam notification.

Hello Greetings

The BITSAT 2025 application form is expected to be released in the second week of January 2025, with the last date to submit the application form being in April 2025 .

Here are the *Important Dates* for BITSAT 2025:

- _Application Form Release_: Second week of January 2025

- _Last Date to Apply_: April 2025

- _Application Form Correction_: April 2025

- _Exam Centre Allotment_: May 2025

- _Test Date and Slot Reservation_: May 2025

- _Admit Card Release_: May 2025

- _Exam Date Session 1_: May 2025

- _Exam Date Session 2_: June 2025

As for Manipal, I couldn't find any specific information on the exam notification. However, I recommend checking the official website of Manipal University for the latest updates on exam notifications and application deadlines.

It's essential to stay updated with the latest information, so keep checking the official websites for BITSAT and Manipal University.

Have a great day

Read Complete Answer

Sakshi Badoni 04 January,2025

how to check jee mains and bitsat score cards after 3 years ? i wrote exam in 2021

Every year JEE mains and BITSAT portal was updated with New registration process and notification for upcoming student. There was no options to check previous year Result

However you can get it from your coaching institute they saved it during result time you can ask them for it.

its Must to keep All Documents safe and saved to avoid relevant problems.

HOpe this will helpful to you..

Read Complete Answer

sadafreen356 13 December,2024

Is cracking bitsat easy than jee mains

Hello,

Cracking BITSAT and JEE Mains both require dedication and hard work. However, the difficulty level and competition vary between the two exams.

BITSAT is considered relatively easier than JEE Mains for several reasons:

1. Less competition: BITSAT is a standalone exam for admission to BITS Pilani, whereas JEE Mains is a national-level exam with a much larger candidate pool.

2. Syllabus: BITSAT syllabus is similar to JEE Mains, but it has a stronger focus on logical reasoning and English proficiency.

3. Exam pattern: BITSAT has a unique exam pattern, with a mix of multiple-choice questions and a section on logical reasoning.

4. Cut-off: BITSAT cut-offs are generally lower than JEE Mains cut-offs.

However, it's essential to remember that BITSAT is still a competitive exam, and cracking it requires:

1. Strong fundamentals: A solid grasp of physics, chemistry, and mathematics concepts.

2. Practice and strategy: Familiarity with the exam pattern, time management, and practice with sample papers and mock tests.

3. Logical reasoning and English skills: BITSAT places significant emphasis on logical reasoning and English proficiency.

In conclusion, while BITSAT might be considered relatively easier than JEE Mains

https://www.careers360.com/exams/bitsat

Read Complete Answer

Tanya Gupta 09 December,2024

My son is in class 12 ..in class 12. pCM..wanted to take a break and appear for bitsat

Dear aspirant !

Hope you are doing well ! If he is confident and sure that after taking a break he can crack the entrance exam of bitsat then let him give the exam and give him a chance . Drop year has been proved beneficial for many students in the past . Taking a year drop doesn't mean it's the end of the road for you. It only means that you have taken an extra year to give a better chance to your career . Don't listen to people who tell you that dropping your year means you are risking your career.

Hope it helps you !

Thank you

Read Complete Answer

Shivanshu 26 November,2024

View All

Popular BITSAT Questions

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

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

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

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

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

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)
XO_{3 ,}acidic
Option 3)
X₂O₃ , amphoteric
Option 4)
X₂O_3,Basic

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

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.

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.

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)
i and iv

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5 Jobs Available

Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available

Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.

2 Jobs Available

Safety Manager

2 Jobs Available

Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available

Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available

Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available

Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials.

2 Jobs Available

Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available

Geologist

A geologist attempts to comprehend the historical backdrop of the planet we live on, all the more likely to anticipate the future and clarify current events. He or she analyses the components, deployments, results, physical characteristics, and past of the planet. A geologist examines the landforms and landscapes of the earth in relation to the geology, climatic, and human processes that have shaped them.

A geologist studies earth procedures, for example, seismic tremors, avalanches, floods, and volcanic eruptions to review land and draw up safe structure plans. When he or she researches earth materials, explores metals and minerals, yet in addition search for oil, petroleum gas, water, and strategies to extricate these.

2 Jobs Available

Energy Performance Engineer

Energy efficiency engineering is a broad field of engineering which deals with energy efficiency, energy services, facility management, plant engineering, and sustainable energy resources. Energy efficiency engineering is one of the most recent engineering disciplines to emerge. The field combines the knowledge and understanding of physics, chemistry, and mathematics, with economic and environmental engineering practices. The main job of individuals who opt for a career as an energy performance engineer is to find the most efficient and sustainable path to operate buildings and manufacturing processes.

Individuals who opt for a career as energy performance engineers apply their understanding and knowledge to increase efficiency and further develop renewable sources of energy. The energy efficiency engineers also examine the use of energy in those procedures and suggest the ways in which systems can be improved.

2 Jobs Available

Petroleum Engineer

A career as a Petroleum engineer is concerned with activities related to producing petroleum. These products can be in the form of either crude oil or natural gas. Petroleum engineering also requires the exploration and refinement of petroleum resources. Therefore, a career as a petroleum engineer comes up with oil and gas onshore jobs. There are also desk jobs in the petroleum industry. In layman’s terms, a petroleum engineer is a person who finds the best way to drill and extract oil from oil wells. Individuals who opt for a career as petroleum engineer also tries to find new ways to extract oil in an efficient manner.

2 Jobs Available

Transportation Planner

3 Jobs Available

Civil Engineer

A career as a civil engineer is of great importance for the infrastructural growth of the country. It is one of the most popular professions and there is great professional as well as personal growth in this civil engineering career path. There is job satisfaction in this civil engineering career path, but it also comes with a lot of stress, as there are multiple projects that need to be handled and have to be completed on time. Students should pursue physics, chemistry and mathematics in their 10+2 to become civil engineers.

2 Jobs Available

Transportation Engineer

A career as a Transportation Engineer is someone who takes care of people's safety. He or she is responsible for designing, planning and constructing a safe and secure transportation system. The transportation sector has seen a huge transformation and is growing day by day and improving every day.

As a Transport Engineer, he or she needs to solve complex problems such as accidents, costs, traffic flow, and statistics. A Transport Engineer also collaborates for projects with some other companies.

1 Jobs Available

Loco Pilot

A loco pilot or locomotive pilot is a professional responsible for operating trains. He or she starts, stops, or controls the speed of the train. A locomotive pilot ensures that the train operates according to time schedules and signals. These loco pilots are responsible for carrying people and products to distinct destinations.

A loco pilot has thorough knowledge and understanding of the railway operations, rules, regulations, protocols, and measures to take in times of emergency. Their role is crucial in ensuring passenger and freight trains' smooth and safe operation. Here, in this article, we will discuss everything on how to how to become a loco pilot.

2 Jobs Available