100+ Free IISER IAT Practice Questions

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2026 Statistics

Key Facts: IISER IAT Exam

60 MCQs

Total questions in the IAT exam (15 per subject)

Official IAT Exam Pattern

180 mins

Time limit allowed to complete the online test

Official IAT Guidelines

240 marks

Maximum possible score (+4/-1 marking scheme)

Official IAT Marking Scheme

₹2,000

Application fee for General/OBC/EWS candidates

Official IAT Fee Schedule

9 Campuses

Accepting institutions (7 IISERs + IISc + IITM)

Official IAT Intake Information

100% Free

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The IISER Aptitude Test (IAT) is a 3-hour computer-based exam consisting of 60 MCQs (15 questions each in Physics, Chemistry, Mathematics, and Biology) based on the Class 11/12 NCERT syllabus. Correct answers earn +4 marks, while incorrect answers lose 1 mark. The registration fee is ₹2,000 for General/OBC/EWS and ₹1,000 for reserved categories.

Sample IISER IAT Practice Questions

Try these sample questions to test your IISER IAT exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A body of mass 5 kg is acted upon by a constant force of 20 N. If it starts from rest, what is its kinetic energy after 4 seconds?

A.320 J

B.640 J

C.160 J

D.1280 J

Explanation: The acceleration of the body is a = F/m = 20/5 = 4 m/s^2. The velocity after 4 seconds is v = u + at = 0 + (4)(4) = 16 m/s. The kinetic energy is KE = 0.5 * m * v^2 = 0.5 * 5 * 16^2 = 0.5 * 5 * 256 = 640 J.

2A ball is thrown vertically upwards with a speed of 20 m/s from the top of a tower of height 25 m. Find the time taken by the ball to hit the ground. (Take g = 10 m/s^2).

A.3 s

B.4 s

C.5 s

D.6 s

Explanation: Using the equation of motion s = ut + 0.5 * a * t^2 with upward direction positive: -25 = 20t - 0.5 * 10 * t^2. Rearranging gives 5t^2 - 20t - 25 = 0, which simplifies to t^2 - 4t - 5 = 0. Factoring this quadratic equation yields (t - 5)(t + 1) = 0, giving the physical time t = 5 seconds.

3A block of mass 2 kg is placed on a rough horizontal surface. The coefficient of static friction is 0.4 and kinetic friction is 0.3. If a horizontal force of 6 N is applied on the block, what is the frictional force acting on the block? (Take g = 10 m/s^2).

A.8 N

B.6 N

C.5.88 N

D.0 N

Explanation: The maximum static frictional force is f_s(max) = mu_s * N = mu_s * m * g = 0.4 * 2 * 10 = 8 N. Since the applied force of 6 N is less than the maximum static friction, the block remains at rest. Therefore, the static frictional force must exactly balance the applied force, meaning it is 6 N.

4A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an inclined plane of height 7 m. What is its linear velocity when it reaches the bottom of the incline? (Take g = 10 m/s^2).

A.7 m/s

B.10 m/s

C.14 m/s

D.12 m/s

Explanation: By conservation of mechanical energy, Mgh = 0.5 * M * v^2 + 0.5 * I * w^2. For a solid sphere, I = 0.4 * M * R^2, and for rolling without slipping, w = v/R. Substituting these values gives Mgh = 0.5 * M * v^2 + 0.2 * M * v^2 = 0.7 * M * v^2, which reduces to gh = 0.7 * v^2. Thus, v = sqrt(10 * gh / 7) = sqrt(10 * 10 * 7 / 7) = 10 m/s.

5A satellite is revolving around the Earth in a circular orbit of radius R. If its kinetic energy is E, what is its total mechanical energy?

A.-E

B.E

C.-2E

D.2E

Explanation: For a satellite in a stable circular orbit under a central gravitational force, the gravitational potential energy is U = -2 * KE. Since the kinetic energy is E, the potential energy is -2E. The total mechanical energy is the sum of kinetic and potential energies: TE = KE + U = E - 2E = -E.

6A uniform rod of length L and mass M is free to rotate in a vertical plane about a frictionless hinge at one end. The rod is released from rest in the horizontal position. What is the angular acceleration of the rod immediately after release?

A.3g / (2L)

B.g / L

C.2g / (3L)

D.3g / L

Explanation: The torque about the hinge immediately after release is due to gravity acting at the center of mass: torque = Mg * (L/2). The moment of inertia of a uniform rod of length L about one of its ends is I = (1/3) * M * L^2. Using torque = I * alpha, we get Mg * (L/2) = (1/3) * M * L^2 * alpha. Solving for alpha gives alpha = 3g / (2L).

7A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a_c is varying with time t as a_c = k^2 * r * t^2, where k is a constant. What is the power delivered to the particle by the forces acting on it?

A.m * k^2 * r^2 * t

B.m * k * r^2 * t^2

C.Zero

D.2 * m * k^2 * r^2 * t

Explanation: The centripetal acceleration is a_c = v^2/r = k^2 * r * t^2, which gives velocity v = k * r * t. The tangential acceleration is a_t = dv/dt = k * r. Since the centripetal force is perpendicular to the velocity, it does no work and delivers zero power. Power is delivered only by the tangential force: P = F_t * v = (m * a_t) * v = (m * k * r) * (k * r * t) = m * k^2 * r^2 * t.

8An ideal gas undergoes an isothermal expansion at temperature T from volume V to 2V. What is the work done by the gas?

A.nRT ln(2)

B.nRT

C.Zero

D.nRT ln(0.5)

Explanation: The work done by an ideal gas during an isothermal process is given by W = nRT * ln(V_f / V_i). Since the volume expands from V to 2V, V_f / V_i = 2. Thus, the work done is W = nRT * ln(2).

9A Carnot engine operates between two reservoirs at temperatures 500 K and 300 K. If the engine receives 1000 J of heat from the hot reservoir in each cycle, how much heat is discarded to the cold reservoir?

A.400 J

B.600 J

C.800 J

D.200 J

Explanation: The efficiency of a Carnot engine is eta = 1 - T_C / T_H = 1 - 300/500 = 0.40 (or 40%). The work done per cycle is W = eta * Q_H = 0.40 * 1000 J = 400 J. By conservation of energy, the heat discarded to the cold reservoir is Q_C = Q_H - W = 1000 J - 400 J = 600 J.

10One mole of a monoatomic ideal gas (Cv = 1.5 R) is mixed with one mole of a diatomic ideal gas (Cv = 2.5 R). What is the molar specific heat at constant volume (Cv) of the mixture?

A.1.5 R

B.2.0 R

C.2.5 R

D.3.0 R

Explanation: The molar specific heat at constant volume of a mixture of gases is given by Cv = (n1 * Cv1 + n2 * Cv2) / (n1 + n2). Substituting the given values: Cv = (1 * 1.5 R + 1 * 2.5 R) / (1 + 1) = 4.0 R / 2 = 2.0 R.

About the IISER IAT Exam

The IISER Aptitude Test (IAT) is an entrance examination conducted for admission to the 5-Year BS-MS Dual Degree and 4-Year BS degree programs at the seven IISER campuses, IISc Bangalore, and IIT Madras. It is a computer-based test comprising 60 multiple-choice questions across Physics, Chemistry, Mathematics, and Biology (15 questions each). The exam syllabus is aligned with Class 11 and 12 NCERT curriculum. This practice bank offers 100 high-quality questions spanning all four subjects to help you master the core concepts.

Questions

60 scored questions

Time Limit

3 hours (180 minutes)

Passing Score

Admissions are merit-based. Scores are out of 240, and cutoffs vary by year, category, and preferred IISER campus.

Exam Fee

₹2,000 for General/OBC-NCL/EWS; ₹1,000 for SC/ST/PwD/Third Gender/Kashmiri Migrants. (Indian Institutes of Science Education and Research (IISER))

IISER IAT Exam Content Outline

25%

Physics

Mechanics, Thermodynamics, Electrodynamics, Optics, and Modern Physics

25%

Chemistry

Physical, Inorganic, and Organic Chemistry

25%

Mathematics

Algebra, Calculus, Coordinate Geometry, Trigonometry, and Probability

25%

Biology

Cell Biology, Genetics, Human Physiology, Ecology, and Evolution

How to Pass the IISER IAT Exam

What You Need to Know

Passing score: Admissions are merit-based. Scores are out of 240, and cutoffs vary by year, category, and preferred IISER campus.
Exam length: 60 questions
Time limit: 3 hours (180 minutes)
Exam fee: ₹2,000 for General/OBC-NCL/EWS; ₹1,000 for SC/ST/PwD/Third Gender/Kashmiri Migrants.

Keys to Passing

Complete 500+ practice questions
Score 80%+ consistently before scheduling
Focus on highest-weighted sections
Use our AI tutor for tough concepts

IISER IAT Study Tips from Top Performers

1Thoroughly study NCERT textbooks for Class 11 and 12 as they form the core syllabus for all four subjects.

2Practice time management since you need to solve 60 complex questions across four distinct subjects within 180 minutes.

3Pay close attention to negative marking: avoid guessing on questions where you cannot eliminate options.

4Solve previous years' IAT papers and take mock tests to get accustomed to the online computer-based format.

5Dedicate equal study time to Physics, Chemistry, Mathematics, and Biology to ensure a balanced overall performance.

Frequently Asked Questions

What is the marking scheme for the IISER Aptitude Test (IAT)?

IAT features a total of 240 marks. Each correct response is awarded +4 marks, each incorrect response results in -1 mark (negative marking), and unanswered questions receive 0 marks.

Which institutions accept IISER Aptitude Test (IAT) scores?

IAT scores are accepted for admission to the BS-MS and BS programs at all seven IISERs (Berhampur, Bhopal, Kolkata, Mohali, Pune, Tirupati, Thiruvananthapuram), IISc Bangalore, and IIT Madras.

What is the eligibility criteria for IAT 2026?

Candidates must have passed Class 12 or equivalent in the science stream in 2025 or 2026 from a recognized board, with at least three subjects out of Physics, Chemistry, Mathematics, and Biology.

Do I have to attempt all four sections (Physics, Chemistry, Math, Biology) in the IAT?

Yes, the IAT contains questions from all four subjects (15 questions each), and all are compiled in a single paper. While you are not forced by the system to answer every section, attempting all subjects maximizes your potential score out of 240.

What is the application fee for the IAT exam?

The fee is ₹2,000 for General, OBC-NCL, and EWS candidates. For candidates under SC, ST, PwD, Third Gender, and Kashmiri Migrants, the application fee is ₹1,000.