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100+ Free CSIR NET Practice Questions

Pass your CSIR UGC National Eligibility Test exam on the first try — instant access, no signup required.

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If you toss a fair die once, what is the probability of rolling a prime number?

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Key Facts: CSIR NET Exam

180 minutes

Single continuous session, no sectional time limits

csirnet.nta.nic.in

5 subjects

Chemical, Earth, Life, Mathematical, Physical Sciences

CSIR HRDG

INR 1150

General-category fee for CSIR NET 2026

NTA Information Bulletin

25-33%

Negative marking varies by subject and part

CSIR NET syllabus and pattern

100

Free practice questions here

OpenExamPrep

CSIR NET is a 3-hour computer-based MCQ test across 5 science subjects with three parts: Part A (20 common aptitude questions, choose 15), Part B (subject MCQs), Part C (HOTS MCQs). Total ~75 MCQs per paper, negative marking 25-33% by subject. Conducted by NTA on behalf of CSIR for JRF and Assistant Professor eligibility.

Sample CSIR NET Practice Questions

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

1If 15 men can complete a piece of work in 24 days, how many days will 20 men take to complete the same work, assuming all work at the same rate?
A.18 days
B.20 days
C.16 days
D.22 days
Explanation: Work is constant: men × days = 15 × 24 = 360 man-days. With 20 men: 360 / 20 = 18 days. This is the inverse-proportionality rule (M1·D1 = M2·D2).
2The mean of 10 numbers is 25. If one number which was incorrectly recorded as 32 is corrected to 23, what is the new mean?
A.24.1
B.24.5
C.25.9
D.23.7
Explanation: Original sum = 10 × 25 = 250. Corrected sum = 250 − 32 + 23 = 241. New mean = 241 / 10 = 24.1. Adjusting the total by the error (−9) then dividing by n gives the corrected mean directly.
3A train travels the first half of a journey at 60 km/h and the second half at 90 km/h. What is its average speed over the entire trip?
A.72 km/h
B.75 km/h
C.70 km/h
D.78 km/h
Explanation: Average speed for equal distances = 2·v1·v2 / (v1 + v2) = 2(60)(90) / 150 = 10800/150 = 72 km/h. This is the harmonic mean, not the arithmetic mean, because equal distances mean unequal times.
4In a class of 60 students, 35 study Physics, 30 study Chemistry, and 12 study both. How many students study neither subject?
A.7
B.5
C.10
D.8
Explanation: By inclusion–exclusion: |P ∪ C| = 35 + 30 − 12 = 53. Students studying neither = 60 − 53 = 7. The formula |A ∪ B| = |A| + |B| − |A ∩ B| avoids double-counting students in both sets.
5A solution contains 20% salt by mass. How much water must be added to 500 g of this solution to reduce the salt concentration to 8%?
A.750 g
B.600 g
C.1000 g
D.500 g
Explanation: Salt mass = 0.20 × 500 = 100 g (constant). New solution mass = 100 / 0.08 = 1250 g. Water added = 1250 − 500 = 750 g. Diluting a solution conserves solute mass while increasing total mass.
6A sum of money doubles itself in 8 years at simple interest. In how many years will it triple at the same rate?
A.16 years
B.12 years
C.20 years
D.24 years
Explanation: If P doubles in 8 years, SI = P, so rate R = (100 × P)/(P × 8) = 12.5% per year. To triple, SI = 2P, time = (2P × 100)/(P × 12.5) = 16 years. SI is linear in time, so doubling and tripling timescales relate proportionally.
7Find the next number in the series: 2, 6, 12, 20, 30, ?
A.42
B.40
C.44
D.38
Explanation: The differences are 4, 6, 8, 10, 12 — increasing by 2 each step. So next = 30 + 12 = 42. Equivalently, the nth term is n(n+1), giving 1·2, 2·3, 3·4, 4·5, 5·6, 6·7 = 42.
8A bag contains 4 red and 6 blue balls. Two balls are drawn at random without replacement. What is the probability that both are blue?
A.1/3
B.3/10
C.2/9
D.1/2
Explanation: P(both blue) = (6/10) × (5/9) = 30/90 = 1/3. Without replacement reduces both the favourable and total counts by one in the second draw.
9The price of a commodity rose by 25% and then fell by 20%. What is the net percentage change in price?
A.0%
B.5% increase
C.5% decrease
D.1% decrease
Explanation: Successive change: 1.25 × 0.80 = 1.00, so net change is 0%. The rise (+25%) and fall (−20%) are inverse multipliers because 1/0.80 = 1.25.
10In a row of children facing north, Ravi is 7th from the left and Suresh is 12th from the right. After they exchange positions, Ravi becomes 22nd from the left. How many children are in the row?
A.33
B.32
C.34
D.31
Explanation: Suresh's original position from left = 22 (Ravi took Suresh's spot, which is 22nd from left). Total = (position from left) + (position from right) − 1 = 22 + 12 − 1 = 33.

About the CSIR NET Exam

CSIR NET (Council of Scientific & Industrial Research National Eligibility Test) is conducted by the National Testing Agency (NTA) on behalf of CSIR to determine eligibility for Junior Research Fellowship (JRF) and Lectureship/Assistant Professor positions in Indian universities and research institutes. The exam is held in five science subjects: Chemical Sciences, Earth Atmospheric Ocean and Planetary Sciences, Life Sciences, Mathematical Sciences, and Physical Sciences. Each paper is a 3-hour Computer-Based Test with three parts — Part A (General Aptitude, common across subjects), Part B (subject MCQs), and Part C (Higher Order Thinking Skills). Negative marking ranges from 25% to 33% depending on subject. The June 2026 session is the next scheduled cycle.

Questions

100 scored questions

Time Limit

180 minutes (3 hours, single continuous session)

Passing Score

Subject-specific cutoffs released by CSIR; typically ~33-50% for Lectureship, higher for JRF

Exam Fee

INR 1150 (General); INR 600 (EWS/OBC); INR 325 (SC/ST/PwD) (National Testing Agency (NTA) on behalf of CSIR)

CSIR NET Exam Content Outline

~50%

Part A — General Aptitude

20 common questions across all 5 subjects: numerical ability, reasoning, data interpretation, graphical analysis, percentages, ratios, basic geometry, probability, and scientific aptitude. Candidates attempt 15 of 20.

~10%

Chemical Sciences fundamentals

Acid-base equilibria, kinetics, ideal gas law, organic functional groups, atomic structure, and SI units for the chemistry subject paper.

~10%

Earth Sciences fundamentals

Rock types, atmospheric composition, planetary gravity, altitude pressure, and Coriolis effect for the Earth/Atmospheric/Ocean/Planetary Sciences paper.

~10%

Life Sciences fundamentals

Cell organelles, DNA structure and replication, Mendelian genetics, photosynthesis, bacterial growth, and evolution basics for the Life Sciences paper.

~10%

Mathematical Sciences fundamentals

Set theory, calculus, linear algebra, determinants, eigenvalues, quadratic equations, and function evaluation for the Mathematical Sciences paper.

~10%

Physical Sciences fundamentals

Kinetic energy, electromagnetic waves, thermodynamics, kinetic theory degrees of freedom, optics refraction, and SI units of pressure for the Physical Sciences paper.

How to Pass the CSIR NET Exam

What You Need to Know

  • Passing score: Subject-specific cutoffs released by CSIR; typically ~33-50% for Lectureship, higher for JRF
  • Exam length: 100 questions
  • Time limit: 180 minutes (3 hours, single continuous session)
  • Exam fee: INR 1150 (General); INR 600 (EWS/OBC); INR 325 (SC/ST/PwD)

Keys to Passing

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

CSIR NET Study Tips from Top Performers

1Master Part A General Aptitude first — it is common across all 5 subjects and offers 30 marks with relatively predictable patterns
2Practice 3-hour mock tests under exam conditions to build the stamina needed for a single continuous session without sectional breaks
3Build a formula notebook by subject — kinetics, thermodynamics, set theory, organelle functions — and revise it daily in the final month
4Track negative-marking discipline: in subjects with 33% negative marking, attempt only when you can eliminate at least two distractors
5Use previous-year CSIR NET papers from csirhrdg.res.in for the last 5-10 years — pattern repetition in Part C is significant
6For Part A graph and DI questions, learn to read pie/bar/line/histogram charts quickly — speed here frees up time for HOTS Part C problems

Frequently Asked Questions

What is the CSIR NET exam pattern in 2026?

CSIR NET 2026 is a 180-minute (3-hour) computer-based test with a single continuous session. Each subject paper has three parts: Part A (20 General Aptitude MCQs, attempt 15), Part B (subject MCQs), and Part C (Higher Order Thinking Skills MCQs). Total is approximately 75 MCQs per paper, with negative marking of 25% (Life/Earth/Chemical/Physical) or 33% on some sections; Mathematical Sciences Part C has no negative marking.

Which five subjects are offered in CSIR NET?

CSIR NET is conducted in five science subjects: (1) Chemical Sciences, (2) Earth, Atmospheric, Ocean and Planetary Sciences, (3) Life Sciences, (4) Mathematical Sciences, and (5) Physical Sciences. Candidates choose one subject at application based on their post-graduate background.

What is the difference between JRF and Lectureship qualification?

JRF (Junior Research Fellowship) requires a higher score and provides a monthly stipend for pursuing PhD research at CSIR-recognized institutions. Lectureship/Assistant Professor qualification has a lower cutoff and certifies eligibility to apply for teaching positions. JRF candidates are also automatically Lectureship-qualified.

What is the application fee for CSIR NET 2026?

The CSIR NET 2026 application fee is INR 1150 for General/Unreserved candidates, INR 600 for EWS and OBC-NCL candidates, and INR 325 for SC/ST/PwD candidates. The fee is paid online via the official NTA portal at csirnet.nta.nic.in during the registration window.

How is Part A General Aptitude scored?

Part A contains 20 common General Aptitude questions across all subjects. Candidates choose any 15 to attempt; only the first 15 attempted (or all 15 if exactly 15 are attempted) count for scoring. Each correct answer carries 2 marks (total 30 for Part A). Negative marking applies as per subject rules.

When is CSIR NET 2026 conducted and when do registrations open?

CSIR NET 2026 is scheduled for the June 2026 session. The official notification, registration window, and exam city slip release dates are published on csirnet.nta.nic.in. Registrations typically open 6-8 weeks before the exam date. Admit cards are released about a week before the test.

What is the eligibility for CSIR NET 2026?

Candidates need an MSc or equivalent post-graduate degree in a relevant science subject with at least 55% marks (50% for SC/ST/PwD/OBC-NCL/EWS). Final-year students may apply as Result Awaited. Age limit applies only to JRF: maximum 28 years (with relaxations for reserved categories); no upper age for Lectureship/Assistant Professor.