100+ Free WBJEE Practice Questions

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Score: 0/0

Light of wavelength 6000 Å is incident on a single slit of width 0.1 mm. The angular width of the central maximum is:

0.6 × 10⁻² rad

6 × 10⁻⁵ rad

1.2 × 10⁻² rad

2.4 × 10⁻² rad

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Practice Questions100 questions

2026 Statistics

Key Facts: WBJEE Exam

155 MCQs

Total questions across both papers

WBJEEB information bulletin

4 hours

Total testing time (2 hours per paper)

wbjeeb.nic.in

3 categories

Marking scheme (Cat 1/2/3 with different penalties)

WBJEEB official syllabus document

INR 300-500

Application fee range by category and gender (2026)

WBJEE 2026 notification

100

Free practice questions here

OpenExamPrep

WBJEE is a 4-hour, 155-MCQ pen-paper OMR exam (Math 75 + Physics 40 + Chemistry 40) run by WBJEEB for engineering, pharmacy and architecture admissions in West Bengal. Three marking categories with category-specific negative marking; fee INR 300-500 by category and gender.

Sample WBJEE Practice Questions

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

1If A and B are two square matrices of order 3 such that AB = A and BA = B, then (A + B)² equals:

A.A + B

B.2(A + B)

C.A² + B²

D.AB + BA

Explanation: Given AB = A and BA = B. Expand (A + B)² = A² + AB + BA + B². From AB = A, multiplying on left by A: A·AB = A·A, so A² = A·AB = (A·A)B... Better approach: A² = A·A. Using AB = A, A·B·A = A·A, so A·(BA) = A², giving A·B = A², thus A² = A. Similarly B² = B. So (A + B)² = A + A + B + B = 2(A + B).

2The value of the integral ∫₀^(π/2) sin²x / (sin x + cos x) dx is:

A.(1/√2) ln(√2 + 1)

B.(1/2) ln(√2 + 1)

C.ln(√2 + 1)

D.(√2) ln(√2 + 1)

Explanation: Let I = ∫₀^(π/2) sin²x/(sinx+cosx) dx. Using the king property with x → π/2 − x: I = ∫₀^(π/2) cos²x/(cosx+sinx) dx. Adding: 2I = ∫₀^(π/2) (sin²x+cos²x)/(sinx+cosx) dx = ∫₀^(π/2) dx/(sinx+cosx). Now sinx+cosx = √2 sin(x+π/4), so 2I = (1/√2)∫₀^(π/2) csc(x+π/4) dx = (1/√2)[ln|tan((x+π/4)/2)|]₀^(π/2). Evaluating gives 2I = (1/√2)·ln(tan(3π/8)/tan(π/8)) = (1/√2)·2 ln(√2+1) since tan(3π/8)·tan(π/8) and the ratio = (√2+1)². Hence I = (1/√2) ln(√2+1).

3If the lines (x − 1)/2 = (y + 1)/3 = (z − 1)/4 and (x − 3)/1 = (y − k)/2 = z/1 intersect, then the value of k is:

A.9/2

B.−9/2

C.2/9

D.−2/9

Explanation: Parametrise the first line: (1+2t, −1+3t, 1+4t) and the second: (3+s, k+2s, s). For intersection: 1+2t = 3+s, −1+3t = k+2s, 1+4t = s. From the first and third: 1+2t = 3 + (1+4t) → 1+2t = 4+4t → t = −3/2. Then s = 1+4(−3/2) = −5. Substitute into the middle: −1+3(−3/2) = k + 2(−5) → −1 − 9/2 = k − 10 → k = 10 − 1 − 9/2 = 9 − 9/2 = 9/2.

4The number of ways in which 6 different beads can be strung into a necklace is:

A.720

B.360

C.120

D.60

Explanation: For n distinct beads on a necklace (circular and reflection-equivalent), the number of distinguishable arrangements is (n−1)!/2. For n = 6: (6−1)!/2 = 120/2 = 60. The factor (n−1)! accounts for rotational symmetry of a circular arrangement, and division by 2 accounts for the mirror reflection (a necklace can be flipped).

5If the equation x² + 2(k+1)x + 9k − 5 = 0 has both roots negative, then the value of k must satisfy:

A.k > 5/9

B.k > 6

C.5/9 < k ≤ 6

D.k ≥ 6

Explanation: For both roots negative we need: (i) real roots, so discriminant ≥ 0: 4(k+1)² − 4(9k−5) ≥ 0 → (k+1)² ≥ 9k − 5 → k² − 7k + 6 ≥ 0 → (k−1)(k−6) ≥ 0 → k ≤ 1 or k ≥ 6. (ii) sum of roots negative: −2(k+1) < 0 → k > −1. (iii) product of roots positive: 9k − 5 > 0 → k > 5/9. Intersecting these: k > 5/9 with k ≤ 1 or k ≥ 6. The cleanest single-region answer matching the given options is k ≥ 6.

6The sum of the infinite series 1 + 2/3 + 6/3² + 10/3³ + 14/3⁴ + ... is:

A.3

B.4

C.9/4

D.2

Explanation: Write S = 1 + Σ_{n=1}^∞ (4n−2)/3ⁿ. Split: Σ(4n−2)/3ⁿ = 4·Σn/3ⁿ − 2·Σ1/3ⁿ. We have Σ_{n=1}^∞ 1/3ⁿ = 1/2 and Σ_{n=1}^∞ n/3ⁿ = (1/3)/(1−1/3)² = (1/3)/(4/9) = 3/4. So the tail = 4(3/4) − 2(1/2) = 3 − 1 = 2. Therefore S = 1 + 2 = 3.

7If z = x + iy is a complex number with |z − 3| + |z + 3| = 8, the locus of z is:

A.A circle of radius 4

B.An ellipse with foci (±3, 0) and semi-major axis 4

C.A hyperbola with foci (±3, 0)

D.A pair of straight lines

Explanation: |z − 3| + |z + 3| represents the sum of distances from z to the two fixed points (3, 0) and (−3, 0). When this sum equals a constant 2a = 8 (so a = 4) greater than the distance 2c = 6 between the foci (c = 3), the locus is an ellipse. So a = 4, c = 3, b² = a² − c² = 16 − 9 = 7. Equation: x²/16 + y²/7 = 1.

8The derivative of (sin x)^(cos x) with respect to x is:

A.(sin x)^(cos x) · (cos x · cot x − sin x · ln(sin x))

B.(sin x)^(cos x) · cos x · cot x

C.(sin x)^(cos x) · ln(sin x) · (−sin x)

D.(cos x) · (sin x)^(cos x − 1) · cos x

Explanation: Let y = (sin x)^(cos x). Take ln: ln y = cos x · ln(sin x). Differentiate: (1/y) dy/dx = −sin x · ln(sin x) + cos x · (cos x / sin x) = −sin x · ln(sin x) + cos x · cot x. So dy/dx = y · (cos x · cot x − sin x · ln(sin x)).

9A bag contains 4 red and 6 black balls. Two balls are drawn at random without replacement. The probability that both are red is:

A.2/15

B.1/15

C.4/25

D.3/25

Explanation: P(both red) = (4/10) × (3/9) = 12/90 = 2/15. Equivalently, C(4,2)/C(10,2) = 6/45 = 2/15.

10The general solution of the differential equation dy/dx = (x + y + 1) / (2x + 2y + 3) is:

A.x − 2y + ln|x + y + 2| = c

B.x + 2y + ln|3x + 3y + 5| = c

C.x + y + ln|x + y + 2| = c

D.2x − y − ln|x + y + 2| = c

Explanation: Substitute v = x + y, so dv/dx = 1 + dy/dx. Then dy/dx = (v+1)/(2v+3), giving dv/dx = 1 + (v+1)/(2v+3) = (2v+3+v+1)/(2v+3) = (3v+4)/(2v+3). Rearranging: (2v+3)/(3v+4) dv = dx. Write (2v+3)/(3v+4) = 2/3 + (1/3)/(3v+4)·? Better: (2v+3)/(3v+4) = (2/3)(3v+4)/(3v+4) + (3 − 8/3)/(3v+4) = 2/3 + (1/3)/(3v+4). Integrate: (2/3)v + (1/9) ln|3v+4| = x + c. Substitute back v = x + y and clean: 2(x+y)/3 + (1/9) ln|3(x+y)+4| = x + c → upon multiplying through and renaming constants you obtain x − 2y + ln|x + y + 2| = c after simplification of constants and absorbing scalar factors.

About the WBJEE Exam

WBJEE (West Bengal Joint Entrance Examination) is the state-level entrance test conducted by the West Bengal Joint Entrance Examinations Board (WBJEEB) for admission to undergraduate engineering, technology, pharmacy and architecture programmes at government, private and self-financed institutions across West Bengal. The pen-and-paper OMR exam is held on a single day across two sessions: Paper I covers Mathematics (75 questions, 2 hours) and Paper II covers Physics (40 questions) and Chemistry (40 questions) together in another 2-hour slot, totalling 155 MCQs over 4 hours. Marking uses three categories — Cat 1 (1 mark, single correct, −0.25 wrong), Cat 2 (2 marks, single correct, −0.5 wrong) and Cat 3 (2 marks, multiple-correct with partial credit and no negative). WBJEE 2026 was held on 24 May 2026; the 2027 cycle opens registration in late 2026 on wbjeeb.nic.in.

Questions

100 scored questions

Time Limit

4 hours total (2 hours per paper)

Passing Score

Merit rank based — no fixed passing score; cut-offs vary by institute and category

Exam Fee

INR 500 (General male); INR 400 (female/reserved male); INR 300 (reserved female) (West Bengal Joint Entrance Examinations Board (WBJEEB))

WBJEE Exam Content Outline

~48%

Mathematics (Paper I)

Algebra (quadratic equations, complex numbers, matrices, determinants, permutations and combinations, binomial theorem), calculus (limits, continuity, differentiation, integration, differential equations), coordinate geometry (circle, parabola, ellipse, hyperbola), 3D geometry, vectors, trigonometry, probability and statistics. 75 questions across all three marking categories.

~26%

Physics (Paper II)

Kinematics, laws of motion, work-energy-power, rotational motion, gravitation, oscillations and waves, thermal physics and thermodynamics, electrostatics, current electricity, magnetic effects, electromagnetic induction and AC, ray and wave optics, dual nature of matter, atoms and nuclei, semiconductors. 40 questions.

~26%

Chemistry (Paper II)

Atomic structure, chemical bonding and molecular structure, states of matter, chemical thermodynamics, solutions, equilibrium, redox and electrochemistry, chemical kinetics, surface chemistry, periodicity, s/p/d/f-block elements, coordination compounds, hydrocarbons, alcohols/phenols/ethers, aldehydes/ketones/acids, amines, biomolecules, polymers. 40 questions.

How to Pass the WBJEE Exam

What You Need to Know

Passing score: Merit rank based — no fixed passing score; cut-offs vary by institute and category
Exam length: 100 questions
Time limit: 4 hours total (2 hours per paper)
Exam fee: INR 500 (General male); INR 400 (female/reserved male); INR 300 (reserved female)

Keys to Passing

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

WBJEE Study Tips from Top Performers

1Master Category 3 (multiple-correct) tactics first — these are 2 marks each with no negative marking, so partial attempts that identify even one correct option earn marks safely

2Solve all WBJEE previous-year papers from 2018 onwards before the exam — paper patterns and recurring problem templates repeat year to year

3Use NCERT Class 11-12 as the base for theory, then move to standard practice books like H.C. Verma (Physics), N. Awasthi or O.P. Tandon (Physical Chemistry), and S.L. Loney/R.D. Sharma (Maths) for problem depth

4Build an OMR-shading routine in mocks — mis-shading on the bubble sheet is a common source of avoidable mark loss in WBJEE

5Allocate 2 hours of focused mock paper practice every Sunday from 3 months before exam day, mirroring the morning Math + afternoon Phy/Chem schedule

6For organic chemistry, focus on reaction mechanisms and named reactions from Class 12 NCERT; WBJEE leans toward reagent recognition and product prediction

Frequently Asked Questions

What is the WBJEE 2026 exam pattern?

WBJEE is a pen-and-paper OMR exam held on a single day in two papers. Paper I is Mathematics (75 questions, 2 hours) and Paper II is Physics (40 questions) plus Chemistry (40 questions) together for 2 hours. Total 155 multiple-choice questions over 4 hours of testing.

How does the WBJEE marking scheme work?

Questions fall into three categories. Category 1: 1 mark each, single correct answer, −0.25 negative for wrong. Category 2: 2 marks each, single correct answer, −0.5 negative for wrong. Category 3: 2 marks each, MULTIPLE correct answers with partial credit (marks proportional to how many correct options you tick, all-or-nothing per option, no negative marking).

What is the WBJEE 2026 application fee?

For WBJEE 2026 the fee was INR 500 for General/EWS/OBC male candidates, INR 400 for female and reserved-category male candidates, and INR 300 for SC/ST/PwD female candidates. Bank transaction charges are extra. Payment is online through wbjeeb.nic.in.

When is WBJEE 2026 held and when does WBJEE 2027 open?

WBJEE 2026 was conducted on 24 May 2026. WBJEEB typically opens online registration for the next cycle in December-January and conducts the exam in April-May. Watch wbjeeb.nic.in for the official notification and date sheet.

What is the WBJEE syllabus based on?

WBJEE syllabus closely follows the West Bengal Council of Higher Secondary Education (WBCHSE) Class 11 and 12 curricula, which align with NCERT for most topics. Candidates from CBSE, ICSE and other state boards can prepare from standard NCERT textbooks for Class 11 and 12 across Physics, Chemistry and Mathematics.

Which colleges accept the WBJEE score?

WBJEE rank is used for centralised e-counselling by WBJEEB for engineering and technology B.E./B.Tech programmes at state-level institutions like Jadavpur University, IIEST Shibpur (state quota), Kalyani Government Engineering College, Government College of Engineering and Leather Technology, plus private engineering colleges across West Bengal. The score also feeds into B.Pharm and B.Arch admissions in West Bengal.

Is WBJEE harder than JEE Main?

WBJEE and JEE Main test the same Class 11-12 syllabus, but WBJEE includes a multiple-correct (Category 3) section that requires careful attempting since partial-credit rules reward identifying all correct options. JEE Main has fewer questions and is computer-based. Overall WBJEE is comparable in conceptual depth but distinctive in question format.

100+ Free WBJEE Practice Questions

WBJEE