Preparing for JEE Main B.E./B.Tech means mastering not just what to solve, but how to think through each problem. Below are 10 questions — 3 from Physics, 3 from Chemistry, and 4 from Mathematics — each with a detailed solution that shows the reasoning, not just the final answer. Use these to sharpen your approach before the exam.
Physics
Question 1: Kinematics
A particle moves along the x-axis with velocity v(t) = 3t² - 6t m/s. If it starts at x = 2 m at t = 0, find its position at t = 2 s.
Solution: Position is the integral of velocity: x(t) = ∫ v(t) dt = ∫ (3t² - 6t) dt = t³ - 3t² + C. At t = 0, x = 2 → C = 2. So x(t) = t³ - 3t² + 2. At t = 2: x = 8 - 12 + 2 = -2 m.
Question 2: Work and Energy
A block of mass 2 kg is pushed up a 30° incline with a constant force of 20 N parallel to the incline. The coefficient of kinetic friction is 0.2. Find the work done by friction when the block moves 3 m up the incline. (g = 10 m/s²)
Solution: Normal force = mg cosθ = 2 × 10 × cos30° = 20 × √3/2 = 17.32 N. Friction force = μ N = 0.2 × 17.32 = 3.464 N. Work done by friction = -friction force × distance = -3.464 × 3 = -10.39 J.
Question 3: Electrostatics
Two point charges +4 μC and -2 μC are placed 0.3 m apart in air. Find the electric field at the midpoint of the line joining them. (k = 9 × 10⁹ N m²/C²)
Solution: Midpoint distance from each charge = 0.15 m. Field due to +4 μC: E₁ = k × (4 × 10⁻⁶) / (0.15)² = 9 × 10⁹ × 4 × 10⁻⁶ / 0.0225 = 1.6 × 10⁶ N/C away from the charge. Field due to -2 μC: E₂ = k × (2 × 10⁻⁶) / (0.15)² = 9 × 10⁹ × 2 × 10⁻⁶ / 0.0225 = 0.8 × 10⁶ N/C toward the charge (same direction as E₁). Net field = 1.6 × 10⁶ + 0.8 × 10⁶ = 2.4 × 10⁶ N/C.
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Chemistry
Question 4: Thermodynamics
Calculate the enthalpy change for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l). Given bond energies: H-H = 436 kJ/mol, O=O = 498 kJ/mol, O-H = 464 kJ/mol.
Solution: Bonds broken: 2 H-H + 1 O=O = 2×436 + 498 = 1370 kJ. Bonds formed: 4 O-H = 4×464 = 1856 kJ. ΔH = bonds broken - bonds formed = 1370 - 1856 = -486 kJ.
Question 5: Equilibrium
The pH of a 0.1 M acetic acid solution is 2.87. Find its Ka. (log 1.35 = 0.13)
Solution: pH = 2.87 → [H⁺] = 10⁻²·⁸⁷ = 1.35 × 10⁻³ M. For weak acid, [H⁺] = √(Ka × C) → Ka = [H⁺]² / C = (1.35 × 10⁻³)² / 0.1 = 1.8225 × 10⁻⁶ / 0.1 = 1.82 × 10⁻⁵.
Question 6: Organic Chemistry
Identify the product when propene reacts with HBr in the presence of peroxide.
Solution: Peroxide causes anti-Markovnikov addition. H adds to the carbon with more H atoms (C1), Br adds to C2. Product: 1-bromopropane (CH₃CH₂CH₂Br).
Mathematics
Question 7: Calculus
Find the area bounded by the curve y = x² and the line y = x + 2.
Solution: Intersection: x² = x + 2 → x² - x - 2 = 0 → (x-2)(x+1) = 0 → x = -1, 2. Area = ∫ from -1 to 2 of (x+2 - x²) dx = [x²/2 + 2x - x³/3] from -1 to 2. At 2: 2 + 4 - 8/3 = 6 - 8/3 = 10/3. At -1: 1/2 - 2 + 1/3 = -1 + 1/6 = -5/6. Area = 10/3 - (-5/6) = 10/3 + 5/6 = 20/6 + 5/6 = 25/6 sq units.
Question 8: Vectors and 3D
Find the angle between the vectors a = 2i - j + k and b = i + 2j - k.
Solution: Dot product: a·b = 2×1 + (-1)×2 + 1×(-1) = 2 - 2 - 1 = -1. |a| = √(4+1+1) = √6, |b| = √(1+4+1) = √6. cos θ = (a·b)/(|a||b|) = -1/6 → θ = cos⁻¹(-1/6).
Question 9: Probability
A bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Find the probability that both are red.
Solution: P(first red) = 3/8. After one red, remaining: 2 red, 5 blue → P(second red) = 2/7. P(both red) = (3/8) × (2/7) = 6/56 = 3/28.
Question 10: Matrices
If A = [[1, 2], [3, 4]], find A² - 5A + 2I.
Solution: A² = [[1×1+2×3, 1×2+2×4], [3×1+4×3, 3×2+4×4]] = [[7, 10], [15, 22]]. 5A = [[5, 10], [15, 20]]. 2I = [[2, 0], [0, 2]]. A² - 5A + 2I = [[7-5+2, 10-10+0], [15-15+0, 22-20+2]] = [[4, 0], [0, 4]].
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