JEE Main 2025 previous year paper practice · PYQ-pattern set

JEE Main 2025 PYQ-pattern questions

Showing 10 of the 13 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · Kinematics

   A ball is thrown vertically upward with a speed of 25 m/s from the top of a tower 40 m high. What is the maximum height attained by the ball from the ground? (Take g = 10 m/s²)

   • A. 71.25 m
   • B. 65 m
   • C. 80 m
   • D. 75 m
   Answer and solution

   Answer: A

   Maximum height above the tower h = u²/(2g) = 625/20 = 31.25 m. Total maximum height from ground = 40 + 31.25 = 71.25 m.

2. Q2 · Electrostatics

   Two identical metallic spheres A and B have charges +6Q and -2Q respectively. They are brought into contact and then separated. What is the charge on sphere A after separation?

   • A. +Q
   • B. +2Q
   • C. +3Q
   • D. +4Q
   Answer and solution

   Answer: B

   When identical conductors touch, charge redistributes equally. Total charge = +6Q + (-2Q) = +4Q. After separation, each sphere has +4Q/2 = +2Q.

3. Q3 · Oscillations and Waves

   A transverse wave travels along a string with a speed of 80 m/s. If the frequency of the wave is 200 Hz, the wavelength (in cm) is

   • A. 20
   • B. 40
   • C. 60
   • D. 80
   Answer and solution

   Answer: B

   Wave speed v = fλ, where f is frequency and λ is wavelength. Thus λ = v/f = 80/200 = 0.4 m = 40 cm.

4. Q4 · Dual Nature of Matter and Radiation

   Light of wavelength 220 nm is incident on a metal surface with work function 3.5 eV. What is the maximum kinetic energy of the ejected photoelectrons? (Use h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

   • A. 1.2 eV
   • B. 2.1 eV
   • C. 5.6 eV
   • D. 0.8 eV
   Answer and solution

   Answer: B

   Energy of incident photon E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(220×10⁻⁹) = 9.04×10⁻¹⁹ J = 5.65 eV. Maximum KE = E - φ = 5.65 - 3.5 = 2.15 eV ≈ 2.1 eV.

5. Q5 · Some Basic Concepts in Chemistry

   A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. The empirical formula of the compound is:

   • A. CH₂O
   • B. C₂H₄O₂
   • C. CHO
   • D. C₃H₆O₃
   Answer and solution

   Answer: A

   Mole ratio: C: 40/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33. Simplest ratio = 1:2:1, giving empirical formula CH₂O.

6. Q6 · Coordination Compounds

   The complex [Co(NH₃)₄Cl₂]NO₃ exhibits geometrical isomerism. How many geometrical isomers are possible for this complex?

   • A. 2
   • B. 3
   • C. 4
   • D. No geometrical isomers
   Answer and solution

   Answer: A

   The complex is octahedral with formula [Ma₄b₂]⁺. It can have cis (Cl atoms adjacent) and trans (Cl atoms opposite) isomers, giving 2 geometrical isomers.

7. Q7 · Purification and Characterisation of Organic Compounds

   Which of the following techniques is most suitable for the purification of a solid organic compound that sublimes readily at atmospheric pressure?

   • A. Steam distillation
   • B. Sublimation
   • C. Crystallisation
   • D. Fractional distillation
   Answer and solution

   Answer: B

   Sublimation is the process where a solid directly converts to gas without passing through the liquid phase, making it ideal for purifying compounds like camphor, naphthalene, and benzoic acid that sublime at atmospheric pressure.

8. Q8 · Sequences and Series

   If the sum of the first n terms of an A.P. is given by Sₙ = 3n² + 5n, then the 15th term of this A.P. is:

   • A. 86
   • B. 89
   • C. 92
   • D. 95
   Answer and solution

   Answer: C

   aₙ = Sₙ - Sₙ₋₁. S₁₅ = 3(225)+5(15) = 750, S₁₄ = 3(196)+5(14) = 658. a₁₅ = 750 - 658 = 92.

9. Q9 · Trigonometry

   If tan α and tan β are the roots of the equation x² - px + q = 0, then the value of sin²(α + β) is:

   • A. p² / (p² + (1-q)²)
   • B. p² / (p² + q²)
   • C. (1-q)² / (p² + (1-q)²)
   • D. q² / (p² + q²)
   Answer and solution

   Answer: A

   Using tan(α+β) = (tan α + tan β)/(1 - tan α tan β) = p/(1-q). Then sin²(α+β) = tan²(α+β)/(1 + tan²(α+β)) = (p²/(1-q)²) / (1 + p²/(1-q)²) = p²/(p² + (1-q)²).

10. Q10 · Laws of Motion

    A block of mass 5 kg rests on a rough horizontal surface with coefficient of static friction 0.4. A horizontal force F is applied on the block such that it just begins to move. If the angle between the applied force and horizontal is 30°, the magnitude of F is (take g = 10 m/s²)

    • A. 20 N
    • B. 25 N
    • C. 30 N
    • D. 35 N
    Answer and solution

    Answer: A

    With force at 30° above horizontal, normal force N = mg - F sin30. At onset of motion: F cos30 = μN = μ(mg - F sin30). So F(0.866) = 0.4(50 - 0.5F) = 20 - 0.2F, giving 1.066F = 20, F ≈ 18.8 N, closest to 20 N.