JEE Advanced Practice Set 2
Original pattern-based JEE Advanced full-length mock with 36 questions: 12 Mathematics, 12 Physics, and 12 Chemistry for practical two-paper strategy practice.
Exam Pattern & Rules
Negative Marking: May apply for some questions
Note: Both papers are compulsory. Pattern is not fixed in the same way as JEE Main.
Practice Set Details
JEE Advanced Practice Set 2: all 36 questions with answers
Try each question first, then open “Show answer” to check the correct option and explanation.
Q1.Solve for x: 7x + 10 = 66.
- A.9
- B.7
- C.10
- D.8
Show answer
Correct answer: D. 8
Subtract 10 from both sides: 7x = 56. Divide by 7: x = 8.
Q2.One root of x² − 18x + 80 = 0 is 8. The other root is:
- A.9
- B.18
- C.10
- D.11
Show answer
Correct answer: C. 10
Product of roots = 80. Other root = 80 ÷ 8 = 10.
Q3.The 8th term of an AP with first term 7 and common difference 7 is:
- A.64
- B.56
- C.49
- D.63
Show answer
Correct answer: B. 56
aₙ = a + (n−1)d = 7 + 7×7 = 56.
Q4.Slope of the line through (6, 7) and (9, 28) is:
- A.7
- B.8
- C.6
- D.9
Show answer
Correct answer: A. 7
Slope = (28−7) ÷ (9−6) = 21 ÷ 3 = 7.
Q5.Value of sin 90° is:
- A.0
- B.−1
- C.1/2
- D.1
Show answer
Correct answer: D. 1
sin 90° = 1.
Q6.A bag contains 8 red and 10 blue balls. Probability of drawing a red ball is:
- A.8/17
- B.11/19
- C.8/18
- D.10/19
Show answer
Correct answer: C. 8/18
P(red) = 8/18.
Q7.Determinant of matrix [[7, 1], [3, 8]] is:
- A.23
- B.53
- C.58
- D.50
Show answer
Correct answer: B. 53
det = ad − bc = 7×8 − 1×3 = 53.
Q8.Evaluate: log₂ₐₑₒₓ3(3^7) [log base 3].
- A.7
- B.8
- C.6
- D.21
Show answer
Correct answer: A. 7
logₐ(a^n) = n, so log_3(3^7) = 7.
Q9.Find 11P4.
- A.1980
- B.7928
- C.7912
- D.7920
Show answer
Correct answer: D. 7920
11P4 = 11!/(11−4)! = 11×10×9×8 = 7920.
Q10.Find 12C2.
- A.69
- B.63
- C.66
- D.24
Show answer
Correct answer: C. 66
nC2 = n(n−1)/2 = 12×11/2 = 66.
Q11.Simple interest on Rs 1500 at 9% p.a. for 3 years is:
- A.Rs 810
- B.Rs 405
- C.Rs 675
- D.Rs 270
Show answer
Correct answer: B. Rs 405
SI = P×R×T/100 = 1500×9×3/100 = Rs 405.
Q12.Area of a triangle with base 11 cm and height 9 cm is:
- A.49.5 cm²
- B.99 cm²
- C.55.5 cm²
- D.43.5 cm²
Show answer
Correct answer: A. 49.5 cm²
Area = (1/2) × base × height = (1/2) × 11 × 9 = 49.5 cm².
Q13.A body starts with velocity 7 m/s and has acceleration 7 m/s². Its velocity after 3 s is:
- A.21 m/s
- B.35 m/s
- C.10 m/s
- D.28 m/s
Show answer
Correct answer: D. 28 m/s
v = u + at = 7 + 7×3 = 28 m/s.
Q14.For u = 6 m/s, a = 2 m/s², t = 5 s, displacement is:
- A.65 m
- B.53 m
- C.55 m
- D.30 m
Show answer
Correct answer: C. 55 m
s = ut + ½At² = 6×5 + 0.5×2×5² = 55 m.
Q15.Net force needed to give 7 kg mass an acceleration of 8 m/s² is:
- A.15 N
- B.56 N
- C.64 N
- D.48 N
Show answer
Correct answer: B. 56 N
F = ma = 7×8 = 56 N.
Q16.Work done when force 35 N moves a body 8 m in its direction is:
- A.280 J
- B.315 J
- C.245 J
- D.43 J
Show answer
Correct answer: A. 280 J
W = F×d = 35×8 = 280 J.
Q17.A body does 350 J of work in 10 s. Its power is:
- A.70 W
- B.45 W
- C.30 W
- D.35 W
Show answer
Correct answer: D. 35 W
P = W/t = 350/10 = 35 W.
Q18.A body of mass 8 kg moves at 9 m/s. Its momentum is:
- A.63 kg·m/s
- B.17 kg·m/s
- C.72 kg·m/s
- D.80 kg·m/s
Show answer
Correct answer: C. 72 kg·m/s
p = mv = 8×9 = 72 kg·m/s.
Q19.Kinetic energy of 7 kg body moving at 9 m/s is:
- A.276.5 J
- B.283.5 J
- C.63 J
- D.346.5 J
Show answer
Correct answer: B. 283.5 J
KE = ½mv² = 0.5×7×9² = 283.5 J.
Q20.Voltage across a 8 Ω resistor carrying 7 A current is:
- A.56 V
- B.64 V
- C.48 V
- D.17 V
Show answer
Correct answer: A. 56 V
V = IR = 7×8 = 56 V.
Q21.Two resistors 9 Ω and 11 Ω connected in series. Total resistance is:
- A.99 Ω
- B.2 Ω
- C.22 Ω
- D.20 Ω
Show answer
Correct answer: D. 20 Ω
Series: R = R₁ + R₂ = 9 + 11 = 20 Ω.
Q22.Power of a lens with focal length 70 cm is:
- A.2.43 D
- B.0.7 D
- C.1.43 D
- D.70 D
Show answer
Correct answer: C. 1.43 D
P = 1/f = 1/0.7 m = 1.43 D.
Q23.Wave speed for frequency 150 Hz and wavelength 7 m is:
- A.157 m/s
- B.1050 m/s
- C.1200 m/s
- D.980 m/s
Show answer
Correct answer: B. 1050 m/s
v = fλ = 150×7 = 1050 m/s.
Q24.Standard acceleration due to gravity at Earth's surface is:
- A.9.8 m/s²
- B.6.7 m/s²
- C.11.2 m/s²
- D.3.7 m/s²
Show answer
Correct answer: A. 9.8 m/s²
g ≈ 9.8 m/s² at Earth's surface.
Q25.If [H⁺] = 10⁻6 mol/L, the pH of the solution is:
- A.7
- B.8
- C.5
- D.6
Show answer
Correct answer: D. 6
pH = −log[H⁺] = −log(10⁻6) = 6.
Q26.Molar mass of HCl is:
- A.37 g/mol
- B.18 g/mol
- C.36.5 g/mol
- D.35.5 g/mol
Show answer
Correct answer: C. 36.5 g/mol
Add atomic masses: molar mass of HCl = 36.5 g/mol.
Q27.Oxidation state of oxygen in H₂O is:
- A.−1
- B.−2
- C.+2
- D.0
Show answer
Correct answer: B. −2
H is +1 (×2), so O = −2.
Q28.Formula of iron(III) chloride is:
- A.FeCl₃
- B.FeCl₂
- C.Fe₃Cl
- D.FeCl
Show answer
Correct answer: A. FeCl₃
Fe(+3) needs 3 Cl(−1).
Q29.Gas produced when Na reacts with water is:
- A.oxygen
- B.sodium oxide
- C.hydroxide gas
- D.hydrogen
Show answer
Correct answer: D. hydrogen
2Na + 2H₂O → 2NaOH + H₂↑.
Q30.Blue litmus paper turns red in:
- A.neutral salt solution
- B.distilled water
- C.acid
- D.base
Show answer
Correct answer: C. acid
Acids turn blue litmus red.
Q31.IUPAC name of CH₃OH is:
- A.methanal
- B.methanol
- C.ethanol
- D.propanol
Show answer
Correct answer: B. methanol
CH₃OH is methanol.
Q32.Which element has electronic configuration 2,8,7?
- A.Chlorine
- B.Fluorine
- C.Bromine
- D.Neon
Show answer
Correct answer: A. Chlorine
2+8+7=17 protons → Cl.
Q33.Number of moles in 40 g of NaOH (M = 40 g/mol) is:
- A.2
- B.3
- C.2.5
- D.1
Show answer
Correct answer: D. 1
n = mass/M = 40/40 = 1 mol.
Q34.Electron configuration of Na⁺ is:
- A.1s²2s²2p⁶ 3s²
- B.1s²2s²2p⁵
- C.1s²2s²2p⁶
- D.1s²2s²2p⁶ 3s¹
Show answer
Correct answer: C. 1s²2s²2p⁶
Na loses 1e⁻ → 10 electrons (Ne config).
Q35.Hybridisation of carbon in methane (CH₄) is:
- A.dsp²
- B.sp³
- C.sp²
- D.sp
Show answer
Correct answer: B. sp³
CH₄ is tetrahedral; C is sp³ hybridised.
Q36.Rust is chemically:
- A.Fe₂O₃·xH₂O
- B.FeO
- C.Fe₃O₄
- D.FeSO₄
Show answer
Correct answer: A. Fe₂O₃·xH₂O
Rust = hydrated iron(III) oxide.