JEE Advanced Practice Set 1

JEE Advanced · 2026 · with answers and explanations

questions
30
Total marks
90
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 3 marks · −1

A projectile is fired at an angle θ with the horizontal with speed u. At the highest point, its speed is u/2. The angle of projection θ is:

  1. A.30°
  2. B.45°
  3. C.60°
  4. D.75°

Correct answer: C

Explanation: At the highest point, only the horizontal component remains: ucosθ = u/2, so cosθ = 1/2, giving θ = 60°.

Question 2

Physics · 3 marks · −1

A particle undergoes uniform circular motion with angular velocity ω in a circle of radius R. If the centripetal acceleration is 4π²R and the time period is T, then T equals:

  1. A.0.5 s
  2. B.1 s
  3. C.2 s
  4. D.π s

Correct answer: B

Explanation: Centripetal acceleration a = ω²R = (2π/T)²R = 4π²R/T². Given a = 4π²R, so 4π²R/T² = 4π²R, giving T² = 1, hence T = 1 s.

Question 3

Physics · 3 marks · −1

A uniform rod of mass M and length L is pivoted at one end and released from a horizontal position. The angular velocity of the rod when it becomes vertical is:

  1. A.√(g/L)
  2. B.√(2g/L)
  3. C.√(3g/L)
  4. D.√(6g/L)

Correct answer: C

Explanation: The centre of mass falls through L/2. By energy conservation: Mg(L/2) = ½Iω², where I = ML²/3 (about the pivot). So MgL/2 = ½(ML²/3)ω², giving ω² = 3g/L, hence ω = √(3g/L).

Question 4

Physics · 3 marks · −1

An ideal gas undergoes an isothermal expansion to twice its original volume. If the initial pressure is P, the work done by the gas is:

  1. A.PV
  2. B.PV ln 2
  3. C.2PV
  4. D.PV/ln 2

Correct answer: B

Explanation: For isothermal process, W = nRT ln(V₂/V₁) = PV ln(2), since PV = nRT. The work done is PV ln 2.

Question 5

Physics · 3 marks · −1

In a double-slit experiment, the fringe width is β when the wavelength used is 600 nm. If the wavelength is changed to 400 nm, the new fringe width is:

  1. A.β/2
  2. B.2β/3
  3. C.3β/2
  4. D.3β/4

Correct answer: B

Explanation: Fringe width β = λD/d. Since D and d are constant, β ∝ λ. New fringe width = β × (400/600) = 2β/3.

Question 6

Physics · 3 marks · −1

A particle executes SHM with amplitude A. At what displacement from the mean position is its kinetic energy equal to its potential energy?

  1. A.A/2
  2. B.A/√2
  3. C.A/4
  4. D.A√3/2

Correct answer: B

Explanation: KE = ½ mω²(A² - x²) and PE = ½mω²x². Setting KE = PE: A² - x² = x², so 2x² = A², giving x = A/√2.

Question 7

Physics · 3 marks · −1

An ideal gas at temperature T has rms speed v. If the temperature is increased to 4T, the new rms speed is:

  1. A.v/2
  2. B.v
  3. C.2v
  4. D.4v

Correct answer: C

Explanation: v_rms = √(3kT/m). When temperature becomes 4T, v'_rms = √(3k(4T)/m) = 2√(3kT/m) = 2v.

Question 8

Physics · 3 marks · −1

Two charged conducting spheres of radii R and 2R are connected by a wire. If the total charge is Q, the charge on the smaller sphere after equilibrium is:

  1. A.Q/2
  2. B.Q/3
  3. C.Q/4
  4. D.2Q/3

Correct answer: B

Explanation: When connected, potentials equalise: q₁/(4πε₀R) = q₂/(4πε₀(2R)), so q₁/R = q₂/(2R), giving q₂ = 2q₁. Since q₁ + q₂ = Q: q₁ + 2q₁ = Q, so q₁ = Q/3.

Question 9

Physics · 3 marks · −1

In the Bohr model of hydrogen, the ratio of the kinetic energy to the total energy of an electron in the nth orbit is:

  1. A.1
  2. B.-1
  3. C.2
  4. D.-2

Correct answer: B

Explanation: In the Bohr model, KE = -E_total (total energy is negative). The ratio KE/E_total = -1. The kinetic energy equals the magnitude of total energy but is positive while total energy is negative.

Question 10

Physics · 3 marks · −1

A copper wire of length L and area of cross-section A has resistance R. If the wire is stretched uniformly to length 3L, the new resistance is:

  1. A.3R
  2. B.6R
  3. C.9R
  4. D.27R

Correct answer: C

Explanation: Volume remains constant: AL = A'(3L), so A' = A/3. New resistance R' = ρ(3L)/(A/3) = 9ρL/A = 9R.

Question 11

Chemistry · 3 marks · −1

For a first-order reaction, the time required for 99% completion is approximately how many times the time required for 90% completion?

  1. A.1.5
  2. B.2
  3. C.5
  4. D.10

Correct answer: B

Explanation: For first-order: t = (2.303/k) log(a/(a-x)). t₉₉ = (2.303/k)log(100/1) = (2.303/k)(2). t₉₀ = (2.303/k)log(100/10) = (2.303/k)(1). So t₉₉/t₉₀ = 2.

Question 12

Chemistry · 3 marks · −1

The boiling point of a solution containing 18 g of glucose (molar mass 180 g/mol) in 100 g of water is (Kᵇ for water = 0.52 K·kg/mol, boiling point of water = 100°C):

  1. A.100.26°C
  2. B.100.52°C
  3. C.101.04°C
  4. D.100.78°C

Correct answer: B

Explanation: Moles of glucose = 18/180 = 0.1 mol. Molality = 0.1/0.1 = 1 m. ΔTᵇ = Kᵇ × m = 0.52 × 1 = 0.52 K. Boiling point = 100.52°C.

Question 13

Chemistry · 3 marks · −1

The common ion effect causes a decrease in the solubility of a sparingly soluble salt. If the Kₛₚ of AgCl is 1.6 × 10⁻¹⁰, the solubility of AgCl in 0.1 M NaCl solution is:

  1. A.1.6 × 10⁻⁵ M
  2. B.1.6 × 10⁻⁹ M
  3. C.1.6 × 10⁻¹¹ M
  4. D.1.26 × 10⁻⁵ M

Correct answer: B

Explanation: In 0.1 M NaCl, [Cl⁻] ≈ 0.1 M. Kₛₚ = [Ag⁺][Cl⁻] = s × 0.1. So s = 1.6×10⁻¹⁰/0.1 = 1.6×10⁻⁹ M.

Question 14

Chemistry · 3 marks · −1

According to molecular orbital theory, the bond order of O₂⁺ is:

  1. A.1.5
  2. B.2.0
  3. C.2.5
  4. D.3.0

Correct answer: C

Explanation: O₂ has 16 electrons (BO = 2). O₂⁺ has 15 electrons — one electron removed from the antibonding π* orbital. Bond order = (10-5)/2 = 2.5.

Question 15

Chemistry · 3 marks · −1

The d-block element that shows the maximum number of oxidation states is:

  1. A.Chromium
  2. B.Manganese
  3. C.Iron
  4. D.Cobalt

Correct answer: B

Explanation: Manganese (Mn) shows oxidation states from +2 to +7 (and 0), the maximum among 3d elements. This is because it has 5 unpaired d-electrons plus 2 s-electrons, allowing a wide range of oxidation states.

Question 16

Chemistry · 3 marks · −1

The number of geometrical isomers possible for the complex [Ma₂b₂] in a square planar geometry is:

  1. A.1
  2. B.2
  3. C.3
  4. D.4

Correct answer: B

Explanation: The square planar complex [Ma₂b₂] can exist as cis (same ligands adjacent) and trans (same ligands opposite) isomers. So there are 2 geometrical isomers.

Question 17

Chemistry · 3 marks · −1

The major product formed when phenol reacts with bromine water is:

  1. A.2-Bromophenol
  2. B.4-Bromophenol
  3. C.2,4-Dibromophenol
  4. D.2,4,6-Tribromophenol

Correct answer: D

Explanation: Phenol is highly activated towards electrophilic aromatic substitution. With bromine water (excess), all three ortho/para positions are brominated to give 2,4,6-tribromophenol as a white precipitate.

Question 18

Chemistry · 3 marks · −1

Aniline can be distinguished from N-methylaniline by which of the following tests?

  1. A.Hinsberg test
  2. B.Carbylamine test
  3. C.Litmus test
  4. D.Biuret test

Correct answer: B

Explanation: The carbylamine (isocyanide) test is given by primary amines only. Aniline is a primary amine and gives a positive carbylamine test, while N-methylaniline is a secondary amine and does not.

Question 19

Chemistry · 3 marks · −1

The reaction of chlorobenzene with sodium in dry ether to form biphenyl is known as:

  1. A.Friedel-Crafts reaction
  2. B.Wurtz reaction
  3. C.Fittig reaction
  4. D.Sandmeyer reaction

Correct answer: C

Explanation: The Wurtz-Fittig reaction involves the coupling of an aryl halide with an alkyl halide (or two aryl halides) using sodium in dry ether. When two moles of chlorobenzene react with sodium, biphenyl is formed — this specific case is the Fittig reaction.

Question 20

Chemistry · 3 marks · −1

For an exothermic reaction at equilibrium, if the temperature is increased, which of the following is true?

  1. A.Equilibrium shifts forward, K increases
  2. B.Equilibrium shifts backward, K decreases
  3. C.Equilibrium shifts forward, K decreases
  4. D.No change in equilibrium or K

Correct answer: B

Explanation: By Le Chatelier's principle, increasing temperature for an exothermic reaction shifts equilibrium towards reactants (backward direction). The equilibrium constant K decreases with increase in temperature for exothermic reactions.

Question 21

Mathematics · 3 marks · −1

The solution of the differential equation dy/dx = y/x, with y(1) = 2, is:

  1. A.y = x + 1
  2. B.y = 2x
  3. C.y = x² + 1
  4. D.y = 2x²

Correct answer: B

Explanation: Separating variables: dy/y = dx/x. Integrating: ln|y| = ln|x| + C, so y = kx. Using y(1) = 2: k = 2. The solution is y = 2x.

Question 22

Mathematics · 3 marks · −1

The value of ∫₀⁻ x² e⁻ˣ dx is (where the integral is from 0 to infinity, using the Gamma function):

  1. A.1
  2. B.2
  3. C.π
  4. D.√π

Correct answer: B

Explanation: Using the substitution and Gamma function: ∫₀⁻ x²e⁻ˣdx = Γ(3) = 2! = 2.

Question 23

Mathematics · 3 marks · −1

The eccentricity of the ellipse x²/25 + y²/16 = 1 is:

  1. A.3/5
  2. B.4/5
  3. C.5/3
  4. D.3/4

Correct answer: A

Explanation: For the ellipse x²/a² + y²/b² = 1 with a > b: a² = 25, b² = 16. e = √(1 - b²/a²) = √(1 - 16/25) = √(9/25) = 3/5.

Question 24

Mathematics · 3 marks · −1

The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:

  1. A.π/6
  2. B.π/4
  3. C.π/3
  4. D.π/2

Correct answer: C

Explanation: The normals are n₁ = (2,-1,1) and n₂ = (1,1,2). cosθ = |n₁·n₂|/(|n₁||n₂|) = |2-1+2|/(√6 × √6) = 3/6 = 1/2. So θ = 60° = π/3.

Question 25

Mathematics · 3 marks · −1

If vectors a⃗ = 2î + 3ĵ + k̂ and b⃗ = î - ĵ + 2k̂, then a⃗ × b⃗ is:

  1. A.7î - 3ĵ - 5k̂
  2. B.7î + 3ĵ - 5k̂
  3. C.-7î + 3ĵ + 5k̂
  4. D.7î - 3ĵ + 5k̂

Correct answer: A

Explanation: a⃗ × b⃗ = |î ĵ k̂; 2 3 1; 1 -1 2| = î(6-(-1)) - ĵ(4-1) + k̂(-2-3) = 7î - 3ĵ - 5k̂.

Question 26

Mathematics · 3 marks · −1

Five persons are to be seated in a row. The probability that two particular persons are always adjacent is:

  1. A.1/5
  2. B.2/5
  3. C.3/5
  4. D.1/10

Correct answer: B

Explanation: Total arrangements = 5! = 120. Treating the two persons as one unit: 4! × 2! = 48 arrangements (2! for internal arrangement). Probability = 48/120 = 2/5.

Question 27

Mathematics · 3 marks · −1

If A is a 3×3 matrix with |A| = -2, then |adj(A)| is:

  1. A.-4
  2. B.4
  3. C.-8
  4. D.8

Correct answer: B

Explanation: For a square matrix of order n, |adj(A)| = |A|^(n-1). Here n = 3, so |adj(A)| = (-2)² = 4.

Question 28

Mathematics · 3 marks · −1

The value of lim(x→0) (eˣ - 1 - x)/x² is:

  1. A.0
  2. B.1/2
  3. C.1
  4. D.Does not exist

Correct answer: B

Explanation: Using L'Hôpital's rule (0/0 form): lim(x→0) (eˣ - 1)/(2x), still 0/0, apply again: lim(x→0) eˣ/2 = 1/2.

Question 29

Mathematics · 3 marks · −1

If z = (1 + i)/(1 - i), then z⁴ equals:

  1. A.1
  2. B.-1
  3. C.i
  4. D.-i

Correct answer: A

Explanation: z = (1+i)/(1-i) × (1+i)/(1+i) = (1+2i+i²)/(1-i²) = (1+2i-1)/(1+1) = 2i/2 = i. So z⁴ = i⁴ = 1.

Question 30

Mathematics · 3 marks · −1

The number of ways in which 5 boys and 5 girls can be seated in a row so that no two girls sit together is:

  1. A.86400
  2. B.14400
  3. C.7200
  4. D.120 × 120

Correct answer: A

Explanation: First arrange 5 boys in 5! ways. This creates 6 gaps (including ends). Choose 5 of these 6 gaps for the girls: C(6,5) ways, then arrange the 5 girls in 5! ways. Total = 5! × 6 × 5! = 120 × 6 × 120 = 86400.

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