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

JEE Advanced 2025 PYQ-pattern questions

Showing 10 of the 22 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 · General

   The dimensions of the physical quantity (ε₀E²), where ε₀ is the permittivity of free space and E is the electric field, are the same as:

   • A. Energy
   • B. Power
   • C. Pressure
   • D. Force
   Answer and solution

   Answer: C

   ε₀ has dimensions [M⁻¹L⁻³T⁴A²] and E has [MLT⁻³A⁻¹]. Therefore ε₀E² has dimensions [M⁻¹L⁻³T⁴A²][M²L²T⁻⁶A⁻²] = [ML⁻¹T⁻²], which is pressure (force per unit area).

2. Q2 · States of Matter: Gases and Liquids

   A closed vessel contains an ideal gas at 27°C. If the root mean square speed of gas molecules is 400 m/s, what will be the temperature (in °C) at which the root mean square speed becomes 600 m/s?

   • A. 402°C
   • B. 402 K
   • C. 675°C
   • D. 405°C
   Answer and solution

   Answer: A

   vrms proportional to sqrt(T). T1 = 300 K. T2 = 300*(600/400)^2 = 300*2.25 = 675 K = 675-273 = 402°C.

3. Q3 · Electrochemistry

   A current of 2.5 A is passed through an aqueous solution of CuSO₄ for 40 minutes. The mass of copper deposited at the cathode is approximately: (Atomic mass of Cu = 63.5, Faraday constant F = 96500 C/mol)

   • A. 1.98 g
   • B. 3.96 g
   • C. 0.99 g
   • D. 7.92 g
   Answer and solution

   Answer: A

   Using Faraday's law: m = (Q × M)/(n × F). Q = I × t = 2.5 × 2400 = 6000 C. For Cu²⁺, n = 2. m = (6000 × 63.5)/(2 × 96500) ≈ 1.97 g.

4. Q4 · Classification of Elements and Periodicity in Properties

   Consider four consecutive elements in the third period with atomic numbers Z, Z+1, Z+2, and Z+3. The first ionization energies follow the order I₁(Z) < I₁(Z+1) < I₁(Z+3) < I₁(Z+2). Which of the following represents element Z?

   • A. Mg
   • B. Al
   • C. Si
   • D. Na
   Answer and solution

   Answer: B

   For Z=Al: Al(577)<Si(786)<S(1000)<P(1012). This gives I₁(Z)<I₁(Z+1)<I₁(Z+3)<I₁(Z+2) since Z+2=P has the highest due to its half-filled 3p³ stability exceeding S. Matches exactly.

5. Q5 · p-Block Elements

   Among the following, the number of species that are planar is: XeF₄, SF₄, BrF₅, ClF₃

   • A. 0
   • B. 1
   • C. 2
   • D. 3
   Answer and solution

   Answer: C

   XeF₄ is square planar (all atoms in one plane). ClF₃ is T-shaped which is also planar (all 4 atoms lie in a single plane). SF₄ (seesaw) and BrF₅ (square pyramidal) are non-planar. Thus 2 species are planar.

6. Q6 · Coordination Compounds

   A coordination compound with formula [M(NH₃)₄Br₂]Cl exhibits geometrical isomerism but not optical isomerism. The geometry and magnetic nature of the complex if M is Co³⁺ are respectively:

   • A. Octahedral, diamagnetic
   • B. Tetrahedral, paramagnetic
   • C. Square planar, paramagnetic
   • D. Octahedral, paramagnetic
   Answer and solution

   Answer: A

   Co³⁺ with strong field NH₃ ligands forms low-spin octahedral complex with d⁶ configuration (all electrons paired), hence diamagnetic. [MA₄B₂] octahedral complexes show cis-trans isomerism.

7. Q7 · Alkanes

   When 2-methylbutane undergoes monochlorination in the presence of sunlight, how many distinct monochlorinated products (excluding stereoisomers) are formed?

   • A. 3
   • B. 4
   • C. 5
   • D. 6
   Answer and solution

   Answer: B

   2-methylbutane has four distinct types of hydrogen atoms: at C-1, C-2, C-3, and C-4 (methyl). Chlorination at each position gives four different monochlorinated products.

8. Q8 · Carboxylic Acids

   Which of the following carboxylic acids is the MOST acidic?

   • A. CH₃CH₂COOH
   • B. ClCH₂CH₂COOH
   • C. CH₃CHClCOOH
   • D. Cl₂CHCOOH
   Answer and solution

   Answer: D

   Acidity of carboxylic acids increases with electron-withdrawing groups closer to the –COOH group. Cl₂CHCOOH has two chlorine atoms on the α-carbon, making it the most acidic due to maximum stabilization of the carboxylate anion.

9. Q9 · Biomolecules

   A linear hexasaccharide (molar mass = 918 g mol⁻¹) on complete hydrolysis produces two monosaccharides: glucose and fructose. The amount of glucose formed is 64.71% (w/w) of the total monosaccharides produced. How many glucose units are present in one molecule of the hexasaccharide? (Given: Molar mass in g mol⁻¹: glucose = 180, fructose = 180; Atomic mass in amu: H = 1, O = 16)

   • A. 3
   • B. 4
   • C. 2
   • D. 5
   Answer and solution

   Answer: B

   On hydrolysis, hexasaccharide + 5H₂O → 6 monosaccharides. Total mass of monosaccharides = 918 + 90 = 1008 g. If glucose is 64.71%, mass = 652 g, giving 652/180 ≈ 3.62 ≈ 4 units of glucose.

10. Q10 · Algebra

    If α, β are the roots of x² - 3x + 1 = 0, then the value of α⁷ + β⁷ is:

    • A. 322
    • B. 843
    • C. 643
    • D. 987
    Answer and solution

    Answer: B

    Using Newton's identity S_n = 3·S_(n-1) - S_(n-2) with S_0=2, S_1=3: S_2=7, S_3=18, S_4=47, S_5=123, S_6=322, S_7=843.