MHT-CET 2025 previous year paper practice · PYQ-pattern set

MHT-CET 2025 PYQ-pattern questions

Showing 10 of the 19 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 · Units and Measurement (Class 11 — ~20% weight)

   The dimensions of coefficient of viscosity are:

   • A. [ML⁻¹T⁻¹]
   • B. [MLT⁻²]
   • C. [ML²T⁻¹]
   • D. [M⁰L⁰T⁰]
   Answer and solution

   Answer: A

   Coefficient of viscosity (η) is defined as force per unit area per unit velocity gradient. Its dimensional formula is [ML⁻¹T⁻¹].

2. Q2 · Thermal Properties of Matter (Class 11 — ~20% weight)

   A metallic rod of length 1 m has its one end at 100°C and the other end at 0°C. The temperature at a point 40 cm from the hot end in steady state is:

   • A. 40°C
   • B. 60°C
   • C. 50°C
   • D. 70°C
   Answer and solution

   Answer: B

   In steady state, temperature varies linearly. At 40 cm from hot end, T = 100 – (100/100) × 40 = 100 – 40 = 60°C.

3. Q3 · Mechanical Properties of Solids (Class 12)

   A steel wire of length 3 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by 1.5 mm under a load. If Young's modulus for steel is 2 × 10¹¹ N/m², what is the applied force?

   • A. 100 N
   • B. 200 N
   • C. 300 N
   • D. 400 N
   Answer and solution

   Answer: B

   Y = (F/A) × (L/ΔL). Rearranging, F = Y × A × (ΔL/L) = 2×10¹¹ × 2×10⁻⁶ × (1.5×10⁻³/3) = 200 N.

4. Q4 · Electrostatics (Class 12)

   Two point charges of +3 μC and +5 μC are placed 20 cm apart in air. What is the magnitude of the electrostatic force between them? (k = 9 × 10⁹ N m²/C²)

   • A. 2.7 N
   • B. 3.4 N
   • C. 4.5 N
   • D. 5.0 N
   Answer and solution

   Answer: B

   F = k q₁q₂/r² = (9×10⁹ × 3×10⁻⁶ × 5×10⁻⁶) / (0.2)² = 135×10⁻³ / 0.04 = 3.375 ≈ 3.4 N.

5. Q5 · Capacitance (Class 12)

   A parallel plate capacitor with air as dielectric has capacitance 8 μF. The area is divided into two equal halves and filled with two dielectrics of constants K₁ = 3 and K₂ = 5 respectively. The new capacitance of the system will be

   • A. 24 μF
   • B. 32 μF
   • C. 16 μF
   • D. 40 μF
   Answer and solution

   Answer: B

   When area is halved, each half acts as capacitor with C/2. With dielectrics, capacitances become (C/2)K₁ and (C/2)K₂ in parallel. Total = (C/2)(K₁ + K₂) = (8/2)(3+5) = 32 μF.

6. Q6 · Atoms and Nuclei (Class 12)

   A radioactive element X undergoes decay with a half-life of 4 days. If initially there are 8 × 10²⁰ atoms, how many atoms of X will remain after 12 days?

   • A. 1 × 10²⁰
   • B. 2 × 10²⁰
   • C. 4 × 10²⁰
   • D. 6 × 10²⁰
   Answer and solution

   Answer: A

   After 12 days, three half-lives pass (12/4 = 3). Remaining atoms = N₀/(2³) = 8 × 10²⁰ / 8 = 1 × 10²⁰.

7. Q7 · Some Basic Concepts of Chemistry (Class 11 — ~20% weight)

   How many moles of oxygen atoms are present in 0.5 moles of calcium carbonate (CaCO₃)?

   • A. 0.5 moles
   • B. 1.0 moles
   • C. 1.5 moles
   • D. 2.0 moles
   Answer and solution

   Answer: C

   Each mole of CaCO₃ contains 3 moles of oxygen atoms. Therefore, 0.5 moles contain 0.5 × 3 = 1.5 moles of oxygen.

8. Q8 · Redox Reactions (Class 11 — ~20% weight)

   In the reaction: MnO₄²⁻ + 4H⁺ + 2e⁻ → MnO₂ + 2H₂O, what is the change in oxidation state of manganese?

   • A. +7 to +4
   • B. +6 to +4
   • C. +7 to +2
   • D. +5 to +3
   Answer and solution

   Answer: B

   In MnO₄²⁻, oxidation state of Mn is +6 (let x: x + 4(-2) = -2, x = +6). In MnO₂, Mn is +4. Change is +6 to +4.

9. Q9 · Solutions (Class 12)

   A solution containing 3.42 g of a non-volatile solute in 250 g of water has a boiling point elevation of 0.104 K. What is the molar mass of the solute? (Kb for water = 0.52 K kg mol⁻¹)

   • A. 68 g mol⁻¹
   • B. 85 g mol⁻¹
   • C. 102 g mol⁻¹
   • D. 136 g mol⁻¹
   Answer and solution

   Answer: A

   ΔTb = Kb × m; 0.104 = 0.52 × (3.42/M)/(0.25); M = (0.52 × 3.42)/(0.104 × 0.25) = 68 g mol⁻¹.

10. Q10 · p-Block Elements (Class 12)

    Identify the correct increasing order of boiling points of the hydrides of group 15 elements.

    • A. NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃
    • B. BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃
    • C. PH₃ < AsH₃ < SbH₃ < BiH₃ < NH₃
    • D. NH₃ < BiH₃ < SbH₃ < AsH₃ < PH₃
    Answer and solution

    Answer: C

    NH₃ shows exceptionally high boiling point due to strong hydrogen bonding. For the rest, boiling point increases down the group due to increasing molecular mass (PH₃ < AsH₃ < SbH₃ < BiH₃).