MHT-CET 2023 previous year paper practice · PYQ-pattern set

MHT-CET 2023 PYQ-pattern questions

Showing 10 of the 18 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 · Laws of Motion (Class 11 — ~20% weight)

   A block of mass 8 kg is placed on a rough horizontal surface with coefficient of friction μ = 0.4. A horizontal force of 50 N is applied on the block. The frictional force acting on the block is: (Take g = 10 m/s²)

   • A. 50 N
   • B. 32 N
   • C. 40 N
   • D. 18 N
   Answer and solution

   Answer: B

   Maximum static friction = μN = 0.4 × 8 × 10 = 32 N. Since applied force (50 N) exceeds this, kinetic friction acts, which equals 32 N.

2. Q2 · Waves (Class 11 — ~20% weight)

   A sound wave of frequency 500 Hz travels in air with speed 350 m/s. After the wave completes 250 oscillations, what distance has the wave travelled?

   • A. 125 m
   • B. 175 m
   • C. 200 m
   • D. 250 m
   Answer and solution

   Answer: B

   Wavelength λ = v/f = 350/500 = 0.7 m. Distance for 250 vibrations = 250 × λ = 250 × 0.7 = 175 m.

3. Q3 · Thermodynamics (Class 12)

   An ideal gas undergoes an adiabatic expansion from volume V to 2V. If the initial pressure is P and the adiabatic exponent γ = 1.4, what is the final pressure?

   • A. P / 2.64
   • B. P / 2.00
   • C. P / 1.96
   • D. P / 1.40
   Answer and solution

   Answer: A

   For adiabatic process PV^γ = constant. P₁V₁^γ = P₂V₂^γ ⇒ P × V^1.4 = P₂ × (2V)^1.4, so P₂ = P / 2^1.4 = P / 2.64.

4. Q4 · Magnetic Effects of Electric Current (Class 12)

   A circular coil of radius 10 cm with 200 turns carries a current of 2 A. The magnetic field at the center of the coil is (μ₀ = 4π × 10⁻⁷ T·m/A)

   • A. 2.5 × 10⁻³ T
   • B. 5.0 × 10⁻³ T
   • C. 1.26 × 10⁻³ T
   • D. 6.28 × 10⁻³ T
   Answer and solution

   Answer: A

   Magnetic field at center B = (μ₀NI)/(2R) = (4π×10⁻⁷ × 200 × 2)/(2 × 0.1) = 2.5 × 10⁻³ T.

5. Q5 · Dual Nature of Radiation and Matter (Class 12)

   The work function of a metal is 3.0 eV. When light of wavelength 4000 Å is incident on it, the maximum kinetic energy of the emitted photoelectrons is (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

   • A. 0.1 eV
   • B. 0.5 eV
   • C. 1.0 eV
   • D. 0.2 eV
   Answer and solution

   Answer: A

   Energy of photon E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(4×10⁻⁷) = 4.97×10⁻¹⁹ J = 3.1 eV. KE_max = E - φ = 3.1 - 3.0 = 0.1 eV.

6. Q6 · Communication Systems (Class 12)

   A carrier wave has a peak voltage of 20 V and is used to transmit a message signal with a modulation index of 60%. What is the peak voltage of the modulating signal?

   • A. 8 V
   • B. 10 V
   • C. 12 V
   • D. 15 V
   Answer and solution

   Answer: C

   Modulation index μ = (Aₘ / Aᶜ), so Aₘ = μ × Aᶜ = 0.60 × 20 V = 12 V.

7. Q7 · Periodic Classification (Class 11 — ~20% weight)

   Which of the following elements has the highest electronegativity?

   • A. Oxygen
   • B. Fluorine
   • C. Nitrogen
   • D. Chlorine
   Answer and solution

   Answer: B

   Fluorine is the most electronegative element in the periodic table.

8. Q8 · Basic Principles of Organic Chemistry (Class 11 — ~20% weight)

   What is the hybridization of the carbon atom in a carbanion (CH₃⁻)?

   • A. sp
   • B. sp²
   • C. sp³
   • D. sp³d
   Answer and solution

   Answer: C

   In CH₃⁻, carbon has three bond pairs with hydrogen and one lone pair, totaling four electron pairs requiring sp³ hybridization with pyramidal geometry.

9. Q9 · Alcohols, Phenols and Ethers (Class 12)

   Which from the following reagents is used to distinguish between phenol and ethanol?

   • A. Lucas reagent
   • B. Neutral FeCl₃
   • C. Bromine water
   • D. Both B and C
   Answer and solution

   Answer: D

   Phenol gives a violet coloration with neutral FeCl₃ and forms a white precipitate of 2,4,6-tribromophenol with bromine water. Ethanol does not give these reactions.

10. Q10 · Polymers (Class 12)

    Nylon-6,10 is obtained by the condensation polymerization of:

    • A. Hexamethylenediamine and adipic acid
    • B. Hexamethylenediamine and sebacic acid
    • C. Caprolactam and sebacic acid
    • D. Ethylene glycol and terephthalic acid
    Answer and solution

    Answer: B

    Nylon-6,10 is formed by condensation of hexamethylenediamine (6 carbons) and sebacic acid (10 carbons). The numbers in the name indicate the carbon atoms in the diamine and diacid respectively.