NEET UG 2022 previous year paper practice · PYQ-pattern set

NEET UG 2022 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 · Properties of Bulk Matter

   Water flows through a horizontal pipe of varying cross-section. At section A, the area is 10 cm² and velocity is 2 m/s. At section B, the area is 5 cm². What is the velocity at section B?

   • A. 1 m/s
   • B. 2 m/s
   • C. 4 m/s
   • D. 8 m/s
   Answer and solution

   Answer: C

   By continuity equation, A₁v₁ = A₂v₂. So 10 × 2 = 5 × v₂, giving v₂ = 4 m/s.

2. Q2 · Work, Energy and Power

   A body of mass 2 kg is thrown vertically upward with a kinetic energy of 490 J. What is the maximum height it can reach? (Take g = 10 m/s²)

   • A. 24.5 m
   • B. 49 m
   • C. 12.25 m
   • D. 98 m
   Answer and solution

   Answer: A

   At maximum height, all kinetic energy converts to potential energy: KE = mgh, so 490 = 2 × 10 × h, giving h = 24.5 m.

3. Q3 · Waves

   A string of length 1 m and mass 5 g is under tension of 200 N. The speed of transverse wave in the string is:

   • A. 100 m/s
   • B. 200 m/s
   • C. 400 m/s
   • D. 50 m/s
   Answer and solution

   Answer: B

   Linear mass density μ = m/l = 5×10⁻³/1 = 5×10⁻³ kg/m. Wave speed v = √(T/μ) = √(200/5×10⁻³) = √40000 = 200 m/s.

4. Q4 · Electrostatics

   Two point charges +3 μC and -3 μC are placed 10 cm apart. What is the electric potential at the midpoint of the line joining them?

   • A. Zero
   • B. 5.4 × 10⁵ V
   • C. 2.7 × 10⁵ V
   • D. 1.08 × 10⁶ V
   Answer and solution

   Answer: A

   At the midpoint, distances from both charges are equal (5 cm each). Potential V = kq₁/r + kq₂/r = k(+3μC)/0.05 + k(-3μC)/0.05 = 0.

5. Q5 · Work, Energy and Power

   A body of mass 2 kg is moving with a velocity of 6 m/s. The kinetic energy of the body is:

   • A. 18 J
   • B. 24 J
   • C. 36 J
   • D. 72 J
   Answer and solution

   Answer: C

   Kinetic Energy KE = ½mv² = ½ × 2 × (6)² = ½ × 2 × 36 = 36 J.

6. Q6 · Physical World and Measurement

   The least count of a vernier caliper is 0.01 cm. If the main scale reading is 3.4 cm and 7th vernier division coincides with a main scale division, the measurement is:

   • A. 3.47 cm
   • B. 3.41 cm
   • C. 3.07 cm
   • D. 3.74 cm
   Answer and solution

   Answer: A

   Total reading = Main scale reading + (Vernier coinciding division × Least count) = 3.4 + (7 × 0.01) = 3.4 + 0.07 = 3.47 cm.

7. Q7 · Motion of System of Particles and Rigid Body

   The moment of inertia of a thin uniform rod of mass M and length L about an axis perpendicular to its length passing through its center is:

   • A. ML²/3
   • B. ML²/12
   • C. ML²/6
   • D. ML²/4
   Answer and solution

   Answer: B

   The moment of inertia of a uniform rod about its center perpendicular to length is ML²/12 (standard result from integration).

8. Q8 · Electrostatics

   Two point charges +4 μC and –4 μC are separated by a distance of 6 cm in air. The electric potential at the midpoint of the line joining them is:

   • A. Zero
   • B. 1.2 × 10⁶ V
   • C. 2.4 × 10⁶ V
   • D. 3.6 × 10⁶ V
   Answer and solution

   Answer: A

   At the midpoint, both charges are at equal distance (3 cm). The potential due to +4 μC is positive and due to –4 μC is negative; they cancel exactly, making net potential zero.

9. Q9 · Behaviour of Perfect Gases and Kinetic Theory

   A gas at 27°C has a pressure of 2 atm in a container. If the temperature is increased to 327°C while keeping the volume constant, the new pressure will be:

   • A. 2 atm
   • B. 3 atm
   • C. 4 atm
   • D. 6 atm
   Answer and solution

   Answer: C

   At constant volume, P/T = constant. P₂/P₁ = T₂/T₁ = 600/300 = 2. Therefore P₂ = 2P₁ = 2 × 2 = 4 atm.

10. Q10 · Magnetic Effects of Current and Magnetism

    A straight conductor of length 0.5 m carrying a current of 10 A is placed perpendicular to a uniform magnetic field of 0.2 T. The force experienced by the conductor is:

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

    Answer: B

    Force F = BIL sinθ = 0.2 × 10 × 0.5 × sin90° = 0.2 × 10 × 0.5 × 1 = 1.0 N.