NEET UG 2024 previous year paper practice · PYQ-pattern set

NEET UG 2024 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 · Physical World and Measurement

   The least count of a vernier caliper is 0.01 cm. When the jaws are closed, the 5th division of the vernier scale coincides with a main scale division. If while measuring the diameter of a sphere, the main scale reading is 3.2 cm and the 8th vernier division coincides with a main scale division, what is the correct diameter?

   • A. 3.23 cm
   • B. 3.28 cm
   • C. 3.33 cm
   • D. 3.18 cm
   Answer and solution

   Answer: A

   Zero error = +5 × 0.01 = +0.05 cm. Reading = 3.2 + (8 × 0.01) = 3.28 cm. Corrected reading = 3.28 - 0.05 = 3.23 cm.

2. Q2 · Kinematics

   A stone is dropped from a height h. It covers a distance h/2 in the last second of its fall. What is the total time of fall?

   • A. (1 + √2) s
   • B. 2 s
   • C. (2 + √2) s
   • D. √2 s
   Answer and solution

   Answer: C

   Distance in first (t-1) seconds equals h/2, so ½g(t-1)² = ½·(½gt²), giving (t-1)²=t²/2, thus t-1=t/√2. Solving: t=√2/(√2-1)=2+√2 s.

3. Q3 · Gravitation

   At what height above the Earth's surface will the acceleration due to gravity be 25% of its value on the surface? (R = radius of Earth)

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

   Answer: A

   g' = g(R/(R+h))². For g' = g/4, we get (R/(R+h))² = 1/4, so R/(R+h) = 1/2, giving R+h = 2R, thus h = R.

4. Q4 · Magnetic Effects of Current and Magnetism

   A circular coil of radius 5 cm has 100 turns and carries a current of 2 A. The magnetic field at the center of the coil is: (μ₀ = 4π × 10⁻⁷ T·m/A)

   • A. 2.51 × 10⁻³ T
   • B. 5.02 × 10⁻³ T
   • C. 1.26 × 10⁻³ T
   • D. 7.54 × 10⁻³ T
   Answer and solution

   Answer: A

   Magnetic field B = (μ₀nI)/(2r) = (4π×10⁻⁷ × 100 × 2)/(2 × 0.05) = (8π×10⁻⁵)/(0.1) = 2.51 × 10⁻³ T.

5. Q5 · Behaviour of Perfect Gases and Kinetic Theory

   An ideal gas undergoes an isothermal process at 300 K. If the volume of the gas is doubled, what is the work done by the gas? (Given: initial pressure = 2 × 10⁵ Pa, initial volume = 0.01 m³, and ln 2 = 0.693)

   • A. 1386 J
   • B. 693 J
   • C. 2772 J
   • D. 346.5 J
   Answer and solution

   Answer: A

   For isothermal process, W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = 2×10⁵ × 0.01 × ln(2) = 2000 × 0.693 = 1386 J.

6. Q6 · Kinematics

   A particle moves along a straight line such that its position is given by x = 6t² - t³ (in metres, t in seconds). At what time does the particle achieve maximum velocity along the positive x-direction?

   • A. 2 s
   • B. 4 s
   • C. 6 s
   • D. 8 s
   Answer and solution

   Answer: A

   Velocity v = dx/dt = 12t - 3t². Maximum velocity when dv/dt = 0: 12 - 6t = 0, so t = 2 s. Checking options: at t=2, v = 24-12 = 12 m/s (positive); at t=4, v = 48-48 = 0. So t=2 s gives maximum velocity in positive direction.

7. Q7 · Laws of Motion

   Two blocks of masses 4 kg and 6 kg are connected by a light inextensible string passing over a smooth pulley. The acceleration of the system is (g = 10 m/s²):

   • A. 1 m/s²
   • B. 2 m/s²
   • C. 4 m/s²
   • D. 6 m/s²
   Answer and solution

   Answer: B

   For an Atwood machine, acceleration a = (m₂ - m₁)g/(m₁ + m₂) = (6 - 4) × 10/(4 + 6) = 20/10 = 2 m/s².

8. Q8 · Thermodynamics

   An ideal gas undergoes an isothermal process at temperature T. If the volume is doubled, the change in internal energy is:

   • A. RT
   • B. 2RT
   • C. Zero
   • D. RT ln2
   Answer and solution

   Answer: C

   For an ideal gas in an isothermal process, temperature remains constant, so change in internal energy ΔU = 0 (since U depends only on temperature for ideal gas).

9. Q9 · Oscillations

   A simple pendulum oscillates with a time period of 2 s in air. If the same pendulum is taken to a planet where acceleration due to gravity is one-fourth that of Earth, what will be its new time period?

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

   Answer: C

   Time period T = 2π√(L/g). Since T ∝ 1/√g, when g becomes g/4, T becomes T√4 = 2T = 2 × 2 = 4 s.

10. Q10 · Electromagnetic Induction and Alternating Currents

    A coil of 200 turns and area 0.02 m² is placed in a uniform magnetic field of 0.5 T. If the coil is rotated by 90° in 0.1 s, the average induced emf is:

    • A. 10 V
    • B. 20 V
    • C. 40 V
    • D. 80 V
    Answer and solution

    Answer: B

    Change in flux Δφ = BA = 0.5 × 0.02 = 0.01 Wb. Induced emf = N(Δφ/Δt) = 200 × (0.01/0.1) = 200 × 0.1 = 20 V.