NEET UG 2023 previous year paper practice · PYQ-pattern set

NEET UG 2023 PYQ-pattern questions

Showing 10 of the 20 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

   Two blocks of masses 3 kg and 5 kg are connected by a light inextensible string passing over a smooth pulley. If the system is released from rest, what is the acceleration of the blocks? (Take g = 10 m/s²)

   • A. 1.5 m/s²
   • B. 2.5 m/s²
   • C. 3.75 m/s²
   • D. 5.0 m/s²
   Answer and solution

   Answer: B

   For Atwood's machine, a = (m₂ - m₁)g/(m₁ + m₂) = (5 - 3) × 10/(3 + 5) = 20/8 = 2.5 m/s².

2. Q2 · Properties of Bulk Matter

   A metal wire of length 2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by 2 mm when a load of 160 N is applied. What is the Young's modulus of the material?

   • A. 4 × 10¹⁰ N/m²
   • B. 8 × 10¹⁰ N/m²
   • C. 1.6 × 10¹⁰ N/m²
   • D. 3.2 × 10¹⁰ N/m²
   Answer and solution

   Answer: B

   Young's modulus Y = (F/A)/(ΔL/L) = (F × L)/(A × ΔL) = (160 × 2)/(2 × 10⁻⁶ × 2 × 10⁻³) = 320/(4 × 10⁻⁹) = 8 × 10¹⁰ N/m².

3. Q3 · Thermodynamics

   An ideal gas undergoes an adiabatic expansion. If its volume doubles, what is the ratio of final to initial temperature? (Given γ = 5/3)

   • A. 1/2^(2/3)
   • B. 1/2
   • C. 2^(2/3)
   • D. 1/4
   Answer and solution

   Answer: A

   For adiabatic process, TV^(γ-1) = constant. So T₁V₁^(γ-1) = T₂V₂^(γ-1). With V₂ = 2V₁ and γ = 5/3, T₂/T₁ = (V₁/V₂)^(2/3) = (1/2)^(2/3) = 1/2^(2/3).

4. Q4 · Electromagnetic Induction and Alternating Currents

   An AC voltage V = 200 sin(100πt) volts is applied to a pure inductor of inductance 0.5 H. The maximum current in the circuit is:

   • A. 1.27 A
   • B. 2.54 A
   • C. 4.00 A
   • D. 0.64 A
   Answer and solution

   Answer: A

   V₀ = 200 V, ω = 100π rad/s. Inductive reactance XL = ωL = 100π × 0.5 = 50π Ω. Maximum current I₀ = V₀/XL = 200/(50π) = 4/π = 1.27 A.

5. Q5 · Oscillations

   A simple pendulum has a time period of 2 s on Earth. What will be its time period on a planet where the acceleration due to gravity is one-fourth that of Earth?

   • A. 1 s
   • B. 4 s
   • C. 8 s
   • D. 0.5 s
   Answer and solution

   Answer: B

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

6. Q6 · Laws of Motion

   A block of mass 3 kg rests on a horizontal surface with coefficient of static friction 0.4. The minimum horizontal force required to just move the block is (g = 10 m/s²):

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

   Answer: C

   The minimum force required to just move the block is equal to the maximum static friction: F = μₛ × m × g = 0.4 × 3 × 10 = 12 N.

7. Q7 · Work, Energy and Power

   A force F = (3x² + 2) N acts on a particle along the x-axis. The work done by the force in displacing the particle from x = 0 to x = 2 m is:

   • A. 6 J
   • B. 8 J
   • C. 10 J
   • D. 12 J
   Answer and solution

   Answer: D

   W = ∫₀² (3x²+2) dx = [x³ + 2x]₀² = (8 + 4) - 0 = 12 J.

8. Q8 · Gravitation

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

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

   Answer: A

   g' = gR²/(R+h)². Given g' = g/4, so R²/(R+h)² = 1/4, giving (R+h)² = 4R², so R+h = 2R, hence h = R.

9. Q9 · Waves

   A sound wave travels in air with a speed of 340 m/s. If the wavelength of the wave is 68 cm, what is the frequency of the wave?

   • A. 200 Hz
   • B. 500 Hz
   • C. 680 Hz
   • D. 1000 Hz
   Answer and solution

   Answer: B

   Using v = fλ, frequency f = v/λ = 340/(0.68) = 500 Hz.

10. Q10 · Electromagnetic Waves

    The frequency of an electromagnetic wave in free space is 6 × 10¹⁴ Hz. The wavelength of this wave is:

    • A. 500 nm
    • B. 600 nm
    • C. 700 nm
    • D. 800 nm
    Answer and solution

    Answer: A

    Using c = fλ, wavelength λ = c/f = (3×10⁸)/(6×10¹⁴) = 5×10⁻⁷ m = 500 nm.