NEET 2023 Previous Year Question Paper

Section A

1. Two forces of 3 N and 4 N act on a body at right angles to each other. The magnitude of their resultant is

   1. 5 N
   2. 12 N
   3. 1 N
   4. 7 N
   Answer and explanation

   Answer: A (5 N)

   R = √(3² + 4²) = √25 = 5 N.

2. Statement I: The electric field inside a uniformly charged thin spherical shell is zero. Statement II: The electric potential everywhere inside such a shell is zero.

   1. Both Statement I and Statement II are true
   2. Both Statement I and Statement II are false
   3. Statement I is true but Statement II is false
   4. Statement I is false but Statement II is true
   Answer and explanation

   Answer: C (Statement I is true but Statement II is false)

   By Gauss's law no charge is enclosed inside the shell, so E = 0 (Statement I true). Since E = 0, V is constant inside and equal to the surface value kQ/R, which is not zero (Statement II false).

3. A heater is rated 1100 W, 220 V. Its resistance is

   1. 5 Ω
   2. 44 Ω
   3. 0.2 Ω
   4. 242 Ω
   Answer and explanation

   Answer: B (44 Ω)

   R = V²/P = 220²/1100 = 48 400/1100 = 44 Ω.

4. The energy of the electron in the n = 3 orbit of a hydrogen atom is

   1. −13.6 eV
   2. −0.85 eV
   3. −3.4 eV
   4. −1.51 eV
   Answer and explanation

   Answer: D (−1.51 eV)

   Eₙ = −13.6/n² eV = −13.6/9 ≈ −1.51 eV.

Section B

Attempt any 2 of these 3 questions (real paper: any 10 of 15).

5. A block slides down a rough incline of 30° with constant velocity. The coefficient of kinetic friction is

   1. √3
   2. √3/2
   3. 1/√3
   4. 1/2
   Answer and explanation

   Answer: C (1/√3)

   Constant velocity: mg sin θ = μ mg cos θ, so μ = tan 30° = 1/√3 ≈ 0.577.

6. If the linear momentum of a body is doubled, its kinetic energy becomes

   1. half
   2. the same
   3. four times
   4. double
   Answer and explanation

   Answer: C (four times)

   K = p²/2m, so doubling p multiplies K by 4.

7. A Zener diode used as a voltage regulator is operated in

   1. reverse bias below breakdown
   2. reverse bias in the breakdown region
   3. forward bias
   4. zero bias
   Answer and explanation

   Answer: B (reverse bias in the breakdown region)

   In reverse breakdown the Zener voltage stays nearly constant over a wide range of current, which gives the regulation.

Section A

8. Statement I: The boiling point of H₂O is much higher than that of H₂S because of hydrogen bonding in water. Statement II: H₂S forms stronger hydrogen bonds than H₂O.

   1. Both Statement I and Statement II are true
   2. Both Statement I and Statement II are false
   3. Statement I is true but Statement II is false
   4. Statement I is false but Statement II is true
   Answer and explanation

   Answer: C (Statement I is true but Statement II is false)

   O is small and highly electronegative, so water molecules are strongly hydrogen-bonded (Statement I true). S is larger and less electronegative, so H₂S shows negligible hydrogen bonding (Statement II false).

9. Which of the following molecules is paramagnetic?

   1. O₂
   2. F₂
   3. H₂
   4. N₂
   Answer and explanation

   Answer: A (O₂)

   By MO theory O₂ has two unpaired electrons in π*2p orbitals; N₂, F₂ and H₂ have all electrons paired.

10. What volume of 0.1 M H₂SO₄ will exactly neutralise 20 mL of 0.2 M NaOH?

    1. 20 mL
    2. 10 mL
    3. 40 mL
    4. 80 mL
    Answer and explanation

    Answer: A (20 mL)

    Moles NaOH = 0.2 × 0.020 = 0.004 mol. H₂SO₄ is diprotic, so 0.002 mol is needed: V = 0.002/0.1 = 0.020 L = 20 mL.

11. Clemmensen reduction converts a ketone C=O group into

    1. CH₂
    2. CH–OH
    3. COOH
    4. C=C
    Answer and explanation

    Answer: A (CH₂)

    Zn–Hg/conc. HCl reduces >C=O to >CH₂ (e.g. acetophenone → ethylbenzene).

Section B

Attempt any 2 of these 3 questions (real paper: any 10 of 15).

12. The amount of electricity needed to deposit 27 g of aluminium from Al³⁺ ions is (Al = 27 g/mol)

    1. 9 F
    2. 2 F
    3. 3 F
    4. 1 F
    Answer and explanation

    Answer: C (3 F)

    Al³⁺ + 3e⁻ → Al. 27 g = 1 mol Al needs 3 mol electrons = 3 F (≈ 2.9 × 10⁵ C).

13. In maltose, the two glucose units are linked by

    1. α-1,4 glycosidic linkage
    2. β-1,4 glycosidic linkage
    3. α-1, β-2 glycosidic linkage
    4. α-1,6 glycosidic linkage
    Answer and explanation

    Answer: A (α-1,4 glycosidic linkage)

    Maltose = two α-D-glucose units joined C1 to C4 (α-1,4). Lactose has a β-1,4 link and sucrose an α-1, β-2 link.

14. In the complex [Fe(CN)₆]⁴⁻, the d-electron configuration of the metal ion is

    1. t₂g³ eg³
    2. t₂g⁶ eg⁰
    3. t₂g⁵ eg¹
    4. t₂g⁴ eg²
    Answer and explanation

    Answer: B (t₂g⁶ eg⁰)

    Fe²⁺ is d⁶. CN⁻ is a strong-field ligand, so Δo > pairing energy and all six electrons pair in t₂g: t₂g⁶ eg⁰ (low spin, diamagnetic).

Section A

15. Statement I: Mitochondria and chloroplasts contain their own DNA. Statement II: Mitochondria and chloroplasts contain 80S ribosomes.

    1. Both Statement I and Statement II are true
    2. Both Statement I and Statement II are false
    3. Statement I is true but Statement II is false
    4. Statement I is false but Statement II is true
    Answer and explanation

    Answer: C (Statement I is true but Statement II is false)

    Both organelles have circular DNA (Statement I true), but their ribosomes are 70S, like bacteria (Statement II false).

16. DNA replication in the eukaryotic cell cycle takes place in

    1. G₂ phase
    2. M phase
    3. S phase
    4. G₁ phase
    Answer and explanation

    Answer: C (S phase)

    DNA content doubles in the S (synthesis) phase of interphase; the chromosome number does not change.

17. Bending of a shoot towards unilateral light is mediated by

    1. auxin
    2. abscisic acid
    3. ethylene
    4. cytokinin
    Answer and explanation

    Answer: A (auxin)

    Auxin moves to the shaded side and stimulates cell elongation there, so the shoot bends towards light.

18. Double fertilisation is a characteristic feature of

    1. pteridophytes
    2. bryophytes
    3. angiosperms
    4. gymnosperms
    Answer and explanation

    Answer: C (angiosperms)

    Syngamy plus triple fusion in the embryo sac occurs only in flowering plants.

Section B

Attempt any 2 of these 3 questions (real paper: any 10 of 15).

19. The net gain of ATP from glycolysis of one glucose molecule (substrate-level only) is

    1. 4
    2. 36
    3. 8
    4. 2
    Answer and explanation

    Answer: D (2)

    4 ATP are produced and 2 ATP are used in glycolysis, so the net gain is 2 ATP (plus 2 NADH).

20. Match List-I with List-II. (a) Ti plasmid (b) Cry proteins (c) Thermostable DNA polymerase (d) EcoRI (i) Bacillus thuringiensis (ii) Escherichia coli (iii) Agrobacterium tumefaciens (iv) Thermus aquaticus

    1. a-ii, b-iv, c-i, d-iii
    2. a-iii, b-i, c-iv, d-ii
    3. a-i, b-iii, c-ii, d-iv
    4. a-iii, b-iv, c-i, d-ii
    Answer and explanation

    Answer: B (a-iii, b-i, c-iv, d-ii)

    Ti plasmid is from Agrobacterium tumefaciens (a-iii); Cry toxins from Bacillus thuringiensis (b-i); Taq polymerase from Thermus aquaticus (c-iv); EcoRI from E. coli RY13 (d-ii).

21. Which ecological pyramid is always upright?

    1. Pyramid of biomass
    2. Pyramid of numbers
    3. Pyramid of energy
    4. All of these
    Answer and explanation

    Answer: C (Pyramid of energy)

    Energy is lost at every transfer between trophic levels, so the pyramid of energy can never be inverted.

Section A

22. Statement I: About 97% of oxygen is transported from the lungs to the tissues by RBCs. Statement II: About 20–25% of carbon dioxide is carried by RBCs as carbamino-haemoglobin.

    1. Both Statement I and Statement II are true
    2. Both Statement I and Statement II are false
    3. Statement I is true but Statement II is false
    4. Statement I is false but Statement II is true
    Answer and explanation

    Answer: A (Both Statement I and Statement II are true)

    Both are standard figures: ~97% of O₂ is carried bound to haemoglobin (3% dissolved in plasma), and 20–25% of CO₂ is carried as carbamino-haemoglobin (about 70% as bicarbonate).

23. Which hormone is released from the posterior pituitary?

    1. Prolactin
    2. Oxytocin
    3. TSH
    4. Growth hormone
    Answer and explanation

    Answer: B (Oxytocin)

    The neurohypophysis stores and releases oxytocin and vasopressin made by the hypothalamus; the others come from the anterior pituitary.

24. Which of the following animals has an open circulatory system?

    1. Earthworm
    2. Frog
    3. Cockroach
    4. Rabbit
    Answer and explanation

    Answer: C (Cockroach)

    In arthropods such as the cockroach, haemolymph flows through open spaces (haemocoel). Earthworm and vertebrates have closed systems.

25. The joint between the atlas and axis vertebrae is a

    1. saddle joint
    2. ball and socket joint
    3. hinge joint
    4. pivot joint
    Answer and explanation

    Answer: D (pivot joint)

    The atlanto-axial joint is a pivot joint, allowing rotation of the head.

Section B

Attempt any 2 of these 3 questions (real paper: any 10 of 15).

26. Ringworm in humans is caused by

    1. viruses
    2. bacteria
    3. protozoa
    4. fungi
    Answer and explanation

    Answer: D (fungi)

    Ringworm is caused by fungi of the genera Microsporum, Trichophyton and Epidermophyton.

27. In a population in Hardy–Weinberg equilibrium, the frequency of the dominant allele is 0.7. The frequency of homozygous recessive individuals is

    1. 0.09
    2. 0.49
    3. 0.21
    4. 0.3
    Answer and explanation

    Answer: A (0.09)

    q = 1 − 0.7 = 0.3, so q² = 0.09.

28. The human embryo implants in the uterine wall at the

    1. gastrula stage
    2. morula stage
    3. zygote stage
    4. blastocyst stage
    Answer and explanation

    Answer: D (blastocyst stage)

    The blastocyst's trophoblast attaches to the endometrium and the inner cell mass becomes the embryo.