Free NEET UG mock test — 180 questions, 3 hours, +4/−1 🇮🇳

8 original practice questions written by Myndaq in the NEET UG style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1PhysicsOne correct answer

   A 3 kg block, initially at rest on a rough horizontal floor, is pulled by a constant horizontal force of 18 N and starts sliding at once. The coefficient of kinetic friction between the block and the floor is 0.25 (take g = 10 m/s²). The kinetic energy of the block 4 s after the force is first applied is

   1. A. 210 J
   2. B. 294 J
   3. C. 504 J
   4. D. 864 J
   Show answer and explanation

   Answer: B

   Kinetic friction = μmg = 0.25 × 3 × 10 = 7.5 N. Net force = 18 − 7.5 = 10.5 N, so a = 10.5/3 = 3.5 m/s².

   After 4 s: v = 3.5 × 4 = 14 m/s, so KE = ½ × 3 × 14² = 294 J.

   Check with the work-energy theorem: distance = ½ × 3.5 × 4² = 28 m; work by the pull = 18 × 28 = 504 J; work against friction = 7.5 × 28 = 210 J; 504 − 210 = 294 J. The option 504 J is only the work done by the applied force, 210 J is only the energy lost to friction, and 864 J is what you get by ignoring friction (a = 6 m/s², v = 24 m/s).

2. Question 2PhysicsOne correct answer

   Assertion (A): A thin convex lens made of glass of refractive index 1.5 has a longer focal length when it is immersed in water (refractive index 4/3) than when it is held in air.

   Reason (R): By the lens maker's formula, the power of a thin lens is proportional to (μ₂/μ₁ − 1), where μ₂ is the refractive index of the lens material and μ₁ that of the surrounding medium, and this factor is smaller in water than in air.

   Read the Assertion (A) and the Reason (R) above. Which option describes them correctly?

   1. A. A and R are both correct, and R explains why A holds.
   2. B. A and R are both correct, but R does not explain A.
   3. C. A is correct, but R is incorrect.
   4. D. A is incorrect, but R is correct.
   Show answer and explanation

   Answer: A

   Lens maker's formula: 1/f = (μ₂/μ₁ − 1)(1/R₁ − 1/R₂). The curvature term does not change when the lens is moved into water.

   In air: μ₂/μ₁ − 1 = 1.5 − 1 = 0.5. In water: μ₂/μ₁ − 1 = 1.5/(4/3) − 1 = 9/8 − 1 = 1/8. So f(water)/f(air) = 0.5 ÷ (1/8) = 4. The focal length becomes four times longer and the lens is still converging, so A is correct.

   R states exactly the dependence that causes this change (the relative index falls from 1.5 to 1.125), so R is correct and is the explanation of A. The other options fail because both statements are true and R does account for A.

3. Question 3ChemistryOne correct answer

   At 25 °C, 10 mL of 0.2 M HCl is mixed with 30 mL of 0.1 M NaOH. The pH of the resulting solution is (log 2 = 0.30)

   1. A. 1.60
   2. B. 12.60
   3. C. 12.40
   4. D. 13.00
   Show answer and explanation

   Answer: C

   Moles of H⁺ = 10 mL × 0.2 M = 2 mmol. Moles of OH⁻ = 30 mL × 0.1 M = 3 mmol. After neutralisation, 1 mmol of OH⁻ is left in a total volume of 10 + 30 = 40 mL.

   [OH⁻] = 1/40 = 0.025 M = 2.5 × 10^-2 M. log 2.5 = log 10 − log 4 = 1 − 0.60 = 0.40, so pOH = 2 − 0.40 = 1.60 and pH = 14 − 1.60 = 12.40.

   1.60 is the pOH, not the pH. 12.60 comes from adding the log term instead of subtracting it (pOH taken as 1.40). 13.00 treats the solution as if the 0.1 M NaOH were unreacted and undiluted.

4. Question 4ChemistryOne correct answer

   List I (conversion)

   P. Benzenediazonium chloride → chlorobenzene

   Q. Propanone → propane

   R. Ethanoyl chloride → ethanal

   S. Phenol → 2-hydroxybenzaldehyde (salicylaldehyde)

   List II (reagent or conditions)

   1. CHCl₃ with aqueous NaOH, followed by acidification

   2. Zinc amalgam with concentrated HCl

   3. Cu₂Cl₂ with HCl

   4. H₂ over palladium on barium sulphate

   Pair each conversion in List I with the reagent in List II that brings it about. Which option gives the right pairing?

   1. A. P-3, Q-4, R-2, S-1
   2. B. P-1, Q-2, R-4, S-3
   3. C. P-2, Q-3, R-1, S-4
   4. D. P-3, Q-2, R-4, S-1
   Show answer and explanation

   Answer: D

   P: replacing the diazonium group by chlorine with cuprous chloride in HCl is the Sandmeyer reaction, so P-3.

   Q: reducing the C=O group of a ketone all the way to CH₂ with Zn-Hg and concentrated HCl is the Clemmensen reduction, so Q-2.

   R: an acid chloride is reduced only as far as the aldehyde by hydrogen over Pd on BaSO₄ (a partly poisoned catalyst, Rosenmund reduction), so R-4.

   S: phenol with chloroform and aqueous alkali, then acid, introduces a CHO group ortho to the OH (Reimer-Tiemann reaction), so S-1.

   The pairing P-3, Q-2, R-4, S-1 is option D. Option A swaps the two reductions (Pd/BaSO₄ would not take a ketone to an alkane), and options B and C mismatch the diazonium and phenol reactions.

5. Question 5Biology (Botany)One correct answer

   In a plant species, allele T (tall) is completely dominant over t (dwarf), while flower colour shows incomplete dominance: RR plants have red flowers, Rr plants pink and rr plants white. The two genes assort independently. A tall, pink-flowered plant heterozygous at both loci is crossed with a dwarf, pink-flowered plant. What fraction of the progeny is expected to be tall with red flowers?

   1. A. 1/8
   2. B. 3/16
   3. C. 1/4
   4. D. 3/8
   Show answer and explanation

   Answer: A

   The cross is TtRr × ttRr. Treat each gene separately, since they assort independently.

   Height: Tt × tt gives 1/2 Tt (tall) and 1/2 tt (dwarf). Colour: Rr × Rr gives 1/4 RR (red), 1/2 Rr (pink) and 1/4 rr (white).

   Tall and red = 1/2 × 1/4 = 1/8.

   3/16 wrongly applies the 9:3:3:1 dihybrid ratio, which needs both parents to be double heterozygotes. 1/4 is the fraction of tall pink plants (1/2 × 1/2). 3/8 treats R as fully dominant and counts red and pink together (1/2 × 3/4).

6. Question 6Biology (Botany)One correct answer

   a. In mesophyll cells, PEP carboxylase fixes CO₂ with phosphoenol pyruvate to form oxaloacetic acid, a four-carbon compound.

   b. The bundle sheath cells of C4 plants lack RuBisCO.

   c. In C4 plants, the Calvin cycle runs in the bundle sheath cells.

   d. C4 plants show very little photorespiration because the CO₂ concentration around RuBisCO is kept high.

   Which of the statements a to d about the C4 (Hatch and Slack) pathway are correct?

   1. A. a, b and c only
   2. B. a, c and d only
   3. C. b and d only
   4. D. a, b, c and d
   Show answer and explanation

   Answer: B

   a is correct: the primary CO₂ acceptor in C4 mesophyll cells is the three-carbon PEP, the enzyme is PEP carboxylase, and the first stable product is the four-carbon oxaloacetic acid.

   b is wrong: bundle sheath cells are rich in RuBisCO and lack PEP carboxylase. This is why c is correct: the C4 acid is carried to the bundle sheath, decarboxylated there, and the released CO₂ enters the Calvin cycle in those cells.

   d is correct: decarboxylation raises the CO₂ level around RuBisCO, so its oxygenase activity is minimised and photorespiration is negligible, which helps C4 plants in hot, bright conditions.

   So only a, c and d are correct. Every other option includes the false statement b or leaves out a true one.

7. Question 7Biology (Zoology)One correct answer

   In a resting adult, each ventricle holds 140 mL of blood at the end of diastole and 60 mL at the end of systole, and the heart beats 64 times per minute. The cardiac output of this person is

   1. A. 3.84 L/min
   2. B. 8.96 L/min
   3. C. 12.80 L/min
   4. D. 5.12 L/min
   Show answer and explanation

   Answer: D

   Stroke volume is the blood pumped out by a ventricle in one beat: end-diastolic volume − end-systolic volume = 140 − 60 = 80 mL.

   Cardiac output = stroke volume × heart rate = 80 mL × 64 per minute = 5120 mL/min = 5.12 L/min.

   3.84 L/min multiplies the blood left behind (60 mL) by the heart rate, 8.96 L/min uses the full end-diastolic volume (140 mL) as if the ventricle emptied completely, and 12.80 L/min adds the two volumes instead of subtracting them.

8. Question 8Biology (Zoology)One correct answer

   Each option lists a human digestive enzyme, the substrate it acts on and the product it forms. Which option is INCORRECTLY matched?

   1. A. Enterokinase — trypsinogen — trypsin
   2. B. Pepsin — proteins — proteoses and peptones
   3. C. Nucleosidases — nucleotides — nucleosides
   4. D. Dipeptidases — dipeptides — amino acids
   Show answer and explanation

   Answer: C

   Nucleotides are converted to nucleosides by nucleotidases. Nucleosidases act one step later, splitting nucleosides into pentose sugars and nitrogenous bases, so option C is the mismatch.

   The rest are correct: enterokinase from the intestinal mucosa activates trypsinogen to trypsin, which then activates the other pancreatic proenzymes; pepsin in the stomach breaks proteins into proteoses and peptones; and dipeptidases in the intestinal juice (succus entericus) hydrolyse dipeptides into amino acids.