Answers & Explanations For Free NEET Practice Questions 2026

Free NEET Practice Questions 2026: To help NEET UG aspirants accurately evaluate their performance, we have created this dedicated answers and explanations post for the free practice paper. This section allows students to easily cross-check their responses, understand the correct concepts behind each question, and identify the topics that need further revision. By carefully reviewing the explanations, learners can avoid repeating mistakes and strengthen their conceptual clarity, an essential step toward improving scores in upcoming mock tests and the final NEET examination.

Ans.1. B. Horizontal component of velocity. This component is unaffected by gravity and remains constant throughout the projectile motion because no force acts along the horizontal direction. This is a key feature of ideal projectile motion in uniform gravitational field without air resistance.

Ans.2. B. F/3. The system’s acceleration is $F\mathrm{/}(3m)$, and the tension is the force needed to accelerate mass $m$m at this rate, giving $T=m\cdot F\mathrm{/}(3m)=F\mathrm{/}3$. This approach uses the idea of treating both blocks as a single system to find acceleration first, then analyzing one block to find tension.

Ans.3. C. $v_{\max }=\sqrt{{\mu }_{s}gR}$ We set maximum friction ${\mu }_{s}mg$μsmg equal to centripetal requirement $m{v}^2\mathrm{/}R$ and solve for $v$. Any other factor, such as 2, would imply additional forces or different conditions, which are not present here.

Ans.4. D. 1/2mgh. Because gravitational potential energy is directly proportional to height, falling half the distance reduces the potential energy by half. This lost potential energy appears as kinetic energy, and numerically equals the work done by gravity over that portion of the motion.

Ans.5.B. W=(P1+P2)/2(V1−V2). This formula uses the average pressure during the process multiplied by the change in volume to give the work done in a linear compression or expansion.

Ans.6. C sin i/sin r = c/v. This directly follows from combining Snell’s law with the definition of refractive index as the ratio of light speeds in the two media, establishing that more optically dense media (lower speed) have higher refractive index.

Ans.7. D.  4:1. Because area $A$ is proportional to $R^2$, doubling the radius increases the area fourfold, and thus the stored energy becomes one-fourth, making the thin wire’s energy four times that of the thick wire for the same tension and length.

Ans.8. D. The sum of kinetic and potential energies remains constant in time. In a frictionless SHM system, potential energy is maximum at extremes and kinetic energy is maximum at the mean, but their sum is constant and equal to ${\displaystyle \frac{1}{2}}k{A}^2$.

Ans.9. C. λ/2. Since $\lambda \propto 1\mathrm{/}\sqrt{V}$, quadrupling $V$ multiplies the denominator by 2, thereby halving the wavelength from $\lambda$ to $\lambda \mathrm{/}2$.

Ans.10. A. 5000. For a tenfold increase in voltage, the turns on the secondary must also be ten times the primary, consistent with the transformer ratio relation.

Ans.11. C. It becomes $\beta \mathrm{/}\mu$. Immersing the apparatus in water reduces the effective wavelength to $\lambda \mathrm{/}\mu$, making fringes more closely spaced by the same factor.

Ans.12. A. A0/8. This follows directly from applying the half-life concept three times in succession (10, 20, 30 days).

Ans.13. C.  It remains $B$B. The proportional increase from current is exactly offset by the proportional increase in distance.

Ans.14. D.  Both have the same de Broglie wavelength. This highlights that de Broglie waves are fundamentally associated with momentum, not directly with mass or speed taken alone.

Ans.15. B. Photoelectric current increases, but stopping potential remains the same. Doubling intensity raises the emission rate of electrons, increasing current, but does not change the energy per electron, so stopping potential is unchanged.

Ans.16. D. Virtual, erect, and magnified. This is characteristic of the simple magnifying glass arrangement where the object is within one focal length of a convex lens.

Ans.17. A. Current is maximum and power factor is 1. The maximum current arises from minimal impedance, while unity power factor indicates efficient real power transfer.

Ans.18. A. Solid sphere. Using $a={\displaystyle \frac{g\sin \theta }{1+I\mathrm{/}(m{R}^2)}}$, the smaller $I\mathrm{/}(m{R}^2)$ for the sphere gives greater acceleration.

Ans.19. C.  Volume $V$ versus temperature $T$ at constant pressure (in Kelvin). Charles’s law states $V\propto T$ at constant pressure, which is exactly a straight line through the origin on a $V$–$T$ (Kelvin) plot.

Ans.20. C. H–Cl. The H–Cl bond exhibits a larger electronegativity difference compared to H–Br and H–I and has an optimal bond length that maintains significant charge separation, enhancing ionic character. This is supported by comparative dipole moments and behaviors in chemical reactions. 

Ans.21. A. 0.1 $\text{mol}{\text{kg}}^{-1}$ ${\text{CaCl}}_2$. Calcium chloride dissociates in water into three ions: one ${\text{Ca}}^{2+}$ and two ${\text{Cl}}^{-}$, so its theoretical van’t Hoff factor is $i\approx 3$. For the same molality and solvent, a higher $i$ means a larger $\mathrm{\Delta }{T}_{\text{f}}$, so the solution will exhibit a stronger freezing point depression. 

Ans.22. D. NH3 and ${\text{NH}}_4^{+}$. These two species differ only by the presence or absence of one ${\text{H}}^{+}$ ion. ${\text{NH}}_3$ acts as a base by accepting a proton to form ${\text{NH}}_4^{+}$, its conjugate acid. Recognizing such simple relationships is fundamental for solving acid-base equilibrium questions in NEET UG Chemistry.

Ans.23. C. 4s. With the smallest $n+l$ value (4) among the given options, 4s is energetically the most favorable orbital for electrons, so it fills before 3d, 4p, and 4d according to the Aufbau principle. 

Ans.24. D. Zinc is oxidized and acts as a reducing agent. Zinc changes from oxidation state 0 to +2, losing two electrons, whereas ${\text{H}}^{+}$ ions from the acid gain those electrons and are reduced to ${\text{H}}_2$ gas. 

Ans.25. A. A catalyst decreases the activation energy of both forward and backward reactions. This reduction in activation energy enhances the reaction rate in both directions, helping the system reach equilibrium faster without altering the equilibrium constant or the net enthalpy change.

Ans.26. D. ${\text{BF}}_3$. In boron trifluoride, boron has three bonding pairs and no lone pairs, giving an ${\text{AX}}_3$ type molecule in VSEPR terms. These three electron pairs arrange themselves at ${120}^{\circ }$ angles in a trigonal planar geometry, lying in one plane.

Ans.27. D.  Physical adsorption is due to weak van der Waals forces. In physisorption, the adsorbate is held on the surface of the adsorbent by weak intermolecular attractions. 

Ans.28. D. CH3CH=CHCH3. In this alkene, each carbon of the double bond is attached to two different groups. The left carbon is attached to ${\text{CH}}_3$ and $\text{H}$, and the right carbon is attached to $\text{H}$ and ${\text{CH}}_3$.

Ans.29. A. 0.01 $\text{M}$ NaOH. This is a strong base that fully dissociates into ${\text{Na}}^{+}$ and ${\text{OH}}^{-}$ ions. With $[{\text{OH}}^{-}]\approx 0.01\text{M}$, the solution has $\text{pOH}=2$pOH=2 and $\text{pH}=12$, which is much higher (more basic) than the pH values of the weak acid, strong acid, or salt solutions listed.

Ans.30. B. Electrolytic refining. Blister copper, obtained after smelting and converting, still contains impurities like iron, zinc, silver, and gold. By using it as an anode in an electrolytic cell with a copper sulphate solution, pure copper dissolves and redeposits on the cathode, while impurities either settle as anode mud or remain in solution.

Ans.31. A. Enthalpy of mixing $(\mathrm{\Delta }{H}_{\text{mix}})$ is zero. In an ideal solution, A-B interactions are similar in strength to A-A and B-B interactions, so no heat is absorbed or released on mixing, making $\mathrm{\Delta }{H}_{\text{mix}}=0$. Additionally, there is no significant volume change on mixing ($\mathrm{\Delta }{V}_{\text{mix}}=0$), and the solution obeys Raoult’s law over the entire composition range.

Ans.32. B. Lyophobic sol. Sulphur sol prepared by oxidation methods (such as bubbling ${\text{H}}_2\text{S}$ into ${\text{SO}}_2$ solution) yields colloidal sulphur in water that has little affinity for the dispersion medium. Such sols are relatively unstable, easily coagulated by electrolytes, and do not readily re-form once destroyed.

Ans.33. A. 3,2,3,+1/2. In this set, the value of $m=3$ is not allowed for $l=2$, because $m$ must lie between $-l$ and $+l$, i.e., between −2 and +2.

Ans.34. A. Fehling’s solution. Aldehydes typically reduce Fehling’s solution to give a red precipitate of ${\text{Cu}}_2\text{O}$, while ketones generally do not show this reaction, making Fehling’s solution a reliable qualitative reagent to distinguish these two types of carbonyl compounds.

Ans.35. B. 4. For $n=4$n=4, $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90\text{BM}$ , matching the observed value.

Ans.36. A. It can act as both an oxidizing and a reducing agent. In acidic medium, ${\text{H}}_2{\text{O}}_2$ often oxidizes species like ${\text{Fe}}^{2+}$ to ${\text{Fe}}^{3+}$, while in alkaline medium, it can reduce species like ${\text{KMnO}}_4$ in some cases, and under other conditions, it can itself be reduced or oxidized further. 

Ans.37. A. Phloem of vascular bundles. Phloem is a complex tissue composed of sieve tube elements, companion cells, phloem parenchyma, and sometimes fibers. In a leaf, the sucrose and other organic solutes produced by mesophyll cells during photosynthesis are loaded into the sieve tubes of the phloem.

Ans.38. B. The active site has a precise 3D shape complementary to the substrate. This specific three-dimensional arrangement means only certain substrates can form the correct interactions (hydrogen bonds, ionic bonds, hydrophobic interactions) within the active site. As a result, only those molecules can be converted efficiently to products. 

Ans.39. D.  Arrested at metaphase II and completed at fertilization (not ovulation). Primary oocytes begin meiosis during fetal development and pause in diplotene of prophase I. At puberty, during each menstrual cycle, a few oocytes resume meiosis I, one typically completes it to form a secondary oocyte and a polar body. The secondary oocyte then proceeds to metaphase II and stops again. Only when sperm enters the secondary oocyte does meiosis II finish, producing an ovum and second polar body.

Ans.40. D. It has a large surface area due to villi and microvilli and rich blood supply. These structural adaptations greatly increase the absorptive surface area while minimizing the distance that nutrients must travel to enter the blood or lymph. Each villus contains capillaries and a lacteal, allowing rapid transport of absorbed amino acids, monosaccharides, fatty acids, and glycerol. 

Ans.41. B. They recycle nutrients by breaking down dead organic matter. By decomposing leaf litter, carcasses, and faeces, decomposers release mineral nutrients into the soil and water, maintaining the nutrient pool. 

Ans.42. A. Hemoglobin’s affinity for ${\text{O}}_2$ decreases with increased ${\text{CO}}_2$ and lower pH. The Bohr effect describes how elevated ${\text{CO}}_2$and hydrogen ion concentration in metabolically active tissues shift the oxyhemoglobin dissociation curve to the right. This shift means that at the same partial pressure of ${\text{O}}_2$, hemoglobin releases more oxygen.

Ans.43. C.  Ribosomes. Ribosomes serve as a common feature connecting all domains of life: Bacteria, Archaea, and Eukarya. Their RNA components are so conserved that ribosomal RNA sequences are used to infer evolutionary relationships. The presence of ribosomes in every cell type emphasizes that protein synthesis is a universal requirement of life.

Ans.44. A. Tendency of genes located close together on the same chromosome to be inherited together. Linkage is a physical and statistical phenomenon observed during meiosis, where the proximity of genes on a chromosome influences how often recombination separates them. If two genes are very close, they are strongly linked and show low recombination frequency; if farther apart, they are more likely to recombine.

Ans.45. A. Hydrogen bonding between water molecules absorbs large amounts of heat. This high specific heat means that water in blood, cells, and extracellular fluid can act as a thermal buffer, moderating internal temperature changes even when the environment fluctuates. It is this unique combination of polarity and hydrogen bonding that underlies water’s role in temperature regulation in organisms.

Ans.46. B. Ultrafiltration due to high hydrostatic pressure in glomerular capillaries. In the glomerulus, blood enters via the afferent arteriole and exits via a narrower efferent arteriole, creating high hydrostatic pressure. This pressure forces water and small solutes (like glucose, amino acids, urea, and ions) through the filtration membrane (fenestrated endothelium, basement membrane, and podocyte slit pores) into Bowman’s space, forming glomerular filtrate.

Ans.47. B. A single amino acid can be coded by more than one codon. This redundancy means many amino acids are represented by multiple codons (for example, arginine, leucine, and serine have six codons each). This pattern reflects the structure of tRNA anticodons and the wobble hypothesis, where flexibility in base pairing at the third codon position allows a single tRNA to recognize multiple codons.

Ans.48. D. Adequate water supply, light, and low internal ${\text{CO}}_2$ concentration. Under these conditions, photosynthesis is active, internal ${\text{CO}}_2$ is being utilized, and guard cells accumulate solutes (like ${\text{K}}^{+}$), increasing their turgor. As a result, stomata open to allow more ${\text{CO}}_2$ to enter for continued photosynthesis, while transpiration occurs.

Ans.49. D.  Because random mating and absence of evolutionary forces maintain genetic stability. Hardy-Weinberg equilibrium is based on conditions such as large population size, random mating, no mutation, no migration, and no natural selection. When these conditions are met, allele and genotype frequencies remain constant from generation to generation.

Ans.50. A.  They require a host cell’s machinery to replicate their nucleic acid and synthesize proteins. Without host cells, viruses cannot grow, metabolize, or reproduce, which is why they are called obligate intracellular parasites.

Practice Questions: Free NEET Practice Questions 2026

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