NEET UG Questions and Answers

NEET UG Questions and Answers: Preparing for NEET UG requires more than just attempting questions; it’s about understanding the logic behind every answer. In this article, we provide detailed answers and clear explanations to the latest NEET UG questions so that you can evaluate your performance effectively. Use this section to cross-check your responses, identify weak areas, and strengthen your conceptual clarity in Physics, Chemistry, and Biology. Remember, every mistake corrected today brings you one step closer to your dream medical college.

Ans.1. D. Its velocity is zero, and acceleration is g downward at the highest point. At the topmost point, the velocity is momentarily 0, not g upward or downward. The acceleration remains g downward throughout the motion due to the constant gravitational force mg.

Ans.2. C. Power in $R_1$ is greater than in $R_2$. In a parallel connection, the same voltage $V$ appears across each resistor. Power in each resistor is $P=\frac{V^2}{R}$ . Because $R_2>R_1$, the denominator for $R_2$ is larger, so $P_2=\frac{V^2}{R_2}$ is smaller than $P_1=\frac{V^2}{R_1}$. Thus, the smaller resistance dissipates more power at the same voltage.

Ans.3. D. μ=sini/sinr. Snell’s law for refraction at a boundary between two media is $n_1\sin i=n_2\sin r$ . For air to glass, the refractive index of air is nearly 1, so $\mu =n_2={\displaystyle \frac{\sin i}{\sin r}}$. Because glass is optically denser than air, $\mu >1$, which implies $\sin i>\sin r$ and hence $i>r$. Writing $\mu ={\displaystyle \frac{\sin r}{\sin i}}$would give a value less than 1, which contradicts the known fact that glass has a higher refractive index than air.

Ans.4. D. v/2. For a satellite of mass $m$m in circular orbit, the gravitational force provides the centripetal force: ${\displaystyle \frac{GMm}{r^2}}={\displaystyle \frac{m{v}^2}{r}}$. This simplifies to $v=\sqrt{{\displaystyle \frac{GM}{r}}}$. Thus, orbital speed is proportional to ${\displaystyle \frac{1}{\sqrt{r}}}$. If the new radius is $4R$, the new speed $v^{\mathrm{\prime }}=\sqrt{{\displaystyle \frac{GM}{4R}}}={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{GM}{R}}}={\displaystyle \frac{v}{2}}$. So the speed decreases as the satellite moves to a higher orbit.

Ans.5. D. Temperature. In an isothermal process, the system is in thermal contact with a heat reservoir that maintains its temperature at a fixed value. According to the ideal gas law $PV=nRT$ , if $T$ is constant and $n$does not change, then $PV$ remains constant, which implies that pressure and volume adjust so their product stays the same.

Ans.6. A. Kinetic energy. The key principle is that to bring a moving body to rest, an opposing force must remove all its kinetic energy. Using $K=\frac{1}{2}m{v}^2$ and the work-energy theorem $W=\mathrm{\Delta }K$, the required work in magnitude is $\frac{1}{2}m{v}^2$.

Ans.7. C. Acceleration. The key property of SHM is that the acceleration is directly proportional to displacement but opposite in direction: $a=-{\omega }^{2}x$ , where $\omega$ is the angular frequency. 

Ans.8. B. Convex lens with object beyond $2F$. For a convex (converging) lens, when the object is placed between $F$ and the optic centre, the image is virtual and enlarged, not real and diminished. 

Ans.9. A. It becomes $2v_d$. The current density is given by $J=ne{v}_d$ , where n is the number density of charge carriers and $e$ is the charge of an electron. For a conductor of fixed cross-sectional area, current $I$ is proportional to drift velocity: $I\propto v_d$. 

Ans.10. B. Increases by about 3%. Using $R=\rho {\displaystyle \frac{L}{A}}$ and constant volume $V=LA$, the new length and area satisfy L′=1.01L and $A^{\mathrm{\prime }}={\displaystyle \frac{A}{1.01}}$. Substituting into the resistance expression, $R^{\mathrm{\prime }}=\rho {\displaystyle \frac{L^{\mathrm{\prime }}}{A^{\mathrm{\prime }}}}=\rho {\displaystyle \frac{1.01L}{A\mathrm{/}1.01}}=\rho {\displaystyle \frac{(1.01{)}^{2}L}{A}}\approx 1.0201R$

Ans.11. B. Charge. In a series connection, the same current passes through every component, so the amount of charge that accumulates on each capacitor plate is the same. Let the charge be Q.

Ans.12. D. Becomes half. The fringe width $\beta$ in Young’s double-slit experiment is given by $\beta ={\displaystyle \frac{\lambda D}{d}}$, where $\lambda$ is the wavelength of light, $D$ is the distance between slits and screen, and $d$ is the slit separation. If $d$ is doubled and $\lambda$ and $D$ are unchanged, the new fringe width is ${\beta }^{\mathrm{\prime }}={\displaystyle \frac{\lambda D}{2d}}={\displaystyle \frac{\beta }{2}}$.

Ans.13. C. Zero. The magnetic force is $F=BIL\sin \theta$. The maximum possible value is $F_{\text{max}}=BIL$ when $\theta ={90}^{\circ }$. For $\theta =0^{\circ }$, we have $F=0$, so there is no force at all. 

Ans.14. A.  Increases by a factor of 16. According to Stefan-Boltzmann law, the radiant power per unit area of a black body is $E=\sigma T^4$ , where $T$ is the absolute temperature in kelvin. 

Ans.15. D. Becomes half. From $v=\nu \lambda$, if $v$ is constant and $\nu$ν increases, then $\lambda$ must decrease proportionally. If frequency doubles, $\lambda$ becomes half; if frequency triples, $\lambda$ becomes one-third, and so on. In this question, where speed is explicitly stated to remain constant, an increased frequency necessarily implies a shorter wavelength.

Ans.16. D. Adding an inert gas at constant pressure. Under constant pressure, adding an inert gas necessarily increases the volume, which reduces the partial pressures of all gaseous species.

Ans.17. B. 0.1 M ${\text{CaCl}}_2$. Because boiling point elevation depends on the total number of dissolved particles, ionic solutes that dissociate into multiple ions produce a larger effect.

Ans.18. A. Kc=Kp(RT)−Δn. By rearranging the fundamental relation, you divide by $(RT{)}^{\mathrm{\Delta }n}$ to solve for $K_c$.

Ans.19. B. $\text{Na}<\text{Al}<\text{Mg}<\text{Si}$. This order reflects periodic trends: ionization enthalpy generally increases across a period from left to right, but there are anomalies, such as $\text{Al}$ having slightly lower ionization enthalpy than $\text{Mg}$ because the $3p$ electron in $\text{Al}$ is less tightly held than the paired $3s^2$ electrons in $\text{Mg}$.

Ans.20. B. H2, Anode: ${\text{Cl}}_2$. This outcome is consistent with the standard electrode potentials and the practical chlor-alkali process, where chloride is oxidized and water is reduced.

Ans.21. D. CH3COOH and ${\text{NaCH}}_3\text{COO}$Na. This combination directly provides a weak acid–conjugate base pair at significant concentrations, allowing the solution to resist pH changes by shifting the acid–base equilibrium in response to added ${\text{H}}^{+}$ or ${\text{OH}}^{-}$, which is the defining property of a buffer.

Ans.22. D. sp3d2 hybridization, octahedral. With six equivalent S-F bonds, sulfur must use six hybrid orbitals, which arise from mixing one s, three p, and two d orbitals.

Ans.23. D. The reaction is third order overall and first order with respect to B. Recognizing that the exponents in the rate law directly represent orders is key to predicting how the rate changes with concentration.

Ans.24. C. o-dichlorobenzene and p-dichlorobenzene, where the same substituents are attached at different positions on the same benzene ring framework, fitting the definition of position isomerism.

Ans.25. C.  Physical adsorption involves weak van der Waals forces and is multilayered. In physisorption, the forces involved are universal intermolecular attractions, leading to relatively non-specific adsorption that can form multiple layers, especially at low temperatures and near saturation pressures, which is an important concept in surface chemistry.

Ans.26. C. NH3<CH3NH2<(CH3)2NH<(CH3)3N. This reflects the increasing +I effect of the growing number of methyl groups in the gas phase, where solvation does not play a role.

Ans.27. A. Al2O3, which exhibits both acidic and basic behavior by reacting with both acids and bases to form corresponding salts and water.

Ans.28. C. O2 has two unpaired electrons in ${\pi }_{2p}^{\ast }$ antibonding orbitals. These unpaired electrons produce a net magnetic moment, explaining why liquid ${\text{O}}_2$ is drawn into a magnetic field, a classic demonstration in chemistry.

Ans.29. C.  First order; slope = $-k$. This specific graphical signature helps quickly distinguish reaction order in experimental data, which is a key point in NEET kinetics.

Ans.30. A. MgCl2. Higher charge and smaller ionic radius of ${\text{Mg}}^{2+}$ increase lattice enthalpy relative to ${\text{Na}}^{+}$ or ${\text{K}}^{+}$, leading to a higher melting point for ${\text{MgCl}}_2$ compared with $\text{NaCl}$ and $\text{KCl}$.

Ans.31. B. Dilation of the afferent arteriole. By widening the afferent arteriole, the kidney allows more blood to enter the glomerulus, thereby counteracting moderate decreases in systemic blood pressure. This maintains or increases glomerular hydrostatic pressure and helps sustain an adequate GFR so that nitrogenous wastes such as urea and creatinine continue to be removed from the body efficiently.

Ans.32. C. Scattered vascular bundles in ground tissue. This pattern of numerous, closed vascular bundles distributed irregularly throughout the ground tissue is unique to monocot stems in a typical transverse section. By contrast, dicot stems display vascular bundles arranged in a ring with a distinct cortex and pith, often with cambium present, allowing secondary growth.

Ans.33. A. Natural selection. Repeated pesticide use creates a consistent selection pressure, favoring insects that already carry resistance alleles. These individuals survive, reproduce, and increase the frequency of resistance within the population. Over several generations, the entire population becomes dominated by resistant individuals, illustrating how natural selection drives adaptive evolution.

Ans.34. D. Carbon dioxide is initially fixed into a $4$-carbon compound in mesophyll cells. This 4-carbon compound, oxaloacetate, is the hallmark of the C4 pathway. It is part of a mechanism that pumps ${\text{CO}}_2$ into bundle sheath cells, allowing photosynthesis to proceed efficiently even under high light, high temperature, and low ${\text{CO}}_2$ conditions.

Ans.35. A. Their red blood cells have antigen A and antigen Rh on the surface. Because of these antigens, the person can safely receive A positive and O positive blood (with cross-matching), as these will not present foreign A or Rh antigens that their immune system would attack.

Ans.36. B. Transfer of maternal antibodies to the fetus through the placenta. In this case, the fetus receives preformed IgG antibodies from the mother. These antibodies protect the newborn during early life before its own immune system becomes fully functional, illustrating passive, temporary immunity.

Ans.37. B. Crossing over between non-sister chromatids during prophase I. This event physically exchanges segments of chromatids, generating recombinant chromosomes with novel allele combinations. Together with independent assortment, crossing over is a major source of genetic diversity among offspring.

Ans.38. A. Luteinizing hormone $(\text{LH})$. LH from the anterior pituitary binds to specific receptors on Leydig cells, causing them to synthesize and release testosterone. This hormone is then responsible for driving spermatogenesis in cooperation with FSH and for producing the characteristic male phenotype.

Ans.39. D. It acts as the pacemaker by initiating the heartbeat. The SA node, located in the right atrium, spontaneously generates rhythmic electrical impulses that set the basic heart rate, earning it the title of the natural pacemaker of the heart.

Ans.40. D. Planaria – regeneration and fragmentation. The ability of Planaria to produce new individuals from body fragments via regeneration demonstrates a key form of asexual reproduction tested in exams.

Ans.41. A. Trypsin – proteins. Trypsin is a pancreatic enzyme that continues protein digestion in the small intestine by breaking peptide bonds, particularly after basic amino acids, producing smaller peptides and amino acids.

Ans.42. D. They cannot synthesize their own ATP and replicate only inside host cells. This dependence on the metabolic and synthetic machinery of the host cell defines their parasitic, intracellular lifestyle.

Ans.43. B. Inhibition of lateral bud growth by the apical bud, mainly via auxin. Auxin produced at the shoot tip is transported basipetally (downward), suppressing the outgrowth of axillary buds and maintaining a main central stem.

Ans.44. D. Smooth muscle is non-striated and involuntary, while cardiac muscle is striated and involuntary. This distinction highlights both structural differences (presence or absence of striations) and functional similarities (involuntary control).

Ans.45. D.  Biomass and stability of the community generally increase. The accumulation of organic matter, development of layers (strata), and more efficient nutrient cycling contribute to this greater stability.

Ans.46. A. Increase in ${\text{CO}}_2$ and ${\text{H}}^{+}$ concentration decreases hemoglobin’s affinity for ${\text{O}}_2$, promoting oxygen release in tissues. This adaptation ensures efficient oxygen delivery during conditions of high metabolic activity.

Ans.47. A. Parenchyma – photosynthesis and storage. This pairing accurately reflects the primary roles of parenchyma cells in plants.

Ans.48. D. 9 tall yellow : $3$ tall green : $3$ dwarf yellow : $1$ dwarf green. This is the classical Mendelian ratio for a dihybrid cross with independent assortment and complete dominance for both traits.

Ans.49. D. It has villi and microvilli that greatly increase surface area and contains an extensive blood supply. This combination of surface modifications and vascularization explains its central role in nutrient absorption.

Ans.50. A.  Photosynthesis by green plants and phytoplankton. Primary productivity is the rate at which light energy is converted into chemical energy in the form of organic compounds through photosynthesis.

We hope these answers and explanations helped you analyze your preparation level and clear your doubts. Make it a habit to review solutions carefully and note down important concepts for quick revision. Consistent practice combined with proper analysis is the key to scoring high in NEET UG. Keep practicing, stay confident, and continue working hard; your medical dream is absolutely achievable!

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