Answers & Solutions For Day 30 NEET Practice Set

Answers & Solutions For Day 30 NEET Practice Set: After attempting the NEET UG 2026 Day 30 Practice Questions, it’s time for the most important step: analyzing your performance! The Answers & Solutions for Day 30 are designed not just to reveal the correct options, but to help you understand the logic, concepts, and shortcuts behind each question. Whether you got stuck on a tricky Physics numerical, a conceptual Chemistry reaction, or a statement-based Biology question, this detailed solution set will clear your doubts and strengthen your fundamentals.

Ans.1. A. It stays at rest because friction equals weight. The frictional force acts upward, balancing the downward weight, and because frictional force can adjust up to its maximum value, it just takes the value $20\text{N}$20N needed to hold the block in equilibrium. Thus, net force along the vertical direction is zero, and the block remains stationary on the wall.

Ans.2. D. 0.5 m. The key is equating the initial spring energy to the work done by friction. The spring stores 2J of energy, and friction does work $W_f={\mu }_{k}mgd$. With ${\mu }_k=0.2$, $m=2\text{kg}$, and $g=10{\text{m s}}^{-2}$, the frictional force is 4N, so the work done over distance $d$d is 4d. Since all 2J are dissipated, $2=4d$, giving $d=0.5\text{m}$. 

Ans.3. A. The horizontal range is maximum for $\theta ={45}^{\circ }$ if landing height equals projection height. Gravity acts vertically downward with constant magnitude g. As a result, the vertical component of velocity changes with time according to $v_y=u\sin \theta -gt$. It decreases on the way up, becomes zero only at the highest point, and then becomes negative as the projectile falls. By contrast, the horizontal motion experiences no acceleration (in the ideal case), so $v_x=u\cos \theta$stays constant. Using projectile equations, the range when launching and landing at the same level is $R=\frac{u^2\sin 2\theta }{g}$. This range is maximum when $\sin 2\theta$ is maximum, i.e., equal to 1, which occurs at $2\theta ={90}^{\circ }$ or $\theta ={45}^{\circ }$

Ans.4. C. The capacitance becomes half and the potential doubles. For a parallel-plate capacitor in vacuum, capacitance is given by $C={\epsilon }_{0}A\mathrm{/}d$ where $A$ is the plate area and $d$ is the separation. When the distance doubles to $2d$, the new capacitance becomes $C^{\mathrm{\prime }}={\epsilon }_{0}A\mathrm{/}(2d)=C\mathrm{/}2$. 

Ans.5.D. The car will skid outward because the required centripetal force exceeds the maximum friction. The centripetal force required to keep the car moving in a circle is directed towards the center. Static friction between tyres and road is what actually provides this force. If the maximum frictional force ${\mu }_{s}mg$ is less than the required centripetal force $m{v}^2\mathrm{/}r$, then friction cannot keep the car on a circular path. The car then tends to move in a straight line tangent to the circle, which from the driver’s perspective appears as skidding outward.

Ans.6. B. B/2. The magnetic field due to a long straight current-carrying wire is $B=\frac{\mu I}{2\pi r}$. This means the field is inversely proportional to r. Changing the distance from $r$ to $2r$ makes the field $B^{\mathrm{\prime }}=\frac{\mu I}{2\pi (2r)}=B\mathrm{/}2$.

Ans.7. C. The net heat absorbed by the gas in one cycle is $500\text{J}$500J. In cyclic processes, $\mathrm{\Delta }U=0$ΔU=0. Applying the first law, $\mathrm{\Delta }U=Q-W$, simplifies to $0=Q-W$, so $Q=W$. Since the area enclosed on a clockwise $P\text{–}V$ diagram represents positive work done by the gas, here $W=+500\text{J}$. Consequently, $Q=+500\text{J}$.

Ans.8. A. 35∘. Applying Snell’s law, ${\mu }_1\sin i={\mu }_2\sin r$ with ${\mu }_1=1$ (air), ${\mu }_2=1.5$ (glass), and $i={60}^{\circ }$, we get $\sin r=(\sqrt{3}\mathrm{/}2)\mathrm{/}1.5=\sqrt{3}\mathrm{/}3\approx 0.577$, giving $r\approx {35}^{\circ }$. Keeping the angle equal to ${60}^{\circ }$ would imply no refraction and identical speeds in both media, which contradicts the given refractive index different from 1.

Ans.9. B. 2:1. Activity $A$ is proportional to the number of undecayed nuclei $N$ and is given by $A=\lambda N$, where $\lambda$ is the decay constant. 

Ans.10. C. Real, inverted image at $30\text{cm}$30cm in front of the mirror (or more precisely, at $30\text{cm}$ if calculated with the correct distances). Using the mirror formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where $f$ is focal length, $v$v is image distance and $u$u is object distance (with sign convention), we have $f=-10\text{cm}$ for a concave mirror and $u=-15\text{cm}$.

Ans.11. A. 2.25R. From $R=\rho L\mathrm{/}A$, if the length is increased by a factor of $1.5$ and volume is constant, then area decreases by $1\mathrm{/}1.5$. The new resistance becomes $R^{\mathrm{\prime }}=\rho (1.5L)\mathrm{/}(A\mathrm{/}1.5)=\rho L\mathrm{/}A\cdot ({1.5}^2)=2.25R$.

Ans.12. D.  $U=\frac{1}{2}k{x}^2$. At $x=A\mathrm{/}2$x=A/2, $U=\frac{1}{2}k(A\mathrm{/}2{)}^2=\frac{1}{8}k{A}^2$ So $K=E-U=\frac{1}{2}k{A}^2-\frac{1}{8}k{A}^2=\frac{3}{8}k{A}^2$. Dividing by $E$ yields $K\mathrm{/}E=3\mathrm{/}4$. 

Ans.13. C. [ML2T−1]. Planck’s constant appears in the relation $E=h\nu$, where $E$ is energy and $\nu$is frequency. The dimensional formula of energy is $[M{L}^2{T}^{-2}]$, and frequency has dimensions $[T^{-1}]$. Therefore, $[h]=[E]\mathrm{/}[\nu ]=[M{L}^2{T}^{-2}]\mathrm{/}[T^{-1}]=[M{L}^2{T}^{-1}]$.

Ans.14. C. 200. The key step is converting $10\text{mV}$ to volts: $10\text{mV}=0.01\text{V}$. Then the gain is $A_v=2\mathrm{/}0.01=200$. 

Ans.15. B. Impedance is minimum and equal to resistance. At resonance, the inductive reactance $X_L=\omega L$ equals the capacitive reactance $X_C=1\mathrm{/}(\omega C)$. Hence, the net reactance $X=X_L-X_C=0$.

Ans.16. D. Hydrogen bonding in ice creates an open hexagonal structure. Intermolecular hydrogen bonds arrange water molecules into a spacious lattice in ice, increasing the volume per molecule. Because density equals mass divided by volume, this larger volume lowers the density of ice compared to liquid water.

Ans.17. D. Si. Moving from left to right in period 3, effective nuclear charge increases and atomic radius decreases, raising ionization enthalpy overall. Silicon has configuration $[\text{Ne}]3s^{2}3p^2$ and lies further to the right than $\text{Na}$, $\text{Mg}$, and $\text{Al}$, so its outer electrons experience a stronger attraction to the nucleus.

Ans.18. C. NH4OH and ${\text{NH}}_4\text{Cl}$. Here, ${\text{NH}}_4\text{OH}$ (weak base) and ${\text{NH}}_4^{+}$from ${\text{NH}}_4\text{Cl}$ (its conjugate acid) coexist, providing the capacity to neutralize small amounts of added acid or base. 

Ans.19. D. Octahedral. ${\text{SF}}_6$ corresponds to an ${\text{AX}}_6$ arrangement, leading to an octahedral electron pair and molecular geometry. In this shape, four fluorine atoms occupy equatorial positions in a square plane, and two occupy axial positions above and below the plane.

Ans.20. D. Gas molecules occupy negligible volume compared to container. At high pressure, real gas molecules have appreciable volume, making the free space for movement smaller than the container volume. This violates the ideal gas model and requires corrections like those in the van der Waals equation. 

Ans.21. A. Chemisorption is specific, often forms monolayer, and increases with temperature up to a point. In chemisorption, the process resembles a chemical reaction between surface and adsorbate, requiring activation energy and therefore being favored by moderate temperature increases until bond formation is maximized. After a certain temperature, desorption may dominate. 

Ans.22. C. CH3CH=CHCH3. Here, each double-bonded carbon has a hydrogen and a methyl group attached, making two distinct substituents on each carbon. This allows for $\text{cis}$ and $\text{trans}$ configurations based on the relative positions of the ${\text{CH}}_3$ groups, a standard NEET UG organic chemistry concept.

Ans.23. B. Tollens’ reagent. Aldehydes reduce the diamminesilver(I) complex to metallic silver, giving a characteristic silver mirror, while most ketones fail to do so. 

Ans.24. D. 0.1M ${\text{CaCl}}_2$. Colligative properties like boiling point elevation depend on the total number of solute particles, not their nature. ${\text{CaCl}}_2$ dissociates into three ions (${\text{Ca}}^{2+}$ and $2{\text{Cl}}^{-}$), giving a higher van ‘t Hoff factor $i$i than $\text{NaCl}$, ${\text{KNO}}_3$, or glucose. Therefore, it causes the greatest increase in boiling point for the same molar concentration.

Ans.25. C. F2. Fluorine’s extremely high electronegativity and small atomic size contribute to a very high tendency to gain electrons, making it the strongest oxidizing agent among the halogens in aqueous solution. 

Ans.26. A. They often exhibit variable oxidation states due to involvement of both $s$ and $d$ electrons. Transition metals can lose their outer $ns$ and $n-1d$electrons, leading to multiple stable oxidation states.

Ans.27. D. Reaction of ${\text{CH}}_3\text{Br}$ with ${\text{OH}}^{-}$ to form ${\text{CH}}_3\text{OH}$. This reaction involves displacement of the bromide ion by the nucleophilic hydroxide ion on a saturated carbon, fitting the definition of nucleophilic substitution

Ans.28. A. A catalyst provides an alternative pathway with lower activation energy. This lower energy route increases the reaction rate but leaves $K$ unchanged

Ans.29. C. Teflon is obtained by polymerization of tetrafluoroethene. Teflon (polytetrafluoroethene) is produced via free-radical addition polymerization of the monomer tetrafluoroethene $({\text{CF}}_2={\text{CF}}_2)$, yielding a linear polymer with repeating $\text{–}{\text{CF}}_2\text{–}{\text{CF}}_2\text{–}$ units. NEET UG polymer questions often focus on correctly matching monomers to polymers and recognizing addition versus condensation processes.

Ans.30. B. Presence of a chiral carbon atom or overall chirality. A chiral center is usually a carbon attached to four different substituents, though molecules can also be chiral without a single chiral center (axial chirality, etc.). 

Ans.31. A.  Light reactions in thylakoid membranes; Calvin cycle in stroma. In chloroplasts, the light-dependent reactions occur on the thylakoid membranes where pigment-protein complexes like photosystem II, photosystem I, and ATP synthase are embedded. 

Ans.32. A. Carotenoids and other accessory pigments broaden the range of usable light. Chlorophyll a shows strong absorption peaks in the blue-violet and red regions, with relatively poor absorption in the green region. Accessory pigments help cover some of these gaps, thereby improving efficiency across a wider set of wavelengths.

Ans.33. D. Circular DNA not associated with histone proteins in most groups. In prokaryotes, the nucleoid region contains the genetic material in direct contact with the cytoplasm, without a surrounding nuclear membrane. Their DNA is usually circular and lacks the complex histone-based chromatin organization characteristic of eukaryotes.

Ans.34. A. Ethylene; stimulating fruit ripening and associated enzymatic changes. Ethylene acts as a powerful regulator of the fruit ripening program, especially in climacteric fruits. It triggers the breakdown of chlorophyll, leading to color changes, and it induces the synthesis of enzymes that degrade complex polysaccharides in the cell wall, resulting in softening. Additionally, ethylene influences the accumulation of sugars and organic acids that define the fruit’s sweetness and flavor.

Ans.35. D. Proximal convoluted tubule – major reabsorption of glucose, amino acids, and ions. At Bowman’s capsule, the process is primarily mechanical filtration driven by hydrostatic pressure, with size-selective permeability. Once filtrate enters the PCT, it encounters cells specialized for recovering essential nutrients and ions. This process dramatically reduces the amount of solute and water that would otherwise be lost. 

Ans.36. A. Four sperm cells; crossing over and independent assortment of homologous chromosomes. In spermatogenesis, a diploid primary spermatocyte undergoes meiosis I and meiosis II to produce four haploid spermatids, each of which matures into a sperm. Genetic variation arises mainly during meiosis I. Crossing over occurs during prophase I when homologous chromosomes pair as bivalents and exchange segments, creating new combinations of alleles on each chromatid. 

Ans.37. A. Significant reduction in soil nitrogen due to continuous cropping without replenishment. Primary productivity depends on the availability of sunlight, water, ${\text{CO}}_2$, and essential nutrients. In many grasslands, nitrogen is the main limiting nutrient, so its depletion severely restricts plant growth. Without adequate nitrogen, plants cannot build the proteins and chlorophyll needed for photosynthesis and metabolism. As a result, the rate at which they convert solar energy into biomass decreases, which directly lowers primary productivity in that ecosystem.

Ans.38. A. Innate immunity includes physical barriers and cellular defenses present from birth, while adaptive immunity involves antigen-specific responses that develop after exposure. Innate immunity is the body’s first line of defense, comprising physical barriers like skin and mucous membranes, chemical barriers such as lysozyme and stomach acid, and cellular components like phagocytes (for example, neutrophils and macrophages), natural killer cells, and complement proteins. These responses are rapid and broadly targeted against common features of pathogens. Adaptive immunity, by contrast, is mediated by lymphocytes (B cells and T cells) that recognize specific antigens via unique receptors. 

Ans.39. B.  Opening of voltage-gated potassium channels leading to efflux of ${\text{K}}^{+}$K+. In a typical neuron, an action potential begins when a threshold stimulus opens voltage-gated sodium channels, allowing ${\text{Na}}^{+}$Na+ to rush into the cell and depolarize the membrane toward a positive potential. Soon after, these sodium channels inactivate, and voltage-gated potassium channels open. The resulting efflux of ${\text{K}}^{+}$K+ out of the cell drives the membrane potential back toward the resting level (repolarization) and often slightly beyond (hyperpolarization).

Ans.40. D. Albumin. While fibrinogen contributes somewhat to plasma oncotic pressure, albumin’s higher concentration and specific properties make it the dominant determinant of colloid osmotic pressure. Albumin also has a broad transport role that fibrinogen lacks, carrying hormones, lipids, and other small molecules through the circulation. Without sufficient albumin, individuals can develop edema due to decreased plasma oncotic pressure, even if other proteins like fibrinogen are present.

Ans.41. C. Differential gene expression in cells that all contain essentially the same DNA. During development, cells in the embryo undergo mitotic divisions, so the daughter cells inherit nearly identical DNA. As development proceeds, signaling molecules, cell–cell interactions, and gradients of morphogens cause certain sets of genes to be turned on or off in specific cells. This process, called differential gene expression, leads to distinct patterns of proteins being produced in different cells. 

Ans.42. A. Presence of Kranz anatomy with bundle sheath cells performing the Calvin cycle, and initial ${\text{CO}}_2$ fixation in mesophyll cells via PEP carboxylase. In C4 plants, bundle sheath cells are large, chloroplast-rich cells surrounding veins, and they host the Calvin cycle under conditions of elevated ${\text{CO}}_2$. 

Ans.43. B. The operon is only weakly expressed because the repressor is inactivated by lactose, but cAMP levels are low due to high glucose. This is an example of catabolite repression: the presence of a preferred carbon source (glucose) suppresses the expression of genes needed to metabolize alternative sources (like lactose). Only when glucose levels drop, cAMP rises, CAP–cAMP binds efficiently, and together with lactose-mediated repressor inactivation, the operon becomes strongly expressed.

Ans.44. C.  Attachment and embedding of the blastocyst into the endometrial lining of the uterus. Ovulation provides the oocyte that may be fertilized by sperm in the oviduct. Only after fertilization and several days of cleavage divisions does the embryo reach the uterus as a blastocyst ready for implantation. During implantation, the blastocyst becomes anchored in the endometrium, allowing the exchange of nutrients and gases that supports continued development.

Ans.45. B. They lack cellular machinery for independent metabolism and replication, requiring a host cell to reproduce. Viral simplicity means they carry only a minimal set of genes, typically those needed to enter host cells, replicate their genome, and assemble virions. All other metabolic processes, such as energy production and protein synthesis, are borrowed from the host. This dependence on host cell machinery is what makes them obligate intracellular parasites.

Ans.46. D. Allele and genotype frequencies remain constant from generation to generation in the absence of evolutionary forces. Under Hardy-Weinberg conditions, natural selection is absent, so no genotype has a systematic advantage or disadvantage, and allele frequencies do not change. If selection occurs, alleles associated with higher fitness increase in frequency, causing a departure from HardyWeinberg equilibrium.

Ans.47. A. Vitamin D – rickets in children due to defective bone mineralization. Vitamin D is essential for normal calcium and phosphate metabolism. In children, deficiency impairs the absorption of calcium and phosphate from the gut, leading to inadequate mineralization of growing bones. This results in soft, deformed bones, bowing of the legs, delayed closure of fontanelles, and other skeletal abnormalities collectively known as rickets. In adults, a similar defect in bone mineralization leads to osteomalacia.

Ans.48. D.  IOIO. People with IOIO genotype lack active glycosyltransferase enzymes that would attach A or B sugars to the H antigen on red blood cells. As a result, they display neither A nor B antigens, which defines the O blood group. The O phenotype is recessive, so it is only expressed when both alleles are IO, with no IA or IB present.

Ans.49. C. They have permeable skin and complex life cycles that expose them to both aquatic and terrestrial pollutants. Amphibians such as frogs and salamanders typically have moist, thin skin that allows gases and water-soluble substances to pass easily. This makes them particularly sensitive to contaminants, changes in pH, and other chemical alterations in their environment. Additionally, many amphibians have life cycles that include aquatic larval stages and terrestrial adult stages, so they experience conditions in both habitats. Because they respond quickly to environmental degradation, declines in amphibian populations can serve as early warning signs of ecosystem health problems.

Ans.50. A. They break down dead organic matter, releasing inorganic nutrients back into the environment for reuse by producers. Decomposers, including bacteria and fungi, play a crucial role in nutrient cycling by digesting dead plants, animals, and waste products. Through processes like mineralization, they transform complex organic compounds into simpler inorganic forms that can be taken up again by producers (plants and algae).

Remember, improvement begins with analysis. Reviewing solutions carefully helps you identify weak areas, avoid repeated mistakes, and sharpen your exam strategy. Use Day 30’s answer key as a learning tool, not just a score-checking sheet. Stay consistent, keep revising, and come back tomorrow stronger for the next practice set.

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