Day 29 Practice Question’s Solutions

Day 29 Practice Question’s Solutions: This article contains the complete answers and detailed explanations for the Day 29 question set (NEET UG Practice Questions Day 29). By now, your consistency is building real momentum and that’s exactly what NEET preparation demands. These solutions are not just for checking scores, but for understanding concepts deeply, identifying weak areas, and avoiding similar mistakes in the actual exam. Go through each explanation carefully and turn today’s analysis into tomorrow’s improvement!

Ans.1. C.  A straight line with negative slope, crossing $v=0$ at some $t$t and extending symmetrically above and below the time axis. Under constant acceleration $a=-g$, the relation $v=u+at$ shows that $v$v depends linearly on $t$.

Ans.2. A. 2m s−2. First, calculate the frictional force using $f_k={\mu }_{k}N$ , where ${\mu }_k=0.3$ and $N$ is the normal reaction. On a horizontal surface with no vertical acceleration, $N=mg=2\times 10=20\text{N}$Hence, $f_k=0.3\times 20=6\text{N}$. The applied force is $10\text{N}$, so the net force in the horizontal direction is $F_{\text{net}}=10-6=4\text{N}$ in the direction of pulling. Using Newton’s second law, $F_{\text{net}}=ma$ , we get $a=F_{\text{net}}\mathrm{/}m=4\mathrm{/}2=2{\text{m s}}^{-2}$. Thus the acceleration must be greater than $1{\text{m s}}^{-2}$.

Ans.3. D. 8000N. The correct formula is $F_c={\displaystyle \frac{m{v}^2}{r}}$. With $m=1000\text{kg}$, $v=20{\text{m s}}^{-1}$, and $r=50\text{m}$, the calculation gives $F_c=8000\text{N}$. To get $4000\text{N}$, you would effectively be halving the required inward force, which would not be enough to bend the car’s path into a circle of that radius at the given speed. In reality, insufficient centripetal force would cause the car to drift outward, following a path of larger radius.

Ans. 4. D. k/2. When springs are connected in series, the effective spring constant $k_{\text{eff}}$ is given by ${\displaystyle \frac{1}{k_{\text{eff}}}}={\displaystyle \frac{1}{k_1}}+{\displaystyle \frac{1}{k_2}}+\dots$. For two identical springs with $k_1=k_2=k$, this becomes ${\displaystyle \frac{1}{k_{\text{eff}}}}={\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{k}}={\displaystyle \frac{2}{k}}$. Therefore, $k_{\text{eff}}={\displaystyle \frac{k}{2}}$. Physically, adding springs in series increases the total extension for a given force, effectively reducing the overall stiffness compared to a single spring.

Ans.5. D. 3/4. Using the standard SHM energy relations, total energy is $E={\displaystyle \frac{1}{2}}m{\omega }^2{A}^2$ and potential energy at displacement $x$ is $U={\displaystyle \frac{1}{2}}m{\omega }^2{x}^2$. For $x={\displaystyle \frac{A}{2}}$, we get $U={\displaystyle \frac{1}{2}}m{\omega }^2({\displaystyle \frac{A}{2}}{)}^2={\displaystyle \frac{1}{4}}E$. Therefore, the kinetic energy must be $K=E-U=E-{\displaystyle \frac{1}{4}}E={\displaystyle \frac{3}{4}}E$.

Ans.6. A. 30

Ans.7. D. 30cm. The thin lens formula is ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$ , where $f$ is the focal length, $v$v is the image distance, and $u$ is the object distance (negative for real objects according to the sign convention). Given $f=20\text{cm}$ and $v=60\text{cm}$, we write ${\displaystyle \frac{1}{20}}={\displaystyle \frac{1}{60}}-{\displaystyle \frac{1}{u}}$. Rearranging gives ${\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{60}}-{\displaystyle \frac{1}{20}}={\displaystyle \frac{1-3}{60}}=-{\displaystyle \frac{2}{60}}=-{\displaystyle \frac{1}{30}}$

Ans.8. D. 0.05J. The energy stored in a capacitor is given by $U={\displaystyle \frac{1}{2}}C{V}^2$$C=10\mu \text{F}=10\times {10}^{-6}\text{F}$, and $V=100\text{V}$=100V. Substituting, we get $U={\displaystyle \frac{1}{2}}\times 10\times {10}^{-6}\times {100}^2={\displaystyle \frac{1}{2}}\times 10\times {10}^{-6}\times 10000$

Ans.9. C. $\propto {\displaystyle \frac{1}{r^3}}$In electrostatics, a point charge field falls as $1\mathrm{/}{r}^2$, a dipole as $1\mathrm{/}{r}^3$ and higher-order multipoles fall off even faster . For our configuration, with charges $+Q$+ and $-Q$ separated by $2a$, the far-field behavior ($r\gg a$) is dominated by the dipole term, yielding $E\propto {\displaystyle \frac{p}{r^3}}$, where $p$ is the dipole moment. 

Ans.10. B. Resistance of the conductor. From the expression $F=BIL$ for a conductor perpendicular to the magnetic field, $F$ is directly proportional to $B$. If the magnetic field strength is increased, the force on the conductor increases correspondingly. 

Ans.11. D. 10,000 turns. From ${\displaystyle \frac{V_s}{V_p}}={\displaystyle \frac{N_s}{N_p}}$ , we find ${\displaystyle \frac{2200}{220}}={\displaystyle \frac{N_s}{1000}}$, giving $N_s=10,000$. 

Ans.12. D. Gamma rays. The electromagnetic spectrum is arranged in order of decreasing wavelength (and increasing frequency) from radio waves, microwaves, infrared, visible, ultraviolet, X-rays, to gamma rays . 

Ans.13. B. Increase in number of photoelectrons emitted per unit time. The threshold frequency $f_0$ satisfies $h{f}_0=\varphi$ , where $\varphi$ is the work function. Neither $\varphi$ nor $f_0$ is affected by how bright (intense) the light is. 

Ans.14. C. The acceleration is directed towards the centre and has magnitude ${\displaystyle \frac{v^2}{r}}$. In an inertial frame of reference, the only real acceleration for uniform circular motion is centripetal and points inward. 

Ans.15. A. Wavelength decreases but frequency remains the same. When light enters glass from air, its speed decreases from $c$ to $v={\displaystyle \frac{c}{\mu }}$ . The relationship among speed, wavelength, and frequency is $v=\nu \lambda$. Since $\nu$ (frequency) remains constant, a decrease in speed implies a proportional decrease in wavelength: ${\lambda }^{\mathrm{\prime }}={\displaystyle \frac{v}{\nu }}={\displaystyle \frac{c\mathrm{/}\mu }{\nu }}={\displaystyle \frac{\lambda }{\mu }}$. Thus the wavelength becomes shorter in the denser medium, while the frequency retains the value determined by the source.

Ans.16. C. The azimuthal quantum number (l), also called the angular momentum quantum number, determines the shape of an orbital. It defines the subshell type: l=0 for s (spherical), l=1 for p (dumbbell), l=2 for d (cloverleaf), and l=3 for f orbitals.

Ans. 17. D. ${\text{CH}}_3\text{COONa}$CH3COONa. Only when you have a weak acid like ${\text{CH}}_3\text{COOH}$ and its conjugate base in significant amounts can the solution consume added ${\text{H}}^{+}$ or ${\text{OH}}^{-}$ without large pH changes.

Ans.18. A. Na+ has a larger size and weaker polarizing power, making ${\text{Na}}_2{\text{CO}}_3$ decompose more readily. Actually, ${\text{Na}}_2{\text{CO}}_3$ is more thermally stable than ${\text{Li}}_2{\text{CO}}_3$.

Ans.19. A. 0.1molkg−1 ${\text{MgCl}}_2$. Because ${\text{MgCl}}_2$ dissociates into three ions, it effectively triples the particle concentration, giving a larger $\mathrm{\Delta }{T}_b$ according to $\mathrm{\Delta }{T}_b=i{K}_{b}m$. 

Ans.20. C. $3d$ electron. Shielding increases as you move to higher principal quantum numbers and orbitals that are more diffuse and less penetrating. $3d$ electrons are shielded by all the inner-shell 1s, 2s, 2p, 3s, and 3p electrons.

Ans.21. D. Adsorption occurs on a homogeneous surface forming only a monolayer of adsorbate. According to Langmuir, once all surface sites are occupied, no further increase in adsorption occurs, so the isotherm approaches a limiting value rather than growing indefinitely with pressure .

Ans.22. A. ΔS=0.5Rln2. According to $\mathrm{\Delta }S=nR\ln$, halving the number of moles halves the entropy change for the same volume ratio, so the correct numerical coefficient must match the number of moles given.

Ans. 23. C. ${\text{O}}_2^{2-}$. When ${\text{O}}_2$ gains two electrons to become the peroxide ion, these extra electrons fill the previously half-occupied ${\pi }_{2p}^{\ast }$ orbitals, pairing up all electrons. With no unpaired electrons remaining, ${\text{O}}_2^{2-}$ becomes diamagnetic, meaning it is weakly repelled by a magnetic field .

Ans.24. D.  Ionization enthalpy generally increases from left to right, with small dips at specific elements. This pattern reflects the balance between the strong pull of the nucleus, subshell structure, and electron–electron repulsions, which create predictable but slightly irregular trends .

Ans. 25. A. Cathode: ${\text{H}}_2$; Anode: ${\text{Cl}}_2$. This reflects the dominant half-reactions in concentrated brine, which are responsible for the chlor-alkali process that supplies chlorine gas and sodium hydroxide industrially.

Ans.26. B. 2-Chlorobutane. The parent chain has four carbon atoms (butane), and the chlorine substituent is located on the second carbon when numbered from the end nearest the substituent, giving the name 2-chlorobutane according to IUPAC rules for haloalkanes .

Ans.27. D. Sodium bicarbonate solution. Ethanoic acid reacts with ${\text{NaHCO}}_3$to produce brisk effervescence of ${\text{CO}}_2$: ${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3\to {\text{CH}}_3\text{COONa}+{\text{CO}}_2+{\text{H}}_2\text{O}$. Ethanol does not react with sodium bicarbonate under these conditions, so this test clearly differentiates between the two.

Ans.28.A. Rate becomes 18 times. The rate law exponents are key: order 1 in A, order 2 in B leads to a combined multiplicative effect of $2\times 9$ when their concentrations are changed as specified .

Ans.29. D. Graphite has delocalized $\pi$-electrons that can move freely, while diamond has localized σ-bonds only. The availability of delocalized electrons in graphite is crucial for electrical conduction, unlike in diamond .

Ans.30. C. BaCl2 and ${\text{Na}}_2{\text{SO}}_4$. Their ions form ${\text{BaSO}}_4$, an insoluble sulfate, which precipitates from solution, illustrating selective precipitation based on solubility product considerations .

Ans.31. A. They temporarily bind specific solutes and move them down their concentration gradient without using ATP. Carrier‑mediated facilitated diffusion involves a specific interaction between the carrier and the transported molecule. The carrier changes conformation in a controlled way, exposing the solute first to one side of the membrane and then to the other, always following the solute’s concentration gradient.

Ans.32. A. Auxin. Auxin stimulates stem elongation by altering cell wall plasticity and promoting water uptake into cells, enabling them to expand. In phototropism, auxin’s unequal distribution between the shaded and illuminated sides of a stem causes unequal elongation.

Ans. 33. B. increased ${\text{CO}}_2$ and ${\text{H}}^{+}$ concentration in blood and cerebrospinal fluid. When ${\text{CO}}_2$ levels rise, ${\text{CO}}_2$ diffuses into cerebrospinal fluid and forms carbonic acid, which dissociates into ${\text{H}}^{+}$ and ${\text{HCO}}_3^{-}$. The increased ${\text{H}}^{+}$ lowers pH and stimulates central chemoreceptors, which then signal the respiratory centre in the medulla to increase the rate and depth of breathing.

Ans.34. A. DNA polymerase I. Once primase has created RNA primers and DNA polymerase III has elongated them, the RNA segments must be removed. DNA polymerase I performs this task using its unique 5′ → 3′ exonuclease activity and simultaneously fills in the region with DNA nucleotides. 

Ans.35. D. Proximal convoluted tubule – major site of selective reabsorption of nutrients and ions. In the PCT, glucose and amino acids are reabsorbed almost completely via secondary active transport mechanisms that are coupled to ${\text{Na}}^{+}$ reabsorption.

Ans.36. B. perform the Calvin cycle using ${\text{CO}}_2$ released from a 4‑carbon acid. The decarboxylation of 4‑carbon compounds such as malate inside bundle sheath cells enriches the local ${\text{CO}}_2$ concentration. Rubisco, present in these cells, then uses this ${\text{CO}}_2$ in the Calvin cycle to form sugars via 3‑phosphoglycerate intermediates. 

Ans.37. A. It states that alleles of different genes assort independently only when they are located on different, non‑homologous chromosomes or far apart on the same chromosome. Independent assortment is strictly true for genes on different chromosomes because their orientation on the metaphase plate is random.

Ans.38. D. Intact skin and mucous membranes. These structures form the first line of defence in innate immunity. Skin provides a tough, keratinised barrier and an acidic environment due to secretions like sweat and sebum. Mucous membranes line the respiratory, digestive, and urogenital tracts, secreting mucus that traps microbes. 

Ans.39. B. The operon is strongly activated because the repressor is inactivated and cAMP‑CAP complex enhances transcription. This dual regulation allows the bacterium to prioritize glucose but switch to lactose metabolism when glucose is scarce and lactose is available, a phenomenon known as diauxic growth.

Ans.40. A. Population. Population refers specifically to one species in a given area that can interbreed, making it the basic unit for studying genetic variation and evolutionary processes. For example, a population of tiger in a forest or a population of lotus plants in a pond are typical illustrations used in exam questions.

Ans.41. C. Crossing over between non‑sister chromatids during prophase I. During this stage, homologous chromosomes undergo synapsis and exchange segments. The resulting recombinant chromatids introduce new allele combinations that contribute significantly to genetic diversity among offspring.

Ans.42. D.  It acts as a carboxylase, catalysing the fixation of ${\text{CO}}_2$ to RuBP, but it can also act as an oxygenase under high ${\text{O}}_2$ conditions. Understanding this dual role helps explain why photorespiration occurs and why some plants have evolved strategies to minimize it.

Ans.43. B. Salivary amylase – acts on starch to produce maltose. Salivary amylase (ptyalin), secreted by salivary glands, begins carbohydrate digestion in the mouth. It hydrolyses starch (a polysaccharide) into maltose and small dextrins. Although its activity is limited by the short time food stays in the mouth and the subsequent acidic pH of the stomach, this step marks the start of digestion of complex carbohydrates in humans.

Ans.44. B. A surge in luteinising hormone (LH) secretion from the anterior pituitary. This LH surge leads directly to rupture of the mature follicle and release of the oocyte, and is a key hormonal event tested in NEET questions on the menstrual cycle.

Ans.45. B. They lack their own metabolic machinery and must use the host cell’s biosynthetic systems to replicate. Their dependence on host cells for reproduction distinguishes them from free‑living microorganisms like bacteria and is a fundamental property emphasized in NEET biology.

Ans.46. A. Clownfish living among the tentacles of sea anemones, both gaining protection and food benefits. In this mutualistic association, the clownfish gains protection from predators among the stinging tentacles, which it is immune to, and the anemone benefits from increased water circulation and bits of leftover food from the fish. The fish may also chase away potential anemone predators. Both partners gain from the relationship, which is the hallmark of mutualism.

Ans.47. B. Leukocytes – defend the body against infections and foreign substances. White blood cells (leukocytes) include various types such as neutrophils, lymphocytes, monocytes, eosinophils, and basophils. Together, they perform immune functions like phagocytosis, antibody production, and release of inflammatory mediators, providing defence against pathogens and foreign materials.

Ans.48. A. Sympathetic division of the autonomic nervous system generally prepares the body for ‘fight or flight’ responses. Sympathetic activation increases heart rate, dilates bronchi, mobilises glucose, and diverts blood flow to skeletal muscles, enabling the body to respond rapidly to stressful or emergency situations. This concept frequently appears in NEET questions on nervous system physiology.

Ans.49. D. Primary lymphoid organs are sites of lymphocyte maturation, while secondary lymphoid organs are sites where mature lymphocytes encounter antigens and initiate immune responses. In humans, the primary lymphoid organs are bone marrow (for B cell maturation and origin of all lymphocytes) and thymus (for T cell maturation). Secondary organs include lymph nodes, spleen, tonsils, and Peyer’s patches, where antigens are trapped and presented to lymphocytes.

Ans.50. D. The maximum number of individuals of that species that the environment can support sustainably, given available resources. This concept explains the upper limit of population size imposed by the environment and is central to the logistic growth model in population ecology.

That concludes the answers and explanations for the Day 29 question set. Remember, analyzing solutions is just as important as solving questions. Even a small mistake today can become a strong concept tomorrow if reviewed properly. Stay consistent, stay confident, and keep improving step by step. Get ready for the next challenge – your NEET success story is being written one practice set at a time!

All The Best!