Important NEET UG 2026 Practice Questions (Day 34)

Important NEET UG 2026 Practice Questions (Day 34): After a short pause due to the Holi holidays, we are excited to restart our Daily NEET UG 2026 Practice Questions series. We hope you also had a refreshing break and are now ready to get back into your preparation routine. Regular MCQ practice is essential for mastering concepts, improving accuracy, and building the speed required for the NEET exam. With Day 34, we are back on track and will be posting daily practice questions consistently to help you stay focused on your goal of securing a top score in NEET UG 2026.

PHYSICS

Q.1. A physical quantity P is related to four observations $a, b, c$ and $d$ as follows:$$P = \frac{a^3 b^2}{c \sqrt{d}}$$

The percentage errors of measurement in $a, b, c,$ and $d$ are 1%, 3%, 2% and 4% respectively. The percentage error in the quantity $P$ is:

A. 13%

B. 15%

C. 10%

D. 2%

Q.2. The sun rotates around its centre once in 27 days. What will be the period of revolution if the sun were to expand to twice its present radius without any external influence? Assume the sun to be a sphere of uniform density.

A. 105 days

B. 115 days

C. 108 days

D. 100 days

Q.3. The radius of Martian orbit around the sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 earth days. Then which of the following is the length of 1 year on mercury?

A. 225 earth days

B. 172 earth days

C. 124 earth days

D. 88 earth days

Q.4. A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two are added in series. The net effective resistance of the combination is:

A. $\frac{R}{32}$

B. $\frac{R}{16}$

C. $\frac{R}{8}$

D. $\frac{R}{64}$

Q.5. A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the Y-axis is:

1. $\frac{7}{40}$
   ​
2. $\frac{7}{57}$
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3. $\frac{7}{64}$
   ​
4. $\frac{7}{8}$

CHEMISTRY

Q.6. For an ideal gas undergoing a reversible isothermal expansion at temperature T, which of the following statements is most accurate regarding the relationship between work $W$W, heat $q$q, and change in internal energy ΔU?

A. ΔU>0 and $q>0$ because heat must be absorbed during expansion, but W=0 in a reversible process.

B. ΔU=q and is positive, while W=0 since the process is isothermal.

C. ΔU=0, q=0, and W=0 because temperature is constant.

D. ΔU=0, q=−W, and the magnitude of work depends on the final volume.

Q.7. A sample of a radioactive nuclide has an initial activity of 800Bq. After $30\text{minutes}$, its activity falls to 50Bq. Assuming first-order decay, what is the approximate half-life of the nuclide?

A. 15minutes

B. 10minutes

C. 7.5minutes

D. 5minutes

Q.8. Which of the following combinations correctly matches the type of inhibition with its characteristic effect on enzyme kinetics in a Lineweaver-Burk plot?

A. Uncompetitive inhibition – $K_m$ increased and $V_{\max }$ unchanged.

B. Noncompetitive inhibition – $K_m$ unchanged and $V_{\max }$ decreased.

C. Competitive inhibition – increases apparent $K_m$ and decreases $V_{\max }$.

D. Mixed inhibition – both $K_m$ and $V_{\max }$ remain unchanged.

Q.9. Which of the following statements about colligative properties is most accurate?

A. Vapor pressure lowering depends on the identity of the solute and not just on the number of particles present.

B. Osmotic pressure is given by $\pi =iMRT$, showing dependence on both solute concentration and degree of dissociation.

C. Freezing point depression is directly proportional to molality and independent of the van’t Hoff factor.

D. Boiling point elevation decreases with increasing solute concentration for ideal solutions.

Q.10. Which of the following best explains the term “buffer capacity” of a solution?

A. The ability of a buffer to maintain constant pH regardless of the amount of acid or base added.

B. The concentration of salt present in a buffer solution.

C. The exact pH at which the buffer solution resists any change in pH.

D. The amount of strong acid or base that can be added before the buffer’s pH changes by more than one unit.

BIOLOGY

Q.11. In human physiology, which hormone–function pairing is most accurate?

A. Glucagon – decreases blood glucose by promoting uptake of glucose into skeletal muscle.

B. Parathyroid hormone – increases blood calcium levels by stimulating bone resorption and enhancing renal ${\text{Ca}}^{2+}$reabsorption.

C. Insulin – stimulates glycogen breakdown and gluconeogenesis in the liver during fasting.

D. Aldosterone – decreases ${\text{Na}}^{+}$ reabsorption and promotes ${\text{K}}^{+}$ retention in the distal nephron.

Q.12. In human respiration, which of the following changes occurs when a person transitions abruptly from rest to vigorous exercise?

A. Arterial pH rapidly becomes highly alkaline due to excessive removal of ${\text{CO}}_2$ at exercise onset.

B. Arterial ${\text{CO}}_2$ partial pressure immediately rises sharply and remains very high throughout exercise.

C. Ventilation initially decreases because more ${\text{O}}_2$ is delivered to muscles by increased cardiac output alone.

D. Ventilation increases primarily due to a rapid feed-forward drive from motor cortex and proprioceptors even before ${\text{CO}}_2$ accumulates.

Q.13. Which of the following best explains why myelinated axons conduct nerve impulses faster than unmyelinated axons of the same diameter?

A. Myelin allows action potentials to propagate by continuous conduction along the axon membrane.

B. Myelin permits saltatory conduction, where action potentials jump between nodes of Ranvier, reducing membrane area that must depolarize.

C. Myelin increases the number of voltage-gated ${\text{Na}}^{+}$ channels along the entire length of the axon.

D. Myelin increases the diameter of the axon, reducing internal resistance to ion flow.

Q.14. An enzyme exhibits a sigmoidal velocity-substrate concentration curve. Which of the following most likely explains this observation?

A. The enzyme follows simple Michaelis-Menten kinetics with a single active site

B. The enzyme is allosteric and shows cooperative binding of substrate

C. The enzyme’s Vmax is reduced due to non-competitive inhibition

D. The enzyme is competitively inhibited by a product of the same pathway

Q.15. In a population of flowering plants, a rare recessive allele causes male sterility when homozygous. Assuming random mating and no selection on other traits, what is the long-term effect of this allele on population genetics?

A. Its frequency will decline because homozygous recessive individuals contribute fewer gametes

B. It will increase in frequency due to genetic drift even in a large population

C. It will quickly reach fixation because heterozygotes have a reproductive advantage

D. It will remain at the same frequency because selection does not act on recessive alleles

ANSWERS & EXPLANATIONS

Ans.1. B. 13%

Ans. 2. C. 108

Ans. 3. D. 88 days (solve using Kepler’s law)

Ans. 4. C. R/16

Ans.5. B. 7/57

Ans.6.D. ΔU=0, $q=-W$, and the magnitude of work depends on the final volume. The first law of thermodynamics states $\mathrm{\Delta }U=q+W$. With $\mathrm{\Delta }U=0$, we get $q=-W$. The nonzero work term shows that energy is transferred as work, and an equal amount of heat must flow in to maintain constant temperature, balancing the energy ledger.

Ans.7.C. 5minutes. The decay follows $A_t=A_0{\left(\frac{1}{2}\right)}^{t\mathrm{/}{t}_{1\mathrm{/}2}}$. We have $\frac{50}{800}={\left(\frac{1}{2}\right)}^{30\mathrm{/}{t}_{1\mathrm{/}2}}$, so $\frac{1}{16}={\left(\frac{1}{2}\right)}^4$. This implies $30\mathrm{/}{t}_{1\mathrm{/}2}=4$, giving $t_{1\mathrm{/}2}=\frac{30}{4}=7.5\text{minutes}$. Careful comparison of each option shows that $7.5\text{minutes}$ matches the mathematical result, so a $10\text{minute}$ half-life is inconsistent with the data.

Ans.8. B. Noncompetitive inhibition – $K_m$ unchanged and $V_{\max }$ decreased. In pure noncompetitive inhibition, the inhibitor binds to an allosteric site equally well whether or not substrate is bound, reducing the total number of active enzyme molecules. This lowers $V_{\max }$ but does not affect substrate binding affinity, so $K_m$ remains unchanged. On a Lineweaver–Burk plot, lines for different inhibitor concentrations intersect on the x-axis.

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Ans.9. B. Osmotic pressure is given by $\pi =iMRT$ showing dependence on both solute concentration and degree of dissociation. This relationship explicitly includes $i$ and clearly demonstrates how colligative properties depend on the effective number of solute particles.

Ans.10. D. The amount of strong acid or base that can be added before the buffer’s pH changes by more than one unit. This reflects the realistic, finite nature of buffering action.

Ans.11. B.  Parathyroid hormone – increases blood calcium levels by stimulating bone resorption and enhancing renal ${\text{Ca}}^{2+}$ reabsorption. This pairing accurately reflects the main role of PTH in calcium regulation.

Ans.12. D. Ventilation increases primarily due to a rapid feed-forward drive from motor cortex and proprioceptors even before ${\text{CO}}_2$ accumulates. This coordinated adjustment of both cardiovascular and respiratory systems allows the body to meet the metabolic requirements of exercise efficiently.

Ans.13. B. Myelin permits saltatory conduction, where action potentials jump between nodes of Ranvier, reducing membrane area that must depolarize. Myelinated axons conduct impulses faster than unmyelinated ones primarily because of saltatory conduction. In this process, the action potential “jumps” from one node of Ranvier to the next rather than traveling continuously along the entire membrane. This occurs because the myelin sheath acts as an electrical insulator, preventing ion flow and leakage across the axonal membrane in the insulated regions. Consequently, depolarization is only required at the exposed nodes, which significantly reduces the total membrane area that must be depolarized and increases the overall speed of transmission.

Ans.14. B. The enzyme is allosteric and shows cooperative binding of substrate. Cooperative binding typically arises in oligomeric enzymes where conformational changes in one subunit influence neighboring subunits. This yields an S-shaped curve when velocity is plotted against substrate concentration. The sigmoidal curve conveys that the enzyme is relatively inactive at low substrate levels but becomes highly active as substrate concentration crosses a critical threshold, providing a switch-like regulatory mechanism in metabolic pathways.

Ans.15. A. Its frequency will decline because homozygous recessive individuals contribute fewer gametes. The reduced reproductive output of these individuals means that the recessive allele is systematically passed on less often than the dominant allele. Over many generations, this consistent disadvantage results in a lower frequency of the recessive allele unless some balancing mechanism, such as heterozygote advantage or mutation pressure, counteracts the selection.

Keep practicing, stay curious, and don’t forget that small daily efforts lead to big results. Make sure you revisit your weak areas and revise regularly alongside solving questions. We will now be posting practice questions regularly again, so be sure to visit daily and keep your preparation momentum strong for NEET UG. Stay focused, keep learning, and all the best for your journey toward becoming a future doctor!

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