NEET UG Practice Paper Day 39

Welcome to NEET UG Practice Paper Day 39, a vital milestone in your preparation journey for the NEET UG 2026 exam. This practice paper, meticulously crafted from the Class 11 and 12 NCERT syllabus, features a balanced mix of 15 questions across Biology, Chemistry, and Physics to enhance your conceptual clarity, time management, and exam temperament.

Q.1. In the reaction of (R)-2-bromobutane with aqueous KOH, a mixture of alcohols is formed. If the reaction proceeds through both SN​1 and SN​2 pathways simultaneously such that the SN​2 rate is twice the SN​1 rate, what is the optical purity of the resulting 2-butanol?

A. 0%

B. 50%

C. 66.7%

D. 33.3%

Q.2. A double heterozygous plant (AaBb) is test-crossed. The offspring show: AaBb (42), aabb (44), Aabb (8), and aaBb (6). What is the map distance between genes A and B, and what was the linkage phase in the parent?

A. 14 cM; Repulsion (Trans) phase

B. 7 cM; Coupling (Cis) phase

C. 14 cM; Coupling (Cis) phase

D. 86 cM; Repulsion (Trans) phase

Q.3. A gas undergoes a reversible isothermal expansion from volume $V_1$ to $V_2$ at temperature $T$. The gas has a temperature-dependent equation of state given by $P=\frac{nRT}{V-b(T)}$, where $b(T)$ increases linearly with $T$. For this process, which statement about the work done $W$ and the internal energy change $\mathrm{\Delta }U$ is most appropriate, assuming the gas has non-ideal interactions but constant molar heat capacity at constant volume $C_V$?

A. W depends only on $V_1$ and $V_2$, while $\mathrm{\Delta }U=0$ because the process is isothermal

B. W depends on the exact form of $b(T)$ and $\mathrm{\Delta }U>0$ because expansion increases potential energy

C. W and $\mathrm{\Delta }U$ are both path dependent and cannot be determined for a reversible process

D. W depends only on $V_1$, $V_2$, and $T$, while $\mathrm{\Delta }U$ may be non-zero because intermolecular potential energy can change

Q.4. A radioactive sample contains a mixture of two nuclides, $X$ and $Y$, that decay independently. $X$ has a half-life of 2 hours, and $Y$ has a half-life of 6 hours. Initially, the activities of X and $Y$ are equal. After 6 hours, what is the ratio of the total activity of the sample to its initial total activity?

A. $\frac{1}{4}$

B. 5/16

C. $\frac{5}{12}$

D.$\frac{3}{8}$

Q.5. Which of the following molecules has the maximum number of lone pairs on the central atom?

A. SF4​

B. I3−​

C. XeF2​

D. ClF3​

BIOLOGY

Q.6. In a complex food web of a temperate forest, a pesticide selectively killing predatory insects is introduced. Over time, a marked increase in herbivorous insect population is observed, followed by a significant decline in certain plant species and later a crash in the herbivore population. This pattern best illustrates which ecological concept?

A. Competitive exclusion among herbivores

B. Allee effect in primary producers

C. Trophic cascade with top-down control

D. Bottom-up regulation of trophic levels

Q.7. Which of the following experimental observations most strongly supports the fluid mosaic model of biological membranes rather than a static protein-lipid sandwich model?

A. X-ray diffraction patterns revealing regular, crystalline arrangement of membrane lipids

B. High proportion of cholesterol in animal cell membranes but not in plant membranes

C.Presence of distinct thicknesses of membranes in different organelles under electron microscopy

D. Rapid lateral diffusion of fluorescently labeled membrane proteins within the plane of the membrane

Q.8. A gene encoding an enzyme has a point mutation in its coding region that changes a codon from $GAA$ to $GAG$, but the enzyme’s activity and structure remain unchanged. Later, another mutation occurs in the same gene’s intron, disrupting a key branch point adenine required for splicing. What is the most likely combined effect on the protein product?

A. Both mutations together cause a frameshift in the coding sequence

B. The first mutation is silent, but the second leads to aberrant splicing and likely a nonfunctional protein

C. The first mutation reduces enzyme efficiency, while the second compensates by altering splicing to restore function

D. No change at all, because both mutations are silent

Q.9. In a population of flowering plants, a particular trait is controlled by a single gene with two alleles, $A$ and $a$. Allele $A$ is incompletely dominant over $a$, leading to three distinct phenotypes in $AA$, $Aa$, and $aa$ individuals. If the population is in Hardy-Weinberg equilibrium and the frequency of the intermediate phenotype is 0.48, what is the frequency of allele $A$?

A. 0.52

B. 0.24

C. 0.60

D. 0.40

Q.10. A bacterial operon is regulated by both a repressor protein that binds the operator and an activator protein that binds a site upstream of the promoter. The activator enhances RNA polymerase binding only when a specific small molecule is present. A mutation occurs that prevents the activator from binding DNA but leaves the repressor functional. In the presence of the small molecule and absence of any inducer for the repressor, what is the expected transcriptional state of the operon?

A. Basal-level transcription independent of regulator proteins

B. Constitutively high transcription because the small molecule is present

C. Oscillating transcription levels due to conflicting regulatory signals

D. No transcription because the repressor remains bound and the activator cannot assist RNA polymerase

PHYSICS

Q.11. A parallel-plate capacitor with plate area $A$ and separation $d$ is connected to a battery of emf $V$ and fully charged. While still connected to the battery, a dielectric slab of dielectric constant $K$ and thickness $d$ is fully inserted between the plates, completely filling the space. Which statement about the energy stored in the capacitor before and after insertion is correct?

A. The stored energy remains unchanged because the voltage is constant

B. The stored energy increases by a factor of $K$ because the capacitance increases

C. The stored energy decreases by a factor of $K$ because the voltage is constant

D. The stored energy becomes zero because the electric field inside the dielectric is cancelled

Q.12. A body of mass $m$m executes simple harmonic motion (SHM) along the x-axis under a restoring force $F=-kx$. Its total mechanical energy is $E$. At a displacement $x=\frac{A}{2}$ from equilibrium, where $A$ is the amplitude, what fraction of the total mechanical energy is kinetic?

A. $\frac{1}{4}$

B. $\frac{1}{2}$

C. $\frac{3}{8}$

D. $\frac{3}{4}$

Q.13. In Bohr’s model of the hydrogen atom, the radius of the $n$nth orbit is given by $r_n=n^2{a}_0$, where $a_0$ is the Bohr radius. If an electron makes a transition from $n=4$ to $n=2$, what is the ratio of the initial orbital angular momentum to the final orbital angular momentum according to this model?

A. 4:2

B. 1:2

C.16:4

D. 2:1

Q.14. A projectile is launched from the ground with speed $u$ at an angle $\theta$ to the horizontal in a region where air resistance is not negligible and is proportional to the instantaneous velocity (linear drag). Compared to ideal projectile motion without air resistance, which qualitative change in the trajectory is most accurate?

A. The time of flight increases, but the horizontal range remains the same

B. Both the maximum height and horizontal range decrease, and the trajectory becomes more steeply curved

C. The horizontal range decreases, but the maximum height increases

D. The trajectory remains parabolic but with reduced range

Q.15. A thin convex lens of focal length $f$ forms a real, inverted image of an object placed at a distance $2f$ from the lens. The lens is then cut along a diameter into two equal halves, and one half is laterally displaced by a small distance $d$ parallel to the cut while keeping both halves perpendicular to the principal axis. What will happen to the image of the object?

A. The image will become virtual and closer to the lens

B. The image will split into two separate real images, both inverted and located symmetrically about the original image position

C. The image will disappear because the lens is no longer continuous

D. The image will remain single, real, inverted, and at the same position but with reduced brightness

Congratulations on completing Day 39 of your NEET UG Practice Paper series – your dedication is building the resilience needed for success. Analyze your performance to identify strengths and gaps, revise weak topics promptly, and carry this momentum forward; excellence in NEET comes from consistent, reflective practice like this.

ANSWERS & EXPLANATIONS

Ans.1. C. 66.7% . The reaction produces a mixture of (R)- and (S)-2-butanol due to competing SN1 (racemization) and SN2 (inversion) pathways.

SN2 rate is twice the SN1 rate, so total rate = 3 parts (SN1: 1 part, SN2: 2 parts).​

SN1 gives 50% (R)-2-butanol and 50% (S)-2-butanol (racemic).

SN2 gives 100% (S)-2-butanol (inversion from (R)-starting material).

Enantiomer	From SN1 (1/3 total)	From SN2 (2/3 total)	Total fraction
(R)	(0.5)(1/3) = 1/6	0	1/6 (16.7%)
(S)	(0.5)(1/3) = 1/6	(1)(2/3) = 4/6	5/6 (83.3%)

Optical purity (enantiomeric excess) = |% (S) – % (R)| = |83.3% – 16.7%| = 66.7%.

Ans. 2. C. Total progeny: 42 + 44 + 8 + 6 = 100.

Parental classes (most frequent): AaBb (42) and aabb (44), so dihybrid parent is AB/ab (coupling/cis phase).

Recombinants: Aabb (8) + aaBb (6) = 14.

Map distance = (14/100) × 100 = 14 cM.

$$Ans. 3. D. For reversible isothermal expansion, work $W=-{\int }_{V_1}^{V_2}PdV=-nRT\ln$, depending only on $V_1,V_2,T$(since $b(T)$ is fixed at constant $T$).​

For non-ideal gases, $\mathrm{\Delta }U=\int$ even isothermally due to volume-dependent intermolecular forces (constant $C_V$ assumes kinetic part only).

Ans. 4. B. 5/16. Initial activities equal: $A_{X0}=A_{Y0}=A_0$, so total initial $A_{\text{total},0}=2A_0$

Decay constants: ${\lambda }_X=\frac{\ln 2}{2}$λ, ${\lambda }_Y=\frac{\ln 2}{6}$.

After 6 hours (3 half-lives for X, 1 for Y):
$A_X=A_0(1\mathrm{/}2{)}^3=A_0\mathrm{/}8$
$A_Y=A_0(1\mathrm{/}2{)}^1=A_0\mathrm{/}2$

Total final: $A_0\mathrm{/}8+A_0\mathrm{/}2=(1\mathrm{/}8+4\mathrm{/}8)A_0=5\mathrm{/}8A_0$
Ratio: $(5\mathrm{/}8A_0)\mathrm{/}(2A_0)=5\mathrm{/}16$

Ans.5. B. I₃⁻

Lone pairs on central atom:

• SF₄ (S): 1 lone pair (10 valence e⁻, 4 bonds = 8 e⁻, 1 pair left)​
• I₃⁻ (central I): 3 lone pairs (22 valence e⁻, 2 bonds = 4 e⁻, 18 e⁻ left = 9 pairs total, 3 bonding + 3 lone pairs)​
• XeF₂ (Xe): 3 lone pairs (22 valence e⁻, 2 bonds, 3 lone pairs)​
• ClF₃ (Cl): 2 lone pairs (28 valence e⁻, 3 bonds, 2 lone pairs)​

I₃⁻ has the maximum (3 lone pairs)

Ans. 6. C. Trophic cascade with top-down control. Killing predatory insects removes top-down control, causing herbivores to surge, overexploit plants (decline), then crash due to food scarcity. This classic sequence demonstrates a trophic cascade propagating downward through the food web.

Ans.7. D. Rapid lateral diffusion of fluorescently labeled membrane proteins within the plane of the membrane. The fluid-mosaic model, proposed by Singer and Nicolson, emphasizes that the membrane is a dynamic, two-dimensional fluid. Experiments, such as cell fusion experiments (Frye and Edidin) and Fluorescence Recovery After Photobloaching (FRAP), demonstrated that membrane proteins and lipids can move laterally, which is incompatible with a rigid “sandwich” of protein layers covering the lipid bilayer.

Ans. 8.  B. GAA → GAG is a silent mutation (both code for glutamic acid), leaving the enzyme unchanged. The intron branch point mutation disrupts splicing, causing intron retention, exon skipping, or cryptic splice site use, typically yielding a nonfunctional protein via frameshift or nonsense-mediated decay.

Ans.9. C. 0.60, use 2pq=0.48

Ans.10. D. No transcription because the repressor remains bound, and the activator cannot assist RNA polymerase.

Ans. 11. B. The capacitor is kept connected to the battery at constant emf $V$, so the voltage across the plates remains fixed at $V$ even after inserting the dielectric.​​The capacitance of a parallel‑plate capacitor increases by a factor of K when the space between the plates is completely filled with a dielectric of constant $K$.

Ans. 12. D. For simple harmonic motion under a restoring force $F=-kx$, the total mechanical energy is$$E=\frac{1}{2}k{A}^2$$

At displacement $x=\frac{A}{2}$, the potential energy is$$U=\frac{1}{2}k{x}^2=\frac{1}{2}k{\left(\frac{A}{2}\right)}^2=\frac{1}{2}k\cdot \frac{A^2}{4}=\frac{1}{8}k{A}^2$$

So the kinetic energy is$$K=E-U=\frac{1}{2}k{A}^2-\frac{1}{8}k{A}^2=\frac{4}{8}k{A}^2-\frac{1}{8}k{A}^2=\frac{3}{8}k{A}^2$$

The fraction of total energy that is kinetic is then$$\frac{K}{E}=\frac{\frac{3}{8}k{A}^2}{\frac{1}{2}k{A}^2}=\frac{3\mathrm{/}8}{1\mathrm{/}2}=\frac{3}{8}\times \frac{2}{1}=\frac{3}{4}$$Ans 13. D. n Bohr’s model, orbital angular momentum is $L_n=n\mathrm{\hslash }$.​

So:

• Initial orbit $n_i=4$: $L_i=4\mathrm{\hslash }$.​
• Final orbit $n_f=2$: $L_f=2\mathrm{\hslash }$.​

Therefore, the ratio $L_i:L_f=4\mathrm{\hslash }:2\mathrm{\hslash }=2:1$.​

Ans. 14. B. Both the maximum height and horizontal range decrease, and the trajectory becomes more steeply curved. With linear drag proportional to velocity, air resistance reduces both the maximum height and horizontal range compared to ideal projectile motion, while making the trajectory asymmetric and more steeply curved.

Ans.15. B. The image will split into two separate real images, both inverted and located symmetrically about the original image position accurately describes this effect.

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