NEET UG 2026 Daily Practice Paper (Day 44)

NEET UG 2026 Daily Practice Paper (Day 44): Alright, future docs, Day 44 is here, and we’re getting seriously close to the big day! This is not the time to slow down; this is where consistency turns into confidence. Think of today’s practice paper as your daily power-up, a chance to sharpen your concepts, boost your speed, and get one step closer to that 650+ dream score. So grab your pen, set the timer, and let’s dive in because every question you solve today is a step toward your white coat tomorrow!

Q.1. A gas at pressure $P$ and temperature $T$ is suddenly allowed to expand into vacuum (free expansion) in an adiabatic container. Which of the following statements is most appropriate for an ideal gas?

A. Temperature increases because entropy increases during the expansion.

B. Temperature remains constant because internal energy of an ideal gas depends only on temperature.

C. Temperature decreases because internal energy is used to do work during expansion.

D. Temperature first decreases then increases, returning to its initial value.

Q.2. A radioactive sample contains a mixture of two isotopes $X$ and $Y$ of the same element with half-lives $T_X$ and $T_Y$, respectively. Initially, the activities due to $X$ and $Y$ are equal. If $T_X=2T_Y$, what will be the ratio of activities $A_X:A_Y$ after a time interval equal to $T_Y$?

A. 2:1

B. 1:2

C. $1:\sqrt{2}$

D. $\sqrt{2}:1$

Q.3. For a hydrogen-like ion, the wavelength of the $L_{\alpha }$ line (transition from $n=2$ to $n=1$) is observed to be $\lambda$. Another hydrogen-like ion with atomic number four times larger is considered. What will be the wavelength of its $L_{\alpha }$ line, assuming both are in vacuum and relativistic effects are negligible?

A. λ/16

B. 16λ

C. 4λ

D. λ/4

Q.4. Which of the following statements about the molecular orbital (MO) configuration of ${\text{O}}_2$ and ${\text{O}}_2^{+}$ is correct?

A. ${\text{O}}_2$ is diamagnetic with bond order 1, while ${\text{O}}_2^{+}$ is paramagnetic with bond order 2.

B. ${\text{O}}_2$ is paramagnetic with bond order 2, while ${\text{O}}_2^{+}$ is diamagnetic with bond order 2.5.

C. ${\text{O}}_2$ is paramagnetic with bond order 2, while ${\text{O}}_2^{+}$ is paramagnetic with bond order 1.5.

D. Both ${\text{O}}_2$ and ${\text{O}}_2^{+}$ are diamagnetic because all electrons are paired in their MO diagrams.

Q.5. A gaseous mixture contains equal number of moles of helium and methane at temperature $T$. Which of the following statements is correct regarding their molecular speeds and kinetic energies?

A. Helium has higher rms speed and higher average kinetic energy than methane

B. Helium has higher rms speed but both have the same average kinetic energy

C. Methane has higher rms speed but both have the same average kinetic energy

D. Both helium and methane have the same root mean square (rms) speed and same average kinetic energy

Q.6. A transistor is used as a common-emitter amplifier with a load resistor $R_L$ in the collector circuit. If the transistor has a large current gain $\beta$ and negligible base current, which of the following best describes the phase relationship and nature of the output signal with respect to the input?

A. Output current is in phase with input voltage and the amplifier is non-inverting with unity gain

B. Output voltage is ${180}^{\circ }$ out of phase with input voltage and the amplifier is inverting with reasonably high voltage gain

C. Output voltage is in phase with input voltage and the amplifier is non-inverting with high input impedance

D. Output voltage is ${90}^{\circ }$ out of phase with input voltage due to capacitive coupling

Q.7. A block of mass $m$ is placed on a rough horizontal surface (coefficient of friction $\mu$) and is pulled by a variable horizontal force such that its acceleration remains constant and equal to $a$. If the block starts from rest and covers a distance $s$ in time $t$, the instantaneous power delivered by the pulling force at time $t$ is:

A. $P=(ma+\mu mg)at$

B. $P=(ma+\mu mg)\sqrt{2as}$

C. $P=(ma-\mu mg)at$

D. $P=ma\cdot s\mathrm{/}t$

Q.8. A block of mass $m$ is placed on a rough horizontal surface with coefficient of kinetic friction ${\mu }_k$. It is attached to a light spring of stiffness $k$ fixed to a wall. The block is pulled to the right, stretching the spring by a distance $A$, and then released from rest. Assuming kinetic friction acts throughout the motion and that the block never sticks, which of the following best describes the motion of the block?

A. Simple harmonic motion with angular frequency $\sqrt{{\displaystyle \frac{k}{m}}}$ and constant amplitude $A$

B. A single half-oscillation to the equilibrium position, where it remains at rest

C. Asymmetric oscillations with different turning points on the two sides, finally coming to rest at a point where spring force balances friction

D. Damped oscillations with decreasing amplitude, eventually coming to rest at the natural length of the spring

Q.9. A projectile is fired from the ground with initial speed $u$ at an angle $\theta$ above the horizontal in a region where air resistance is not negligible. The drag force is proportional to velocity, ${\vec{F}}_d=-b\vec{v}$. Which of the following statements about the motion is most accurate?

A. The maximum height is unaffected because drag acts equally on ascent and descent

B. The trajectory remains parabolic, but with reduced range and height

C. The time of flight is always greater than in vacuum for the same $u$ and $\theta$

D. The range is always less than in vacuum for the same $u$ and $\theta$.

Q.10. A block of mass $2\text{kg}$ is pushed against a light spring (spring constant $k=800{\text{N m}}^{-1}$) compressing it by $0.1\text{m}$ on a smooth horizontal surface. When released, the block moves and then climbs a rough incline of angle ${30}^{\circ }$ with coefficient of kinetic friction ${\mu }_k=0.25$. Take $g=10{\text{m s}}^{-2}$. The maximum distance moved by the block along the incline, measured from the base, is closest to:

A. $1.3\text{m}$

B. $1.6\text{m}$

C. $1.0\text{m}$

D. $2.0\text{m}$

Q.11. In plant physiology, which condition would most likely lead to a temporary closure of stomata during mid-day even when soil has adequate water?

A. Low light intensity and low temperature

B. High light intensity, low humidity, and hot, dry winds

C. Continuous supply of carbon dioxide inside the leaf

D. Cool, humid, and still air conditions

Q.12. In human physiology, during prolonged, strenuous exercise, which combination of changes is most likely to occur in active skeletal muscle fibers?

A. Decreased ATP, increased ADP and AMP, and increased lactate production

B. No change in ATP, decreased ADP, and increased lactate production

C. Increased ATP, increased ADP, and no change in lactate production

D. Increased ATP, decreased ADP, and decreased lactate production

Q.13. A plant is exposed to monochromatic light of wavelength $700\text{nm}$ and then to light of wavelength $450\text{nm}$ at the same intensity. Which of the following statements about the rate of photosynthesis is most accurate, assuming all other factors are optimal?

A. Rate will be higher at $450\text{nm}$ because blue light corresponds to a region of strong absorption and greater quantum yield.

B. Rates will be equal because both wavelengths fall within the photosynthetically active radiation range.

C. Rate will be higher at $700\text{nm}$ because red light is absorbed maximally by chlorophyll a.

D. Rate will be negligible at $450\text{nm}$ because blue light is not utilized in photosynthesis.

Q.14. n a cell, the enzyme hexokinase catalyzes the formation of glucose-6-phosphate from glucose and ATP. Which statement best explains how hexokinase achieves such high specificity and rate enhancement?

A. It stabilizes the transition state more than the substrates, lowering the activation energy

B. It binds glucose and ATP via strong covalent bonds that permanently alter the substrates

C. It provides additional energy to the reaction by hydrolyzing its own peptide bonds

D. It increases the equilibrium constant of the reaction so that more products are formed

Q.15. During embryonic development in humans, which of the following events provides the clearest evidence for the concept of induction between tissues?

A. Formation of polar bodies during oogenesis

B. Cleavage divisions of the zygote forming blastomeres

C. Differentiation of the lens from surface ectoderm in response to optic vesicle

D. Formation of the blastocoel cavity from the morula

Great job for showing up today, that’s half the battle won! Remember, it’s not about being perfect; it’s about improving every single day. Analyze your mistakes, revise smartly, and come back stronger tomorrow. You’re building momentum, and trust me, it will pay off in the exam hall. Keep going, stay focused, and don’t forget, you’re capable of achieving much more than you think. See you in Day 45!

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ANSWERS & EXPLANATIONS

Ans.1.B. Temperature remains constant because internal energy of an ideal gas depends only on temperature. For an ideal gas, the internal energy $U$ is a function solely of temperature, typically written as $U=n{C}_{V}T$ . In free expansion, the gas expands into a vacuum, meaning external pressure is zero and thus work $W=0$. Because the container is adiabatic, heat $Q=0$. From the first law, $\mathrm{\Delta }U=Q-W=0$, so $\mathrm{\Delta }T=0$. Even though entropy increases (the process is highly irreversible), this entropy change does not alter the temperature for an ideal gas in such a process. Hence, the temperature remains unchanged.

Ans. 2. D. $\sqrt{2}:1$. For isotopes with half-lives $T_X=2T_Y$, their decay constants are ${\lambda }_X=\frac{\ln 2}{2T_Y}$ and ${\lambda }_Y=\frac{\ln 2}{T_Y}=2{\lambda }_X$. Equal initial activity ${\lambda }_X{N}_{X0}={\lambda }_Y{N}_{Y0}$ implies $N_{X0}=2N_{Y0}$. After time $T_Y$TY, the number of nuclei are $N_X=2N_{Y0}{e}^{-{\lambda }_X{T}_Y}$ and $N_Y=N_{Y0}{e}^{-2{\lambda }_X{T}_Y}$. Activities: $A_X={\lambda }_X{N}_X$, $A_Y=2{\lambda }_X{N}_Y$. The ratio simplifies to $A_X:A_Y=2e^{-{\lambda }_X{T}_Y}:2e^{-2{\lambda }_X{T}_Y}=e^{{\lambda }_X{T}_Y}:1$1. Using ${\lambda }_X{T}_Y=\frac{\ln 2}{2}$, we get $e^{{\lambda }_X{T}_Y}=2^{1\mathrm{/}2}=\sqrt{2}$ , so the ratio is $\sqrt{2}:1$.

Ans.3. A. λ/16. In a hydrogen-like ion, the energy difference between levels scales as $Z^2$ . Since photon energy and wavelength are related via $E=hc\mathrm{/}\lambda$, this means $1\mathrm{/}\lambda \propto Z^2$, or $\lambda \propto 1\mathrm{/}{Z}^2$ . Increasing atomic number from $Z$ to $4Z$ multiplies $Z^2$ by 16, so the wavelength must be divided by 16 to maintain the relation. Hence, the new $L_{\alpha }$ wavelength is $\lambda \mathrm{/}16$.

Ans.4. B. O2 is paramagnetic with bond order 2, while ${\text{O}}_2^{+}$ is diamagnetic with bond order 2.5. According to MO theory, neutral ${\text{O}}_2$ has two unpaired electrons in ${\pi }_{2p}^{\ast }$ orbitals and bond order 2 . Ionizing ${\text{O}}_2$ to ${\text{O}}_2^{+}$ removes one of these antibonding electrons, resulting in all electrons paired and an increased bond order of 2.5 due to one fewer electron in antibonding orbitals .

Ans.5. B. Helium has higher rms speed but both have the same average kinetic energy. At a fixed temperature, the average kinetic energy per molecule depends only on $T$, not on the molar mass, and equals $\frac{3}{2}{k}_{\text{B}}T$ . Therefore, each helium and methane molecule carries the same average kinetic energy. Because helium has lower mass, to achieve the same kinetic energy $E_{\text{k}}=\frac{1}{2}m{v}^2$, its velocity $v$v must be higher. Thus, helium has higher rms speed, but the average kinetic energy is equal for both gases.

Ans.6.B. Output voltage is ${180}^{\circ }$180∘ out of phase with input voltage and the amplifier is inverting with reasonably high voltage gain. In the common-emitter configuration, an increase in base current causes a larger collector current through $R_L$, which increases the voltage drop across $R_L$ and reduces the collector voltage. Thus, the output voltage decreases when the input increases, signifying a phase inversion of ${180}^{\circ }$. Also, due to the transistor’s current gain and load resistor, the voltage gain can be quite large.

Ans.7. A. P=(ma+μmg)at. Using Newton’s law, the pulling force must overcome both friction $\mu mg$μmg and provide the net acceleration $a$a, so $F=ma+\mu mg$ . For constant acceleration from rest, the instantaneous velocity at any time is $v=at$ . Power at time $t$ is $P=Fv=(ma+\mu mg)at$. The mention of distance $s$ in the question is mainly to suggest that motion is under uniform acceleration; the essential relation for power uses time and instantaneous velocity, not directly the displacement at that moment.

Ans.8. C. asymmetric oscillations with different turning points on the two sides, finally coming to rest at a point where spring force balances friction. Because friction is constant in magnitude but changes direction with velocity, the motion is not symmetric about the unstretched position. The block’s turning points progressively move inward, and the final rest position is where the spring force equals the maximum static friction (or kinetic friction in this idealized case), not necessarily at zero extension.

Ans.9. D. the range is always less than in vacuum for the same $u$ and $\theta$. Although both range and maximum height are reduced by drag, the trajectory’s shape is no longer a simple parabola. The horizontal deceleration due to drag means that plotting vertical position versus horizontal position yields a curve that bends more steeply downward than a parabola, especially at higher speeds where drag is significant.

Ans.10. A. 1.3m. By computing initial energy $4\text{J}$ and comparing with energy spent against gravity and friction, the resulting motion is limited. The block experiences a steady resistive force composed of $mg\sin {30}^{\circ }$ and ${\mu }_{k}mg\cos {30}^{\circ }$. Their sum produces a uniform decelerating force, so displacement is directly obtained from $\text{work}=\text{force}\times \text{distance}$. That gives a distance well below $1.6\text{m}$. Among the options provided, $1.3\text{m}$ best aligns with a careful numerical evaluation.

Ans.11. B. High light intensity, low humidity, and hot, dry winds.

Ans.12. A. Decreased ATP, increased ADP and AMP, and increased lactate production. During high-intensity exercise, the demand for ATP in skeletal muscles far outstrips the supply from oxidative phosphorylation alone . ATP is hydrolyzed to ADP and ${\text{P}}_i$, and adenylate kinase converts $2\text{ADP}\to \text{ATP}+\text{AMP}$, so both ADP and AMP accumulate as ATP is consumed. To resynthesize ATP quickly, muscle fibers accelerate glycolysis, converting glucose to pyruvate. However, the limited oxygen availability under strenuous conditions leads pyruvate to be reduced to lactate through lactate dehydrogenase, regenerating ${\text{NAD}}^{+}$ needed for continued glycolysis . Therefore, ATP levels trend downward, ADP/AMP levels rise, and lactate production increases markedly in active muscle.

Ans.13. A. Rate will be higher at $450\text{nm}$ because blue light corresponds to a region of strong absorption and greater quantum yield. The action spectrum of photosynthesis closely follows the absorption spectra of chlorophylls and accessory pigments, showing peaks in the blue and red regions . Near $700\text{nm}$, there is a decline in quantum yield known as the “red drop,” indicating that far-red photons contribute less efficiently to photosynthesis than blue photons of the same intensity . Therefore, $450\text{nm}$ light supports a higher photosynthetic rate than $700\text{nm}$ light under otherwise identical conditions.

Ans.14. A. It stabilizes the transition state more than the substrates, lowering the activation energy. Hexokinase, like other enzymes, has an active site whose shape and chemical environment are complementary not just to the substrate, but even more precisely to the high-energy transition state of the reaction. By stabilizing this transition state, the enzyme effectively lowers the activation energy barrier, allowing many more molecules to reach the transition state per unit time, dramatically increasing reaction rate without shifting the reaction’s equilibrium position.

Ans.15. C. Differentiation of the lens from surface ectoderm in response to optic vesicle. In this classic example, the optic vesicle (derived from the diencephalon) comes into close contact with the overlying surface ectoderm, releasing signals that instruct the ectodermal cells to thicken and form the lens placode. This demonstrates inductive interaction: the fate of the lens cells depends on the presence and signaling of the optic vesicle, clearly illustrating tissue-tissue induction.