Practice Test - NEET MCQ

The change in kinetic energy of a particle moving in a horizontal plane can be determined using the work-energy theorem. The theorem states that the work done by a force on a particle equals the change in its kinetic energy. We Know that, 

At any θ,
T - mgcosθ =

⇒ T = mg cosθ + 
Since v is constant,
⇒ T will be minimum when cos θ is minimum.
⇒ θ = 180° corresponds to T_min.

Young's Modulus of elasticity =stress/strain
Y= [F/a/△l/l] ​ or Y= [Fl​/a△l]
or △l= [Fl/aY] ​= Fl​/ πr²Y
In the given problem, △l∝ l​/ r²
When both l and r are doubled, △l is halved.

Stress = Force per area and strain is unit less.
Therefore in CGS units dyne per cm square.
Hence A is correct.

We know that magnitude of bulk modulus
K=[P/(dV/V)]
Now, percentage change in volume is. 10^-3 %
Therefore, (dV/V)x100=10^-3
So, dV/V=10^-5
Hence, k is,
K=40atm/10^-5=40x1.01x10⁵ /10^-5 =4.04x10¹¹ N/m² 

Force applied on the base is equal to the weight of the liquid, since in both cases the same liquid has been poured in the container, it will exert the same force on the base of the container.
F¹​=F²​

First, let’s concentrate on the force exerted by the liquid of density 3ρ on the cylinder in the horizontal direction. 
 
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line. 
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
 The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ

Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,

For equilibrium, both the forces should be equal, hence solving the above equation, 
h = R √3/2​​

Let say the tank is accelerating by some acceleration a, such that the rest water in the tanks forms shape like this -

Acceleration, a is right wards.
Where let say h is the height from top till which there is no water
Now if we say V is total volume and B is area of its base and S be its height 
We have ½ h X B = V / 3 
Thus we get h = S/3
Thus the angle in this cross section of vacant triangle is tan^-1 ⅓
Also the same triangle relates a and g, which can be seen when we make the block a inertial frame by adding pseudo force of magnitude ma and directing leftwards, thus we get a/g = ⅓
Thus we get a = g/3

Mn²⁺/ Mn have - 1.18SRPvalue.

Ford-block elements, group number is equal to the number of electrons in (n - 1)d subshell + number of electrons in valence shell (nth shell).

[Kr]4d¹⁰5s⁰

Lanthanides, Ce(Z = 58) ñ Lu(Z = 71) and Actinoids comprises of f block elements.

Elements characterized by the filling of 4f orbitals are lanthanides.

The correct answer is Option B.

14 elements after actinium is called Actinides.
 

Modern periodic law is essentially the result of periodic variation and electronic configuration.

This is the correct option as statement 'b' is false. Biosynthetic phase reactions occur in stroma or matrix of chloroplasts and do not require light.

ATP is synthesized by cells (in mitochondria and chloroplasts) is named phosphorylation.

The Calvin cycle can be described under three stages: carboxylation, reduction and regeneration. Pathway operated in cyclic manner.

Jan Ingenhousz's experiments with an aquatic plant showed that in bright sunlight, small bubbles containing oxygen were formed around the green parts of the plant, while in the dark, they did not. This demonstrated that sunlight is essential for the process by which plants release oxygen.

• Electrons from excited chlorphyll molecule of photosystem II are accepted first by Quinone.
• Photosystem II is a photosynthetic pigment system along with some electron carriers that is located in the appressed part of the grana thylakoids.
• Photosystem II has chlorophyll a, b and carotenoids. Other components of PS II are phaeophytin, plastoquinone (PQ), cytochrome complex and blue coloured copper-containing plastocyanin.

In C₄ plants, Carbon dioxide is fixed in malic acid with the help of enzyme phosphoenel pyruvate (PEP). Malic acid is a four-carbon compound that later changes into oxaloacetic acid.

Dark reaction of Carbon dioxide fixation in CAM plants is called as ossification because it produces Malic acid. These plants open their stomata during night to prevent transpiration.

Pigments are substances that have an ability to absorb light, at specific wavelengths. The chlorophyll a pigment to absorb lights of different wavelengths.

The organism that depends on dead and decaying organic matter is called a saprophyte. Saprophytes are a type of decomposer that obtain their nutrients and energy by breaking down dead plant or animal material. They play a vital role in the ecosystem by recycling nutrients and returning them to the environment for other organisms to use. 

Example: Moulds, Mushroom etc. 

• Autotrophs, on the other hand, are organisms that can produce their own food using inorganic substances and an external energy source, such as sunlight. They convert carbon dioxide and water into organic compounds through the process of photosynthesis (in plants) or chemosynthesis (in some bacteria). Autotrophs do not rely on dead organic matter for their nutrition.
• Carnivores are organisms that primarily consume the flesh of other animals. They obtain their energy and nutrients by hunting and feeding on other living organisms, rather than on dead organic matter.
• Herbivores are organisms that primarily consume plant material. They obtain their energy and nutrients by feeding on plants or plant parts. Herbivores generally do not depend directly on dead and decaying organic matter as their main source of nutrition.

The TCA cycle begins with a condensation reaction. In this step, an acetyl group, derived from the breakdown of glucose or fatty acids, combines with oxaloacetic acid (OAA), a four-carbon molecule, in the presence of water. This condensation reaction forms a six-carbon compound called citric acid or citrate.

The complete reaction is as follows:
Acetyl-CoA + OAA + H₂O → Citric Acid (Citrate)

The condensation step is significant because it marks the entry of the acetyl group into the TCA cycle and the beginning of a series of enzymatic reactions that ultimately lead to the release of energy and the regeneration of OAA.

During anaerobic respiration, less energy is produced compared to aerobic respiration because incomplete oxidation of glucose occurs. Anaerobic respiration occurs in the absence of oxygen, typically in situations where oxygen is not readily available or during strenuous exercise when oxygen demand exceeds supply.

The two molecules of pyruvic acid (produced from one glucose molecule during glycolysis). The acetyl CoA then enters a cyclic pathway, tricarboxylic acid cycle, more commonly called as Krebs’ cycle. Thus, acetyl-CoA interconnecting between two cyclic pathways.

Plants unlike animals have no special systems for breathing or gaseous exchange. Stomata and lenticels allow gaseous exchange by diffusion.

(a) Stomata: Stomata are tiny openings present on the surface of leaves, primarily on the underside. These openings are surrounded by specialized cells called guard cells. Stomata regulate the exchange of gases, including the intake of carbon dioxide (CO₂) for photosynthesis and the release of oxygen (O₂) and water vapor (H₂O) during transpiration. Stomata play a crucial role in the gas exchange process in plants.

(d) Lenticels: Lenticels are small openings or pores in the bark of woody plant stems. They allow for gaseous exchange between the internal tissues of the stem and the external environment. Lenticels facilitate the exchange of gases, such as oxygen (O₂) and carbon dioxide (CO₂), in stems and other woody parts of plants.

The correct statements are:

(A) The oxidation of pyruvic acid molecules formed in glycolysis occurs inside the mitochondria.
(C) Under anaerobic conditions, the pyruvic acid formed during glycolysis is reduced to either ethyl alcohol or lactic acid.

Explanation:

• (A) Correct: Pyruvic acid (a 3-carbon compound) produced in glycolysis enters the mitochondria, where it is oxidatively decarboxylated to form Acetyl CoA (a 2-carbon compound) before entering the Krebs cycle.

• (B) Incorrect: Acetyl CoA is a 2-carbon compound, not 3-carbon.

• (C) Correct: In anaerobic conditions, pyruvate is reduced to:

  • Lactic acid (in animal cells/muscles via lactate fermentation).

  • Ethanol + CO₂ (in yeast/plant cells via alcoholic fermentation).

• (D) Incorrect: Acetyl CoA enters the Krebs cycle (in mitochondria), not the Calvin cycle (which fixes CO₂ in chloroplasts using RuBisCO).

Final Answer:

(A) and (C) are correct

Phosphofructokinase is a kinase enzyme that phosphorylates fructose-6-phosphate in glycolysis. The enzyme catalysed transfer of a phosphoryl group from ATP is an important reaction in a wide variety of biological processes.