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Practice Test - NEET MCQ


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30 Questions MCQ Test 4 Months Preparation for NEET - Practice Test

Practice Test for NEET 2024 is part of 4 Months Preparation for NEET preparation. The Practice Test questions and answers have been prepared according to the NEET exam syllabus.The Practice Test MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test below.
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Practice Test - Question 1

A particle experiences a variable force in a horizontal x-y plane. Assume distance in meters and force in newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane; then kinetic energy changes by

Detailed Solution for Practice Test - Question 1

Practice Test - Question 2

A stone of mass m tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

Detailed Solution for Practice Test - Question 2


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 Tminimum.

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Practice Test - Question 3

The length of the wire is increased by 1 mm on the application of a given load. In a wire of the same material but of length and radius twice that of the first, on application of the same force, extension produced is

Detailed Solution for Practice Test - Question 3

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.

Practice Test - Question 4

The volume of a spherical body is decreased by 10-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 105 N/m2).

Detailed Solution for Practice Test - Question 4

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/m

Practice Test - Question 5

Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let F2 be the force applied by the liquid on the bottom of this new vessel. Then :

Detailed Solution for Practice Test - Question 5

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

Practice Test - Question 6

The vertical limbs of a U shaped tube are filled with a liquid of density r upto a height h on each side. The horizontal portion of the U tube having length 2h contains a liquid of density 2r. The U tube is moved horizontally with an accelerator g/2 parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs, at steady state will be

Detailed Solution for Practice Test - Question 6

Given:
a=g/2​ 
Pressure at A
PA​=Po​+ρgh+(2ρ)g(h−x)=Po​+3ρgh−2ρgx
Pressure at B
PB​=Po​+ρgx
Using
PA​−PB​=[2ρ(h+x)+ρ(h−x)]a
∴ (Po​+3ρgh−2ρgx)−(Po​+ρgx)=[3ρh+ρx]×g/2​
OR 3ρgh−3ρgx=3​ρgh/2+1​ρgx/2
OR 3​ρgh/2=7​ρgx/2  ⟹x=3​h/7
∴ Difference in the heights between two columns ΔH=(2h−x)−x=2h−2x 
⟹ ΔH=2h−6h/7​=8h/7

 

Practice Test - Question 7

Only One Option Correct Type

Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

Q. 

Atomic number of second transition series lies from

Detailed Solution for Practice Test - Question 7

Since,1st transition series belongs to 3d orbital and 2nd transition series belongs to 4d-orbital. According to the given option, atomic number 39 to 48 (5s24d1-10) belongs to second transition series.

Practice Test - Question 8

An element has configuration 4d55s2. The element belongs to

Detailed Solution for Practice Test - Question 8

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).

Practice Test - Question 9

The electronic configuration of palladium is

Detailed Solution for Practice Test - Question 9

[Kr]4d105s0

Practice Test - Question 10

Which has lowest and highest first ionisation enthalpy in 3d series?

Detailed Solution for Practice Test - Question 10

Sc (3d4s2)has 631 kJ mol-1 and Zn (3d10 4s2) has 906 kJ mol-1 

Practice Test - Question 11

f block of the Periodic Table consists of

Detailed Solution for Practice Test - Question 11

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

Practice Test - Question 12

The elements charecterised by the filling of 4 f-orbitals, are:

Detailed Solution for Practice Test - Question 12

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

Practice Test - Question 13

An element belongs to Group 15 and third period of the periodic table. Its electronic configuration will be

Detailed Solution for Practice Test - Question 13

group is 15 so total valence electrons will be 5 while period is 3rd so n value of valence shell will be 3.

Practice Test - Question 14

Consider the following statements and choose the correct answer.

a. Photochemical phase occurs inside the thylakoids, especially those of grana region.
b. Biosynthetic phase reactions occur in stroma or matrix of chloroplasts and are dependent upon light.

Detailed Solution for Practice Test - Question 14

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.

Practice Test - Question 15

In cells, ATP is synthesized in:

Detailed Solution for Practice Test - Question 15

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

Practice Test - Question 16

Photorespiration does NOT occur in:

Detailed Solution for Practice Test - Question 16
  • In the photorespiratory pathway, there is no synthesis of ATP or NADPH. Therefore, photorespiration is a wasteful process.
  • In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO2 at the enzyme site.
Practice Test - Question 17

What did Jan Ingenhousz's experiments reveal about the role of sunlight in plant processes?

Detailed Solution for Practice Test - Question 17

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.

Practice Test - Question 18

Electrons from exited chlorophyll molecules of photosystem II are accepted first by:

Detailed Solution for Practice Test - Question 18
  • 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.
Practice Test - Question 19

In sugarcane, CO2 is fixed in malic acid with the help of enzyme:

Detailed Solution for Practice Test - Question 19

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.

Practice Test - Question 20

Dark fixation of CO2 in CAM plants is called ossification because it produces:

Detailed Solution for Practice Test - Question 20

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.

Practice Test - Question 21

The organism which depend on the dead and decaying organic matter is:

Detailed Solution for Practice Test - Question 21

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.
Practice Test - Question 22

Energy is stored in the form of:

Detailed Solution for Practice Test - Question 22

Energy is a fundamental requirement for all living organisms to carry out their various biological processes. Within cells, energy is stored and transferred in different molecules. The molecule that primarily stores and transfers energy in living systems is called adenosine triphosphate, or ATP.

A: FAD (Flavin adenine dinucleotide) is another molecule involved in energy transfer within cells, but it is not the primary molecule for energy storage.

B: NADH (Nicotinamide adenine dinucleotide) is an important molecule in cellular respiration, which is involved in the transfer of electrons and hydrogen ions during energy production. While it plays a role in energy transfer, it is not the primary molecule for energy storage.

D: ADP (Adenosine diphosphate) is the molecule that forms when one phosphate group is removed from ATP. It is the "lower energy" form of ATP and can be converted back into ATP by adding a phosphate group. However, it is not the primary molecule for energy storage.

Practice Test - Question 23

The TCA cycle starts with:

Detailed Solution for Practice Test - Question 23

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.

Practice Test - Question 24

During anaerobic respiration less energy is produced than aerobic respiration because:

Detailed Solution for Practice Test - Question 24

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.

Practice Test - Question 25

The enzyme that interconnects the glycolysis and kreb cycle is:

Detailed Solution for Practice Test - Question 25

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.

Practice Test - Question 26

In plants, the gaseous exchange take place in:
(a) Stomata
(b) Roots
(c) Stems
(d) Lenticels

Detailed Solution for Practice Test - Question 26

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.

Practice Test - Question 27

Which of the following statements are correct?

(A) The oxidation of pyruvic acid molecules formed in glycolysis occurs inside the mitochondria.

(B) Acetyl CoA is a 3-carbon compound.

(C) Under anaerobic conditions, the pyruvic acid formed during glycolysis is reduced to either ethyl alcohol or lactic acid.

(D) Acetyl CoA molecules enter into cyclic reactions during Calvin cycle.

Detailed Solution for Practice Test - Question 27

Acetyl CoA is a 2-carbon compound and 3-phosphoglycerate molecule enter into cyclic reactions during Calvin cycle.

Practice Test - Question 28

Aerobic respiration is common in:

Detailed Solution for Practice Test - Question 28

Aerobic respiration is the process that leads to a complete oxidation of organic substances in the presence of oxygen, and releases carbon dioxide, water and a large amount of energy present in the substrate. This type of respiration is most common in higher organisms.

Aerobic Respiration:

Practice Test - Question 29

Dough kept overnight in warm place becomes soft and spongy due to:

Detailed Solution for Practice Test - Question 29

Fermentation is a kind of anaerobic respiration in absence of oxygen to produce ethanol and carbon dioxide gas. Dough kept overnight in warm place becomes soft and spongy due to carbon dioxide gas released during fermentation.

Practice Test - Question 30

Which of the following reactions is catalysed by the enzyme phosphofructokinase?

Detailed Solution for Practice Test - Question 30

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.

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