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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

According to Maxwell’s equations

Detailed Solution for Practice Test - Question 1



 

Explanation:

 


  • Maxwell's equations describe how electric and magnetic fields interact with each other.

  • One of Maxwell's equations states that a time-varying magnetic field acts as a source of electric field.

  • This means that when a magnetic field changes over time, it induces an electric field in the surrounding space.

  • Similarly, another Maxwell's equation states that a time-varying electric field acts as a source of magnetic field.

  • Therefore, time-varying magnetic fields and time-varying electric fields are interconnected and can generate each other.


  •  



 

Practice Test - Question 2

The electromagnetic waves ranging in frequencies between 1 GHz and 300 GHz are called _______

Detailed Solution for Practice Test - Question 2

Option 2 : Microwaves

Concept:

  • Frequencies range of Radio waves:  3KHz to 300GHz
  • Frequencies range of Micro waves:  1GHz to 300GHz
  • Frequencies range of Infrared waves:  300GHz to 400THz
  • Frequencies range of light waves:  430THz to 750THz
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Practice Test - Question 3

Plane electromagnetic wave travels in vacuum along z-direction. If the frequency of the wave is 30 MHz, its wavelength is

Detailed Solution for Practice Test - Question 3

The electromagnetic wave travels in vacuum along z-direction. Then, the electric field vector and magnetic field vector will be in the XY plane.
Frequency of the wave = 30 MHz
Converting into Hz, v = 30× 106 Hz.
speed of light, c = 3 × 108 m/s
Wavelength of the wave can be calculated using the following relation:
λ = c/v
⇒ λ = 3 × 108/30 × 106
⇒λ = 10 m

Practice Test - Question 4

Light rays from a point object

Detailed Solution for Practice Test - Question 4

Light from each point on a luminous object travels outward in all directions in straight lines. Light travels at very high speeds, but is not instantaneously everywhere. Light from a point object travels indefinitely until it collides with matter in its path to be partially absorbed and reflected.

Practice Test - Question 5

The graph drawn with object distance along abscissa & image as ordinate for a convex lens is

Detailed Solution for Practice Test - Question 5

In a convex lens,  (1/v)−(1/u)=(1/f)​
Or (1/v)​−(1/f)​=(1/u)​ where u is always negative and f is always positive.
(1/v)​=(1/f)​−(1/u)​
So, as u is increased 1/u will decrease and  (1/f)​−(1/u)​  will increase, so, 1/v​ increases or v decreases.
So, its graph will be a hyperbola

Practice Test - Question 6

Is light a particle or a wave?

Detailed Solution for Practice Test - Question 6

Explanation:explanation none

Practice Test - Question 7

The angle of incidence at which the reflected beam is fully polarized is called

Detailed Solution for Practice Test - Question 7

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.

Practice Test - Question 8

The refractive index of glass is 1.5 which of the two colors red and violet travels slower in a glass prism?

Detailed Solution for Practice Test - Question 8

(a) Speed of light in glass is c/μ=2×108m/s

(b) The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

Practice Test - Question 9

In a young’s double slit experiment, the central bright fringe can be identified by

Detailed Solution for Practice Test - Question 9

When we use white light, the central bright will have light from all wavelengths as none of them cancel out. Hence the central bright fringe appears white.
For other bright fringes, depending on the wavelength of light constructive interference will not take place for certain wavelengths. Hence they will not be white, rather will be coloured, hence differentiated from central fringe.

Practice Test - Question 10

Direction (Q. Nos. 1-21) This section contains 21 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

Q. For zeroth order reaction,

A → B

[A]0 = 0.01 M, [A]t = 0.008Matter 10 min.

Thus, half-life is

Detailed Solution for Practice Test - Question 10

For zeroth order reaction

When reaction is 50% completed. 

Practice Test - Question 11

A Certain Zeroth Order reaction has k = 0.025 Ms-1 for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?

Detailed Solution for Practice Test - Question 11

x(product) formed after 15 s
= 0.025 Ms-1 x 15s
= 0.375 M
Then, reactant left
= 0.500 - 0.375
= 0.125 M

Practice Test - Question 12

For Zeroth Order Reaction, variation of x with time is shown as 

Q. At initial concenration of the reactant as 17.32 min dm-1 , half - life period is 

Detailed Solution for Practice Test - Question 12

Practice Test - Question 13

In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M is 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

Detailed Solution for Practice Test - Question 13

For first order reaction,

time = 15 min to change its concentration from 0.8 m to 0.4 m Thus, 50% reaction.
Thus,  t50 = 15 min



Practice Test - Question 14

For the First order reaction, concentration of the reactant after two averagelives is reduced to

Detailed Solution for Practice Test - Question 14

For first order reaction,




Practice Test - Question 15

For the first order reaction with, C0 as the initial concentration and C at time  t, (C0 - C) =

Detailed Solution for Practice Test - Question 15




Practice Test - Question 16

Select the option that correctly identifies the chemical bonds present in the given biomolecules. 
Polysaccharides - A, Proteins - B, Fats - C, Water - D

Detailed Solution for Practice Test - Question 16

In polysaccharides, monosaccharides are linked together by glycosidic bond (C−O−C). It is formed by dehydration between two carbon atoms of two adjacent monosaccharides. Water has hydrogen bonding. Amino acids in a protein are linked through peptide bond whereas ester bond is present in fats.

Practice Test - Question 17

An alpha helix is a:

Detailed Solution for Practice Test - Question 17

Ans: 2
Sol: An alpha helix is a secondary structure of a protein because it involves the local coiling of the polypeptide chain into a helix, stabilized by hydrogen bonds between the backbone atoms. It is one of the common structural patterns found within proteins, distinct from the primary, tertiary, and quaternary structures.
4o

Practice Test - Question 18

In a DNA molecule, the phosphate group is attached to ____carbon of the sugar residue of its own nucleotide and _____ carbon of the sugar residue of the next nucleotide by ____bonds.

Detailed Solution for Practice Test - Question 18

A deoxyriboncleotide of DNA is formed by cross-linking of three chemicals- deoxyribose sugar (C5H10O4), a nitrogen base and phosphoric acid (H3PO4). The backbone of a DNA strand is built up of alternate deoxyribose and phosphoric acid grounds. The phosphate group is connected to carbon 5' of the sugar residue of its own nucleotide and cerbon 3' of the sugar residue of the next nucleotide by phosphodiester bonds.

Practice Test - Question 19

Which of the following statements is not correct regarding chitin?

Detailed Solution for Practice Test - Question 19

Chitin is the second most abundant polysaccharide (after cellulose). Chitin is not a storage polysaccharide rather it is a structural homopolysaccharide which forms the structure component of fungal walls and exoskeleton of artropods. It is an unbranched polysaccharide formed of N- acetylglucosamine (NAG) monomers joined together by β−1,4 linkages.

Practice Test - Question 20

B-DNA which is right-handed double helix contains ______ base pairs per turn of the helix and each turn is _______ long.

Detailed Solution for Practice Test - Question 20

Correct option is B)

B-DNA is the most commonly found form of DNA in nature. This form of DNA is very narrow and elongated.
B-DNA contains 10 base pairs per turn and each turn is about 34 Å long.

Practice Test - Question 21

The correct order of chemical composition of living tissues/cells in term of percentage of the total cellular mass is

Detailed Solution for Practice Test - Question 21

The average composition of a living cell is water 70−90%, proteins 10−15%, nucleic acids 5−7%, carbohydrates 3%, lipids 2%, ions 1%.

Practice Test - Question 22

Which bonds are indicated by X and Y in the given diagram?

Detailed Solution for Practice Test - Question 22

Given figure represents a part of DNA molecules, in which phosphoric acid groups and deoxyribose sugars are linked together by means of phosphodiester bonds. The two DNA chains are antiparallel, i.e., they run parallel but in opposite directions. In one chain the direction is 5'→3' while in the opposite one it is 3'→5'. The chains are held together by hydrogen bonds between their bases. Adenine (A), a purine of one chain lies exactly opposite to thymine (T), a pyrimidine of the other chain . Similarly, cytosine (C), a primidine lies opposite to guanine (G) a purine. Three hydrogen bonds occur between cytosine and guanine (C≡G) at positions 1'−1',2'−6' and 6'−2'. There are two such hydrogen bonds between adenone and thymine (A=T) which are formed at positions 1'−3' and 6'−4'. Hydrogen bonds occur between hydrogen of one base and oxygen or nitrogen of the other base.

Practice Test - Question 23

What will be the molecular formula of a polypeptide consisting of 10 glycine molecules when the formula of glycine is    C2H5​NO2?

Detailed Solution for Practice Test - Question 23

A polypeptide is formed by the condensation of glycine monomers, i.e., by the eliminetion of one water molecule during the formation of each peptide bond. When 10 glycine (C2H5O2N), molecules participate, 9H2O molecules are eliminated. Thus resulting polypeptide will prossess the molecular formula C20H32O11N10.

Practice Test - Question 24

Plants show mitotic divisions in

Detailed Solution for Practice Test - Question 24

Mitosis is the type of cell division that ensures equal distribution of genetic material in daughter cells. Mitosis can occur both in diploid and haploid cells.
The main function of mitosis is to make copies of cells for growth and regeneration.

Practice Test - Question 25

Plant Cytokinesis differ from animals Cytokinesis in having

Detailed Solution for Practice Test - Question 25

In plants, cytokinesis or division of cytoplasm occurs due to the formation of the cell plate.

In animals, cytokinesis occurs due to the formation of constriction in the middle cell membrane.

Cytokinesis- Definition and Process (in animal and plant cells)

Practice Test - Question 26

Cell growth results in disturbing the ratio between

Detailed Solution for Practice Test - Question 26

Cell growth results in disturbing the ratio between the nucleus and the cytoplasm.
Therefore, it becomes essential for the cell to divide to restore the nucleo-cytoplasmic ratio.

Practice Test - Question 27

Chromosome appeared beaded during

Detailed Solution for Practice Test - Question 27

Chromosomes appeared beaded during the leptotene stage of the prophase of meiosis I, also the longest stage of meiosis cell division.
Chromatin materials start condensing after that stage.

Practice Test - Question 28

The transition period between M- phase I and M- phase II without DNA replication

Detailed Solution for Practice Test - Question 28

The transition period between the M-phase of Meiosis I and M-phase II during which no replication of DNA occurs is called interkinesis.

Practice Test - Question 29

During which stage do the chromatids of a bivalent become distinct?

Detailed Solution for Practice Test - Question 29

Bivalent or tetrad formation takes place during the zygotene stage of prophase I, after synapsis has occurred and homologous chromosomes form pairs. However, the bivalent is distinct only in the next stage, pachytene.

Practice Test - Question 30

In some lower plants and social insects, the haploid cells are divided by

Detailed Solution for Practice Test - Question 30

In some lower plants and in some social insects haploid cells also divide by mitosis.

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