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VITEEE PCBE Mock Test - 8 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE PCBE Mock Test - 8

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VITEEE PCBE Mock Test - 8 - Question 1

An irreversible or permanent increase in size, mass or volume of a cell, organ or organism is called as ________.

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 1

Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell organ or organism. Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at curtain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new als are always being added to the plant body by the activity of the meristem is called the open form of growth.

VITEEE PCBE Mock Test - 8 - Question 2

Read the following statements regarding arithmetic growth and select the correct answer.
(i) Rate of growth is constant.
(ii) One daughter cell remains meristematic while the other one differentiates and matures.
(iii) Mathematical expression is L=L0 +rt.

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 2

Increase in growth per unit time is called as growth rate. The growth rate may be arithmetic or geometrical. Arithmetic Growth is a type of growth in which the rate of growth is constant and increase in growth occurs in arithmetic progression-- 2, 4, 6, 8, 10,12. Meristematic cells at the growing point divide in such a fashion that one daughter cell remains meristematic while the other grows and differentiates. the process continues. Mathematically, arithmetic growth is expressed as
L=L0 +rt.
where Lt = length after time t, L0   = length at the beginning, and r = growth rate. On plotting growth against time, a linear graph is obtained.

VITEEE PCBE Mock Test - 8 - Question 3

Growth at cellular level, is principally a consequence of increase in the amount of

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 3

Growth, at a cellular level, is principally a consequence of increase in the amount of protoplasm. Growth is measured by a variety of parameters such as  (i) increase in length, e.g, stem,root, pollen tube (ii) increase in volume eg., fruits. (iii) Increase in area, eg.leaves (iv) increase in diameter, eg, tree trunks, fruits (v) Increase in fresh or dry weight. One single maize root apical meristem can given to more than 17,500 new cells per hour, wheres as cell in watermelon may increase in size by upto 3, 50,000 the former, growth is expressed as increase the cell number latter expresses growth as increase in the size of cell.

VITEEE PCBE Mock Test - 8 - Question 4

Growth in plants is

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 4

The meristem which is consumed in the formation of an organ is called determinate meristem. The meristem which continues its activity throughout life of the plant is called indeterminate meristem. Root apical meristem, shoot apical meristem, intercalary meristem (e.g., grass) and lateral meristems are all Indeterminate meristems. Plant growth is generally indeterminate, i.e., plants retain the capacity for unlimited growth throughout their life, whereas it is determinate in the meristem which is consumed in the formation of an organ.

VITEEE PCBE Mock Test - 8 - Question 5

Vascular cambium and cork cambium are

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 5

Lateral meristem occurs on the sides and takes part in increasing girth of the plant. Intra fascicular cambium is the primary lateral meristem which lies in vascular bundles of dicot and gymnosperm stems in between phloem and xylem. Examples of  secondary lateral meristems are vascular cambium of the root, inter fascicular cambium of stem,cork cambium etc. that  take part in the secondary growth.

VITEEE PCBE Mock Test - 8 - Question 6

The pre-hypertension blood pressure value is a measurement between:

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 6

Prehypertension is defined as a systolic pressure from 120–139 millimeters of mercury (mm Hg) or a diastolic pressure from 80–89 mm Hg.

VITEEE PCBE Mock Test - 8 - Question 7

 Photophosphorylation is the process in which

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 7

Photophosphorylation is the conversion of ADP to ATP using the energy of sunlight by activation of PSII. This involves the splitting of the water molecule in oxygen and hydrogen protons (H+), a process known as photolysis.

VITEEE PCBE Mock Test - 8 - Question 8

Bt toxin is obtained from

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 8

Bt toxin is obtained from Bacillus thuringiensis bacterium that can kill certain insects like tobacco budworm, armyworm, beetles and flies.

VITEEE PCBE Mock Test - 8 - Question 9

Double helix model of DNA was proposed by proposed by

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 9

Double helix model of DNA was proposed by James Watson and Francis Crick in 1953 based on X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin.

VITEEE PCBE Mock Test - 8 - Question 10

Drosophila flies with XXY genotype are females, but human beings with such genotype are abnormal males. It shows that

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 10

Sex in Drosophila is a function of the ratio of the number ofX chromosomes to the number of autosomal sets. Therefore a Drosophila with a X/A =1.0 will be a female whereas the one with a X/Aratio=0.5 will be male However, in humans the presence or absence of the Y chromosome determines sex.

VITEEE PCBE Mock Test - 8 - Question 11
A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing
Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 11
The image can be formed on the screen if it is real. Real image of reduced size can be formed by a conave mirror or a convex lens.
Let u = 2f + x.
Then,




v =
It is given that u + v = 1 m.
2f + x + = (2f + x)
Or
Or (2f + x)2 < (f + x)
This will be true only when f < 0.25 m.
VITEEE PCBE Mock Test - 8 - Question 12
If a screw gauge moves 1 mm in two rotations, then the pitch of the screw gauge is
Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 12
Pitch = distance covered by screw head/number of rotations
= (1 mm)/2
= 0.5 mm
VITEEE PCBE Mock Test - 8 - Question 13

The activity of a radioactive sample is measured as N₀ counts per minute at t = 0, and N₀e⁻ᵦt counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is:

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 13

Number of nuclei left undecayed after time t is given by,

Where n is number of half lives in time t.

According to question, N = N0/e

Substituting in (i), we get


Taking log on both sides, we get

Activity of the sample reduces to half at its half life.

VITEEE PCBE Mock Test - 8 - Question 14

 

A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in the figure. The force F exerted by the sphere on a particle of mass mlocated at a distance r from the centre of sphere varies as (0 ≤ r ≤ ∞)

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 14

F = 0 when 0 ≤ r ≤ R₁
Because intensity is zero inside the cavity.

F increases when R₁ ≤ r ≤ R₂
As there will be more mass enclosed if the radius of the enclosure is increased.

F ∝ 1/r² when r > R₂
As all mass is enclosed, it will behave as a point mass.

VITEEE PCBE Mock Test - 8 - Question 15

The correct pair of orbitals involved in π-bonding between metal and CO in metal carbonyl complexes is

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 15

VITEEE PCBE Mock Test - 8 - Question 16

The most suitable reagent for the conversion of 2-phenylpropanamide into 1-phenylethylamine is:

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 16

The most suitable reagent for the conversion of 2-phenylpropanamide into 1-phenylethylamine is Br2, NaOH.

VITEEE PCBE Mock Test - 8 - Question 17

For a 1st order chemical reaction,

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 17

For a 1st order chemical reaction,


Arrhenius equation:

[A = Pre-exponential factor]
In a 1st order reaction, the unit of the pre-exponential factor is reciprocal second.
Because the pre-exponential factor depends on the frequency of collisions, it is related to collision theory and transition state theory.

VITEEE PCBE Mock Test - 8 - Question 18
Which one of the following pairs of solution can we expect to be isotonic at the same temperature?
Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 18
'i' for Ca(NO3)2 and Na2SO4 is 3 each. Their colligative molarities are also the same. Hence, 0.1 M Ca(NO3)2 and 0.1 M Na2SO4 solutions are isotonic.
VITEEE PCBE Mock Test - 8 - Question 19

Directions: In the following question, two statements are given. One is assertion and the other is reason. Examine the statements carefully and mark the correct answer according to the instructions given below.
Assertion: [Co(NO2)3(NH3)3] does not show optical isomerism.
Reason: It has a plane of symmetry.

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 19

Optical isomerism is shown by chiral molecules, due to the absence of any plane of symmetry, axis of symmetry or centre of symmetry in the molecule. If the molecules possess any of these elements of symmetry, the compound will not be optically active.
[Co(NO
2)3(NH3)3] possesses a plane of symmetry and can be superimposed on its mirror image, so it is an optically inactive complex.

VITEEE PCBE Mock Test - 8 - Question 20

Directions: In the following question, two statements are given. One is Assertion, and the other is Reason. Examine the statements carefully and mark the correct answer according to the instructions given below.
Assertion: The oxidation of ketone by perbenzoic acid gives esters.
Reason: Perbenzoic acid oxidises because of the release of nascent oxygen on dissociation.

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 20


The reaction does not occur because of nascent oxygen. Hence, reason is false.

VITEEE PCBE Mock Test - 8 - Question 21

Which is non-reducing sugar

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 21

The carbohydrates or sugar where free aldehyde or ketonic group is absent (utilized in glycosidic bond formation) can not reduce the above reagents are called non-reducing sugar i.e., Sucrose, glycogen, starch.

VITEEE PCBE Mock Test - 8 - Question 22

Substances which can be stretched to couse large strain are called

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 22

Substances which can be stretched to couse large strain are called elastomers. e.g. tissue of aorta, rubber etc.

VITEEE PCBE Mock Test - 8 - Question 23

The IUPAC name of compound  is

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 23

If atom or group of higher priority are on opposite direction at the double bond of each carbon atom then the configuration is known as E and if they are in same direction then the configuration is known as Z configuration.

VITEEE PCBE Mock Test - 8 - Question 24

Which of the following statements are correct?
(i) Fullerenes have dangling bonds
(ii) Fullerenes are cage-like molecules
(iii) Graphite is thermodynamically most stable allotrope of carbon
(iv) Graphite is slippery and hard and therefore used as a dry lubricant in machines

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 24

(ii) Fullerenes are cage-like molecules.
(iii) Graphite is thermodynamically most stable allotrope of carbon.
(iv) Graphite is slippery and hard and therefore used as a dry lubricant in machines.

VITEEE PCBE Mock Test - 8 - Question 25

Compare vitamin List I with its deficiency disease List II.

Codes :

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 25

VITEEE PCBE Mock Test - 8 - Question 26

The shortest O−O bond can be seen in which of the following?

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 26

Bond order is calculated as:
Bond order = (Nb - Na) / 2

Where:

  • Nb = Number of electrons in bonding molecular orbitals
  • Na = Number of electrons in antibonding molecular orbitals

For the following molecules:

  • O2 → Bond order = 2 (Molecular orbital theory)
  • O3 → Bond order = 1.5 (Bond order calculated based on the resonating structures)
  • O2²⁻ → Bond order = 1 (Molecular orbital theory)
  • O⁻₂ → Bond order = (10 - 7) / 2 = 1.5 (Molecular orbital theory)

As bond order increases, bond length decreases.

VITEEE PCBE Mock Test - 8 - Question 27

What is the role of a broad-spectrum antibiotic?

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 27

Antibiotics which kills or inhibit a wide range of gram-positive and gram-negative bacteria are said to be broad spectrum antibiotics.
So Broad-spectrum antibiotics act on different antigens.

VITEEE PCBE Mock Test - 8 - Question 28

The key step in Cannizaro's reaction is the intermolecular shift of

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 28

Cannizaro reaction,
HCHO + NaOH → CH3OH + HCOONa
This reaction takes place by those compounds which has no α−H atom.
Inter molecular shift of hydride ion is key step of Cannizaro reaction

VITEEE PCBE Mock Test - 8 - Question 29

In the given figure, AB || CD and CD || MN.

Find the value of x + y + z.

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 29

AB || MN and AM is the transversal.
∠1 + (x + 65°) = 180° (Co-interior angles)
90° + x + 65° = 180°
Or x = 25°
Now, CD || MN and BM is the transversal.
y + 65° = 180° (Co-interior angles)
y = 115°
Now, AB || CD and BM is the transversal.
z = y (Corresponding angles)
z = 115°
Now, x + y + z = 25° + 115° + 115° = 255°

VITEEE PCBE Mock Test - 8 - Question 30

The difference between the age of Guri and Shuri is 4 years. Five years ago, the sum of their ages was 36 years. What will be the ratio of the age of Guri to that of Shuri 7 years from now, if Shuri is younger than Guri?

Detailed Solution for VITEEE PCBE Mock Test - 8 - Question 30

Let the age of Guri and Shuri five years ago be x and y, respectively.
According to the question,
x + y = 36
x = 36 - y ... (i)
Also,x - y = 4
Putting the value of x as given in (i) we get,
36 - y - y = 4
36 - 2y = 4
-2y = 4 - 36
-2y = -32
2y = 32
y = 32 ÷ 2
y = 16
Therefore, the age of Shuri, 5 years ago = 16 years
Putting y = 16 in the equation x - y = 4,
x - 16 = 4
x = 4 + 16
x = 20
Hence, age of Guri, 5 years ago = 20 years
Present age of Guri = 20 + 5 = 25 years
Present age of Shuri = 16 + 5 = 21 years
Age of Guri and Shuri, 7 years from now, will be 32 years and 28 years, respectively.
Required ratio = 32 : 28
= 8 : 7

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