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VITEEE PCBE Mock Test - 9 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE PCBE Mock Test - 9

VITEEE PCBE Mock Test - 9 for JEE 2025 is part of VITEEE: Subject Wise and Full Length MOCK Tests preparation. The VITEEE PCBE Mock Test - 9 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE PCBE Mock Test - 9 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE PCBE Mock Test - 9 below.
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VITEEE PCBE Mock Test - 9 - Question 1

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

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 1

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.

VITEEE PCBE Mock Test - 9 - Question 2

As they release hydrolase that digest old andd damaged cells, the term suicide bags is aptly used by cell biologists for

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 2

The lysosomes may be called "suicide bags’ of the cell in view of their autolytic role, or "disposal units’ of the cell because they digest the incoming food materials and remove the foreign bodies, toxic molecules, and debris, or "recyling centres" as they break down worn out cells cell organelles to component molecules for building organelles and cells.

VITEEE PCBE Mock Test - 9 - Question 3

Ribosomes are synthesized in

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 3

Nucleolus is a dense, rounded, dark-staining, granular structure without a limiting membrane. It consists largely of RNAs and proteins. Nucleolus syntesizes and stores RNA. It also receives ribosomal proteins from the cytoplasm for storage. It forms ribosomal subunits by wrapping the RNAs with nitosomal proteins. The ribosomal subunits later have the nucleus through the nuclear pores.

VITEEE PCBE Mock Test - 9 - Question 4

Deposition of uric acid crystals with in the synovial joint causes:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 4

In gout, there occurs a defect in uric acid metabolism resulting into its elevated level in blood (hyperuricemia). This is followed by precipitation of excessive uric acid which gets deposited in the joint spaces. These deposited crystals of uric acid causes pain in different bony joints.

VITEEE PCBE Mock Test - 9 - Question 5

The number of floating ribs, in the human body, is

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 5

The last two pairs i.e. 11th and 12th pairs ribs remain free anteriorly, hence, they are called as floating ribs.

VITEEE PCBE Mock Test - 9 - Question 6

What is the correct sequence for parturition to occur?

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 6
  • Parturition starts with the signal released by the fetus and the placenta.
  • These act on the pituitary to release oxytocin.
  • Oxytocin acts as a stimulant leading to contractions of the uterine muscles.
  • The uterine contractions feedbacks to release more oxytocin from the pituitary.
  • This results in more powerful contractions of the uterus until the baby is delivered in the process of parturition.
VITEEE PCBE Mock Test - 9 - Question 7

The process of adaptive radiation in Australian marsupials resulted in:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 7

Adaptive radiation in Australian marsupials refers to the process where several distinct species evolved from a common ancestor to occupy various ecological niches. This diversification was a result of adaptation to different environments, not due to migration or extinction.

VITEEE PCBE Mock Test - 9 - Question 8

Which of the following is not a typical feature of Down’s syndrome?

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

Gynecomastia in males is not a typical feature of Down’s syndrome. Typical features include short stature, mental retardation, and characteristic physical traits like a broad palm with a single crease.

VITEEE PCBE Mock Test - 9 - Question 9

How does RNA interference (RNAi) protect plants from nematode infestation?

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

RNA interference (RNAi) protects plants from nematode infestation by silencing specific mRNA of the nematode through the formation of complementary double-stranded RNA. This prevents translation, thereby disrupting the nematode's survival.

VITEEE PCBE Mock Test - 9 - Question 10

Select the statement which explains best parasitism:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 10

Parasitism is a type of symbiotic relationship where one organism benefits at the expense of the other, which is harmed.

VITEEE PCBE Mock Test - 9 - Question 11

What is the role of 23s rRNA in the process of translation in prokaryotes?

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 11

23S rRNA in the large subunit of the ribosome is catalytically active and plays a crucial role in the formation of peptide bonds during protein synthesis in prokaryotes.

VITEEE PCBE Mock Test - 9 - Question 12
A closely-wound solenoid of 1000 turns, having area of cross-section 1.5 × 10–4 m2 and carrying a current of 2 A, is suspended through its centre, allowing it to turn in a horizontal plane. What is the force on the solenoid if a uniform horizontal magnetic field of 5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?
Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 12
The net force experienced by a magnetic dipole in a uniform magnetic field is zero. The magnetic field exerts equal and opposite force on the north and south poles. Hence, the correct choice is (4).
VITEEE PCBE Mock Test - 9 - Question 13

If the velocity of light 'c', gravitational constant 'G' and Planck's constant 'h' are chosen as fundamental units, then the dimensions of length 'L' in the new system will be

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 13

Let, L = [hacbGd]
[L]1 = [M1L2T -1]a[LT -1]b[M-1L3T -2]d
[L]1 = [Ma-dL2a+b+3dT -a-b-2d]
Comparing the dimensions of M, L and T on both sides, we have
a - d = 0 ..............1
2a + b + 3d = 1 ..........2
-a - b - 2d = 0 ............3
From equation 1, a = d
and from equation 1 and 3, b = -3a
Then, substituting the value of b and d in 2,

VITEEE PCBE Mock Test - 9 - Question 14

A force of 1200 N acts on a 0.5 kg steel ball as a result of a collision lasting 25 ms. If the force is in a direction opposite to the initial velocity of 14 m/s-1, then the final speed of the steel ball would be:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 14

Initial velocity of steel ball, u = 14 m/s-1
Mass of steel ball, m = 0.5 kg
Force acting on the steel ball from the opposite direction, F = -1200 N
Time duration for collision, t = 25 ms = 25 × 10⁻³ s

Let the final velocity of the ball be v.

Applying Newton's second law of motion to the steel ball, i.e.

[−ve sign shows that now ball will move in the direction of force]

VITEEE PCBE Mock Test - 9 - Question 15

For the wave shown in figure, the equation for the wave, travelling along +x-axis with velocity 350 m s−1 when its position is at t = 0, is, (All variables are in SI unit.)

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 15

At t = 0,
y = 0.05 sin(kx),

2.5λ = 20 cm,
λ = 8 cm.

⇒ λ = 0.8 cm.

⇒ k = 2π/λ = 2π/0.8 = 5π/2.

V = 350 m/s.

From y = A sin(k(x - vt)),
y = 0.05 sin(5π/2(x - 350t))
= 0.05 sin(5π/2 x - 350t)
= 0.05 sin(5π/2 x - (5π/2)(350)t).

VITEEE PCBE Mock Test - 9 - Question 16

Which of the following statement is true?

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 16

From the law of equipartition of energy law, the total kinetic energy of the gas is KE = f/2nRT, which is proportional to the temperature.
At absolute zero temperature, kinetic energy of the gas molecules becomes zero, but they may possess some potential energy, so we cannot say that absolute zero degree temperature is zero energy temperature.
The RMS speed of a gas is given by .
For different gases or at different temperature, it will be different. But it does not depend on the pressure.
From the ideal gas equation, PV = nRT, two different gases at same temperature, pressure and volume will have the same number of moles and same number of molecules.

VITEEE PCBE Mock Test - 9 - Question 17

For an alternating current

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 17

sForinusoidal alternating current or triangualr and square wave alternating current the average value for a period is always zero.

But in case of following currnet peak value will not be equal to rms value.

If AC is the square wave then all these three options are possible. i.e.,

Irms = Iav = I0, here I0 is peak value

VITEEE PCBE Mock Test - 9 - Question 18

A unit less quantity

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 18

For a quantity to be unit less, its dimensions= 1 = M0L0T0
Hence, it has to be dimensionless.
Some examples are refractive index and relative density, these are unit less and dimensionless.
Quantities like angles are dimensionless, but they have a unit.
Hence, the correct answer is never has a non-zero dimension.

VITEEE PCBE Mock Test - 9 - Question 19

Coefficient of restitution during the collision is changed to 1/2, keeping all other parameters unchanged. What is the velocity of the ball B after the collision?

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 19

Let the velocity of A and B be vA & vB along the line of impact after collision.

Now, net force along the line of impact is 0.
So, applying conservation of momentum

On solving eq 1 & eq 2

VITEEE PCBE Mock Test - 9 - Question 20

The reaction 2A → 2B + C obeys the second order rate law. When the partial pressure of A is doubled, the rate would

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 20

When the partial pressure of A is doubled, its concentration gets doubled. Since rate = k(A)2, the rate becomes four times the original.

VITEEE PCBE Mock Test - 9 - Question 21

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: Deoxyribose (C5H10O4) is not a carbohydrate.
Reason: Carbohydrates are hydrates of carbon. So, the compounds which follow Cx(H2O)y formula are carbohydrates.

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 21

Deoxyribose is a carbohydrate. Hence, the assertion is wrong.
Deoxyribose does not have the formula Cx(H2O)y, but is very much a carbohydrate. Hence, the definition of carbohydrates given in the reason is not valid.
The correct definition is that carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones.

VITEEE PCBE Mock Test - 9 - Question 22
Electronic configurations of different elements are given below in the options. Based on their electronic configurations, choose the one having the highest ionisation energy.
Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 22
[Ar] 3d10 4s2 is the electronic configuration of zinc.
Zinc has the highest ionisation energy among the elements of the 3d transition series as it has the highest effective nuclear charge and a stable electronic configuration of fully filled orbitals.
VITEEE PCBE Mock Test - 9 - Question 23

Freundlich adsorption isotherm is a graph between

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 23

Freundlich adsorption isotherm is a graph between extent of adsorption and pressure at a constant temperature.

x/m = Kρ1/n (at constant T)

VITEEE PCBE Mock Test - 9 - Question 24

Chile salt petre is an ore of

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 24

Chile salt petre (NaNO3) sodium nitrate is the nitrate ore of sodium.
It is a deliquescent crystalline sodium salt that is found chiefly in northern Chile. It is a white crystalline solid, odourless and has a slightly bitter taste. It is non-combustible and is highly soluble in water, hydrazine and ammonia. It dissolves in alcohol, slightly dissolves in pyridine, and completely insoluble in acetone.

VITEEE PCBE Mock Test - 9 - Question 25

Find out the incorrect match from the following:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 25

In the above molecule, the central atom xenon undergo sp3 hybridisation, hence, the pi bonds are due to p orbitals of oxygen and d orbitals of xenon.

In the above molecule, the central atom sulphur undergo sp3 hybridisation, hence, the pi bond is pπ−dπ bond.

In the above molecule, the central atom sulphur undergo sp2 hybridisation, hence, out of three pi bonds one pi bond is pπ−pπ bond and other two are pπ−dπ bonds.

In the above molecule, the central atom chlorine undergo sp3 hybridisation, hence, the two pi bonds are pπ−dπbonds.

VITEEE PCBE Mock Test - 9 - Question 26

In a covalent bond formation:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 26

A covalent bond exist between two similar atoms or the atoms which has less electronegative difference.
Equal sharing of electrons between two atoms takes place forming covalent bond to complete the octet.
For example, in hydrogen molecules the bond is covalent, as both the atoms are similar.

VITEEE PCBE Mock Test - 9 - Question 27

Daily requirement of vitamin-A for adult women is

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 27

The recommended daily amount of vitamin-A for adult women is 700 micrograms. Vitamin-A is a nutrient important for good vision, growth, cell division, reproduction and immunity. It is found in spinach, dairy products, etc.

VITEEE PCBE Mock Test - 9 - Question 28

P, Q, R, S and T are standing in a line. Q is taller than T, but shorter than P. S is shorter than T, but taller than R. Who is the shortest?

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 28

Correct Answer: R
P > Q > T

T > S > R
P > Q > T > S > R

VITEEE PCBE Mock Test - 9 - Question 29

The number of 3×3 matrices A , whose entries are either 1 or −1 and for which the system  has exactly distinct solutions, is

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 29

It is not possible that three planes intersect exactly at 3  points. So, no such matrix is possible. Therefore, the answer is 0 (zero).
Thus, option A, that is, 0, is the correct answer.

VITEEE PCBE Mock Test - 9 - Question 30

Let f(x) = ax² + bx + c, g(x) = ax² + qx + r, where a, b, c, q, r ∈ R and a < 0. If α and β are the roots of f(x) = 0, and α + δ and β + δ are the roots of g(x) = 0, then:

Detailed Solution for VITEEE PCBE Mock Test - 9 - Question 30

Roots of the quadratic equation f(x) = ax² + bx + c are α and β.

Let D₁ be the discriminant of the above quadratic.

The difference of the roots |α - β| = D₁ / √a ...(i)

The maximum value of the above quadratic is fmax = -D₁ / 4a.

Roots of the quadratic equation g(x) = ax² + qx + r are α + δ and β + δ.

Let D₂ be the discriminant of the above quadratic.

The difference of the roots |(α + δ) - (β + δ)| = |α - β| = D₂ / √a ...(ii)

The maximum value of the above quadratic is gmax = -D₂ / 4a.

From (i) and (ii), D₁ = D₂
Thus, fmax = gmax.

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