VITEEE PCBE Mock Test - 4 - JEE MCQ

• Mosses develop into a stage known as protonema from spores before developing into thallus, which resembles leaves. In contrast, this transitional stage is not present in liverworts, therefore the spores grow into a thallus without a protonema.
• Mosses are tiny, flowerless plants that belong to the Bryophyta category, along with liverworts and hornworts. They lack xylem and phloem-like circulatory systems and primarily absorb water and nutrients through their leaves. They often grow in bunches or mats on the forest floor in moist, shaded areas. They often only reach a height of 10 cm, although the unusual species Dawsonia may reach heights of 50 cm.

Mosses have protonema; liverworts develop directly into thallus.

The Genus ranks directly above Species in the taxonomic hierarchy. It groups together species that are closely related and share a more recent common ancestor than those grouped at higher taxonomic levels. This classification reflects evolutionary relationships and morphological similarities among species. An interesting fact is that the genus name is always capitalized and, together with the specific epithet (species name), forms the binomial nomenclature of an organism, such as Homo sapiens for humans.

The correct answer is :
Option D : Both A and R are true and R is the correct explanation of A

• Statement A is true. A flower is indeed a modified shoot wherein the shoot apical meristem changes to a floral meristem.
• Statement R is also true. In the process of flower formation, the internodes of the shoot get condensed, and instead of leaves, different floral appendages (such as sepals, petals, stamens, and carpels) are produced laterally at successive nodes.
• Moreover, Statement R is indeed the correct explanation for Statement A. The transition from the shoot apical meristem to the floral meristem involves the condensation of internodes and the lateral production of floral appendages at nodes, which leads to the formation of a flower.

The correct option is 2 i.e., conversion of Succinyl-CoA to Succinic acid.
During the conversion of succinyl-CoA to succinic acid a molecule of GTP is synthesised. This is a substrate level phosphorylation.
The other 3 steps involve oxidation-reduction reaction where NAD⁺ is reduced to NADH⁺ H⁺ and one point where FAD⁺ is reduced to FADH2.

Topic in NCERT: Citric acid cycle

Line in NCERT: "during the conversion of succinyl-coa to succinic acid a molecule of gtp is synthesised. this is a substrate level phosphorylation."

The cardiac output of a person having 72 heart beats per minute and a stroke volume of 50 mL would be 72 x 50 = 3600 mL. Thus, the correct answer is 3600 mL, i.e Option D.

A typical flower with superior ovary and other floral parts inferior is. A typical flower that has an ovary placed superior along with the other floral organs is known as hypogynous. The other floral organs are also attached under the ovary to the receptacle.

Hypogynous flower: superior ovary.

When the stamens get binded in a bunch or a single bundle, it is called Monadelphous.
The stamens are binded through the filaments of the anthers.
Stamen is composed of several anthers. And filament is a thread like structure which works as a connecting medium for stamen and thalamus.

• When there is a decrease in glomerular blood flow, the JG (Juxtaglomerular) cells in the kidneys respond by releasing a hormone called renin.
• Renin plays a crucial role in the renin-angiotensin mechanism. It converts angiotensinogen, a protein present in the blood, into angiotensin I, which is further converted to angiotensin II.
• Angiotensin II acts as a potent vasoconstrictor, causing the blood vessels to narrow, which increases the glomerular blood pressure and subsequently the glomerular filtration rate (GFR).
• The release of renin is part of a complex regulatory system that helps regulate blood pressure and maintain renal function.

Line in NCERT: "The JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II."

Solution: During inspiration, the diaphragm contracts and moves downward, increasing the volume of the thoracic cavity and decreasing the pressure inside the lungs, allowing air to flow in. This is a crucial mechanism in the breathing process, demonstrating the diaphragm's essential role in respiratory mechanics.

The correct order from smallest to largest is virus, PPLO (Mycoplasma), typical bacteria, and typical eukaryotic cell. Viruses are the smallest, usually only visible with an electron microscope. PPLOs are slightly larger than the largest viruses but smaller than typical bacteria. Typical bacteria are larger than viruses and PPLOs and are visible under light microscopes. Eukaryotic cells are the largest, significantly bigger than viruses, PPLOs, and typical bacteria, making Option A the correct choice.

Topic in NCERT: PROKARYOTIC CELLS

Line in NCERT: "Typical bacteria (1-2 µm) PPLO (about 0.1 µm) A typical eukaryotic cell (10-20 µm) Viruses (0.02-0.2 µm)"

Figure shown above represent translation process in which protein is produced. Ribosome provide site for protein synthesis and t-RNA brings the amino acids. The ‘x’ is the polypeptide chain produced.

I =
=

n = = 0.625 x 10¹⁹ x 1/4

When a charged particle enters a uniform magnetic field at a right angle to the direction of the field, the following changes occur:

• A: Plane of motion of the particle: The particle will follow a circular or helical path due to the magnetic force, which acts perpendicular to the velocity of the particle. The plane of motion changes as the particle moves in a circular trajectory.
• B: Velocity of the particle: The magnitude of the velocity remains constant (since the magnetic force does not do work), but the direction of the velocity changes because the particle moves in a circular path. Therefore, the velocity vector changes direction.
• C: Direction of motion: The direction of motion changes continuously because the charged particle moves in a circular path due to the Lorentz force exerted by the magnetic field.

Conclusion: All of the above aspects change in the scenario described.
Thus, the correct answer is: D: All of these.

The value of g at a height R_e from Earth's surface is .


So,

By the geometric addition of vectors, the sum of the three vectors shown in the figure is zero, since they form the three sides of a triangle.

That is, all the force vectors add up together and get nullified, and thus, the particle is in equilibrium. 

Since the net force on the particle is zero,  remains unchanged.

The resultant intensity of two periodic waves is given by:

I = I₁ + I₂ + 2√(I₁I₂) cos(δ)

Where δ is the phase difference between the waves.

For maximum intensity, δ = 2nπ; n = 0, 1, 2, ...

Therefore, for zero-order maxima, cos(δ) = 1.

I_max = I₁ + I₂ + 2√(I₁I₂)

= (√I₁ + √I₂)²

= I₁ + I₂ + 2√(I₁I₂)

For minimum intensity, δ = (2n - 1)π; n = 1, 2, ...

Therefore, for first-order minima, cos(δ) = -1.

I_min = I₁ + I₂ - 2√(I₁I₂)

= (√I₁ - √I₂)²

= I₁ + I₂ - 2√(I₁I₂)

Therefore,
I_max + I_min = (√I₁ + √I₂)² + (√I₁ - √I₂)²
= 2(I₁ + I₂)

The current will be equally divided at junction P. The field at the centre due to wires PQ and SR will be equal in magnitude but opposite in the direction, so its effective field will be zero. Similarly, net field due to wires PS and QR is zero. Therefore, the net field at the centre of the loop is zero.

Electrode potential of the cell must be +ve for spontaneous reaction.
Zn²⁺ → Zn; E⁰ = -0.76 V
Cu²⁺ → Cu; E⁰ = -0.34 V
Redox reaction is:

E_cell = E⁰_cathode - E⁰_anode
= -034 - (-0.76)
= +0.42 V
E_cell is positive, so the above reaction is feasible.
Hence, option (2) is the correct answer.

[IO₂F₅]^2- ion hybridisation is sp³d³ and shape is pentagonal bipyramidal.
Double bond causes more repulsion, so they would be on axial position at 180^o angle to each other, so shape is

The given information describes a straight-line plot where:

• The y-axis represents 1/(a - x), where a is the initial concentration and x is the concentration at time t.
• The x-axis represents time.
• The slope of the line is equal to the rate constant (k).
• The y-intercept is 1/a.

This setup matches the integrated rate law for a second-order reaction. For a second-order reaction, the equation is:

This is a straight-line equation of the form y = mx + b, where:

m (slope) is the rate constant k,

b (y-intercept) is 1/a.

Conclusion: This graph suggests that the reaction is second-order.
So, the correct answer is: B: 2.

Given Chemical Reaction:
2AgCl(g) + H₂(g) → 2HCl(aq) + 2Ag(s)

In this reaction:
AgCl (solid) is reduced to Ag (solid).
H₂ (gas) is oxidized to H⁺ (aq) which forms HCl (aqueous).

Understanding the Galvanic Cell Notation:
In the galvanic cell notation, we represent the anode and cathode reactions as well as their respective phases:
The anode (oxidation site) is where the oxidation reaction occurs.
The cathode (reduction site) is where the reduction reaction occurs.
The phase boundary is represented by a slash (/).
The double vertical line (||) separates the two half-cells.

Step 1: Oxidation (Anode)
At the anode, H₂(g) gets oxidized to H⁺. This happens in the presence of HCl.
So, the anode half-cell will be represented as: Pt(s), H₂(g), 1 bar | 1 M HCl(aq)

Step 2: Reduction (Cathode)
At the cathode, AgCl(s) is reduced to Ag(s).

So, the cathode half-cell will be represented as: AgCl(s) | Ag(s)

Step 3: Final Cell Notation
Putting the anode and cathode together with the appropriate phase boundaries:
The final cell notation is:
Pt(s), H₂(g), 1 bar / 1 M HCl(aq) || AgCl(s) / Ag(s)

Conclusion: The correct notation for the galvanic cell is:
C: Pt(s), H₂(g), 1 bar / 1 M HCl(aq) || AgCl(s) / Ag(s)

Zone refining:-

It is based upon fractional crystallization as the impurity prefer to stay in the melt and on solidification, only pure metal solidifies on the surface of melt.

The principle in this process is solubility of the impurity in the metal in melt state is greater than in solid state.

Tertiary amines have low boiling points due to absence of hydrogen bonding among isomeric amines. Primary amine have high boiling point due to the presence of extensive hydrogen bonding among the amines. 
The order is 1^∘ > 2^∘ >3^∘.
Boiling point ∝ force of attraction.
Primary amine has only 2 active hydrogen for Hydrogen bonding, whereas secondary amine has 1 active hydrogen and tertiary amine has 0 active hydrogen required for hydrogen bonding.

So, ​PhLi react readily with 1 but does not add to 2.

Fluorine cannot form multiple bonds with the central metal atom but oxygen atom can do the same.

Last paragraph of the passage says, "While cross-country comparisons are somewhat difficult given that the reporting of equity ownership data is not uniform ..." So, option 3 is correct.

The firms with the concentrated ownership and insider control are the firms with majority of the holding lying in the hands of owners. Here, the equity is not shared among many shareholders. So, option 2 is correct.

AD = Height of the tower.

C is the top of the platform.

BC = DE = 40√3 m

Calculating:

tan 30° = AB / BC
1 / √3 = AB / 40√3

AB = 40 m

Height of the tower = AB + Height of the platform
AD = 40 + 5 = 45 m

Thus, the tower is 45 metres tall.