VITEEE PCBE Mock Test - 3 - JEE MCQ

Statements ii, iii, and iv are correct. Birds' forelimbs are indeed modified into wings, they have a highly efficient four-chambered heart to support their high metabolism, and they practice internal fertilization. Statement i is incorrect as not all birds can fly; some species like ostriches, emus, and penguins are flightless.

• Assertion: True. Algae are classified into Chlorophyceae, Phaeophyceae, and Rhodophyceae, primarily based on differences in their pigment composition and type of stored food, such as starch, mannitol, or laminarin.
• Reason: False. While vegetative reproduction like fragmentation is a common method among algae, it is incorrect to state that algae reproduce exclusively through such methods. Algae can also reproduce asexually through spores and sexually through various types of gametes and reproductive strategies (isogamous, anisogamous, oogamous).
• Therefore, the Assertion is correct, but the Reason is false, making Option C the correct choice.

Coronary heart disease (CHD) is a disease in which a waxy substance called plaque builds up inside the coronary arteries. These arteries supply oxygen-rich blood to your heart muscle. When plaque builds up in the arteries, the condition is called atherosclerosis. The buildup of plaque occurs over many years.

• The respiratory or gas-exchange surface consists of millions of small sacs, or alveoli, lined by a simple squamous epithelium.
• This epithelium is exceedingly thin to facilitate the diffusion of oxygen and CO₂. 
• The alveolar walls also contain cuboidal surfactant-secreting cells.
  

Hence, the correct option is A

NCERT Reference: Topic- EXCHANGE OF GASES” of chapter "Breathing and Exchange of Gases" of NCERT 

S-shaped growth curve is also called Verhulst-Pearl logistic curve and is represented by the following equation :

where dN/dt= rate of change in population size, r= intrinsic rate of natural increase, N= population density, K= carrying capacity and 

Topic in NCERT: Logistic Growth

Line in NCERT: "This type of population growth is called Verhulst-Pearl Logistic Growth and is described by the following equation: dN/dt = rN(K-N)"

Hormone receptors are specific to the hormones they bind, allowing for precise regulation of physiological processes.

A mature ovarian follicle is called Graafian follicle. After ovulation, the empty Graafian follicle shows deposition of lutein and forms corpus luteum that ultimately degenerates. If fertilisation occurs, the corpus luteum secretes progesterone, a hormone that causes further change in the endometrium, allowing it to provide a good milieu in which a zygote (fertilised ovum) can grow through the stages of gestation to become a fetus.
Sweat, and not sebum, is produced by sweat glands.
Niacin is also called vitamin B₃.
SA Node is called pacemaker of the heart.

By the end of sixth months of pregnancy, eye-lashes are formed in the foetus.

In eukaryotes, the coding sequences or expressed sequences are defined as exons. Exons appear in mature or processed RNA. The exons are interrupted by introns, they do not appear in mature or processed RNA.

NCERT Topic: Structural genes in eukaryotes

NCERT Line: "The coding sequences or expressed sequences are defined as exons. Exons are said to be those sequence that appear in mature or processed RNA. The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA."

As the electric potential of both the spheres is same,

If σ be the surface charge density,

Let l₀ be the length of the solenoid, r be the radius and n be the number of turns per unit length.
Length of the wire l = l₀n(2πr) and
Self Inductance  L = μ₀n²l₀πr²
where l₀ is length of solenoid =1 m, n is the number of turns per unit length.

 (as ω will be same in both cases)



= 6
The correct answer is 6.

To prove this, we will calculate the work done by the electric force in a closed loop ABCD.

During displacement AD and BC, the work done by the electric force is 0 because the electrostatic force is perpendicular to displacement.

Work done during AB = R α E₁

Work done during CD = −r α E₂

Total work done = R α E₁ − r α E₂

Since the work done in a closed loop should be zero:

∴ E₁ / E₂ = r / R

Given h_5th = 2 × h_6th. By solving we get u = 65 m s⁻¹

The electron microscope is based on the de-Broglie wavelength of electron. The de-Broglie wavelength of the electron is given by,

 where V is the potential difference applied.

Hence, the ratio of the wavelength will be,

In the presence of friction, acceleration, a = (gsinθ − μgcosθ)

∴ Time taken to slide down the incline,

In the absence of friction, time taken to slide down the incline, 

According to the question, t₁ = 2t₂

sinθ = 4sinθ − 4μcosθ

On heating the system  x, r, d all increases, since, the expansion of isotropic solids is similar to true photographic enlargement.

The given reaction is S_N1 type.
For alkyl group, the reactivity decreases in the order 3° > 2° > 1°.
So, (2) is correct.

Hexagonal close-packing is a three-dimensional close packing.
HCP arrangement of atoms→→ 
The unit cell consists of three layers of atoms.
In the bottom (A) layer, each sphere is in direct contact with six others spheres. After packing, two types of voids are formed, i.e., b and c.

A new layer of spheres is placed in such a manner that the triangular b voids of the first layer are fully occupied, but voids c are unoccupied hollows.

Now the third layer (A) is again placed over the (B) layer in such a way that the bottom (A) is aligned with top (A).

Hence, the arrangement is, ABABABAB. In the hexagonal close packing structure the coordination number of each sphere is 12. The packing efficiency is 74%.

The manganate and permanganate ions are tetrahedral due to the π− bonding involves overlap of p− orbitals of oxygen with d− orbitals of manganate.

Saccharin is an artificial sweetener that effectively has no food energy.
It is about 300−400 times as sweet as sucrose but has a bitter or metallic by taste, especially at the higher concentrations.

The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics. The rate of reaction depends upon the concentration of only one reactant, which is tertiary butyl bromide.
Rate ∝ [Reactant]¹

This reaction is called HVZ reaction.

A metal complex consists of a central metal atom or ion that is bonded to one or more ligands, which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal.
Higher the charge and smaller the size of the ligand, more stable is the complex formed, as the ligand can approach the metal more closely. In case of neutral monodentate ligands, high dipole moment and small size favours the more stable complex.
A metal ion with a high charge and a small size forms more stable complexes as it can attract the ligands more closely. As the size will increase, the ionisation potential will decrease, which means that a greater ionisation potential favours the more stable complex.

Here, 'sustenance' means a requirement for survival of the continuous market (it already exists, but public-float is essential). 'Maintenance' means to preserve a condition and here, it says a public-float is essential if the continuous market for listed securities is to continue. So, 'maintenance' is the correct option. 'Subsistence' means the state of having the bare minimum you need for survival or existence. 'Presence' means 'to be present', 'occurrence' means 'to happen', and 'evidence' means 'proof'. So, option 2 is the correct answer.

There are two circles in the figure.

Given:

Area of the inner circular lawn = 154 m²

⇒ πr² = 154
⇒ (22/7) × r² = 154
⇒ r² = (154 × 7) / 22
⇒ r² = 49
⇒ r = √49 = 7 m

Given: Width of the path = 7 m

Radius of the circular lawn including the path (R) =
Radius of the circular lawn (r) + Width of the path
⇒ 7 + 7 = 14 m

Area of the circular lawn including the path
= πR²
= (22/7) × 14²
= (22/7) × 196
= 616 m²