SRMJEE Mock Test - 2 (Medical) - JEE MCQ

The situation is shown in the figure. Plate 1 has surface charge density σ and plate 2 has surface charge density σ. The electric field at point P due to two charged plates add up, giving


Given, σ = 2.64 × 10⁻¹² C/m²,
ε₀​ = 8.85 × 10⁻¹² C²/N − m²
Hence,

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens.
Hence, μ₁ < μ₂.

The refractive index of glass prism is:
= =
, where v is the speed of the light in glass.
v =

Benzaldehyde condenses with N, N-dimethyl aniline in presence of anhydrous ZnCI₂ to give malachite green.

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

x/m = Kρ^1/n (at constant T)

The structures of pyrrole and pyridine are:

Pyrrole is a weaker base than pyridine because when pyrrole accepts proton it loses its aromatic properties. When it accepts a proton, the sp² hybridised N-atoms are converted to a sp³ hybridised N-atom, which makes the molecule non-aromatic.

Neoprene is synthetic rubber. It is a polymer of chloroprene and is resistant to the action of oils, gasoline and other solvents.

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.

Iron is extracted from iron ore in a huge container called a blast furnace. 
Reactions in which oxygen is removed are called reduction reactions.
CO is a strong reducing agent.
The Iron oxide is reduced by CO at 850°C

This problem includes a conceptual mix of Ionization Energy (IE) and Electron Affinity (EA). This problem can be solved by understanding the exact concepts involved in IE and EA.

The amount of energy involved in losing an electron from the outermost shell is known as IE, while the amount of energy involved in accepting one extra electron is known as EA. We know that:

IE = EA

In process (ii), IE is required to remove the electron:
Cl⁻ → Cl + e⁻

While in process (i), EA is involved:
Cl + e⁻ → Cl⁻

The same amount of energy is released when an electron is absorbed.

So, |IE of Process (ii)| = |EA of Process (i)|.

Since, in mitosis the number of chromosomes in the parent and progeny cells is the same, it is also called as equational division.

Prawn, scorpion and locusta belong to phylum Arthropoda. Arthropoda is the largest phylum of animalia, and includes the insects arachnids, crustaceans and others. It is a group of animals with features including bilaterally symmetrical, triploblostic tube within tube plan, organ system level of organization and metemerically segmented body.

Laminaria (kelp) and Fucus (rock weed) are marine algae. They are the members of the class Phaeophyceae, which is commonly known as brown algae. Brown algae are characterized by their brown pigment, fucoxanthin, which gives them their color.

Phylum Hemichordata consists of a small group of worm-like marine animals with an organ-system level of organization. These animals are bilaterally symmetrical, triploblastic, and coelomate. The excretory function in Hemichordata is carried out through the proboscis gland.

The correct matching of respiratory capacities with their respiratory volumes are:

(A) In human beings, the menstrual cycle ceases around 50 years of age; this is termed as "menopause." Menopause is a natural biological process that marks the end of a woman's reproductive years. During menopause, the ovaries stop releasing eggs, and the levels of certain hormones, such as estrogen and progesterone, decrease. 

(B) The milk produced during the initial few days of lactation is called "colostrum." Colostrum is a specialized form of breast milk that is rich in essential nutrients and antibodies. It contains several bioactive compounds absolutely essential to develop resistance for newborn babies. These antibodies help protect the infant from infections and provide important nutrients for their early growth and development.

(C) At the completion of the "meiosis I" division, the primary oocyte divides into the "secondary oocyte" and a smaller cell called the "first polar body." Meiosis is the process by which germ cells (eggs and sperm) are formed. In meiosis I, the primary oocyte undergoes a reduction in chromosome number, resulting in the formation of the secondary oocyte and the first polar body. The secondary oocyte is the cell that can potentially be fertilized by a sperm, while the first polar body usually degenerates and does not play a direct role in reproduction.

Since the woman is a carrier of the colour blindness allele (XᴄX), and the man is normal (XY), there is a 50% chance that a son will inherit the Xᴄ chromosome and be colour blind, and a 50% chance for a daughter to be a carrier like the mother.

Statement I:
"Dryopithecus was more man-like, and Ramapithecus was more ape-like."

• This is incorrect.

• In reality:

  • Dryopithecus (a Miocene ape, ~12-9 mya) was clearly ape-like, with no significant human-like features.

  • Ramapithecus (now considered part of Sivapithecus, ~12-7 mya) showed some dental features initially thought to be hominin-like (more human), though later research showed it was likely an orangutan ancestor.

Statement II:
"Homo erectus ate more meat than Homo habilis."

• This is correct.

• Evidence shows:

  • Homo habilis (~2.4-1.4 mya) had a mixed diet but primarily ate plants, with some meat (evidenced by simple stone tools and occasional cut marks on bones).

  • Homo erectus (~1.9 mya-110 kya) showed:

    • More advanced hunting tools (Acheulean hand axes)

    • Archaeological evidence of systematic hunting

    • Biological adaptations (smaller gut size) suggesting higher meat consumption

    • Higher protein requirements for their larger brain size

Key Points:

1. Statement I reverses the actual evolutionary relationships - Dryopithecus was an ape, while Ramapithecus showed some (initially overinterpreted) human-like dental features.

2. Statement II correctly reflects the dietary shift between these two Homo species, with erectus being more carnivorous due to:

   • Technological advances

   • Increased brain size

   • Migration to environments requiring higher-calorie foods

Final Answer:

• Statement I is false

• Statement II is true

Plasmids are circular DNA molecules that replicate independently of the bacterial chromosome and typically contain an origin of replication and one or more antibiotic resistance genes. However, plasmids are primarily used in prokaryotic cells (such as bacteria) for gene cloning and expression, not just in eukaryotic cells. Therefore, statement 3 is incorrect.

B cells primarily function in humoral immunity by secreting antibodies that target and neutralize pathogens like viruses. Cytotoxic T cells, on the other hand, are involved in cell-mediated immunity, directly killing virus-infected cells. Therefore, the correct description is that B cells secrete antibodies, while cytotoxic T cells kill virus-infected cells.

Transgenic animals are genetically modified primarily to produce substances that are useful for medical applications, such as proteins, hormones, or other therapeutic agents. This has led to the creation of animals that can produce human proteins like insulin or clotting factors in their milk or blood.

The other options are not correct as transgenic animals are not typically created to increase disease susceptibility, decrease growth rates, or reproduce asexually for uniformity

As shown in the figure, you initially face east, i.e. the direction OA. On turning 100° clockwise, you face the direction OB. On further turning 145° anticlockwise, you face the direction OC. Clearly, OC makes an angle of (145° - 100°), i.e. 45° with OA.

Area of square ABCD = 64 cm²
CD × CD = 64
CD = 8 cm
InΔDCE,
Area of ΔDCE = 1/2× DC × CE = 1/2 × 8 × CE = 4 CE ... (i)
Also,
tan ∠EDC = tan 45 =
Or CE = 8 × tan 45^o
CE = 8 (As tan 45^o = 1)
From (i),
Area of DCE = 4 × 8 = 32 sq. cm

 If 10 sin θ^{_{= 6,}}
sinθ = 6/10 = 3/5

sinθ =
Base = = 4k
tanθ = 3/4
cotθ = 4/3
tanθ+ cotθ = 3/4 + 4/3 =

Required pair of letters are as follows:
(i) BA−  No letter between B and A

(ii) G(A)I− 1 letter between G and I

(iii) R(GAI)N - 3 letter between R and N

(iv) ED− No letter between E and D

Clearly, there are 4 such pairs.

Hence, D is the correct option. 

Hope against hope means cling to a mere possibility.
Hope or wish for with little reason or justification, as in I'm hoping against hope that someone will return my wallet. This expression, based on the biblical “Who against hope believed in hope”
Use of this Phrase as given  -
"they were hoping against hope that he would find a way out"
The most suitable option will be option C , That's why  correct answer is (c).

A long time ago, on a big tree in the lap of the mountain, lived a bird named Sindhuka. It was a rather special bird because its droppings turned into gold as soon as they hit the ground.
The following is possibly the most appropriate title for the story : 
The Bird with the Gold Dropping