SRMJEE Mock Test - 5 (Medical) - JEE MCQ

Equivalent circuit of the given circuit is

Between points C and D resistors 2Ω, 2Ω and 2Ω are in series, therefore their equivalent resistance,
R' = 2 + 2 + 2 = 6Ω
Resistors R` and 6Ω are In parallel, therefore their equivalent resistance is given by

R" = 3Ω
Now between points A and B, 1Ω, 3Ω and 1Ω are in series
Therefore, resultant resistance is
R = 1 + 3 + 1 = 5Ω

For the parallel plate capacitor, the capacitance is given by:

When the battery is disconnected, the charge remains constant. As the separation between the plates is increased, the capacitance decreases, and voltage across the capacitor will increase as we have:

Force Diagram on the Body

As shown in the figure, when the lower block of mass m₂ = 6 kg is subjected to a horizontal constant force while the upper block of mass m₁ = 2 kg is attached to a string, both blocks experience a friction force as shown in the figure.

Since the blocks are in motion, we need to apply the equation of motion by balancing the forces acting on the block. The free-body diagram (FBD) of the block is given below:

Let the applied force be F, and it is given that μ = 0.42. Using Newton’s second law of motion, the net force equation is:

m₂ a = F - f₁ - f₂ ...(1)

Here:

f₁ = μ N₁ = μ m₁ g = 0.42 × 2 × 10 = 8.2 N
This is the frictional force between the two blocks.

f₂ = μ N₂ = μ (m₁ + m₂) g
f₂ = 0.42 × (2 + 6) × 10 = 33.6 N
This is the friction force between the 6 kg block and the ground.

From equation (1):

6 × 1.5 = F - 8.2 - 33.6
⇒ F = 50.8 N

By using of dimensional analysis = n₂u₂ = n₁u₁

= x^-2

Let the initial projectile velocity at A be u, and the velocity when it falls on the same horizontal surface at B be v.

Applying energy conservation between A and B:

(1/2) * m * u² + 0 = (1/2) * m * v² + 0

⇒ u² = v²

⇒ v = u

Thus, the velocity remains the same at B.

From the symmetry of projectile motion, the projectile strikes point B at an angle of 45°.

The angular velocity at point B with respect to the starting point is:

ω = V⊥ / R = V⊥ / AB

Here, V⊥ = u sin 45°, and AB is the range of the projectile motion.

The range of projectile motion is given by:

AB = (u² sin(2θ)) / g

For θ = 45°:

AB = (u² sin(2 × 45°)) / g = u² / g

Thus, the angular velocity is:

The point source and its image due to the reflecting surface create two coherent sources. Interference fringes on a distant screen are circular, as the wavefronts from point sources form circular patterns.
Correct answer: A: circular.

For a zero-order reaction, - (d[A]/dt) = k
Thus, [A]_t – [A]₀ = –kt

HBr reacts the fastest with 2-methyl propan-2-ol due to the formation of stable tertiary carbocation.

Organometallic compounds are those compounds which have metal linked directly with the carbon atom of an organic compound/group. These are formed either by σ-bond or π-bond, or both. Cis-platin is not an organometallic compound as it is an example of coordination complex compound. It is clear from the structures that cis-platin does not contain any metal-carbon bond (as shown):

A: Na⁺, SO₄²⁻
Sodium sulfate (Na₂SO₄) is highly soluble in water because sodium salts are generally soluble. Na⁺ and SO₄²⁻ will not form a precipitate.

B: NH₄⁺, CO₃²⁻
Ammonium carbonate (NH₄)₂CO₃ is soluble in water. Therefore, the ions NH₄⁺ and CO₃²⁻ will not form a precipitate under normal conditions.

C: Na⁺, S²⁻
Sodium sulfide (Na₂S) is soluble in water. However, S²⁻ (sulfide ion) can form insoluble compounds with many metal cations, but Na⁺ is not one of them. Therefore, Na⁺ and S²⁻ will not form a precipitate.

D: Fe³⁺, HPO₄²⁻
Iron(III) phosphate (FePO₄) is insoluble in water, so when Fe³⁺ (iron(III) ion) and HPO₄²⁻ (hydrogen phosphate ion) are mixed in dilute solutions, they will form a precipitate.

Conclusion: The pair that will form a precipitate is Fe³⁺ and HPO₄²⁻.
Correct Answer: (D) Fe³⁺, HPO₄²⁻.

Hydrolysis is any chemical reaction in which a molecule of water breaks one or more chemical bonds. The term is used broadly for substitution, elimination, and solvation reactions in which water is the nucleophile. Thus hydrolysis adds water to break down, whereas condensation builds up by removing water.

1. Understanding Spermatogenesis:
- Spermatogenesis is the process of sperm formation that occurs in the male reproductive system, specifically in the seminiferous tubules of the testes.

2. Identifying Key Hormones:
- The process of spermatogenesis is regulated by several hormones. The main hormones involved are Follicle Stimulating Hormone (FSH), Luteinizing Hormone (LH), and testosterone.

3. Role of FSH:
- FSH is secreted by the anterior lobe of the pituitary gland. Its primary role in spermatogenesis is to stimulate the Sertoli cells in the testes, which support and nourish developing sperm cells.

4. Role of LH:
- LH, also secreted by the anterior pituitary gland, stimulates the Leydig cells in the testes to produce testosterone. This hormone is crucial for the overall process of spermatogenesis.

5. Role of Testosterone:
- Testosterone, produced by the Leydig cells, is essential for the development of male secondary sexual characteristics and also plays a significant role in the maturation of sperm.

6. Conclusion:
- The correct answer to the question is that spermatogenesis and sperm differentiation are under the control of both FSH and testosterone. Therefore, the correct option is "FSH and testosterone."

Final Answer: FSH and testosterone.

The structural gene (z) of the lac operon codes for β-galactosidase. It is responsible for the hydrolysis of polysaccharides. The ‘y’ genes code for permease. It increases the permeability of a cell to β-galactosidase. The ‘a’ genes code for transacetylase.

In the lac operon, gene y codes for permease, which increases cell permeability to β-galactosides. Gene z codes for β-galactosidase, and gene a for transacetylase.

Heterogeneous nuclear RNA or hnRNA is a term that encompasses various types and sizes of RNAs found in the eukaryotic cell nucleus. When this undergoes modifications and processing it will give rise to the mature mRNA that has the inherent code for the protein synthesis (amino acid code). The mRNA in eukaryotes is synthesized in the nucleus and hence should be transported to the cytoplasm and into the ribosome for protein synthesis by the process of translation. 

NCERT Topic: The fully processed hnRNA is called mRNA and is transported out of the nucleus into the cytoplasm for translation.

NCERT Line: "It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation."

Genotypes of different individuals in the given pedigree chart can be illustrated as:

The right granted to the company or organization of which others cannot use it is called biopatent. This right is given as intellectual right to the company.

Topic in NCERT: Biotechnology and its applications

Line in NCERT: "the patent extends to functional equivalents, implying that other people selling basmati rice could be restricted by the patent."

Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like the cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin protein. Each of the peripheral fibrils is a triplet. The adjacent triplets are also linked.

Option 3: Diploid somatic cells.

During mitotic division, also known as cell division, the genetic material in a cell's nucleus is replicated and then the cell divides into two daughter cells. In animal cells, this process occurs in diploid somatic cells, which are cells that contain two copies of each chromosome (one from each parent). Haploid cells, on the other hand, only contain one copy of each chromosome. Haploid cells can be either germ cells, which are cells that give rise to gametes (sperm or eggs), or they can be cells that have undergone meiosis, a type of cell division that reduces the chromosome number by half.

In the given scenario, the laboratory population of fruit flies consists of 40 fruit flies, and during a specified time interval, 4 individuals died. To calculate the death rate, we divide the number of deaths by the total population.

Number of deaths = 4
Total population = 40

Death rate = Number of deaths / Total population
= 4 / 40
= 0.1 individuals per fruitfly per week

Option a) "0.1 individuals per fruitfly per week" is the correct answer because it accurately represents the death rate calculated based on the given information.

Death rate is calculated as deaths divided by population: 4/40 = 0.1 individuals per fruit fly per week.

Topic in NCERT: Population Dynamics

Line in NCERT: "If 4 individuals in a laboratory population of 40 fruitflies died during a specified time interval, say a week, the death rate in the population during that period is 4/40 = 0.1 individuals per fruitfly per week."

Clarias gariepinus, the African catifish has been illegally introduced for aquaculture in India. It is threatening native catfishes of Indian rivers.

According to Oparin-Haldane theory, life originated through bio-chemical processes of atoms that combined to form molecules. Molecules in turn reacted to produce inorganic and organic compounds. Organic compounds interacted to produce all types of macromolecules which organised to form the first form of life or cells.

Topic in NCERT: Origin of life

Line in NCERT: "oparin of russia and haldane of england proposed that the first form of life could have come from pre-existing non-living organic molecules (e.g. rna, protein, etc.) and that formation of life was preceded by chemical evolution, i.e., formation of diverse organic molecules from inorganic constituents."

Systemic circulation involves the flow of oxygenated blood from the heart to the body’s tissues and the return of deoxygenated blood to the heart. In option D, blood flows from the left atrium to the left ventricle, through the aorta and arteries to the tissues, and returns via veins to the right atrium, accurately representing systemic circulation. Options A, B, and C include incorrect pathways involving the pulmonary artery or illogical heart chamber sequences.

• Statement a: Correct. The collecting duct extends from the cortex to the medulla, as per NCERT.
• Statement b: Incorrect. Columns of Bertini are renal columns, not tubular structures. Each kidney has about one million nephrons, which are the complex tubular structures.
• Statement c: Correct. The collecting duct reabsorbs water, especially under antidiuretic hormone (ADH) influence, to produce concentrated urine.

Thus, only statement b is false, making option B the correct answer.

Topic in NCERT: Collecting duct

Line in NCERT: "Collecting Duct: This long duct extends from the cortex of the kidney to the inner parts of the medulla. Large amounts of water could be reabsorbed from this region to produce a concentrated urine."

Calvin used the radioactive isotope of carbon (14C) of Chlorella. He observed that after second exposure of Chlorella in presence of 14 CO2, the radioactive isotope of carbon was found mainly in 3-phosphoglyceric acid (PGA). This clearly indicated that the three carbon compound 3- phosphoglyceric acid was the first stable product photosynthesis and the carbon from the CO2 is fixed in the preparation of the photosynthates.
Hence, the correct answer is 'PGA'.

Topic in NCERT: Photosynthesis in higher plants

Line in NCERT: "the first product identified was 3-phosphoglyceric acid or in short pga."

∠CAB + ∠ABC + ∠BCA = 180°
80° + 60° + ∠BCA = 180°
∠BCA = 180° – 140° = 40°
∠BCA = 40°
x = 40/2= 20°
∠CBD = 60°/2 = 30°
y = 180° – 50° = 130°
So, 20° and 130° are the correct values.

∠PQR = 50° (Alternate Segment Theorem)
∠PQO + ∠OQR = 50°
30° + ∠OQR = 50°
∠OQR = 20°
Since PO = OQ, (Radii of the same circle)
∠OPQ = ∠OQP = 30°
∠OQR = ∠ORQ = 20°

According to the given information in the question, diagram can be drawn as,

B is the mother of H. C is the nephew of E. A is the husband of H. A is third to the left of H. Both the neighbours of C are females. F and G are daughters of H.
Thus, option D is the correct answer.

Let the cost price of the tablet = x Rupees

So, the cost price of the mobile = 1.5x Rupees

Total cost price of both articles = x + 1.5x = 2.5x

Selling Prices:
The tablet is sold at an 8% loss, so its selling price = (92/100) × x = 0.92x
The mobile phone is sold at a 10% profit, so its selling price = (110/100) × 1.5x = 1.65x
Total Selling Price:
Total selling price = 0.92x + 1.65x = 2.57x

Gain Calculation:
Gain = Selling Price - Cost Price
= (2.57x - 2.5x) = 0.07x

Gain Percentage:
(0.07 / 2.5) × 100 = 2.8%

In the passage, “deferred” refers to purchase plans delayed to the future. “Postponed” means delayed, making option C the correct synonym.

In the passage, “ramp up” refers to increasing production. The opposite is to decrease, and “dampen” means to reduce or make less, making it the correct choice. “Escalate” and “intensify” are synonyms, and “dole out” refers to distributing, not reducing.