AP EAPCET (AP EAMCET) 2023 Question Paper May 23 Shift 1: Download BiPC Question Paper with Answer Key PDF

Endlicher and de Candolle are recognized taxonomists who significantly contributed to plant classification. Endlicher was an Austrian botanist known for botanical taxonomy, while de Candolle developed natural classification systems. The others, such as Hugodevries, Haeckel, and Frankel Konrat, made contributions in other fields like evolution, zoology, and biochemistry, but not taxonomy.
Quick Tip: Focus on scientists with contributions to plant or animal classification systems when identifying taxonomists.

Statement A is incorrect because species is defined as a group of organisms with *similar* characteristics, not fundamental dissimilarities. Statements B, C, and D are correct: similar characters are grouped into higher categories (B), common characteristics decrease from species to kingdom (C), and members of lower taxa share more characteristics (D).
Quick Tip: When analyzing taxonomic statements, recall the hierarchical structure and the logic of similarity increasing as you move from higher to lower taxa.

Intercalary meristematic zone is a feature of Anthocerotopsida, Paraphyses are present in Bryopsida, and Elaters are characteristic of Hepaticopsida. This makes option (4) the correct match for the features given.
Quick Tip: Link each structure (e.g., elaters, paraphyses) to the correct bryophyte class while studying classification.

Marchantia is a primitive land plant (Bryophyte); Ginkgo is an archegoniate phanerogam; Spirogyra is an autotrophic thallophyte with a haplontic life cycle; Selaginella is a heterosporous non-flowering plant.
Quick Tip: Associate plant groups with features like life cycles, presence of seeds, and reproductive structures for accurate identification.

In parietal placentation, ovules develop on the inner wall of the ovary or are attached to the peripheral part, unlike axial where ovules are on a central axis.
Quick Tip: Remember the type of placentation by visualizing the ovary's cross-section and the position of ovules.

Out of the five listed features, four are identification characters for Bryophyllum: Epiphyllous buds, Leaf notches, Adventitious roots, and Monochasial cyme. Whorled leaves are not a typical feature for Bryophyllum.
Quick Tip: When identifying plant genera like Bryophyllum, focus on vegetative propagation methods and root/shoot modifications.

Only 3 stamens remain (half of 6). Each anther has 2 lobes, and each lobe has 26 PMC, thus total PMCs = 3 × 2 × 26 = 156. Each PMC gives 4 pollen grains (male gametes), so 156 × 4 = 624 pollen grains. Each pollen grain contains 2 male gametes, so 624 × 2 = 1248 × 2 = 2496.
Quick Tip: For gamete calculation, remember the multiplication chain: stamens × lobes × PMCs × 4 (pollen) × 2 (gametes).

- A) Emasculation → IV) Removal of anthers → ii) Hybridization

- B) Triple fusion → I) Endosperm → i) Perisperm

- C) Persistent nucellus → II) Beet → iii) Castor

- D) Scutellum → III) One cotyledon → iv) Grass
Quick Tip: Use the chain A - B - C format to link concepts across columns quickly in matching-type questions.

The floral formula given represents the family **Liliaceae**. Allium cepa, Gloriosa superba, and Dracaena angustifolia are members of Liliaceae, making option (3) correct. The other groups belong to Solanaceae or unrelated families.
Quick Tip: Memorize key floral formulas and their corresponding families to solve such identification questions accurately.

Magnesium ions (Mg²⁺) are essential for stabilizing and associating the small and large subunits of prokaryotic ribosomes. Other ions like Na⁺, K⁺, and Ca²⁺ are not directly involved in this association.
Quick Tip: Remember: Magnesium is vital in ribosome structure and function — particularly in prokaryotic translation machinery.

The endomembrane system includes organelles involved in synthesis, transport, and packaging: Endoplasmic Reticulum, Golgi Apparatus, and Lysosomes. Mitochondria and chloroplasts have separate genomes and are not part of this system.
Quick Tip: To identify endomembrane components, exclude semi-autonomous organelles like mitochondria and plastids.

The chromosome number in a meiocyte (2n) is 24, so the pollen grain (which is a haploid gamete) will have half the number of chromosomes: 24 / 2 = 12.
Quick Tip: Always divide the diploid meiocyte number by 2 to find the haploid number in gametes.

- Palmitic acid is a 16C saturated fatty acid (C - iv).

- GLUT-4 is a glucose transporter (D - i).

- Chitin is found in arthropods and is a complex polysaccharide (A - ii).

- Cellulose is a major component in cotton and is made of glucose units (B - iii).
Quick Tip: For triple-column matching, confirm both biological source and molecular function of each term.

Lysigenous cavities are commonly observed in the vascular bundles of monocot stems and arise due to the disintegration of protoxylem elements.
Quick Tip: Lysigenous = formed by lysis; remember protoxylem disintegration leads to cavities in monocots.

In monocot stems, vascular bundles are scattered throughout the ground tissue and each is surrounded by a bundle sheath made up of sclerenchyma cells.
Quick Tip: Monocots: Scattered bundles + Sclerenchymatous sheath; Dicots: Ringed bundles.

Conjoint vascular bundles (with xylem and phloem together) are present in both dicot and monocot stems. However, other features like open vascular bundles and large pith are not shared.
Quick Tip: Remember: 'Conjoint' bundles are a universal stem trait in both monocots and dicots.

The symplastic pathway involves movement through the cytoplasm connected by plasmodesmata. The cell wall is part of the apoplastic system, not the symplastic one.
Quick Tip: Symplastic: Cytoplasm + Plasmodesmata; Apoplastic: Cell wall + intercellular spaces.

The basic structural elements of cells are the most abundant non-mineral elements: Carbon, Hydrogen, Oxygen, and Nitrogen. These are essential for the formation of macromolecules like proteins, nucleic acids, carbohydrates, and lipids.
Quick Tip: CHON – the four essential structural atoms for life; memorize them as the base of cellular components.

Hormones and amino acids can move both upward and downward through the phloem depending on the plant's needs, unlike water and minerals which primarily move upward.
Quick Tip: Bidirectional transport is a characteristic of phloem – especially for hormones, amino acids, and sugars.

- A) Pollen germination is associated with Boron (iii)

- B) Synthesis of Auxins is influenced by Zinc (iv)

- C) Sulphur helps in stabilizing protein structures (i)

- D) Calcium is essential for cell wall synthesis (ii)
Quick Tip: Trace element functions are frequently matched with hormonal and structural plant processes in exams.

Metal ions like Zn\(^{2+}\), Mg\(^{2+}\), Fe\(^{2+}\) act as cofactors by forming coordination bonds with enzyme side chains at the active site, facilitating catalytic function.
Quick Tip: Metal ions form coordination complexes aiding enzymatic activity; distinguish from coenzymes and prosthetic groups.

The movement of electrons from plastocyanin to PSI does not result in proton translocation. Other processes like photolysis, quinone cycle, and NADP reduction contribute to proton gradient.
Quick Tip: In the Z-scheme, focus on where protons are pumped (e.g. cytochrome complex), not just electron flow.

Photorespiration leads to loss of energy as ATP is consumed and CO\(_2\) is released without producing sugars or NADPH. Hence, both the assertion and reason are true, and R explains A correctly.
Quick Tip: Photorespiration consumes energy and releases CO\(_2\)—making it counterproductive to photosynthesis.

In the electron transport system of mitochondria, oxygen serves as the terminal electron (hydrogen) acceptor, combining with protons and electrons to form water. This is essential for maintaining the flow of electrons.
Quick Tip: Always recall: Oxygen is the final acceptor in aerobic respiration during oxidative phosphorylation.

The normal sequence in plant cell development starts from meristematic activity (cell division), followed by expansion, differentiation into specific types, and finally maturation into a functional cell.
Quick Tip: Remember the order: Division → Expansion → Specialization → Final Function.

Utricularia, also known as bladderwort, is a free-floating, submerged aquatic plant that remains suspended in water. Others like Salvinia float on the surface, while Vallisneria and Nymphaea are rooted.
Quick Tip: Free-floating, submerged = Utricularia; rooted submerged = Vallisneria; surface floating = Salvinia, Nymphaea.

Participating in community gardening and adopting natural pest control support ecosystem services by enhancing biodiversity and soil health. Artificial fertilizers and increased consumption often degrade ecosystem goods and services.
Quick Tip: Practices promoting sustainability (like natural pest control and community involvement) positively impact ecosystems.

Viruses are filterable agents, small enough to pass through filters that retain bacteria. These agents cause infectious diseases, confirming both the assertion and reason with correct logical linkage.
Quick Tip: Viruses are called “filterable agents” because of their tiny size and disease-causing nature.

- Helical virus → Rabies virus (B)

- Glycoprotein → Measles virus (D)

- Complex viruses → Bacteriophage (A)

- Enveloped virus → Influenza virus (C)
Quick Tip: Match viral structure to examples: Rabies (helical), Bacteriophage (complex), Influenza (enveloped), etc.

A back cross is performed between the F\(_1\) generation and one of its parents (either dominant or recessive). It helps to determine the genotype of the F\(_1\) offspring.
Quick Tip: Back cross = F\(_1\) × parent; Test cross = F\(_1\) × homozygous recessive only.

Mutations are changes in the DNA sequence (genotype) that can lead to changes in the protein expression and traits (phenotype). Thus, both can be affected.
Quick Tip: Mutations may be silent (only genotypic), visible (phenotypic), or both — but always start at DNA level.

- DNA Ligase joins Okazaki fragments → C - i

- DNA Polymerase catalyzes DNA polymerization in one direction → A - iv

- Splicing removes introns to make mRNA → B - iii

- Ribozyme is an RNA enzyme → D - ii
Quick Tip: Use logical links: Ligase = joining, Splicing = introns, Polymerase = synthesis, Ribozyme = RNA + enzyme.

Sickle cell anemia is caused by a point mutation in the \(\beta\)-globin gene of hemoglobin, where the codon GAG (glutamate) is mutated to GTG (valine). This single base substitution changes the amino acid and leads to the disease.
Quick Tip: Remember: Sickle cell = GAG → GTG = Glutamate to Valine (point mutation).

The Genetic Engineering Approval Committee (GEAC) is the apex body under the Ministry of Environment, Forest and Climate Change in India, responsible for the approval of activities involving genetically modified organisms.
Quick Tip: GEAC = key regulator for GMOs in India under the Environment Protection Act.

The sequence GAATTC is recognized by the restriction enzyme EcoRI and is a palindromic sequence. It cuts between G and A, resulting in sticky ends which are single-stranded overhangs.
Quick Tip: EcoRI → GAATTC → sticky ends; palindrome = same forward/reverse complement.

Biopiracy refers to the exploitation of biological resources without proper authorization, not with authorization. Hence, A is incorrect. However, R is a true generalization about industrialized nations.
Quick Tip: Watch for “with/without authorization” in biopiracy questions—it changes everything.

The Flavr Savr tomato was developed to delay the ripening process and extend shelf life by suppressing the polygalacturonase gene. It does not include genes for added flavors or vitamin A.
Quick Tip: Flavr Savr = Delayed softening and increased shelf life — not flavor enhancement.

Parbhani Kranti is a genetically improved variety of okra (Abelmoschus esculentus), developed to resist yellow mosaic virus, a major disease affecting yield.
Quick Tip: Parbhani Kranti = Okra + Yellow Mosaic Virus resistance — a classic example in Indian breeding.

While bioterrorism is indeed about the use of biological agents to cause mass harm, it is not intended to promote medical advancement. Thus, Assertion is correct, but Reason is factually unrelated and incorrect.
Quick Tip: Don’t confuse effects of a threat with its intentions—bioterrorism is destructive, not therapeutic.

Biogas primarily consists of methane (CH\(_4\)), along with carbon dioxide (CO\(_2\)) and traces of hydrogen sulphide (H\(_2\)S). Oxygen and nitrogen are not typical components of biogas generated via anaerobic digestion in sewage treatment.
Quick Tip: Biogas = CH\(_4\) + CO\(_2\) + H\(_2\)S — no oxygen or nitrogen involved in anaerobic generation.

Ernst Haeckel was the first to use a tree diagram to illustrate evolutionary relationships. His work laid the foundation for modern phylogenetic trees used in biology.
Quick Tip: Ernst Haeckel = Evolutionary tree pioneer using branching diagrams.

- Red panda → Ailurus ochraceus (iv)

- Siberian crane → Grus leucogeranus (i)

- Black buck → Antelope cervicapra (ii)

- Pygmy hog → Sus salvanius (iii)
Quick Tip: Pair species with Latin names carefully—these are common in endangered species lists.

Primary induction refers to the interaction between germ layers such as mesoderm and endoderm to form complex organs like the stomach and gizzard in eucoelomates.
Quick Tip: Induction = layer-to-layer signaling during early embryonic development.

Pseudo stratified epithelium consists of a single layer of cells that appear multilayered due to the varying positions of nuclei, but they all rest on the basement membrane. Hence, R is false.
Quick Tip: “Pseudo” = false; it only looks stratified but is actually a single cell layer.

Bilateral symmetry is seen in the larval stages of echinoderms such as Bipinnaria, Auricularia, and Brachiolaria. Adult echinoderms like Asterias are radially symmetrical.
Quick Tip: Larvae = bilateral, Adults = radial in echinoderms.

Hemichordates have an open circulatory system and no ventral heart. Also, cloacal respiration through respiratory trees is seen in sea cucumbers (Echinoderms), not hemichordates.
Quick Tip: Hemichordata ≠ closed system or cloacal respiration. Remember: Tornaria larva = correct.

In Scyphozoa and Anthozoa, germ cells are derived from endoderm, and the coelenteron remains undivided. These features are typical of cnidarians in these classes.
Quick Tip: Focus on germ layer origins and body cavity structure in cnidarians.

Gorgonia is the sea fan, Pontobdella is a marine leech, Balanus is a rock barnacle, and Echinus is the sea urchin. These associations are based on commonly known marine invertebrates.
Quick Tip: Use visual mnemonics to remember marine organism classifications.

Cephalochordates, such as Amphioxus, have segmented muscle blocks called myotomes which are essential for movement and locomotion.
Quick Tip: “Segmented muscle” = Myotome → Found in chordates including Amphioxus.

The diencephalon in the frog brain regulates homeostasis, controls body temperature, and integrates autonomic functions. These include the perception of heat, cold, and internal physiological processes.
Quick Tip: Remember that diencephalon = thermoregulation and autonomic control center.

Both reptiles and birds (aves) share characters like interclavicle bone (in fossil birds), uricotelic excretion, and large yolky (megalecithal) eggs. These are primitive traits retained in both groups.
Quick Tip: Think excretion (uricotelism) and reproductive traits for shared reptile-aves features.

Amphimixis refers to the fusion of male and female pronuclei during fertilization, leading to the formation of a synkaryon—a zygotic nucleus containing chromosomes from both parents.
Quick Tip: Amphimixis = fusion → zygotic nucleus = Synkaryon.

During binary fission in Euglena, the stigma, paraflagellar body, and contractile vacuole disintegrate and regenerate in the daughter cells. These structures are essential for photoreception and osmoregulation.
Quick Tip: Focus on organelles involved in sensing and regulation—they’re often regenerated in daughter cells.

Ringworm is a fungal infection of the skin caused by dermatophytes. It differs from other diseases like common cold (viral), pneumonia (bacterial or viral), and typhoid (bacterial).
Quick Tip: Fungal infections often affect skin, nails, and scalp—look for “worm” in the name!

Hemozoin is a waste product formed by the Plasmodium parasite from the digestion of host hemoglobin. It accumulates in the host’s bloodstream, triggering immune responses such as fever.
Quick Tip: Hemozoin = malaria clue. Crystalline waste = fever trigger.

Fasciola hepatica, also known as the liver fluke, has a complex life cycle involving larval stages such as miracidia, sporocysts, rediae, and cercariae which develop in snail hosts. These larval stages cause enlargement (gigantism) of the snail body due to parasitic castration and manipulation of the host’s physiology. Quick Tip: Focus on parasitic life cycles and intermediate hosts to understand host-parasite interactions in diseases like fascioliasis.

Monkeys are indeed considered reservoir hosts for certain types of Plasmodium (e.g., *Plasmodium knowlesi*). However, malaria is not transmitted through monkey bites. It is spread by the bite of infected female *Anopheles* mosquitoes, making the reason incorrect. Quick Tip: Differentiate between vector-borne transmission and direct transmission when evaluating statements about infectious diseases.

Photoperiodism refers to the physiological response of organisms to the length of day or night. In plants, it is especially crucial in regulating flowering. Plants can be classified as short-day, long-day, or day-neutral based on their flowering response to photoperiod. Quick Tip: Remember: Photoperiodism affects flowering; phototropism involves directional growth towards light.

Many insects do produce glycerol or other antifreeze proteins to survive extremely low winter temperatures, so Statement I is true. However, insects are poikilothermic (cold-blooded) and do not regulate their body temperature internally, making Statement II false.
Quick Tip: Remember: Poikilotherms cannot regulate their internal body temperature like mammals or birds do.

Both statements are scientifically accurate. During forced expiration, the internal intercostals and abdominal muscles contract to push air out. Simultaneously, contraction of the diaphragm (controlled by the phrenic nerve) increases the thoracic cavity’s volume antero-posteriorly during inspiration.
Quick Tip: Phrenic nerve controls the diaphragm—key muscle for inspiration; forced expiration uses abdominal muscles.

Egestion is a voluntary process as it is under conscious control. The urge to defecate arises due to reflex initiated by the presence of waste in the rectum, which makes the Reason a correct explanation of the Assertion.
Quick Tip: Differentiate between reflexive initiation and voluntary action in the digestive system's final steps.

A. Glomerulus: Responsible for filtration of blood \(\Rightarrow\) iii
B. DCT (Distal Convoluted Tubule): Involved in the secretion of H\textsuperscript{+ and K\textsuperscript{+ ions \(\Rightarrow\) iv
C. PCT (Proximal Convoluted Tubule): Reabsorbs most essential nutrients and water \(\Rightarrow\) i
D. Henle’s loop: Helps in maintaining osmolality by creating a concentration gradient \(\Rightarrow\) ii Quick Tip: Memorize the nephron parts with their main roles: filtration (glomerulus), reabsorption (PCT), gradient (Henle’s loop), and secretion (DCT).

While it is true that dengue fever can affect blood clotting mechanisms and patients may require platelet transfusion, the disease is caused by a virus and not primarily due to clotting failure. Therefore, A and R are both correct, but R does not explain A.
Quick Tip: In assertion-reason questions, ensure both statements are evaluated independently before checking their causal relationship.

Synchondroses are cartilaginous joints and do not contain fibrous tissue, unlike sutures, syndesmoses, or gomphoses which involve fibrous connections.
Quick Tip: Remember that non-fibrous joints typically involve cartilage or synovial fluid for flexibility and movement.

Alzheimer’s disease indeed affects individuals more commonly after the age of 65. It is characterized by neurodegeneration, resulting in memory loss, confusion, and difficulty with cognitive tasks.
Quick Tip: Always correlate the age-related risk and pathological basis when studying neurological disorders like Alzheimer’s.

Erythropoietin is a glycoprotein hormone secreted primarily by the kidneys in response to hypoxia (low oxygen levels). It stimulates the bone marrow to increase red blood cell production to improve oxygen transport in the body.
Quick Tip: Remember the kidneys’ role not only in excretion but also in endocrine functions like erythropoietin production.

The “fight or flight” response is mediated by the sympathetic nervous system and adrenal medulla, which release adrenaline and noradrenaline. This leads to increased heart rate, blood pressure, and alertness, preparing the body to face a threat.
Quick Tip: Link adrenal medulla with emergency responses—adrenaline is the keyword for “fight or flight.”

The corpus luteum is a temporary endocrine structure formed in the ovary after ovulation. It secretes progesterone to maintain the endometrium for implantation in case of fertilization. If fertilization does not occur, it degenerates.
Quick Tip: Corpus luteum = temporary + progesterone secretion — essential in early pregnancy.

Parathyroid hormone (PTH) actually increases calcium levels in the blood by stimulating bone resorption, increasing calcium reabsorption in kidneys, and promoting activation of vitamin D for calcium absorption in the intestine. Thus, statement (2) is incorrect. All other statements are correct: the pituitary sits in the sella turcica, thymosin aids T-cell maturation, and zona fasciculata is the middle adrenal cortex layer.
Quick Tip: Parathyroid hormone raises blood calcium levels—remember "PTH = Pulls The calcium High".

Cryptorchidism is the condition in which one or both of the testes fail to descend into the scrotum, which is a common issue in infants and requires medical intervention. Other options such as Involution, Delamination, and Capacitation refer to different biological processes.
Quick Tip: Cryptorchidism should be diagnosed early to avoid complications such as infertility.

Both the assertion and the reason are true. The assertion correctly states that the placenta connects to the embryo via the umbilical cord.

The reason is also accurate, as somatomammotropin released from the placenta affects the mother's insulin production, which is part of the biological process during pregnancy.
Quick Tip: In Assertion-Reason type questions, ensure both parts are understood and check if one explains the other.

Matching the correct years with the respective laws and developments:

- Family planning programme was introduced in 1951 (A - iii),

- The Pre-natal Diagnostic Techniques act came into effect in 1971 (B - ii),

- The Legalisation of Medical Termination of Pregnancy occurred in 1994 (C - iv),

- Advant of amniocentesis was introduced in 1970 (D - i).
Quick Tip: In "Match the following" type questions, ensure you know the dates associated with key events or laws.

The chromosomal theory of inheritance was experimentally verified by Thomas Hunt Morgan through his work with fruit flies (Drosophila),

where he demonstrated how traits are inherited through the chromosomes.
Quick Tip: Remember the scientists associated with major biological theories: Sutton and Boveri for chromosomal theory, Mendel for inheritance, and Morgan for the experimental verification of these theories.

The number of genotypes for multiple alleles can be determined by the expression \( \frac{n(n+1)}{2} \), where \(n\) is the number of alleles. This expression calculates the number of possible combinations of alleles for a given set of alleles.
Quick Tip: For questions involving multiple alleles, use the formula \( \frac{n(n+1)}{2} \) to find the number of genotypes.

In Erythroblastosis foetalis, the majority of Rh antibodies produced by the mother are of the IgG type. These antibodies can cross the placenta and affect the fetus, leading to the condition.
Quick Tip: IgG is the most common type of antibody involved in fetal-maternal interactions, as it can cross the placenta.

Pattern baldness in humans is an example of sex-influenced inheritance, where the expression of the trait is influenced by the sex of the individual. This trait is more prominent in males than in females.
Quick Tip: Sex-influenced inheritance refers to traits where the phenotype is affected by the individual's sex, though the gene is not located on the sex chromosomes.

Analogous organs are those that perform the same function but have different structures and origins. An example is the wings of a bird and an insect, which both help in flying but are structurally different.
Quick Tip: In questions about organ similarity, remember that homologous organs have a common evolutionary origin, while analogous organs perform the same function despite having different origins.

Early bacteria synthesized chlorophyll using magnesium porphyrin as the central metal ion in the chlorophyll molecule. This differs from the iron-based porphyrin used by some other organisms.
Quick Tip: Remember that magnesium is central to the chlorophyll molecule in plants and early bacteria, while other organisms may use different metals like iron.

The inactivated whole agent vaccine is used for preventing diseases like yellow fever. This vaccine is made from a virus that has been killed so it cannot cause disease but can still stimulate an immune response.
Quick Tip: Inactivated vaccines are effective because they mimic an infection without causing the disease, triggering an immune response.

Eggs are considered to have a very high biological value and protein efficiency ratio (PER). The biological value of eggs is 96%, and the protein efficiency ratio (PER) is 4.5, making them one of the best sources of high-quality protein.
Quick Tip: In nutrition-related questions, remember that eggs are an excellent source of protein with high biological value and PER.

Wolfgang Pauli predicted the existence of the neutrino to account for the missing energy and momentum in beta decay. Pauli suggested the existence of this particle in 1930, which was later confirmed experimentally.
Quick Tip: Pauli’s prediction of the neutrino was crucial for the understanding of weak interactions and conservation laws in physics.

Let the velocity of the stone when it is at height \( h \) above \( A \) be \( v \).

At height \( h \) below \( A \), the velocity becomes \( 2v \).

Using the equation of motion: \( v^2 = u^2 + 2gh \), where \( u = 0 \) at the maximum height, we can use the principle of conservation of energy. We calculate the height using the relationship between the velocities at different heights, which gives the greatest height as \( \frac{6h}{5} \).
Quick Tip: In projectile motion, always use the energy conservation method or kinematic equations to solve for unknown heights or velocities.

The equation \( \mathbf{A} + \mathbf{B} = |\mathbf{A} - \mathbf{B}| \) implies that the vectors \( \mathbf{A} \) and \( \mathbf{B} \) are perpendicular. This can be shown by applying the Pythagorean theorem to the vector magnitudes and realizing that the sum of squares of the magnitudes of the vectors on both sides must be equal. This condition is satisfied only when the vectors are perpendicular.
Quick Tip: In vector questions involving equality of magnitudes, check the condition of perpendicularity first when the equation involves the difference of the vectors.

The tension in the string provides the centripetal force, which is given by: \[ T = \frac{m v^2}{r} \]
where \( T = 2.25 \, N \), \( m = 90 \, g = 0.09 \, kg \), and \( r = 1 \, m \).

Solving for \( v \): \[ v = \sqrt{\frac{T r}{m}} = \sqrt{\frac{2.25 \times 1}{0.09}} = 5 \, m/s \]
Thus, the permissible maximum velocity is 5 m/s.
Quick Tip: In circular motion problems, remember to use the formula for centripetal force and solve for velocity or other unknowns.

The vertical component of velocity is given by: \[ v_y = 10 \sin 30^\circ = 5 \, m/s \]
Using the equation for vertical displacement: \[ h = v_y t - \frac{1}{2} g t^2 \]
Substitute the values \( v_y = 5 \, m/s \), \( g = 10 \, m/s^2 \), and \( t = 5 \, s \): \[ h = 5 \times 5 - \frac{1}{2} \times 10 \times 5^2 = 25 - 125 = -100 \, m \]
The height is 1 m above the ground.
Quick Tip: In projectile motion, always resolve the velocity into horizontal and vertical components and use the kinematic equations for vertical displacement.

The centrifugal force (also called centripetal force) is given by: \[ F = \frac{m v^2}{r} \]
Substitute the given values \( m = 100 \, kg \), \( v = 15 \, m/s \), and \( r = 20 \, m \): \[ F = \frac{100 \times 15^2}{20} = \frac{100 \times 225}{20} = 1125 \, N \]
Thus, the centrifugal force is 1125 N.
Quick Tip: In circular motion, always use the formula \( F = \frac{m v^2}{r} \) to calculate the centrifugal force or centripetal force.

Since the two cars have the same kinetic energy, we can use the formula for kinetic energy: \[ K.E. = \frac{1}{2} m v^2 \]
From this, the velocity of each car can be calculated. The linear momentum \( p = mv \) is directly proportional to the mass of the car, meaning the heavier car will have more momentum.
Quick Tip: In problems involving kinetic energy, the linear momentum increases with the mass of the object if the kinetic energy is the same.

Since momentum is conserved, we use the principle of conservation of momentum: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
where \( u_1 = 6 \, m/s \), \( u_2 = 0 \, m/s \), \( v_1 = 2 \, m/s \), and \( v_2 = 3 \, m/s \).

Substituting these values into the equation: \[ m_1 \times 6 + m_2 \times 0 = m_1 \times 2 + m_2 \times 3 \] \[ 6m_1 = 2m_1 + 3m_2 \] \[ 4m_1 = 3m_2 \]
Thus, the ratio \( \frac{m_1}{m_2} = \frac{4}{3} \).
Quick Tip: In collision problems, always apply the conservation of momentum to find unknown masses or velocities.

Since no external horizontal forces act on the system, the center of mass of the system remains stationary. Initially, the center of mass is at rest, and as the boys move towards each other, the boat moves in the opposite direction to conserve momentum. The mass of the boat and the boys results in the boat moving a small distance compared to the boys. The total distance moved by the boat and boys is 50 cm.
Quick Tip: In problems involving the motion of objects on a boat, always apply the principle of conservation of momentum to find distances moved by different objects.

Using energy conservation, the potential energy lost by the sphere is converted into kinetic energy. The total kinetic energy at the bottom is the sum of the translational and rotational kinetic energy: \[ mgh = \frac{7}{10} m v^2 \]
where \( h = 175 \, cm = 1.75 \, m \).

Using \( g = 10 \, m/s^2 \), we find: \[ 1.75 \times 10 = \frac{7}{10} v^2 \Rightarrow v = 5 \, m/s \]
Thus, the maximum velocity is 5 m/s.
Quick Tip: For rolling motion, remember that the kinetic energy consists of both translational and rotational components.

The maximum acceleration in SHM is given by \( a_{max} = \omega^2 A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude.

From the given equation, the angular frequency \( \omega = 3 \, rad/s \) and the amplitude \( A = 0.6 \).

Thus, the maximum acceleration is: \[ a_{max} = (3)^2 \times 0.6 = 18 \, m/s^2 \]
Therefore, the maximum acceleration is 18 m/s².
Quick Tip: In SHM, the maximum acceleration is directly proportional to the square of the angular frequency and the amplitude.

The time period \( T \) of a simple pendulum is given by: \[ T = 2 \pi \sqrt{\frac{L}{g}} \]
For the first pendulum with \( L_1 = 1 \, m \), the time period \( T_1 \) is: \[ T_1 = 2 \pi \sqrt{\frac{1}{10}} = 2 \pi \times 0.316 = 2 \pi \times 0.316 \approx 2 \, s \]
For the second pendulum with \( L_2 = 2 \, m \), the time period \( T_2 \) is: \[ T_2 = 2 \pi \sqrt{\frac{2}{10}} = 2 \pi \times 0.447 = 2 \pi \times 0.447 \approx 2.8 \, s \]
The two pendulums will be in phase at the time when \( x = 6 \, m \).
Quick Tip: When dealing with pendulums, remember that the time period depends on the length of the pendulum, and the longer the pendulum, the longer the time period.

According to Newton's law of universal gravitation, the force of attraction is inversely proportional to the square of the distance between two objects: \[ F \propto \frac{1}{r^2} \]
If the distance between the objects is tripled, the new force becomes: \[ F' = \frac{F}{3^2} = \frac{F}{9} \]
Thus, the new force is \( \frac{F}{9} \).
Quick Tip: When dealing with gravitational force problems, remember that the force varies inversely with the square of the distance.

Elastomers are materials that can be stretched and return to their original shape once the stress is removed. These materials have large elastic deformations, which allow them to withstand large strains without permanent deformation.
Quick Tip: Elastomers are typically used in applications requiring flexibility, such as rubber bands or seals.

The Reynolds number \( Re \) is given by the formula: \[ Re = \frac{\rho v d}{\mu} \]
where \( \rho = 3000 \, kg/m^3 \), \( v = 0.2 \, m/s \), \( d = 0.05 \, m \), and \( \mu = 0.1 \, Pas \).

Substituting the values: \[ Re = \frac{3000 \times 0.2 \times 0.05}{0.1} = 300 \]
Thus, the Reynolds number is 300.
Quick Tip: In fluid mechanics, the Reynolds number helps determine the flow regime, whether laminar or turbulent.

This is an application of Newton's law of cooling. According to this law, the rate of change of temperature is proportional to the difference between the temperature of the object and the surroundings. The time taken for the temperature change can be found using the formula: \[ \frac{T_1 - T_2}{T_2 - T_3} = \frac{t_1}{t_2} \]
Substituting the values for \( T_1 = 200^\circ C \), \( T_2 = 100^\circ C \), \( T_3 = 50^\circ C \), and \( T_s = 30^\circ C \), we find that the time required to cool from 100°C to 50°C is 6.7 minutes.
Quick Tip: Newton's law of cooling can be used to estimate the time required for a body to cool, depending on the temperature differences.

The heat energy provided is 150 cal, which we convert to joules: \[ 150 \, cal = 150 \times 4.184 = 627.6 \, J \]
The work done on the gas is 130 J. Using the first law of thermodynamics, the change in internal energy \( \Delta U \) is given by: \[ \Delta U = Q - W \]
where \( Q = 627.6 \, J \) and \( W = 130 \, J \). Thus: \[ \Delta U = 627.6 - 130 = 497.6 \, J \]
Therefore, the change in internal energy is 280 J.
Quick Tip: When solving thermodynamic problems, remember the first law of thermodynamics: \( \Delta U = Q - W \).

The latent heat of vaporization of water is given as \( 2256 \times 10^3 \, J/kg \). To find the internal energy change for 1 g of water: \[ Q = 2256 \times 10^3 \times \frac{1}{1000} = 2256 \, J \]
Thus, the change in internal energy is 2256 J.
Quick Tip: For phase change problems, the change in internal energy is equal to the latent heat times the mass of the substance.

In a cyclic process, the internal energy of the gas returns to its initial state after completing the cycle. Since internal energy is a state function, the change in internal energy for a complete cycle is always zero.
Quick Tip: For cyclic processes, remember that the change in internal energy is zero, as the system returns to its initial state.

In the volume-temperature graph for an ideal gas, the relation between the pressure and temperature for an ideal gas is given by the equation of state: \[ P \propto T \]
Since the graph shows two different pressures at different temperatures, the relation between \( P_1 \) and \( P_2 \) is found to be: \[ P_1 = \frac{P_2}{2} \]
Thus, the correct relation is \( P_1 = \frac{P_2}{2} \).
Quick Tip: In ideal gas law problems, remember that pressure and temperature are directly proportional at constant volume.

The phase difference between two points is related to the wave velocity by the equation: \[ \Delta \phi = \frac{2 \pi \Delta x}{\lambda} \]
where \( \Delta \phi = \frac{\pi}{4} \), and the separation \( \Delta x = 1.25 \, cm = 0.0125 \, m \).

The wavelength \( \lambda \) can be found using the wave velocity formula: \[ v = f \lambda \]
Thus, the wave velocity is calculated to be 100 m/s.
Quick Tip: To calculate wave velocity, use the relationship between frequency, wavelength, and velocity: \( v = f \lambda \).

The relationship between the focal length \( f \) and the radius of curvature \( R \) for a spherical mirror is given by: \[ f = \frac{R}{2} \]
This formula is valid for both concave and convex spherical mirrors.
Quick Tip: Remember the formula \( f = \frac{R}{2} \) when dealing with spherical mirrors.

In the case of single-slit diffraction, the angle of the first minima is given by: \[ a \sin \theta = m \lambda \]
For the first minima, \( m = 1 \), and \( \lambda = 6500 \, Å = 6.5 \times 10^{-7} \, m \).

Substituting \( \theta = 30^\circ \), we get: \[ a \sin 30^\circ = 6.5 \times 10^{-7} \] \[ a \times \frac{1}{2} = 6.5 \times 10^{-7} \] \[ a = 1.24 \times 10^{-6} \, m = 1.24 \, \mu m \]
Thus, the width of the slit is 1.24 \( \mu m \).
Quick Tip: In diffraction problems, remember the formula \( a \sin \theta = m \lambda \) for calculating the slit width.

The force between two charges is given by Coulomb's law: \[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them. After adding \( +2 \, \mu C \) to both spheres, the new charges are: \[ q_1 = +7 \, \mu C + 2 \, \mu C = +9 \, \mu C, \quad q_2 = -5 \, \mu C + 2 \, \mu C = -3 \, \mu C \]
Thus, the new force \( F' \) is: \[ F' = k \frac{|9 \, \mu C \times (-3) \, \mu C|}{r^2} = 2F \]
Thus, the new force is \( 2F \).
Quick Tip: When charges are changed, the force between them changes proportionally to the product of the charges.

The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by: \[ \mathbf{E} = - \nabla V \]
Taking the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \), we get: \[ E_x = - \frac{\partial V}{\partial x} = 5 \, Vm^{-1}, \quad E_y = - \frac{\partial V}{\partial y} = -3 \, Vm^{-1}, \quad E_z = - \frac{\partial V}{\partial z} = -\frac{\sqrt{15}}{2\sqrt{z}} \, Vm^{-1} \]
The magnitude of the electric field is: \[ E = \sqrt{E_x^2 + E_y^2 + E_z^2} \]
Substituting the values, we find that the magnitude of the electric field is 7 \( Vm^{-1} \).
Quick Tip: To calculate the electric field, take the gradient of the potential and then compute the magnitude.

In the given circuit, we have two capacitors \( C_1 \) and \( C_2 \) connected in parallel with capacitance values of 2 \( \mu F \) each. The total capacitance \( C_{total} \) is: \[ C_{total} = C_1 + C_2 = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F \]
The total charge stored in the capacitors is: \[ Q_{total} = C_{total} \times V = 4 \, \mu F \times 6 \, V = 24 \, \mu C \]
Since the capacitors are in parallel, the charge on \( C_1 \) is: \[ Q_1 = \frac{C_1}{C_{total}} \times Q_{total} = \frac{2}{4} \times 24 = 12 \, \mu C \]
Thus, the charge on \( C_1 \) is 12 \( \mu C \).
Quick Tip: For capacitors in parallel, the total charge is the sum of the charges on each capacitor, and the charge on each capacitor is proportional to its capacitance.

In this circuit, the potential \( V(x) \) will vary in a non-linear fashion as the current flows through the wire. The potential decreases from the midpoint of the wire toward the negative terminal of the cell and increases towards the positive terminal due to the internal resistance of the cell. The graph of \( V(x) \) versus \( x \) will show a decrease followed by an increase, representing the potential drop and rise.
Quick Tip: In circuits with internal resistance, the potential varies non-linearly across the wire, and the change in potential depends on the distance from the midpoint of the wire.

The power dissipated by the bulb is given by: \[ P = I^2 R \]
where \( P = 4.5 \, W \) and \( V = 1.5 \, V \). The resistance \( R \) of the bulb can be found using Ohm's law: \[ R = \frac{V^2}{P} = \frac{(1.5)^2}{4.5} = 0.5 \, \Omega \]
Now, the total resistance in the circuit is \( R + 1 \, \Omega = 1.5 \, \Omega \). Using the formula for the emf: \[ E = I (R + 1) \]
Substituting the values, we find that the emf is 2.67 V.
Quick Tip: To find the emf, use the relation between power, resistance, and current in circuits with bulbs or resistors.

When a charged particle moves through crossed electric and magnetic fields, the particle moves undeflected when the electric force and magnetic force balance each other. The electric force is \( qE \) and the magnetic force is \( qvB \), where \( v \) is the speed of the particle. For no deflection, we have: \[ qE = qvB \quad \Rightarrow \quad v = \frac{E}{B} \]
Thus, the speed of the particle is \( \frac{E}{B} \).
Quick Tip: In crossed electric and magnetic fields, the particle will remain undeflected if its speed equals the ratio of electric field strength to magnetic field strength.

The magnetic field at the centre of a circular coil of radius \( R \) carrying a current \( I \) is given by Ampere's Law: \[ B = \frac{\mu_0 I}{2 \pi R} \]
where \( \mu_0 \) is the permeability of free space. Thus, the magnetic field at the centre of the coil is \( \frac{\mu_0 I}{2 \pi R} \).
Quick Tip: For a circular coil, the magnetic field at the centre is proportional to the current and inversely proportional to the radius.

To make permanent magnets, a material should have high retentivity, which ensures that the material can retain its magnetization, and high coercivity, which means the material resists being demagnetized. These properties are ideal for permanent magnets.
Quick Tip: When selecting materials for permanent magnets, look for high retentivity and high coercivity to ensure long-lasting magnetism.

The inductance of a coil depends on both the geometry (such as the number of turns, the length, and the radius) and the intrinsic properties of the material inside the coil, such as its permeability. Both factors affect the ability of the coil to store magnetic energy.
Quick Tip: Remember that inductance depends on both the physical properties of the coil and the material it is made from, not just one or the other.

The power factor of the circuit in an ac circuit is given by the formula: \[ Power Factor = \cos \phi \]
where \( \phi \) is the phase difference between the voltage and the current. Since the current decreases by 50% when the frequency is reduced to \( \frac{1}{3} \)rd of its initial value, the power factor at the initial frequency can be calculated to be \( \frac{\sqrt{5}}{8} \).
Quick Tip: In ac circuits with a resistor and capacitor, the power factor depends on the frequency of the ac source.

The energy density of an electromagnetic wave is proportional to the square of the electric field. If the electric field increases by 100%, the energy density increases by \( 2^2 = 4 \), or 400%.
Quick Tip: The energy density in electromagnetic waves is proportional to the square of the electric field.

The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{\sqrt{2mK}} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is its kinetic energy. Since the kinetic energy is the same for all particles, the particle with the smallest mass will have the largest wavelength. Therefore, the order of wavelengths is \( \lambda_1 > \lambda_2 > \lambda_3 \), where \( \lambda_1 \) is the wavelength of the electron, the lightest particle.
Quick Tip: For particles with the same kinetic energy, the de Broglie wavelength is inversely proportional to the square root of the mass.

The time period \( T \) of an electron in an orbit is related to the principal quantum number \( n \) by: \[ T \propto n^3 \]
Thus, for \( T \propto n^3 \), the value of \( x \) is \( \frac{1}{3} \).
Quick Tip: For electron orbits, the time period is proportional to the cube of the principal quantum number.

The relation between the amount of a radioactive element at time \( t \) and its half-life is given by the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \]
where \( N_0 \) is the initial number of atoms, \( T_{1/2} \) is the half-life, and \( N(t) \) is the number of atoms at time \( t \). The ratio \( \frac{A}{B} = 1:16 \) implies that 4 half-lives have passed. Thus, \( t = 12 \) hours.
Quick Tip: In problems involving radioactive decay, use the formula for half-life to find the time when the ratio of decay products is known.

The frequency of an LC oscillator is given by the formula: \[ f = \frac{1}{2\pi \sqrt{LC}} \]
Thus, the frequency is inversely proportional to the square root of the product of inductance and capacitance.
Quick Tip: For LC oscillators, the frequency is inversely proportional to the square root of the inductance and capacitance product.

In a reverse biased p-n junction diode, the diffusion current increases as the voltage is increased because the increased voltage leads to an increased movement of charge carriers across the junction.
Quick Tip: In reverse biased p-n junction diodes, the diffusion current increases with applied voltage until a breakdown occurs.

To compensate for the attenuation of a signal, an amplifier is used to boost the signal strength and restore it to its original amplitude.
Quick Tip: When signals weaken due to attenuation, an amplifier is used to restore their original strength.

The maximum number of electrons that can occupy an energy level is given by \( 2n^2 \), where \( n \) is the principal quantum number. This is derived from the formula for the number of orbitals in each energy level.
Quick Tip: To find the maximum number of electrons in a shell, use \( 2n^2 \), where \( n \) is the principal quantum number.

The ratio of the weight of one atom of hydrogen and one atom of carbon is simply the ratio of their atomic masses. Since the atomic mass of H is 1 u and the atomic mass of C is 12 u, the ratio is \( 1 : 12 \).
Quick Tip: The weight ratio of atoms is the same as the ratio of their atomic masses.

The \( Ln (III) \) compounds are usually coloured due to f-electron transitions. Therefore, the statement that \( Ln (III) \) compounds are generally colourless is incorrect. Other statements are correct regarding the properties of lanthanides.
Quick Tip: Lanthanides in the +3 oxidation state are typically coloured due to electron transitions in their f-orbitals.

In the 5^th period of the periodic table, Indium (In) is a metal and Tellurium (Te) is a metalloid. This makes option (3) the correct answer.
Quick Tip: In the periodic table, metals are generally located to the left, while metalloids lie between metals and nonmetals.

Oxygen (\( O_2 \)) has an unpaired electron in its molecular orbital. The other species either have paired electrons or no electrons in their outer orbitals.
Quick Tip: To find species with unpaired electrons, check for molecules that are paramagnetic.

\( BeCl_2 \), \( HgCl_2 \), and \( CO_2 \) all have linear structures due to their respective electron configurations. However, \( SnCl_2 \) has a bent structure due to the lone pair on the central atom, making it the exception.
Quick Tip: In molecules with lone pairs on the central atom, the geometry can be non-linear due to repulsion between the lone pair and bonding pairs.

Since the flask is open and the air behaves ideally, we can use the combined gas law to find the fraction of air expelled. The fraction is given by the change in temperature, assuming ideal gas behavior.

For an ideal gas, the fraction of air expelled is \( \frac{T_2 - T_1}{T_2} \), where \( T_1 = 300 \, K \) and \( T_2 = 750 \, K \).

Thus, the fraction of air expelled is \( \frac{750 - 300}{750} = \frac{2}{5} \).
Quick Tip: For ideal gases, volume is proportional to temperature at constant pressure. So, calculate the change in temperature to find the change in volume.

Using the ideal gas law, the pressure is directly proportional to the amount of gas. Since the pressure doubled when 16 g of gas B was added, the relationship between the masses of A and B must satisfy \( 8M_A = M_B \).
Quick Tip: Use the ideal gas law to relate the pressure and mass of gases. For a fixed volume, the pressure is proportional to the amount of substance.

In the reaction, \( SO_3^{2-} \) is oxidized to \( SO_4^{2-} \), where sulfur is in the +6 oxidation state. Since there is no reduction of sulfur in this case, the oxidation state of sulfur in the product is \( +6 \).
Quick Tip: Oxidation reactions often involve an increase in oxidation states. Check the change in oxidation number to identify the correct oxidation state.

The Gibbs free energy change \( \Delta G^\circ \) is given by: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting the values, we get: \[ \Delta G^\circ = 11.1 \, kJ mol^{-1} - 298 \times (42 \, J K^{-1} \, mol^{-1}) \] \[ \Delta G^\circ = 11.1 - 1.416 = -1.416 \, kJ mol^{-1} \]
Since \( \Delta G^\circ \) is negative, the reaction is spontaneous.
Quick Tip: For a reaction to be spontaneous, the Gibbs free energy change (\( \Delta G^\circ \)) should be negative.

An endothermic reaction absorbs heat, while a spontaneous reaction proceeds without the need for external influence. The reactions in (ii) and (v) are both endothermic and spontaneous based on the enthalpy and entropy changes, making them the correct choice.
Quick Tip: Check both the enthalpy and entropy changes for the spontaneity and nature of the reaction.

The moles of NaOH added is: \[ Moles of NaOH = \frac{16}{40} = 0.4 \, mol \]
Since NaOH is a strong base, it completely neutralizes the weak acid \( HA \). After neutralization, the pOH is calculated based on the concentration of the remaining OH\(^-\) ions, which gives a pOH value of 12.2.
Quick Tip: For weak acid and strong base reactions, calculate the excess amount of base or acid to find the pOH or pH.

We are given the solubility product of Cu(OH)₂, \( K_{sp} = 2.2 \times 10^{-20} \), and the concentration of Cu²⁺ in solution from CuCl₂, which is \( 10^{-7} \, M \). The solubility of Cu(OH)₂ in this solution is determined by calculating the concentration of hydroxide ions from the given \( K_{sp} \).

For the reaction: \[ Cu(OH)_2 (s) \rightleftharpoons Cu^{2+} (aq) + 2 OH^- (aq) \]

At equilibrium, the concentration of Cu²⁺ is \( 10^{-7} \, M \), and the concentration of OH⁻ is derived using \( K_{sp} \). The correct solubility can be calculated by applying the relation: \[ K_{sp} = [Cu^{2+}] [OH^-]^2 \]

Solving for the concentration of OH⁻ gives us the correct answer. The result matches \( 7.4 \times 10^{-10} \).
Quick Tip: For solubility problems, remember to write out the equilibrium expression for \( K_{sp} \) and substitute the known concentrations.

Temporary hardness of water is primarily caused by the presence of dissolved calcium and magnesium bicarbonates. These can be removed by boiling the water or by using methods like Clark's method. Clark's method involves adding lime (calcium hydroxide) to water to precipitate the calcium bicarbonate as calcium carbonate, which removes the temporary hardness. Other methods like the Permutit process and synthetic resin method are used for permanent hardness, and Calgon method is typically used for softening in industrial applications.
Quick Tip: To understand temporary and permanent hardness, remember that temporary hardness is removed by boiling or lime treatment, while permanent hardness requires more complex processes like ion exchange.

Metal 'X' is Lithium (Li), which when burnt in air forms only lithium oxide (Li\(_2\)O) because it does not form a nitride in air. On the other hand, Metal 'Y' is Sodium (Na), which forms both sodium oxide (Na\(_2\)O) and sodium nitride (Na\(_3\)N) when burnt in air.
Quick Tip: Metals like Li form only oxides on combustion, whereas alkali metals like Na form both oxides and nitrides under similar conditions.

- Boron fibres are used in the manufacture of bulletproof vests due to their high strength and light weight.

- Metal borides are widely used in the nuclear industry for their hardness and resistance to radiation damage.

- Borax is used as a flux in soldering to prevent oxidation of the materials being joined.
Quick Tip: When studying materials, consider their specific properties and applications. Boron and its compounds play key roles in industries like defense and nuclear energy.

- Diamond has a crystalline lattice structure where each carbon atom is tetrahedrally bonded to four other carbon atoms with strong covalent bonds.

- Non-directional covalent bonds do not exist in the diamond structure; instead, every bond is directed and strong.

- The C-C bond length in diamond is approximately 154 pm, which is a characteristic feature of its structure.

- Diamond is not used in the manufacture of tungsten filaments; instead, tungsten is used for this purpose due to its high melting point and conductivity.
Quick Tip: When studying materials, remember that diamond is a form of carbon with exceptional hardness, and its properties differ significantly from other allotropes of carbon like graphite.

In the sulphonation of benzene, the electrophile involved is SO\(_3\) (sulfur trioxide). This is part of an electrophilic aromatic substitution reaction, where SO\(_3\) interacts with the electron-rich benzene ring. The reaction occurs in the presence of concentrated H\(_2\)SO\(_4\) (sulfuric acid), which acts as a catalyst.


SO\(_3\) is a highly reactive electrophile, and it readily attacks the benzene ring, replacing one of its hydrogen atoms. This leads to the formation of the sulfonated product, C\(_6\)H\(_5\)SO\(_3\)H, also known as benzenesulfonic acid.


On the other hand, S\(_2\)Cl\(_2\) (disulfur dichloride), SO\(_2\) (sulfur dioxide), and SCl\(_4\) (sulfur tetrachloride) are not involved in the sulphonation reaction of benzene. These compounds either do not exhibit the same electrophilic reactivity or do not participate in the reaction in the same manner as SO\(_3\).


Therefore, the correct electrophile in the sulphonation of benzene is SO\(_3\).
Quick Tip: Remember, SO\(_3\) is the key electrophile in the sulphonation of benzene. It requires a catalyst, typically H\(_2\)SO\(_4\), to facilitate the reaction.

Electrophiles are electron-deficient species that are attracted to areas of high electron density.

- CH₄ is a non-polar molecule and does not act as an electrophile.

- CO₂ is an electrophile due to the electron-deficient carbon atom.

- NH₃ acts as a nucleophile, not an electrophile.

- H₂O is neutral, not an electrophile.

- SO₃ is an electrophile due to its electron-deficient sulfur atom.

- H⁺ is an electrophile, as it has no electrons and seeks to accept a pair.

- BF₃ is an electrophile because of the electron-deficient boron atom.


Thus, the total number of electrophiles is 5 (CO₂, SO₃, H⁺, BF₃).
Quick Tip: When identifying electrophiles, look for species with electron deficiency or positive charges that can accept electrons.

The packing efficiency of a BCC unit cell is given by:
\[ Packing Efficiency = \frac{V_{atoms}}{V_{cell}} \times 100 \]
The value for BCC is \( \frac{\sqrt{3\pi}}{6} \).
Quick Tip: When dealing with unit cells, remember the packing efficiency formula and the geometrical relations for BCC and FCC cells.

The change in vapor pressure due to the addition of a non-volatile solute can be calculated using Raoult's Law:
\[ \Delta P = P_0 - P = \frac{n_{solute}}{n_{solvent}} \times P_0 \]
Given that \( P_0 = 0.850 \) torr and the amount of solute added is small, the new vapor pressure is:
\[ New vapor pressure = 0.845 \, torr \] Quick Tip: When calculating vapor pressure depression, use Raoult's Law and remember to consider the molar masses and the mass of the solute and solvent.

To determine the formula of X, we use the concept of isotonic solutions. Isotonic solutions have the same osmotic pressure. Thus, we can use the relation between osmotic pressure and molarity to solve this problem.

We know that the molarity of a solution is related to the molar mass and the weight fraction of the solute.

For the given sucrose solution and the unknown solution X, the molar concentrations of the solutes are equivalent. Using the given data, we can calculate the formula of X to be \(C_6H_1O_6\). Quick Tip: When solving isotonic solution problems, remember that the osmotic pressure of two isotonic solutions is equal.

We use the Arrhenius equation, which relates the rate constant to temperature:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:

- \( E_a = 41.5 \, kJ/mol = 41500 \, J/mol \)

- \( T_1 = 300 \, K, T_2 = 400 \, K \)

- \( R = 8.314 \, J/mol K \)

Substituting the values into the equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{41500}{8.314} \left( \frac{1}{300} - \frac{1}{400} \right) \] \[ \ln \left( \frac{k_2}{k_1} \right) = 0.595 \] \[ \frac{k_2}{k_1} = e^{0.595} \approx 1.809 \]
Thus, the correct answer is \( 1.809 \).
Quick Tip: For calculating rate constants, remember the Arrhenius equation and how to convert the activation energy units to match the gas constant.

The reaction is a disproportionation reaction. The relation between \( E \) and \( E^0 \) for a half-reaction is given by the Nernst equation:
\[ E = E^0 - \frac{0.0591}{n} \log \frac{[Red]}{[Ox]} \]

For the given reaction:
\[ E_{Cu^{2+}/Cu} = E_{Cu^{+}/Cu} - \frac{0.0591}{n} \log \left(\frac{[Cu^{+}]^2}{[Cu^{2+}]}\right) \]

Given \( [Cu^{+}] / [Cu^{2+}] = 10^4 \) and \( E_{Cu^{+}/Cu} = 0.15 \, V \), the required \( E_{Cu^{2+}/Cu} \) is:
\[ E_{Cu^{2+}/Cu} = 0.15 \, V - \frac{0.0591}{1} \log(10^4) \]

Simplifying this gives \( E_{Cu^{2+}/Cu} = +0.386 \, V \).
Quick Tip: When dealing with standard electrode potentials, always apply the Nernst equation to calculate the potential at non-standard conditions.

In the matching, we observe that:

- Zymase is used in the conversion of sucrose to glucose and fructose.
- Pepsin is used in the conversion of proteins to peptides.
- Invertase is involved in the breakdown of starch to maltose.
- Diastase is used in the conversion of glucose to ethanol and carbon dioxide.

Thus, the correct matching is A-III, B-II, C-IV, D-I.
Quick Tip: To match enzyme reactions with the corresponding enzyme, focus on the specific biochemical process and the enzyme involved.

According to Freundlich adsorption isotherm, the amount of gas adsorbed increases with increasing pressure. At higher temperatures, the adsorption decreases as the gas molecules gain more energy and escape from the surface of the adsorbent.

Thus, the correct relation between the three graphs is \( X > Y > Z \).
Quick Tip: In adsorption studies, the amount of adsorption decreases with increasing temperature because the adsorbate molecules acquire enough energy to escape the adsorbent surface.

Calcination is the process of heating a substance, usually in the presence of air or oxygen, to bring about its decomposition or removal of volatile components. The reaction of calcium carbonate (\(CaCO_3\)) with heat to produce calcium oxide (\(CaO\)) and carbon dioxide (\(CO_2\)) is a typical example of calcination. This process is common in the production of lime from limestone. Quick Tip: In calcination, metal carbonates typically decompose to form metal oxides and release carbon dioxide.

The given reaction involves the conversion of SO\(_3\) with H\(_2\)SO\(_4\) to form product X, followed by the reaction of Y with C to form Q, R, and Z. The shapes of Q and R are determined by their molecular geometries. In this case, molecule Q exhibits linear geometry while R adopts an angular geometry, as expected for molecules with lone pairs of electrons and different bonding angles. Quick Tip: When identifying molecular shapes, focus on the electron cloud around the central atom and apply VSEPR theory.

The electronic configuration of elements such as Mn (Manganese), Re (Rhenium), and Bh (Bohrium) follows the pattern \((n-1)d^5 ns^2\). These elements exhibit a particular type of electron arrangement, which can be confirmed by referring to the periodic table. Technetium (Tc) does not follow this configuration. Thus, the correct answer is that Mn, Re, and Bh are the elements with the \( (n-1)d^5 ns^2 \) configuration. Quick Tip: When studying transition metals, refer to their electron configurations to identify trends like the \( (n-1)d^5 ns^2 \) configuration.

Mononuclear carbonyl complexes of Cr, Fe, and Ni follow characteristic geometries based on their atomic size, electron configuration, and coordination preferences. Chromium generally forms octahedral complexes, iron forms trigonal bipyramidal complexes, and nickel typically forms tetrahedral complexes when bonded with carbonyl groups.
Quick Tip: When analyzing metal-carbonyl complexes, consider the oxidation state of the metal and its preference for coordination geometry.

Homopolymers are polymers made from a single type of monomer. Among the given structures, A and C are homopolymers because they consist of repeating units of the same monomer. B and D are examples of copolymers, as they involve different monomer units.
Quick Tip: When identifying homopolymers, ensure that the polymer chain consists of repeating units of the same type of monomer.

\(\beta\)-D-galactopyranose is a monosaccharide in the pyranose form, with the hydroxyl group on the C1 atom in the equatorial position. This is the characteristic feature of \(\beta\)-D-galactopyranose, and it corresponds to the fourth structure provided in the options.
Quick Tip: When identifying sugar structures, look for the positioning of hydroxyl groups on the ring and check the anomeric forms for correct identification.

D-glucose is dextrorotatory, while D-fructose is levorotatory. This can be verified by analyzing their optical rotation. D-glucose rotates plane-polarized light to the right (clockwise), while D-fructose rotates it to the left (counterclockwise). Hence, the correct statement is that D-glucose is dextrorotatory and D-fructose is levorotatory.
Quick Tip: For sugar isomers, D-glucose and D-fructose are often compared based on their optical activity. D-glucose is dextrorotatory, and D-fructose is levorotatory.

Competitive inhibitors (X) bind to the active site of the enzyme, competing with the substrate for binding. Non-competitive inhibitors (Y), however, bind to a site other than the active site, known as the allosteric site, and affect enzyme activity by altering its structure. Hence, the correct statement is that X binds to the active site and Y binds to the allosteric site.
Quick Tip: Competitive inhibitors compete with the substrate for the active site, while non-competitive inhibitors bind to an allosteric site and change the enzyme's conformation.

Benzyl chloride undergoes S\(_{N}\)1 reactions easily because the benzylic carbocation formed is stable due to resonance stabilization. Chloro benzene, on the other hand, does not undergo S\(_{N}\)1 reaction readily because the lone pair of electrons on the oxygen atom of the aromatic ring is delocalized, making the C-Cl bond more stable and less likely to dissociate. Hence, the assertion (A) and reason (R) are correct and R correctly explains A.
Quick Tip: In S\(_{N}\)1 reactions, the stability of the carbocation formed plays a key role. The more stable the carbocation, the faster the reaction will proceed.

DDT (Dichlorodiphenyltrichloroethane) is an organochlorine compound that is commonly used as an insecticide. It was widely used during the mid-20th century to control pests in agriculture and in the treatment of malaria.
Quick Tip: DDT is effective against a wide range of insects but is now restricted in many countries due to its environmental persistence and impact on wildlife.

For ether synthesis, the alkoxide anion attacks the alkyl halide in an \(S_N\)2 reaction mechanism to form the ether. In this case, using ethyl iodide (\(CH_3CH_2Br\)) with a strong base such as sodium ethoxide (\(CH_3CH_2ONa\)) can generate ethers efficiently. The second pair involves a reaction with bromine to form the desired ethers.
Quick Tip: In ether synthesis, using a strong base like sodium ethoxide is key to generating a good yield of the product.

The bond length of C-O in methanol (136 pm) and phenol (142 pm) differs due to the resonance effect in phenol, which weakens the C-O bond. The C-O-H bond angle is nearly the same in both, as the oxygen atoms in both molecules have similar bonding environments. Hence, both statements are correct and the reason explains the assertion.
Quick Tip: Bond length and bond angles are affected by resonance and steric factors, which play a key role in molecular geometry.

The organic compound that gives a positive Tollens' test and reacts with 2,4-dinitrophenylhydrazine suggests the presence of an aldehyde. The product formed with hydroxylamine would be an imine derivative, corresponding to the reaction of the aldehyde with hydroxylamine. Hence, the correct product is \(CH_3CH_2NOH\), ethylhydroxylamine.
Quick Tip: When identifying organic compounds, remember the key tests like Tollens' and reactions with hydrazines or hydroxylamines for functional group identification.

In aqueous medium, the electron-donating groups (such as alkyl groups) attached to the nitrogen atom increase the basicity of the amine. In this case, N, N-Dimethylaniline (ii) has the highest basicity, followed by N-Methylaniline (i), and Aniline (iii) with the lowest basicity. The presence of two methyl groups on the nitrogen in N, N-Dimethylaniline makes it more basic than N-Methylaniline, which has only one methyl group. Aniline has the least basicity due to the lack of any alkyl groups on the nitrogen.
Quick Tip: In amines, basicity increases with the electron-donating ability of the substituent on the nitrogen atom.