Grand Mdcat Mock 6th April 2025 — Solved Past Paper with Answers

Q1. End-stage renal failure is called:

• A. Kidney failure stage
• B. Hyperoxaluria
• C. Uremia✓
• D. Hypercalcemia

Explanation: Option C is correct.Uremia is the term for high levels of urea in the blood. The uremic syndrome can be defined as the terminal clinical manifestation of kidney failure (also called renal failure), uremia is mostly a consequence of kidney failure. It can be defined as an excess of amino acid and protein metabolism end products, such as urea and creatinine, in the blood that would be normally excreted in the urine.

Q2. The disease in which patients passed urine that rapidly turned black on exposure to air is called:

• A. Phenylketonuria
• B. Alkaptonuria✓
• C. Sickle cell anemia
• D. Hemophilia
• E. Anuria

Explanation: Alkaptonuria is an inherited condition that results in the buildup of homogentisic acid, causing urine to turn black upon exposure to air. This distinctive symptom differentiates it from other disorders. Phenylketonuria involves amino acid metabolism but does not affect urine color. Sickle cell anemia and hemophilia are blood disorders, not related to urine. Anuria is the absence of urine production, not a change in color.

Q3. The excretion of hypertonic urine in humans is associated best with the:

• A. Glomerular capsule
• B. Proximal convoluted tubule
• C. Loop of Henle✓
• D. Distal convoluted tubule

Explanation: Hypertonic Urine means there is a higher solute concentration than the body fluid (the solution has more osmotic pressure; water moves in by osmosis). This means maximum water has been reabsorbed. Despite Option B, Option C, and Option D, playing a role in water reabsorption, the main site where water reabsorption occurs is the Loop of Henle. Glomerular Capsule (or the Bowman capsule, is a cup-like sac at the beginning of the nephron that is involved in the filtration of the blood coming into the kidney. The proximal Convoluted Tubule is present after the Bowman's capsule while the Distal convoluted tubule is present after the loop of Henle. The proximal Convoluted Tubule is the site where Sodium ions are reabsorbed while the Distal convoluted tubule is where the body carries our processes to maintain the body acid-base concentration. On the other hand, the Loop of Henle is associated with both the solute and water reabsorption using the counter-current multiplier, which involves the usage of energy to generate an osmotic gradient that enables the reabsorption of water from the filtrate and into the blood, producing concentrated urine.

Q4. Transgenic plants are the ones:

• A. Generated by introducing foreign DNA into a cell and regenerating a plant from that cell✓
• B. Produced after protoplast fusion in artificial medium
• C. Grown in artificial medium after hybridization in the field
• D. Produced by a somatic embryo in artificial medium

Explanation: The plants produced through genetic engineering contain genes, usually from an unrelated organism. Such genes are called transgenes and the plants having transgenes are called transgenic plants. Recombinant DNA techniques are being used to improve crop plants by increasing their productivity, making them more nutritious, and by developing disease resistance. Transgenic plants have a natural resistance to herbicides and pests. In the future, plants may have an ability to fix atmospheric nitrogen and an increased ability to grow in arid and salty soils.

Q5. The contractile protein of skeletal muscle involving ATPase activity is:

• A. Troponin
• B. Tropomyosin
• C. Myosin✓
• D. Actinin

Explanation: The correct answer is Myosin. Myosin is a contractile protein with intrinsic ATPase activity, essential for muscle contraction. Its role involves the hydrolysis of ATP, which provides the necessary energy for muscle movement by interacting with actin filaments. On the other hand, Troponin and Tropomyosin are regulatory proteins that do not possess ATPase activity but are crucial in regulating the interaction between actin and myosin. Alpha-actinin is a structural protein that helps maintain the organization of the muscle structure rather than directly participating in contraction.

Q6. Unpaired facial bones are:

• A. Maxilla, Zygomatic
• B. Palatine, Inferior concha
• C. Nasal, Lacrimal
• D. Mandible, Vomer✓

Explanation: The maxillae are the largest bones in the face and form the upper jaw. The zygomatic bones are the cheekbones. The palatine bones form the hard palate. The inferior nasal conchae are the small, scroll-shaped bones that project into the nasal cavity. The nasal bones form the bridge of the nose. The lacrimal bones are small, teardrop-shaped bones that form part of the eye socket. The facial bones are the bones that form the face of the skull. There are 14 facial bones, of which 6 are paired and 2 are unpaired. The unpaired facial bones are the vomer and the mandible. The vomer is a small, triangular bone that forms the posterior part of the nasal septum. It is the only unpaired bone in the nasal cavity. The mandible is the largest and strongest bone in the face. It forms the lower jaw and is the only movable bone in the skull.

Q7. During contraction, which statement is correct?

• A. I-band increases
• B. A-band decreases
• C. H-zone disappears, A-band constant✓
• D. Z-line increases

Explanation: In sliding filament theory, A-band remains same, H-zone shortens and may disappear.

Q8. Which bones meet at the elbow joint, and what kind of movement do they allow?

• A. Option A
• B. Option B
• C. Option C
• D. Option D✓

Explanation: At the elbow joint, the ulna and humerus meet, and they cause back-and-forth movement of the forearm. At the shoulder joint, the humerus and scapula meet, and there they cause full axial movement.

Q9. Muscle fiber has many _ and in turns these have many _.

• A. Myofibrils; Sarcomeres✓
• B. Myofibrils; Thin filaments
• C. Myosin; Thin filaments
• D. Myosin; Sarcomeres

Explanation: Each skeletal muscle fibre is a single cylindrical muscle cell. An individual skeletal muscle may be made up of hundreds, or even thousands, of muscle fibres bundled together and wrapped in a connective tissue covering.

Q10. It increases the excitability of neurons and results in loss of sensation.

• A. Cramp
• B. Muscle fatigue
• C. Tetany✓
• D. Tetanus

Explanation: Tetany is the correct answer as it results from increased excitability of neurons caused by low calcium levels, leading to both involuntary muscle contractions and potential sensory disturbances. Cramp and muscle fatigue are primarily physical and do not involve the nervous system in the same way. Tetanus, while causing muscle spasms, is due to a bacterial infection and does not involve loss of sensation.

Q11. Endorphins are produced in:

• A. Brain✓
• B. Adrenal gland
• C. Stomach
• D. Thymus

Explanation: Endorphins are neurotransmitters that are produced in various parts of the brain, including the hypothalamus and the pituitary gland. They act as natural painkillers and mood elevators.

Q12. Which hormone induces labor pains?

• A. Estrogen
• B. Oxytocin✓
• C. Progesterone
• D. LH

Explanation: Oxytocin is often referred to as the "love hormone" or "cuddle hormone." In the context of childbirth, it plays a crucial role in initiating and regulating uterine contractions during labor. These contractions are essential for the progression of labor and eventually lead to the delivery of the baby.

Q13. The estrogen hormone secretion during oogenesis is stimulated by:

• A. Luteinizing Hormone
• B. Inhibin
• C. Follicle-Stimulating Hormone✓
• D. Testosterone

Explanation: FSH is responsible for stimulating the growth and development of ovarian follicles in the ovaries. These developing follicles, as they mature, produce estrogen.

Q14. Causative agent of a sexually transmitted disease that affects mucous membrane of the urogenital tract is:

• A. Staphylococcus aureus
• B. Treponema pallidum
• C. Neisseria gonorrhoeae✓
• D. Escherichia coli

Explanation: Sexually transmitted diseases (STDs), or sexually transmitted infections (STIs), are infections that are passed from one person to another through sexual contact. They are usually spread during vaginal, oral, or anal sex. Gonorrhea is a sexually transmitted disease caused by bacterial infection. The agent for Gonorrhea is Neisseria Gonorrhoeae. It invades the mucus membranes of reproductive tracts including the cervix, vagina, uterus, fallopian tubes in women, and urethra in both men and women.

Q15. During luteal phase of menstrual cycle, the hormone at its peak:

• A. Progesterone✓
• B. Estrogen
• C. LH
• D. GnRH

Explanation: Rising levels of progesterone from the corpus luteum act on endometrium, causing the arteries to elaborate and converting the functional layer into secretory layer.

Q16. Menstrual flow occurs due to lack of:

• A. Oxytocin
• B. Vasopressin
• C. Progesterone✓
• D. FSH

Explanation: The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometrium. Such as endometrium is necessary for implantation of the fertilized ovum and other events of pregnancy. In the absence of fertilization, the corpus luteum degenerates. This causes disintegration of the endometrium leading to menstruation. The menstrual flow results due to breakdown of endometrial lining of the uterus and its blood vessels which forms liquid that comes out through vagina.

Q17. Select the correct option describing gonadotropin activity in a normal pregnant female.

• A. High level of FSH and LH stimulates the thickening of endometrium
• B. High level of FSH and LH facilitates implantation of the embryo
• C. High level of hCG stimulates the synthesis of estrogen and progesterone✓
• D. High level of hCG stimulates the thickening of endometrium

Explanation: The trophoblastic cells secrete human chorionic gonadotropin hormone which has properties similar to those of lutenizing hormone (LH) of the pituitary gland. It takes over the job of pituitary LH during pregnancy. The hCG maintains the corpus luteum and stimulates it to secrete progesterone. The latter maintains the endometrium of the uterus and causes it to grow throughout pregnancy. This also prevents menstruation. Progesterone also causes increased secretion of mucus in the cervix of the uterus that forms a protective plug during pregnancy.

Q18. Pairing of homologous chromosomes during zygotene in meiosis is:

• A. Non specific but exactly pointed
• B. Highly specific and exactly pointed✓
• C. Non specific and non pointed
• D. Highly specific but fused

Explanation: The first essential phenomenon of meiosis i.e., pairing of homologous chromosomes called synapsis starts. This pairing is highly specific and exactly pointed, but with no definite starting point (s.) Each paired but not fused, complex structure is called a bivalent or tetrad.

Q19. The Leydig’s cells as found in the human body are the secretory source of:

• A. Progesterone
• B. Intestinal mucus
• C. Glucagon
• D. Androgens✓

Explanation: Interstitial cells or Leydig cells are the cells interspersed between the seminiferous tubules of the testis. They secrete androgens (e.g., testosterone) in response to stimulation by luteinizing hormone from the anterior pituitary gland.

Q20. Seed dormancy is due to the :

• A. Ethylene
• B. Abscisic Acid✓
• C. IAA
• D. Starch

Explanation: .ABA induces dormancy and prevents premature germination.

Q21. In the higher plants, growth is confined to certain regions called the:

• A. Meristems✓
• B. Epistems
• C. Peristems
• D. Mesostems

Explanation: Meristems = actively dividing cells at root/shoot tips or sides.

Q22. The increase in thickness of stems and roots due to the activity of lateral meristems is called:

• A. Primary Growth
• B. Secondary Growth✓
• C. Tertiary Growth
• D. Both A and B

Explanation: Secondary growth = thickening by vascular & cork cambium.

Q23. Which of the following is not true about viruses?

• A. Contain DNA
• B. Can replicate on their own✓
• C. Can infect bacteria
• D. They have a sub-cellular structure

Explanation: Viruses cannot replicate on their own. They lack the cellular machinery needed for reproduction and must infect a host cell to replicate. Once inside a host cell, a virus hijacks the cell's machinery to produce viral components, which are then assembled into new virus particles. This dependence on a host cell is a defining characteristic of viruses.

Q24. Antibiotics can be used against:

• A. Herpes simplex
• B. Influenza
• C. Polio
• D. Salmonella typhi✓

Explanation: Caused by the bacterium Salmonella typhi, which is the agent of typhoid fever. Antibiotics are effective here and are used to treat the infection.

Q25. Your neighbour has a flower garden in which there are red flowers and white flowers. These flowers are diploid organisms, and the flower colour is an autosomal trait. The gene for red flowers (R) is dominant, while the gene for white flowers (r) is recessive. Which of the following could be the genotype of a red flower?

• A. Rr
• B. RR, Rr, or rr
• C. rr
• D. RR or Rr✓

Explanation: Option A : If there is only the Rr allele, the flower will be red instead of white. Option B : If the alleles are Rr and RR, both will produce red flowers. Incase, if no 'R' is present and 'rr' are present, the flower will be white instead of red. Option C : If alleles are 'rr' the flower will be white instead of red. Option : D A red flower must have an 'R' allele which conveys the red phenotype. Since the red allele is dominant, the flower remains as long as red as one of the copies of this allele is present. Thus, both the 'Rr' and 'RR' will produce red flowers. In case, if no 'R' allele is present and 'rr' alleles are present, the flower will be white instead of red.

Q26. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are the same as 1 : 2 : 1 and offspring are intermediate of both parents. It represents a case of:

• A. Co­dominance
• B. Dihybrid cross
• C. Monohybrid cross with complete dominance
• D. Monohybrid cross with incomplete dominance.✓

Explanation: The F2 generation in a Mendelian cross with a 1:2:1 genotypic and phenotypic ratio indicates a monohybrid cross with incomplete dominance. Let's explain why the other options are incorrect: 1. Codominance: Codominance results in a 1:2:1 phenotypic ratio, but the genotypic ratio would be different. In codominance, both alleles are fully expressed in the heterozygous condition, leading to a 1:2:1 phenotypic ratio, but a 1:2:1 genotypic ratio would not be expected.2. Dihybrid cross: A dihybrid cross typically involves two pairs of contrasting traits and would result in a 9:3:3:1 phenotypic ratio, not the 1:2:1 ratio observed in this case.3. Monohybrid cross with complete dominance: In a monohybrid cross with complete dominance, you would expect a 3:1 phenotypic ratio (3 dominant: 1 recessive) in the F2 generation, which is different from the 1:2:1 ratio observed here. So, the only option that matches the 1:2:1 genotypic and phenotypic ratio is a monohybrid cross with incomplete dominance, where the heterozygous individuals exhibit an intermediate phenotype between the two homozygous parents. The inheritance of flower color in the dog flower (snapdragon or Antirrhinum sp -Codominance results in a 1:2:1 phenotypic ratio, but the genotypic ratio would be different. In codominance, both alleles are fully expressed in the heterozygous condition, leading to a 1:2:1 phenotypic ratio, but a 1:2:1 genotypic ratio would be expected.

Q27. A test cross is carried out to:

• A. Determine the genotype of a plant at F2✓
• B. Predict whether two traits are linked
• C. Assess the number of alleles of a gene
• D. Determine whether two species or varieties will breed successfully or not

Explanation: A test cross is performed to determine the genotype of the F2 plant. In a typical test cross, an organism showing dominant phenotype and whose genotype is to be determined is crossed with one that is homozygous recessive for the allele being investigated, instead of self­crossing. The progenies of such a cross can easily be analyzed to predict the genotype of the test organism. Given ahead is an illustration of a test cross:

Q28. Which traits are more likely to affect men than women?

• A. X-linked recessive✓
• B. X-linked dominant
• C. Autosomal dominant
• D. Autosomal recessive

Explanation: The correct answer is X-linked recessive because these conditions predominantly affect males due to their single X chromosome. If a male inherits an X chromosome with a recessive mutation, he will express the disorder since there is no second X chromosome to counteract the effect. In contrast, females would need two copies of the recessive mutation (one on each X chromosome) to exhibit the disorder, making it less common in females.Other options are incorrect because:X-linked dominant: This disorder can affect both genders, and males can show symptoms, but it is not more likely to affect them than females.Autosomal dominant: This pattern impacts both sexes equally, so it does not favor males over females.Autosomal recessive: Similar to autosomal dominant, this pattern affects both sexes equally, requiring two copies of the gene to express the disorder.

Q29. An Rh-negative woman is married to an Rh-positive man whose father was also Rh-negative. What are the chances that their child will have Rh negative blood group?

• A. 25%
• B. 50%✓
• C. 75%
• D. 100%

Explanation: Given that the man’s father was Rh-negative and the man is Rh-positive, the man must be heterozygous for Rh-positive. A negative woman should be homozygous as a recessive allele only expresses itself in homozygous conditions. When a heterozygous dominant individual is crossed with a homozygous recessive individual, the offspring show the dominant and recessive phenotypes in a ratio of 1:1. Hence, there is a fifty percent chance of their child being “affected” with Rh negative phenotype.

Q30. ABO blood group system is due to:

• A. Multifactor inheritance
• B. Incomplete dominance
• C. Multiple allelism✓
• D. Epistasis

Explanation: The correct answer is c) Multiple allelism. The ABO blood group system is determined by the presence of three alleles: A, B, and O. These alleles dictate the antigens present on the surface of red blood cells. An individual inherits two alleles, one from each parent, resulting in combinations that define their blood type. Unlike epistasis or multifactor inheritance, multiple allelism involves variations within a single gene locus. Therefore, other options like multifactor inheritance, incomplete dominance, and epistasis do not apply to the ABO system.

Q31. Which of the following reduces the chance of recombination and variation?

• A. Genetic recombination
• B. Mutation
• C. Meiosis
• D. Gene linkage✓

Explanation: Gene linkage is the tendency of DNA sequences that are close together on a chromosome to be inherited together during the meiosis phase of sexual reproduction. Linked genes are found on the same chromosome and are inherited together. Since they don't independently assort, the gametes produced show less genetic variation than unlinked genes. Genetic recombination, meiosis, and mutation all increase the chances of variation.

Q32. The history of life is like a tree, with multiple branching and rebranching from a common trunk, in view of:

• A. Darwin✓
• B. Lamarck
• C. Cuvier
• D. Lyell

Explanation: The correct answer is Darwin. Charles Darwin used the 'tree of life' metaphor to describe how species evolve from common ancestors, branching out like a tree. This illustrates the diversity of life and evolutionary relationships. Lamarck did not use this metaphor and focused on acquired characteristics. Cuvier was known for his work in palaeontology and catastrophism, not evolutionary trees. Lyell was a geologist whose work influenced Darwin but did not propose the tree analogy.

Q33. Homologous structures show _ evolution?

• A. Parallel
• B. Divergent✓
• C. Convergent
• D. Adaptive

Explanation: Homologous structures demonstrate evidence of divergent evolution. These structures originate from a common ancestor but have evolved to serve different functions in different species due to adaptations to various environments. For example, the forelimbs of humans, bats, and whales have different functions but share a common structural origin. This indicates divergent evolution, where species evolve different adaptations from a shared ancestral trait. In contrast, parallel evolution involves related species developing similar traits independently. Convergent evolution involves unrelated species independently developing similar traits. Adaptive evolution refers broadly to changes that enhance survival and reproduction and is not specific to homologous structures.

Q34. The severe reduction in population size is due to which of the following?

• A. Genetic drift
• B. Outbreeding
• C. Inbreeding
• D. Bottleneck effect✓

Explanation: The correct answer is the bottleneck effect. This effect occurs when a population experiences a sharp reduction in size due to sudden environmental events, such as natural disasters. The survival of only a few individuals leads to a loss of genetic diversity and can drastically change allele frequencies. In contrast, genetic drift is a slower process that does not necessarily involve a reduction in population size. Outbreeding increases genetic diversity and does not reduce population size. Inbreeding affects genetic health over time but does not result in sudden population size reduction.

Q35. According to Lamarckism, the basis of evolution is:

• A. Inheritance of acquired characteristics✓
• B. Mutation
• C. Natural selection
• D. Survival of the fittest

Explanation: According to Lamarck, these acquired traits could be passed on to offspring. This means that traits developed during an organism's lifetime could be inherited by the next generation.The inheritance of acquired characteristics is the idea that traits a parent organism develops during its lifetime can be passed on to its offspring, a concept famously associated with Jean-Baptiste Lamarck. For example, a giraffe stretching its neck to reach higher leaves would pass that longer neck to its young. This theory was largely discredited by modern genetics, which shows that acquired traits (like a broken bone or a suntan) do not change the DNA in reproductive cells and therefore cannot be inherited.

Q36. Where are the enzymes required for the replication of HIV found?

• A. Surrounding the viral core
• B. Inside the capsid✓
• C. In the protein spikes
• D. Outside the capsid

Explanation: The enzymes required for HIV replication, such as reverse transcriptase and integrase, are indeed found within the capsid of the virus. The capsid is a protein shell that encases the viral RNA and these essential enzymes. Once HIV enters a host cell, the capsid disassembles, releasing the viral RNA and enzymes, which are necessary for the replication process.

Q37. The best definition of natural selection is:

• A. Survival of the fittest
• B. The process by which individuals with advantageous traits are more likely to survive and reproduce✓
• C. Those who eat better, are healthier, and live longer are the most fit within a population
• D. Preservation of traits that lead to increased survival and reproduction

Explanation: Natural selection is a fundamental mechanism of evolution, where individuals with traits that enhance their ability to survive and reproduce are more likely to pass those traits to subsequent generations. The correct option accurately describes this process by emphasizing the role of advantageous traits in increasing reproductive success. Other options either oversimplify the concept to mere survival or fail to encompass the full scope of evolutionary fitness, which includes both survival and successful reproduction.

Q38. The Nissl’s granules of nerves cell are made up of :

• A. DNA
• B. RNA
• C. Ribosome✓
• D. Protein

Explanation: Nissl's granules (or Nissl bodies) are clusters of rough endoplasmic reticulum and ribosomes found in neurons. They are rich in RNA and are responsible for protein synthesis, essential for nerve cell maintenance and repair.

Q39. Which of the following pairs of hormones are not antagonistic (having opposite effects) to each other?

• A. Aldosterone: Atrial Natriuretic Factor
• B. Relaxin: Inhibin✓
• C. Parathormone: Calcitonin
• D. Insulin: Glucagon

Explanation: Relaxin hormone is secreted by ovary and placenta during pregnancy, which relaxes ligaments in pelvis and softens and widens cervix during childbirth. Inhibin secreted by granulosa cells in the ovaries inhibits secretion of FSH by anterior pituitary. Thus, relaxin and inhibin have different functions and are not antagonistic.

Q40. The soluble shield of macromolecules which surrounds bacteria is called

• A. Pili
• B. Slime✓
• C. Capsule
• D. Flagella

Explanation: Slime is a loosely attached, easily removable layer that provides some protection but does not form a well-defined structure around the bacterium.

Q41. The following form of circular double stranded, self-replicating DNA molecules present in many bacteria in addition to the chromosomes are not essential for bacterial growth and metabolism:

• A. Nucleoid
• B. Plasmids✓
• C. Mesosomes
• D. Nuclear region

Explanation: Plasmids are small, circular DNA molecules that exist independently of the chromosome and carry non-essential but often beneficial genes, such as antibiotic resistance.

Q42. During cell growth, DNA synthesis takes place in the_ phase.

• A. S­-phase✓
• B. G-1 phase
• C. G-2 phase
• D. M-phase

Explanation: In S-phase, DNA replicates, doubling the genetic material for division.

Q43. The first step of central dogma is the transfer of information from:

• A. DNA to Protein
• B. DNA to mRNA✓
• C. RNA to Protein
• D. DNA to tRNA

Explanation: Transcription is the first step, where DNA is used to synthesize messenger RNA (mRNA).

Q44. In Gram -ve bacteria, the peptidoglycan layer is:

• A. Condensed
• B. Thick
• C. Moderate
• D. Thin✓
• E. Entangled

Explanation: In gram-negative bacteria, the peptidoglycan layer is relatively thin compared to gram-positive bacteria.

Q45. Arrange the following events of meiosis in correct sequence:(i) Crossing over (ii) Synapsis (iii) Terminalisation of chiasmata (iv) Disappearance of nucleolus

• A. (i), (ii), (iii), (iv)
• B. (ii), (iii), (iv), (i)
• C. (ii), (i), (iv), (iii)
• D. (ii), (i), (iii), (iv)✓

Explanation: Prophase ­I of meiosis has been divided into five sub­stages which occur in the following sequence:Leptotene ➡ Zygotene ➡ Pachytene ➡ Diplotene ➡ Diakinesis. Synapsis i.e., the pairing of homologous chromosomes occurs during zygotene. Crossing over i.e., exchange of chromatid segments occurs during pachytene. Terminalisation of chiasmata i.e., shifting of chiasmata towards the ends of chromosomes and complete disappearance of nucleolus take place during diakinesis.

Q46. The interphase of meiosis lacks:

• A. G1 phase
• B. G2 phase✓
• C. Interphase
• D. None of these

Explanation: Interphase of first meiotic division lacks G2 phase, while interphase of second meiotic division lacks S-phase.

Q47. The hereditary material is:

• A. DNA✓
• B. Protein
• C. Both DNA and Protein
• D. Neither DNA nor Protein

Explanation: DNA carries genetic information and is responsible for inheritance.

Q48. DNA polymerase enzyme for PCR is isolated from bacteria, Thermus aquaticus, because:

• A. It can work at high speed
• B. It can withstand high denaturation temperature✓
• C. It can withstand low denaturation temperature
• D. It can be used again and again

Explanation: DNA polymerase from Thermus aquaticus (Taq polymerase) is thermostable, so it does not denature at the high temperatures (-95 °C) used during the PCR denaturation step.

Q49. Highly condensed proteins of chromatin are called:

• A. Homochromatin
• B. Heterochromatin✓
• C. Semi-heterochromatin
• D. Semi-homochromatin

Explanation: Heterochromatin is the tightly packed form of DNA, making it less transcriptionally active.

Q50. Name the protein complex that stabilizes the separated strand of DNA during replication:

• A. Double Stranded binding proteins
• B. Double-stranded binding enzymes
• C. Single-stranded binding proteins✓
• D. Single-stranded binding enzyme

Explanation: Single-stranded (SSBs) are proteins that bind to single-stranded DNA during processes such as DNA replication and repair. They stabilize the unwound DNA strands, preventing them from re-annealing or forming secondary structures. By keeping the strands separate, SSBs allow other enzymes, like DNA polymerases, to access the DNA for replication and repair.

Q51. The longest stage of interphase in the human cell cycle is:

• A. M-phase
• B. S-phase✓
• C. G2-phase
• D. Gl-phase

Explanation: In case of human cells, the average cell cycle is about 24 hours and the length of each phase is variable. For example:G1-phase = 9 hoursS-phase = 10 hoursG2-phase = 4.5 hoursM-phase MMM = 30 minutes

Q52. What type of genetic transfer was displayed in the Frederick Griffith experiment?

• A. Transduction
• B. Transformation✓
• C. Binary fission
• D. Conjugation

Explanation: Option A: Transduction is a method of genetic transfer in bacteria mediated by bacteriophages (viruses that infect bacteria). It involves the transfer of genetic material from one bacterium to another through the infection of a bacteriophage. This process was not demonstrated in the Frederick Griffith experiment. Option B: The Frederick Griffith experiment demonstrated bacterial transformation, where genetic material from one bacterium is taken up and incorporated by another bacterium, leading to a change in its characteristics. Option C: Binary fission is a form of asexual reproduction in which a single cell divides into two identical daughter cells. It does not involve the transfer of genetic material between different bacteria. Option D: Conjugation is a process of genetic transfer in bacteria that involves the direct physical contact between two bacterial cells. Conjugation was not observed in the Frederick Griffith experiment.

Q53. _ is the stage of mitosis characterized by the physical separation of _ chromatids.

• A. Interphase, daughter
• B. Telophase, genetic
• C. Metaphase, sister
• D. Anaphase, sister✓

Explanation: Anaphase is the correct answer because it is the stage in mitosis where sister chromatids are physically separated and move towards opposite poles of the cell. This separation is facilitated by the shortening of spindle fibers. In contrast, during Metaphase, chromosomes only align at the cell's equator without separation. Telophase involves the reformation of the nuclear envelope around the separated chromatids, and Interphase is the preparatory phase for cell division, not part of mitosis itself.

Q54. Crossing over occur between _ of homologous chromosomes:

• A. Sister pair of chromatid
• B. Sister chromatids of homologous
• C. Non-sister chromatids✓
• D. All of them

Explanation: During crossing over, homologous chromosomes come together in order to form a tetrad. This close contact allows the non-sister chromatids of homologous chromosomes to attach to one another and exchange nucleotide sequences.Crossing over is the biological process during meiosis (sexual reproduction) where homologous chromosomes exchange genetic material, leading to the creation of recombinant chromosomes and increased genetic variation in gametes like sperm and eggs. This process involves the physical swapping of segments between non-sister chromatids of paired homologous chromosomes, resulting in unique combinations of maternal and paternal genes in the resulting sex cells, which is crucial for the diversity and adaptation of a species.

Q55. Which of the following statements is true about metaphase?

• A. Chromosomes are thickest during this phase.
• B. Chromosomes are shortest during this phase.
• C. Chromosomes are longest during this phase.
• D. Chromosomes are shortest and thickest during this phase.✓

Explanation: During the metaphase of mitosis, the chromosomes align themselves in the equator forming a metaphase plate. The chromosomes are the shortest and thickest during this time.Metaphase is the third stage of mitosis, during which replicated chromosomes are aligned along the metaphase plate, an imaginary line across the center of the cell. This precise alignment is achieved as spindle fibers from opposite poles of the cell attach to the kinetochores of each chromosome, creating balanced tension that prepares the chromosomes for separation in the next stage, anaphase.

Q56. Which one of the following is the main cause of cancer?

• A. Mutation✓
• B. Controlled cell division
• C. Regulated mitosis
• D. Haploid division

Explanation: Cancer is caused by changes (mutations) to the DNA within cells. The DNA inside a cell is packaged into a large number of individual genes, each of which contains a set of instructions telling the cell what functions to perform, as well as how to grow and divide.Cancer arises from genetic mutations that disrupt a cell's normal growth and division, leading to uncontrolled proliferation and potential spread. These mutations can occur in proto-oncogenes, turning them into oncogenes that drive growth, or in tumor suppressor genes that lose their ability to slow or stop cell growth. DNA repair genes can also be affected, preventing the correction of other mutations, which accumulate over time.

Q57. During meiosis I, the chromosomes start pairing at the:

• A. Zygotene✓
• B. Pachytene
• C. Diplotene
• D. Leptotene

Explanation: During zygotene or zygonema of meiotic prophase I the chromosomes become shorter and thicker. The homologous chromosomes come to lie side ­by ­side in pairs. This pairing of homologous chromosomes is known as synapsis or syndesis. A pair of homologous chromosomes lying together is called a bivalent.

Q58. Which hormone possesses an anti ­insulin effect?

• A. Cortisol✓
• B. Calcitonin
• C. Oxytocin
• D. Aldosterone

Explanation: Option A is correct.Insulin decreases the level of glucose in the blood while cortisol (secreted by the middle region of the adrenal cortex) increases the blood glucose level by converting proteins and fats into carbohydrates, which are, in turn, converted to glucose.

Q59. Myelinated, single and long fiber that takes message towards the cell body is:

• A. Dendron✓
• B. Axon
• C. Dendrite
• D. Soma

Explanation: Dendron is a single, long protoplasmic extension of a neuron that carries impulses toward the cell body. Smaller branches from dendron are called dendrites.

Q60. Which of the following is involved in sleeping and waking?

• A. Thalamus
• B. Brain stem
• C. Hypothalmus✓
• D. Cerebellum

Explanation: C is correct, as along with other functions like regulating body temperature and appetite, the hypothalamus is also responsible for regulating the sleep-wake cycle. The thalamus is a relay station for sensory information, such as touch, taste, smell, sight, and hearing. It sends this information to the cerebrum, which is the largest part of the brain and is responsible for processing information and making decisions. The brain stem is the lowest part of the brain and is responsible for many vital functions, such as breathing, heart rate, and blood pressure. It also controls reflexes, such as the knee-jerk reflex. The cerebellum is responsible for balance, coordination, and movement. It also helps to regulate emotions.

Q61. The abundant inhibitory neurotransmitter found in the CNS is called?

• A. Gamma-aminobutyric acid (GABA)✓
• B. Glutamate
• C. Acetylcholine
• D. Dopamine

Explanation: GABA, or gamma-aminobutyric acid, is the primary inhibitory neurotransmitter in the central nervous system, responsible for reducing neuronal excitability and creating a calming effect by blocking nerve signals.

Q62. The diseases that arise due to age are:

• A. Alzheimer's disease✓
• B. Cystic fibrosis
• C. Sickle cell anemia
• D. Phenylketonuria

Explanation: Alzheimer's disease is a neurodegenerative disorder that is most commonly associated with aging. It is characterized by progressive memory loss, cognitive decline, and behavioral changes.

Q63. Which of the following neurotransmitters is lying outside the central nervous system?

• A. Acetylcholine✓
• B. Endorphins
• C. Gamma-aminobutyric acid
• D. Dopamine

Explanation: Acetylcholine is a neurotransmitter involved in both the central and peripheral nervous systems. It plays a key role in muscle contraction, memory, learning, and attention. In the central nervous system, acetylcholine is important for cognitive functions, while in the peripheral nervous system, it enables communication between nerves and muscles. Low levels of acetylcholine are associated with conditions like Alzheimer's disease.

Q64. Depolarization during conduction of nerve impulse is due to:

• A. Inward movement of Na+✓
• B. Inward movement of K+
• C. Outward movement of K+
• D. Outward movement of Na+

Explanation: The inward movement of Na+ (sodium ions) occurs during the depolarization phase of a nerve impulse. When a neuron is stimulated, voltage-gated sodium channels open, allowing Na+ to rush into the cell. This influx of positively charged sodium ions causes the inside of the neuron to become less negative (or more positive), leading to the generation of an action potential. This rapid change in membrane potential is what propagates the nerve impulse along the axon.

Q65. Which of the following is NOT a feature of the autonomic nervous system?

• A. Regulate response of visceral organs
• B. Regulate response of skeletal muscles✓
• C. Regulate response of glands
• D. Regulate response of smooth muscles

Explanation: The autonomic nervous system (ANS) does not directly regulate the response of skeletal muscles. Skeletal muscles are controlled by the somatic nervous system, which governs voluntary movements. However, the ANS can indirectly affect skeletal muscles by regulating blood flow, oxygen delivery, and metabolic processes during physical activity, especially through the sympathetic division during "fight or flight" responses.

Q66. The children suffering from SCIDS lack an enzyme called:

• A. Adenosine deaminase✓
• B. Homogentisic and hydroxylase
• C. Phenylalanine hydroxylase
• D. Homogentisic and dehydrogenase

Explanation: Children suffering from severe combined immunodeficiency syndrome (SCID) due to adenosine deaminase (ADA) deficiency lack this critical enzyme, which is essential for the maturation of T and B lymphocytes. Without ADA, these immune cells cannot function properly, leading to life-threatening infections. The other options refer to different enzymes that are involved in metabolic pathways unrelated to SCID: homogentisic acid and its associated enzymes deal with the metabolism of phenylalanine and tyrosine, while phenylalanine hydroxylase is specifically involved in converting phenylalanine to tyrosine. Therefore, they do not relate to the immune deficiency seen in SCID.

Q67. Stimulation of a muscle fiber by a motor neuron occurs at:

• A. The neuromuscular junction✓
• B. The transverse tubules
• C. The myofibril
• D. The sarcoplasmic reticulum

Explanation: The correct answer is the neuromuscular junction, where a motor neuron communicates with a muscle fiber. This junction includes the axon terminals of the neuron and the motor end plate of the muscle fiber. The transverse tubules, myofibrils, and sarcoplasmic reticulum have different roles in muscle function. Transverse tubules aid in action potential propagation, myofibrils are the contractile units, and the sarcoplasmic reticulum stores calcium ions, but none of these are directly involved in the initial neuron-muscle fiber interaction.

Q68. Which of the following statements is true about savannah?

• A. Dry season is very long and temperature ranges more than 18°C throughout the year✓
• B. Dry season is very short and temperature ranges less than 18°C throughout the year
• C. Dry season is very long and temperature ranges less than 18°C throughout the year
• D. Dry season is very short and temperature ranges more than 18°C throughout the year

Explanation: This is the correct answer. Savannas are characterised by a distinct dry season, which can last for several months. The temperatures in savannas are generally warm to hot throughout the year, with average temperatures often exceeding 18°C. Therefore, the statement that is true about savannas is that the dry season is very long and the temperature ranges more than 18°C throughout the year.

Q69. When a charged particle is projected perpendicular to a magnetic field:

• A. Its path is circular in a plane perpendicular to the plane of magnetic field✓
• B. The speed and kinetic energy of the particle not remains constant
• C. The velocity and momentum of the particle changes in direction and magnitude
• D. The time period of revolution, angular and frequency of revolution is independent of velocity of the particle and radius of circular path

Explanation: When a charged particle is projected perpendicular to a magnetic field,The path of particle is circular in a plane perpendicular to the plane of magnetic field.The correct option is A.

Q70. A parallel plate capacitor has a capacitance C. If the distance between the plates halves, the capacitance will be?

• A. 0.5 C
• B. 4 C
• C. 0.25 C
• D. 2 C✓

Explanation: The capacitance is inversely proportional to distance between the plates. Halving the distance would double the capacitance giving us 2C.

Q71. Which statement describes the electric potential difference between two points in the electric field of charge Q?

• A. The difference of electric field between the points per unit charge.
• B. The ratio of the power dissipated between the points to the mass of charge.
• C. The work done in moving a test charge between points divided by the magnitude of the test charge.✓
• D. The force required to move a unit positive charge between the points per unit charge.

Explanation: This question is simply revising how the electric potential is defined as it is work done per unit charge in moving that charge from one place to the other in the electric field.All other options are not the definitions of electric potential.

Q72. Two capacitors C1 = 2µ and C2 = 4µ F are connected in series across in a 100V supply. Find the effective capacitance.

• A. ½ µF
• B. 3/2 µF
• C. 5/2 µF
• D. 4/3 µF✓
• E. 2 µF

Explanation: Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)Therefore,1/Ceffective = 1/C1 + 1/C21/Ceffective = ½ +¼ 1/Ceffective = ¾C effective= 4/3 µF Hence option D is the correct answer.

Q73. Two charges of -2µC and +4µC are a fixed distance apart. As shown below: A is +4µC and B is -2µC CAt which position can the electric field be zero?

• A. at A only
• B. at B only
• C. at C only✓
• D. at A, B and C

Explanation: Point C has no charge thus the electric field can be zero. (Note that when the charge is positive, the electric field is away from the point and vice versa so point A and B can never be zero).

Q74. Electric potential energy per unit charge is:

• A. Electric flux
• B. Electric field
• C. Electric potential✓
• D. Electric intensity

Explanation: OPTION A: Electric flux is defined as: The number of lines of force that pass through the area placed in the electric field.OPTION B: An electric field is defined as any region around a charge in which an electric test charge would experience an electric force.OPTION C: Energy per unit charge is called electric potential.V=E/qHence, C is the correct option.OPTION D: The intensity of an electric field at any point is the force per unit positive test charge placed at that point.E=F/qo

Q75. The value of K depends upon _.

• A. Charges
• B. System of units and nature of medium✓
• C. The distance between charges
• D. Nature of medium

Explanation: K depends on •System of units •Nature of medium separating charges

Q76. A 50 Ampere-hour battery can supply a current of 50 A for 1 hour, 25 A for 2 hour, and so on. Then the total energy stored in the 12V-50 Ampere-hour battery is:

• A. 600 J
• B. 600 Wh✓
• C. Depends on for how much time it is used
• D. 3.6× 104 J

Explanation: The energy stored in a battery can be calculated using the formula: Energy (Wh) = Voltage (V) × Capacity (Ah). In this case, the battery has a voltage of 12V and a capacity of 50Ah. Therefore, the total energy stored is:Energy = 12V × 50Ah = 600Wh.Option A (600 J) is incorrect because it uses joules instead of watt-hours as the unit of energy for this context. Option C is misleading because, although the time affects how long the energy can be drawn, it does not change the total stored energy. Option D (3.6 × 104 J) is also incorrect as it represents a much lower value, equivalent to 10 Wh, which does not reflect the actual energy capacity of the battery. Thus, the correct answer is Option B (600 Wh).

Q77. In the given circuit R1 > R2 . Find which of the following statements is correct.

• A. Potential difference across R1 is higher
• B. Current through R1 is greater than in R2
• C. Power consumed in R1 is greater than in R2
• D. None of these✓

Explanation: In a parallel circuit, all resistors have the same potential difference across them, regardless of their resistance. So, the voltage across R1 and R2​ is equal.The current through each resistor in parallel depends on its resistance. The resistor with less resistance allows more current. Since R1>R2R_1 > R_2R1​>R2​, the current through R1 is less than through R2Similarly, power dissipated in a resistor is inversely proportional to resistance when voltage is constant. So, the resistor with lower resistance (i.e., R2) will consume more power than R1.

Q78. An electric heater is designed to operate in a 100 V main with a power output of 1000 W. When it is connected to a 25 V source, the power output is:

• A. 40 W
• B. 100 W
• C. 62.5 W✓
• D. 250 W

Explanation: To solve this problem, we need to determine the resistance of the heater at its rated conditions. The heater is rated for 100 V and 1000 W. Using the formula for power, P = V²/R, we can find the resistance:R = V²/P = 100×100/1000 = 10Ω.Now, using the same resistance with a new voltage of 25 V, we calculate the new power output:P = V²/R = 25×25/10 = 62.5 W.Therefore, the correct answer is 62.5 W.The other options are incorrect because they either use incorrect resistance or do not apply the formula correctly under the changed voltage conditions.

Q79. In the circuit shown, the potential drop across the 6 Ω resistor is 12 V. The emf of the ideal battery is:

• A. 8 V
• B. 16 V
• C. 24 V✓
• D. 32 V

Explanation: To solve this problem, we use Ohm's Law, V = IR. Given that the potential drop across the 6 Ω resistor is 12 V, we can calculate the current through it: I = V/R = 12 V / 6 Ω = 2 A. Assuming this is the only resistor in the circuit, and since the battery is ideal (no internal resistance), the emf of the battery must equal the potential drop required to maintain this current, which is 24 V. Thus, Option C is correct. All other options either underestimate or overestimate the necessary emf based on the given potential drop across the resistor.

Q80. The current in a resistor is 8.0 mA. What charge flows through the resistor in 0.020 s?

• A. 0.16 mC✓
• B. 40 mC
• C. 1.6 mC
• D. 0.40 mC

Explanation: The correct answer is 0.16 mC. To find the charge (Q) that flows through the resistor, we use the formula Q = It, where I is the current (8.0 mA or 8 x 10-3 A) and t is the time (0.020 s). Substituting the given values, Q = 8 x 10-3 x 0.020 = 0.16 mC. This calculation is straightforward if we correctly convert the current from milliamps to amps and apply the formula. The other options result from various miscalculations, such as incorrect multiplication or misunderstanding of unit conversion.

Q81. Resistance of tungsten wire at 150 oC is 133Ω. Its resistance temperature coefficient is 0.0045/oC. The resistance of this wire at 500oC will be:

• A. 180Ω
• B. 225Ω
• C. 342Ω✓
• D. 317Ω

Explanation: To find the resistance of the tungsten wire at 500 oC, we apply the resistance-temperature relationship formula:R = R0(1 + α(T - T0))Where:- R0 = 133Ω (resistance at T0 = 150 oC)- α = 0.0045/oC (temperature coefficient)- T0 = 150 oC- T = 500 oCSubstituting these values into the formula gives:R = 133Ω(1 + 0.0045(500 - 150))Calculating the temperature difference: 500 - 150 = 350 oCNow, calculating the coefficient effect: 0.0045 × 350 = 1.575So, R = 133Ω(1 + 1.575) = 133Ω × 2.575 ≈ 342Ω.Thus, the correct answer is 342Ω. The other options (180Ω, 225Ω, and 317Ω) are incorrect because they do not accurately apply the formula or account for the significant increase in resistance due to the temperature change.

Q82. SI unit of resistivity is _

• A. Ω-m2
• B. (Ω-m)-1
• C. Ω-m✓
• D. (Ω-m)-2

Explanation: Resistivity (ρ) is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is given by the equation ρ = RA/L, where R is resistance, A is cross-sectional area, and L is length. Substituting the respective units, we get ρ = Ω × m²/m = Ωm, making Ωm the correct SI unit for resistivity.Option A (Ω-m2) is incorrect because it is not a recognized unit for resistivity. Option B ((Ω-m)-1) is incorrect as it represents the unit of conductivity, the reciprocal of resistivity. Option D ((Ω-m)-2) is incorrect since it does not correspond to any meaningful electrical property.

Q83. A charged particle is moving in a circular orbit of radius 6 cm with a uniform speed of 3x106 m/s under the action of a uniform magnetic field 2x10-4 Wb/m2 at right angles to the plane of the orbit. The charge-to-mass ratio of the particle is:

• A. 5x109 C / kg
• B. 2.5x1011 C / kg✓
• C. 5x1011 C / kg
• D. 5x1012 C / kg

Explanation: Following is the solution to this question:

Q84. A charged particle moves through a magnetic field directed perpendicular to its direction of motion. Which of the following quantities of the particle will not change?

• A. Momentum
• B. Speed✓
• C. Velocity
• D. None of these

Explanation: Following is the solution to this question:

Q85. A coil of 600 turns is threaded by a flux of 8x10-5 webers if this flux is reduced to 3x10-5 webers in 0.015 seconds. The average induced e.m.f. is:

• A. -2.0 volts
• B. -3.0 volts
• C. +2.0 volts✓
• D. +2.5 volts
• E. +3.0 volts

Explanation: Emf = - N (dФ/dt)where number of turns are denoted by N, dФ is the change in flux and dt is the time interval.Change in flux = Final Flux - Initial FluxChange in flux = (3 × 10⁻⁵) - (8 × 10⁻⁵) WbChange in flux = - (5 × 10⁻⁵) Wbtime interval = 0.015 sectime interval = 15 × 10⁻³ secNumber of turns in coil = 600Putting all the values in formula.Emf = - 600 × (-(5 × 10⁻⁵)/15 × 10⁻³) VoltsEmf = 600 × (5 × 10⁻²/15 ) VoltsEmf = 600 × (10⁻²/3 ) VoltsEmf = 200 × 10⁻² VoltsEmf = 2 Volts

Q86. A transformer is based on the theory of:

• A. Self inductance only
• B. Mutual inductance✓
• C. Capacitive effect
• D. All of these options are correct

Explanation: This is a fact to be memorized. A transformer is based on the theory of mutual inductance. Hence, Option B is correct.

Q87. The emf induced in a circuit according to Faraday's law depends on the:

• A. Maximum magnetic flux
• B. Initial magnetic flux
• C. Change in magnetic flux
• D. The rate of change of magnetic flux✓

Explanation: According to Faraday's laws of electromagnetic induction, induced emf in a coil depends upon the number of turns and the rate of change of flux linking the coil. The larger the rate of change of current, the greater is emf induced in the coil.

Q88. Which of the following phenomenon proves that light waves are transverse waves?

• A. Polarization✓
• B. Refraction
• C. Interference
• D. Diffraction

Explanation: Only transverse waves can undergo polarisation. Longitudinal waves cannot undergo polarization. Since light can undergo polarisation, it is a transverse wave.In summary, while refraction, interference, and diffraction are all phenomena that demonstrate the wave nature of light, they do not specifically prove that light waves are transverse waves. Each of these phenomena can be understood within the broader framework of wave optics, which encompasses both transverse and longitudinal waves.

Q89. A star moving away from earth shows?

• A. Green shift
• B. Red shift✓
• C. Blue shift
• D. None of these options is correct

Explanation: When a star emits light, the color of its light as observed on earth depends on its motion relative to earth. If a star is moving towards the earth then its light is shifted to higher frequencies on the color spectrum (towards the gamma-ray end of the spectrum). This shift towards higher frequency is called a "blue shift". The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum (towards the orange/red/infrared/microwave/radio end of the spectrum). A lower frequency shift is called a "redshift". The faster a star moves away from the earth, the more its light is shifted to lower-frequency colors. This effect is known as the "Doppler shift".

Q90. Which one of the following requires no medium?

• A. Beta rays
• B. Alpha rays
• C. Gamma rays✓
• D. All of these

Explanation: Out of all these, Gamma rays are an example of electromagnetic waves, which require NO medium to travel and propagate from one place to the other or even through a vacuum.On the other hand, mechanical waves require a medium to travel and propagate from one place to the other and can NOT pass through the vacuum. Examples of mechanical waves are sound waves, alpha rays, beta rays, etc.The picture attached represents the 'Electromagnetic Spectrum', of which 'Gamma Rays' are a part of.

Q91. The region that lies between IR and UV regions is called:

• A. Radio waves
• B. None of these
• C. X ray region
• D. Visible light✓
• E. All of these

Explanation: The visible light region falls between the ultraviolet (UV) and infrared (IR) regions on the electromagnetic spectrum. The UV region has a wavelength in the range of 100-400 nm whereas the IR region has a wavelength in the range of 760-1 million nm. The narrow range of 400-700 nm represents rays in the visible light range; rays of wavelengths visible to the human eye.

Q92. An electromagnetic wave travels in a straight line through a vacuum. The wave has a frequency of 6.0 THz. What is the number of wavelengths in a distance of 1.0 m along the wave?

• A. 5.0 × 10–5
• B. 2.0 × 101
• C. 2.0 × 104✓
• D. 5.0 × 107

Explanation: Explanation is given in image.

Q93. Mutual inductance has a practical role in performance of:

• A. AC generator
• B. Radio choke
• C. DC generator
• D. Transformer✓

Explanation: In a transformer, mutual induction occurs when voltage is induced in the secondary coil with the help of change in flux in the primary coil. Hence, we can say that mutual induction has a practical role in the working of a transformer, and D option is the correct answer.

Q94. In a half-wave rectifier, the diode conducts during:

• A. One half of the input cycle✓
• B. A portion of the positive half of the input cycle
• C. A portion of the negative half of the input cycle
• D. Both halves of the input cycles

Explanation: The answer is A since the positioning of the diode would ultimately decide whether the positive half is allowed to flow or the negative half. Since we aren’t aware of the position/direction of the diode, we would go with A.

Q95. In a step up transformer _.

• A. Vs > Vp while Is > Ip
• B. Vs < Vp while Is > Ip
• C. Vs = Vp while Is > Ip
• D. Vs > Vp while Is < Ip✓

Explanation: A step-up transformer increases the secondary voltage (Vs) compared to the primary voltage (Vp) while decreasing the secondary current (Is) compared to the primary current (Ip). This is consistent with the conservation of energy, where the power (P = V × I) remains constant (minus losses). Therefore, as voltage increases, current must decrease. Option A is incorrect because both voltage and current cannot increase simultaneously due to conservation of energy. Option B is incorrect as it describes a step-down transformer. Option C is incorrect because it suggests no voltage change, which does not apply to step-up transformers.

Q96. In electronic circuits, a PN junction diode is used to convert alternating current into direct current. This process is called:

• A. Rectification✓
• B. Amplification
• C. Doping
• D. Integrated Circuit

Explanation: Rectification is the conversion of alternating current to direct current. Rectification is performed by a diode that allows current to flow in one direction but not in the opposite direction.

Q97. An A.C emf of E = 100 sin (100 π t) volt is concerned with a choke of negligible resistance. In order to produce current of amplitude 1 A, the inductance of the choke should be:

• A. 200 H
• B. 2 π H
• C. 1 / π H✓
• D. 2 / π H

Explanation: This is the following solution.

Q98. The principle of an AC generator is.

• A. Lenz's Law
• B. Faraday's Law✓
• C. Self-Induction
• D. Ampere's Law

Explanation: The correct answer is Faraday's Law. AC generators operate on the principle of electromagnetic induction, where an emf is induced in a coil as it rotates in a magnetic field, causing a change in the magnetic flux through the coil. This principle is succinctly described by Faraday's Law, which states that the induced emf in a circuit is proportional to the rate of change of magnetic flux through the circuit.Lenz's Law, while related to electromagnetic induction, describes the direction of the induced emf and current. Self-induction refers to emf induced by a change in current in the same coil, not by external magnetic fields. Ampere's Law deals with the relationship between current and magnetic fields in closed loops, which is not directly applicable to the operation of generators.

Q99. In case of half wave rectification the resistance of diode during negative half of A.C is:

• A. Very high✓
• B. Very low
• C. A few ohms
• D. Negative

Explanation: A half-wave rectifier converts an AC signal to DC by passing either the negative or positive half-cycle of the waveform and blocking the other. Half-wave rectifiers can be easily constructed using only one diode, but are less efficient than full-wave rectifiers. During negative half of half wave rectifier, the diode is reverse biased i.e the current of few ohms and hence very high resistance.

Q100. Transformer is used in rectification to _ the supply voltage.

• A. Step up
• B. Step down✓
• C. Equalize
• D. None of these options is correct

Explanation: The purpose of the transformer is to step down the AC voltage from the mains supply and feed it to the rectifier.

Q101. If a substance can undergo plastic deformation, until it breaks, it is:

• A. Ductile substance✓
• B. Brittle substance
• C. Crystalline substance
• D. Polymeric substance

Explanation: Substances that elongate considerably and undergo plastic deformation before they break are known as ductile substances (A).

Q102. Hooke's law states that under normal conditions:

• A. Stress is inversely proportional to strain till elastic limit
• B. Stress is directly proportional to strain till elastic limit✓
• C. Stress is independent of strain
• D. Stress is proportional to elastic modulus

Explanation: Hooke's law is a principle in physics that relates to the elasticity of materials. It states that, under normal conditions, the stress applied to a solid material is directly proportional to the strain that results as long as the material remains within its elastic limit. This means that the amount of deformation or stretching of the material is directly proportional to the force applied until a point is reached where the material can no longer return to its original shape and permanent deformation occurs. The constant of proportionality between stress and strain is called the modulus of elasticity or Young's modulus.

Q103. Strain energy in deformed energy is stored in the form of:

• A. Elastic Energy
• B. Potential Energy✓
• C. Plastic Energy
• D. Kinetic Energy

Explanation: For an object that is deformed energy as stored as the potential energy and is known as the Strain energy.Elastic energy is the energy stored in a deformed object when it undergoes elastic deformation. This energy is stored in the form of potential energy within the object's molecular or atomic bonds.Potential energy is a type of energy associated with the position or configuration of an object. In the case of a deforming train, potential energy is stored as the train's shape changes. This potential energy can be released or converted into other forms of energy, such as kinetic energy, when the train returns to its original shape or undergoes further deformation.Plastic energy Is not a correct term.Kinetic energy is the energy of motion. While a deforming train may have some kinetic energy due to its motion.

Q104. The stress-strain graph deduced the following limits successively:

• A. Proportional limit, yield limit, elastic limit
• B. Yield limit, elastic limit, proportional limit
• C. Proportional limit, elastic limit, yield limit✓
• D. Elastic limit, proportional limit, yield limit

Explanation: The following limits are deduced in the order; Proportional limit, elastic limit, yield limit.

Q105. The ratio of applied stress to volumetric strain is called:

• A. Bulk Modulus✓
• B. Shear Modulus
• C. Tensile modulus
• D. Young’s Modulus

Explanation: Within the elastic limit, the ratio of volumetric stress to the corresponding volumetric strain in a body is always constant, which is called the bulk modulus of elasticity. In simpler words, the bulk modulus is nothing but a numerical constant that is used to measure and describe the elastic properties of a solid or fluid when pressure is applied to all the surfaces.

Q106. A completely filled band is called:

• A. Valence band✓
• B. Conduction band
• C. Forbidden band
• D. None of these

Explanation: In certain specific materials like insulators and semiconductors, the valence band is typically not fully filled at normal ambient conditions. It contains electrons occupying various energy states, but there are still available energy levels within the valence band for electrons to move around and participate in bonding. However, in some cases, particularly with certain elements or compounds, the valence band can be nearly full or completely filled. This configuration usually contributes to their insulating properties because there's limited availability of energy states for electrons to move and conduct electricity.

Q107. Near absolute zero temperature pure semi-conductors behave like:

• A. Conductors
• B. Metals
• C. Insulators✓
• D. None of these

Explanation: Near absolute zero temperature, pure semiconductors behave more like insulators. At extremely low temperatures, most of the electrons are bound tightly in the valence band, and very few are able to overcome the energy gap to move into the conduction band. This lack of available electrons in the conduction band results in minimal conductivity, similar to how insulators behave due to the absence of free charge carriers at normal temperatures.

Q108. In full wave rectifier, diodes are used:

• A. Two✓
• B. One
• C. Three
• D. Five

Explanation: In a full-wave rectifier, either two diodes or four diodes can be used depending on the specific configuration. A center-tapped full-wave rectifier uses two diodes and a center-tapped transformer. On the other hand, a bridge rectifier configuration uses four diodes arranged in a bridge circuit. Both setups achieve full-wave rectification, but they utilize different numbers of diodes in their designs.

Q109. As the conduction in a P-type substances takes place due to holes, so they are known as_.

• A. Majority carriers✓
• B. Minority carriers
• C. P-type
• D. N-type

Explanation: In a P-type semiconductor, holes are considered the majority carriers. Majority carriers are the dominant charge carriers in a semiconductor material, meaning they are the most prevalent type of mobile charge carriers in that material. In P-type semiconductors, where holes are the primary carriers, they outnumber the electrons, which are the minority carriers in this context.

Q110. Minimum energy required to eject an electron from metal surface is called:

• A. Stopping potential
• B. Electromotive force
• C. Work function✓
• D. Threshold frequency

Explanation: The work function is the minimum energy needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface. Stopping potential is defined as the potential required to stop ejection of electrons from a metal surface when an incident beam of energy greater than the work potential of metal is directed on it. Threshold frequency is the minimum frequency of incident light that can cause emission of electrons from the metal surface.

Q111. If frequency of incident light is increased beyond threshold frequency, keeping light intensity constant, then:

• A. Stopping potential increases✓
• B. Stopping potential decreases
• C. Stopping potential remains constant
• D. Stopping potential becomes infinite

Explanation: The stopping potential is the voltage required to completely stop photoelectrons. The energy of the electrons are directly related with the magnitude of the stopping potential. The greater the frequency of the incident photon, the higher the energy of the electron and the higher will be the stopping potential.

Q112. The energy of an X-ray quantum of wavelength 1.0 x 10-10 m is:

• A. 1.99 x 10-15 J✓
• B. 3 x 108 J
• C. 6.6 x 10-34 J
• D. 19.89 x 10-26 J
• E. 1.99 J

Explanation: ->v=fλ3 x 108=f x (1.0 X 10-10)f=3 x 108/1.0 X 10-10=3 x 1018->E=hfE=(6.62607004 × 10-34 )*(3 x 1018)=1.99 x 10-15

Q113. As the temperature of the black body is raised, the black body radiations become richer in:

• A. Intermediate wavelengths
• B. Longer wavelengths
• C. Shorter wavelengths✓
• D. Low frequencies

Explanation: According to Wien's Law, as the temperature of the blackbody increases, the wavelength of emitted radiations decreases.Hence the answer is CThe intensity (or flux) at all wavelengths increases as the temperature of the blackbody increases. The total energy being radiated (the area under the curve) increases rapidly as the temperature increases (Stefan–Boltzmann Law).

Q114. An electron moving with velocity v has momentum 3 x 10 -26 Kg.m/s. The de Broglie wavelength associated with it is _. Value of h = 6.63 x 10-34 is.

• A. 24.1 nm
• B. 22.14 m
• C. 22.1 nm✓
• D. 22.1 mm

Explanation: The de Broglie wavelength formula is λ = h/p, where h is Planck's constant and p is the momentum of the electron. Given the momentum p = 3 × 10-26 Kg.m/s and h = 6.63 × 10-34 J.s, substitute these values into the formula to find λ:λ = 6.63 × 10-34 / 3 × 10-26 = 22.1 nm.Option A is incorrect due to a calculation error. Option B is incorrect because it suggests a much larger and incorrect unit (meters). Option D is incorrect due to the wrong unit (millimeters). Thus, Option C is the correct answer with the correct units and calculation.

Q115. Calculate the energy of a photon, with a frequency of 3.0 x 1018 Hz.(h = 6.63 x 10-34)

• A. 19.89 x 10-16 J✓
• B. 11.89 x 10-16 J
• C. 1.89 x 10-16 J
• D. 19.89 x 10-18 J

Explanation: The correct option is A.E=hf = 6.63×10-34 × 3×10¹⁸ = 19.89×10-16 J

Q116. The reverse process of pair production is known as:

• A. Annihilation of matter✓
• B. Anti-pair production
• C. Materialization of matter
• D. Annihilation of practical into its anti-particle

Explanation: When a particle and its antiparticle collide, they annihilate each other, converting their mass into energy according to Einstein's famous equation E=mc2. This process is known as the annihilation of matter and it is the reverse of pair production.

Q117. Which of the following physical phenomena connot be described only by the wave theory of the electromagnetic radiation?

• A. Diffraction
• B. Interference
• C. Photoelectric Effect✓
• D. Polarization

Explanation: The correct answer is the Photoelectric Effect. This phenomenon cannot be adequately described by the wave theory of light alone because it demonstrates that light can behave as a stream of particles (photons), with each photon having a discrete energy that can be transferred to electrons. This is in contrast to wave-based phenomena like diffraction, interference, and polarization, which can be fully explained by the wave theory of electromagnetic radiation.

Q118. Find the longest wavelength in the Paschen series, given R = 1.097 x 107m-1.

• A. 11097 A
• B. 18750 A✓
• C. 17884 A
• D. 19001 A

Explanation: Option B is correct according to the calculations.

Q119. Which of the following series lie in the visible region?

• A. Lyman
• B. Paschen
• C. Balmer✓
• D. Pfund

Explanation: Four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths between 400nm−700nm.The Lyman, Balmer, Paschen, and Pfund series are terms used to describe different series of spectral lines in the emission spectrum of hydrogen and other elements. These series correspond to transitions of electrons between different energy levels within the atom.1. **Lyman Series:** The Lyman series is a series of spectral lines in the ultraviolet region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 1 energy level (ground state) of the hydrogen atom. The lines in this series are named after the scientist Theodore Lyman who studied them.2. **Balmer Series:** The Balmer series is a series of spectral lines in the visible region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 2 energy level of the hydrogen atom. The lines in this series were discovered by Johann Balmer.3. **Paschen Series:** The Paschen series is a series of spectral lines in the infrared region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 3 energy level of the hydrogen atom. The lines in this series were studied by Friedrich Paschen.4. **Pfund Series:** The Pfund series is a series of spectral lines in the infrared region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 4 energy level of the hydrogen atom. The lines in this series were observed by August Herman Pfund.These series of spectral lines occur because of the quantized nature of energy levels in atoms. When electrons transition from higher energy levels to lower energy levels, they emit photons of specific energies, which correspond to specific wavelengths or frequencies of light. Each series corresponds to transitions ending at a specific lower energy level, and the lines within each series represent different possible paths for electrons to take while transitioning to that level.These spectral series provided crucial insights into the quantization of energy in atoms and played a significant role in the development of quantum mechanics.

Q120. The ratio of the wavelength of the last line of the Balmer series and the last line of the Lyman series is:

• A. 2
• B. 1
• C. 4✓
• D. 0.05

Explanation: Option C is correct. The following is the correct solution:

Q121. A radioactive substance has a half-life of four months. 3/4 of the substance will decay in:

• A. 6 months
• B. 12 months
• C. 8 months✓
• D. 16 months

Explanation: For each half-life, the quantity of the substance will be half. Starting from 100%3/4 of 100% = 75% which means 25% should remain.100% ---> 50% ---> 25%It took 2 half-lives to reach 25% which is 3/4th of the original value hence, the time taken is 4 x 2 = 8 months.

Q122. Half-life of radioactive elements depends upon:

• A. Amount of sample present
• B. Pressure
• C. Temperature
• D. None✓

Explanation: Half-life is a natural process that does not depend upon the amount of sample present, pressure, and temperature.

Q123. 92U238nucleus emits two α-particles and two β-particles and transforms into a thorium nucleus. Which of the following is the mass number and atomic number of the thorium nucleus so produced?

• A. 230, 90✓
• B. 234, 90
• C. 230. 88
• D. 234, 88

Explanation: The equation can be written as:23892U → 23090Th + 242He + 2 0-1e

Q124. Carbonates of sodium are stable but carbonates of lithium are less stable because of?

• A. Low electronegativity
• B. High charge density✓
• C. Low charge density
• D. Not known yet

Explanation: Lithium ion (Li⁺) is very small with a high charge density. This high charge density polarizes the carbonate ion (CO₃²⁻) strongly, making lithium carbonate less stable. Sodium ions (Na⁺) are larger with lower charge density, so their carbonates are more stable.

Q125. Homocyclic hydrocarbons are further divided into:

• A. Alicyclic and aromatic✓
• B. Carbocyclic and heterocyclic
• C. Acyclic and heterocyclic
• D. Heterocyclic and aromatic

Explanation: Homocyclic compounds are divided into two compounds. Alicyclic compounds and aromatic compounds.

Q126. Which contains sp2 hybridization?

• A. Methane
• B. Ethane
• C. Alkyne
• D. None of these options are correct✓

Explanation: The hybridization of methane molecules occurs by mixing one orbital with three p orbitals. In ethane (CH3CH3), both carbons are sp3-hybridised. Alkynes undergo sp hybridization.

Q127. Which of the following compounds will give a secondary alcohol after reaction with NaBH4?

• A. CH3COCH3✓
• B. CH3COOCH3
• C. CH3CH2CHO
• D. CH3CH2COOH

Explanation: Option A is a ketone, and its reduction (NaBH4 is a reducing agent) results in a secondary alcohol. This is because the carbonyl functional group is present in the middle of the chain so, the hydroxyl group is formed in the same place after reduction, giving a secondary alcohol.Option B is an ester, which does not undergo reduction with NaBH4.Option C is an aldehyde, that forms a primary alcohol, upon reduction, as its carbonyl group is towards the terminal end of the chain, thereby forming a terminal hydroxyl group.Option D is a carboxylic acid, which does not undergo reduction with NaBH4.

Q128. The test used to distinguish among primary, secondary, and tertiary alcohols is:

• A. 2,4 - DNPH test
• B. Tollens test
• C. Lucas Test✓
• D. Fehling solution test

Explanation: The Lucas test is used to differentiate primary, secondary, and tertiary alcohols. It is based on the rate of turbidity formation due to alkyl chloride formation. Tertiary alcohols react fastest, secondary slower, and primary show little or no reaction.

Q129. Ketones can be made by oxidation of:

• A. Aldehydes
• B. Primary Alcohols
• C. Secondary Alcohols✓
• D. Tertiary Alcohols

Explanation: Option A; aldehydes oxidize into carboxylic acids.Option B; primary alcohols oxidize into aldehydes and then to carboxylic acids.Option C; secondary alcohols oxidize into ketones.Option D; tertiary alcohols are resistant to oxidation.

Q130. Alkylbenzene is formed when benzene is treated with an alkyl halide In the presence of anhydrous aluminum chloride, Identify the type of reaction.

• A. Halogenation
• B. Friedel-Crafts acylation reaction
• C. Friedel-Crafts alkylation reaction✓
• D. Sulphonation

Explanation: The type of reaction where benzene is treated with an alkyl halide in the presence of anhydrous ammonium chloride to form an alkylbenzene, is known as Friedal-Crafts alkylation reaction.

Q131. Three alternate single and double bonds in benzene are called?

• A. Conjugate bonds✓
• B. Coordinate covalent bonds
• C. Fixed bonds
• D. Ionic bonds

Explanation: Three alternate single and double bonds in benzene are called conjugate bonds. In benzene, the six carbon atoms form a planar ring structure with alternating single and double bonds.

Q132. Consider the chlorination of methane, the attack of chlorine free radical on methane form methyl free radical occurs in?

• A. Initiation step
• B. Propagation step✓
• C. Termination step
• D. Last step

Explanation: In the chlorination of methane, the attack of a chlorine free radical on methane to form a methyl free radical occurs in the propagation step.

Q133. Which of the following is the correct name of CH3CH2CH2COCH2CHO?

• A. 3-oxo hexanal✓
• B. 3-one hexanal
• C. 2-oxo hexanol
• D. 3-one hexanol

Explanation: In the given compound, there is a carbonyl group (C=O) located at the third carbon atom (counting from the left) of the carbon chain. The prefix "oxo" is used to indicate the presence of a carbonyl group. The compound is a hexanal derivative, as it contains six carbon atoms. Therefore, the correct name for the compound is 3-oxo hexanal. The compound is: CH3-CH2-CH2-CO-CH2-CHO

Q134. Which of the most suitable reagent for the conversion of R-CH2OH -->RCHO?

• A. KMnO4/NaOH
• B. Pyridinium chlorochromate✓
• C. CrO3
• D. Cr2O4/H2SO4 (Conc.)

Explanation: RCH2OH→RCHO its an oxidation reaction. Pyridinium chlorochromate is amild oxidising agent that oxidises primary alcohol to aldehydes and secondary alcohols to ketones. It does not effect any other functional group and therefore have high selectivity for oxidation of alcohols. Whereas acidic permanganate, acidic dichromate, chromic anhydride in glacial acetic acid are strong oxidising agents and can do further oxidation to carboxylic acid. The most suitable reagent for RCH2OH→RCHO is pyridinium chlorochromate (PCC)

Q135. Which among the following have least pH?

• A. CH3CH2COOH
• B. CH2CICH2COOH
• C. CH3CHCI2COOH✓
• D. CH3CH2CH2COOH

Explanation: The compound with the least pH (highest acidity) is CH3-CH-Cl2-COOH (2,2-dichloropropanoic acid) due to the presence of two chlorine atoms, which significantly enhances its electron-withdrawing effect and acidity.

Q136. Which one of the following methods is used to prepare acid anhydride?

• A. Dehydration of carboxylic acid with P2O5✓
• B. Reaction of carboxylic acid with SOCl2
• C. Reaction of carboxylic acid with NH3
• D. Reaction of carboxylic acid with alcohol in the presence of cone. H2SO4

Explanation: P₂O5 acts as a strong dehydrating agent and removes water from two molecules of a carboxylic acid, forming an acid. anhydride. Other options lead to acid chloride (SOCI2), amide (NH3) or ester (alcohol+ conc. H2SO4).

Q137. Organic acid which cannot obtained by hydrolysis of fats:

• A. Succinic acid✓
• B. Acetic acid
• C. Butyric acid
• D. Propionic acid

Explanation: All Carboxylic acids are not obtained by hydrolysis of fats because the hydrolysis of fats produces fatty acids which are monocarboxylic acids.The hydrolysis of fats is a process that breaks down fats into their constituent components, which include glycerol and fatty acids.Fatty acids are long-chain carboxylic acids that contain an even number of carbon atoms.Therefore, the hydrolysis of fats does not produce dicarboxylic acids. Aliphatic monocarboxylic acid is called fatty acid.Fats on hydrolysis give saturated fatty acids. Succinic acid is a dicarboxylic acid so it is not a fatty acid.

Q138. Which of the following compounds shows maximum Hydrogen Bonding?

• A. CH3OH
• B. CH3CH2OH
• C. C6H5OH✓
• D. C6H11OH

Explanation: Phenol has a benzene ring that can participate in pi-bonding (π-bonding) with other electron-rich aromatic systems in addition to forming hydrogen bonds. This makes it more likely to show stronger hydrogen bonding compared to the other options.

Q139. Which of the following solvents favor SN2 reactions?

• A. Water
• B. Ammonia
• C. Carbon tetrachloride✓
• D. Acetic acid

Explanation: The correct answer is Carbon tetrachloride, which is a non-polar solvent that does not stabilize ions and thus allows for a more favorable environment for SN2 reactions. In SN2 mechanisms, the nucleophile attacks the substrate in a single concerted step, and polar protic solvents like water, ammonia, and acetic acid can hinder this process by stabilizing the transition state or the leaving group, making them less favorable for such reactions.In contrast, polar aprotic solvents (like carbon tetrachloride) do not solvate the nucleophile as strongly, allowing it to remain reactive and effectively participate in the nucleophilic substitution process.

Q140. Au3+ has _ configuration.

• A. 5d7
• B. 4d9
• C. 5d8✓
• D. 5d10

Explanation: C is the correct answer. The electronic configuration of Au in its neutral state is [Xe] 4f14 5d10 6s1. When Au loses three electrons to form Au3+, it loses the 6s1 and two 5d electrons, leading to the configuration [Xe] 4f14 5d8.

Q141. In a reaction having both alkyl halide and base, the base will attack on the _.

• A. Electrophilic carbon
• B. Nucleophilic carbon
• C. Beta-hydrogen✓
• D. None of these options is correct

Explanation: In a reaction having both alkyl halide and base, the base will attack the beta-hydrogen.

Q142. Enzymes which bring about exchange of a functional group are called:

• A. Oxidoreductase
• B. Hydrolases
• C. Ligases
• D. Transferases✓

Explanation: Transferases are a class of enzymes that facilitate the transfer of a specific functional group, such as a phosphate group, a methyl group, or an amino group, from one molecule (the donor) to another molecule (the acceptor). The name of the transferase usually indicates the nature of the functional group being transferred.

Q143. Which of the following compounds will react with Tollens reagent to give metallic silver?

• A. Methanal✓
• B. Acetic acid
• C. Ethyl alcohol
• D. Acetore

Explanation: Methanal is an aldehyde that can undergo oxidation with Tollen's reagent to produce metallic silver. The rest of the compounds won't react with Tollen's reagent so option A is correct.

Q144. What a catalyst work by:

• A. Increasing the frequency of collisions
• B. Increasing the proportion of particles with energy greater than the activation✓
• C. Increasing the activation energy
• D. Increasing the concentration

Explanation: A catalyst primarily functions by increasing the proportion of particles with energy greater than the activation energy required for the reaction to occur, allowing the reaction to proceed at a faster rate. Thus, option B is the correct answer.

Q145. What is the value of C-C bond length in benzene?

• A. 154.0 pm
• B. 120.0 pm
• C. 134.0 pm
• D. 139.7 pm✓

Explanation: The correct answer is 139.7 pm. The typical C-C bond length in benzene is around 140 pm. This value may vary slightly depending on experimental conditions. The unique structure of benzene with its alternating single and double bonds contributes to its stability and aromaticity. Options A, B, and C are incorrect because they provide values that are not in line with the typical C-C bond length in benzene.

Q146. The reactivity of alkenes is due to the presence of?

• A. Sigma bond
• B. Two pi bond
• C. One pi bond✓
• D. Due to electrophilic nature

Explanation: The reactivity of an alkene is primarily due to the presence of a pi bond. This pi bond, formed by the overlapping of p orbitals, allows alkenes to undergo addition reactions and form new compounds. The characteristic reactivity of alkenes arises from the ability of the pi bond to interact with other molecules, leading to diverse chemical transformations. Therefore, the correct answer is Option C: One pi bond.While sigma bonds are essential for the structure of alkenes, they are not directly responsible for their reactivity. Two pi bonds are not typically found in alkenes, as they generally contain a single pi bond. Additionally, while alkenes can exhibit electrophilic behavior, it is the presence of pi bonds that primarily governs their reactivity.

Q147. Which of the following has the maximum number of unpaired d electrons?

• A. Mg2+
• B. Ti3+
• C. V3+
• D. Fe2+✓

Explanation: The correct answer is Fe2+. The electron configuration for Fe2+ is [Ar] 3d6, with the 3d subshell holding six electrons: four of these are unpaired due to Hund's rule, which states that electrons fill degenerate orbitals singly first. Mg2+ lacks d electrons entirely, Ti3+ has only one unpaired d electron ([Ar] 3d1), and V3+ has two unpaired d electrons ([Ar] 3d2).

Q148. Cyanohydrins can be synthesized from ketones through:

• A. Nucleophilic addition reaction✓
• B. Unimolecular Nucleophilic substitution reaction
• C. Electrophilic substitution reaction
• D. Bimolecular Nucleophilic substitution reaction
• E. Nucleophilic elimination reaction

Explanation: A cyanohydrin reaction is an organic chemical reaction by an aldehyde or ketone with a cyanide anion or a nitrile to form a cyanohydrin.This nucleophilic addition is a reversible reaction but with aliphatic carbonyl compounds equilibrium is in favor of the reaction products. The cyanide source can be potassium cyanide, sodium cyanide or trimethylsilyl cyanide.

Q149. The increasing stability order of the following compounds (1 )CH3 CH2 CH2+(2) (CH₃)₃C+(3) CH3 CH2 CH+ CH3 are ?

• A. 1 > 2 > 3
• B. 3 > 2 > 1
• C. 2> 3 > 1✓
• D. 3 > 1 > 2

Explanation: 1. CH₃CH₂CH₂⁺ (Primary carbocation):This is a primary carbocation, meaning it has one alkyl group attached directly to the positively charged carbon atom. Primary carbocations are the least stable among the three options because they lack electron-donating alkyl groups to help stabilize the positive charge.2. (CH₃)₃C⁺ (Tertiary carbocation):This is a tertiary carbocation, meaning it has three alkyl groups attached to the positively charged carbon atom. Tertiary carbocations are the most stable among the three options because they benefit from the electron-donating nature of the three alkyl groups, which helps stabilize the positive charge.3. CH3CH2CH+CH3 (Secondary carbocation):This is a secondary carbocation, meaning it has two alkyl groups attached to the positively charged carbon atom. Secondary carbocations are intermediate in stability between primary and tertiary carbocations.

Q150. When 2-bromopropane reacts with sodium ethoxide, the major product is/are?

• A. Propane
• B. Propene✓
• C. Ethyl isopropyl ether
• D. All are formed

Explanation: The correct major product when 2-bromopropane reacts with sodium ethoxide is B. Propene (propylene).

Q151. Methyl ketones can be characterized by performing:

• A. Iodoform test✓
• B. Schiff’s test
• C. Benedict’s reagent test
• D. Tollen’s test
• E. Cannizzaro’s test

Explanation: When methyl ketone is heated with iodine in the presence of sodium hydroxide, yellow precipitate of iodoform is obtained.Hence, methyl ketones are characterized by iodoform test. In this reaction, methyl ketone is converted to the sodium salt of carboxylic acid.

Q152. Formaldehyde polymerizes to form?

• A. Bakelite
• B. Paraldehyde
• C. Metaformaldehyde✓
• D. All of these

Explanation: Formaldehyde (CH2O) can undergo polymerization to form metaformaldehyde, which is a cyclic trimer consisting of repeating units of formaldehyde. This compound has the chemical formula (CH2O)3and is a direct result of the polymerization process. In contrast, Bakelite is a synthetic resin derived from phenolic compounds and formaldehyde, but it does not result from the polymerization of formaldehyde alone.Paraldehyde, while related to aldehydes in structure, is a cyclic compound formed from acetaldehyde, not from formaldehyde. Therefore, the only correct answer is metaformaldehyde.

Q153. The only o, p−directing group which is deactivating in nature is:

• A. -NH2
• B. -OH
• C. -X (halogens)✓
• D. -R (alkyl groups)

Explanation: Halides are ortho-para- directing groups but unlike most ortho-para- directors halides tend to deactivate benzene. This unusual behaviour can be explained by two properties:1. Since the halogens are very electronegative they cause inductive withdrawal (withdrawal of electrons from the carbon atom of benzene).2. Since the halogens have non-bonding electrons they can donate electron density through pi bonding (resonance donation).Option C is correct.

Q154. When an unsymmetrical alkene undergoes addition reactions, the negative part of attacking reagent is added to that double-bonded carbon atom which contains:

• A. Highest number of chloride atoms
• B. Lesser number of hydrogen atoms✓
• C. Highest number of hydrogen atoms
• D. Moderate number of hydrogen atoms
• E. Lesser number of chloride atoms

Explanation: This is based on Markovnikov's Rule, which states that in the addition of a protic acid (such as HCl, HBr, etc.) to an unsymmetrical alkene, the hydrogen atom (the positive part of the reagent) attaches to the carbon with the greater number of hydrogen atoms, and the negative part (such as Cl⁻, Br⁻, etc.) attaches to the carbon with the fewer hydrogen atoms.

Q155. According to Raoult's law, the relative lowering of vapor pressure is equal to:

• A. Mole fraction of solute✓
• B. Mole fraction of solvent
• C. Molarity
• D. Molality
• E. Parts per million

Explanation: In Raoult's Law, the lowering of vapor pressure of a solvent in a solution is directly related to the mole fraction of the solute.

Q156. It is used to distinguish between primary, secondary, and tertiary alcohol:

• A. Tollens Reagent
• B. Grignard Reagent
• C. Benedict Reagent
• D. Bloor Reagent
• E. Lucas Reagent✓

Explanation: Lucas reagent, a mixture of concentrated hydrochloric acid (HCl) and zinc chloride (ZnCl₂), is primarily used to distinguish between primary, secondary, and tertiary alcohols through a process known as the Lucas test. When an alcohol is treated with Lucas reagent, the reaction proceeds via the formation of an alkyl chloride. Tertiary alcohols react rapidly, often within minutes, to form insoluble alkyl chlorides, leading to an immediate cloudiness. Secondary alcohols take longer, while primary alcohols react very slowly, if at all, under these conditions. This differential reactivity helps in identifying the alcohol type based on the time taken for the reaction to occur.

Q157. Ketones, when treated with LiAlH4, reduce to:

• A. Primary alcohol
• B. Tertiary alcohol
• C. Alcohol
• D. Dihydric alcohol
• E. Secondary alcohol✓

Explanation: Ketones reduce to form secondary alcohols in the presence of lithium aluminum hydride (LiAlH₄). LiAlH₄ is a strong reducing agent that donates hydride ions (H⁻) to the carbonyl carbon of the ketone, resulting in the formation of the corresponding alcohol. For example, if you reduce a ketone like acetone (2-propanone), it will be converted to isopropanol (2-propanol).

Q158. Which of the following catalyst is used in the Friedel craft reaction?

• A. ZnCl2
• B. KMnO4
• C. AlCl3✓
• D. V2O5

Explanation: Anhydrous Aluminum Chloride is the catalyst used in the Friedel-craft reaction.

Q159. Reduction of aldehydes & ketones by Zn-Hg amalgam and concentrated HCl results in conversion to an alkane. This reaction is known as:

• A. Down Reduction
• B. Cope Reduction
• C. Wolff-Kishner Reduction
• D. Clemmensen reduction✓
• E. Sodium tetrahydroboride

Explanation: Clemmensen reduction is a chemical reaction used to convert carbonyl compounds (such as aldehydes and ketones) into alkanes. This reduction is typically carried out using zinc amalgam (Zn/Hg) in the presence of hydrochloric acid (HCl).

Q160. Deficiency causes loss of weight, appetite, and taste.

• A. K
• B. P
• C. Fe
• D. Zn✓

Explanation: Zinc plays a crucial role in the functioning of taste and smell receptors. A deficiency can result in altered taste perception (dysgeusia) or a diminished sense of smell (hyposmia), making food less appealing. zinc deficiency can lead to loss of appetite. This symptom is part of a broader range of effects that zinc deficiency can have on the body.

Q161. The rate of SN1 mechanism depends upon:

• A. Concentration of nucleophile
• B. Concentration of both substrate and nucleophile
• C. Polar non-polar solvent
• D. Concentration of substrate only✓
• E. Concentration of reactant

Explanation: The rate of an S_N1 reaction (unimolecular nucleophilic substitution) depends solely on the concentration of the substrate (the molecule undergoing substitution). This is because the rate-determining step is the formation of the carbocation after the leaving group departs. Once the carbocation is formed, the nucleophile can quickly react with it.

Q162. Aldehydes & ketones can be converted to alkane. This reaction is called:

• A. Ozonolysis
• B. Wolff-Kishner reaction✓
• C. Grignard reaction
• D. Friedel-Craft reaction
• E. None of these

Explanation: The Wolff-Kishner reaction is a chemical reaction used to reduce aldehydes or ketones to alkanes. This reaction involves the treatment of the carbonyl compound with hydrazine (NH₂NH₂) in the presence of a strong base (like potassium hydroxide, KOH) and heat. During the reaction, the carbonyl group (C=O) is converted into a methylene group (CH₂), effectively removing the oxygen atom. The reaction is performed under strongly basic conditions and is often used when acidic conditions are not suitable, unlike the Clemmensen reduction.

Q163. An example of a bidentate ligand among the following is:

• A. Br
• B. CN-
• C. C204-✓
• D. OH-
• E. CI

Explanation: Oxalate ion (C₂O₄²⁻): Bidentate ligand, anionic, coordinates through both oxygen atoms, forming chelate complexes with metal ions.

Q164. Which one of the following is more soluble in water?

• A. Be(OH)2
• B. Ca(OH)2
• C. Mg(OH)2
• D. Ba(OH)2✓

Explanation: Group II metal hydroxides become more soluble in water as you go down the column(group). This trend can be explained by the decrease in the lattice energy of the hydroxide salt and by the increase in the coordination number of the metal ion as you go down the column.

Q165. The carbon atom carrying positive charge and attached to three other atoms or group is called:

• A. Carbocation✓
• B. Carbanion
• C. Oxonium ion
• D. Carbene

Explanation: A carbocation is a positively charged carbon atom that is attached to three other atoms or groups. It has an electron deficiency, leading to a positive charge. Therefore, option a is correct.

Q166. Nylon-6,6 is also called:

• A. Poly styrene
• B. Polyester
• C. Polyamide✓
• D. Polyvinyl alcohol

Explanation: Nylon-6,6 is a synthetic polymer called a polyamide because of the characteristic monomers of amides in the backbone chainPolystyrene is a plastic. It is made up of many styrenes. Styrene is a small organic compound with the chemical formula C6H5CH=CH2.Polyester is a synthetic fabric that's usually derived from petroleum. This fabric is one of the world's most popular textiles, and it is used in thousands of different consumer and industrial applications. Chemically, polyester is a polymer primarily composed of compounds within the ester functional group.Poly is a water-soluble synthetic polymer. It has the idealized formula [CH₂CH]ₙ.

Q167. The electrophile which is considered to be an active agent in nitration of benzene is:

• A. NO2+✓
• B. NO2-
• C. NO+
• D. HNO2+

Explanation: Benzene reacts with nitric acid at 50-55°C in the presence of sulphuric acid to form nitrobenzene. This reaction is known as the nitration of Benzene.

Q168. Which of the following alkyl halides has the highest boiling point.

• A. n-butyliodide✓
• B. isobutyl iodide
• C. isopropyl bromide
• D. n-propyl bromide

Explanation: The given compounds are simply hydrocarbons substituted with halogens. Thus they are called alkyl halides. Some of them are branched. Generally, molecules that do not have branching increase their boiling point with respect to their increase in molecular weight.The boiling point is dependent on the forces between the molecules. Moreover, the boiling point increases when the number of carbon atoms is increased. When the number of carbon atoms increases, the surface area and the size of the molecule also increase. There are van der Waals forces between them. This is also increased with the increase in the surface area. When the force is more, more energy, i.e., more temperature, is needed for boiling. As it is obvious that the branched compounds have less surface area. Thus the force is less. So branched alkyl halides have less boiling point.

Q169. The oxidation number of cobalt in the given coordination complex is:[Co(H2NCH2CH2NH2)3]2(SO4)3

• A. III✓
• B. II
• C. IV
• D. VI

Explanation: In the given complex, [Co(H2NCH2CH2NH2)3]2(SO4)3 Cobalt is bonded with three molecules of ethylenediamine which is a bidendate ligand having neutral charge. Moreover the coordination sphere is bonded with three molecules of sulphate ions each having -2 charge. Using formula: Charge of Metal + Charge on Ligand + Charge on coordination sphere 2(x + 3(0) ) = 3(-2) 2x = -6 x = -3

Q170. Pyrole belongs to which class of compounds?

• A. Hydrocarbons
• B. Homocyclic
• C. Alicyclic
• D. Heterocyclic✓

Explanation: Pyrole is a heterocyclic compound. Heterocyclic compounds are organic compounds with a ring structure that contains in the cycle at least one carbon atom and at least one other element, such as N, O, or S.

Q171. Ethyl alcohol in the presence of H2SO4 at 170°C produces

• A. Ether
• B. Ethene✓
• C. Ester
• D. Ethane

Explanation: When ethyl alcohol reacts with H2SO4 at high temperature, beta elimination takes place resulting in formation of ethene.

Q172. Among the following hydrocarbons which one has acidic hydrogen?

• A. C2H6
• B. C2H4
• C. C2H2✓
• D. C3H6

Explanation: In alkynes,terminal alkyne has acidic hydrogen due to the high level of s character in the sp hybrid orbital, which bonds with the s orbital of the hydrogen atom to form a single covalent bond.

Q173. Greater number of Alkyl groups on substrate favours:

• A. Substrate reaction
• B. Elimination reaction✓
• C. Oxidation reaction
• D. Free radical reaction

Explanation: A greater number of alkyl groups on a substrate favors elimination reactions because it leads to the formation of more stable and substituted double bonds, reduces steric hindrance, enhances hyperconjugation, and stabilizes the transition state of the elimination reaction.

Q174. When phenol reacts with excess of bromine in aqueous solution it results in the formation of:

• A. Ortho/para bromophenol
• B. Meta bromophenol
• C. 2,4,6-Tribromophenol✓
• D. 3,5-Dibromophenol

Explanation: An aqueous solution of phenol reacts with bromine water to give white ppt. of 2,4,6- tribromophenol.

Q175. Propylene glycol and trimethylene glycol are:

• A. Functional group isomers
• B. Metamers
• C. Position isomers✓
• D. Tautomers

Explanation: Propylene glycol and trimethylene glycol are positional isomers. In positional isomers, the same molecular formula is retained, but the functional groups or substituents are attached at different positions on the carbon chain. Both of these compounds are diols (containing two hydroxyl groups), but the positioning of the hydroxyl groups differs, which gives them distinct properties.

Q176. Aldehyde and ketone on reaction with hydroxylamine form:

• A. Hydrazine
• B. Hydrazone
• C. Oxime✓
• D. Imine

Explanation: When aldehydes and ketones react with hydroxylamine (NH₂OH), they form a class of compounds known as oximes. The reaction involves the nucleophilic addition of hydroxylamine to the carbonyl group (C=O) of the aldehyde or ketone, resulting in the formation of an oxime, which contains the functional group C=N-OH.

Q177. Demonstrate control of tenses and sentence structure.It _ (rained) since last night, and it looks as if it may rain all day.

• A. Raining
• B. Is raining
• C. It has been raining✓
• D. Rained

Explanation: The perfect present form of rain, i.e., "has been raining,” will be used, as the fact that it has been raining is an action that occurred in the past and is given importance in the present.

Q178. Fill in the blanks with the most appropriate option. When we went to cinema, the film _.

• A. Already started
• B. Had already started✓
• C. Would already start
• D. Started already

Explanation: The other options are wrong as 'when we went' indicates the use of the past tense. The correct answer is B) had already started.

Q179. Choose the correct spelling:

• A. Entripreneur
• B. Enterpreneur
• C. Entrepreneur✓
• D. Entreprineur

Explanation: The correct spelling is entrepreneur. It refers to a person who sets up a business or businesses, taking on financial risks in the hope of profit.

Q180. Choose the correctly structured sentence:

• A. Had he lived in England he would miss his family.
• B. Had he lived in England he would have missed his family.✓
• C. Had he lived in England he had missed his family.
• D. Had he live in England he will missed his family.

Explanation: Option B is correct because it is the only sentence that adheres perfectly to grammar rules. The first part of the sentence is in the past perfect tense (indicated by the verb form "had he"), so the next part of the sentence should also conform to the same tense so the correct context should be “would have”.

Q181. Spot the error.Even the doctor put in his best efforts, he could not succeed in saving the patient.

• A. Even the doctor✓
• B. Put in his best efforts, he
• C. Could not succeed in
• D. Saving the patient.

Explanation: The phrase “Even the doctor” is incorrect here because it changes the meaning. “Even the doctor” implies surprise that the doctor, specifically, tried, which is not the intended conditional sense. The sentence needs “Even if the doctor” to correctly express a hypothetical situation: regardless of the doctor’s efforts, the patient could not be saved.

Q182. Which way shall we go?The underlined word is:

• A. Demonstrative adjective
• B. Interrogative pronoun
• C. Interrogative adjective✓
• D. Exclamatory adjective

Explanation: The underlined word 'which' is an interrogative adjective because it modifies the noun 'way' and directly precedes it. Interrogative adjectives are used to ask questions about nouns. The other options are incorrect: a demonstrative adjective would indicate position or distance, which 'which' does not; an interrogative pronoun would stand alone and be followed by a verb, not a noun; and an exclamatory adjective is used to express strong feelings, which is not applicable in this question context.

Q183. Synonym of "Juxtaposition" is

• A. Contrast✓
• B. Wit
• C. Image
• D. Freedom

Explanation: Option A is correct because "contrast" is the most suitable synonym for the word "juxtaposition." Juxtaposition refers to placing things side by side for comparison or contrast.

Q184. I _ (am) working all afternoon and have just finished the assignment.

• A. Had been✓
• B. Have been
• C. Am
• D. Shall be

Explanation: The sentence, 'been working all afternoon' and 'just finished' shows both past and present efforts. "Have" is a present form while "had" is the past form of ‘to have'. "Am" is for the first person singular (I am), which in this case is grammatically incorrect when said out loud. When two things that happened in the past and one event started and finished before the other one started, "Had" is placed before the main verb for the event that happened first.

Q185. Demonstrate the correct use of article:Gold is _ precious metal.

• A. A✓
• B. An
• C. The
• D. No article

Explanation: In the phrase "Gold is a precious metal," the article "a" is used because it precedes the noun "precious metal" to indicate that gold belongs to a category or class of substances known as precious metals. "A" is an indefinite article used before singular nouns to refer to any member of a general category. In this case, "precious metal" refers to a broad category of metals that are rare, valuable, and often used in jewelry or currency.

Q186. Use an appropriate article:Can you pass me _ salt?

• A. A
• B. An
• C. The✓
• D. No article

Explanation: The noun in the question is "salt" and the tense is present simple. We use "the" in a sentence when we are referring to a specific or particular noun. Hence “the” will be used.

Q187. Spot the error:Now the time was (A) to escape (B) and he opened the window (C) and jumped out(D).

• A. Now the time was✓
• B. To escape
• C. And he opened the window
• D. And jumped out

Explanation: The error is in Option A, 'Now the time was.' The phrase should be 'Now is the time,' which is a fixed expression in English indicating that a particular moment has arrived. The other options are grammatically correct.

Q188. Demonstrate correct use of articles and prepositions. My friend has been living _ Karachi for two years.

• A. On
• B. At
• C. In✓
• D. Across

Explanation: "In" is a preposition used when something is enclosed or located within a defined space, e.g., Karachi.The preposition "on" is used to indicate a surface or a position that is physically above or attached to something.The preposition "at" is used to indicate a specific location or point in time, but it does not convey the meaning of living in a place.The preposition "in" is used to indicate the location of a place where someone lives, for example, KarachiThe preposition "across" is used to indicate movement from one side of something to the other.

Q189. What type of sentence is this?When I go swimming, I have to keep my eyes closed underwater.

• A. Complex✓
• B. Simple
• C. Compound
• D. None

Explanation: This sentence is an example of complex sentence.It has an independent clause (main clause): "I have to keep my eyes closed underwater" and a dependent clause (subordinate clause): "When I go swimming".The dependent clause "When I go swimming" cannot stand alone as a complete sentence, and it relies on the main clause to make sense. The word "when" is a subordinating conjunction that introduces the dependent clause and indicates the condition or circumstance under which the main action takes place.

Q190. Choose the correct option. I prefer fruits _ sweets.

• A. Then
• B. On
• C. Over✓
• D. From

Explanation: We use prefer to say we like one thing or activity more than another. Over is used when there are two clear choices in the phrase. Think of over as setting a list of preferences and putting one over top the other. I prefer jogging over running and walking or as in this case, I prefer fruits over sweets.

Q191. From each of the following, choose the sentence with the correct comma placement.

• A. She finished her work, and then took a long lunch.✓
• B. She finished her work, and then, took a long lunch.
• C. She finished her work and then took a long lunch.
• D. She finished her work and then, took a long lunch.

Explanation: The correct sentence is: 'She finished her work, and then took a long lunch.' This is because the sentence consists of two independent clauses: 'She finished her work' and 'then took a long lunch'. According to punctuation rules, a comma should be placed before the conjunction 'and' when it connects two independent clauses. Options B and D are incorrect because they include unnecessary commas that disrupt the sentence flow. Option C is incorrect because it lacks the necessary comma before 'and'.

Q192. Choose the correct passive voice:How did she defraud him of his savings?

• A. How had he been defrauded of his savings?
• B. How had he been defrauded by her?
• C. How was he defrauded of his savings?✓
• D. How was he defrauded by her of his savings?

Explanation: This option correctly transforms the sentence into the passive voice by using the verb “was defrauded” indicating the action being done to the subject “he”. The prepositional phrase “of his savings” remains intact as it represents what was taken from him.

Q193. Choose the correct indirect speech:The teacher said, “Amna, watch your steps.”

• A. The teacher ordered Amna that she should watch her steps.
• B. The teacher ordered Amna to watch your steps.
• C. The teacher ordered Amna to watch her steps.✓
• D. The teacher requested Amna to watch your steps.

Explanation: The correct answer is: 'The teacher ordered Amna to watch her steps.' In indirect speech, the verb 'said' is typically changed to match the speaker's intention—in this case, 'ordered' is appropriate as the teacher is giving an instruction. Additionally, the possessive adjective should match the subject (Amna), so 'her' is used instead of 'your'. Other options either use incorrect pronouns or verbs that do not accurately reflect the speaker's intention.

Q194. Read the following passage to answer the given question:Right after the Civil War, many distraught soldiers made their way West to find fame and fortune. Some could not go home because there were no homes to go to. The war had devastated them. One young man, Will Goodlad, made his fortune in the hills of Colorado. He found gold in a little river near Grand Junction. His fortune was short lived, however. In 1875, he declared bankruptcy and returned to the land of his birth- the Piedmont of South Carolina. For which side did Will fight during the War?

• A. East
• B. West
• C. North
• D. South✓

Explanation: The passage indicates that Will Goodlad was born in the Piedmont of South Carolina, a state that was part of the Confederacy during the Civil War. Therefore, it is likely that he fought for the South. The terms East and West do not correspond to sides in the Civil War, and while it is possible for a Southerner to have fought for the North, the passage suggests loyalty to his birthplace.

Q195. Read the passage and the following statements below. Then choose the correct option, basing your answer only on the information provided.Queen Elizabeth II's Platinum Jubilee, celebrating her 70 years on the British throne, was above all a tribute to one of history's great acts of constancy. Her reign spanned virtually the entire post-World War II era, making her a witness to cultural upheavals from the Beatles to Brexit.STATEMENTS:I. There has been another queen of the British throne named Elizabeth before her.II. Brexit is a normal occurrence.III. Elizabeth was Queen of the British during World War II.

• A. I, II and III; all are correct
• B. Only llI is correct
• C. Only I is correct✓
• D. Only I and III are correct

Explanation: In the passage queen Elizabeth has a II with it, this shows that there has been another queen with Elizabeth in her name, Brexit is not a normal occurrence and the passage clearly states that she rules post world war 2 so only statement 1 is correct.Based on the passage, the correct option is:I. There has been another queen of the British throne named Elizabeth before her.The passage refers to Queen Elizabeth II, indicating that there was a Queen Elizabeth I before her.The passage does not support statements II and III:- II. Brexit is not described as a normal occurrence, but rather as one of the cultural upheavals that Queen Elizabeth II witnessed during her reign.- III. The passage states that Queen Elizabeth II's reign spanned "virtually the entire post-World War II era", implying that she was not queen during World War II itself.

Q196. Statement:According to reports, the childrens below age 10 are being aggressive because of mobile phones.Course of Action:I. Parents needs to be strict with their child.II. Parents need to be concerned and limit their screening time.

• A. Both of the courses of action follow
• B. None of the courses of action follow
• C. Only courses of action I follows
• D. Only courses of action II follows✓

Explanation: The root cause of children getting aggressive is their mobile phones hence limiting the time that children use their phones can solve this problem. Lenient parentage is not the problem plus strict parenting sometimes causes rebellious behavior.

Q197. Read the statement given, assume it is true, and select the correct course of action:A public sector television channel is worried about the quality of its programs and, considering the competition from several private sector television channels, has decided to provide some incentives in order to attract talent for its programs.Course of action:a) Public sector television channel has decided to revise its fee structure for artists.b) It should not revise its fee structure until the private sector channels do so.

• A. Both A and B follow
• B. Only A follows✓
• C. Only B follows
• D. Neither A nor B follows
• E. Either A or B follows

Explanation: By revising its fee structure, the public sector television channel can attract better talent, which can help improve the quality of its programs and make it more competitive against private sector channels.

Q198. Parveen, Qadir, Rehan, Salim, and Tehmina are five people in a family. If Parveen is the daughter of Qadir, Qadir is the son of Rehan, Rehan is the father of Salim, while Tehmina is the daughter of Parveen, then which of the following is true?

• A. Rehan is the uncle of Parveen
• B. Qadir is the grandfather of Parveen
• C. Qadir is the daughter of Salim
• D. Parveen is the sister of Tehmina
• E. None of them✓

Explanation: Let’s break down the relationships based on the information given:Parveen is the daughter of Qadir: This means Qadir is Parveen's father.Qadir is the son of Rehan: This means Rehan is Qadir's father.Rehan is the father of Salim: This means Rehan has two children—Qadir and Salim.Tehmina is the daughter of Parveen: This means Parveen is Tehmina's mother.

Q199. Look at the numbers below. Identify the pattern and logically deduce the correct option that should follow: 1, 4, 9, 16,

• A. 54
• B. 36
• C. 49
• D. 25✓

Explanation: These numbers are perfect squares of natural numbers.The next number should logically be 5^2 = 25

Q200. All hammers are tools. Some tools are useless things. All useless things are trash. Which of the following is NECESSARILY TRUE given only the information above?ConclusionsI. Some hammers are trash. II. Some tools are trash.III. All useless things are tools.

• A. I and III
• B. I and II
• C. II and III
• D. II✓

Explanation: Now look at the conclusions provided, For I, if we look at the diagram some hammers are not trash hence I is wrong. For II, Some tools are trash as indicated in the yellow-shaded region hence II is correct. For III, Not ALL useless things are tools rather some useless things are tools so III is also wrong. The only option correct according to the above analyses is D ( only conclusion II).