Uhs Mdcat 2012 — Solved Past Paper with Answers

Q1. He had a heart attack and all attempts to _ him failed.

• A. Renew
• B. Resuscitate✓
• C. Revise
• D. Refurnish

Explanation: Resuscitate means to revive from unconsciousness or apparent death.Renew : to resume after an interruption Revise: to examine and make corrections Refurnish: to furnish again or differently

Q2. The _ stench of dead animals and plants made Mumtaz ill.

• A. Putrid✓
• B. Purified
• C. Perturbed
• D. Purchase

Explanation: Putrid is a fetid smell decaying or rotting organic matterPurified: cleansedPerturbed: unsettled, anxious or concerned Purchase: to buy

Q3. While going up the hills by bus, she felt _ inside.

• A. Fishy
• B. Itchy
• C. Queasy✓
• D. Squeezy

Explanation: Queasy: feeling sick or nauseous Fishy: feeling of doubt or suspicion Itchy: causing an itchSqueezy: flexible

Q4. The craft statesman manipulated the situation by making false promises and declaring sport festivities a _ to fool the public.

• A. Red-Herring✓
• B. Red-Feather
• C. Red-Heiring
• D. Red-Haring

Explanation: Red herring is a clue or piece of information which is intended to be misleading or distracting.The remaining options are incorrect because they can not be used in this situation.

Q5. Select the incorrect segment:

• A. Discarded
• B. As
• C. There
• D. For✓

Explanation: The correct answer is Option D: For. The preposition "for" is incorrectly used in this context. The sentence requires "in" to accurately indicate the relationship between the actions and their reasons. The other options, "Discarded", "As", and "There", are correctly used in their respective contexts. "Discarded" properly describes the rejection of the theory, "As" correctly introduces the reason, and "There" serves as an appropriate expletive construction.

Q6. Choose the correct option:

• A. Why does not Nomana remained true to her husband?
• B. Why did not Nomana remain true to her husband?✓
• C. Why had not Nomana remain true to her husband?
• D. Why did not Nomana remained true to her husband?

Explanation: The sentence is in past tense so it would be “did” (the past tense of do). "Nomana" is a singular noun, so “remain” will be used.

Q7. Identify the correct option:

• A. All my childhood, I longed desperately in for a tricycle.
• B. All my childhood, I longed desperately to a tricycle.
• C. All my childhood, I longed desperately for a tricycle.✓
• D. All my childhood, I longed desperately at a tricycle.

Explanation: Adverbial Phrase: "All my childhood" is an adverbial phrase that provides additional information about the duration or time frame during which the action occurred. In this case, it indicates that the longing for a tricycle persisted throughout the speaker's childhood. Main Clause: "I longed desperately for a tricycle" is the main clause. It expresses the speaker's strong desire for a tricycle. Adverb "Desperately": The word "desperately" modifies the verb "longed," emphasizing the intensity or urgency of the longing. Together, the sentence effectively communicates the speaker's prolonged and intense desire for a tricycle during their entire childhood. The structure is grammatically sound and conveys the intended meaning.

Q8. Choose the correct option:

• A. Bill Gates is one of the wealthiest person in the world.
• B. Bill Gates is one of the wealthy persons in the world.
• C. Bill Gates is one of the wealthiest persons in the world.✓
• D. Bill Gates is one of the more wealthy person in the world.

Explanation: Since he is included in among the most wealthy people of the world and since there are many of those, “persons” (plural) and “wealthiest” will be used.

Q9. Choose the correct option:

• A. Her father is a SP in the Punjab Police.
• B. Her father was a SP in the Punjab Police.
• C. Her father is an SP in the Punjab Police.✓
• D. Her father are a SP in the Punjab Police.

Explanation: When an indefinite article is needed immediately before an abbreviation, the pronunciation of the abbreviation determines whether 'a' or 'an' should be used. The trick here is to use your ears (how the acronym is pronounced), not your eyes (how it's spelled). HIV (pronounced "aitch eye vee") begins with a vowel sound, so an HIV patient is correct.

Q10. Choose the correct option:

• A. Hand up the answer sheet to me.
• B. Hand over the answer sheet to me.✓
• C. Hand down the answer sheet to me.
• D. Hand for the answer sheet to me.

Explanation: "Hand over" is the correct phrasal verb.

Q11. Select the nearest correct meaning of the given word:Dissonance

• A. Inconsistency✓
• B. Expansion
• C. Perception
• D. Warp

Explanation: Dissonance is the lack of harmony.

Q12. Select the nearest correct meaning of the given word:Trifle

• A. Pudding
• B. Minor✓
• C. Deluge
• D. Treble

Explanation: Trifle means something of little importance or value.

Q13. Select the nearest correct meaning of the given word:Murky

• A. Dusty
• B. Squeamy
• C. Clear
• D. Unclear✓

Explanation: Murky means obscured. Clear would be the opposite.

Q14. Select the nearest correct meaning of the given word:Faux

• A. Blunder
• B. Mistake
• C. Indiscrete
• D. False✓

Explanation: Faux is something fake or not genuine, artificial.

Q15. Select the nearest correct meaning of the given word:Myriad

• A. Countable
• B. Multitude✓
• C. Measured
• D. Blurred

Explanation: Myriad means countless or in extremely great number.

Q16. Select the nearest correct meaning of the given word:Facile

• A. Fallacy
• B. Depict
• C. Delicate
• D. Superficial✓

Explanation: Facile is synonymous with simplistic or effortless.

Q17. Select the nearest correct meaning of the given word:Magnum

• A. Masterpiece✓
• B. Magnanimity
• C. Modest
• D. Magnetic

Explanation: Magnum is a Latin word meaning "great". "Magnum opus" means a masterpiece.

Q18. Select the NEAREST CORRECT MEANING of the given word: SIDLE

• A. Sneak✓
• B. Sift
• C. Siege
• D. Sieve

Explanation: To sidle means to walk in a furtive, unobtrusive, or timid manner, especially sideways or obliquely.

Q19. Select the nearest correct meaning of the given word:Plethora

• A. Plastic
• B. Superfluity✓
• C. Measure
• D. Malleable

Explanation: "Plethora" means a large or excessive amount of something.

Q20. Select the nearest correct meaning of the given word:Vertex

• A. Poetry
• B. Depth
• C. Zenith✓
• D. Diminish

Explanation: "Vertex" means the highest point; the top or apex. It is also synonymous with "zenith" or "peak".

Q21. The part of the neuron fibre that conducts nerve impulses away from the cell body is:

• A. Dendron
• B. Dendrites
• C. Axon✓
• D. Peripheral Branch

Explanation: Simply, nerve impulses travel through the dendron towards the cell body and are carried away from the cell body by the axon.

Q22. The number of cranial nerves in human is:

• A. 31 pairs
• B. 12 pairs✓
• C. 24 pairs
• D. 62 pairs

Explanation: Humans have 12 pairs of cranial nerves, which connect the brain to different parts of the body.

Q23. The part of brain which controls breathing, heart rate and swallowing is:

• A. Cerebrum
• B. Cerebellum
• C. Medulla✓
• D. Hypothalamus

Explanation: The medulla is responsible for, among other things, the control of involuntary activities, such as heartbeat, respiration, etc.

Q24. Syphilis is a sexually transmitted disease which is caused by:

• A. HIV / AIDS
• B. Pseudomonas Pyogenes
• C. Treponema Pallidum✓
• D. Neisseria

Explanation: The bacteria responsible for causing syphilis is treponema pallidum.

Q25. Discharge of ovum or secondary oocyte from the ovary or Graffian follicle is called:

• A. Fertilization
• B. Pollination
• C. Follicle formation
• D. Ovulation✓

Explanation: Ovulation is the process of release or discharge of ovum or secondary oocyte from the Graffian follicle after maturation of the Graffian follicle.This should not be confused with fertilization which is the fusion of male and female gametes for zygote formation.

Q26. Second meiotic division in the secondary oocyte proceeds as far as :

• A. Metaphase✓
• B. Prophase
• C. Anaphase
• D. Telophase

Explanation: Until fertilization, the secondary oocyte stops at the metaphase. After fertilization, it completes meiosis and forms a mature oocyte and secondary polar body.

Q27. Which one of the following differentiates directly into mature sperm?

• A. Primary spermatocyte
• B. Secondary spermatocyte
• C. Spermatogonia
• D. Spermatid✓

Explanation: Spermatids are haploid cells that result from the second meiotic division of secondary spermatocytes. They undergo a process called spermiogenesis, which involves extensive structural and morphological changes to transform into mature sperm.

Q28. Uterus opens into the vagina through:

• A. Cervix✓
• B. Fallopian tube
• C. External genitalia
• D. Vulva

Explanation: The cervix is the lower part of the uterus, which opens into the vagina.

Q29. Each muscle fibre is surrounded by membrane which is called:

• A. Sarcomere
• B. Sarcolemma✓
• C. Twitch fibre
• D. Capsule

Explanation: Each muscle fiber is surrounded by the sarcolemma, which also contains the cytoplasm of the cell.

Q30. When calcium ions are released from the sarcoplasmic reticulum, they bind with _ during muscle contraction.

• A. Tropomyosin
• B. Sarcolemma
• C. Cytosol’s ions
• D. Troponin✓

Explanation: Calcium ions, which are released, bind to troponin, which in turn allows the tropomyosin to give way, thus allowing the muscle to stay contracted.

Q31. The human and mammalian skeleton can be divided into two parts, the axial skeleton and:

• A. Appendicular skeleton✓
• B. Exoskeleton
• C. Endoskeleton
• D. Hydrostatic skeleton

Explanation: The human skeleton consists of the axial skeleton, which comprises the cranium, backbone, and ribs, and the appendicular skeleton, which consists of the upper and lower skeleton.

Q32. The last four vertebrae in humans are fused to form a structure called:

• A. Sacrum
• B. Cervical vertebrae
• C. Pubis
• D. Coccyx✓

Explanation: The last 4 vertebrae in humans are fused into a triangular structure known as the coccyx.

Q33. How many bones are involved in the formation of each half of the pelvic girdle?

• A. 3 bones✓
• B. 4 bones
• C. 2 bones
• D. 1 bone

Explanation: Each half of the pelvic girdle comprises 3 bones: the ilium, the pubis, and the ischium.

Q34. Ductless glands are known as:

• A. Endocrine gland✓
• B. Exocrine gland
• C. Salivary glands
• D. Bile glands

Explanation: The endocrine system is made up of ductless glands, which help control several aspects of the body.

Q35. Gastrin is the hormone which is produced by the:

• A. Liver
• B. Pyloric region of stomach✓
• C. Adrenal gland
• D. Mucosal lining of intestine

Explanation: Gastrin hormone is released by the pyloric region of the stomach. It is a peptide hormone that stimulates HCl secretion by parietal cells during and after a meal.

Q36. β-cells of the pancreas (changed from the liver) secrete a hormone that is called :

• A. Insulin✓
• B. Glucagon
• C. Antidiuretic hormone
• D. Gastrin

Explanation: Insulin is released by Beta-cells in the islets of Langerhans in the pancreas and helps maintain the blood sugar level in the body.

Q37. Vasopressin (ADH) and oxytocin are released from the:

• A. Placenta
• B. Ovary
• C. Anterior pituitary
• D. Posterior pituitary✓

Explanation: Vasopressin (also known as ADH) and Oxytocin are the only two hormones released from the posterior lobe of the pituitary gland. It must be noted that these two hormones are actually produced by the hypothalamus but stored and released from the posterior pituitary gland.

Q38. Antigen is a foreign protein or any other molecule which stimulates the formation of :

• A. MHC complex
• B. Immunogen
• C. Mucus
• D. Antibodies✓

Explanation: d) Antibodies: Antibodies, also known as immunoglobulins, are proteins produced by the immune system in response to the presence of antigens. Antibodies specifically bind to antigens, marking them for destruction by immune cells. While antibodies are produced in response to antigens, they are not responsible for stimulating their own formation.

Q39. Antibodies are produced by which of the following lymphocytes?

• A. B lymphocytes✓
• B. A lymphocytes
• C. T lymphocytes
• D. B and T lymphocytes

Explanation: B lymphocytes are a type of white blood cells which produce antibodies, which protect the body against antigens.

Q40. T-lymphocytes become mature and competent under the influence of:

• A. Liver
• B. Bursa of fabricius
• C. Thymus gland✓
• D. Spleen

Explanation: T-lymphocytes are a type of immune cells. They originate from stem cells in the bone marrow and are sent to the thymus gland for maturation which makes them immunocompetent.

Q41. Mucous membranes are part of body defense system and they offer:

• A. Physical Barriers✓
• B. Mechanical Barriers
• C. Chemical Barriers
• D. Biological Barriers

Explanation: Mucous membranes act as physical barriers by trapping pathogens and preventing them from entering the body.

Q42. Snake bite is treated with which type of immunization?

• A. Active
• B. Passive✓
• C. Humoral
• D. Specific

Explanation: Passive immunity provides immediate protection by injecting immunoglobulins, and this protection lasts for a short period of time, which is perfect for a snake bite.

Q43. The product(s) of cyclic photophosphorylation is/are:

• A. ATP✓
• B. NADP
• C. NADP and ATP
• D. NADP, ATP, and O2

Explanation: The primary product of cyclic photophosphorylation in photosynthesis is ATP (adenosine triphosphate). Cyclic photophosphorylation is one of the two types of light reactions that occur in the thylakoid membrane of the chloroplasts. The overall result is the generation of ATP without the production of NADPH or the release of oxygen. Cyclic photophosphorylation is particularly important in certain photosynthetic bacteria and in the light reactions of photosynthesis in the chloroplasts of some plants.

Q44. The total NADH formed by one glucose molecule during the Krebs cycle is

• A. 6✓
• B. 3
• C. 8
• D. 4

Explanation: The total NADH formed by one glucose molecule during the Krebs cycle is 6. Each glucose is converted into two acetyl-CoA molecules, and each acetyl-CoA produces 3 NADH in one turn of the cycle, giving a total of 6 NADH per glucose. The 3 NADH option is incorrect because it accounts for only one acetyl-CoA, not both. The 8 NADH option is too high and does not match the actual NADH generated in the Krebs cycle, though the total NADH, including glycolysis and the link reaction, may be higher. The option of 4 NADH underestimates the amount, as two acetyl-CoA molecules produce 6 NADH in total.

Q45. The terminal electron acceptor in electron transport chain is:

• A. Hydrogen
• B. Iron
• C. Cytochrome
• D. Oxygen✓

Explanation: The terminal electron acceptor in the electron transport chain is oxygen (O2). In aerobic respiration, which occurs in the presence of oxygen, the final stage of the electron transport chain involves the transfer of electrons to molecular oxygen.

Q46. The end product of glycolysis is:

• A. ADP
• B. Reduced FAD
• C. Citric acid
• D. Pyruvate✓

Explanation: The end product of glycolysis is two molecules of pyruvate(3C compound). Glycolysis is the first stage of cellular respiration, and it takes place in the cytoplasm of the cell. During glycolysis, one molecule of glucose (a six-carbon sugar) is broken down into two molecules of pyruvate (each containing three carbon atoms).

Q47. One molecule of FADH₂ is produced in the Krebs cycle during the conversion of:

• A. Fumarate to Malate
• B. Succinate to Fumarate✓
• C. Malate to Oxaloacetate
• D. α-Ketoglutarate to Succinate

Explanation: The conversion of succinate to fumarate is the correct step where one molecule of FADH₂ is produced in the Krebs cycle. This reaction is catalyzed by the enzyme succinate dehydrogenase, which is located on the inner mitochondrial membrane. During this oxidation process, two hydrogen atoms are removed from succinate and transferred to flavin adenine dinucleotide (FAD), reducing it to FADH₂. This reduced coenzyme later donates electrons to the electron transport chain, contributing to ATP production. The reason FAD rather than NAD⁺ is used here is that the oxidation of succinate is not sufficiently energetic to reduce NAD⁺, but it is enough to reduce FAD.The conversion of fumarate to malate involves the addition of water, not the reduction of FAD. The conversion of α-ketoglutarate to succinate produces NADH and CO₂, not FADH₂. The conversion of malate to oxaloacetate also produces NADH, not FADH₂.

Q48. In recombinant DNA technology _ are tools for manipulating DNA.

• A. Viruses
• B. Chromosomes
• C. Enzymes✓
• D. Genes

Explanation: Enzymes are essential to DNA recombinant technology and help modify and change DNA molecules as needed. In recombinant DNA technology, enzymes like restriction enzymes and DNA Ligase are used to cut and join DNA, while cloning vectors (like plasmids) are used to carry the DNA into a host cell. Other tools, like the PCR are used to amplify DNA fragments.

Q49. In the DNA finger printing process, the use of _ produces a distinctive pattern on autoradiography or X-ray film.

• A. Restriction enzyme
• B. DNA Ligase
• C. Macrosatellites
• D. Probes for genetic markers✓

Explanation: In DNA fingerprinting, probes for genetic markers are used to identify markers such as microsatellites and thus produce distinctive patterns on autoradiography and X-ray films.A single-stranded DNA or RNA fragment that is complementary to a specific target DNA or RNA sequence. Labeled with a radioactive or fluorescent tag, these probes are used to detect and identify that specific sequence in a sample through a process called hybridization.

Q50. In the recombinant DNA technology plasmids are used as:

• A. Genetic material
• B. Enzymes
• C. Vectors✓
• D. Probes

Explanation: Plasmids are small, circular, double-stranded DNA molecules, which are used in DNA recombinant technology by modifying the plasmid and then introducing them into cells.A single-stranded DNA or RNA fragment that is complementary to a specific target DNA or RNA sequence. Labeled with a radioactive or fluorescent tag, these probes are used to detect and identify that specific sequence in a sample through a process called hybridization.

Q51. In which process, multiple copies of the desired genes are produced?

• A. Polymerase chain reaction✓
• B. Gene sequencing
• C. Analyzing DNA
• D. DNA finger printing

Explanation: It is used to replicate a few or even a singular DNA molecule into millions of copies. It is very specific the target DNA sequence can be less than one part in a million of the total DNA sample.It is widely used in molecular biology, medical diagnostics, and forensics because it can amplify a tiny amount of DNA into a large enough quantity for detailed study.

Q52. The enzyme adenosine deaminase is missing in a person suffering from:

• A. Cystic fibrosis
• B. Hypercholesterolemia
• C. Severe combined immunodeficiency syndrome✓
• D. Parkinson’s disease

Explanation: Adenosine deaminase deficiency is an inherited disorder, which causes severe combined immunodeficiency syndrome, and it causes the afflicted person to have virtually no immunity to bacteria, viruses or fungi.It is treated with the help of gene therapy in which a virus is used to deliver the normal gene to bone marrow cells.

Q53. What is the niche of an organism in an ecosystem?

• A. Role played by many organisms in an ecosystem
• B. Role played by a dead organism in an ecosystem
• C. Role played by community of microorganisms in their ecosystem
• D. Role played by an organism in its ecosystem✓

Explanation: The niche of an organism is the role that it plays in a community, especially its living conditions and its interaction with other organisms in the community, and also its behavior and influence.

Q54. The distinct levels or links of food chain are called

• A. Trophic level✓
• B. Food web
• C. Energy pyramid
• D. Food chain

Explanation: A trophic level refers to a level or a position in a food chain, a food web, or an ecological pyramid.

Q55. A relationship between two or more organisms of different species in which all partners get benefit is called

• A. Symbiosis✓
• B. Parasitism
• C. Commensalism
• D. Predation

Explanation: Symbiosis is the interaction of two dissimilar organisms, in which both organisms benefit from the interaction.

Q56. Bacteria and fungi are examples of

• A. Producers
• B. Decomposers✓
• C. Consumers
• D. Denvers

Explanation: Bacteria and fungi are examples of decomposers, and gain energy from breaking down dead organisms or waste from other organisms.

Q57. The cause of acid rain is

• A. Oxides of carbon
• B. Oxides of nitrogen and Sulphur✓
• C. Oxides of Sulphur
• D. Oxides of nitrogen

Explanation: Nitrogen dioxide acts as a catalyst and reacts with sulphur dioxide to form sulphur trioxide, which is the main culprit of acid rain.

Q58. When the presence of a gene at one locus suppresses the effect of a gene at another locus, the phenomenon is called

• A. Hypostasis
• B. Pleiotropy
• C. Epistasis✓
• D. Epitrop

Explanation: Epistasis is the interaction of genes, which are not alleles, leading to the suppression of the effect of one gene due to the presence of another.

Q59. The gene for ABO-blood group systems in humans is represented by symbol

• A. X
• B. I✓
• C. Y
• D. O

Explanation: The gene for ABO blood group systems is represented by I, with the three alleles being IA, IB, and i.

Q60. When a single gene affects two or more traits, the phenomenon is called :

• A. Epistasis
• B. Pleiotropy✓
• C. Dominance
• D. Over dominance

Explanation: Pleitropy is a phenomenon, when a single gene affects apparently unrelated phenotypic traits.

Q61. The comparative embryology of all vertebrates shows the development of :

• A. Hairs
• B. Gill pouches✓
• C. Scales
• D. Fins

Explanation: Gill pouches were found to be present in the embryos of all vertebrates, even in air breathing reptiles and mammals.

Q62. In men, sex-determination depends upon the nature of

• A. Heterogametic male✓
• B. Homogametic female
• C. Heterogametic female
• D. Homogametic male

Explanation: In humans, the sex is determined by the heterogametic male, who have the XY chromosome, while the female have the XX chromosome.

Q63. Population of different species (plants and animals) living in the same habitat form a:

• A. Community✓
• B. Ecosystem
• C. Biosphere
• D. Microhabitat

Explanation: A community is a group of interacting species, which live in the same location.

Q64. The part of the body which forms a structural and functional unit and is composed of more than one tissue is called

• A. Organ✓
• B. Organelle
• C. Organ system
• D. Whole organism

Explanation: An organ is defined as a collection of tissues, which combine to form a functional unit within the body, specialized at carrying out a specific task.

Q65. A method in which pests are destroyed by using same living organisms or natural enemies is called

• A. Pasteurization
• B. Integrated disease management
• C. Biological control✓
• D. Genetic engineering

Explanation: Control of pests by means of introducing natural predators or diseases affecting the pests comes under the umbrella of biological control.

Q66. Chemicals produced by microorganisms which are capable of destroying the growth of microbes are called:

• A. Antigen
• B. Biocidal
• C. Antiseptics
• D. Antibiotics✓

Explanation: Antibiotics are chemicals which are produced by micro-organisms, and are particularly adept at destroying bacteria

Q67. Plastids are only found in the:

• A. Animals and Plants
• B. Animals
• C. Plants✓
• D. Viruses

Explanation: Plastids are double membraned organelles, which contain pigments used in photosynthesis, and also manufacture and store compounds in plant cells.

Q68. Plasma membrane is chemically composed of:

• A. Phospholipids only
• B. Lipids and proteins✓
• C. Lipids and carbohydrates
• D. Glycoproteins

Explanation: The plasma membrane is mainly composed of lipids and proteins, which form the structural and functional basis of the membrane. Lipids, primarily phospholipids, provide the membrane with a flexible matrix, while proteins serve various functions, including transport, communication, and structural support. While carbohydrates are present, they are typically attached to lipids and proteins, enhancing their functions rather than forming the main structural components. Thus, 'Lipids and proteins' is the correct answer. Options suggesting only lipids, lipids and carbohydrates, or glycoproteins alone do not account for the essential role of proteins in the membrane's composition.

Q69. The endoplasmic reticulum contains a system of flattened membrane-bounded sacs, which are named as

• A. Cristae
• B. Marks
• C. Cisternae✓
• D. Tubules

Explanation: Cisternae are the membrane-bound sacs that make up the endoplasmic reticulum and the Golgi body. While cristae are folded inner membrane structures found in mitochondria, not in the endoplasmic reticulum. "Marks" is not a term associated with the structure of the endoplasmic reticulum. While the endoplasmic reticulum does contain tubular structures, they are not the main focus of this question. The flattened sacs are the primary structural feature being referred to in this context. Therefore, the only correct option is Cisternae.

Q70. Lipid synthesis/organisation takes place in which of the following organelle?

• A. Mitochondria
• B. Vacuoles
• C. Rough endoplasmic reticulum
• D. Smooth endoplasmic reticulum✓

Explanation: SER is involve in lipid synthesis and also it's organization such as phospholipids and cholesterol. It also involve in the maintaining of structure of ER

Q71. Ribosomes exist in two forms, either attached with RER or freely dispersed in the:

• A. Tonoplast
• B. Golgi bodies
• C. Cytoplasm✓
• D. SER

Explanation: Ribosomes are essential for protein synthesis and can be found in two key locations within the cell: attached to the rough endoplasmic reticulum (RER) or freely dispersed in the cytoplasm. In the cytoplasm, ribosomes synthesize proteins that are essential for various cellular functions. The other options are incorrect: the tonoplast and Golgi bodies are involved in different cellular activities unrelated to ribosome function, and the smooth endoplasmic reticulum (SER) does not support ribosome attachment or dispersion.

Q72. Exchange of segments between homologous chromosomes is called:

• A. Segregation
• B. Independent assortment
• C. Crossing over✓
• D. Mutation

Explanation: During synapsis, when genes on chromatids of homologous chromosomes are tightly align, genetic material can transfer between non-sister chromatids and is known as crossing over.Crossing over is the biological process during sexual reproduction in which homologous chromosomes exchange genetic material, creating new combinations of genes on the resulting chromosomes. This "recombination" occurs in prophase I of meiosis and is vital for generating genetic diversity within a species, which helps with adaptation to changing environments and overall species survival.

Q73. If a person has 44 autosomes + XXY, he will suffer from:

• A. Klinefelter’s syndrome✓
• B. Down’s syndrome
• C. Turner’s syndrome
• D. Edward’s syndrome

Explanation: The correct answer is Klinefelter's syndrome. Klinefelter's syndrome is characterized by the presence of an additional X chromosome in males, resulting in a karyotype of 47,XXY. This condition can lead to various physical, hormonal, and developmental differences in affected individuals.

Q74. The ribosomal RNA is synthesized and stored in :

• A. Endoplasmic reticulum
• B. Nucleolus✓
• C. Golgi complex
• D. Chromosomes

Explanation: Ribosomal RNA is synthesized in the dense region of the nucleus as it is the site for formation of ribosomes and is called the nucleolus, which contains the genes relevant to formation of ribosomal RNA.

Q75. In which stage of Interphase, there is an increase in cell size and many biochemical are formed:

• A. G2 phase
• B. G1 phase✓
• C. S phase
• D. C phase

Explanation: As the cell grows and expands after cell division, that is the point when the nucleus is most active, i.e. during the first stage of interphase.G1(Gap 1) is the period of extensive metabolic activity, in which cell normally grows in size, specific enzymes, are synthesized and DNA base units are accumulated for theDNA synthesis.

Q76. In Down’s syndrome, which one of the following pair of chromosome fails to segregate?

• A. 7
• B. 18
• C. 21✓
• D. 19

Explanation: Down’s syndrome is a genetic condition in which the 21st chromosome fails to segregate, leading to physical and mental developmental delay and disabilities. Down syndrome symptoms include distinctive physical features like upward slanting eyes, a flattened face, a short neck, and small ears, hands, and feet, along with low muscle tone and loose joints. Individuals also experience intellectual and developmental delays, and may face other health issues such as heart problems, hearing loss, gastrointestinal issues, and vision problems.

Q77. Carbohydrates are organic molecules and contain three elements

• A. Carboan, water and oxygen
• B. Carbon, Sulphur and hydrogen
• C. Carbon, calcium and hydrogen
• D. Carbon, ahydrogen and oxygen✓

Explanation: Carbohydrates are organic molecules, which are built out of carbon, hydrogen and oxygen molecules

Q78. Which one are intermediates in respiration and photosynthesis both?

• A. Ribose and heptolose
• B. Glyceraldehyde and dihydroxyacetone✓
• C. Glucose and galactose
• D. Fructose and ribulose

Explanation: Glyceraldehyde and dihydroxyacetone are intermediates in both glycolysis (respiration) and the Calvin cycle (photosynthesis).

Q79. Which of the following is a peptide bond?

• A. –C–N✓
• B. -C–O
• C. –C–P
• D. –C–S

Explanation: The peptide bond is the bond formed when the carboxyl group reacts with an amide group forming a -C-N- bond.

Q80. Which one of the following is an example of unsaturated fatty acid?

• A. Butyric Acid
• B. Oleic Acid✓
• C. Palmitic Acid
• D. Acetic Acid

Explanation: Correct answer is Oleic Acid as it is a monounsaturated fatty acid.The example of an unsaturated fatty acid among the options provided is Oleic acid.Unsaturated fatty acids are a type of fatty acid that contains one or more double bonds in their carbon chain.Butyric acid is a saturated fatty acid. It is a four-carbon chain fatty acid with no double bonds.Oleic acid is an unsaturated fatty acid. It is an eighteen-carbon chain fatty acid with one double bond located between the ninth and tenth carbon atoms from the methyl (CH3) end of the chain.Palmitic acid is a saturated fatty acid. It is a sixteen-carbon chain fatty acid with no double bonds.Acetic acid is not a fatty acid. It is a simple organic compound known as a carboxylic acid, with the chemical formula CH3COOH. It is the primary component of vinegar.

Q81. Which of the following combination of base pair is absent in DNA?

• A. A–T
• B. C–G
• C. A–U✓
• D. T–A

Explanation: In RNA, the two pyrimidine bases are uracil and cytosine. While in DNA, the pyrimidine bases are thymine and cytosine. Thus, uracil is absent in DNA and is replace by thymine.

Q82. The type of inhibition in which the inhibitor has no structural similarity to the substrate and combines with the enzyme at other than the active site is called:

• A. Irreversible inhibition
• B. Competitive inhibition
• C. Non-competitive inhibition✓
• D. Reversible inhibition

Explanation: Non-competitive inhibition occurs when an inhibitor binds to another part of an anzyme, denatures the active site, and inhibits the reaction. The body frequently uses this to monitor and manage enzyme-controlled reactions.

Q83. The inhibitors that bind tightly and permanently to enzymes and change their globular structure and catalytic activity are:

• A. Reversible inhibitors
• B. Irreversible inhibitors✓
• C. Competitive inhibitors
• D. Non-competitive inhibitors

Explanation: Irreversible inhibitors are characterized by their ability to bind permanently to enzymes, often through covalent bonds, leading to an alteration in the enzyme's globular structure and a permanent loss of catalytic activity. Unlike reversible inhibitors, which can detach and allow the enzyme to regain function, irreversible inhibitors permanently inactivate the enzyme. This distinguishes them from both competitive and non-competitive inhibitors, which are types of reversible inhibitors that do not permanently change the enzyme's structure.

Q84. Enzyme succinate dehydrogenase converts succinate into:

• A. Malate
• B. Malonic acid
• C. Citrate
• D. Fumarate✓

Explanation: In Krebs cycle, an enzyme called succinate dehydrogenase acts on its substrate, succinate, and converts it into product fumarate.

Q85. If the detachable cofactor is an inorganic ion, then it is designated as:

• A. Coenzyme
• B. Prosthetic group
• C. Holoenzyme
• D. Activator✓

Explanation: The detachable cofactor is known as an activator if it is an inorganic ion. If the non-protein part is covalently bonded, it is known as a prosthetic group. If it is loosely attached to the protein part, it is known as a coenzyme. The enzyme with its cofactor is called holoenzyme.

Q86. In HIV viruses, reverse transcriptase converts single-stranded RNA into double-stranded viral DNA. This process is called

• A. Translation
• B. Duplication
• C. Replication
• D. Reverse Transcription✓

Explanation: Reverse transcription is a process by which the DNA molecule is formed from the RNA template and the process is catalyzed by the enzyme, reverse transcriptase.

Q87. Mesosomes are infoldings of the cell membrane and are involved in

• A. DNA replication✓
• B. RNA synthesis
• C. Protein synthesis
• D. Metabolism

Explanation: The cell membrane invaginates into the cytoplasm forming a structure called a mesosome. Mesosomes are in the form of vesicles, tubules, or lamellae. Mesosomes are involved in DNA replication and cell division whereas some mesosomes are also involved in the export of exocellular enzymes. Respiratory enzymes are also present in the mesosomes.

Q88. Most widespread problem of the antibiotics misuse is the

• A. Rapid cure
• B. Increased resistance in pathogen✓
• C. Disturbance of metabolism
• D. Immunity

Explanation: Misuse and overuse of antibiotics gives the bacteria more chances to develop resistance to that antibiotic, and this resistance can be shared between different species.

Q89. Which of the following components is found in the cell wall of fungi?

• A. Cellulose
• B. Chitin✓
• C. Proteins
• D. Glycerol

Explanation: Chitin is a structural polysaccharide found in cell walls of fungi and made up of N-acetyl glucosamine subunits, while cell walls of plants are made up of cellulose.

Q90. The male reproductive parts of the flower are called:

• A. Gynoecium
• B. Calyx
• C. Androecium✓
• D. Corolla

Explanation: The androecium is the male reproductive part of a flower, consisting of one or more stamens. Each stamen typically consists of two main parts:Anther: This is the part of the stamen that produces and releases pollen. Pollen contains the male gametes (sperm cells) necessary for fertilization.Filament: This is the slender stalk or thread-like structure that supports the anther and positions it for optimal pollen dispersal.

Q91. Fasciola is the name given to

• A. Tapeworm
• B. Planaria
• C. Liver fluke✓
• D. Earthworm

Explanation: Fasciola is a type of parasitic flatworm that is commonly known as the liver fluke. It is found in the liver and bile ducts of various mammals, including humans. Infection occurs by ingesting contaminated water or plants. The life cycle of the liver fluke involves a snail as an intermediate host and a mammal as a definitive host. The disease caused by Fasciola is called fascioliasis, which can lead to liver damage and other serious health problems if left untreated.

Q92. Ascaris is

• A. Diploblastic
• B. Triploblastic✓
• C. Haploid
• D. Acoelomate

Explanation: Ascaris is a genus of parasitic nematode worms known as the "small intestinal roundworms" that can infect the human small intestine. Ascaris worms are triploblastic, which means they have three germ layers during embryonic development: the ectoderm, mesoderm, and endoderm. They are also pseudocoelomates, meaning they have a body cavity that is not completely lined by mesodermal tissue.

Q93. During development, in an animal, mesoderm layer gives rise to

• A. Nervous System
• B. Alimentary canal lining
• C. Muscular and skeletal system✓
• D. Mouth

Explanation: During development, the mesoderm layer gives rise to the muscular system, skeletal system, circulatory system, and other organs such as the kidneys and reproductive system. The ectoderm gives rise to the nervous system and the skin, while the endoderm gives rise to the lining of the digestive and respiratory tracts and other internal organs such as the liver and pancreas.

Q94. Polymorphism is characteristic feature of

• A. Porifera
• B. Cnidaria✓
• C. Annelida
• D. Nematodes

Explanation: Polymorphism refers to occurrence of more than one individual in a single organism, and is a characteristic feature of Cnidarians

Q95. The muscles of the stomach walls thoroughly mix up the food with gastric juices and the resulting semi-solid/semi-liquid material is called:

• A. Bolus
• B. Bolus or chyme
• C. Mucus
• D. Chyme✓

Explanation: After food enters the stomach, the muscular walls of the stomach contract and mix the food with gastric juices, breaking it down into a semi-solid/semi-liquid substance called chyme. The chyme is then slowly released from the stomach into the small intestine, where further digestion and absorption of nutrients occur. A bolus is a term used to describe a small, rounded mass of food that has been chewed and is ready to be swallowed. Mucus is a slimy substance secreted by the mucous membranes that line various organs, including the stomach and intestines, to protect and lubricate them.

Q96. Trypsinogen is converted into trypsin by the activity of

• A. Goblet cells
• B. Absorptive cells
• C. Enterokinase✓
• D. Peptidase

Explanation: Enterokinase, also known as enteropeptidase, converts inactive trypsinogen into active trypsin in the duodenum, where it helps in protein digestion

Q97. In large intestine, vitamin K is formed by the activity of:

• A. Symbiotic Bacteria✓
• B. Obligate Bacteria
• C. Parasitic Bacteria
• D. Facultative Bacteria

Explanation: [A] Symbiotic bacteria are bacteria that live in a symbiotic relationship with other organisms. It benefits both the organisms like Escherichia coli live in symbiotic association in human intestine and produce large quantities of vitamin K and in human intestine. [B] An obligate aerobe is an organism that requires oxygen to grow. Through cellular respiration, these organisms use oxygen to metabolise substances, like sugars or fats, to obtain energy. [C] parasite bacteria live on or in a host organism and get their food from or at the expense of its host. [D] Bacteria that can use dissolved oxygen (DO) or oxygen obtained from food materials such as sulphate or nitrate ions, or some can respire through glycolysis

Q98. Goblet cells secrete:

• A. HCl
• B. Mucus✓
• C. Enzymes
• D. Amylase

Explanation: Goblet cells are unicellular glands that are present in the lining of the respiratory and digestive tracts. These cells secrete mucus, a sticky and slimy substance that helps to lubricate and protect the lining of the organs from mechanical damage and harmful pathogens. In the respiratory tract, mucus traps dust, bacteria, and other foreign particles and prevents them from entering the lungs. In the digestive tract, mucus lubricates the food bolus and facilitates its movement through the gut.

Q99. Mature mammalian red blood cells do not have

• A. Nucleus✓
• B. Red color
• C. Fluids
• D. Haemoglobin

Explanation: Mature mammalian red blood cells, also known as erythrocytes, lack a nucleus. During their development in the bone marrow, erythrocytes have a nucleus, but they extrude it before entering the bloodstream. This allows them to have more space to carry the oxygen-carrying protein, hemoglobin. Without a nucleus, erythrocytes have a biconcave shape, which increases their surface area and allows for more efficient oxygen exchange. The lack of a nucleus also means that erythrocytes cannot divide or repair themselves, which is why their lifespan is typically around 120 days.

Q100. In a normal person plasma constitutes about _ by volume of blood:

• A. 50%
• B. 60%
• C. 45%
• D. 55%✓

Explanation: Plasma is the liquid component of blood and makes up the largest proportion of blood volume. It is a yellowish fluid that consists of water, proteins, electrolytes, and other substances. In a normal person, plasma constitutes about 55% by volume of blood. The remaining 45% is composed of red blood cells, white blood cells, and platelets.

Q101. Which vein has oxygenated blood?

• A. Renal vein
• B. Subclavian vein
• C. Pulmonary vein✓
• D. Jugular vein

Explanation: The pulmonary vein is the only vein in the human body that carries oxygenated blood, while all other veins carry deoxygenated blood. It carries blood from the lungs to the left atrium of the heart, where it is then pumped out to the rest of the body via the aorta.

Q102. What is the residual volume of air which always remains inside the lungs of human?

• A. 3.5 Liters
• B. 0.5 Liters
• C. 5.0 Liters
• D. 1.5 Liters✓

Explanation: Residual volume is the amount of air that always remains inside the lungs even after a person has forcefully exhaled as much air as possible. This air is important to keep the lungs open and prevent collapse. The residual volume of air in human lungs is typically around 1.5 liters, which is almost one-third of the total lung capacity.

Q103. In nephron, most of the reabsorption takes place in the:

• A. Distal tubule
• B. Proximal tubule✓
• C. Ascending limb
• D. Descending limb

Explanation: The proximal convoluted tubule (PCT) is the site where ~65– 80% of filtered water, electrolytes, glucose, and amino acids are reabsorbed back into the bloodstream due to its highly permeable epithelium and extensive surface area.

Q104. Detection of change and signalling for the effector’s response to the control system is a:

• A. Negative feedback
• B. Positive feedback
• C. Inter-coordination
• D. Feedback mechanism✓

Explanation: The feedback mechanism involves a response to a stimulus by means of an effector, which impacts conditions to bring them back to the normal range.

Q105. What are the three components of mechanism of homeostatic regulations?

• A. Receptors, control centre and effectors✓
• B. Sensory, motor and associative neurons
• C. CNS, PNS and diffused nervous system
• D. Cerebrum, cerebellum and pons

Explanation: Option A is correct.The three components of the mechanism of homeostatic regulation are:Receptors:These are specialized cells or structures that detect changes in the internal or external environment. They can be located in various parts of the body, such as the skin, organs, and blood vessels.Control center:This is a part of the nervous system or endocrine system that receives information from the receptors and determines the appropriate response to maintain homeostasisEffectors:These are cells or organs that carry out the response directed by the control center. They can be muscles, glands, or organs such as the heart or kidneys.Together, these three components work to maintain the internal environment of the body within a narrow range of conditions, despite changes in the external environment or internal conditions. This is essential for the proper functioning of cells, tissues, organs, and for the overall health of the organism.

Q106. Blood enters the glomerulus through

• A. Efferent arteriole
• B. Afferent arteriole✓
• C. Renal artery
• D. Renal vein

Explanation: The glomerulus receives blood from the afferent arteriole, which carries blood into the glomerular capillary network for filtration. The efferent arteriole carries blood away from the glomerulus.

Q107. Which portion of nephron is under the control of ADH?

• A. Bowman’s capsule
• B. Ascending arm
• C. Distal and collecting ducts✓
• D. Descending arm

Explanation: Option C is correct.ADH (antidiuretic hormone), also known as vasopressin, regulates water balance in the body by controlling the permeability of water in the distal tubules and collecting ducts of the nephron in the kidney. When the body needs to conserve water, ADH is released from the pituitary gland and acts on the cells in the distal tubules and collecting ducts to increase their permeability to water, allowing more water to be reabsorbed from the urine back into the bloodstream. This results in the production of more concentrated urine and helps conserve water in the body.

Q108. The cause of Parkinson’s disease is the death of brain cells that produce:

• A. Dopamine✓
• B. Acetylcholine
• C. ADH hormone
• D. Oxytocin

Explanation: Parkinson's disease is a neurodegenerative disorder that affects movement. It is caused by the death of brain cells in a region of the brain that produces dopamine, a neurotransmitter that helps to regulate movement and emotional responses. The death of these cells leads to a deficiency of dopamine in the brain, resulting in the characteristic symptoms of Parkinson's disease, such as involuntary tremors, diminished motor power, and rigidity, which causes difficulty with balance and coordination. While the exact cause of the death of these brain cells is not known, it is thought to be related to a combination of genetic and environmental factors, like head trauma.

Q109. In the below reaction the nucleophile which attacks the carbon atom of acid is:

• A. OH-
• B. P
• C. Cl-✓
• D. H-

Explanation: In the given reaction, the nucleophile which attacks the carbon atom of acid (CH3COOH) is CI (chloride ion). The reaction involves the substitution of the -OH group of the carboxylic acid (CH3COOH) with a chloride (-CI) group. resulting in the formation of an acyl chloride (CH3COCI) and the generation of HCI and POCI3

Q110. When ethanoic acid reacts with methylamine, an amide is formed. What is the structure of the amide formed?

• A. A
• B. B
• C. C
• D. D✓

Explanation: When ethanoic acid reacts with methylamine, it undergoes a condensation reaction to form an amide. The reaction can be represented as:CH3COOH + CH3NH2 → CH3CONHCH3 + H2OThe product formed is methyl ethanamide, which has the structure CH3-CO-NH-CH3. It is also commonly known as N-methyl acetamide. In the amide, the carbonyl group (C=O) of ethanoic acid reacts with the nitrogen atom of the methylamine, and a peptide bond is formed between them. Amides are important functional groups in biochemistry and are present in many biomolecules such as proteins and nucleic acids.

Q111. Organic compound containing both amino and carboxyl group is known as:

• A. Amino acid✓
• B. Fatty acid
• C. Saccharide
• D. Amide

Explanation: An amino acid is an organic molecule that is made up of a basic amino group (−NH2), an acidic carboxyl group (−COOH), and an organic R group (or side chain) that is unique to each amino acid.

Q112. Alanine is an aminoacid which showsneutral effect on litmus paper,the formula of alanine may be

• A. A✓
• B. B
• C. C
• D. D

Explanation: Alanine is a non-polar, aliphatic, and alpha-amino acid that contains both an amine (-NH2) and a carboxyl (-COOH) group attached to the alpha-carbon atom. It has a simple structure with a methyl (-CH3) group as its side chain. Alanine is one of the 20 common amino acids that make up proteins, and its neutral effect on litmus paper is due to its zwitterionic form at physiological pH.

Q113. Which of the following structures is not an alpha amino acid?

• A. A
• B. B✓
• C. C
• D. D

Explanation: An amino acid is a molecule containing both amine and carboxyl functional groups. Amino acids are the building blocks of proteins. The α-Amino acid is the most important type. α-Amino acid consists of an amino, a carboxyl, an R group, and a hydrogen atom that attaches to the same a-carbon atom

Q114. The skeletal formula of dipeptide formed between aspartic acid and phenylalanine is given below: How many functional groups are present in its formula?

• A. 5✓
• B. 2
• C. 4
• D. 3

Explanation: There are five functional groups present in the given formula. of the dipeptide: Carboxylic acid [COOH) group: one at the N-terminus of aspartic acidAmine (NH2) group: one at the C-terminus of phenylalanine Amide (O-C-NH) group: one linking the two amino acids Ester (O=C-O-) group: one in the side chain of aspartic acid Phenyl (benzene ring) group: one in the side chain of phenylalanine

Q115. In basic conditions, amino acid exists in which of the following forms?

• A. A
• B. B
• C. C
• D. D✓

Explanation: In basic conditions, the amino acid exists in the form of its conjugate base called the zwitterion or dipolar ion. The amino group (-NH2) acts as a base and accepts a proton (H+) to form an ammonium ion (-NH3+). The carboxyl group (-COOH) acts as an acid and donates a proton to form a carboxylate ion (-COO-). These two groups are present in the same molecule and balance each other's charges, resulting in the formation of a dipolar ion. This dipolar ion has no net charge and is represented as:H2N-CH2-COO-Thus, option D (H2N-CH2-COOH) is the correct form of an amino acid under basic conditions.

Q116. The following dipeptide is called:

• A. Glycyl glycine
• B. Glycyl alanine✓
• C. Alaninyl alanine
• D. Alaninyl glycine

Explanation: Left residue is glycine (-CH2-NH2), right residue is alanine (α- C bears a CH3). They are joined by a peptide bond from glycine’s carbonyl to alanine’s NH → glycyl-alanine. Left residue is glycine (-CH2-NH2), right residue is alanine (α-C bears a CH3). They are joined by a peptide bond from glycine’s carbonyl to alanine’s NH → glycyl-alanine.

Q117. The principal energy storage carbohydrate in animals is:

• A. Glucose
• B. Starch
• C. Protein
• D. Glycogen✓

Explanation: Glycogen is the principal energy storage carbohydrate in animals. It is a highly branched polymer of glucose molecules and is stored in the liver and muscle cells. Glycogen is synthesized from glucose by a process called glycogenesis and broken down into glucose by a process called glycogenolysis. When blood glucose levels are low, glycogen is broken down to release glucose into the bloodstream to maintain a constant supply of energy for the body

Q118. Starch is a polymer of:

• A. β–D–glucose
• B. α––glucose✓
• C. γ–D–glucose
• D. α–L–glucose

Explanation: Starch is a polysaccharide composed of repeating units of alpha-glucose molecules joined together through glycosidic bonds. It is the principal energy storage carbohydrate in plants, and is commonly found in grains, potatoes, and other starchy vegetables. The alpha-glucose molecules in starch can be arranged in two forms: amylose, which is a linear chain of alpha-glucose molecules linked by alpha-1, 4-glycosidic bonds, and amylopectin, which is a branched chain of alpha-glucose molecules linked by alpha-1, 4-glycosidic bonds and alpha-1, 6-glycosidic bonds.

Q119. The reaction between fats and caustic soda is called:

• A. Hydrogenolysis
• B. Fermentation
• C. Carboxylation
• D. Saponification✓

Explanation: Saponification is a process that involves the conversion of fat, oil, or lipid (fats in this case) into soap and alcohol by the action of aqueous alkali (caustic soda in this case).

Q120. Adipic acid and hexamethylene diamine both of which have _ carbon atoms:

• A. 7
• B. 8
• C. 6✓
• D. 4

Explanation: Nylon 6,6 is made of two monomers each containing 6 carbon atoms, hexamethylenediamine, and adipic acid, which give nylon 6,6 its name.

Q121. Lactose is a sugar present in milk. It is an example of:

• A. Disaccharides✓
• B. Monosaccharides
• C. Polysaccharides
• D. Starch

Explanation: It consists of one unit of galactose plus one of glucose linked together by β-(1→4) glycosidic bond, then it is a disaccharide.

Q122. Macromolecules are described as large molecules built up from small repeating units called:

• A. Monomers✓
• B. Isomers
• C. Metamers
• D. Tautomers

Explanation: A monomer is a molecule that “can undergo polymerization thereby contributing constitutional units to the essential structure of a macromolecule.

Q123. The increase in the concentration of oxidizing agents in smog like H2O2, HNO3, PAN, and ozone in the air is called:

• A. Carbonated smog
• B. Nitrated smog
• C. Photochemical smog✓
• D. Sulphonated smog

Explanation: Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight.

Q124. Which is the metal, whose elevated concentration is harmful to fish as it clogs the gills thus causing suffocation?

• A. Na
• B. Pb
• C. Zn
• D. Al✓

Explanation: Elevated levels of aluminum can affect some species' ability to regulate ions, like salts, and inhibit respiratory functions, like breathing. Aluminum can accumulate on the surface of a fish's gill, leading to respiratory dysfunction, and possibly death.

Q125. An organic compound has the empirical formula C3H3O if the molar mass of the compound is 110.15 gmol-1. The molecular formula of this organic compound is: (A, of C=12, H=1.008and O=16)

• A. C6H6O2✓
• B. C3H3O
• C. C9H9O3
• D. C6H6O3

Explanation: The empirical formula mass can be calculated as follows:Empirical formula mass = (3 x atomic mass of C) + (3 x atomic mass of H) + (1 x atomic mass of O) Empirical formula mass = (3 x 12.011) + (3 x 1.008) + 16.00 Empirical formula mass = 55.05 g/molThe molecular formula mass can be calculated using the following equation:Molecular formula mass = n x empirical formula masswhere n is the number of empirical formula units in the molecular formula.n can be calculated as follows:n = Molecular formula mass / Empirical formula massn = 110.15 / 55.05 n = 2Therefore, the molecular formula is 2 times the empirical formula, and the molecular formula is:2(C3H3O) = C6H6O2So, the correct option is A) C6H6O2.

Q126. When 8 grams (4 moles) of H2 react with 2 moles of O2 , how many moles of water will be formed?

• A. 5
• B. 4✓
• C. 6
• D. 3

Explanation: The balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water is:2H2 + O2 → 2H2OFrom the equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water. Therefore, 4 moles of hydrogen gas will react with (1/2) x 4 = 2 moles of oxygen gas to form (2 x 4) = 8 moles of water.So, the answer is 8/2 = 4 moles of water.

Q127. The number of molecules in 22.4 dm3 of H2 gas at 0°C and 1 atm are:

• A. 60.2 x 1023
• B. 6.02 x 1022
• C. 6.02 x 1025
• D. 6.02 x 1023✓

Explanation: This is Avogrado's number of molecules, 6.02×1023

Q128. Correct order of boiling points of the given liquids is:

• A. H2O > HF > HCl > NH3
• B. HF > H2O > HCl > NH3
• C. H2O > HF > NH3 > HCl✓
• D. HF > H2O > NH3 > HCl

Explanation: H2O has a greater boiling point than HF because H2O can form 4 hydrogen bonds while HF has 2 hydrogen bonds and to break all the hydrogen bonds more energy is needed by H2O than HF. Due to the strong force of intermolecular hydrogen bonding. Due to the high difference in the electronegativity of hydrogen and fluorine as compared to that of hydrogen and nitrogen, more amount of energy is required to break the hydrogen-fluorine bond. Because of the high electronegativity of nitrogen than chlorine, the bond between N−H is stronger than a bond between H−Cl. Therefore the boiling point of NH3 is more than HCl.

Q129. The relative energies of 4s, 4p and 3d orbitals are in the order:

• A. 3d < 4p <4s
• B. 4s < 3d < 4p✓
• C. 4p < 4s < 3d
• D. 4p < 3d < 4s

Explanation: The Aufbau principle states that electrons fill lower-energy atomic orbitals before filling higher-energy ones (Aufbau is German for "building-up"). By following this rule, we can predict the electron configurations for atoms or ions. To find the energy of orbitals, we use the (n+l) principle where n= energy level, while l= orbital level.For 4s: n+l= 4 + 0 = 4For 3d: n+l= 3 + 2= 5For 4p: n+l = 4+ 1= 5 (but since the distance of 4p is greater than that of 3d, 3d gets filled first)Final order: 4s<3d<4p

Q130. With increase in the value of Principal Quantum Number (n), the shape of the s-orbitals remains the same although their sizes:

• A. Decrease
• B. Increase✓
• C. Remain the same
• D. May or may not remain the same

Explanation: As the principal quantum number (n) increases, the size of the s-orbitals also increases. The shape of the s-orbitals remains spherical, with a probability distribution that becomes more spread out from the nucleus as n increases. So, while the shape (spherical) remains the same, the size of the orbital increases with higher values of n ( as shown in figure). This means that s-orbitals with larger principal quantum numbers have a higher probability of finding electrons farther away from the nucleus. Therefore, option B is correct.

Q131. The angle between unhybridized p-orbital and three sp2 hybrid orbitals of each carbon atom in ethene is:

• A. 120°
• B. 90°✓
• C. 109.5°
• D. 180°

Explanation: Option B is correct. A 90-degree angle is observed between the unhybridized p-orbital and the sp2 hybrid orbitals in ethene.

Q132. In ‘H-F’ bond electronegativity difference is ‘1.9’. What is the type of this bond?

• A. Polar covalent bond✓
• B. Non-polar covalent bond
• C. Pi (π) bond
• D. Co-ordinate covalent bond

Explanation: Fluorine is the most electronegative element with an electronegativity of 4.0 on the Pauling scale, while hydrogen has an electronegativity of 2.1. Due to this significant difference in electronegativity, the electron pair in the H-F bond is pulled towards fluorine, creating a partial negative charge on fluorine and a partial positive charge on hydrogen. Thus creating a polar covalent bond.

Q133. ‘∆H’ will be given a negative sign in:

• A. Exothermic reactions✓
• B. Decomposition reactions
• C. Dissociation reaction
• D. Endothermic reactions

Explanation: In exothermic reactions, heat is released to the surroundings as a product. This means that the enthalpy change (∆H) is negative because the system loses energy in the form of heat. The negative sign indicates that the reaction releases heat and is exothermic.In contrast, endothermic reactions absorb heat from the surroundings, which means that the enthalpy change (∆H) is positive. Decomposition and dissociation reactions can be either endothermic or exothermic, depending on the specific reaction and the energy involved.

Q134. The lattice energy of an ionic crystal is the enthalpy of:

• A. Combustion
• B. Dissociation
• C. Dissolution
• D. Formation✓

Explanation: The lattice energy of an ionic crystal is the energy released when separate gaseous ions are combined to form a solid ionic compound. It is a measure of the strength of the ionic bonds in the crystal. The formation of an ionic compound is an exothermic process, meaning that it releases energy, which is reflected by the negative sign of the lattice energy. The higher the magnitude of the lattice energy, the stronger the ionic bonds, and the more stable the crystal.

Q135. As the number of solute particles increases, the freezing point of the solution:

• A. Remains the same
• B. Increases
• C. First increases, then decreases
• D. Decreases✓

Explanation: When a solute is dissolved in a solvent, the freezing point of the solution decreases compared to the pure solvent. This happens because the solute particles disrupt the crystal lattice structure of the solvent, making it more difficult for the solvent to freeze. The extent to which the freezing point is lowered depends on the concentration of the solute particles in the solution.According to the Colligative property, the change in freezing point of a solution is directly proportional to the molality of the solution, which is the number of moles of solute per kilogram of solvent. Thus, as the number of solute particles increases, the molality of the solution increases, and so does the degree of depression of the freezing point. As a result, the freezing point of the solution lowers.

Q136. Boiling point constants help us to determine:

• A. Molar masses✓
• B. Volumes
• C. Pressures
• D. Masses

Explanation: Boiling point elevation is a colligative property that depends on the number of solute particles in a solution. The boiling point constant, Kb, is a proportionality constant that relates the molality of a solution to the elevation in boiling point.The equation for boiling point elevation is:ΔTb = Kb x molalitywhere ΔTb is the change in boiling point, Kb is the boiling point constant, and molality is the concentration of the solute in mol/kg of the solvent.By measuring the change in boiling point and knowing the boiling point constant of the solvent, we can calculate the molality of the solution. Then, using the molality and mass of the solute, we can calculate the molar mass of the solute. Therefore, boiling point constants help us to determine molar masses of solutes in solutions.

Q137. In electrolysis of aqueous CuCl2, the metal deposited at the cathode is:

• A. Na
• B. Al
• C. Pb
• D. Cu✓

Explanation: Copper deposits at cathode and chlorine is liberated at anode.During the electrolysis of aqueous CuCl2, Cu²⁺ ions and H⁺ ions are present in the solution. At the anode, Cl⁻ ions are oxidized to Cl₂ gas:2Cl⁻ → Cl₂ + 2e⁻At the cathode, Cu²⁺ ions are reduced to Cu metal:Cu²⁺ + 2e⁻ → CuTherefore, the metal deposited at the cathode is copper (Cu).

Q138. In MgCl2, the oxidation state of ‘Cl’ is :

• A. 0
• B. +2
• C. -2
• D. -1✓

Explanation: In MgCl2, the oxidation state of Mg is +2 since it is a Group 2 element and tends to lose two electrons to attain a noble gas configuration. Since MgCl2 is a neutral compound, the sum of the oxidation states of all the atoms in the compound should be zero.Let the oxidation state of Cl be x. Since there are two Cl atoms, the total contribution of Cl to the oxidation state will be 2x. Therefore, we can write:+2 + 2x = 0Solving for x, we get:x = -1Thus, the oxidation state of Cl in MgCl2 is -1.

Q139. The formation of NH3 is a reversible and exothermic process, what will happen on cooling?

• A. More reactant will form
• B. More N2 will be formed
• C. More H2 will be formed
• D. More product (NH3) will be formed✓

Explanation: According to Le Chatelier's principle, when a system at equilibrium is subjected to a change in temperature, pressure, or concentration of reactants or products, the equilibrium shifts in a direction that tends to counteract the imposed change.In the given scenario, the formation of NH3 is exothermic, meaning that heat is released when NH3 is formed. If the reaction mixture is cooled, the equilibrium will shift in the direction that releases more heat. Thus, the equilibrium will shift towards the forward direction (towards the formation of (NH3) to counteract the decrease in temperature. Therefore, more NH3 will be formed.Hence, the correct option is D) More product (NH3) will be formed.

Q140. A buffer solution is that which resists/minimizes the change in:

• A. pOH
• B. pH✓
• C. pKa
• D. pKb

Explanation: A buffer solution is a solution that can resist a change in pH upon addition of an acidic or basic component. It contains a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer solution minimizes changes in pH because the weak acid or base reacts with any added acid or base, respectively, to maintain the solution's pH within a relatively narrow range. The buffer solution can only resist a change in pH within a certain range, known as the buffer capacity.

Q141. In some reactions, a product formed acts as a catalyst. The phenomenon is called:

• A. Negative catalysis
• B. Activation of catalyst
• C. Heterogeneous catalysis
• D. Autocatalysis✓

Explanation: Autocatalysis is the phenomenon in which a product formed in a reaction acts as a catalyst for the same reaction. As the concentration of the product increases, the rate of the reaction also increases. This can lead to a positive feedback loop, where the reaction rate increases exponentially until other factors limit the process.

Q142. The reaction rate in forwarding direction decreases with the passage of time because:

• A. Concentration of reactants decrease✓
• B. Concentration of product decreases
• C. The order of reaction changes
• D. Temperature of the system changes

Explanation: The rate of a chemical reaction depends on the concentration of reactants. As the reaction proceeds, the concentration of reactants decreases while the concentration of products increases. This leads to a decrease in the rate of the reaction in the forward direction with time. The rate of the reverse reaction, on the other hand, increases with time as the concentration of products increases. This is known as a reversible reaction and the rates of the forward and reverse reactions become equal at equilibrium.

Q143. Which one remains the same throughout a period?

• A. Atomic radius
• B. Melting point
• C. Number of shells (orbits)✓
• D. Electrical conductivity

Explanation: Across a period, the atomic number increases, which means the number of protons increases in the nucleus. This results in the increased positive charge of the nucleus, which attracts the electrons more strongly towards the nucleus. This results in a decrease in atomic size (atomic radius) and an increase in the ionization energy and electron affinity of the elements.Melting point and electrical conductivity vary across a period due to the different bonding patterns and arrangements in the elements. In contrast, the number of shells or orbits remains the same across a period since the elements in a period have the same number of electron shells. Therefore, correct option is number of shells.

Q144. More the ionization energy of an element:

• A. More the electropositivity
• B. More the reducing power
• C. Less the metallic character✓
• D. Bigger the atomic radius

Explanation: Ionization energy is the energy required to remove an electron from a neutral gaseous atom, and it is a measure of the tendency of an element to form a cation. Elements with high ionization energies tend to have low electropositivity and reducing power because they are less likely to give up electrons.On the other hand, metallic character refers to the ability of an element to lose electrons and form positive ions. Elements with low ionization energies tend to have high metallic character, while those with high ionization energies tend to have low metallic character. Therefore, as the ionization energy of an element increases, its metallic character decreases.

Q145. Alkaline earth metal hydroxides decompose on heating. Which of the following reactions is a correct representation of this decomposition?

• A. A✓
• B. B
• C. C
• D. D

Explanation: When heated, alkaline earth metal hydroxides undergo thermal decomposition to produce the corresponding metal oxide and water. The reaction can be represented as:M(OH)2(s) → MO(s) + H2O(l)where M represents an alkaline earth metal.For example, when magnesium hydroxide is heated, it decomposes to produce magnesium oxide and water:Mg(OH)2(s) → MgO(s) + H2O(l)

Q146. Carbon has the unique ability to form long chains by bonding with other carbon atoms. This property of self-linking in carbon is known as:

• A. Condensation
• B. Polymerization
• C. Cyclization
• D. Catenation✓

Explanation: Catenation is the ability of an element to bond with itself to form long chains or rings. Carbon has a very strong tendency to catenate due to the strength and versatility of its covalent bonds. Carbon atoms can form multiple bonds with each other to create long chains, branched chains, and rings. This ability of carbon to form long chains is the basis of organic chemistry, where the vast majority of compounds are based on carbon chains or rings.

Q147. Oxidation state of ‘Mn’ in KMnO4, K2MnO4, MnO2 and MnSO4 is in the order:

• A. +7, +6, +2, +4
• B. +6, +7, +2, +4
• C. +7, +6, +4, +2✓
• D. +4, +6, +7, +2

Explanation: In KMnO4, we know that oxygen has a -2 oxidation state, and potassium has a +1 oxidation state. Let the oxidation state of manganese be x.Then, using the fact that the compound has a neutral charge:(+1) + x + 4(-2) = 0 x - 6 = 0 x = +6In K2MnO4, we have two potassium ions, each with a +1 oxidation state. Again, let the oxidation state of manganese be x.Then:2(+1) + x + 4(-2) = 02 + x - 8 = 0 x = +6In MnO2, we know that oxygen has a -2 oxidation state, so:2(-2) + x = 0x = +4In MnSO4, we know that oxygen has a -2 oxidation state, and sulfur has a +6 oxidation state. Let the oxidation state of manganese be x. Then: x + (-2)4 + (+6) = 0x - 8 + 6 = 0x = +2Thus, the correct order of increasing oxidation state of Mn is +7, +6, +4, +2, which corresponds to option (C).

Q148. Which pair of transition elements shows abnormal electronic configuration?

• A. Sc and Zn
• B. Cu and Sc
• C. Zn and Cu
• D. Cu and Cr✓

Explanation: Both Cu and Cr show abnormal electronic configurations.Cu (Copper) has the electronic configuration of [Ar] 3d10 4s1 instead of [Ar] 3d9 4s2. This is because a filled d-subshell (d10) and a half-filled s-subshell (4s1) are more stable configurations than a partially filled d-subshell (d9) and a filled s-subshell (4s2).Cr (Chromium) has the electronic configuration of [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2. This is because a half-filled d-subshell (d5) is more stable than a partially filled d-subshell (d4) due to the exchange energy.

Q149. The acid rain water has pH:

• A. Below 5✓
• B. 7
• C. Between 5 and 7
• D. Between 7 and 14

Explanation: It results from high levels of atmospheric nitric and sulfuric acids that get washed down to earth. Oxidation of sulfur and nitrogen in coal or other fossil fuels leads to the generation of acidic pollutants in the atmosphere.

Q150. In the Contact Process for manufacturing sulphuric acid, Sulphur trioxide (SO3) is not absorbed in water because:

• A. The reaction does not go to completion
• B. The reaction is highly exothermic✓
• C. The reaction is quite slow
• D. SO3 is insoluble in water

Explanation: In the Contact Process, sulphur trioxide (SO3) is produced by the oxidation of sulphur dioxide (SO2) in the presence of a vanadium(V) oxide catalyst. The reaction is highly exothermic and produces a lot of heat. This means that if SO3 were to come into contact with water, it would react vigorously with it, releasing a large amount of heat and forming sulphuric acid (H2SO4) through the following reaction:SO3(g) + H2O(l) → H2SO4(l) (exothermic)This reaction is highly exothermic and can cause the mixture to boil and even explode due to the sudden release ofheat. Therefore, in the Contact Process, SO3 is not allowed to come into contact with water and is instead absorbed in concentrated sulphuric acid to form oleum, which can be further diluted with water to produce sulphuric acid.

Q151. In modern Haber Process Plants, the temperature maintained during the process is:

• A. 670 – 770 K (400 °C – 500 °C)✓
• B. 270 – 370 K (0 °C – 100 °C)
• C. 370 – 470 K (100 °C – 200 °C)
• D. 570 – 600 K (300 °C – 380 °C)

Explanation: The Haber Process is an important industrial process that is used to produce ammonia (NH3) from nitrogen and hydrogen gases. It is carried out at high pressure and temperature in the presence of an iron catalyst. The optimal temperature range for the Haber process is between 400 °C to 500 °C, and the pressure used in modern plants is about 200 atmospheres. The temperature range used in the process is carefully chosen because it provides a good balance between the kinetics of the reaction and the thermodynamics of the reaction. At lower temperatures, the reaction is too slow, while at higher temperatures, the equilibrium of the reaction shifts towards the formation of nitrogen and hydrogen instead of ammonia, resulting in a lower yield of ammonia.

Q152. In the Haber process for manufacturing of ammonia, Nitrogen is taken from:

• A. Proteins occurring in living bodies
• B. Ammonium salts obtained industrially
• C. Air✓
• D. Minerals containing nitrates

Explanation: In the Haber process, Nitrogen is obtained from the air, which contains about 78% nitrogen gas. The air is compressed and then passed through a catalyst bed along with hydrogen gas at a high temperature and pressure. The nitrogen gas reacts with hydrogen gas in the presence of an iron catalyst to form ammonia gas. Haber process is an important industrial process for the production of ammonia, which is a key component in the manufacture of fertilizers, explosives, and other chemical compounds.

Q153. Ethene on polymerization gives the product polyethylene. This reaction may be called as:

• A. Addition✓
• B. Condensation
• C. Substitution
• D. Pyrolysis

Explanation: Polymerization of ethene involves the addition of many ethene molecules to form a long-chain hydrocarbon called polyethylene. During the polymerization process, the carbon-carbon double bond in ethene is broken, and the carbons in the double bond form single bonds with other ethene molecules. As a result, the polymer chain grows longer and longer with each addition, ultimately forming a long-chain polymer. This reaction is called an addition polymerization reaction because the monomers add together to form the polymer.Condensation reactions, on the other hand, typically involve the elimination of a small molecule, such as water or alcohol, as the monomers react to form a polymer. Substitution and pyrolysis reactions are different types of reactions that do not apply to the polymerization of ethene to form polyethylene.

Q154. In the following, which one is a free radical?

• A. Cl-
• B. Cl+
• C. Cl2
• D. Cl0✓

Explanation: A free radical is an atom, molecule or ion that has an unpaired electron. In the given options, CI° has only one electron in its valence shell, making it an unpaired electron and a free radical. CI- and CI+ have an equal number of electrons and protons, and therefore they are not free radicals. Cl2 is a molecule, and although it contains two chlorine atoms, they are bonded together by a covalent bond, and not unpaired electrons, making it not a free radical.

Q155. See the question image and answer:

• A. Acylation✓
• B. Carbonyl reduction
• C. Alkylation
• D. Formylation

Explanation: The introduction of an RC(O)+ group (also called an acyl group) in benzene is called acylation. This reaction is typically carried out using a catalyst such as aluminium chloride (AlCl3) or iron chloride (FeCl3). During the reaction, the catalyst activates the acylating agent such as acetyl chloride (CH3COCl) by removing a halide ion, creating an acylium ion (RC(O)+) that is then attacked by the benzene ring. The final product is an aromatic ketone.

Q156. The alkaline hydrolysis of bromoethane shown below gives alcohol as the product H3C―CH2―Br → H3C―CH2―OH. The reagent and the condition used in this reaction may be:

• A. H2O at room temperature
• B. Ethanol, heat
• C. KOH in alcohol
• D. Dilute NaOH(aq) warm✓

Explanation: The given reaction is an example of alkaline hydrolysis or nucleophilic substitution reaction in which a halogen atom in an alkyl halide is replaced by a hydroxyl (-OH) group from a nucleophilic hydroxide ion (OH-).The reaction conditions required for this reaction are basic and involve the use of a strong nucleophile, which is provided by the hydroxide ion. Dilute NaOH(aq) is commonly used as a reagent for this reaction, as it provides the necessary conditions for hydrolysis to occur. The reaction is usually carried out under reflux conditions (heating with constant boiling) to drive the reaction to completion.

Q157. In the reaction of ethene with bromine the intermediate formed is:

• A. A✓
• B. B
• C. C
• D. D

Explanation: The reaction of ethene with bromine is a typical example of electrophilic addition reaction. In this reaction, the first step involves the formation of an intermediate called a bromonium ion. This is formed due to the addition of bromine molecule across the double bond of ethene. The bromonium ion is a cyclic intermediate having a positive charge on the bromine atom. The correct representation of the intermediate formed in the reaction is given in (A) option. This intermediate is highly reactive and can react with a nucleophile to form the final product

Q158. In substitution reactions, dihaloalkane or secondary halogenoalkane give / show:

• A. SN1 Mechanism
• B. SN2 Mechanism
• C. Both E1 and E2
• D. Both SN1 and SN2✓

Explanation: In the case of dihaloalkanes or secondary halogenoalkanes, both SN1 and SN2 mechanisms can occur. However, SN1 mechanism is favored in polar solvents like water or alcohol, while SN1 mechanism is favored in non-polar solvents like ether or hydrocarbons.In the SN1 mechanism, the halogenoalkane undergoes a heterolytic bond cleavage, forming a carbocation intermediate, which is then attacked by the nucleophile. In the SN2 mechanism, the nucleophile attacks the halogenoalkane at the same time as the leaving group departs, leading to inversion of configuration at the carbon center.It is worth noting that the primary halogenoalkanes typically undergo SN2 reactions exclusively, while tertiary halogenoalkanes typically undergo SN1 reactions exclusively.

Q159. The dehydration of ethyl alcohol with concentrated H2SO4 at 140°C gives:

• A. Ethene (ethylene)
• B. Diethyl ether✓
• C. Alcohol
• D. Carboxylic acid

Explanation: When ethanol is heated with conc. H2SO4 at140∘C (413K), the product obtained is diethyl ether.2C2H5OH + H2SO4 at 140∘C(413K) → C2H5−O−C2H5 + H2O

Q160. Ethanol can be converted in to ethanoic acid by:

• A. Oxidation✓
• B. Fermentation
• C. Hydration
• D. Hydrogenation

Explanation: Ethanol can be oxidized to ethanoic acid by using oxidizing agents such as potassium permanganate (KMnO4), chromium trioxide (CrO3) or acidified potassium dichromate (K2Cr2O7). The reaction involves the loss of hydrogen atoms from the ethanol molecule, resulting in the formation of an aldehyde (acetaldehyde) which is then further oxidized to form the carboxylic acid (ethanoic acid).For example, the oxidation of ethanol using acidified potassium dichromate can be represented by the following equation:CH3CH2OH + 2[O] → CH3CHO + H2OCH3CHO + [O] → CH3COOHWhere [O] represents the oxidizing agent.

Q161. The following structure is of a:

• A. Secondary alcohol
• B. Primary alcohol
• C. Tertiary alcohol✓
• D. Carboxylic acid

Explanation: A tertiary alcohol has the hydroxyl group on a carbon atom, which is bonded to two other carbon atoms.

Q162. When ethanol is warmed with ethanoic acid in the presence of a strong acid catalyst, an ester ethyl ethanoate is formed.CH3CH2OH + CH3CO2H → CH3CO2CH2CH3During this reaction:

• A. Alcohol is reduced
• B. O-H bond in ethanoic acid is broken
• C. O-H bond in ethanol is broken✓
• D. Acid is oxidised

Explanation: In the given reaction, ethanol and ethanoic acid react in the presence of a strong acid catalyst (usually concentrated sulphuric acid) to form an ester, ethyl ethanoate. This reaction is an example of an esterification reaction.The reaction involves the breaking of the O-H bond in the ethanol molecule and the O-H bond in the carboxylic acid molecule. The hydrogen from the -OH group of the ethanol combines with the hydroxyl group (-OH) of the carboxylic acid, leading to the formation of a water molecule. The remaining carbon-oxygen bond in both the ethanol and the carboxylic acid forms the ester linkage (-COO-) of the ester product.Therefore, option (C) is the correct answer.

Q163. Primary alcohols usually give us aldehydes when oxidized in the presence of Na2Cr3O7, what will the product be, when the secondary alcohols are oxidized in the same conditions?

• A. Alkenes
• B. Alkynes
• C. Alkyl halides
• D. Ketones✓

Explanation: When secondary alcohols are oxidized with Na2Cr2O7 in the presence of H2SO4, they are converted to ketones. This process does not go beyond the ketone stage as further oxidation of ketones is not possible under these conditions. In contrast, primary alcohols are oxidized to aldehydes, and further oxidation leads to the formation of carboxylic acids. The oxidation of secondary alcohols does not lead to the formation of alkenes, alkynes, or alkyl halides because these functional groups do not contain an carbonyl group that is oxidized functional group of -OH group.

Q164. Formaldehyde reacts with HCN (NaCN + HCl) to give a compound:

• A. A
• B. B✓
• C. C
• D. D

Explanation: The reaction of formaldehyde with HCN (sodium cyanide and hydrochloric acid) is an example of a nucleophilic addition reaction. The reaction takes place as follows:H2C=O + HCN → H2C(OH)(CN)The product formed is called hydroxymethyl cyanide, which is also known as formaldehyde cyanohydrin. In this product, the HCN has added to the carbonyl carbon of formaldehyde, forming a new carbon-carbon bond and creating an alcohol and nitrile group. Therefore, the correct option is B, which represents the structure of hydroxymethyl cyanide.

Q165. Iodoform test will not be positive with:

• A. A✓
• B. B
• C. C
• D. D

Explanation: Iodoform test is a test for the presence of a methyl ketone or a compound containing methyl group is adjacent to the functional group. In this test, the compound is treated with iodine in the presence of a mild base such as sodium hydroxide (NaOH), and the formation of a yellow precipitate of iodoform (CHI3) indicates the presence of a methyl ketone or a compound containing CH3 adjacent to functional group.Out of the given options, only option A (H₂C-CH₂-CO-CH₂-CH3) does not have a methyl group adjacent to the carbonyl group. Therefore, iodoform test will not be positive for option A. Options B (H3C--CH2-CO-CH3), C (C2H5OH), and D (H3C-CO-H) have a methyl group adjacent to the functional group, and hence, will give a positive iodoform test.

Q166. When CH3-CH2-OH is oxidized in the presence of K2Cr2O7 and H2SO4, the product formed is:

• A. A✓
• B. B
• C. C
• D. D

Explanation: When CH3-CH2-OH is oxidized in the presence of K2Cr2O7 and H2SO4, the product formed is ethanal (acetaldehyde), which can further oxidize to give ethanoic acid (acetic acid) as follows:CH3-CH2-OH + [O] → CH3-CHO + H2OCH3-CHO + [O] → H3C-CO-OHThe oxidation of alcohol to aldehyde is a partial oxidation, and the further oxidation of aldehyde to carboxylic acid is a complete oxidation. The presence of the K2Cr2O7 and H2SO4 provide the necessary oxidizing conditions.

Q167. The diagram shows a small magnet hanging on a thread near the end of a solenoid carrying a steady current ‘I’. What happens to the magnet as the iron core is inserted into the solenoid?

• A. It moves towards solenoid and rotates through 180˚
• B. It moves towards the solenoid✓
• C. It moves away from solenoid
• D. It moves away from solenoid and rotates through 180˚

Explanation: According to the right-hand grip rule, a north pole is induced at the end of the solenoid which faces the south pole of the magnet. Since opposite poles attract, the magnet will move towards the solenoid.

Q168. A 10 cm long solenoid has 100 turns. What will be the magnetic field inside it along its axis if one-microampere current is passed through it?

• A. 4π x 10-13 tesla
• B. 4π x 10-17 tesla
• C. 4π x 10-10 tesla✓
• D. 4π x 10-16tesla

Explanation: Given that,Length = L= 10cm which is 0.1mNumber of turns = n = 100Current passed = 1μ which is 1 x 10-6AUsing the formula for magnetic induction,𝐵=𝜇0 n I / LWhere,𝜇0 is the magnetic permeability of free space, and mu naught value : µ0 = 4π × 10-7 H/m.n is the number of turns of wire per metreI is the current through the wireAnd L is its length,B=4π × 10-7 x 100 x 1 x 10-6/ 0.1B = 4π x 10-10tesla

Q169. Two long straight parallel wires held vertically have equal but opposite currents as shown in the figure. Which of the following effects will be observed?

• A. Magnetic field at ‘X’ is stronger than that at ‘Y’ and ‘Z’
• B. Magnetic field at ‘X’ is weaker than that at ‘Y’ and ‘Z’
• C. Magnetic field at ‘X’, ‘Y’ and ‘Z’ is same
• D. Magnetic field at ‘X’ is weaker than that at ‘Y’ but stronger than that at ‘Z’.
• E. Magnetic field at ‘Y’ is stronger than that at ‘X’ and ‘Z’✓

Explanation: When two parallel wires carry equal and opposite currents, they create a magnetic field around them. The direction of the magnetic field at any point is given by the right-hand rule. Now, let's consider the magnetic field at points X, Y, and Z. At point Y, the magnetic fields created by the two wires are in the same direction and add up, resulting in a stronger magnetic field than at either X or Z.At point X, the magnetic field due to the current in the left wire is in the upward direction, and the magnetic field due to the current in the right wire is in the downward direction. These two magnetic fields oppose each other, resulting in a weaker magnetic field than at point Y.At point Z, the magnetic field due to the current in the right wire is in the upward direction, and the magnetic field due to the current in the left wire is in the downward direction. These two magnetic fields also oppose each other, resulting in a weaker magnetic field than at point Y.Therefore, the magnetic field at Y is stronger than that at X and Z.

Q170. The kinetic energy K.E. with which the electron strikes the target is given by:

• A. K.E. = e2V
• B. K.E. = hc/λ
• C. K.E. = hf2
• D. K.E. = eV✓

Explanation: An electron cannot travel at the speed of light, unless the wave properties are being studied it's wavelength cannot be observed; however eV is a frequent way to describe energy of an electron. Where ‘e’ is the charge of the electron and ‘V’ is the potential difference between the filament and the positive accelerating electrode

Q171. LASER is an acronym for:

• A. Light amplification by stimulated emission of radiation✓
• B. Light annihilation by stimulated emission of radiation
• C. Light amplitude of stimulated emission of radiation
• D. Light amplification by stimulated emission of radio

Explanation: Light amplification by stimulated emission of radiation

Q172. X-rays can be produced by bombardment of _ on target metal:

• A. Protons
• B. Electrons✓
• C. Neutrons
• D. Alpha particles

Explanation: X rays are produced due to electrons. When a heated filament emits electrons by thermionic emission in a glass envelope.These electrons are then accelerated using a high voltage. These accelerated electrons strike a metal target at an angle which produces X rays. This can further be explained by the diagram of a simple X Ray tube.

Q173. Laser light is monochromatic which means:

• A. It consists of one ray of light
• B. It consists of one wavelength✓
• C. It consists of carbon monoxide gas
• D. It consists of photons having 1 eV energy

Explanation: Monochromatic light refers to light that has only one wavelength or frequency. In other words, all the photons in monochromatic light have the same energy and travel in the same direction. Laser light is an example of monochromatic light because it is produced by a process called stimulated emission, which produces light waves of a very specific wavelength.

Q174. If an electron in the ‘K’ shell is removed and an electron from the ‘L’ shell jumps to occupy the hole in the ‘K’ shell, it emits a photon of energy:

• A. hfKα = EL – EK✓
• B. hc = EL – EK
• C. h/λKα = EL – EK
• D. hfKα = EK – EL

Explanation: When an electron is removed from the K-shell, leaving a hole, an electron from a higher energy level (L-shell in this case) can jump down to fill the gap, emitting a photon of energy equal to the difference in energy between the two levels. This is known as the characteristic X-ray emission.The energy of the emitted photon is given by the equation:hfKa = EL - Ek

Q175. Which of the following properties must be there in a substance so that it can be used as a target in an X-ray tube?

• A. It must have low melting point
• B. It must have low atomic number
• C. It must have high reflecting ability
• D. It must have high atomic number✓

Explanation: An X-ray tube is a device that produces X-rays by accelerating electrons and directing them towards a target material. When high-speed electrons collide with the target material, X-rays are produced through a process called bremsstrahlung radiation.The efficiency of X-ray production is directly proportional to the atomic number of the target material. This is because higher atomic number elements have more electrons that can be excited by the high-speed electrons, leading to a greater number of X-rays produced.Therefore, a substance with a high atomic number, such as tungsten, is an ideal target material for an X-ray tube. Additionally, the target material should have a high melting point and good thermal conductivity to withstand the high heat generated during X-ray production.

Q176. Which of the following can be used to produce population inversion for the emission of a Laser?

• A. Optical pumping✓
• B. Optical fibre
• C. Optical instrument
• D. Optical polarization

Explanation: Optical pumping is the process of photon emission at an element, since amplification of photons is needed to produce a laser, photon pumping is done to cause excitation of electrons to help with it.

Q177. What is the charge on alpha particles emitted during the phenomenon of radioactivity?

• A. +e
• B. –e
• C. –2e
• D. +2e✓

Explanation: Alpha particles are emitted during the process of alpha decay, which is a type of radioactive decay. An alpha particle consists of two protons and two neutrons bound together, which is equivalent to a helium nucleus.Since a proton has a positive charge of +e and there are two protons in an alpha particle, the total charge on an alpha particle is +2e. The neutrons in an alpha particle have no charge, so they do not contribute to the particle's overall charge.

Q178. A radioactive nuclide decays by emitting an alpha particle, a beta particle, and a gamma ray photon, the change in the nucleon number will be:

• A. -4✓
• B. -1
• C. -2
• D. -3

Explanation: An alpha particle is a helium nucleus with two protons and two neutrons which means it has a mass number of 4. (42α)A beta particle is a fast-moving electron. An excess neutron transforms into a proton and an electron.The proton stays in the nucleus and the electron is ejected energetically. Therefore in beta emission, the nucleon number remains the same. (01β)In gamma radiation, no particle is emitted instead there is the emission of radiation hence no change in nucleon number occurs during gamma emission(00γ)Therefore, the change in nucleon number will only be caused by the emission of an alpha particle hence the change will be -4.

Q179. The half-life of sodium-24 is _ which is used to estimate the volume of blood in a patient.

• A. 6 hours
• B. 15 hours✓
• C. 8 hours
• D. 15 days

Explanation: Sodium-24 is a radioactive isotope of sodium with a half-life of approximately 15 hours. This means that if a sample of sodium-24 contains 1000 atoms, after 15 hours, only 500 atoms will remain.The radioactive decay of sodium-24 can be used to estimate the volume of blood in a patient. This involves injecting a small amount of sodium chromate, which contains sodium-24, into the patient's bloodstream. As the sodium-24 decays, it is eliminated from the body through the urine. By measuring the rate at which sodium-24 is excreted in the urine, the volume of blood in the patient's body can be estimated.

Q180. Which of the following is the unit of absorbed dose?

• A. Sievert
• B. Gray✓
• C. Roentgen
• D. Curie

Explanation: Absorbed dose is a measure of the amount of energy deposited in a material by ionizing radiation per unit mass. The unit of absorbed dose is the gray (Gy), which is defined as the absorption of one joule of radiation energy per kilogram of the material.The sievert (Sv) is a unit of dose equivalent, which takes into account the biological effect of different types of ionizing radiation on human tissue. The roentgen (R) is a unit of exposure, which measures the ionization produced by X-rays or gamma rays in air. The curie (Ci) is a unit of activity, which measures the rate of radioactive decay of a substance.

Q181. In a radioactive phenomenon observation is shown in the figure where α deviates less than β in some electric or magnetic field (not shown in the figure). What is the reason for less deviation?

• A. α is charged particle
• B. α is neutral particle
• C. α is heavier particle✓
• D. α is lighter particle

Explanation: In a magnetic or electric field, deviation occurs to the charge of a particle and how much a particle deviates depends on the mass of a particle. An alpha particle is identical to the nucleus of the Helium-4 atom. It consists of two protons and two neutrons bound together having a mass of four units and a positive charge of two. A gamma particle has no charge and no mass, hence it is undeflected. A beta particle has a negative charge and lower mass as compared to an alpha particle.Alpha particles deflect the least because it is heavier than beta and gamma.42α ,0-1β ,00γ .

Q182. The isotope of Iodine-131 is used in the treatment of:

• A. Blood cancer
• B. Bone cancer
• C. Lung tumour
• D. Thyroid cancer✓

Explanation: Thyroid glands absorb nearly all of the iodine in your body. Because of this, radioactive iodine ‘I-131’ can be used to treat thyroid cancer. Where the radiation can destroy the thyroid gland and any other thyroid cells (including cancer cells) that take up iodine, with little effect on the rest of your body.

Q183. Which of the following effects is observed due to the emission of β- during the phenomenon of radioactivity?

• A. A increases by 1 and Z remains same
• B. Z increases by 1 and A remains same✓
• C. Z decreases by 1 and A remains same
• D. A decreases by 1 and Z remains same

Explanation: When a beta particle is emitted, a neutron in the nucleus is converted into a proton, which increases the atomic number Z by 1. However, the mass number A of the atom remains the same, since the mass of an electron or positron is negligible compared to that of a proton or neutron.For example:Therefore, the atomic number Z increases by 1, but the mass number A remains the same when a beta particle (B-) is emitted during radioactive decay.

Q184. Electric charge on an object is measured as 5 micro-coulombs. The value of this charge can be expressed in terms of base units as:

• A. 5000 ampere second
• B. 5 x 10-6 ampere second✓
• C. 5 x 106 coulomb second
• D. 500 coulomb second

Explanation: The following is the solution:Electric charge is a derived quantity, and its base unit is the coulomb (C), which is defined as the charge carried by a current of one ampere (A) in one second (s).So, 1 C = 1 A s.The prefix "micro" (μ) represents a factor of 10-6. Therefore, 5 microcoulombs (5 μC) can be expressed as:5 μC = 5 x 10-6 CTo express this value in terms of base units, we need to substitute the base unit of charge (coulomb) into the equation:5 x 10-6 C = (5 x 10-6 A) x 1 sSo, the value of 5 microcoulombs can be expressed in terms of base units as 5 x 10-6 ampere second, which is option B.

Q185. If ‘m’ is the mass, ‘c’ is the velocity of light and x = mc2, then dimensions of ‘x’ will be:

• A. [LT-1]
• B. [ML2T-2]✓
• C. [MLT-1]
• D. [MLT-2]

Explanation: The equation x = mc² represents the relationship between mass and energy, where x is the energy equivalent of a mass m, and c is the speed of light.To determine the dimensions of x, we need to consider the dimensions of each variable in the equation. The dimensions of mass are [M], the dimensions of the speed of light are [LT-1].Using the rules of dimensional analysis, we can find the dimensions of x by substituting the dimensions of m and c into the equation and simplifying:x = mc²= [M][LT-1]²= [M][L2T-2]Therefore, the dimensions of x are [ML2T-2], which is option B.

Q186. A force ‘F’ is acting at point ‘P’ of a uniform rod capable of rotating about ‘O’. What is the torque about ‘O’?

• A. (OP)(F tanϴ)
• B. (OP)(F)
• C. (OP)(F sinϴ)
• D. (OP)(F cosϴ)✓

Explanation: Torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector,It is important that the force being applied is perpendicular to the rotational axis therefore we will use a component of force F which is perpendicular to the rotational axis. Since the angle θ is between the force F and the perpendicular. The force applied on the rotational axis for torque will be Fcosθ.The distance given here is OP and the force applied is Fcosθ.Torque = d x FTorque = (OP)(Fcosθ)

Q187. An object of mass ‘m’ is suspended in an elevator moving downward with acceleration equal to acceleration due to gravity. What is the apparent weight of object?

• A. Zero✓
• B. 2mg
• C. mg
• D. mg/2

Explanation: The apparent weight of the object is the force exerted on it by the elevator floor, which is equal to the net force acting on the object.Using the formula Fnet = mg - ma Where,m is the mass of the object g is the acceleration due to gravity a is the acceleration of the elevatorIn this case, the elevator is moving downward with an acceleration equal to the acceleration due to gravity, which means a = g.Therefore,Fnet = mg - ma = m(g - g) = 0.So, the apparent weight of the object is zero. Option A is the correct answer.

Q188. Stokes’ Law for steady motion in a fluid of infinite extent is given by:

• A. F = 6πηrv✓
• B. F = (4/3)πr3ρg
• C. F = 6πηr2ρ
• D. F = 2gr2ρ/9η

Explanation: Stokes' Law describes the drag force experienced by a small spherical object moving through a fluid at a low velocity. According to the law, the drag force is proportional to the velocity of the object, the radius of the object, and the viscosity of the fluid.The formula for the drag force, F, can be written as:F = 6πηгv, Where,η is the viscosity of the fluid г is the radius of the object v is the velocity of the object

Q189. If the speed of efflux through a small hole in a large tank is 9.8 m/s. Find the height at the fluid above the hole:

• A. 1 m
• B. 9.8 m
• C. 4.9 m✓
• D. 19.6 m

Explanation: The speed of efflux through a small hole in a large tank is given by the Torricelli's Law, which is given as:v = √(2gh)where v is the speed of efflux, g is the acceleration due to gravity, h is the height of the fluid above the hole.We are given that the speed of efflux through the hole is 9.8 m/s, which is equal to the acceleration due to gravity, i.e., v = g = 9.8 m/s².Substituting these values in the above equation, we get:9.8 = √(2gh)Squaring both sides of the equation, we get:96.04 = 2ghh = 96.04 / (2g)h = 96.04 / (2 x 9.8)h = 4.9 metersTherefore, the height of the fluid above the hole is 4.9 meters. Option C is the correct answer.

Q190. Flow speed of the fluid through a non-uniform pipe increases from 1 m/sec to 3 m/sec. If change in P.E. is zero, then pressure difference between two points will be: (density of the fluid = 1000 kg/m3).

• A. 1000 N/m2
• B. 9000 N/m2
• C. 8000 N/m2
• D. 4000 N/m2✓

Explanation: According to Bernoulli's equation, the pressure difference between two points in a fluid is related to the change in velocity of the fluid and the density of the fluid.The equation is given as:P1 + (1/2)ρv1² = P2 + (1/2)ρv2²where P1 and P2 are the pressures at two different points in the pipe, v1 and v2 are the velocities at those points, and ρ is the density of the fluid.In this problem, we are given that the change in potential energy is zero, which means that we can neglect the potential energy term in Bernoulli's equation.Also, the density of the fluid is given as 1000 kg/m³.Using Bernoulli's equation, we can write:P1 + (1/2)ρv1² = P2 + (1/2)ρv2²Since the change in potential energy is zero, the pressure difference between the two points is given by:ΔP = P1 - P2 = (1/2)ρ(v2² - v1²)Substituting the given values, we get:ΔP = (1/2) x 1000 x (3² - 1²)ΔP = 4000 N/m²Therefore, the pressure difference between the two points is 4000 N/m². Option D is the correct answer.

Q191. Polarization of light exhibited the nature of light as:

• A. Longitudinal wave
• B. Compressional wave
• C. Transverse wave✓
• D. Electromagnetic wave

Explanation: Polarization of light is a phenomenon that occurs when a transverse wave such as light is restricted to vibrating in a particular plane. This means that the electric field and magnetic field vectors of the light wave oscillate perpendicular to the direction of propagation of the wave.

Q192. The concentration of a sugar solution can be determined by:

• A. Un-polarized light
• B. Plane polarized light✓
• C. Interference of light
• D. Diffraction of light

Explanation: The concentration of a sugar solution can be determined using a technique called polarimetry, which involves the use of plane-polarized light. When plane-polarized light is passed through a sugar solution, the plane of polarization of the light is rotated by an amount that depends on the concentration of the sugar in the solution. The plane of polarization of the light is rotated by an angle θ as it passes through the sugar solution. By measuring the angle of rotation and knowing the specific rotation of the sugar, the concentration of the sugar in the solution can be determined. Therefore, the correct option is B) Plane polarized light.

Q193. The information from one place to another can be transmitted very safely and easily by:

• A. Copper wire
• B. Aluminium wire
• C. Photodiode
• D. Optical fibre✓

Explanation: Optical fiber is a technology that allows the transmission of information from one place to another using light signals. An optical fiber is a thin, flexible, and transparent fiber made of glass or plastic that can carry light signals over long distances with minimal loss of signal.The principle of optical fiber communication is based on the phenomenon of total internal reflection, where light traveling through a medium with a higher refractive index is reflected when it encounters a medium with a lower refractive index at an angle greater than the critical angle.Compared to Copper wire or Aluminium wire, optical fiber has several advantages, including higher bandwidth, lower attenuation, greater immunity to electromagnetic interference, and greater security since it is difficult to tap into the signal without breaking the fiber.Therefore, the correct option is Optical fiber.

Q194. The image of an object placed inside the focal length of a convex lens will be largest and clearest when it is at the:

• A. Less than 25 cm
• B. Near point✓
• C. Greater than 25 cm
• D. Infinity

Explanation: When an object is placed inside the focal length of a convex lens, a virtual, upright, and magnified image is formed on the same side of the lens as the object. The size and clarity of the image depend on the distance between the lens and the object. As the object is moved closer to the lens, the image becomes larger and more magnified. However, if the object is moved too close to the lens, the image becomes blurred and distorted. The distance at which the image is largest and clearest is called the "near point" or "least distance of distinct vision". For a normal eye, the near point is about 25 cm from the eye. Therefore, when an object is placed inside the focal length of a convex lens, the image will be largest and clearest when it is at the near point, which is about 25 cm from the eye. The image will be smaller and less clear if the object is placed closer or further away from the near point.

Q195. A simple harmonic oscillator has a time period of 10 seconds. Which equation rotates its acceleration ‘a’ and displacement ‘x’?

• A. a = -2 x
• B. a = -(20π)x
• C. a = -(2π/10)2 x✓
• D. a = -(20π)2 x

Explanation: The motion of a simple harmonic oscillator can be described by the following equation:a = -ω2 * xwhere 'a' is the acceleration, 'x' is the displacement, and ω is the angular frequency of the oscillator.The time period of the oscillator is given by:T = 2π/ωwhere 'T' is the time period.In this case, the time period is given as 10 seconds. Therefore, we can find the angular frequency as:ω = 2π/ T = 2π /10 =π/5Substituting this value of ω in the equation of motion, we get:a = - (π /5)2 * xSimplifying, we get:a = - (2π /10)2 * xTherefore, the correct option is C) a = - (2π / 10)2 * x

Q196. When the length of a simple pendulum is doubled, the ratio of the new frequency to the old frequency is:

• A. 1/4
• B. 1/2
• C. √2
• D. 1/√2✓

Explanation: f = 1/2π × √(g/L)where 'g' is the acceleration due to gravity and 'L' is the length of the pendulum.When the length of the pendulum is doubled, the new length becomes 2L. Therefore, the new frequency can be calculated using the same formula:f' = 1/2π × √(g/2L)To find the ratio of the new frequency to the old frequency, we can divide f' by f:f'/f = [1/2π × √(g/2L)] / [1/2π × √(g/L)]f'/f = √(L/2L)f'/f = √(1/2)f'/f = 1/√2Therefore, the correct option is D) 1/√2

Q197. In the diagram below, the displacement of an oscillating particle is plotted against time. What does the length ‘PR’ on the time axis represent?

• A. Twice the frequency
• B. Half the period✓
• C. Half the frequency
• D. Twice the period

Explanation: The time period of the wave is the time between two consecutive troughs or two consecutive crests. In the given diagram you can see the distance between a trough and a crest which means its half a period.

Q198. When the source of sound moves towards the stationary observer, the value of apparent frequency ‘fo’ is:

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

Explanation: The formula to calculate the apparent frequency (f°) of a sound wave due to the Doppler effect is given by:f°= (V± Uo/ V± Ui) × fwhere V is the speed of sound, Uo is the speed of the observer, Ui is the speed of the sound source, and f is the actual frequency of the sound wave.When a source of sound moves towards a stationary observer, the observer perceives the frequency of the sound to be higher than the actual frequency. Therefore the sign in the denominator would be negative to get higher apparent frequency than actual frequency. The observer is stationary, so the speed of the observer (Uo) is zero. Thus the formula in this case would be:f°= (V/V-Ui) × f

Q199. The ratio of tensile stress to tensile strain is called:

• A. Modulus of elasticity
• B. Bulk Modulus
• C. Young’s Modulus✓
• D. Shear Modulus

Explanation: Young's modulus is a measure of the stiffness of a material. It is defined as the ratio of tensile stress to tensile strain within the elastic limit of a material. In other words, Young's modulus is the ratio of the applied stress to the resulting strain, within the elastic limit of the material.Young's modulus is denoted by the symbol "E" and has the unit of Pascals (Pa) or Newtons per square meter (N/m²). It is a measure of how much a material will stretch (i.e. its elongation) under a given load. The higher the value of Young's modulus, the stiffer the material is.

Q200. A wire is stretched by a force ‘F’ which causes an extension ∆l, the energy stored in the wire is:

• A. F∆l
• B. 2F∆l
• C. ½ F∆l2
• D. ½ F∆l✓

Explanation: Given that, force applied = F . extension = ∆l and we assume the length of the wire to be = L.If the elastic limit is not exceeded then the stress is directly proportional to strain.Where stress is the amount of force applied per unit area (σ = F/A)And strain is extension per unit length (ε = ∆l/l)Hence,Energy stored = ½ x stress x strain x volumeEnergy stored = ½ x (F/A) x (∆l / L) x A x LEnergy stored = ½ F∆l

Q201. H2 and O2 both are at thermal equilibrium at temperature 300 K. Oxygen molecule is 16 times massive than hydrogen. Root mean square speed of hydrogen is:

• A. 4 root mean square of oxygen✓
• B. ¼ root mean square of oxygen
• C. 1/16 root mean square of oxygen
• D. 1/6 root mean square of oxygen

Explanation: According to the kinetic theory of gas,Crms=√(3RT / M )Where, ‘Crms’ is the root-mean-square of the velocity, ‘M’ is the molar mass of the gas in kilograms per mole, ‘R’ is the molar gas constant, and T is the temperature in Kelvin.At a constant temperature the Crms is directly proportional to the squareroot of 1/M shown as,Crms α (√1 / M )Equation 1:Crms of H2/ Crms of O2 = √ (MO2/ MH2)Since MO2 = 16 x MH2MO2/ MH2 = 16Then replacing MO2/ MH2 in equation 1 with 16 will give,Crms of H2/ Crms of O2 = √ 16Crms of H2/ Crms of O2 = 4Crms of H2 = 4 x Crms of O2

Q202. Which of the following is the expression of mean square speed of ‘N’ gas molecules contained in a cylinder?

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

Explanation: The mean square speed of N gas molecules is the average of the squares of their individual speeds. To calculate this, we square the speed of each molecule, sum them up, and then divide by the total number of molecules N. So the expression for mean square speed is:(v12 + v22 + ... + vx2) / NOption B is the expression that matches this formula, which squares each velocity term and averages them over N molecules.

Q203. If ‘Q’ is the amount of heat supplied to a system and ‘W’ is the work done, then change in internal energy can be defined as:

• A. Q/W
• B. Q – W✓
• C. W/Q
• D. 1 + Q/W

Explanation: The 1st law of thermodynamics is given by the equation:∆Q=∆U+∆W Where,∆Q is the heat supplied to the system ∆U is the change in internal energy of the system ∆W is the work done by the system. To derive the equation for change in internal energy, we can rearrange the above equation as follows:∆U = ∆Q - ∆WThis equation tells us that the change in internal energy of the system (∆U) is equal to the heat supplied to the system (∆Q) minus the work done by the system (∆W).

Q204. A heat engine operating according to the second law of thermodynamics rejects one fourth of the heat taken from a high temperature reservoir. What is the percentage efficiency of a heat engine?

• A. 100%
• B. 25%
• C. 50%
• D. 75%✓

Explanation: The percentage efficiency of a heat engine is given by:Efficiency = (Useful work output / Heat input) x 100%According to the second law of thermodynamics, some heat must be rejected to a low temperature reservoir during the operation of a heat engine. Let Qh be the heat taken from a high temperature reservoir, and Qc be the heat rejected to a low temperature reservoir. The net work output of the engine is given by:W = Qh - QcGiven that the engine rejects one-fourth of the heat taken from the high temperature reservoir, we can write:Qc = Qh/4Substituting this in the above equation for W, we get:W = Qh - Qh/4 = 3Qh/4The efficiency of the engine is therefore:Efficiency = (W / Qh) x 100%Efficiency = [(3Qh/4) / Qh] x 100%Efficiency = 75%Therefore, the correct option is D) 75%.

Q205. First law of thermodynamics under adiabatic conditions can be mathematically written as:

• A. Q = W
• B. Q = ∆U
• C. Q= U + W
• D. W = -∆U✓

Explanation: The First law of thermodynamics is written as:∆Q = ∆U + ∆WIn adiabatic conditions, ∆Q = 0This is because adiabatic conditions imply that there is no heat exchange between the system and its surroundings. Therefore, the heat transferred (∆Q) is zero.So equation will become:0 = ∆U + ∆WRearranging this equation, we get:∆U = -∆WTherefore, the correct option is D) W = -∆U.

Q206. What is the logic symbol for a NOT Gate?

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

Explanation: In the following diagrams, A is a NOT gate. B is a OR gate, C is AND gate and D is a symbol for PN junction diode.

Q207. The voltage that is applied across X-plates is provided by a circuit called:

• A. Audio generator
• B. Time base generator✓
• C. Signal generator
• D. Linear generator

Explanation: The circuit that provides voltage to X-plates is called a "time base generator". It is a type of electronic circuit used in electronic displays such as oscilloscopes and television sets. The time base generator generates a sawtooth waveform which is used to control the horizontal deflection of the electron beam in a CRO (Cathode Ray Oscilloscope). This allows the display of time-varying signals such as waveforms and images.

Q208. What will be the effect on the capacitance of a capacitor if area of each plate is doubled while separation between the plates is halved?

• A. Capacitance remains same
• B. Capacitance becomes double
• C. Capacitance becomes four times✓
• D. Capacitance reduces to half

Explanation: The formula for capacitance is𝐶=ε𝐴 / 𝑑where,A is the area of the plate, d is the distance between the two parallel plates.If area is doubled, it becomes 2Aif distance is halved it becomes d/2So the new capacitance will be,𝐶2=ε2𝐴÷ 𝑑/2 = 4ε𝐴/𝑑Since 𝐶1 = ε𝐴/𝑑𝐶2= 4𝐶1Hence capacitance will become 4 times.

Q209. 10 V potential difference is applied across the plate of the 1 µF capacitor. What is the energy stored in the capacitor?

• A. 0.5 mJ
• B. 0.05 mJ✓
• C. 5 mJ
• D. 50 mJ

Explanation: The energy stored in a capacitor can be calculated using the formula:E = (1/2) * C * V2Where,E is the energy storedC is the capacitanceV is the potential difference applied across the capacitor.Substituting the given values into the formula, we get:E = (1/2) * 1 uF * (10 V)2Since 1 uF = 10-6 F, we can rewrite the expression as:E = (1/2) * 10-6 F * (10 V)2Simplifying the expression, we get:E = 0.5 * 10-6 * 100E = 0.05 * 10-3 JE = 0.05 mJTherefore, the energy stored in the capacitor is 0.05 mJ. Hence, option (B) is correct.

Q210. Which one of the following is I-V curve of a junction diode?

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

Explanation: Option B shows the I-V curve of a junction diode. This curve represents the behavior of the diode under both forward and reverse bias conditions.Under forward bias, the diode allows current to flow in the forward direction, with increasing current for increasing voltage until it reaches a certain voltage threshold called the forward voltage. Beyond this threshold, the current increases rapidly as the diode enters into saturation.Under reverse bias, the diode blocks the flow of current until the reverse voltage reaches a certain value called the breakdown voltage, beyond which the current increases rapidly.Therefore, option B is the correct choice for the I-V curve of a junction diode among the following diagrams.

Q211. Choose the word that best fits in with the meaning of the sentence as a whole. Indolence gives vent to _ disposition in human life.

• A. Static✓
• B. Enthusiastic
• C. Energetic
• D. Filthy

Explanation: Static means to be stationary/to stop. It fits in the question as indolence (laziness) means to not move and not do work.Enthusiastic and energetic are antonyms and do not fit in.Filthy means dirty which also does not fit in.

Q212. Choose the word that best fits in with the meaning of the sentence as a whole. The Quaid’s _ enthusiasm led the Muslims Indo-Pak to independence.

• A. Simplified
• B. Latent
• C. Onerous✓
• D. Threatening

Explanation: The correct answer is 'Onerous.' The word 'onerous' conveys a sense of hard work and significant effort, which fits the context of enthusiasm leading a movement to independence. 'Simplified' does not convey the weight of effort. 'Latent' means hidden, suggesting inactivity, which is the opposite of active enthusiasm. 'Threatening' implies danger, not a positive driving force for independence.

Q213. Choose the word that best fits in with the meaning of the sentence as a whole. He _ the incident to the back of his mind.

• A. Revered
• B. Regulated
• C. Reagitated
• D. Relegated✓

Explanation: Relegated means to assign an inferior rank or position to. It can be used here as the meaning of pushing the incident back to his mind, and thus lowering its importance/rank. Revered means respected/honouredReagitated is not a word.

Q214. Choose the word that best fits in with the meaning of the sentence as a whole. She managed to _ a ticket for the cricket match.

• A. Procure✓
• B. Obscure
• C. Improvise
• D. Preclude

Explanation: Obscure means unclear.Improvise means to produce or make (something) from whatever is available. Preclude means to prevent from happening; make impossible.

Q215. Spot the error:Only twice my father did stop to wipe the sweat from his eyes

• A. Twice
• B. Did✓
• C. To wipe
• D. From

Explanation: - The sentence should read "Only twice did my father stop to wipe the sweat from his eyes".- The word "did" is incorrectly placed in the original sentence. It should be placed immediately after "twice" to form the correct sentence structure.

Q216. Spot the error.A tree slid down the slope, leaned against and settled heavily to the ground.

• A. Down
• B. Leaned against
• C. Heavily
• D. To the ground✓

Explanation: The correct statement will be 'A tree slid down the slope, leaned against and settled heavily on the ground.' The error lies in the phrase 'to the ground.' The preposition 'on' is appropriate here to indicate where the tree settled. Options A, B, and C are all correct phrases within the context of the sentence and do not contain any errors.

Q217. Spot the error. The latter continued: “You see how is it- Ralston filled the place up with young men.”

• A. Latter
• B. How is it✓
• C. Up
• D. With

Explanation: The correct phrasing should be 'You see how it is - Ralston filled the place up with young men.'The error lies in the phrase 'how is it,' which disrupts the expected structure of the sentence. The correct form, 'how it is,' maintains proper syntax and clarity.Option A, 'latter,' is not the error since it correctly indicates the second person being discussed. Option C, 'up,' is used correctly and does not contain an error, as does Option D, 'with.' Both are valid phrases in this context.

Q218. Spot the error. The delegates in Anatolia could transfer their activities to Istanbul, put Mustafa Kamal’s ideas into practice and yet not longer stand in opposition to the Padishah.

• A. In
• B. To
• C. Into
• D. Not longer✓

Explanation: "No longer" would be correct form of expression in this sentence. The correct sentence is, "The delegates in Anatolia could transfer their activities to Istanbul, put Mustafa Kamal’s ideas into practice and yet no longer stand in opposition to the Padishah."

Q219. Female workers are entitled 56 days rest on full salary before giving birth to children.

• A. Entitled✓
• B. 56 days rest
• C. On full salary
• D. Giving birth

Explanation: "Entitled to" will be the correct expression. The correct sentence is Female workers are entitled to 56 days rest on full salary before giving birth to children."

Q220. Fill in the blank with appropriate option.There is nothing interesting in mathematics. I _ read grammar.

• A. Could
• B. Can
• C. Might as well✓
• D. Must

Explanation: The correct answer is 'might as well' because it suggests that, given the lack of interest in mathematics, the speaker sees reading grammar as a reasonable alternative. The phrase implies a decision made due to the circumstances, rather than a necessity or ability. The other options — 'could', 'can', and 'must' — do not fit the context of this choice. 'Could' and 'can' imply ability rather than preference, and 'must' implies obligation rather than a casual decision.