Etea Mdcat 2019 — Solved Past Paper with Answers

Q1. Sadia wore her rain boots; _ her feet stayed dry during the storm.

• A. However
• B. Therefore✓
• C. On the other hand
• D. Still

Explanation: This is the correct answer. "Therefore" is used to show a logical consequence or conclusion. In this context, wearing rain boots logically leads to the result that Sadia's feet stayed dry during the storm.

Q2. Anum asked me, "Did you see the drama on television last night?" Choose the correct indirect speech:

• A. Anum asked me whether I saw the drama on television the earlier night.
• B. Anum asked me whether I had seen the drama on television the earlier night.✓
• C. Anum asked me did I see the drama on television the last night.
• D. Anum asked me whether I had seen the drama on television the last night.

Explanation: Interrogative sentences beginning with an auxiliary verb are changed into indirect speech using the connective if or whether. The reporting verb said (or any other word used as the reporting verb) changes to asked, queried, questioned, demanded of, or enquired of in the indirect speech. Note that of is used after enquired and demanded only when the reporting verb has an object. So B will be the correct option.

Q3. Do not make so much noise, Farrah _ to study for her ESL test!

• A. Try
• B. Tries
• C. Tried
• D. Is trying✓

Explanation: This is the correct answer. "Is trying" is the present progressive (continuous) tense form of the verb "try," used to indicate an action that is currently happening. In this context, it shows that Farrah is currently making an effort to study for her ESL test, which is in line with the instruction not to make noise.

Q4. Choose the correct voice. Zara changed the flat tire.

• A. The flat tire was changed by Zara.✓
• B. The flat tire is changed by Zara.
• C. The flat tire has been changed by Zara.
• D. The flat tire had changed by Zara.

Explanation: The passive voice is used here, indicating that the subject (the flat tire) is having the action of being changed performed on it by the agent (Zara). The verb "was changed" is in the simple past tense, indicating a completed action.

Q5. The simplest form of learning is:

• A. Latent learning
• B. Imprinting
• C. Insight
• D. Habituation✓

Explanation: A learned behavior in which an individual’s response to stimuli decreases over time is called habituation. It is learning not to respond to some unimportant stimuli. It is an extremely simple form of learning, in which an animal, after a period of exposure to a stimulus, stops responding. Hence, D is the correct option.

Q6. Sorry she can't come to the phone. She _ a bath!

• A. Is having✓
• B. Having
• C. Have
• D. Has

Explanation: This option is correct. "Is having" is the present progressive (continuous) tense form of the verb "have," used to indicate an action that is currently happening. In this context, it means that the person is currently in the process of taking a bath, which prevents them from coming to the phone.

Q7. Choose the word nearest in meaning to ENIGMA"

• A. Evaluation
• B. Puzzle✓
• C. Answer
• D. Account

Explanation: Enigma means a person or thing that is mysterious or difficult to understand So the answer should be PUZZLE as it is difficult to understand.

Q8. When I went back to my hometown three years ago, I found that a lot of changes _

• A. Had taken place✓
• B. Have taken place
• C. Are taken place
• D. Were taken place

Explanation: This option is incorrect because "have taken place" is the present perfect tense, which indicates actions that started in the past and have relevance to the present. It does not fit the context of the sentence, which is discussing a specific past event (going back to the hometown three years ago).

Q9. Choose the correct sentence indicating facts and habits.

• A. He is clever and he lacks experience.✓
• B. He is clever but he is lacking experience
• C. He is clever but he is lacked experience
• D. He is clever but he is lack experience

Explanation: The simple present tense is used to describe facts and habits. Option A is correct as it is a fact that he is clever and he lacks experience.

Q10. Look! A hamster _ by cat

• A. Has been chased
• B. Was being chased
• C. Is being chased✓
• D. Is chased

Explanation: The word LOOK with an exclamation mark tells that the cat is chasing the hamster right now so the present continuous tense will be used.

Q11. Choose the word opposite in meaning to "VOCIFEROUS"

• A. Silent✓
• B. Boisterous
• C. Blatant
• D. Noisy

Explanation: Vociferous means expressing or characterized by vehement opinions; loud and forceful.So the correct option is A which is "silent", as it is the opposite of Vociferous.

Q12. I'm sorry the house is not available any longer, It _ to timber tycoon

• A. Was being sold
• B. Will be sold
• C. Is sold
• D. Has been sold✓

Explanation: Has been sold” will be used here as the event took place in the past.

Q13. Choose the correct option. I always liked to lean _ the side of mercy

• A. Over
• B. On✓
• C. Towards
• D. About

Explanation: "Lean on" implies relying or depending on something, which fits the context of favoring or supporting the side of mercy.

Q14. They already _ some of the old ones and _ them more comfortable

• A. Repair, make
• B. Repaired, made✓
• C. Repaired, make
• D. Repair, made

Explanation: The sentence involves two actions that have already been completed. Therefore, both verbs should be in the past tense."Repaired" is the past tense of "repair.""Made" is the past tense of "make."

Q15. I was born in Peshawar but I _ most of my childhood in the Mardan

• A. Spends
• B. Spent✓
• C. Was spending
• D. Is spending

Explanation: The sentence talks about a completed period in the past (childhood) with relevance to the present, which is best expressed using the present perfect tense "have spent."The present perfect is often used for an action that started at some time in the past and is still continuing now. So the correct option will be B.

Q16. _ you win first place, you will receive a prize.

• A. Whenever
• B. If✓
• C. Unless
• D. So forth

Explanation: You use if in conditional sentences to introduce the circumstances in which an event or situation might happen, might be happening, or might have happened.As this sentence is conditional IF is used.

Q17. The train was _

• A. Halt
• B. Halted✓
• C. Had halted
• D. Has halted

Explanation: The past is used to describe things that have already happened. So B is the correct answer

Q18. Choose the correct option. He confided _ me.

• A. About
• B. In✓
• C. On
• D. Of

Explanation: The phrasal verb "confide" usually goes with the preposition "in", and it is actually an idiom

Q19. He said. ''you need not wait". Choose the correct indirect speech:

• A. He said that I need not wait.✓
• B. He said you needed no wait.
• C. He said that wait was not need by you
• D. He said that you must not wait

Explanation: In the direct speech, "He said" is the reporting verb indicating that someone is conveying information or a statement. The subject "He" remains the same in indirect speech, but the verb "said" changes according to the context and tense. The direct statement "you need not wait" is directly quoted but needs to be transformed into reported speech. In indirect speech, the pronoun "you" changes according to who is reporting the statement. Here, it changes to "I" because it's now the narrator reporting what was said. The verb "need" remains the same.

Q20. Choose the correct option.It is natural for us to exert _ our own success

• A. In
• B. At✓
• C. Against
• D. Regarding

Explanation: The preposition “at” is used with “exert” to indicate the target or focus of effort. Saying “exert at our own success” means directing effort toward achieving success. Other prepositions like “on” or “for” do not correctly convey the sense of aiming or applying effort toward a goal in this context.

Q21. The vagus nerve is part of which system?

• A. Sympathetic nervous system
• B. Parasympathetic nervous system✓
• C. Somatic nervous system
• D. None of the above options are correct

Explanation: The vagus nerve is primarily involved in the parasympathetic nervous system, which is responsible for regulating various involuntary bodily functions, including digestion, heart rate, and certain aspects of the immune response. It helps to promote 'rest and digest' activities, counterbalancing the 'fight or flight' responses of the sympathetic nervous system. The sympathetic nervous system focuses on immediate responses to stressors, while the somatic nervous system handles voluntary movements. Therefore, the correct answer is the parasympathetic nervous system.

Q22. The genome of influenza virus is made up of _.

• A. Single strand RNA✓
• B. Double strand DNA
• C. Single strand DNA
• D. Double stranded RNA

Explanation: Influenza viruses, belonging to the Orthomyxoviridae family, have a segmented, negative-sense, single-stranded RNA genome. The genome consists of multiple RNA segments, typically eight in the case of influenza A and B viruses. The negative-sense RNA requires the virus to carry its own RNA-dependent RNA polymerase to transcribe it into positive-sense RNA, which can then be used for protein synthesis.

Q23. Galantamine hydrobromide is a compound derived from:

• A. Cannabis
• B. Coca
• C. English yew
• D. Daffodil✓

Explanation: Galantamine Hydrobromide is derived from plants like the daffodil, specifically from the bulbs and flowers of Galanthus nivalis and Narcissus. It is used in the treatment of mild to moderate Alzheimer’s disease. The other options, cannabis, coca, and English yew, are incorrect as they are not sources of Galantamine hydrobromide.

Q24. Mark the correct match:

• A. Haemophilia - Blood cancer
• B. SA node - Pacemaker✓
• C. ECG - Brain
• D. Alpha cell - Insulin

Explanation: The SA node is a crucial component of the cardiac conduction system, acting as the natural pacemaker of the heart by generating electrical pulses. This makes Option B the correct match. Haemophilia is a genetic disorder related to blood clotting, not blood cancer. An ECG is a test for heart activity, not brain function. Alpha cells produce glucagon, not insulin. Therefore, Option B is the only accurate match.

Q25. Cells that kill cells that display foreign motifs on their surface are:

• A. Platelets
• B. Cytotoxic T cells✓
• C. Antigens
• D. Red blood cell

Explanation: Cytotoxic T cells are a type of T lymphocyte that specifically targets and kills cells displaying foreign motifs on their surface. They play a crucial role in the immune response against infected or abnormal cells. Platelets, antigens, and red blood cells do not possess this specific function of directly killing cells with foreign motifs.

Q26. Chitin is a

• A. Lipoprotein
• B. Polysaccharides✓
• C. Glycoprotein
• D. Phospholipids

Explanation: Chitin is a modified polysaccharide that contains nitrogen. It is a structural carbohydrate found in the exoskeleton of arthropods and the cell walls of fungi.Option A, Lipoprotein, is incorrect as it is a combination of lipids and proteins, not a form of carbohydrate.Option C, Glycoprotein, is incorrect as it is a combination of carbohydrates and proteins but is not specifically related to chitin.Option D, Phospholipids, is incorrect as these are lipid molecules, not carbohydrates like chitin.

Q27. Organization of photosynthetic pigment into clusters is:

• A. Photosynthesis
• B. Photosystem✓
• C. Photosynthetic cluster arrangement
• D. Calvin system

Explanation: The organization of photosynthetic pigments into clusters is referred to as a 'photosystem.' A photosystem is a complex of pigments and proteins that work together to capture light energy and initiate the process of photosynthesis. There are two main types of photosystems in photosynthetic organisms: Photosystem I (PSI) and Photosystem II (PSII). Options A, C, and D are incorrect as they do not accurately describe the organization of photosynthetic pigments into clusters.

Q28. Amphibians are poikilotherm. Therefore, they use to hibernate in:

• A. Winter✓
• B. Summer
• C. Autumn
• D. Spring

Explanation: Option A is correct.Correct! Amphibians, as poikilothermic organisms, hibernate in winter when the environment gets too cold for them to regulate their body temperature and avoid the risk of death due to cold or freezing. The other options are incorrect because they do not align with the characteristic behavior of poikilothermic animals like amphibians.

Q29. All of the following are macronutrient except

• A. Cu ions✓
• B. Ca ions
• C. Mg ions
• D. K ions

Explanation: The correct answer is Cu ions. Macronutrients are elements required in large quantities for plant growth. Copper, calcium, magnesium, and potassium are all essential nutrients for plants, but copper is classified as a micronutrient, not a macronutrient. The other options, Ca ions, Mg ions, and K ions, are all macronutrients.

Q30. Purkinje fibres are connected with the impulse-conducting system of:

• A. Heart✓
• B. Brain
• C. Skin
• D. Nephron

Explanation: The correct answer is 'Heart.' Purkinje fibers are specialized conducting fibers located in the heart's ventricular walls and are part of the impulse conducting system of the heart. The other options (Brain, Skin, Nephron) are incorrect as they are not directly connected with Purkinje fibers and their function within the heart.

Q31. The alveoli represent total surface area of

• A. 10-30m
• B. 30-60m
• C. 70-90m✓
• D. 90-110m

Explanation: The alveoli form the gas exchange surface in the lungs, with about 700 million alveoli contributing to a total surface area of 70-90m2. Options A, B, and D provide ranges that are either too low or too high, making them incorrect.

Q32. Some marine fishes possess salt-excreting organs known as:

• A. Thyroid gland
• B. Pituitary gland
• C. Adrenal gland
• D. Rectal gland✓

Explanation: The correct answer is the rectal gland, which is a specialized salt-secreting organ found in cartilaginous fishes like sharks. This gland helps these marine fishes maintain salt balance in their bodies. The other options, including the thyroid, pituitary, and adrenal glands, are not involved in salt excretion in marine fishes.

Q33. Tetanus is an infection of

• A. Respiratory system
• B. Nervous system✓
• C. Circulatory system
• D. Bones and muscles

Explanation: Tetanus is an infection of the nervous system caused by Clostridium tetani bacteria. The correct answer is the nervous system because the bacteria affect nerve cells and can lead to muscle stiffness and spasms. The other options (respiratory system, circulatory system, bones, and muscles) are incorrect because they are not the primary systems affected by tetanus infection.

Q34. _ regulate the body temperature

• A. Hypothalamus✓
• B. Thalamus
• C. Hippocampus
• D. Amygdala

Explanation: The correct answer is the hypothalamus, as it is specifically designed to regulate body temperature, among other vital functions. The other options, while important for various brain functions, do not have a direct role in regulating body temperature.

Q35. A man has to face an interview, but during his first five minutes before the interview, he experiences sweating, increased heart rate, and respiration. Which hormone is responsible for his restlessness?

• A. Adrenocorticotropic hormone
• B. Insulin and glucagon
• C. Epinephrine and norepinephrine✓
• D. Aldosterone

Explanation: Epinephrine and norepinephrine are responsible for the physical symptoms described because they trigger the 'fight-or-flight' response in stressful situations. These hormones increase heart rate, respiration, and other responses to prepare the body for action. The other options, such as ACTH, insulin, glucagon, and aldosterone, do not directly cause the specific symptoms mentioned in the question.

Q36. Hypothalamus connected to pituitary gland via:

• A. Nerves
• B. Infundibulum✓
• C. Blood
• D. No connection

Explanation: The hypothalamus is connected to the pituitary gland via the infundibulum, which is a funnel-shaped structure that physically links the two. This connection allows the hypothalamus to regulate the pituitary gland's hormone secretion. The other options are incorrect as nerves, blood, or no connection do not directly link these two structures in the central nervous system.

Q37. Second meiotic division in oocyte is completed:

• A. When oocyte is fertilized by sperm✓
• B. When ovum is discharged from the ovary
• C. Just before fertilization
• D. Before the onset of menstruation

Explanation: The correct answer is when the oocyte is fertilized by a sperm. This marks the completion of the second meiotic division, leading to the formation of an ovum. The other options do not accurately depict the specific event that signifies the end of this division process, making them incorrect choices.

Q38. The enzyme which is found in saliva accelerates the conversion of starch into sugar is:

• A. Pepsin
• B. Thrombin
• C. Ptyalin✓
• D. Fumarase

Explanation: Ptyalin is a form of amylase found in the saliva of humans and some other animals. The ptyalin enzyme breaks down the carbohydrates into simple sugars, which the body can absorb.

Q39. A pure breeding tall plant was crossed to a dwarf plant. What would be the probability of "Tt" genotype n F2?

• A. 0
• B. 0.25
• C. 0.5✓
• D. 0.75

Explanation: In Mendelian inheritance, when a pure breeding tall plant (TT) is crossed with a dwarf plant (tt), the F1 generation will all be heterozygous (Tt). When these F1 plants are crossed to produce the F2 generation, the expected genotypic ratio is 1 TT : 2 Tt : 1 tt. Therefore, the probability of obtaining a 'Tt' genotype in the F2 generation is 2 out of 4, or 0.5. Options A and D are incorrect because they do not reflect Mendel's expected genotypic ratio. Option B is incorrect because it underestimates the probability of the 'Tt' genotype.

Q40. The number of human spinal nerves is

• A. 60
• B. 62✓
• C. 64
• D. 66

Explanation: The human spinal cord gives rise to 31 pairs of spinal nerves, each pair emerging from a specific segment of the spinal cord. Thus, the total number of individual spinal nerves is 62 (31 pairs of 2 nerves each). Option B is correct because it accurately reflects this count. Options A, C, and D are incorrect as they do not match the actual number of spinal nerves in the human body.

Q41. Diphtheria vaccines are an example of:

• A. Inactivated vaccine
• B. Toxoid vaccine✓
• C. Subunit vaccine
• D. Live, attenuated vaccine

Explanation: Toxoid vaccines, such as those for diphtheria and tetanus, contain inactivated toxins produced by bacteria. These vaccines do not target the bacteria but rather the harmful effects of the toxins they produce. Inactivated vaccines, however, use killed pathogens, primarily viruses, to elicit an immune response. Subunit vaccines use fragments of the pathogen, while live, attenuated vaccines use weakened forms of the whole virus or bacteria. The diphtheria vaccine specifically uses a toxoid to neutralize the effects of the toxin produced by the bacteria.

Q42. Which one of the following items gives its correct total number?

• A. Cervical vertebrae -7✓
• B. Floating ribs in human - 3
• C. Auditory ossicles - 8
• D. Cranium bones - 4

Explanation: There are two pairs of floating ribs. The ossicles are three bones. There are eight cranial bones. The cervical spine is comprised of seven cervical vertebrae. Thus, A is the correct option.

Q43. Find mismatch

• A. Thyroid gland-T3 and T4
• B. Parathyroid gland-Calcitonin✓
• C. Pancreas-Insulin
• D. Gonads-Testes and ovaries

Explanation: The correct answer is Option B: 'Parathyroid gland-Calcitonin'. This is a mismatch because the parathyroid glands produce parathyroid hormone (PTH), not calcitonin. Calcitonin is produced by the thyroid gland and works to reduce blood calcium levels, counteracting PTH. The other options correctly match the glands with their primary hormones or functions: the thyroid produces T3 and T4, the pancreas produces insulin, and the gonads refer to testes and ovaries in the reproductive system.

Q44. To the end of the first trimesters, the embryo can now technically describe as a:

• A. Zygote
• B. Infant
• C. Toddler
• D. Fetus✓

Explanation: At the conclusion of the first trimester, the embryo undergoes significant development and is officially termed a fetus. This transition marks the end of the embryonic stage, as critical structures have begun forming, preparing for further growth and development. Other options such as zygote, infant, and toddler refer to different stages of development, either too early in pregnancy or post-birth, making them incorrect in this context.

Q45. How many pairs of homologous chromosomes are present in Pisum sativum:

• A. Seven pairs✓
• B. Eight pairs
• C. Nine pairs
• D. Ten pairs

Explanation: Pisum sativum, commonly referred to as the pea plant, contains a total of 14 chromosomes, organized into 7 pairs of homologous chromosomes. This diploid configuration is fundamental to understanding Mendelian inheritance, as each pair consists of chromosomes that carry the same genes in identical order but may have different alleles. The options suggesting more than 7 pairs imply a higher total number of chromosomes, which does not pertain to this species and would suggest a different organism altogether.

Q46. Ozone layer in upper atmosphere is being destroyed by:

• A. Chlorofluorocarbon
• B. Freon
• C. Smog
• D. Both A and B✓

Explanation: The correct answer is Both A and B. Chlorofluorocarbons (CFCs) and Freon are both recognized as significant ozone-depleting substances. When these compounds are released into the atmosphere, they slowly rise to the stratosphere. There, they are broken down by ultraviolet (UV) radiation, releasing chlorine atoms. These chlorine atoms act as catalysts in the destruction of ozone molecules, which leads to the thinning of the ozone layer.Option C, Smog, is not responsible for the destruction of the ozone layer. Smog is a type of air pollution that occurs at the Earth's surface, primarily involving ground-level ozone and particulate matter, rather than affecting the stratospheric ozone layer.

Q47. The percentage of fresh water on earth is

• A. 1%✓
• B. 3%
• C. 5%
• D. 7%

Explanation: Water is crucial for life and covers about 70% of Earth's surface. However, 97% of this is saltwater in oceans and seas, which is not directly usable for most human needs without desalination. Of the remaining 3% that is freshwater, approximately 2% is locked in glaciers and ice caps, leaving only around 1% available in liquid form in rivers, lakes, and underground aquifers. This small percentage is what supports most terrestrial life and human activities. Therefore, the correct answer is 1%. The other options overestimate the amount of accessible freshwater, ignoring the significant portion that is trapped in ice.

Q48. Recombinants contains DNA from

• A. Two different sources✓
• B. A single source
• C. Two identical sources
• D. Three identical sources

Explanation: Recombinant DNA technology is the process of joining together DNA from two different species, resulting in a new genetic combination that can be used in various fields such as science, medicine, agriculture, and industry. The key aspect is that the DNA must come from two distinct sources to create the recombinant DNA molecule. Options B, C, and D are incorrect because they involve DNA from the same source or multiple identical sources, which does not meet the criteria for recombination.

Q49. The inner surface of kidney has a deep notch called

• A. Renal pelvis
• B. Hilum✓
• C. Medulla
• D. Pyramid

Explanation: Option B is correct.The hilum is the indentation on the medial side of the kidney through which the renal artery enters and the renal vein and ureter exit. It serves as the entry and exit point for vessels and nerves. The other options do not describe a notch: the Renal Pelvis is a cavity, the Medulla is a tissue layer, and the Pyramid is a structure within the medulla.

Q50. _ is considered as chief structural and functional unit of nervous system.

• A. Cell
• B. Neuron✓
• C. Nephron
• D. Brain

Explanation: The neuron is the chief structural and functional unit of the nervous system. It transmits information through electrical and chemical signals and is responsible for sensations, thoughts, and movements.

Q51. The bacteriophage replicates only inside a _.

• A. Animal cell
• B. Bacterial cell✓
• C. Fungal cell
• D. Animal and bacterial cell

Explanation: Bacteriophages are viruses that specifically infect and replicate within bacterial cells. They are composed of genetic material, either DNA or RNA, encased in a protein coat known as a capsid. Some bacteriophages may also have a lipid envelope. They utilize the host bacterium's cellular machinery for replication. While they are highly effective against bacteria, they do not infect or replicate inside animal or fungal cells. Therefore, the correct answer is a bacterial cell. The other options are incorrect because they do not represent the specific host cell type that bacteriophages target.

Q52. _ is stored in animal cells

• A. Starch
• B. Cellulose
• C. Sucrose
• D. Glycogen✓

Explanation: Animal cells store food in the form of glycogen. The excess of glucose in the blood is converted to glycogen in the muscle cells. The process of formation of glycogen is known as glycogenesis. During the muscle exercise when extra energy is needed the glycogen is broken down to glucose and channelized into the bloodstream by glycogenolysis.

Q53. A bacterium which has a group of two or more flagella inserted at one pole of the cell:

• A. Monotrichous
• B. Peritrichous
• C. Lophotrichous✓
• D. Amphitrichous

Explanation: Bacteria exhibit different flagellar arrangements based on the number and placement of flagella. A lophotrichous bacterium has a cluster of flagella at one end, which aids in directional movement. This distinguishes it from monotrichous (single flagellum), peritrichous (flagella around the entire cell), and amphitrichous (flagella at both ends) arrangements. Therefore, the correct answer is 'Lophotrichous' because it matches the description of a group of flagella at one pole.

Q54. The gametophyte of lycopsida is mainly:

• A. Aerial
• B. Partly aerial and partly underground
• C. Underground✓
• D. Photosynthetic

Explanation: In the plant group Lycopsida, which includes clubmosses, spike mosses, and quillworts, the gametophyte generation is predominantly underground. This generation is typically small and non-photosynthetic, obtaining nutrients through symbiosis with fungi. This contrasts with the sporophyte generation, which is the dominant, visible phase that carries out photosynthesis and produces spores. Thus, the correct answer is that the gametophyte of Lycopsida is 'Underground'. The other options are incorrect because they do not accurately describe the typical habitat or characteristics of the Lycopsida gametophyte.

Q55. Opossum and koala bear belongs to the sub-class:

• A. Prototheria
• B. Eutheria
• C. Metatheria✓
• D. Monotremata

Explanation: Metatheria includes marsupials that possess a pouch and give birth to partially-developed young ones. Well-known marsupials include kangaroos, wallabies, koalas, opossums, woinbats, Tasmanian devils, and the extinct Thylacme.

Q56. The form of immunity which inherit from mother is:

• A. Active immunity
• B. Passive immunity✓
• C. Acquired immunity
• D. Innate immunity

Explanation: Passive immunity is the correct answer because it involves the transfer of antibodies from the mother to the child, offering immediate protection. This occurs naturally during the last months of pregnancy when antibodies pass through the placenta, and also through breast milk after birth. Innate immunity is incorrect as it refers to the natural defense mechanisms present from birth and not specifically derived from maternal antibodies. Active immunity is incorrect because it involves the body generating its own immune response. Acquired immunity is incorrect as it is developed through exposure to pathogens over time.

Q57. The least toxic excretory product is:

• A. Ammonia
• B. Urea
• C. Uric acid✓
• D. Fatty acid

Explanation: Option C is correct.The correct answer is uric acid, which is the least toxic excretory product. It is excreted in a solid or semi-solid form and conserves water, making it ideal for animals like birds and reptiles living in arid environments. Ammonia, on the other hand, is highly toxic and requires a significant amount of water to be safely excreted, which is why it is typically excreted by aquatic animals. Urea is less toxic than ammonia and is excreted by mammals and amphibians, requiring moderate amounts of water. Fatty acids are not directly involved in nitrogenous waste excretion.

Q58. Chemically hormones are:

• A. Carbohydrates
• B. Proteins
• C. Steroids
• D. Both B and C✓

Explanation: Option D is correct. Chemically, hormones are classified into two main types: Protein hormones. Protein hormones are made up of amino acids and are the most common type of hormone. These hormones are produced by the endocrine glands, such as the thyroid gland, the pituitary gland, and the adrenal glands. Steroid hormones: Steroid hormones are made up of cholesterol and are produced by the adrenal glands and the gonads (ovaries and testes).

Q59. DNA polymerase III works always in:

• A. 5 - 2' direction
• B. 5'- 3' direction✓
• C. 3'- 5 direction
• D. 2 - 5 direction

Explanation: DNA Polymerase III always works in the 5' to 3' direction.Explanation:DNA Polymerase III is the primary enzyme responsible for DNA replication in prokaryotes (such as E. coli).It adds nucleotides to the growing DNA strand only in the 5' to 3' direction because it can only add new nucleotides to the 3' end of the growing strand.This means:The leading strand is synthesized continuously in the 5' to 3' direction.The lagging strand is synthesized discontinuously in short fragments (Okazaki fragments) because replication must still proceed in the 5' to 3' direction.

Q60. The biogas plant is tank which is

• A. 5-10 ft deep
• B. 10-15 ft deep✓
• C. 15-20 ft deep
• D. 20-25 ft deep

Explanation: The correct answer is option B: 10-15 ft deep. This depth is generally sufficient for the anaerobic digestion process to occur efficiently, allowing organic waste to be converted into biogas. Option A, 5-10 ft deep, is often too shallow for effective biogas production on a commercial scale. Options C and D, with depths of 15-20 ft and 20-25 ft respectively, suggest depths that exceed typical requirements, making them less efficient and unnecessarily costly for standard applications.

Q61. Which wave lengths are mainly absorbed by chlorophyll?

• A. Violet blue and red✓
• B. Green and blue
• C. Violet and orange
• D. Red and indigo

Explanation: Chlorophyll primarily absorbs light in the blue and red regions of the electromagnetic spectrum. The violet-blue and red wavelengths are most efficiently absorbed by chlorophyll, while green wavelengths are less effectively absorbed and are, instead, reflected, which is why chlorophyll-containing plants appear green to our eyes.

Q62. For hepatitis B the incubation period is between _.

• A. 4 and 20 weeks✓
• B. 6 and 20 weeks
• C. 2-26 weeks
• D. 2-6 weeks

Explanation: The incubation period of the hepatitis B virus ranges from 30 to 180 days, which translates to approximately 4 to 20 weeks. This is why option A is correct. Option B incorrectly suggests a later start to the incubation period, option C offers a range that is too broad, and option D underestimates the period significantly.

Q63. Sulphur bacteria belongs to which group of bacteria called

• A. Beta-proteobacteria
• B. Alpha-proteobacteria
• C. Gamma-proteobacteria✓
• D. Delta-proteobacteria

Explanation: The correct answer is Gamma-proteobacteria. This group includes sulfur bacteria that oxidize hydrogen sulfide (H2S) to obtain energy, engaging in sulfur cycling. Beta-proteobacteria are involved in nitrogen recycling, Alpha-proteobacteria are known for their symbiotic relationships with eukaryotic hosts, and Delta-proteobacteria form complex structures and are not primarily involved in sulfur oxidation.

Q64. Nuclear mitosis occurs in

• A. Plants
• B. Animals
• C. Fungi✓
• D. Monera

Explanation: Fungi are unique in that their mitosis occurs within an intact nuclear envelope, a process known as nuclear mitosis. This distinguishes them from plants and animals, where the nuclear envelope disintegrates during mitosis. Monera, on the other hand, reproduce asexually through binary fission, which is distinct from mitosis entirely.

Q65. Excess glucose is converted in the liver to glycogen in response to the hormone:

• A. Glucagon
• B. Insulin✓
• C. Bile
• D. Both A and B

Explanation: Insulin promotes glycogenesis in the liver and muscles, storing excess glucose as glycogen when blood sugar levels are high.

Q66. During muscle relaxation, the calcium ions are:

• A. Released from the sarcoplasmic reticulum into the sarcoplasm
• B. Transported back from the sarcoplasm into the sarcoplasmic reticulum✓
• C. Further released from the sarcoplasmic reticulum into the sarcoplasm
• D. Remain constant without any movement

Explanation: During muscle relaxation, calcium ions are actively transported back from the sarcoplasm into the sarcoplasmic reticulum, which decreases their concentration in the sarcoplasm. This process allows the muscle fibers to relax as the removal of calcium ions leads to the detachment of the myosin heads from the actin filaments. Option B is correct because it accurately describes this process. The other options incorrectly describe the movement or state of calcium ions during muscle relaxation.

Q67. In male luteinizing hormone also known as

• A. ACTH
• B. ICSH✓
• C. TRF
• D. MSH

Explanation: In males, luteinizing hormone (LH) is also known as interstitial cell stimulating hormone (ICSH). It plays a crucial role in stimulating the testes to produce testosterone, which is essential for male reproductive function. The other options—ACTH, TRF, and MSH—are hormones with entirely different roles in the body and do not influence testosterone production.

Q68. Particular amino acid and tRNA molecule binds together by the action of an enzyme named;

• A. RNA Polymerase
• B. tRNA synthetase
• C. Aminocyle-tRNA synthetase✓
• D. tRNA ligase

Explanation: So, when it comes to linking a specific amino acid with its corresponding tRNA molecule during protein synthesis, there's a special enzyme called aminoacyl-tRNA synthetase that does the job. It acts like a matchmaker, making sure that the right amino acid is paired with the right tRNA molecule. This process is crucial for building proteins correctly.

Q69. Lipid bilayer makes the membrane differently permeable barrier that allows the transport of:

• A. Ionic materials
• B. Polar materials
• C. Non-polar materials✓
• D. Glycoproteins

Explanation: The lipid bilayer of the plasma membrane is composed mainly of phospholipids, which have hydrophobic tails and hydrophilic heads. This structure creates a barrier that is selectively permeable. Non-polar materials can dissolve in the hydrophobic core of the bilayer, allowing them to pass through easily. In contrast, ionic and polar materials, such as ions and polar molecules, are repelled by the hydrophobic core and require specific transport proteins to facilitate their movement across the membrane. Glycoproteins, being large and complex, also require specialized transport mechanisms.

Q70. The following are sexual reproduction methods in bateria except

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

Explanation: Binary fission is the correct answer because it is an asexual reproduction method. In binary fission, a single bacterial cell divides into two identical daughter cells without the exchange of genetic material.Transformation, transduction, and conjugation are methods of genetic recombination in bacteria, which are considered primitive forms of sexual reproduction. These processes involve the transfer or exchange of genetic material between bacteria, leading to genetic variation.

Q71. Lichen is the symbiotic association of fungus with

• A. Bacteria
• B. Algae✓
• C. Other fungus
• D. Animals

Explanation: Symbiosis: Fungi develop symbiotic associations with other organisms e.g. Lichens and Mycorrhizae Lichen: It is a symbiotic association of a fungus with algae.

Q72. The possible reason(s) for cyanosis, one of congenital heart disease is:

• A. Formation of carboxy-hemoglobin✓
• B. The high concentration of oxyhemoglobin
• C. Low level of carbon monoxide (CO)
• D. Low level of hemoglobin

Explanation: The correct answer is the formation of carboxy-hemoglobin. This occurs when carbon monoxide binds to hemoglobin with a much higher affinity than oxygen, preventing adequate oxygen transport in the blood, leading to cyanosis. Option B is incorrect because a high concentration of oxyhemoglobin suggests well-oxygenated blood. Option C is incorrect because low levels of CO do not contribute to cyanosis. Option D is incorrect as low hemoglobin levels lead to anemia, not cyanosis.

Q73. The deficiency of which micronutrient causes goiter formation?

• A. Iron
• B. Zinc
• C. lodine✓
• D. Sodium

Explanation: Goiter is a condition in which the thyroid in neck enlarges, this happens mainly due to the lack of iodine in diet.

Q74. Phosphatases belong to which group of the following:

• A. Lyases
• B. Ligases
• C. Hydrolases✓
• D. None

Explanation: Phosphatases belong to the group of enzymes known as hydrolases. Specifically, they are a type of hydrolase called phosphohydrolases. Phosphatases catalyze the removal of phosphate groups from molecules through hydrolysis reactions. The removal of phosphate groups is a crucial regulatory step in various cellular processes, as it can activate or deactivate proteins and other molecules.

Q75. The ribosomes responsible for protein synthesis are present in the cell _.

• A. Floating in the cytosol
• B. Localized in the nucleus
• C. Bound to rough endoplasmic reticulum
• D. Both floating in the cytosol and bound to rough endoplasmic reticulum✓

Explanation: Ribosomes are essential organelles for protein synthesis and exist in two primary forms within the cell. Free ribosomes float in the cytosol and are responsible for synthesizing proteins that will function within the cytosol. In contrast, ribosomes bound to the rough endoplasmic reticulum (RER) typically synthesize proteins destined for secretion or incorporation into cellular membranes. Therefore, the correct answer is that ribosomes are present both floating in the cytosol and bound to the rough endoplasmic reticulum.Option A is incorrect because while ribosomes do float in the cytosol, this answer does not account for the ribosomes on the RER. Option B is incorrect because ribosomes are not active in protein synthesis within the nucleus. Option C is incorrect because while ribosomes are bound to the RER, this answer does not account for those that float freely in the cytosol.

Q76. _ enzyme needs a primer for the initiation of its function.

• A. RNA polymerase
• B. DNA polymerase✓
• C. Primase
• D. Ligase

Explanation: The synthesis of a primer is necessary because DNA polymerases, can only attach new DNA nucleotides to an existing strand of nucleotides. The primer therefore serves to prime and lay a foundation for DNA synthesis.

Q77. The following histone proteins form a nucleosome complex except:

• A. H1✓
• B. H2A
• C. H2B
• D. H3

Explanation: The nucleosome is the fundamental structural unit of chromatin in eukaryotic cells. It comprises a histone octamer around which DNA is wrapped. This octamer consists of two copies each of four core histone proteins: H2A, H2B, H3, and H4. Histone H1, on the other hand, is not part of the core nucleosome particle. Instead, H1 is associated with the linker DNA between nucleosomes, playing a role in compacting the chromatin structure into higher-order formations. Therefore, the correct answer is H1, as it does not form part of the nucleosome core.

Q78. The bond that is formed b/w two monosaccharide units is called

• A. lonic bond
• B. Hydrogen bond
• C. Peptide bond
• D. Glycosidic bond✓

Explanation: A glycosidic bond is a type of covalent bond which is formed between two units of monosaccharides

Q79. Which statement about chlorophyll is not true?

• A. It contains terminal carbonyl group✓
• B. It contains phyto tail
• C. It contains porphyrin ring
• D. It contains magnesium

Explanation: Chlorophyll is a complex molecule involved in photosynthesis, and it does contain a terminal carbonyl group. The carbonyl group is a functional group consisting of a carbon atom double-bonded to an oxygen atom. In chlorophyll, the carbonyl group is typically present in a phytol side chain. Chlorophyll molecules consist of a porphyrin ring system with a magnesium ion at the center. The phytol side chain, which contains the terminal carbonyl group, is hydrophobic and anchors the chlorophyll molecule in the thylakoid membrane of chloroplasts. The terminal carbonyl group is involved in the overall structure and function of chlorophyll, contributing to its interaction with light and its role in capturing energy during the light-dependent reactions of photosynthesis. Chlorophyll absorbs light energy, and the carbonyl group is part of the chromophore responsible for the absorption of specific wavelengths of light.

Q80. In humans the disease symptoms develop during the:

• A. Log phase✓
• B. Lag phase
• C. Growth phase
• D. Decline phase

Explanation: Bacterial cells divide at exponential rate during log phase, hence disease symptoms become prominent during this phase.

Q81. Independent gametophyte and sporophyte are found in:

• A. Selaginella
• B. Polytrichum
• C. Ectocarpus✓
• D. Liverworts

Explanation: The correct answer is Ectocarpus. In Ectocarpus, both the gametophyte and sporophyte stages are independent of each other, which is characteristic of its alternation of generations. This is unlike most non-vascular plants like Polytrichum (mosses) and Liverworts, where the sporophyte is dependent on the gametophyte. Selaginella, although having a complex life cycle with heterospory, does not have fully independent gametophytes.

Q82. Tmesipteris is an example of:

• A. Horsetail
• B. Club mosses
• C. Psilopsida✓
• D. Pteropsida

Explanation: Tmesipteris is a genus of plants belonging to the class Psilopsida, which is commonly referred to as whisk ferns. Psilopsida is a class of primitive vascular plants that are characterized by their lack of true leaves and roots. They have simple, dichotomously branching stems and reproduce by spores.

Q83. The larva formed during the life cycle of annelida is

• A. Glochidium larva
• B. Bipinnaria larva
• C. Trochophore larva✓
• D. Tornaria larva

Explanation: The trochophore larva is a distinctive stage in the life cycle of marine annelids and is also seen in some mollusks. This larva is free-swimming and characterized by its translucent body and ciliary bands that aid in locomotion. The other options, such as glochidium, bipinnaria, and tornaria, belong to other phyla—Mollusca, Echinodermata, and Hemichordata, respectively—and are not associated with Annelida.

Q84. Ebner's gland on the dorsal surface of the tongue secrete an enzyme:

• A. Amylase
• B. Ptyalin
• C. Lingual lipase✓
• D. Both Amylase and Ptyalin

Explanation: The von Ebner's glands are minor salivary glands located on the dorsal surface of the tongue, specifically near the circumvallate papillae. They secrete lingual lipase, an enzyme that begins the digestion of lipids in the mouth. This enzyme is particularly important for infants, as it helps in the digestion of milk fats. Amylase and ptyalin, on the other hand, are involved in carbohydrate digestion and are not secreted by the von Ebner's glands. Therefore, lingual lipase is the correct answer.

Q85. Antibodies consists of _ polypeptide chains:

• A. 2
• B. 4✓
• C. 6
• D. 8

Explanation: Antibodies, also known as immunoglobulins, are Y-shaped proteins produced by the immune system to identify and neutralize foreign objects like bacteria and viruses. Each antibody molecule consists of four polypeptide chains: two heavy chains and two light chains. These chains are linked by disulfide bonds. The heavy chains are longer and form the backbone of the Y shape, while the light chains are shorter and attached to the outer sides of the Y. Options A, C, and D are incorrect because they suggest a different number of chains than the actual four: two heavy and two light chains.

Q86. Platyhelminthes are

• A. Bilaterally symmetrical and diploblastic
• B. Bilaterally symmetrical and triploblastic✓
• C. Radially symmetrical and triploblastic
• D. Radially symmetrical and diploblastic

Explanation: Platyhelminthes, commonly known as flatworms, are a phylum within the kingdom Animalia. They are characterized by their bilateral symmetry and triploblastic body structure, meaning they are composed of three primary germ layers: ectoderm, mesoderm, and endoderm. This triploblastic feature differentiates them from diploblastic organisms, which only have two germ layers. Additionally, Platyhelminthes are distinct from radially symmetrical organisms like cnidarians; they have a head and tail, leading to a definite direction of movement. The absence of an anus and circulatory system, along with respiration via diffusion, further distinguishes them within the animal kingdom.

Q87. The scientific name of freshwater mussel is

• A. Mytilus edulis
• B. Loligo peali
• C. Anodonta grandis✓
• D. Anodonta bairdi

Explanation: Freshwater mussels are an essential component of our rivers and streams. By their siphoning actions, mussels filter bacteria, algae, and other small particles, which make them one of the few animals that improve water quality. Mussels also serve as a food source to many species of fish, reptiles, birds, and mammals. Freshwater mussel is a common for ‘Anodonta grandis’.

Q88. Potamogeton is an example of _

• A. Xerophytes
• B. Mesophytes
• C. Hydrophytes✓
• D. Halophytes

Explanation: Option C is correct.Potamogeton is a genus of aquatic plants commonly known as pondweed. These plants are classified as hydrophytes because they are adapted to grow in water or very moist environments. This distinguishes them from xerophytes, which thrive in dry conditions; mesophytes, which prefer moderate water availability; and halophytes, which are adapted to saline environments.

Q89. _ and cytokinins stimulates fruit ripening.

• A. Option A: Cytokinins
• B. Option B: Abscisic acid
• C. Option C: Ethylene✓
• D. Option D: Auxin

Explanation: Ethylene is the correct answer because it is the primary plant hormone responsible for promoting fruit ripening. It accelerates the changes in fruit texture, color, and flavor, making it edible. While cytokinins, abscisic acid, and auxin have important roles in plant growth and development, they do not specifically trigger the ripening process like ethylene does.

Q90. The bulbourethral glands produce:

• A. Acidic fluid
• B. Alkaline fluid✓
• C. Semen
• D. Mucus

Explanation: The bulbourethral glands produce an alkaline fluid that helps neutralize any remaining acidity in the male urethra, which could otherwise harm sperm. This fluid is released in response to sexual stimulation and is part of the pre-ejaculatory fluid. The other options are incorrect: Acidic fluid is not produced by these glands; semen is a combination of various fluids from different glands, not solely from the bulbourethral glands; and while the fluid has mucus-like properties, its key characteristic is being alkaline.

Q91. A condition in which an abnormally large volume of urine is produced is

• A. Polydipsia
• B. Polyuria✓
• C. Polyphagia
• D. Polyanypsida
• E. Both A and B

Explanation: Option B is correct.Polyuria is defined as the production of an abnormally large volume of urine. It is a key symptom in conditions such as diabetes insipidus and diabetes mellitus.Option A, Polydipsia, pertains to excessive thirst, which often accompanies polyuria but is not the same thing.Option C, Polyphagia, relates to excessive eating.Option D, Polyanypsida, is not a recognised medical term.Option E suggests both A and B, but the condition described in the question specifically relates to urine production, not thirst.

Q92. HIV destroys a type of defense cell in the body called _ helper lymphocyte.

• A. TD4
• B. T4
• C. C4
• D. CD4✓

Explanation: HIV (Human Immunodeficiency Virus) primarily targets and infects CD4-positive T cells, which are crucial for coordinating the immune response. The virus binds to the CD4 receptor, enters the cell, and integrates its genetic material into the host's DNA. This leads to the progressive loss of these cells, weakening the immune system. The incorrect options refer to terms or components not directly related to the action of HIV on immune cells: 'TD4' is not a recognized term, 'T4' is incorrect as it does not specifically denote CD4 cells, and 'C4' is part of the complement system and not relevant to lymphocyte targeting by HIV.

Q93. Acetabularia crenulata has a _ shaped cap:

• A. Irregular✓
• B. Umbrella
• C. Regular
• D. Disc

Explanation: Acetabularia is a genus of unicellular green algae, notable for its large size and distinctive cap shapes. Specifically, Acetabularia crenulata features an irregular-shaped cap, which is a key characteristic for identifying this species. In contrast, Acetabularia mediterranea has an umbrella-shaped cap. Understanding these differences is crucial for distinguishing between species in the Acetabularia genus. The incorrect options represent shapes that do not correspond with the known morphology of Acetabularia crenulata.

Q94. The safranin stain is usable for:

• A. Fungal hyphae
• B. Cytoplasm/Cellulose
• C. Blood cells
• D. Lignin✓

Explanation: Safranin is used extensively in plant histology to stain lignin in xylem tissues, where it imparts a red color. This is because safranin has a high affinity for lignified cell walls. Other stains like aniline blue, eosin, and Leishman's stain are more suited to their specific applications: aniline blue for fungal hyphae, eosin for cytoplasm, and Leishman's stain for blood cells, respectively, due to their unique binding properties with different cellular components.

Q95. In the human skull the unpaired bones are:

• A. Frontal, occipital, ethmoid and sphenoid✓
• B. Frontal, ethmoid, sphenoid and zygomatic
• C. Ethmoid, sphenoid, zygomatic and frontal
• D. Temporal, sphenoid, frontal and ethmoid

Explanation: The skull consists of 8 bones. Unpaired Bones Paired Bones. Frontal, Parietal, Occipital, Temporal, Sphenoid, Ethmoid.

Q96. Functionally _ pairs of cranial nerves are sensory in nature and _ pairs are mixed in nature and _ are motor in nature

• A. 3,4 and 5✓
• B. 4,5 and 3
• C. 3,5 and 4
• D. 4,3 and 5

Explanation: The 12 pairs of cranial nerves are divided based on their functionality: 3 pairs are sensory in nature (olfactory, optic, and vestibulocochlear), 4 pairs are mixed (trigeminal, facial, glossopharyngeal, and vagus), and 5 pairs are motor (oculomotor, trochlear, abducens, accessory, and hypoglossal). Understanding these classifications can help in correctly identifying their functions and roles in the nervous system. The correct option, 3, 4, and 5, reflects this accurate distribution.

Q97. DNA finger printing refer to:

• A. Techniques used to identify physical fingerprints of individuals
• B. Molecular analysis of DNA profiles to identify individuals✓
• C. Use of imprinting devices to analyze DNA samples
• D. Combination of physical fingerprint identification and DNA analysis using imprinting devices

Explanation: DNA fingerprinting, also known as DNA profiling, is a molecular technique used to identify individuals based on unique patterns within their DNA. This method analyzes specific regions of the DNA, such as short tandem repeats (STRs), which vary greatly among individuals but are inherited. This variation allows for the creation of a unique genetic profile for each person. It is widely used in forensic science for solving crimes, as well as in paternity testing and genetic research.Option A is incorrect because it relates to physical fingerprint analysis, not DNA. Option C is incorrect because imprinting devices are not used in DNA fingerprinting. Option D is incorrect because it inaccurately combines unrelated processes, whereas DNA fingerprinting focuses solely on molecular DNA analysis.

Q98. Oleic acid is a fatty acid with 18 carbon atoms. It breaks down into 9 acetyl groups. It is estimated that these nine acetyl groups would generate _ ATP molecules.

• A. 81
• B. 98
• C. 101
• D. 108✓

Explanation: The breakdown of oleic acid into acetyl-CoA units involves beta-oxidation, where each acetyl-CoA enters the citric acid cycle. Each acetyl-CoA contributes approximately 10 ATP through both the citric acid cycle and oxidative phosphorylation. Therefore, 9 acetyl-CoA molecules result in 9 x 12 = 108 ATP. The other options are incorrect due to miscalculations or incorrect assumptions about the ATP yield per acetyl-CoA.

Q99. Horsetails are included in class:

• A. Pteropsida
• B. Lycopsida
• C. Psilopsida
• D. Sphenopsida✓

Explanation: Horsetails are classified under the class Sphenopsida, also known as Equisetopsida. This class is characterized by jointed stems, scale-like leaves arranged in whorls, and the presence of silica in their tissues, giving them a rough texture. This group comprises a greater number of fossilized plants compared to living species. Currently, only a single surviving genus, Equisetum, is found within this sub-division. These plants develop from rhizomes, which give rise to two types of aerial stems i.e., fertile and sterile.

Q100. Which one of the following bones is the only movable portion of the skull?

• A. Maxilla
• B. Frontal bone
• C. Mandible✓
• D. Zygomatic

Explanation: The correct answer is the mandible. It is the only bone in the skull that moves, allowing for the opening and closing of the mouth, which is essential for chewing and speaking. The other options, such as the maxilla, frontal bone, and zygomatic bone, are part of the fixed structure of the skull and do not have independent movement.

Q101. Which one of the following is the correct oxidation state of Mn in KMnO4?

• A. -6
• B. 7✓
• C. 9
• D. 10

Explanation: To determine the oxidation state of Mn in KMnO4, we use the known oxidation states: potassium (K) is +1 and oxygen (O) is -2. The compound is neutral, so the sum of the oxidation states must be zero:+1 (K) + Mn + 4(-2) (O) = 01 + Mn - 8 = 0Mn - 7 = 0Mn = +7Thus, the oxidation state of Mn in KMnO4 is +7. The other options are incorrect because they do not satisfy the charge balance required for the compound to be neutral.

Q102. What will be the geometrical shape of a molecule that contains two lone pairs and two bond pairs of electron in valence shell of central atom?

• A. Tetrahedral
• B. Trigonal pyramidal
• C. Angular✓
• D. Linear

Explanation: If there are two bond pairs and two lone pairs of electrons the molecular geometry is angular or bent (e.g. H2O).

Q103. Quantum number which describes the orientation of orbitals in three dimensional space is:

• A. Spin quantum number
• B. Azimuthal quantum number
• C. Magnetic quantum number✓
• D. Principal quantum number

Explanation: Magnetic Quantum Number (m) : Gives the orientation of the orbital in space; in other words, the value of m describes whether an orbital lies along the x-, y-, or z-axis on a three-dimensional graph, with the nucleus of the atom at the origin. m can take on any value from −ltol. Hence option C is correct.

Q104. Which one of the following gas has the highest rate of diffusion at the same temperature and pressure?

• A. HCL
• B. CO2
• C. C2H2✓
• D. C2H6

Explanation: Graham Law states Rate of Diffusion ∝ 1/√(MolecularMass) Option A: HCl has molecular mass 36g Option B: CO2 has molecular mass 44g Option C: C2H2 has molecular mass 26g Option D: C2H6 has molecular mass 30g As C2H2 has lowest molecular mass, it has highest rate of diffusion.

Q105. At higher altitude, the boiling point of water is less than 100°C. This is because of:

• A. Higher atmospheric pressure
• B. Weak hydrogen bonding
• C. No change in atmospheric pressure
• D. Lower atmospheric pressure✓

Explanation: Boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure

Q106. Substance that has sharp melting point in the following is:

• A. Gemstone
• B. Coal tar
• C. Glass
• D. Diamond✓

Explanation: Each carbon atom is covalently bonded to four other carbon atoms in diamond. A lot of energy is needed to separate the atoms. This is because covalent bonds are strong. This is the reason why diamond has a high melting point.

Q107. Which of the following pairs is an example of completely immiscible liquids?

• A. Alcohol and water
• B. Alcohol and ether
• C. Water and ether
• D. Carbon disulphide and water✓

Explanation: CS2 is insoluble in water and more dense (10.5 lb/gal) than water. Hence sinks in water.

Q108. How many elements are there in the third period of the periodic table?

• A. 18
• B. 8✓
• C. 32
• D. 10

Explanation: Second and third period are called short periods and contain eight elements.The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulphur, chlorine, and argon

Q109. The number of isomers of pentane is:

• A. 2
• B. 4
• C. 5
• D. 3✓

Explanation: Pentane (C5H12) is an organic compound with five carbon atoms. Pentane has three isomers that are n-pentane, Iso-pentane (methyl butane) and neopentane (dimethylpropane). Therefore three isomers can be drawn from pentane.

Q110. When ammonium cyanide (NH4CN) salt is dissolved in water the solution will be

• A. Neutral
• B. Acidic
• C. Basic✓
• D. Both acidic and basic

Explanation: Smaller the pKa , the stronger the acid. Very strong acids have pKa less than 1, moderately strong acids have pKa in between 1 to 5 and weak acids have pKa above 5.Ka is directly proportional to the H+ concentration and Kb is directly proportional to the OH- concentration.

Q111. Consider the reversible reaction, N2+3H2 = 2NH3 + Heat. The yield of NH3 will be maximum at:

• A. High temperature and low pressure
• B. High temperature and high pressure
• C. Low temperature and low pressure
• D. Low temperature and high pressure✓

Explanation: Exothermic reaction is a reaction in which heat is given out. For exothermic reaction, by increasing temperature reaction move backwards and value of Kc decreases and by decreasing temperature reaction moves forward and value of Kc increases.Reactions involving only gases and having unequal number of moles,by increasing pressure reaction move to that side where number of moles are less. Since the reaction is exothermic, low temp will favour the product’s (NH3) side. High pressure favours the side with lower no of moles which in this case is the product’s side. So lower temperature and higher pressure will ensure maximum lead of NH3

Q112. In the complex, potassium hexacyanoferrate (III) K3 [Fe(CN)6] the coordination of the Fe is

• A. 9
• B. 3
• C. 6✓
• D. 5

Explanation: Coordination number, also known as ligancy, is the number of atoms, ions, or molecules that a central atom or ion carries in a complex or coordination compound or a crystal as its closest neighbours coordination number of Fe in K3[Fe(CN)6] is 6. Since the Fe atom is surrounded by six cyanide ligands which is mono-dentate, the coordination number is 6.

Q113. The compound which has the higher boiling point in the following is:

• A. Methyl chloride
• B. Methyl iodide✓
• C. Methyl bromide
• D. Both a and b

Explanation: For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Waal forces of attraction increases. Hence the order of boiling points is Methyl chloride (CH3Cl) < methyl bromide (CH3Br) < methyl iodide (CH3I).

Q114. Which one of the following is addition polymer:

• A. Nylon
• B. PVC
• C. Polythene
• D. Both PVC and Polythene✓

Explanation: In addition polymerization (sometimes called chain-growth polymerization), a chain reaction adds new monomer units to the growing polymer molecule one at a time through double or triple bonds in the monomerOn the other hand, condensation polymerization is a process that involves repeated condensation reactions between tri-functional or bi-functional monomers with elimination of small molecules.

Q115. Photochemical smog is primarily caused by:

• A. O3
• B. NO2✓
• C. SO3
• D. CO2

Explanation: Photochemical smog appears to be initiated by nitrogen oxides that are emitted into the air as pollutants mainly from internal combustion engines. Absorbing the visible or ultraviolet energy of sunlight, it forms nitric oxide (NO) to free atoms of oxygen (O), which then combine with molecular oxygen (O2) to form ozone (O3). In the presence of hydrocarbons (other than methane), certain other organic compounds, and sunlight, various chemical reactions take place to form photochemical smog.

Q116. Which of the following is not the major source of organic compound?

• A. Natural gas
• B. Petroleum
• C. Coal
• D. Ammoniacal Liquor✓

Explanation: Ammoniacal liquor is an organic compound. It is a concentrated solution of ammonia, ammonium compounds, and sulfur compounds, obtained as a by-product in the destructive distillation of bituminous coal.

Q117. Which of the following concentration unit is temperature dependent:

• A. Molality
• B. Mole Fraction
• C. Molarity✓
• D. Both Molality and Molarity

Explanation: Molarity of a solution depends upon temperature because volume of a solution is temperature dependent. What happens is that when the temperature is increased the distance between the molecules in a liquid increase leading to volume expansion. In turn leading to decrease in molarity. Molarity = Moles of solute / volume of solution

Q118. Which one is more reactive?

• A. HCHO✓
• B. CH3 CHO
• C. (CH3)2CO
• D. Have equal reactivity

Explanation: Aldehydes are usually more reactive toward nucleophilic substitutions than ketones because of both steric and electronic effects. In aldehydes, the relatively small hydrogen atom is attached to one side of the carbonyl group, while a larger R group is affixed to the other side. In ketones, however, R groups are attached to both sides of the carbonyl group. Thus, steric hindrance is less in aldehydes than in ketones. Electronically, aldehydes have only one R group to supply electrons toward the partially positive carbonyl carbon, while ketones have two electron‐supplying groups attached to the carbonyl carbon. The greater amount of electrons being supplied to the carbonyl carbon, the less the partial positive charge on this atom and the weaker it will become as a nucleus.

Q119. Tertiary alcohols are not oxidized into carbonyl compounds because

• A. They contain more alkyl group
• B. They have no alpha-hydrogen✓
• C. Suitable oxidizing agent is not available
• D. None of the above

Explanation: Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. During oxidation,oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom is attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.

Q120. Which compound shows the highest boiling point

• A. CH3COOH✓
• B. C2H5OH
• C. C2H2O-C2H5
• D. (CH3CH2)3N

Explanation: CH3COOH (acetic acid) has the highest boiling point among the options at 118°C, primarily due to its ability to form strong hydrogen bonds, which require more energy (in the form of heat) to break during the phase change from liquid to gas. In contrast, C2H5OH (ethanol) and the ether compound have boiling points of 78°C and 82°C, respectively, due to weaker hydrogen bonding. Although (CH3CH2)3N (triethylamine) has a higher molecular mass of 101g, it does not engage in hydrogen bonding as effectively as acetic acid, resulting in a boiling point of only 89°C. Thus, the strength of intermolecular forces, particularly hydrogen bonding, plays a critical role in determining boiling points.

Q121. Which of the following pollutant decolorizes the skin?

• A. Mercury
• B. Arsenic✓
• C. Lead
• D. Cadmium

Explanation: Air pollutants damage the skin by inducing oxidative stress. Although human skin acts as a biological shield against pro-oxidative chemicals and physical air pollutants, prolonged or repetitive exposure to high levels of these pollutants may have profound negative effects on the skin.

Q122. Which contains more atoms?

• A. 7 gram Mg
• B. 8 gram Na✓
• C. 9 gram Al
• D. All same

Explanation: As we know, to calculate the number of atoms, [Mass / Molar Mass] x [6.022 x 1023] Hence, Na has the most atoms by putting the values.

Q123. Which contains the highest percentage of nitrogen?

• A. NO
• B. NO2
• C. N2O✓
• D. N2O5

Explanation: N2O contains the highest percentage of nitrogen i.e, 28/44 × 100 = 63.6 %

Q124. Fe+2 will form the most ionic bonds with:

• A. N-3✓
• B. S-2
• C. P-3
• D. F-1

Explanation: The stronger the charges on the cation and anion, the stronger the electrostatic attraction.Smaller ions can get closer → stronger attraction.Nitride(N3−) is smaller than phosphide (P3−)So Fe²⁺ + N³⁻ → stronger electrostatic attraction than Fe²⁺ + P³⁻Highest charge difference: 2+ and 3–Smallest anion radius: N³⁻

Q125. For exothermic reversible reaction, the activation energy for forward direction depends upon:

• A. Temperature
• B. Nature of reactant✓
• C. Nature of product
• D. Both Temperature and Nature of reactant

Explanation: For exothermic reversible reaction activation energy for forward direction depends upon nature of reactant.

Q126. As the polarizing power of cation increases thermal stability of carbonates:

• A. Increases
• B. Decreases✓
• C. Not dependent
• D. Depends upon pressure

Explanation: As the polarizing power of the cation increases, it polarizes the carbonate (anion). The nucleus of the cation attracts electrons of anion causing a decrease in its thermal stability.

Q127. Which of the following elements has lower first ionization energy?

• A. N
• B. O
• C. C
• D. B✓

Explanation: The first ionization decreases from beryllium to boron, and from magnesium to aluminum, as electrons from the p-block start to come into play. In the case of boron, which has an electron configuration of 1s2 2s2 2p1, the 2s electrons shield the higher-energy 2p electron from the nucleus, making it slightly easier to remove. A similar effect occurs in aluminum, which has an electron configuration of 1s2 2s2 2p6 3s2 3p1. Even though oxygen is to the right of nitrogen in period 2, its first ionization energy is slightly lower than that of nitrogen. Nitrogen has an electron configuration of 1s2 2s2 2p3, which puts one electron in each p orbital, making it a half-filled set of orbitals.Half-filled sets of p orbitals are slightly more stable than those with 2 or 4 electrons, which makes it slightly harder to ionize a nitrogen atom. Oxygen has an electron configuration of 1s2 2s2 2p4, which puts another electron in one p orbital; since this is one electron away from being half-filled, it is slightly easier to remove this additional electron.

Q128. The anhydride of HClO4 is

• A. ClO3
• B. ClO2
• C. Cl2 O7✓
• D. CI2 O7

Explanation: Anhydride literally means 'without water. It can be defined as the chemical compound formed by eliminating water from another compound 2HClO4 → Cl2O7 + H2O ​Thus, Anhydride of 2HClO4 is Cl2O7

Q129. A gas diffuses 1/2 times as fast as hydrogen, its molecular mass is:

• A. 50 amu
• B. 25 amu
• C. 16 amu
• D. 8 amu✓

Explanation: Graham found experimentally that the rate of diffusion of a gas is inversely proportional to the square root of the mass of its particles. v1/v2=√(M2/M1) where M1=H2 v1/(1/2)v1 =√(M2/2) 2²=M2/2 => M2 = 4×2 = 8

Q130. Which one of the following ions has more electrons than protons and more protons than neutrons?

• A. D
• B. d-
• C. H-✓
• D. He

Explanation: H is Hydrogen, it’s proton number is 1 so it has 1 proton, 1 electron and no neutrons. The ion of H gains an electron to attain the negative charge so now it has 2 electrons, 1 proton and 0 neutrons

Q131. Ice and water are in equilibrium with each other. By increasing the pressure, the equilibrium will shift in:

• A. Forward direction✓
• B. Reverse direction
• C. To all system at equilibrium
• D. None of the above

Explanation: By increasing the pressure, the equilibrium between ice and water will shift in the direction of the phase with the lower volume, which is the forward direction.

Q132. Steam causes severe bums than boiling water. It is due to

• A. Absence of hydrogen bonding
• B. High latent heat of vaporization✓
• C. Freely moving molecules
• D. Statement is incorrect

Explanation: Steam has more warmth and vitality than water. Due to its dormant warmth of vaporization, it will produce more burn than boiling water. Steam contains the heat energy of boiling water as well as the latent heat of vaporization

Q133. Which oxides of "K" contain more oxygen than its normal oxide?

• A. Peroxide
• B. Super oxide✓
• C. Both contain equal quantity
• D. None of the above

Explanation: Potassium form normal oxide ( in which oxygen has oxidation state -2) K2O2. KO is peroxide (oxygen has oxidation state -1) has same no.of oxygen as normal oxide. KO2 is super oxide (oxygen has oxidation state -½) has more no. of oxygen than normal oxide.

Q134. A gas decolorized alkaline KMnO4 solution but does not give any PPT with ammoniacal AgNO3

• A. Methane
• B. Ethylene✓
• C. Ethane
• D. None of the above

Explanation: Option d) "None of the above" is incorrect because ethylene (option b) decolorizes KMnO4 but does not give a precipitate with ammoniacal AgNO3. Methane (option a) and ethane (option c) are both saturated hydrocarbons with only single bonds and are not reducing agents, so they would not decolorize KMnO4 or react with ammoniacal AgNO3 to form a precipitate. Therefore, the correct answer is option b) Ethylene.

Q135. Why ethanoic acid is a stronger acid in the liquid ammonia than in water

• A. Ammonia is strong base than water✓
• B. Ethanoic acid molecules form H-bonding with water
• C. Ethanoic acid is more soluble in liquid ammonia than in water
• D. None of the above

Explanation: Both water and ammonia can form hydrogen bond, and this is a very strong intermolecular force. In order for ethanoic acid to dissociate, the attraction between the protons and the surrounding solvent must be more favorable then the strength of the hydrogen bond to the ethanoic acid which ammonia provides because it is more basic than water.

Q136. Which ions are used as catalysts in the reaction between persulfate ions and iodide ions?

• A. Lead
• B. Iron✓
• C. Copper
• D. Chromium

Explanation: Fe 2+ and Fe3+ ions are used as catalysts in this reaction since both the reactants are negatively charged, they repel each other, leading to a high activation energy.

Q137. Which one is stronger nucleophile?

• A. C2H5O
• B. C2H5S✓
• C. Both are equally strong
• D. None of the above

Explanation: C2H5S- is the stronger nucleophile due to the positive inductive effect from the alkyl group and since sulfur has more electrons than oxygen, it is attracted toward the center of a positive charge.

Q138. Which one of the following elements has the largest second ionization energy:

• A. O
• B. F
• C. Na✓
• D. N

Explanation: Na has the largest second ionization energy as the second electron is removed from an inner shell, thus the nuclear charge acting on it is stronger and more energy is needed to remove an electron. After losing 1st electron, sodium gains electronic configuration of Neon (noble gas) which has completely filled orbital so it has the highest second ionization energy.

Q139. Which of the following species has the maximum number of unpaired electrons?

• A. O2✓
• B. O+2
• C. O-2
• D. O2-2

Explanation: O2 has 2 unpaired electrons, O+2 has no unpaired electrons, O-2 has paired electrons while O2-2 also has no unpaired electrons hence the correct option is A The number of unpaired electrons in an atom or ion can be determined by examining its electron configuration. Let's analyze the given species: 1. **O2**: Molecular oxygen (O2) has a total of 16 electrons (8 electrons per oxygen atom). Its electron configuration is 1s² 2s² 2p⁴. There are two unpaired electrons in the 2p orbital. 2. **O+2**: Oxygen cation (O+2) is formed by losing two electrons from O2. Its electron configuration becomes 1s² 2s² 2p³. There is one unpaired electron in the 2p orbital. 3. **O-2**: Oxygen anion (O-2) is formed by gaining two electrons by O2. Its electron configuration becomes 1s² 2s² 2p⁶. There are no unpaired electrons. 4. **O2-2**: This is a hypothetical species, as it would mean adding two more electrons to O-2, which is not energetically favorable. If you meant O-2, please refer to the explanation for O-2 above. Based on the analysis, **O2** has the maximum number of unpaired electrons with **two unpaired electrons**.

Q140. A mixture of 10cm3 of oxygen and 50cm3 of hydrogen is sparked continuously. Calculate volume of non limiting reagent left after reaction completion:

• A. 10cm3
• B. 15cm3
• C. 20cm3
• D. 30cm3✓

Explanation: We'll begin by writing the balanced equation for the reaction. This is given below: O2 + 2H2 —> 2H2O From the balanced equation above, we can say that: 1 cm³ of O2 reacted with 2 cm³ of H2 to produce 2 cm³ of H2O. Next, we shall determine the excess reactant. This can be obtained as follow: From the balanced equation above, 1 cm³ of O2 reacted with 2 cm³ of H2. Therefore, 10 cm³ of O2 will react with = (10 × 2)/1 = 10 × 2 = 20 cm³ of H2. From the calculations made above, we can see that only 20 cm³ out of 50 cm³ of H2 given is needed to react completely with 10 cm³ of O2. Therefore, O2 is the limiting reactant and H2 is the excess reactant (non limiting reactant). Finally, we shall determine the volume of the non limiting reactant (excess react) that is remaining after the reaction. This can be obtained as follow: Volume of non limiting reactant (H2) = 50 cm³ Volume of non limiting reactant (H2) that reacted = 20 cm³ Volume of non limiting reactant (H2) remaining = (Volume of non limiting reactant) – (Volume of non limiting reactant that reacted) Volume of non limiting reactant (H2) remaining = 50 – 20 = 30 cm³

Q141. The oxidation states of Nitrogen in NH4NO3 are:

• A. -3 and +5✓
• B. +5 and -3
• C. -3 & -3
• D. Zero

Explanation: In NH₄NO₃, nitrogen is present in two different ions: NH₄⁺ and NO₃⁻. For NH₄⁺, the sum of oxidation states is +1. Since each hydrogen has an oxidation state of +1, the nitrogen must have an oxidation state of -3 to balance it (+1 = -3 + 4). For NO₃⁻, the sum of oxidation states is -1. Oxygen typically has an oxidation state of -2, so the nitrogen must be +5 to account for the charge (-1 = +5 - 6). Therefore, the correct oxidation states of nitrogen in NH₄NO₃ are -3 in NH₄⁺ and +5 in NO₃⁻.

Q142. Which equation relates to the first ionization energy of bromine?

• A. Br(g) → Br-(g) + 1e
• B. Br(g) → Br+(g) + 1e-✓
• C. 1/2 Br2(g) → Br-(g) + 1e-
• D. 1/2 Br2(g) → Br+(g) + 1e-

Explanation: Refer to folllowing explanation

Q143. Co-ordination number of [Co(en)2Cl2]is

• A. -2
• B. 6✓
• C. 4
• D. None

Explanation: In [Co(en)2​Cl2] No. of monodentate ligand (Cl) =2 No. of bidentate ligand,(Ethylene diamine) =2 ∴ Co-ordination no. of the metal =2+2×2=6

Q144. An olefin "X" on ozonolysis gives CH3CH2COCH3 and CH3COCH3. The IUPAC name of X is:

• A. 2 - Butene
• B. 2,3 di Methyl-2 Pentene✓
• C. 2 - Pentene
• D. 1 - Hexene

Explanation: The longest carbon chain is of 5 carbons so the name ends with pentene and since there are two methyl groups as branches so option B is correct

Q145. Which one is more soluble in water?

• A. Secondary amines✓
• B. Tertiary amines
• C. Quatemary amines
• D. All are insoluble

Explanation: Lower amines can form hydrogen bonds with water thus making them more soluble.

Q146. The number of peaks given by ethane thiol in the H-NMR spectrum is:

• A. 2
• B. 3✓
• C. 4
• D. None

Explanation: Ethane thiol has the formula CH3- CH2- SH so the number of peaks of proton NMR are going to be 3

Q147. C4H11N gives the type of isomerism:

• A. Metamerism✓
• B. Optical isomerism
• C. Tautomerism
• D. None

Explanation: It gives metamerism which is when compounds having the same molecular formula but different alkyl groups on either side of functional groups.

Q148. From the following, identify the incorrect statement regarding gas having high value of coefficient of attraction.

• A. Easy to be liquified✓
• B. Having higher critical temperature
• C. Less soluble in water
• D. None

Explanation: A gas having high value of coefficient of attraction is tough to be liquified. It is because, the repulsive forces become more significant and the attractive forces become less dominant. Hence these gases are difficult to be condensed.

Q149. Which one can form more acidic oxide?

• A. Sc
• B. Mn✓
• C. V
• D. Ti

Explanation: The acidic strength of oxides of different elements will depend on the tendency of oxygen atoms to accept electrons. When the oxygen atom is attached to some electronegative element its acidic character increases as the electronegative element pulls the electrons towards its side. In the periodic table, the electropositive character decreases from right to left across a period and increases down the group. Since Mn appears after Sc, V, and Ti in the periodic table it is comparatively more electronegative, and hence Mn is the most acidic oxide among these transition elements.

Q150. Hydration of hydrocarbon give carbonyl compound, the general formula of that hydrocarbon is:

• A. CnH2n+2
• B. CnH2n
• C. CnH2n-2✓
• D. Both CnH2n and CnH2n-2

Explanation: Alkynes on hydration in the presence of H2SO4 and HgSO4 yield carbonyl compounds through enolAll alkynes except ethyne give ketones while ethyne gives aldehyde.

Q151. Consider reversibility in free radical substitution reaction of alkane then Kc value is smallest for:

• A. Initiation step
• B. Propagation step
• C. Termination step✓
• D. All same

Explanation: In a free radical substitution reaction of an alkane, the reaction can proceed through several steps involving free radicals (species with unpaired electrons). For this type of reaction, reversibility is more likely to occur in the termination step.The termination step involves the combination of two free radicals, leading to the formation of a stable molecule. This step can be reversible in certain cases, resulting in the reformation of free radicals. For example:Initiation step: Formation of free radicals (R·) by the homolytic cleavage of a bond in the alkane due to the presence of an initiator (e.g., light or heat). Propagation steps: Free radicals react with the alkane to form new free radicals and alkyl radicals, continuing the chain reaction.Termination step: Two free radicals combine to form a stable molecule.Since the termination step involves the recombination of free radicals and is less likely to be fully irreversible, the equilibrium constant (Kc) for this step would likely be smaller compared to the propagation steps, which involve the generation of new free radicals.It's important to note that the overall reaction is usually considered irreversible, but individual steps within the reaction mechanism may exhibit different degrees of reversibility. The termination step is generally considered to have a smaller Kc value due to its potential for reversibility.

Q152. Ethylenediamine diacetate is

• A. Didentate
• B. Tridentate
• C. Tetradentate✓
• D. Hexadentate

Explanation: Ethylenediamine diacetate is a chemical compound with tetradentate coordination properties, not bidentate. A tetradentate ligand can form four coordination bonds with a central metal ion, while a bidentate ligand can form two coordination bonds. In the case of ethylenediamine diacetate, it has two ethylenediamine groups, each of which can bind to a metal ion through two coordination bonds, totaling four bonds and making it tetradentate.

Q153. Epoxide obtained from isobutylene is further hydrolyzed in the presence of acid The final product is

• A. 2,3 - Butanediol
• B. 1,2 - Butanediol
• C. 2-Methyl-1,2-Propandiol✓
• D. All of them

Explanation: Epoxide obtained from isobutylene is further hydrolyzed in the presence of acid The final product is 2-Methyl-1,2-Propandiol

Q154. In the detection of nitrogen in an organic compound The appearance of Prussian blue colouration is due to the formation of:

• A. Fe4[Fe(CN6)]3✓
• B. Na3[Fe(CN)6]
• C. K3[Fe(CN)6]
• D. None

Explanation: In the Lassaigne’s test, nitrogen in an organic compound forms sodium cyanide (NaCN) when fused with Na. On adding Fe²⁺ ions, it reacts with CN⁻ to form ferric ferrocyanide (Fe₄[Fe(CN)₆]₃). This compound is Prussian blue, giving the characteristic blue coloration.

Q155. The bond angle in H2S is less than H2O. it is due to

• A. Small size of oxygen atom
• B. Greater E.N of oxygen atom✓
• C. Oxygen contain two bone pairs of electrons
• D. All of the above

Explanation: Bond angle of H2S(92° )<H2O(104°). As the electronegativity of the central atom decreases, bond angle decreases. In the present case, S is less electronegative than oxygen. Thus bond pairs in H2S are more away from the central atom than in H2O and thus repulsive forces between bond pairs are smaller producing smaller bond angle

Q156. The auxochrome not concern with Metanil yellow dye:

• A. -SO3H✓
• B. -OH
• C. -NH2
• D. Both -SO3H and -NH2

Explanation: Auxochromes are groups that enhance color by altering the electron distribution in a chromophore. Metanil yellow dye contains –SO₃H (solubility) and –NH₂ (color intensification) as auxochromes. The –OH group is not present and thus does not contribute to this dye’s color.

Q157. Molecular theory for metals is also called as:

• A. Electron pool theory
• B. Valence bond theory
• C. Band Theory✓
• D. None of these options are correct

Explanation: The molecular theory for metals is called band theory. Band theory is a quantum mechanical model that explains the behavior of electrons in solids by grouping them into energy bands. In metals, the delocalization of valence electrons allows them to move freely, offering an explanation for various metallic properties such as electrical conductivity, reflectivity, and melting points.Other options, like electron pool theory and valence bond theory, fail to encompass the full range of metallic properties. Electron pool theory is more simplistic and less comprehensive, while valence bond theory is tailored to covalent bonding in molecules rather than metals.

Q158. The general trend of ionic radii from left to right is:

• A. Increase
• B. Decrease✓
• C. Increase then decrease
• D. Decrease then increase

Explanation: Ionic radius increases as you move from top to bottom on the periodic table. Ionic radius decreases as you move across the periodic table, from left to right.

Q159. Actual yield is a/an:

• A. Theoretical term
• B. Experimental term✓
• C. Stoichiometric term
• D. Supposed term

Explanation: The actual yield is the quantity of a product that is obtained from a chemical reaction. In contrast, the calculated or theoretical yield is the amount of product that could be obtained from a reaction if all of the reactant converted to product. Theoretical yield is based on the limiting reactant.

Q160. The angular acceleration of second hand of watch is

• A. π rad/sec2
• B. 2π rad/sec2
• C. π/2 rad/sec2
• D. None of the above✓

Explanation: Angular acceleration (α) is defined as the change in angular velocity (ꞷ) per unit time. The second-hand watch as a constant angular velocity of rad/s. Therefore, change in (ꞷ) will be zero and α will be zero as well. Option A, B and C are all hence wrong.

Q161. Which one of the following is not a state function?

• A. Work✓
• B. Enthalpy
• C. Internal energy
• D. Pressure

Explanation: The thermodynamic state of a system refers to the temperature, pressure and quantity of substance present. Examples of state functions include density, internal energy, enthalpy, entropy. State functions only depend on these parameters and not on how they were reached, So Work is NOT a state function and hence is the correct option.

Q162. The viscous drag on small spherical body (moving with slow speed v) is propotional to

• A. v✓
• B. Square root of v
• C. 1/square root of v
• D. v2

Explanation: Formula for viscous drag force is Fs = 6πrηv. Form the given formula it can be seen that keeping r and η constant for a specific object in a fluid, F is proportional to v. Option A is hence the only correct relation. Option B, C & D are wrong relations.

Q163. The transverse nature of light is shown by

• A. Interference of light
• B. Refraction of light
• C. Polarization of light✓
• D. Dispersion of light

Explanation: A light wave is an electromagnetic wave which travels through the vacuum of outer space and transparent materials like glass. Electromagnetic waves are transverse waves in the sense that the plane of the wave vibrations is different from the plane of wave propagation. The electromagnetic waves in the true sense have the direction of propagation perpendicular to the direction of its electric and magnetic fields. In the case of polarized waves, the electric and magnetic fields are also at right angles to each other. Polarized light waves are electromagnetic waves in which the vibrations occur in a single plane. Light can be polarized by transmission, reflection, refraction or scattering. Using a polaroid filter, different planes of vibration of polaroid filters can be blocked due to the special material used there. Due to the polarization of light, it is evident that light is a transverse wave.

Q164. An electron is moving along the axis of a solenoid carrying a current. Which of the following is a correct statement about the magnetic force acting on the electron?

• A. The force acts radially inwards
• B. The force acts radially downwards
• C. The force acts in the direction of motion
• D. No force acts✓

Explanation: Consider the diagram of magnetic field shown and observe the magnetic field lines.Option D) When an electron moves along the axis of magnetic field in the solenoid the force acting on it will be given by F= qvB. As the electron is moving along the axis, θ will be zero and is zero.Hence no force will act on the electron.Option A, B & C all talk about forces and hence are wrong.

Q165. The motional E.M.F depends upon:

• A. Strength of magnetic field
• B. Speed of the conductor
• C. Length of conductor
• D. All answers are correct✓

Explanation: The factors on which motional emf depends are the magnetic field and velocity and length of the rod. Thus option D is the correct answer.

Q166. When zinc electrode is coupled with copper electrode in a galvanic cell:

• A. Reduction takes place at zinc electrode
• B. Oxidation takes place at copper electrode
• C. Reduction takes place at copper electrode✓
• D. Both Reduction takes place at zinc electrode and Oxidation takes place at copper electrode

Explanation: A galvanic cell is made by joining two half cells. Cu2+ has a higher reduction potential so it is reduced and reduction talks place at Cu electrode. Consequently, oxidation takes place at Zn electrode as it has lower reduction potential (higher oxidation potential).

Q167. Which one of the following physical quantity does not have the dimension of force per unit area:

• A. Stress
• B. Strain✓
• C. Young modulus
• D. Pressure

Explanation: The strain unit is dimensionless. It is the ratio between the length shift and the initial length, so it is unit-less. Thus B, strain, is the correct option

Q168. In the case of germanium, the value of potential barrier develops across the depletion region is

• A. 0v
• B. 0.3V✓
• C. 0.7V
• D. 0.9V

Explanation: 0.3 V for Germanium and 0.7 V for silicon.

Q169. Electron microscope makes practical use of:

• A. Particle nature of electron
• B. Wave nature of electron✓
• C. Dual nature of electron
• D. None of the above

Explanation: The electron microscope is a powerful scientific instrument that employs a beam of electrons to illuminate an object and magnify its image. By harnessing the wave-like properties of electrons, this technology allows researchers to capture high-resolution images of tiny samples and gain unprecedented insights into their structure and composition.

Q170. Projectile is thrown in such a way that its maximum height equals to its range, the angle of projection is

• A. Tan-1 (45)
• B. Tan-1 (60)
• C. Tan 1 (30)
• D. None✓

Explanation: Maximum Height (H): H = (u²sin²θ) / 2g, where u is the initial velocity, θ is the angle of projection, and g is acceleration due to gravity.Range (R): R = (u²sin2θ) / g2. Set up the Given ConditionThe problem states that the maximum height equals the range:H = R3. Substitute the Formulas(u²sin²θ) / 2g = (u²sin2θ) / g4. SimplifyCancel out 'u²' and 'g' from both sides: sin²θ / 2 = sin2θUse the trigonometric identity sin2θ = 2sinθcosθ: sin²θ / 2 = 2sinθcosθRearrange the equation: sin²θ = 4sinθcosθDivide both sides by sinθ (assuming θ is not 0 or 180 degrees): sinθ = 4cosθDivide both sides by cosθ: tanθ = 45. Solve for θθ = tan⁻¹(4) θ ≈ 75.96 degreesAnswer:The angle of projection is approximately 75.96 degrees.

Q171. Car "X" is travelling at half the speed of car "Y" and mass of car X is twice as compared to the mass of car 'Y". Which of the following statements is correct?

• A. Car "X' has half of the kinetic energy of car "Y'✓
• B. Car "X" has one quarter of the K.E of car "Y"
• C. Car 'X" has twice the K.E of car "Y'
• D. The two cars have the same KE

Explanation: Car X speed = v/2 amd mass = 2m Car Y speed = v and mass = m

Q172. If the wavelength of a transverse is 2cm and the period is 2 seconds, then the wave speed in CGS is:

• A. 0.1 cms-1
• B. 0.2 cms-1
• C. 11 cms-1
• D. 1 cms-1✓

Explanation: Option A, B & C are all numerically wrong.

Q173. A car battery has EMF of 12 volts and internal resistance 5 × 10-2 ohm. If it draws 60 ampere current, then terminal voltage of the battery will be:

• A. 5 volts
• B. 3 volts
• C. 15 volts
• D. 9 volts✓

Explanation: E= V+IrV = E–Ir = 12–(60)(5×10-2) = 9 volts

Q174. The cyclotron frequency of an electron projected with velocity perpendicular to a magnetic field B is given by:

• A. f= mB/πC
• B. f= 2πeB/m
• C. f= eB/2πm✓
• D. f= 2πc/mB

Explanation: For a cyclotron Centripetal force is provided by magnetic force.By derivation, A, B & C are wrong.

Q175. If A, B = 1/2 AB then angle between A and B is

• A. Zero
• B. 30
• C. 60✓
• D. 90

Explanation: In this case dot product is equal to scalar product of two vector A.B= A x B x cos θ In this case on right hand side of equation, ½ will be Hence θ will be 60. Option A, B and D give wrong value of angle

Q176. A train is 200m long and is moving with uniform velocity of 36 km/h the time it will take to cross a bridge of 1 km is

• A. 100 sec
• B. 120 sec✓
• C. 60 sec
• D. 50 sec

Explanation: Consider one point on the train as your frame of reference , so at any given point the train needs to travel 1200m ( 1km +200 m) to reach the other side of the bridge.And converting 36kmph into m/s gives us 10m/s. So by using time = displacement/velocity Time = 1200/10= 120 seconds

Q177. The escape velocity of body from a planet does not depend upon;

• A. The mass of a body✓
• B. The mass of the planet
• C. The average radius of the planet
• D. The density of the planet

Explanation: The escape velocity 'v' of a body depends upon the acceleration due to gravity of the planet and the radius of the planet but not on mass of planet.

Q178. In order to increase the stopping potential, there should be increase in

• A. Intensity of radiation
• B. Wavelength
• C. Frequency of radiation✓
• D. Both wavelength and intensity

Explanation: The energy of incident light is directly proportional to its frequency. Hence, the stopping potential increases when the frequency of the incident light is increased.

Q179. Two meter high tank is full of water. A hole is made in the middle of the tank. The speed of efflux is

• A. 4.9 m/s
• B. 9.8 m/s
• C. 4.42 m/s✓
• D. 3.75 m/s

Explanation: The speed of efflux is equal to the velocity gained by the fluid in falling through the distance (h1-h2) under the action of gravity. Using Bernoulli's equation

Q180. A hail and a rain drop of the same radius are released from the same height, the rain drop will reach

• A. Before hail
• B. After hail✓
• C. At the same time
• D. None of the above

Explanation: The correct answer is option B: 'After hail'. The rain drop, being less dense than the hail, experiences less gravitational force acting on it compared to the denser hail drop. As a result, the hail drop, which is heavier and has a different interaction with air resistance, reaches the ground first. Options A and C are incorrect because they suggest incorrect relationships in falling speed based on density and drag. Option D is also incorrect as it dismisses the correct answer provided in option B.

Q181. Two springs A and B (kA​=2kB​) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

• A. E/2
• B. 2E✓
• C. E
• D. E/4

Explanation: EA = ½ kAxA2 Keeping force constant and using F= kx As kA = 2kB Then 2xA=xB Energy stored in B will be: EB= ½ (kA/2) (2xA)2 EB= ½ (2xA2 kA) Replace EA = ½ kAxA2 EB = 2 EA Hence only B is correct (A, C, D give wrong answer)

Q182. The general form of path difference in young's double slit experiment is "mλ" Its corresponding phase shift (in radians) is

• A. mπ
• B. 2 mπ✓
• C. mπ/2
• D. None

Explanation: Factual. General form for path difference is given as =2𝞹 Option A, C and D are not correct according to this.

Q183. An α-particle is accelerated through a potential difference of 106 volts, K.E is:

• A. 1 MeV
• B. 2 MeV✓
• C. 4 MeV
• D. 8 MeV

Explanation: For a particle that is accelerated, its KE equals VqApplying the formula,KE= Vq AND M=106KE= (106) (2e)KE= 2MeVOption A, C and D give wrong numerical values.

Q184. If there are "n" capacitors each of capacity "c" connected in parallel to "V" volts source then the energy stored is equal to:

• A. CV
• B. ½ nCV2✓
• C. CV2
• D. CV2/2n

Explanation: The energy of a capacitor is calculated using the formula = ½ C V2 multiplied by n i.e number of capacitors

Q185. The electric field strength between a pair of plates is "E". If the separation of the plates is doubled and the potential difference between the plates is increased by a factor of four, the new field strength is:

• A. E
• B. 2E✓
• C. 4E
• D. 8E

Explanation: Electric field is calculated by the formula = Voltage / Distance so doubling the separation of plates decreases the electric field by 2 times and increasing pd between plates by factor of four, increases electric field 4 times so the over all increase is 2E

Q186. Two satellites of masses "3M" and "M" orbit the earth in a circular orbit of radius Y and "3r" respectively, the ratio of their speed is:

• A. 1:1
• B. Square root of 3 :1✓
• C. 3:1
• D. 9:1

Explanation: To move in a circular orbit, gravitational force provides centripetal force. Gravitational force between the two is the same as it is a mutual force. For the same reason, distance between the satellites is the same so speed does not depend on radius.

Q187. The optimum pH of enzyme urease is

• A. 7.8-8.7
• B. 7.0✓
• C. 4.5
• D. 8.0

Explanation: The optimum pH of enzyme urease is 7.0 i.e neutral

Q188. Two wires A and B are made of the same material. Wire A has a length of I, and a diameter of R while wire B has a length of 2L and a diameter of R/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is:

• A. ⅛✓
• B. ¼
• C. 4
• D. 8

Explanation: Wires made of the same material have the same Young Modulus.Hence, option A is the correct choice.

Q189. A wire can sustain the weight of 20kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of

• A. 10kg
• B. 20kg✓
• C. 40kg
• D. 80kg

Explanation: The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same each part can sustain a weight of 20 kg.

Q190. Which of the following is not E.M wave?

• A. Radio waves
• B. X-rays
• C. Light waves
• D. Sound waves✓

Explanation: Factual. Electromagnetic waves include:Gamma rays, X rays, Ultra violet, Visible light, Infra-Red, Microwaves, RADIO WAVES.So, D is not a EM wave.

Q191. A shell of mass m moving with velocity v suddenly breaks into two pieces. The part having mass m/4 remains stationary. The velocity of the other shell will be:

• A. v
• B. 2v
• C. 3v/4
• D. 4v/3✓

Explanation: Using the law conservation of mass, Total momentum Before = Total momentum After Let x be the new speed of remaining mass (3m/4) mv = m1v1 + m2v2 mv = (m/4) (0) + (3m/4) (x) mv = (3m/4) x 4v/3 = x Options A, B and C cannot be derived from the given situation as they donot follow law of conservation of momentum.

Q192. Two blocks "A" and "B" having masses 3kg and 4kg are raised to the same height from earth surface. The ratio of gravitational potential of "A" to that of "B" is

• A. 3:4
• B. 4:3
• C. 1:1✓
• D. None

Explanation: Factual. Gravitational potential for an equipotential surface (one that is at same height) is same for all masses. Hence ratio will be 1:1

Q193. Heat and work are equivalent. This means

• A. When we supply heat to a body we do work on it
• B. When we do work on a body we supply heat to it
• C. The temperature of a body can be increased by doing work on it✓
• D. Heat and work are not inter convertible

Explanation: From the given statement it can be derived Q= W and hence Internal energy change must me zero. (Following from first law of thermodynamics Q= W + U) the temperature of a body can be increased by doing work on it. When work is done on a body, it can lead to an increase in the body's internal energy, which, in turn, can result in a rise in its temperature.Work done on a body involves transferring energy to the body through mechanical means. This energy is converted into the body's internal energy, which is the sum of the kinetic energy and potential energy of its particles. As the internal energy increases, the average kinetic energy of the particles also increases, leading to a rise in temperature.One common example is compressing a gas in a container. When you compress a gas, you are doing work on it by reducing its volume against an external force. The work done on the gas increases its internal energy, and since temperature is a measure of the average kinetic energy of the gas particles, the temperature of the gas increases as well.It's important to note that the increase in temperature due to work done on the body is a result of the transfer of energy and is not a direct heating process like supplying heat through a temperature difference. Nonetheless, work done on a body can indeed raise its temperature by increasing its internal energy.

Q194. The velocity time plot for a particular moving on a straight line is shown in the figure:

• A. The particle has a constant acceleration.✓
• B. The particle has never turned around.
• C. The particle has zero displacement.
• D. The data is insufficient.

Explanation: The particle has a constant acceleration because the slope of the curve is constant.The particle turns around at t=10 seconds.The particle has non-zero displacement because the area above x-axis is not equal to area below the x-axis.Hence, A is correct.

Q195. Mark out the correct option.

• A. The energy of any small part of a string remains constant in a travelling wave
• B. The energy of any small part of a string remains constant in standing wave✓
• C. The energies of all small parts of equal length are equal in a travelling wave
• D. The energies of all small parts of equal length are equal in a travelling wave

Explanation: Option A, C and D are wrong as in a travelling wave energy does not remain constant and is transferred from one point to another.However, in a standing wave as mentioned in option B, Energy of small part remains constant or STANDING in that part.

Q196. A system can be taken from the initial state P1V1 to the final state P2V2 by two different methods. Let Q and W represent the heat given to the system and the work done by the system.Which of the following must be the same in both the methods?

• A. Q
• B. W
• C. Q + W
• D. Q - W✓

Explanation: Internal energy would be the same irrespective of the path.U (internal energy) = Q - W

Q197. At what angle two forces 2F and square root of 2F must act sc that their resultant is F * square root of 10

• A. π/4✓
• B. π/2c
• C. 2π
• D. None of them

Explanation: Explaination for this question will be added shortly.

Q198. When 20J of work was done on a gas, 40J of heat energy was released If the initial internal energy of the gas was 70J. what is the final internal energy?

• A. 50J✓
• B. 60J
• C. 90J
• D. 110J

Explanation: The change in internal energy is 20-40 = -20 J so the final internal energy is x where x-70= -20X=50

Q199. The time required by the projectile to reach the summit point is:

• A. T = square root of (2H/g)✓
• B. T = square root of (3H/g)
• C. T= square root of (4H/g)
• D. T = square root of (H/g)

Explanation: T = square root of (2H/g) is the time required to reach maximum height, this formula has to be memorized

Q200. If energy of activated complex is close to energy of reactants, it means that the reaction is:

• A. Fast✓
• B. Moderate
• C. Slow
• D. Very slow

Explanation: If energy of activated complex is close to energy of reactant, then reaction is fast, and spontaneous. Activated state is the state when reactants gain sufficient energy to be converted into product hence, if energy of reactants is already close to the energy of activated complex, so they will not take time and energy to get converted to activated state. Thus, they will react spontaneously and fast.