Kpk Mdcat Mock Paper 3 — Solved Past Paper with Answers

Q1. Choose the correct spelling:

• A. Exantuated
• B. Axantuated
• C. Accenchuated
• D. Accentuated✓

Explanation: To "accentuate" means to “make more noticeable and prominent” and its correct spelling is given in Option D.

Q2. The headmaster _ to speak to you.

• A. Wants✓
• B. Is wanting
• C. Was wanting
• D. Want

Explanation: “The headmaster wants to speak to you” in Option A sounds most appropriate because "headmaster" is a singular noun so we should complement it with a singular verb - eliminating Option D. The tense is simple present indicated by the verb "speak", hence the present continuous form "is wanting" in Option B is also incorrect. while the past tense form in Option C is also incorrect.

Q3. Identify the errors and choose the correct option: I hope this letters finds in the best of your spirits.

• A. I hope this letter will find you in good of high spirits.
• B. I hope this letter finds you in the best of your spirits.✓
• C. I hope letter finds you in the best of spirits.
• D. I hope the letter found in greatest of sprite.

Explanation: There are two grammatical errors in the sentence above. An object (pronoun/noun) should be used to correctly identify the entity to whom the letter is sent.Option A is incorrect since it makes use of the future tense – indicated by “will find” – while the person will be reading the letter in the present. Option C and Option D are incorrect since they lack the determiner “this” which is necessary to indicate that the letter being mentioned is specific in nature (i.e. the letter that is being read currently). By elimination, Option B is correct.

Q4. It has been raining continuously _ last night.

• A. Since✓
• B. For
• C. From
• D. With

Explanation: We use "for" with a period in the past, present, or future. For refers to periods, e.g. 3 years, 4 hours, ages, a long time, months, years.We use "since" with a point in time in the past. For example: "They’ve lived in Oxford for a couple of months." "They’ve lived in Oxford since 2004."

Q5. I keep the butter in the fridge. Select the correct passive voice:

• A. In the fridge the butter is kept by me.
• B. By me is the butter dept in the fridge.
• C. The butter is kept by me in the fridge.✓
• D. Dept in the fridge by me is the butter.

Explanation: A verb is in the passive voice when the subject of the sentence is acted on by the verb. For example, in “The ball was thrown by the pitcher,” the ball (the subject) receives the action of the verb, and was thrown is in the passive voice.

Q6. Fire destroyed the top floor of the building:

• A. The top floor of the building got destroyed by fire
• B. By fire was destroyed the top floor of the building.
• C. Destroyed by fire was the top floor of the building.
• D. The top floor of the building was destroyed by fire.✓

Explanation: Option D is correct because it adheres to the rules of forming passive voice: the object of the active sentence ('the top floor of the building') becomes the subject of the passive sentence. The verb 'destroyed' is correctly paired with the auxiliary verb 'was' to form the past tense passive voice. Options A, B, and C are incorrect due to improper sentence structure and verb usage. Option A uses an informal construction, Option B misplaces the subject and verb, and Option C incorrectly orders the elements of the sentence.

Q7. An 'elegy‘ is a poem written:

• A. In the memory of a little child
• B. On the death of someone dear✓
• C. On the sighting of an old tutor
• D. In the love of a dear sweetheart

Explanation: An "elegy" is a poem written to express sorrow, grief, or lamentation, typically in response to the death of a loved one or to reflect on the passing of something significant. It is a form of poetry that is meant to evoke feelings of mourning and remembrance.ELEGY, poem lamenting the death of a public personage or of a friend or loved one; by extension, any reflective lyric on the broader theme of human mortality.

Q8. (Fix this) Select the correct sentence:

• A. Certainly she is the best person for the job.
• B. She is the best person for the job certainly.
• C. She is certainly the best person for the job.✓
• D. The best person certainly she is for the job.

Explanation: This sentence is grammatically correct and has a natural flow. "Certainly" is placed directly before "the best person", which makes it clear that it's modifying that phrase.

Q9. The culpable child _ some words to her mother for pardoning his delinquency.

• A. Mumbled✓
• B. Showy
• C. Rude
• D. Crazy

Explanation: 'Mumbled' is to say something indistinctly and quietly, making it difficult for others to hear. So, the correct sentence is, "The culpable child mumbled some words to her mother for pardoning his delinquency."

Q10. "Foraminifers" helps to determine the,

• A. Generation time
• B. Geological age✓
• C. Ecological time
• D. Physiological age

Explanation: The geological time scale relates stratigraphy rock layers) to periods. Foraminifera are geologically short-lived and some forms are only found in specific environments. Therefore, a paleontologist can examine the specimens in a small rock sample like those recovered during the drilling of oil wells and determine the geologic age and environment when the rock formed.

Q11. The surgical transplantation of a matching donor's kidney is the only option left for the permanent treatment of :

• A. Kidney Stones
• B. Hypercalcemia
• C. Uremia✓
• D. Hyperoxaluria

Explanation: Uremia results from end-stage kidney failure, where waste products accumulate in the blood due to severely impaired kidney function. Permanent treatment often requires dialysis or a kidney transplant, making transplantation the only option for a long-term solution.

Q12. CSF Is found in between:

• A. Pia mater and dura mater
• B. Pia mater and arachnoid mater✓
• C. Grey matter and pia mater
• D. Dura mater and grey mater

Explanation: Cerebrospinal fluid (CSF) is a clear, colorless liquid that fills and surrounds the brain and the spinal cord and provides a mechanical barrier against shock. It occupies the subarachnoid space (between the arachnoid mater and the pia mater) and the ventricular system around and inside the brain and spinal cord. Formed primarily in the ventricles of the brain, the cerebrospinal fluid supports the brain and provides lubrication between surrounding bones and the brain and spinal cord. When an individual suffers a head injury, the fluid acts as a cushion, dulling the force by distributing its impact.

Q13. It is adapted to conserve water by over 99.5% reabsorption of glomerular filtrate:

• A. Mammalian body including human
• B. Mammalian kidney including human✓
• C. Mammalian skin including human
• D. Mammalian liver including human

Explanation: The mammalian kidney, including that of humans, is highly adapted to conserve water through the efficient reabsorption of glomerular filtrate.

Q14. Incubation period of ―"HCV" is:

• A. 2-6 weeks
• B. 4-10 weeks
• C. 4-20 weeks✓
• D. 4-26 weeks

Explanation: Hepatitis C is a viral infection that affects the liver. It causes inflammation and swelling, which damages the liver tissues over time. Hepatitis C is caused by a virus that spreads through contact with blood. Transmission occurs when the blood of an infected person enters the body of an uninfected person. Worldwide, transmission commonly occurs in medical settings with unsterilized equipment. An incubation period is the time between infection or contact with the agent and the onset of symptoms or signs of infection. The hepatitis C (HCV) window period is usually 4–20 weeks from the time of exposure. After 6 months, most people will have developed enough antibodies for an HCV test to detect. In rare cases, however, antibodies can take up to 9 months to develop.

Q15. Osteopenia starts at the age of:

• A. 30-40
• B. 30-35
• C. 40-45
• D. 50-60✓

Explanation: Osteopenia is a clinical term used to describe a decrease in bone mineral density (BMD) below normal reference values. Lower BMD indicates we have fewer minerals in our bones than we should, which makes bones weaker. Osteopenia is more common in people older than 50, especially women. Osteopenia isn’t as severe as osteoporosis, a disease that weakens bones so much that they can break more easily. Up until about age 30, a healthy person builds more bone than he or she loses. But after age 35, bones begin to break down faster than they build up. Some things can make bone loss happen more quickly, leading to osteopenia, such as medical conditions such as hyperthyroidism, hormonal changes during menopause, poor nutrition, especially a diet too low in calcium or vitamin D, and unhealthy lifestyle choices, such as smoking, drinking too much alcohol or caffeine, and not exercising.

Q16. Kidney stones are formed due to :

• A. Infectious Diseases
• B. Metabolic Diseases✓
• C. Genetic Disease
• D. Congenital Diseases

Explanation: Kidney stones are formed due to metabolic diseases, which result in abnormalities in the body's normal chemical processes. This disruption can lead to the accumulation of substances that form stones in the kidneys. The other options, such as infectious diseases, genetic diseases, and congenital diseases, are not directly associated with the formation of kidney stones.

Q17. A complex cartilaginous structure surrounding the upper end of the trachea:

• A. Larynx✓
• B. Esophagus
• C. pharynx
• D. Epiglotis

Explanation: The larynx is a cartilaginous structure located above the trachea and contains the vocal cords.

Q18. Select the test used for the estimation of glucose in blood and urine?

• A. Tollen's reagent test
• B. Fehling's solution test
• C. Benedict solution test
• D. All of the above✓

Explanation: The correct answer is Option D: All of the above. Tollen's reagent test, Fehling's solution test, and Benedict solution test are all used for estimating glucose in blood and urine. Each test has its specific characteristics to detect the presence of glucose, making them suitable for this purpose. Options A, B, and C are incorrect as they represent individual tests, whereas Option D encompasses all applicable tests for glucose estimation.

Q19. Trachea lies --- to the esophagus.

• A. Dorsal
• B. Medial
• C. Ventral✓
• D. Lateral

Explanation: 'Ventral' refers to the front or anterior side of the body. The trachea is indeed located ventrally to the esophagus.

Q20. Any DNA molecule having foreign DNA is called:

• A. Mutant
• B. Recombinant✓
• C. Crossing over
• D. All of the above

Explanation: Recombinant DNA is the hybrid DNA made by two or more DNA molecules joined together. So B is the correct option.This technology is used to create new genetic combinations that are then inserted into a host organism, like bacteria, to produce desired traits or products, such as human insulin or genetically modified crops.

Q21. In Eukaryotes, DNA replication proceeds at the rate of:

• A. 50 base pairs per second✓
• B. 40 base pairs per second
• C. 20 base pairs per second
• D. 30 base pairs per second

Explanation: Eukaryotic DNA replication typically occurs at a rate of approximately 50 base pairs per second. This rate reflects the complex nature of eukaryotic cells, which have larger genomes and more intricate regulatory mechanisms than prokaryotic cells. The process is meticulously controlled to ensure accuracy, with proofreading and repair mechanisms in place. The other options, 40, 30, and 20 base pairs per second, underestimate the replication capability of eukaryotic cells, highlighting a lack of alignment with the generally accepted data on replication speeds in these organisms.

Q22. Fatty acids are:

• A. Unsaturated dicarboxylic acid
• B. Long chain alkanoic acid✓
• C. Aromatic carboxylic acid
• D. Aromatic dicarboxylic acid

Explanation: Alkanoic acids are alkane-derived organic compounds that contain the carboxylate (-COOH) functional group and fatty acids are carboxylic acids with long hydrocarbon chains.

Q23. During breathing, air from pharynx enters to:

• A. Trachea✓
• B. Bronchioles
• C. Alveoli
• D. Bronchi

Explanation: When you inhale through your nose or mouth, air travels down your pharynx (back of your throat), passes through your larynx (voice box) and into your trachea (windpipe). Your trachea is divided into two air passages called bronchial tubes. One bronchial tube leads to your left lung, the other to your right lung.

Q24. Which of the following is an enzyme lacking disease?

• A. PKU✓
• B. Alkaptonuria
• C. Anuria
• D. Dluria

Explanation: Phenylketonuria (PKU) is the correct answer as it is a genetic disorder caused by a deficiency of the enzyme phenylalanine hydroxylase (PAH). This enzyme is critical for the metabolism of the amino acid phenylalanine. Without this enzyme, phenylalanine accumulates and can lead to neurological damage. Alkaptonuria also involves an enzyme deficiency, but it is not the enzyme referenced in the question, and anuria and dluria are unrelated to enzyme deficiencies.

Q25. At high altitude, the RBCs in the human blood will :

• A. Increase in number✓
• B. Decrease in number
• C. Increase in size
• D. Decrease in size

Explanation: This statement is accurate. At high altitudes, where the oxygen concentration in the air is lower, the body's response includes an increase in the number of red blood cells (RBCs). This is known as polycythemia. It's an adaptive mechanism to improve oxygen-carrying capacity, helping to transport more oxygen to body tissues.

Q26. Lungs are _ in srcin.

• A. Ectodermal
• B. Endodermal✓
• C. Mesodermal
• D. Preformed

Explanation: The correct answer is Endodermal. The endoderm, one of the three primary germ layers formed during embryonic development, is responsible for the formation of the respiratory tract lining, including the lungs. This layer also forms other vital internal organs such as the liver, pancreas, and parts of the digestive system. The ectoderm and mesoderm, while important for forming other bodily structures, do not contribute to the primary functional tissue of the lungs. 'Preformed' is unrelated to germ layer origins and lung development.

Q27. Lungs do not collapse between breaths, and some air always remains in the lungs, which can never be expelled because:

• A. There is a negative pressure in the lungs
• B. There is a negative intrapleural pressure pulling at the lung walls✓
• C. There is a positive intrapleural pressure
• D. Pressure in the lungs is higher than atmospheric pressure

Explanation: Option B is correct.All options are explained below:a)There is a negative pressure in the lungs."This statement is not entirely accurate. Lungs don't naturally have a negative pressure, and they can't be filled with air if there is a consistent negative pressure. Lungs maintain a degree of positive pressure to keep them expanded.b) "There is a negative intrapleural pressure pulling at the lung walls": This option is correct. Between breaths, the lungs do not collapse due to the presence of a negative intrapleural pressure. Intrapleural pressure is the pressure in the pleural cavity, which surrounds the lungs. It is typically lower than atmospheric pressure, creating a "suction" effect that keeps the lungs inflated. This is a key mechanism for lung expansion and prevents lung collapse.c)"There is a positive intrapleural pressure": This statement is not accurate. A positive intrapleural pressure would not be conducive to maintaining lung expansion; it's the negative intrapleural pressure that plays a critical role in preventing lung collapse.d)"Pressure in the lungs is higher than the atmospheric pressure": This is not correct. The pressure in the lungs, during exhalation, is generally close to atmospheric pressure, but it's the intrapleural pressure that is negative, not the pressure in the lungs themselves.Summary: The correct answer is: "There is a negative intrapleural pressure pulling at the lung walls." The negative intrapleural pressure, which is lower than atmospheric pressure, keeps the lungs expanded by exerting a suction effect on the lung walls, preventing lung collapse between breaths.

Q28. The particular array of chromosomes that an individual possessed is called its:

• A. Genotype
• B. Phenotype
• C. Karyotype✓
• D. Genome

Explanation: The correct answer is Karyotype. A karyotype is a visual profile of an individual's chromosomes, showing the number and appearance of chromosomes, including their size, shape, and any abnormalities. It is often used in genetic testing to identify chromosomal disorders.The term Genotype refers to the specific set of genes an organism carries, which influences its traits but doesn't specifically denote the chromosomal structure.Phenotype describes the observable characteristics or traits of an organism, which are influenced by both its genotype and environmental factors.Genome refers to the entire set of genetic material in an organism but doesn't focus on the physical arrangement or appearance of chromosomes as a karyotype does.

Q29. Changes in gene frequencies in small population by chance is called:

• A. Gene pool
• B. Genetic drift✓
• C. Gene mutation
• D. Gene flow

Explanation: Genetic drift is a mechanism of evolution, it refers to random fluctuations in the frequencies of alleles from generation to generation due to chance of events, therefore the answer will be B.

Q30. Number of chromosomes in Tobacco is:

• A. 45
• B. 48✓
• C. 46
• D. 47

Explanation: The correct number of chromosomes in tobacco is 48. Tobacco is an allopolyploid species, which means it has multiple sets of chromosomes originating from different species. Each gamete contains 24 chromosomes, resulting in a total of 48 chromosomes in somatic cells. The other options, 45, 46, and 47, do not reflect the correct chromosome count for tobacco. Option A (45) and Option D (47) are incorrect as they do not represent the known chromosome numbers for tobacco. Option C (46) is the diploid number for humans, not tobacco.

Q31. Urea formation occurs in:

• A. Kidney
• B. Liver✓
• C. Spleen
• D. Lungs

Explanation: b) Liver:This option states that urea formation occurs in the liver. This statement is correct. The liver is the primary site of urea formation in the body. It is an essential part of the process known as the urea cycle, where toxic ammonia (a byproduct of protein metabolism) is converted into urea, a less toxic compound. Urea is then transported to the kidney for excretion in urine.

Q32. Bone is surrounded by a membrane called:

• A. Perichondrium
• B. Prostomium
• C. Perimycium
• D. Periosteum✓

Explanation: The correct answer is periosteum. The periosteum is a dense fibrous membrane covering the surface of bones. It plays a vital role in bone growth and repair and contains blood vessels and nerves that supply nutrients and sensation to the bone.The perichondrium surrounds cartilage, not bone. The prostomium is unrelated to bone, being a part of certain invertebrates. Perimycium pertains to muscle tissue, not bone.

Q33. A condition of excessive thirst due to diabetes is called:

• A. Polyuria
• B. Glycusuria
• C. Polyphagia
• D. Polydipsia✓

Explanation: Polydipsia is a medical name for the feeling of extreme thirstiness. Polydipsia is an early symptom of diabetes mellitus and diabetes insipidus. Diabetes mellitus causes polydipsia because your blood sugar levels get too high and make you feel thirsty, regardless of how much water you drink. Diabetes insipidus occurs when your body’s fluid levels are out of balance.

Q34. Implantation of zygote takes place in the:

• A. 2nd week✓
• B. 3rd week
• C. 7th week
• D. 5th week

Explanation: Implantation is the process of the blastocyst embedding into the endometrial lining of the uterus, which typically occurs in Week 2 of development. For implantation to occur, the blastocyst must completely hatch from the zona pellucida once the conceptus enters the uterine cavity

Q35. The only human disease caused by Viroid is

• A. Hepatitis A
• B. Hepatitis B
• C. Hepatitis C
• D. Hepatitis D✓

Explanation: Viroids are infectious agents that consist only of naked RNA without any protective layer such as a protein coat. Hepatitis D, also known as the hepatitis delta virus, is an infection that causes the liver to become inflamed. This swelling can impair liver function and cause long-term liver problems, including liver scarring and cancer. The condition is caused by the hepatitis D virus (HDV). Hepatitis D only occurs in people who are also infected with the hepatitis B virus. It is the only human disease known to be caused by a viroid.

Q36. The process of replication in plasmid DNA, other than initiation, is controlled by?

• A. Mitochondrial gene
• B. Plasmid gene
• C. Bacterial gene✓
• D. None of these.

Explanation: The DNA plasmid replicates in a semi­ conservative manner. The initiation of replication is controlled by the plasmid gene and elongation and termination are controlled by bacterial genes.

Q37. Heterospory occur in:

• A. Selaginella✓
• B. Equisetum
• C. Lycopodium
• D. Lepidodendron

Explanation: The phenomenon of the development of two types of spores (differing in size, structure, and function) by the same species is known as heterospory. Selaginella is the sole genus of vascular plants in the family Selaginellaceae, the spikemosses or lesser club mosses. All species of Selaginella are heterosporous; that is, they produce spores of two sizes, the larger designated as megaspores and the smaller as microspores. The megaspores develop into female gametophytes and the microspores into male gametophytes.

Q38. The person is over weight of the body mass index is between:

• A. 15 to 24.9
• B. 17.5 to 24.9
• C. 18.5 to 24.9
• D. 25 to 29.9✓

Explanation: Body Mass Index (BMI) is a calculation that considers an individual's weight in relation to their height to estimate body fat. For adults, a BMI of less than 18.5 is underweight, 18.5 to 24.9 is healthy weight, 25 to 29.9 is overweight, 30 to 39.9 is obese, and 40 or above is severely obese. Therefore, a BMI of 25 to 29.9 indicates that a person is overweight. The other options describe BMI ranges that do not fall into the overweight category.

Q39. Saponification of a fat:

• A. Always results in the formation of soaps.✓
• B. Results in the formation of esters.
• C. Results in the formation of waxes.
• D. Results in the formation of glycerol and soap.

Explanation: Saponification is a chemical process where fats or oils react with an alkali, such as sodium hydroxide (NaOH) or potassium hydroxide (KOH), resulting in the formation of glycerol and soap. The soap consists of the salts of fatty acids, which are the main products of interest. Option A accurately describes this outcome, while the other options incorrectly associate saponification with ester or wax formation. Option D is technically correct but doesn't highlight the primary goal of the process, which is soap production.

Q40. Agarose extracted from sea weeds is used in?

• A. Spectrophotometry
• B. Tissue culture
• C. PCR
• D. Gel Electrophoresis✓

Explanation: In gel electrophoresis, DNA fragments separate (resolve) according to their size through the sieving effect provided by the agarose gel. Agarose is a natural polymer extracted from seaweeds and is commonly used as a matrix.

Q41. A special protein carrier in plasma membrane is:

• A. Catalase
• B. Lipase
• C. Permease✓
• D. Arginase

Explanation: The correct answer is Permease, which is a protein carrier in the plasma membrane. Permease functions by specifically binding to molecules on one side of the membrane and facilitating their movement across the membrane through a conformational change. This is essential for processes like nutrient uptake and waste removal.Other options like Catalase, Lipase, and Arginase are enzymes with roles unrelated to membrane transport. Catalase breaks down hydrogen peroxide, Lipase digests lipids, and Arginase participates in the urea cycle, making them incorrect choices for this question.

Q42. Phagocytosis, pinocytosis and autophagy are the functions of:

• A. Golgi-Apparatus
• B. Lysosomes✓
• C. Peroxisomes
• D. Glyoxisomes

Explanation: Lysosomes perform phagocytosis, pinocutosis and autophagy as they contain enzymes to digest foreign material and casue cell death.

Q43. Thalassaemia major is also known as:

• A. Sickle cell anemia
• B. Cooley's anemia✓
• C. Microcytic anemia
• D. Nutritional anemia

Explanation: Thalassaemia major is also known as Cooley's anemia, named after Thomas B. Cooley, who identified the genetic disorder characterized by reduced hemoglobin production. Sickle cell anemia is a separate genetic condition involving abnormally shaped red blood cells. Microcytic anemia is a general term for small red blood cells, often due to iron deficiency, and can be a symptom of thalassaemia but not a synonym. Nutritional anemia results from nutrient deficiencies, unlike the genetic nature of thalassaemia.

Q44. Restriction endonucleases are enzymes which?

• A. Make cuts at specific positions within the DNA molecule.✓
• B. Recognize a specific nucleotide sequence for binding of DNA ligase.
• C. Restrict the action of the enzyme DNA polymerase.
• D. Remove nucleotides from the ends of the DNA molecule.

Explanation: Restriction endonucleases were found by Arber in 1962 in bacteria. They act as “molecular scissors” or chemical scalpels. They recognize the specific base sequence at palindrome sites in DNA duplex and cut its strands. For example, restriction endonuclease EcoRI found in the colon bacteria E.coli recognizes the base sequence GAATTC in DNA duplex and cuts its strands between G and A.

Q45. Pick one mismatched:

• A. Sublingual - below the tongue
• B. Sub Mandibular - behind jaws
• C. Parotid - in front of the ear
• D. Lacrimal - above the lateral end of the eye✓

Explanation: Humans have three paired major salivary glands (parotid, submandibular, and sublingual) While the lacrimal gland is located within the orbit above the lateral end of the eye. It continually releases fluid which cleanses and protects the eye's surface as it lubricates and moistens it. These lacrimal secretions are commonly known as tears.

Q46. An enzyme of the intestine that converts polypeptides into dipeptides

• A. Trypsin
• B. Maltase
• C. Lipase
• D. Amino peptidase✓

Explanation: Aminopeptidase is an enzyme found in the small intestine, specifically on the brush border of the intestinal lining. It acts on small peptides and converts them into dipeptides by removing amino acids from the N-terminal end of the peptide chain.

Q47. Innermost layer of the stomach is also named:

• A. Mucosa✓
• B. Serosa
• C. Muscularis
• D. Adventitia

Explanation: Mucosa is the innermost layer of the stomach that secretes mucus and digestive enzymes.

Q48. Discontinuous feeding is possible because of:

• A. Stomach✓
• B. Small intestine
• C. The large intestine of food
• D. Oesophagus

Explanation: Discontinuous feeding is possible because the stomach stores food for a long time. It has three functions: Temporarily stores food, Contract and relax to mix and break down food, Produce enzymes and other specialised cells to digest food.

Q49. How many sites of digestion are present in the digestive system of man?

• A. 3✓
• B. 4
• C. 5
• D. 6

Explanation: There are 5 main sites of digestion in humans. mouth, stomach, duodenum, small intestine (jejunum & ileum), and large intestine.

Q50. Chewing of the food in the mouth significantly improves the digestibility of food stuff by breaking down into smaller pieces is:

• A. Mastication✓
• B. Peristalsis
• C. Churning
• D. Acidity

Explanation: Mastication is the chewing process that breaks down food, increasing its surface area for enzyme action.

Q51. All of the following protect the body against entrance of germs except:

• A. Mucus membrane
• B. WBCs
• C. Ciliated cells
• D. RBCs✓

Explanation: RBCs have no role in immunity. RBCs (red blood cells) do not protect the body against the entrance of germs. RBCs are responsible for carrying oxygen to the tissues and carbon dioxide away from the tissues. They cannot engulf or kill germs.

Q52. The cells which play very important role in developing immunity are except:

• A. Monocytes
• B. Lymphocytes
• C. Neutrophils
• D. Thrombocytes✓

Explanation: Thrombocytes (platelets) are for clotting, not immunity. Immune defense relies on lymphocytes, neutrophils, and monocytes.

Q53. Antibodies are manufactured in:

• A. T lymphocytes
• B. Platelets
• C. Red blood cells
• D. B lymphocytes✓

Explanation: Antibodies are manufactured in B lymphocytes. B lymphocytes are a type of white blood cell that are responsible for the body's humoral immune response. The humoral immune response is the part of the immune system that fights off infection by producing antibodies. Antibodies are proteins that bind to specific antigens, which are molecules found on the surface of foreign invaders. When an antibody binds to an antigen, it triggers a cascade of events that ultimately leads to the destruction of the foreign invader. B lymphocytes are produced in the bone marrow. When a B lymphocyte encounters an antigen for the first time, it becomes activated. The activated B lymphocyte then divides and produces a clone of cells that all produce the same antibody. These cells are called plasma cells. Plasma cells secrete antibodies into the blood and lymph fluid.

Q54. Lymphocyte B and T have been named due to their:

• A. Relationship with bursa of Fabricius and thymus gland respectively✓
• B. Origin from bursa Fabricius and thymus, respectively
• C. Storage in bursa of Fabricius and thymus
• D. Destruction in bursa of Fabricius

Explanation: B cells were first discovered in the bursa of Fabricius (birds), and T cells mature in thymus.

Q55. Thymus gland provides immunological competence to:

• A. B-lymphocytes
• B. Antibodies
• C. T-lymphocytes✓
• D. Immunoglobulins

Explanation: The thymus gland is crucial for the maturation of T-lymphocytes, which are essential components of the adaptive immune system. In the thymus, T-lymphocytes undergo a process that equips them with the ability to recognize and respond to specific antigens, thus achieving immunological competence. On the other hand, B-lymphocytes are produced and mature in the bone marrow, where they differentiate independently of the thymus. Antibodies and immunoglobulins are produced by B-lymphocyte-derived plasma cells and do not directly involve the thymus in their production or function.

Q56. An antibody molecule consists of:

• A. Two identical light chains only
• B. Two identical heavy chains only
• C. Two identical light and two identical heavy chains✓
• D. Two identical light chains and two non-identical heavy chains

Explanation: An antibody molecule, also known as an immunoglobulin, is a Y-shaped protein that plays a critical role in the immune response by recognizing and binding to antigens. Its structure includes two identical light chains and two identical heavy chains, which are connected by disulfide bonds. This configuration allows antibodies to specifically bind to antigens. Option A is incorrect because it only mentions light chains. Option B is incorrect because it only mentions heavy chains. Option D is incorrect because it suggests the heavy chains are non-identical, which is not the case.

Q57. Given below is the ECG of a normal human. Which one of its components is correctly interpreted below?

• A. Complex QRS - one complete pulse
• B. Peak T initiation of total cardiac contraction
• C. Peak P and peak R together ­systolic and diastolic blood pressures
• D. Peak P­ initiation of left atrial contraction only✓

Explanation: P wave represents atrial depolarization, which initiates contraction of the atria.QRS complex represents ventricular depolarization, not “one complete pulse.”T wave represents ventricular repolarization, not initiation of total cardiac contraction.Blood pressures are not directly represented by peaks on the ECG.

Q58. If due to some injury, the chordae tendinae of the tricuspid valve of the human heart is partially non-­functional. What will be the immediate effect?

• A. The flow of blood into the aorta will be slowed down
• B. The 'pacemaker' will stop working
• C. The blood will tend to flow back into the left atrium
• D. The flow of blood into the pulmonary artery will be reduced✓

Explanation: The tricuspid valve is located on the right side of the heart and prevents backflow of blood from the right ventricle into the right atrium. If the chordae tendinae are partially non-functional, the tricuspid valve may not close properly, leading to regurgitation of blood from the right ventricle into the right atrium. This can result in a reduction of blood flow into the pulmonary artery, which carries blood from the right ventricle to the lungs.

Q59. Which two of the following changes (i-iv) usually tend to occur in the plain dwellers when they move to high altitudes (3,500 m or more)? (i) Increase in red blood cell size (ii) Increase in red blood cell production (iii) Increased breathing rate (iv) Increase in thrombocyte count Changes occurring are?

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

Explanation: (ii) Increase in red blood cell production: When individuals move to high altitudes, there is a decrease in the availability of oxygen. As a response, the body increases the production of red blood cells to enhance oxygen-carrying capacity and improve oxygen delivery to tissues. (iii) Increased breathing rate: At higher altitudes, the partial pressure of oxygen in the air is lower, leading to reduced oxygen uptake in the lungs. To compensate for this, the body increases the breathing rate, allowing more oxygen to be taken in and delivered to the tissues.

Q60. The fastest distribution of some injectable material/medicine and with no risk of any kind, can be achieved by injecting it into the:

• A. Capillaries
• B. Arteries
• C. Veins✓
• D. Lymph vessels

Explanation: Yes, the fastest distribution of an injectable material or medicine is achieved by injecting it into the veins. This is because veins are blood vessels that carry blood back to the heart. The blood in the veins is under pressure, which helps to push the medicine throughout the body quickly. Veins are blood vessels that carry blood back to the heart. They are typically thin-walled and have valves that prevent blood from flowing backwards.

Q61. Given below are four statements (i-­iv) regarding human blood circulatory system. (i) Arteries are thick-­walled and have narrow lumen as compared to veins. (ii) Angina is acute chest pain when the blood circulation to the brain is reduced. (iii) Persons with blood group AB can donate blood to any person with any blood group under ABO system. (iv) Calcium ions play a very important role in blood clotting. Which two of the above statements is/are correct?

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

Explanation: The correct statements (i)and (iv) are correct.(i) Arteries are thick-walled and have narrow lumen as compared to veins. Arteries are blood vessels that carry oxygenated blood away from the heart to various parts of the body. They have thick walls containing smooth muscle and elastic fibers, which allow them to withstand the high pressure generated by the heart's pumping action. Arteries also have a smaller lumen (the central opening through which blood flows) to maintain the blood pressure and ensure proper blood flow to the tissues. (iv) Calcium ions play a very important role in blood clotting.The statement about calcium ions is correct. Calcium ions play a crucial role in the process of blood clotting or coagulation. When there is an injury or damage to blood vessels, a series of complex reactions involving various clotting factors take place, ultimately leading to the formation of a blood clot to prevent excessive bleeding. Calcium ions are essential for the activation of clotting factors and the formation of a stable blood clot.

Q62. In a standard ECG, which one of the following alphabets is the correct representation of the respective activity of the human heart?

• A. P - start of systole
• B. T - end of diastole
• C. P - depolarization of the atria✓
• D. R - repolarization of the ventricles

Explanation: The P wave in a standard ECG represents the depolarization of the atria. This marks the electrical activity that causes the atria to contract and push blood into the ventricles. The P wave is a small, rounded wave at the beginning of the ECG tracing. It precedes the QRS complex, which indicates ventricular depolarization. The R wave is part of this QRS complex. Conversely, the T wave represents ventricular repolarization, occurring after the ventricles have contracted. Understanding the sequence of these waves is crucial for interpreting heart activity on an ECG.

Q63. Compared to blood, our lymph has:

• A. Plasma without proteins
• B. More WBCs and no RBCs✓
• C. More RBCs and less WBCs
• D. No plasma

Explanation: Lymph has more WBCs and no RBCs compared to blood. White blood cells (WBCs): WBCs are the main cells of the immune system. They help the body to fight off infection. Lymph contains a higher concentration of WBCs than blood does. Red blood cells (RBCs): RBCs are the cells that carry oxygen to the tissues. Lymph does not contain any RBCs.

Q64. During exercise the breathing rate may rise to:

• A. 15 times per minute
• B. 20 times per minute
• C. 30 times per minute✓
• D. 25 times per minute

Explanation: A rest, we inhale and exhale 15-20 times per minute. During exercise, the breathing rate may rise to 30 times per minute.

Q65. Which energy is consumed in breathing?

• A. Chemical✓
• B. Physical
• C. Potential
• D. Mechanical

Explanation: Over all process of breathing is active and consumes energy during inspiration in the form of ATP

Q66. Find out the incorrect statement:

• A. Inspired air contains more O2 than exhaled air:
• B. Expired air has 100 times greater CO2 as inspired air
• C. Amount of N2 is equal in both inhaled and exhaled air
• D. Exhaled air is comparatively drier than inhaled air✓

Explanation: Exhaled air has saturated water contents as compared to inhaled air which has variable water contents.

Q67. During inspiration fresh air moves in which has high percentage of:

• A. CO
• B. O2
• C. N2✓
• D. CO2

Explanation: Highest concentration in inhaled air and exhaled air is that of N2, that is 79%.

Q68. Partial pressure of oxygen is maximum in:

• A. Inspired air✓
• B. Alveolar air
• C. Expired air
• D. Oxygenated blood

Explanation: Oxygen is 21% in inhaled air and it is 16% in exhaled air.

Q69. Which of the following play passive role during breathing?

• A. Lungs✓
• B. Intercostal muscles
• C. Diaphragm
• D. Pleura

Explanation: The lungs are passive structures that expand and recoil due to the active contraction of muscles like the diaphragm and intercostal muscles, which create pressure changes for airflow during breathing.

Q70. Which of the following factor does not alter the rate of breathing by influencing the chemoreceptors in medulla oblongata, aorta and carotid artery?

• A. CO2 partial pressures in the blood
• B. H+ concentration in the blood
• C. O2 partial pressures in the blood
• D. Blood glucose level✓

Explanation: •The rate of breathing is primarily regulated by the respiratory centers in the medulla oblongata and pons, which respond to chemical signals from the chemoreceptors in the aorta and carotid arteries. These chemoreceptors are sensitive to changes in the levels of carbon dioxide (CO2) partial pressure, hydrogen ion (H+) concentration (pH), and oxygen (O2) partial pressure in the blood.•Blood glucose level, on the other hand, does not directly influence the rate of breathing through the chemoreceptors in the medulla oblongata, aorta, and carotid artery. Instead, blood glucose levels are primarily regulated by hormones such as insulin and glucagon, and they play a crucial role in maintaining glucose homeostasis in the body. However, glucose levels can indirectly affect breathing in cases of severe hypoglycemia or hyperglycemia, which may lead to altered levels of consciousness or metabolic imbalances that can impact respiratory function. But it is not a direct regulator of the rate of breathing through the chemoreceptors mentioned.

Q71. In passive immunity which of the following component are injected into blood?

• A. Antigens
• B. Immunogens
• C. Serum
• D. Immunoglobulins✓

Explanation: Immunoglobulins provide passive immunity as the body does not have to produce antibodies; instead, the injected antibodies offer a few months of protection.

Q72. Embryonic mass can generate all of the following except:

• A. Amnion
• B. Chorion✓
• C. Yolk sac
• D. Allantois

Explanation: The embryonic mass is the mass of cells inside the embryo that will eventually give rise to the definitive structures of the fetus. It cannot generate the chorion, which forms the placenta.

Q73. Gel electrophoresis is used for?

• A. Construction of recombinant DNA by joining with cloning vectors.
• B. Isolation of DNA molecules.
• C. Cutting of DNA into fragments.
• D. Separation of DNA fragments according to their size.✓

Explanation: Electrophoresis is a technique used for the separation of substances of different ionic properties. Since the DNA fragments are negatively charged molecules, they can be separated by allowing them to move towards the anode. DNA fragments move towards the anode according to the size of their molecules through the pores of agarose gel. Thus, the smaller fragments move farther away as compared to larger fragments.

Q74. The blood flow in milliliters/ minute during exercise to the skin is:

• A. 1500 ml
• B. 1600 ml
• C. 1800 ml
• D. 1900 ml✓

Explanation: Skin blood flow plays a major role in the distribution of thermal energy during exercise. When we exercise, our body temperature increases and carries the blood toward the skin’s surface, causing us to sweat and cool off. This natural body mechanism can lead to a flushed, red face, which can be especially more noticeable in fair-skinned individuals. The blood flow to the skin during exercise is 1900 ml/min. orMore oxygen is required by the cells as the rate of respiration increases, thus the blood flow to the skin increases as well.REFERENCE: KPK BIOLOGY BOOK

Q75. A single strand of nucleic acid tagged with a radioactive molecule is called?

• A. Vector
• B. Selectable marker
• C. Plasmid
• D. Probe✓

Explanation: Probes are single-stranded, radiolabelled molecules of nucleic acids with known sequence. The probes having sequence complementary to the gene to be identified are supplied. They bind with the particular gene segment. Radiation imaging identifies the location of that particular segment that binds with the probe. Probes are used as an identification tool in gene sequencing.

Q76. The enlarged lining epithelium cells connected with groups of developing spermatozoa in testes is:

• A. Somatic cells
• B. Sertoli cells✓
• C. Stem cells
• D. Totipotent cells

Explanation: Sertoli cells are the epithelial supporting cells of the seminiferous tubules. They are derived from the epithelial sex cords of the developing gonads.

Q77. Transgenic plants are the ones:

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

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

Q78. Hypothalamus is a part of:

• A. Diencephalon✓
• B. Myelencephalon
• C. Telencephalon
• D. Metencephalon

Explanation: The hypothalamus is a region of the forebrain located below the thalamus, forming the basal portion of the diencephalon.

Q79. Human arm is homologous with:

• A. Seal's flipper
• B. Octopus tentacle
• C. Bird's wing
• D. Both Seal's flipper and Bird's wing✓

Explanation: Anatomically, the human arm, the flippers of seals, and the wings of the birds are the forelimbs containing similar bones. These forelimbs have a common origin and similar anatomical features but have been modified to perform different functions. Flippers allow the seal to swim, birds fly using their wings and humans perform a variety of functions using the arms. Such structures are homologous to each other.

Q80. What is true about the Bt toxin?

• A. Bt protein exists as active toxin in the Bacillus
• B. The activated toxin enters the ovaries of the pest to sterilise it and thus prevent its multiplication
• C. The concerned Bacillus has antitoxins
• D. The inactive protoxin gets converted into active form in the insect gut✓

Explanation: Soil bacterium Bacillus thuringiensis produces proteins that kill certain insects like lepidopterans (tobacco budworm, armyworm), coleopterans (beetles), and dipterans (flies, mosquitoes). B. thuringiensis forms some protein crystals which contain a toxic insecticidal protein. This toxin does not kill the Bacillus (bacterium) because it exists as inactive protoxins in them. But, once an insect ingests it, it is converted into an active form of toxin due to the alkaline pH of the alimentary canal. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause swelling and lysis and finally cause the death of the insect.

Q81. A specific nudeotide sequence on DNA molecule to which RNA polymerase attaches to initiate transcription of mRNA from a gene is called:

• A. Poly genes
• B. Genome
• C. Promoter✓
• D. Pleiotropy

Explanation: The first step in transcription is initiation, when the RNA pol binds to the DNA upstream (5′) of the gene at a specialized sequence called a promoter.

Q82. The bone dissolving cells are called:

• A. Osteoclast✓
• B. Osteoblasts
• C. Osteocytes
• D. Fibroblast

Explanation: Osteoclasts are the bone-dissolving cells responsible for bone resorption. Derived from monocytes, they are large, multinucleated cells that break down bone tissue by secreting enzymes and acids, creating resorption pits. This process is vital for bone remodeling and calcium release into the bloodstream. Osteoblasts then fill these pits with new bone tissue, maintaining bone health.The other cell types serve different functions: Osteoblasts create new bone tissue, osteocytes maintain the bone matrix, and fibroblasts are involved in the production of connective tissue, not bone resorption.

Q83. Appendix is vestigial in man but may play role in:

• A. Digestion
• B. Excretion
• C. Immunity✓
• D. Movement

Explanation: The appendix has been found to play a role in mammalian mucosal immune function. It is believed to be involved in extrathymically derived T-lymphocytes and B-lymphocyte mediated immune responses and is also said to produce early defenses that help prevent serious infections in humans. Therefore the answer is C.

Q84. It circulates blood through a capsule in a nephron:

• A. Afferent Arteriole
• B. Peritubular Capillaries
• C. Efferent Arteriole
• D. Glomerulus✓

Explanation: The glomerulus is a tiny, ball-shaped cluster of blood vessels located within the renal cortex of the kidney. It is a critical component of the nephron, which is the functional unit of the kidney responsible for filtering blood and producing urine. Overall, the glomerulus plays a central role in the initial filtration of blood to create the glomerular filtrate. This filtrate undergoes further processing in the renal tubules, where essential substances are reabsorbed, and waste products are eliminated to produce urine. The glomerulus is a key component of the nephron's filtration system and is essential for maintaining overall body homeostasis.

Q85. The birds excrete:

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

Explanation: Birds excrete nitrogenous waste primarily in the form of uric acid. This adaptation is crucial for conserving water, as uric acid requires minimal water to be excreted, allowing birds to thrive in environments where water is scarce. In contrast, ammonia and urea require more water to be safely eliminated from the body, making them more suitable for aquatic animals and mammals, respectively. Acetic acid is not relevant to the excretory processes of birds.

Q86. Microvilli are also called:

• A. Leaf veins
• B. Cristae
• C. Capillaries
• D. Brush border✓

Explanation: Microvilli are small, finger-like projections on the surface of epithelial cells, especially in the small intestine, where they are known as the 'brush border'. This term comes from the appearance of microvilli under a microscope. The brush border increases the surface area for absorption. The other options are incorrect: 'Cristae' are related to mitochondria, 'Capillaries' are blood vessels, and 'Leaf veins' and 'Leaf midribs' are plant structures.

Q87. The structure that is specifically instrumental in the production of concentrated urine is:

• A. Cortical Nephron
• B. Juxtamedullary Nephron✓
• C. Counter Current Multiplier
• D. Restricted Supply of Water

Explanation: Juxtamedullary nephrons have distinct features and functions that make them well-suited for the production of concentrated urine. Juxtamedullary nephrons are primarily located in the inner part of the renal cortex and extend deep into the renal medulla. This positioning allows them to play a key role in regulating water balance and the concentration of urine. Juxtamedullary nephrons make up a smaller portion of the total nephrons in the kidney, but they play a critical role in maintaining water balance and regulating the concentration of urine. The combination of long loops of Henle and the countercurrent mechanism allows the kidney to conserve water when necessary, which is particularly important in maintaining overall body homeostasis, especially in situations of dehydration.

Q88. Which one of the following is a sex-linked inheritance?

• A. Baldness✓
• B. Albinism
• C. None of these
• D. Myopia

Explanation: Males with higher testosterone levels have a dominant baldness gene, whereas females with lower testosterone levels have a recessive baldness gene. So baldness has sex linked inheritance.

Q89. The valve between left ventricle and aorta is called:

• A. Semi lunar valve✓
• B. Blcuspid valve
• C. Tricuspid valve
• D. Pulmonary valve

Explanation: The term semilunar refers to the aortic valve and the pulmonary valve, tricuspid is the valve between the right atrium and ventricular and bicuspid between left atrium and ventricular. Hence the answer will be A.

Q90. Each organism of a species has assumed, in evolutionary history a specific set up of _ at various levels of organization suitable to its surroundings.

• A. Internal environment✓
• B. External environment
• C. Intracellular environment
• D. Intercellular environment

Explanation: Internal environment refers to the conditions, variables, and factors within an organism's body that are crucial for maintaining its health and proper functioning. This concept is closely related to the idea of homeostasis, which is the body's ability to maintain a stable and balanced internal environment despite external changes. The internal environment encompasses a wide range of physiological and biochemical parameters, including temperature, pH, blood pressure, nutrient levels, waste product concentrations, and more.

Q91. Select mineral that is considered as a macronutrient.

• A. Phosphorus✓
• B. Zinc
• C. Iron
• D. Iodine

Explanation: Calcium, sodium, magnesium, phosphorus and potassium are sometimes included as macronutrients because they are required in relatively large quantities compared with other vitamins and minerals.

Q92. Which of the following ions can act both as bronsted acid and base in solvent water?

• A. CN-
• B. SO42-
• C. HCO3-✓
• D. PO43-

Explanation: The correct answer is HCO3- because it can function as both a Bronsted acid and a Bronsted base in water. As a Bronsted acid, it donates a proton (H+) to form CO32-. As a Bronsted base, it accepts a proton to form H2CO3. The other options do not have this dual capability:CN-: Functions mainly as a Bronsted base by accepting protons.SO42-: Predominantly acts as a Bronsted base and is the conjugate base of a strong acid.PO43-: Primarily acts as a Bronsted base by accepting protons.

Q93. Molar extinction coefficient (ε) a constant in Beer- Lambert law is the characteristics of the:

• A. Solute
• B. Solvent
• C. Concentration
• D. All of the above✓

Explanation: Under defined conditions of solvent, pH, and temperature the molar absorption coefficient for a particular compound is a constant at the specified wavelength. Furthermore, it is used to calculate concentration using A = εLcWhere A is the amount of light absorbed by the sample for a particular wavelength,ε is the molar extinction coefficient, L is the distance that the light travels through the solution, and c is the concentration of the absorbing species per unit volume. Hence, it is a characteristic of concentration, as well as solute, and solvent.

Q94. The energy difference between adjacent energy levels of the hydrogen atom:

• A. Increases with increasing energy
• B. Decreases with increasing energy✓
• C. First increases and then decreases with increasing energy
• D. First decreases and then increases with increasing energy

Explanation: As energy levels increase, the energy difference between adjacent ones decreases due to the trend of decreasing energy from the energy equation for each Bohr orbit for hydrogen-like species

Q95. Choose reactants whose reaction product is ester:

• A. CH3COOH and CH3OCH3
• B. CH3COOH and C2H5OH✓
• C. CH3COOH and CH3CHO
• D. CH3COOH and CH3COCH3

Explanation: When primary alcohol is treated with a carboxylic acid in the presence of sulphuric acid a compound is formed. This compound has a sweet smell. The compound obtained is called an ester. The chemical reaction occurring in the formation of the ester is known as an esterification reaction.

Q96. Choose the IUPAC name of the following compound: CH3 |CH3 — CH — CH2 — CH = CH2

• A. 4- Methyl-1-Pentene✓
• B. 2- Methyl-3- Pentene
• C. 2- Methyl-2- Pentene
• D. 4,4-Dirnethyl-2-Pentene

Explanation: According to the IUPAC system of naming compounds, the methyl group is the substituent that will have secondary importance because the double bond is of primary importance. The correct IUPAC name is 4-Methyl-1-Pentene, where the double bond is prioritized. All other options do not follow the IUPAC rules for naming alkenes and identifying the parent alkene chain.

Q97. Which is the strongest acid?

• A. CH3COOH
• B. CH2ClCOOH
• C. CHCl2COOH
• D. CCl3COOH✓

Explanation: The acidity of a carboxylic acid is influenced by the presence of substituents that can stabilize the carboxylate ion. Electron-withdrawing groups, such as chlorine atoms, help stabilize the negative charge by dispersing it, thereby increasing acidity.In this question, CCl3COOH is the strongest acid because it has three chlorine atoms, which are highly effective electron-withdrawing groups. This makes the molecule more acidic than the others, which have fewer chlorine atoms or electron-donating groups like the methyl group in CH3COOH.

Q98. Choose the type of hybridization of carbon atoms in cyclopropane and the bond angle C–C–C.

• A. Sp3, 109.5*
• B. Sp3, 60*✓
• C. Sp2, 120*
• D. Sp2, 107*

Explanation: Cyclopropane has carbons with four bonds, hence they are SP3, but with bond distances less than normal alkanes and C-C-C bond angles of 60 degrees and H-C-H angles of 120 degrees. They are a special case as usually SP3 hybridization displays 109.5-degree angles

Q99. Hemiacetal containing both

• A. Alcohol and aldehyde functional groups
• B. Alcohol and ether functional groups✓
• C. Aldehyde and ether functional groups
• D. Alcohol and carboxylic acid functional groups

Explanation: Hemiacetal is a molecule made up of a core carbon atom connected to four groups: –OR, –OH, –R, and –H. Acetal is a molecule made of a core carbon atom that is attached to two –OR groups, a –R group, and a –H group. RHC(OH)OR' is the general formula for a hemiacetal.orHemiacetals are formed from the combination of alcohols and aldehydes and display at least one hydroxyl group from the alcohol and have an ether functional group with oxygen connecting 2 carbons

Q100. Carbylamine test is given by:

• A. Primary amines✓
• B. Secondary amines
• C. Tertiary amines
• D. All of these

Explanation: Only primary amines give the carbylamine test. Hoffmann's carbylamine test/ Isocyanide test:Aliphatic or aromatic primary amines on heating with chloroform and alcoholic KOH give foul smelling alkyl isocyanides or carbylamines. This test is not given by secondary or tertiary amines.

Q101. Which of the following compounds does not give iodoform test on reaction with I2 and NaOH?

• A. Propanone
• B. Ethanol
• C. Butanone
• D. Methanol✓

Explanation: It does not have methyl group attached to alpha carbon so it will not give idoform test

Q102. Which of the following substitutents is an Ortho and Para director and ring deactivating?

• A. –OH
• B. –NH2
• C. –Cl✓
• D. –OCH3

Explanation: In electrophilic aromatic substitution, the nature of the substituent on the benzene ring plays a crucial role in determining the position of new substituents and the reactivity of the ring. Ortho-para directors are groups that direct incoming electrophiles to the 2, 4 (ortho, para) positions. While most ortho-para directors are activating, -Cl is an exception as it directs due to resonance but deactivates due to a strong inductive effect. This means it reduces the overall reactivity of the benzene ring but still allows for substitution at ortho and para positions.In contrast, –OH and –NH2 groups are strong activators, and –OCH3 is moderately activating; all increase the reactivity of the benzene ring and direct electrophiles to ortho and para positions without deactivating the ring.

Q103. Which of the following compounds undergo nitration most readily?

• A. Benzene
• B. Toluene✓
• C. Benzoic acid
• D. Nitrobenzene

Explanation: The correct answer is Toluene. Toluene undergoes nitration much faster than benzene due to the presence of a methyl group. This group donates electron density to the ring, particularly enhancing the reactivity at the ortho and para positions, making the electrophilic aromatic substitution process more favorable.Why the other options are incorrect:Benzene: While benzene can undergo nitration, it lacks activating groups, making it less reactive compared to toluene.Benzoic Acid: The carboxyl group is a deactivating group, reducing electron density in the ring and hindering electrophilic attack.Nitrobenzene: The nitro group is a strong deactivator, making further nitration extremely difficult due to the electron-deficient ring.

Q104. N2(g) + 3H2 (g) ------> 2NH3(g), △H= 46.1 kj/moleFor the reaction above which statement is true about the equilibrium constant (Keq):

• A. Keq increases with an increase in temperature✓
• B. Keq decreases with an increase in temperature
• C. Keq decreases with an increase in pressure
• D. Keq increases with a decrease in pressure

Explanation: The value of △H is positive which means the reaction is endothermic and heat is absorbed. So if we increase the temperature, the equilibrium will favour the right hand side i.e more products will be formed.

Q105. Which of following functional groups are deactivating and not ortho, para directing?

• A. –R
• B. –COR✓
• C. –NH2
• D. NR2

Explanation: Ortho and para directing grps release electrons to the benzene ring increasing the chemical reactivity of the benzene ring towards electrophiles. Meanwhile meta directing grps withdraw the electrons from the benzene ring towards themselves decreasing their availability towards electrophiles. Therefore the answer will be B.

Q106. Primary amines on treatment with alkyl halide yield;

• A. Secondary amine
• B. Tertiary amine
• C. Quaternary ammonium salt
• D. Mixture of (a), (b) & (c)✓

Explanation: When primary amines are treated with alkyl halides, they undergo a series of nucleophilic substitution reactions. Initially, secondary amines are formed. If the reaction continues, tertiary amines can be generated, and with further reaction, quaternary ammonium salts are produced. Thus, the process often results in a mixture of secondary, tertiary amines, and quaternary ammonium salts, depending on the reaction conditions and stoichiometry. The mixture is due to the stepwise alkylation of the amine.Option A is incorrect because the reaction doesn't necessarily stop at the secondary amine stage. Option B is incorrect as it represents an intermediate, not the complete product mixture. Option C is also incorrect as it is just one of the potential products, not accounting for the others formed during the reaction progression.

Q107. Choose group that cause solubility of the dye in acids.

• A. –OH
• B. –NH2✓
• C. –SO2H
• D. –COOH

Explanation: Amines are used in making azo dyes and nylon apart from medicines and drugs. They are widely used in developing chemicals for crop protection, medication, and water purification. They also find use in products of personal care. Ethanol amines are the most common type of amine used in the global market.

Q108. What is the number of hydrogen atoms in 5 moles of water?

• A. 3.0115 × 1024
• B. 6.023 × 1024✓
• C. 6.023 x 1023
• D. 5.0 x 1023

Explanation: 5 moles of water contains 5 moles of H2, therefore contains 10 moles of H atoms.The number of atoms in one mole of a substance is given by Avogadro's constant, which is 6.02x1023.Thus the number of hydrogen atoms in 5 moles of water = 10x(6.02x1023) = 6.02x1024 H atoms.

Q109. In the main postulates of Bohr atomic theory the angular momentum of an electron in the hydrogen atom is given by the relationship.

• A. mvr = nh
• B. mvr = nh/2π✓
• C. mvr = n/2π
• D. mvr = nh/π

Explanation: The correct relationship for the angular momentum of an electron in a hydrogen atom according to Bohr's atomic theory is mvr = nh/2π, where:m is the mass of the electron,v is the velocity of the electron,r is the radius of the nth orbit,n is the principal quantum number, andh is Planck's constant.Option A is incorrect because it omits the division by 2π. Option C is correct as it matches Bohr's quantization condition. Option D is incorrect because it omits Planck's constant. Option D is incorrect as it uses π instead of 2π, altering the formula.

Q110. The reduction of aldehydes and ketones in the presence of zinc amalgam and HCl is termed as:

• A. Grignard reduction
• B. Clemmenson reduction✓
• C. Wolf-kishner reduction
• D. Friedel-craft reduction

Explanation: The Clemmensen reduction is a reaction that is used to reduce aldehydes or ketones to alkanes using hydrochloric acid and zinc amalgam. The Clemmensen reduction is named after a Danish chemist, Erik Christian Clemmensen.

Q111. Aiman in laboratory dissolve 4g of NaOH in 250ml of water. The molarity of this solution is:

• A. 0.4M✓
• B. 4M
• C. 0.2M
• D. 0.1M

Explanation: M = 4/40 / 250/1000 = 0.4,where 4g is the amount of NaOH, 40 is the grams per mole of NaOH, 250mL is the amount of water and 1000 is the number of ml in a liter.The most common way to express solution concentration is molarity (M), which is defined as the amount of solute in moles divided by the volume of solution in liters: M = moles of solute/liters of solution.

Q112. Which of the following is not a ferromagnetic material?

• A. Iron
• B. Nickel
• C. Cobalt
• D. Barium✓

Explanation: Barium is not a ferromagnetic material. Ferromagnetic materials have a strong magnetic response to an external magnetic field, and they can retain their magnetization even after the external field is removed. Iron, Nickel, and Cobalt are examples of ferromagnetic materials. Barium is a diamagnetic material, which means it is weakly repelled by both poles of a magnet. It does not exhibit the strong magnetic properties characteristic of ferromagnetic materials. Therefore, option d is correct.

Q113. Milk sugar is pasteurized by heating for 15 seconds at the temperature of:

• A. 60 oC
• B. 71 oC✓
• C. 50 oC
• D. 80 oC

Explanation: Pasteurization of milk, requires temperatures of about 63 °C (145 °F) maintained for 30 minutes or, alternatively, heating to a higher temperature, 72 °C (162 °F), and holding for 15 seconds.

Q114. Which one of the following is most ionic?

• A. NaCl
• B. MgCl2
• C. KCl✓
• D. AlCl3

Explanation: The correct answer is KCl because it forms the most ionic bond among the options provided. Potassium (K) is larger and has a lower charge density than sodium (Na), making it easier to lose its valence electron and form a K+ cation. According to Fajan's rules, larger cations with a lower charge density form more ionic bonds, which is why KCl is more ionic than NaCl. NaCl is ionic, but sodium's smaller size compared to potassium leads to a slightly higher charge density, resulting in less ionic character compared to KCl.MgCl2 has a +2 charge on magnesium, which increases the covalent character of the bond due to polarization effects.AlCl3 is more covalent due to aluminum's +3 charge, which causes significant polarization of the chloride ions. The ability of AlCl3 to form a dimer also supports its covalent nature.

Q115. The compound used in borax bead test for the detection of basic redicals to form colored bead is:

• A. H2BO2
• B. (C2H5)3BO3
• C. Ca2B6O115H2O
• D. Na2B4O710H2O✓

Explanation: Per one molecule of borax, there are 10 molecules of water of crystallization. Hence, metaborate is formed in the borax bead test.

Q116. Milk of magnesia is used for treatment of acidity in stomach, its formula is:

• A. Mg(OH)2✓
• B. MgSO4
• C. Ca(OH)2
• D. CaSO4

Explanation: Milk of magnesia is a suspension of magnesium hydroxide, with the chemical formula Mg(OH)2. It is an effective antacid that neutralizes stomach acid, providing relief from heartburn and indigestion. The reaction with stomach acid produces magnesium chloride and water, which reduces acidity. The other options, MgSO4, Ca(OH)2, and CaSO4, either have different uses or are not suitable for treating stomach acidity.

Q117. Which of the following ions play important role in the transportation of carbon dioxide?

• A. Sodium
• B. Potassium
• C. Bicarbonate✓
• D. Chloride

Explanation: The majority of carbon dioxide is transported as bicarbonate ions. This is because erythrocytes have a high activity of carbonic anhydrase, an enzyme that catalyzes the reaction between carbon dioxide and water.

Q118. Which of the following compound is assigned the octane number of 100?

• A. n-heptane
• B. n-octane
• C. 2,3,3-trimethylpentane
• D. 2,2,4-trimethylpentane✓

Explanation: The correct answer is 2,2,4-trimethylpentane, also known as isooctane. It is assigned an octane number of 100 because it has excellent knock resistance in engines. This compound serves as a standard reference point in the octane rating scale, indicating that it can withstand high compression without detonating prematurely. In contrast, n-heptane, with an octane number of 0, has very poor resistance to knocking. n-Octane has an even lower resistance, indicated by its negative octane number. 2,3,3-Trimethylpentane, while having a high knock resistance, does not serve as the standard reference and has a different octane rating.

Q119. The major product of acid catalysed dehydration of 3- pentanol is:

• A. 1-Pentene
• B. 3-Pentene✓
• C. 2-Methylbutane
• D. 3-Methylbutane

Explanation: The major product is 2-pentene because it is the most stable alkene formed after carbocation rearrangement and follows Zaitsev's rule. The reaction proceeds through an E1 mechanism involving a carbocation intermediate and rearrangement.

Q120. Which of the following compound will react most readily with bromine in CCl4?

• A. 1-pentene
• B. 2-pentene
• C. 2-Methyl-1-butene
• D. 3-Methyl-1-butene✓

Explanation: The reaction of alkenes with bromine proceeds through the formation of a carbocation intermediate. The stability of these carbocations is determined by the number of alkyl substituents; more substituents lead to greater stability. Among the given options, 3-methyl-1-butene forms the most stable carbocation due to its tertiary structure, resulting in the fastest reaction with bromine in CCl4. In contrast, 1-pentene forms the least stable primary carbocation, leading to the slowest reaction, while both 2-pentene and 2-methyl-1-butene form secondary carbocations that are more stable than that of 1-pentene but less stable than that of 3-methyl-1-butene, making them less reactive than the correct answer.

Q121. Select the strongest acidThe pKa values are given:

• A. HI, pKa=-10✓
• B. HCN, pKa=9.4
• C. H2SO4, pKa = 1.8
• D. HNO3, pKa = 3.0

Explanation: The strength of an acid is determined by its ability to donate protons, which is quantified by its pKa value. A lower pKa signifies a stronger acid. Hydroiodic acid (HI) with a pKa around -10 is extremely strong, far exceeding the strength of sulfuric acid (H2SO4) with a pKa of 1.8. Hydrocyanic acid (HCN) has a high pKa of 9.4, indicating it is a weak acid. Nitric acid (HNO3) has a pKa of 3.0, making it stronger than HCN but weaker than both H2SO4 and HI. Therefore, considering the correct pKa values, HI is the strongest acid.

Q122. In which of the following pairs are both substances normally crystalline?

• A. Copper and diamond✓
• B. Copper and glass
• C. Copper and rubber
• D. Diamond and glass

Explanation: Diamond is a crystal that is transparent to opaque and generally isotropic, native copper is a polycrystal. Therefore the answer will be A.

Q123. Which one of the following is strongest acid?

• A. CH3COOH
• B. CH3CH2COOH
• C. C6H5CG2COOH
• D. FCH2COOH✓

Explanation: D is the correct option, as it contains a halogen which have strong electron-withdrawing effects.

Q124. The total energy of a hydrogen atom in its ground state is:

• A. Zero
• B. Negative✓
• C. Positive
• D. None of the above

Explanation: The energy of a hydrogen atom in its ground state is negative, approximately -13.6 eV. This negative value indicates that the electron is bound to the nucleus, requiring energy input to free it. The Schrödinger equation and the Bohr model both support this result, with the ground state corresponding to the principal quantum number n = 1. The other options are incorrect: a zero or positive energy would suggest the electron is not bound to the nucleus, contrary to the nature of a ground state electron.

Q125. The smaller the value of pKb:

• A. The weaker the base
• B. The stronger the base✓
• C. The stronger the acid
• D. None of the above

Explanation: The pKb value is the negative logarithm of the base dissociation constant, Kb. It quantifies the strength of a base in a solution. A smaller pKb value reflects a larger Kb, signifying that the base dissociates more completely, thus being stronger. Therefore, Option B is correct. Option A is incorrect because it misinterprets the relationship; Option C is incorrect as it confuses the properties of acids and bases; and Option D is incorrect since Option B provides the accurate explanation.

Q126. Phosphodiester linkage is formed between.

• A. Two nucleotide bases
• B. Amino acid
• C. Two sugar
• D. Nucleotides and phosphates✓

Explanation: In the polynucleotide, adjacent nucleotides are joined by a phosphodiester linkage, which consists of a phosphate group that links the sugars of two nucleotides. This bonding results in a repeating pattern of sugar-phosphate units called the sugar-phosphate backbone. (Note that the nitrogenous bases are not part of the backbone.) Complementary nitrogenous bases are held together by hydrogen bonds.

Q127. The shape of SnCl2 is:

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

Explanation: SnCl2 (Tin dichloride) has a bent or V-shaped geometry, which is an example of an angular or non-linear shape. The molecule has a central tin atom that is bonded to two chlorine atoms. The two bond pairs of electrons repel each other, resulting in a bent shape. The bond angle is approximately 119 degrees. This bent shape is due to the lone pair of electrons present on the tin atom. The lone pair occupies a greater amount of space compared to the bond pair, resulting in the distortion of the molecule. In contrast, the linear shape is a molecular geometry in which the atoms of the molecule are arranged in a straight line, like in HCl or CO2. Tetrahedral shape is a molecular geometry that has four atoms attached to a central atom, like in CH4 or CCl4. Trigonal shape is a molecular geometry that has three atoms attached to a central atom, like in BF3.

Q128. Which is not true about Grignard reagent?

• A. They are highly reactive compounds.
• B. They are very stable compounds and can be isolated easily.✓
• C. They have synthetic importance.
• D. They are represented by the general formula RMgX.

Explanation: The correct answer is They are very stable compounds and can be isolated easily. This statement is false because Grignard reagents are highly reactive and sensitive to moisture and air, which makes them unstable and difficult to isolate. They are usually prepared and used immediately in situ. Conversely, the other options are true: Grignard reagents are highly reactive, important for synthetic applications, and are represented by the formula RMgX.

Q129. Select the true statement about the amorphous solids:

• A. Amorphous substances have a sharp melting point
• B. Amorphous substances do not have a fixed melting point✓
• C. Amorphous substances have proper geometrical shapes
• D. The particles in amorphous substances are arranged in an orderly manner

Explanation: Amorphous solids, unlike crystalline solids, do not have a fixed melting point. This is because they lack the ordered structure that characterizes crystalline materials. Instead, they soften gradually over a range of temperatures. For example, glass softens before melting completely. Amorphous solids have a disordered atomic or molecular structure, which results in their unique properties such as isotropic behavior and lack of sharp melting points. Option B correctly identifies this characteristic, whereas the other options incorrectly attribute properties of crystalline solids to amorphous ones.

Q130. Both NaNO3 an CaCO3 crystallize in Rhombohedral forms therefore they are:

• A. Allotropes
• B. Polymorphous
• C. Isomorphous✓
• D. None of these

Explanation: Two or more substances having the same crystal structure called isomorphous.

Q131. Pure water freezes at 0 oC and boils at 100 oC at standard conditions. Calcium chloride was added to pure water. What do you expect about its freezing point and boiling point?

• A. No change in its freezing point and boiling point
• B. Freezing point increases and boiling point decreases.
• C. Freezing point increases and boiling point increases
• D. Freezing point decreases and boiling point increases✓

Explanation: The presence of impurities in a substance lowers the melting point but increases the boiling point of the substance.

Q132. Equal volume of different gases under same condition of temperature and pressure contain the same number of particles. The above statement is of:

• A. Avogadro's Law✓
• B. Graham's Law
• C. Dalton's Law
• D. Hund's Rule

Explanation: The correct answer is Avogadro's Law. This fundamental principle states that equal volumes of gases, at the same temperature and pressure, will contain an equal number of molecules. This is pivotal for understanding gas behavior in chemistry and physics.Graham's Law is concerned with the diffusion and effusion rates of gases, which depend on molecular weight, not volume and particle number.Dalton's Law explains the pressure contributions of individual gases in a mixture, but does not relate to volume and molecule count.Hund's Rule pertains to electron occupation in atomic orbitals, which is irrelevant to gas laws and molecular volume relationships.

Q133. Sodium chloride crystal structure is:

• A. Hexagonal
• B. Body centered cubic
• C. Face centered cubic✓
• D. Tetragonal

Explanation: Explanation diagram is given below.

Q134. An acid is a substance which accepts:

• A. An electron pair✓
• B. A proton
• C. An electron
• D. Pair of protons

Explanation: The correct answer is that an acid is a substance which accepts an electron pair, according to the Lewis concept. This concept broadens the traditional definition of acids beyond the Bronsted-Lowry theory, which focuses on proton donation. Thus, Option A is correct. Option B is incorrect as it describes the action of acids in the Bronsted-Lowry concept, which involves donating, not accepting, a proton. Option C is incorrect because it misinterprets the Lewis concept, emphasizing that acids accept electron pairs instead of single electrons. Option D is incorrect as no accepted theory or concept describes acids as accepting pairs of protons; the focus remains on single proton donation in the Bronsted-Lowry framework.

Q135. Which is not correct about polyvinyl chloride?

• A. It is used in large scale production of cable insulator
• B. It is a copolymer✓
• C. It is a homopolymer
• D. It is used in the manufacturing of pipes

Explanation: b) It is a copolymer:This option states that polyvinyl chloride is a copolymer. This statement is not correct. PVC is not a copolymer; it is a homopolymer. A homopolymer is a type of polymer that is composed of repeating units of a single monomer. In the case of PVC, the repeating unit is vinyl chloride (CH2=CHCl).

Q136. What are the units of the ratio: specific latent heat of vaporization of water/ specific heat capacity of water

• A. K✓
• B. 1/K
• C. K2
• D. No unit

Explanation: K is the units of the ratio: specific latent heat of vaporization of water/ specific heat capacity of water .

Q137. Moon is revolving around the earth, work done by the earth on the moon is

• A. Positive
• B. Zero✓
• C. Negative
• D. None of these

Explanation: B. The formula for work done on a body is; W = F × d cos θWhere,F = the force of gravitation applied by the earth.d = displacement between the two bodiesθ = the angle between F and dWhen the moon revolves around the earth, there is a 90° angle between the displacement of the moon and the gravitational force exerted by the earth.Putting θ = 90° in the equation,W = F × d cosBut, Cos 90° = 0So, W = 0

Q138. Spectral lines is like a _ of absorbed or emission energy in a spectrum

• A. Charged pattern
• B. Fingerprint pattern✓
• C. Discharged pattern
• D. None of these

Explanation: Fingerprints are unique to each individual, making them excellent for identification. Similarly, spectral lines serve as unique identifiers for elements. Each element has a specific set of spectral lines that act like a 'fingerprint,' allowing scientists to determine the composition of stars and galaxies. Other options, such as 'charged pattern' and 'discharged pattern,' fail to capture the critical aspect of spectral lines being distinct and identifiable for different elements.

Q139. In stationary waves:

• A. There is no net transfer of energy✓
• B. Energy is evenly distributed at all points
• C. Phase is the same for all points
• D. Both (a) & (b)

Explanation: In stationary waves, there is no net transfer of energy because the waves are confined, oscillating back and forth within a fixed region. This causes energy to transfer between nodes and antinodes without progressing through the medium. The correct option, therefore, is that there is no net transfer of energy. Option B is incorrect because energy is not evenly distributed; it varies with position. Option C is incorrect because the phase is not uniform; it changes across the wave. Option D is incorrect because it inaccurately suggests both A and B are correct.

Q140. Michelson interferometer can be used to measure:

• A. Speed of sound
• B. Wavelength of sound
• C. Speed of light
• D. Wavelength of light✓

Explanation: Michelson's Interferometer uses the phenomena of interference to measure the wavelength of light from a source.

Q141. If each vector have unit magnitude than A. A is:

• A. Zero
• B. One✓
• C. Negative One
• D. Two

Explanation: The correct answer is 'One'. When two vectors each have unit magnitude, the dot product is calculated as the cosine of the angle between them. If the vectors are aligned, the angle is 0 degrees, and the cosine of 0 degrees is 1. Thus, the dot product is 1. Options 'Zero', 'Negative One', and 'Two' are incorrect because they do not reflect the properties of the dot product for unit vectors.

Q142. In a purely resistive circuit the current:

• A. Leads the voltage by one-half of a cycle
• B. Leads the voltage by one-fourth of a cycle
• C. Lags the voltage by one-half of a cycle
• D. Is in phase with the voltage✓

Explanation: In a purely resistive circuit, the current and voltage are in phase with each other. This means both reach their maximum and minimum values simultaneously, with no phase shift between them.The reason is that resistance, unlike reactance, does not cause any delay or advancement in the phase of the current relative to the voltage. Therefore, Options A, B, and C are incorrect as they suggest a phase difference, which does not occur in purely resistive circuits. Option D is correct as it accurately describes the in-phase relationship, where the electrical energy is entirely converted into heat without any storage or delay.

Q143. The number of ejected photoelectrons increases with increase.

• A. Increase in intensity of light✓
• B. Increase in wavelength of light
• C. Increase in frequency of light
• D. Never

Explanation: The correct answer is that the number of ejected photoelectrons increases with an increase in the intensity of light. In the photoelectric effect, the intensity of light is proportional to the number of photons hitting a surface per unit time. Therefore, higher intensity means more photons are available to eject more photoelectrons. While frequency and wavelength affect the energy per photon and the ability to eject photoelectrons, they do not directly influence the number of electrons ejected. Thus, increasing the intensity leads to more ejected photoelectrons, assuming the frequency is above the threshold required to overcome the material's work function.

Q144. The number of ejected photoelectrons increases with increase. is strong electrolyte?

• A. Ca(OH)2
• B. SiCl4
• C. KCl✓
• D. SrCl2

Explanation: KCl is a strong electrolyte because it fully dissociates into ions (K+ and Cl-) in water, allowing efficient conduction of electricity. Ca(OH)2 is a weak base and weak electrolyte, which means it does not fully dissociate into ions. SiCl4 does not dissociate at all in water, making it a non-electrolyte. SrCl2 was incorrectly described; it is a strong electrolyte as it also fully dissociates in solution.

Q145. Becquerel is the unit of:

• A. Decay constant
• B. Half life
• C. Mean life
• D. Activity✓

Explanation: The Becquerel (Bq) is the standard unit for measuring radioactivity, specifically the activity of a radioactive substance. It quantifies the number of radioactive decays per second, making it a measure of the substance's activity. This is distinct from the decay constant, half-life, and mean life, which relate to probabilities and time intervals rather than direct decay rates.Option A, the decay constant, is a probability measure indicating decay likelihood; Option B, half-life, is a time measure for decay processes; and Option C, mean life, is an average time measure for decay. Only Option D, activity, correctly identifies what is measured by Becquerels, highlighting the rate of decay events per second.

Q146. A total charge of 100C flows through a 12W bulb in a time of 50 second. What is the potential difference across the bulb during this time?

• A. 0.12V
• B. 2.0V
• C. 6.0V✓
• D. 24V

Explanation: The correct answer is 6.0V, determined by using the formula P = VI, where P is power, V is voltage, and I is current. Given the power of the bulb is 12W and the current is calculated from the charge and time as 2A (100C / 50s), we find V = 12W / 2A = 6V. This satisfies the power equation perfectly.Option A (0.12V) is incorrect because it implies an unrealistically low power. Option B (2.0V) is incorrect because the resulting power equation doesn't match the conditions. Option D (24V) is incorrect because it suggests a current that would only transfer 25C, not the given 100C.

Q147. The changing electric flux in a certain region of space produces:

• A. An electric field
• B. Magnetic field
• C. Both A and B✓
• D. None of the above

Explanation: Faraday's law of electromagnetic induction states that a changing electric flux will result in the induction of a magnetic field. This principle is fundamental in electromagnetism, describing how electric and magnetic fields are interrelated. The correct choice is that both a magnetic and an induced electric field can result from changing fluxes, but not directly from changing electric flux alone. Therefore, the correct answer is that a changing electric flux produces a magnetic field, with the potential to also indirectly influence electric fields through electromagnetic induction processes. Options A and D are incorrect because they don't accurately describe the relationship between electric flux and the fields it influences.

Q148. A hydrogen atom that has lost its electron is moving east in a region where the magnetic field is directed from south to north. It will be deflected:

• A. Up✓
• B. Down
• C. North
• D. South

Explanation: The hydrogen atom, which is effectively a proton in this context since it has lost its electron, is moving east in a magnetic field directed from south to north. To find the direction of the magnetic force acting on the proton, we can use the right-hand rule. For a positive charge like a proton, point your right thumb in the direction of the velocity (east), and your fingers in the direction of the magnetic field (north). The direction in which your palm pushes will indicate the direction of the magnetic force, which is upwards in this case. Therefore, the correct answer is up.Option B (down) is incorrect because the right-hand rule does not support this direction. Option C (north) is incorrect as the force does not act along the direction of the magnetic field but rather perpendicular to it. Option D (south) is also incorrect for the same reason; the magnetic force is not directed south based on the right-hand rule.

Q149. The colour of thin films is a result of:

• A. Dispersion
• B. Absorption of light
• C. Scattering of light
• D. Interference of light✓

Explanation: The correct answer is interference of light. The colors seen in thin films result from the interference of light waves reflected from the top and bottom surfaces of the film. Depending on the thickness of the film and the wavelength of the incoming light, certain wavelengths constructively interfere, while others destructively interfere, creating the vivid colors observed. This is different from dispersion, absorption, or scattering, which do not account for the distinct color patterns seen in thin films.

Q150. Water flows from a 6.0cm diameter pipe into 8.0cm diameter pipe. The speed in the 6.0cm pipe is 5.0m/s. the speed in the 8cm pipe is:

• A. 2.8m/s✓
• B. 3.7m/s
• C. 6.6m/s
• D. 8.8m/s

Explanation: The problem can be solved using the continuity equation, which states that A1V1 = A2V2, where A is the cross-sectional area of the pipe. For circular pipes, A = πr2, so:For the 6.0 cm diameter pipe, r1 = 3 cm, and for the 8.0 cm diameter pipe, r2 = 4 cm.Substituting these into the equation:π(3)2 × 5.0 = π(4)2 × V245 = 16V2V2 = 45/16 = 2.8 m/sThe correct answer is 2.8 m/s. The other options are incorrect as they do not satisfy this calculation based on the continuity equation.

Q151. From the following, which one particle belongs to lepton group:

• A. Neutrinos✓
• B. Protons
• C. Neutrons
• D. Mesons

Explanation: Neutrinos are subatomic particles that belong to the family of elementary particles known as leptons. They are among the most abundant particles in the universe and play a crucial role in various astrophysical and particle physics phenomena.Key characteristics of neutrinos include:1. Electrically neutral: Neutrinos have no electric charge, which means they do not interact with electromagnetic forces. They are only influenced by weak nuclear forces and gravity.2. Very low mass: Neutrinos have extremely low mass compared to other elementary particles. While their exact masses are not yet precisely determined, they are much lighter than electrons and other charged particles.3. Nearly massless and relativistic: Neutrinos are known to travel very close to the speed of light, which makes them nearly massless and relativistic.4. Three types (flavors): There are three known types or flavors of neutrinos: electron neutrinos (νe), muon neutrinos (νμ), and tau neutrinos (ντ). Each type is associated with a specific charged lepton (electron, muon, and tau, respectively).5. Oscillation: Neutrinos can undergo a phenomenon called neutrino oscillation, where they can change from one flavor to another as they travel through space. This discovery proved that neutrinos have mass and provided insight into the fundamental properties of these elusive particles.Neutrinos are produced in various astrophysical processes, such as nuclear reactions in the Sun and other stars, supernovae explosions, and cosmic ray interactions. They can travel vast distances through space without much interaction with matter, making them challenging to detect.Neutrino detectors are specially designed to capture and study neutrinos. They use large volumes of materials, such as water or liquid argon, to detect the very rare interactions of neutrinos with other particles. Neutrino research provides valuable insights into the fundamental nature of matter, the early universe, and astrophysical phenomena that involve high-energy processes.Lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions .There are six leptons in the present structure, the electron, muon, and tau particles and their associated neutrinos.The rest are Hadrons. Hadron is a subatomic particle of a type including the baryons and mesons, which can take part in the strong interaction.

Q152. An electron in a hydrogen atom makes a transition from an energy level with energy E1, to one with energy E2 and simultaneously emits a photon. The wavelength of the emitted photon is:

• A. hc/(E1 – E2)✓
• B. h/(E1 – E2)
• C. h/c(E1 – E2)
• D. (E1 – E2)/hc

Explanation: But as per ptb, the energy change is given by ∆E=E2-E1∆E=hv∆E=hc/λ E2- E1=hc/λ λ =hc/E2- E1 Ans

Q153. A student measures a current as 0.05 A. Which of the following correctly expresses this result?

• A. 50 mA✓
• B. 50 µA
• C. 500 mA
• D. 500 µA

Explanation: The current of 0.05 A can be expressed as 50 mA because 'milli' means one thousandth, making 0.05 A equivalent to 50 milliamperes. Option A is correct. Option B (50 µA) and Option D (500 µA) are incorrect because microamperes are much smaller than milliamperes. Option C (500 mA) is incorrect because it represents a larger current than 0.05 A.

Q154. On the ground, the gravitational force on a satellite is W. What is the gravitational force on the satellite when at a height R/50, where R is the radius of the earth?

• A. 1.04W
• B. 1.02W
• C. 0.50W
• D. 0.96W✓

Explanation: Gravitational force has formula,F=GMm/R2Where G is constant, M is mass of earth ,m is mass of object and R is distance between earth's center and objectOn ground Force= GMm/R2Because distance equals to radius of earthIn given orbit,Force=GMm/(R+R/50)²Taking LCM of R+R/50, we get 51R/50GMm/(51R/50)²GMm/(2601 R²/2500)= GMmx2500/2601R²Equating both these equations:GMm x 2500/2601R²=GMm/R2All the constants will be canceled and we will be left with;2500/2601 = 0.96WSo right option will be D

Q155. In the nuclear reaction shown below what is the value of the coefficient?92U235+ 0n1_ = 56Kr89+γon1 + 200MeV

• A. 0
• B. 1
• C. 2
• D. 3✓

Explanation: In the fission reaction, uranium absorbs a neutron and forms a barium and krypton nucleus which are extremely unstable and they instantaneously release three neutrons between themselves becoming barium144 and krypton89, therefore the answer is option D.Option A is incorrect because a neutron is always produced in fission reactions. Option B is incorrect because fission reactions usually release around 2-3 neutrons, not just one. Option C is incorrect because the fission of U-235 typically releases 2 or 3 neutrons, not just two.

Q156. The vectors A and B are such that |A + B| = |A – B|, then the angle between the two vectors is:

• A. 00
• B. 600
• C. 900✓
• D. 1800

Explanation: Consider the angle between A and B as ∅. The resultant of A + B vector is R = √(A²+B²+2AB cos∅) and the resultant of A - B vector is R = √(A²+B²-2AB cos∅). Given |A+B| = |A-B|, we can equate the expressions and solve for cos∅ to find ∅ = 90°. This is the correct angle between the vectors based on the provided equation.The other options are incorrect because they do not align with the properties of vector magnitudes and the equation given. Option A and D provide angles that are not supported by the equation, and Option B is based on a misinterpretation of the cosine function in this context.

Q157. Which among the following phenomenon shows particle nature of light?

• A. Photoelectric effect✓
• B. Interference
• C. Polarization
• D. Matter waves

Explanation: The photoelectric effect occurs when light ejects electrons from a metal surface, which can only be explained if light consists of particles (photons) with discrete energy packets. This confirms the particle nature of light as explained by Einstein's photon theory.

Q158. For ideal step up transformer Ps_Pp.

• A. Equal to✓
• B. Greater than
• C. Less than
• D. None of these

Explanation: Correct. In an ideal step-up transformer, there is no power loss, meaning the power input (Pp) equals the power output (Ps). This follows from the law of energy conservation, which states that energy cannot be created or destroyed.

Q159. The resistance of a conductor at absolute zero (0K) is:

• A. Almost zero✓
• B. Almost infinite
• C. No prediction at all
• D. May increase or decrease

Explanation: The resistance of a conductor at absolute zero (0K) is almost zero. This phenomenon is known as zero or near-zero resistance, and it occurs in materials with special properties called superconductors.Superconductors are materials that exhibit zero electrical resistance below a critical temperature. At temperatures close to absolute zero, the thermal energy becomes extremely low, and the probability of electron-impurity or electron-phonon scattering is greatly reduced. As a result, the resistance of the superconductor diminishes significantly, approaching zero.

Q160. During adiabatic expansion internal energy decreases by 2J, then work done in this process is:

• A. 2J✓
• B. 1J
• C. -1J
• D. -2J

Explanation: The work done during adiabatic expansion can be calculated using the First Law of Thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). Since this is an adiabatic process, there is no heat added or removed from the system (Q=0). Therefore, ΔU = -2 J (as the internal energy decreases by 2J).Thus, W = -ΔU = -(-2 J) = 2 J. The rest options have incorrect values

Q161. If a system changes from a state P1V1 to P2V2 by two paths then the quantity which remains unchanged is _.

• A. dQ
• B. dW
• C. dQ-dW✓
• D. dQ+dW

Explanation: The quantity that remains unchanged is known as a state function, which depends only on the initial and final states of the system and is independent of the path taken between those states. The only state function among the given options is dQ - dW.dQ is the heat added to the system, and dW is the work done on the system. The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done on the system, i.e., ΔU = dQ - dW. Thus, the quantity dQ - dW is equal to the change in internal energy, which is a state function.

Q162. The radiations all around us are called _ radiations.

• A. Natural
• B. Artificial
• C. Both A or B✓
• D. No radiation is around us

Explanation: Humans come into contact with radiation on a daily basis. These radiations are both natural, such as the rays from the sun, and artificial, such as those from our TV screens or kitchen stoves. This radiation that we experience on a daily basis is known as background radiation

Q163. A charged capacitor stores 10 C at 40 V.. Its stored energy is:

• A. 400 J
• B. 4 J
• C. 0.2 J
• D. 200 J✓

Explanation: The energy stored in a capacitor can be calculated using the formula U = 0.5 * Q * V, where U is the energy, Q is the charge, and V is the voltage. Given Q = 10 C and V = 40 V, the calculation is U = 0.5 * 10 C * 40 V = 200 J. Therefore, the correct answer is 200 J. The other options (400 J, 4 J, and 0.2 J) are incorrect because they do not follow the correct application of the formula.

Q164. A moving charged particle is surrounded by?

• A. 1 field
• B. 3 fields✓
• C. 2 fields
• D. 4 fields

Explanation: A moving charged particle is surrounded by three fields:Electric Field: This field arises due to the charge of the particle itself and affects other charges in its vicinity.Magnetic Field: Generated due to the movement of the charge. The interaction between charge and motion creates this field.Gravitational Field: The mass of the particle generates a gravitational field, albeit usually very weak compared to the other two fields.Option A (1 field) is incorrect because it ignores the presence of the magnetic and gravitational fields. Option C (2 fields) is incorrect as it overlooks the gravitational field. Option D (4 fields) is incorrect because there isn't an additional field present beyond the electric, magnetic, and gravitational fields mentioned.

Q165. Balmer series lies in that region of electromagnetic wave spectrum, which is known as:

• A. Visible region✓
• B. Ultraviolet region
• C. Invisible region
• D. lnfra-red region

Explanation: The Balmer series corresponds to the set of spectral lines in the emission spectrum of hydrogen atoms that result from electron transitions within the atom involving the principal quantum number "n" ranging from 3 to higher levels down to n = 2. These transitions result in the emission of photons with specific energies. The Balmer series is located in the visible region of the electromagnetic spectrum, which ranges from approximately 400 to 700 nanometers (nm). It includes colors that are visible to the human eye.

Q166. Which one of the following pair is of isobars?

• A. 126 C and 146 C
• B. 614 C and 714N✓
• C. 42 He and 31H
• D. Both A and C

Explanation: Isobars have different atomic number but same atomic mass. This option clearly shows elements having same atomic mass but different atomic number.

Q167. If two cars are moving with velocity 10 m/s and 5m/s in opposite direction to each other, then their relative velocity with respect to one another will be:

• A. 5m/s
• B. 10m/s
• C. -5m/s
• D. 15m/s✓

Explanation: The relative velocity of the two cars with respect to one another can be found by adding their velocities. Since they are moving in opposite directions, we need to subtract the velocity of one car from the other: Relative velocity = velocity of first car - velocity of second car = 10 m/s - (-5m/s) = 10 m/s + 5 m/s = 15 m/s

Q168. A.C and D.C have the same:

• A. Effect in charging battery✓
• B. Effect in charging capacitor
• C. Heating effect through a resistance
• D. Effect passing through an inductance

Explanation: The correct option is Effect in charging battery. Both AC and DC can charge a battery, but typically AC is converted to DC for this purpose. The important factor is that the charging mechanism manages the voltage and current appropriately. While AC and DC currents have different characteristics, the end result of charging the battery—restoring its stored energy—can be achieved with both, given the right conditions.Option B is incorrect because capacitors react differently to AC and DC; they charge and discharge continuously with AC. Option C is incorrect because while both AC and DC cause heating, the effect depends on how the AC is measured (RMS value). Option D is incorrect as inductances behave differently under AC, where they induce a current due to the changing magnetic field, unlike the constant field from DC.

Q169. X-rays are widely used as a diagnostic tool in medicine because of its:

• A. Particle property
• B. Cost of X-ray unit is low
• C. High penetrating power✓
• D. It is not electromagnetic waves

Explanation: X-rays are a type of electromagnetic radiation with high frequency and energy, which grants them significant penetrating power. This characteristic is essential for medical diagnostics, as it allows X-rays to pass through the body and produce images of bones and other internal structures. This ability to create detailed images is why they are widely used in medicine.Options A and D are incorrect because X-rays are electromagnetic waves, not particles. Option B is misleading as the cost of X-ray machines is not low; their diagnostic utility justifies their expense.

Q170. A radioactive isotope 'W' decays to 'X' by beta emission which decays to 'Y' by alpha emission and 'Y' decays to 'Z' by alpha emission.What is the change in the atomic number from 'W' to 'Z'?

• A. Increase by 3
• B. Increase by 5
• C. Decrease by 3✓
• D. Decrease by 5

Explanation: When W emits beta radiations, it results in increase of +1 in atomic charge . Further when X emits alpha radiation, it results in decrease of atomic charge by 2 . Then further when Y emits alpha radiations, it results in decrease of atomic charge by further 2. Therefore, from the original atomic charge of W, there was a decrease of a total of 3 in atomic charge.

Q171. Wave nature and particle nature of photon is linked by:

• A. Rest mass of photon
• B. Wavelength of photon
• C. Light speed
• D. Momentum of photon✓

Explanation: Particle nature and wave nature of electromagnetic waves and electrons is called dual nature or wave particle dualism. According to this wave particle dualism, a wave can also exhibit particle behavior and at the same time, a particle can also exhibit wave nature. Momentum of a photon is considered in both the particle and wave nature of a photon.

Q172. In the following question, various terms of an alphabet sequence are given with one or more terms missing as shown by (?). Choose the missing term.Z, ?, T, ?, N, ?, H, ?, B

• A. W, Q, K, E✓
• B. W, R, K, E
• C. X, Q, K, E
• D. X, R, K, E

Explanation: Step 1: Identify the PatternLooking at the given letters:Z → ? → T → ? → N → ? → H → ? → BObserving the position of these letters in the English alphabet:Z (26)T (20)N (14)H (8)B (2)We see that the letters are decreasing in steps of 6:Z (26) → T (20) → N (14) → H (8) → B (2)Each step decreases by 6.Step 2: Fill in the Missing TermsNow, following the pattern:Z (26) → (26 - 6) = T (20)T (20) → (20 - 6) = N (14)N (14) → (14 - 6) = H (8)H (8) → (8 - 6) = B (2)Thus, the missing terms are:Between Z and T: (26 - 3) = W (23)Between T and N: (20 - 3) = Q (17)Between N and H: (14 - 3) = K (11)Between H and B: (8 - 3) = E (5)Final Sequence:Z, W, T, Q, N, K, H, E, BAnswer: W, Q, K, E

Q173. Directions: Read the following information and give answer.Ghulam Rasool is shorter than Ali but taller than Kamran.Naeem is shorter than Kamran.Jameel is taller than Naeem.Ali is taller than Jameel.Who among them is the tallest?

• A. Ali✓
• B. Naeem
• C. Kamran
• D. Jameel

Explanation: Let's analyze the statements:Ghulam Rasool is shorter than Ali but taller than Kamran.Naeem is shorter than Kamran.Jameel is taller than Naeem.Ali is taller than Jameel.From these statements, we can deduce the order:Ali > Ghulam Rasool > Kamran > NaeemAli > Jameel > NaeemThus, Ali is taller than everyone else, making him the tallest.Other options are incorrect because they do not satisfy all the conditions provided in the statements.

Q174. STATEMENTS:I. The dialogue between Jinnah and Gandhi failed.Il. Indian National Congress and Muslim League tried to join hands, so they will be united.

• A. Statement I is the cause and statement II is its effect.
• B. Statement II is the cause and statement I is its effect.
• C. Both statements I and II are independent causes
• D. Both statements I and II are the effects of independent cause.✓

Explanation: Gandhi and Jinnah were unable to reach a decision regarding Pakistan hence the dialogue failed. In order to show a united front to the British as they believe that would help them achieve independence INC and ML tried to join hands(work together). Thus both these statements are effects of independent causes.

Q175. Statements: The increasing population of our nation will lead to the depletion of many essential resources.Conclusions:I. Population of our nation can be controlled.II. The nation will not be able to provide a decent living to its citizens.

• A. Only conclusion I follows
• B. Only conclusion II follows✓
• C. Either conclusion I or II follows
• D. Both conclusions I and II follow

Explanation: The statement directly links population growth to the depletion of essential resources. A lack of essential resources logically implies that providing a decent living for all citizens will become difficult or impossible, making Conclusion II a direct inference. The statement does not provide information about whether the population can be controlled.

Q176. Statements:All flowers are trees. No fruit is tree.Conclusions:I. No fruit is flower.II. Some trees are flowers.Which of the following is/are most appropriate in the above scenario?

• A. Only I
• B. Only II
• C. I and II✓
• D. I and II are inappropriate

Explanation: Statements:All flowers are trees.No fruit is tree.Conclusions:I. No fruit is flower. This follows.Since all flowers are trees and no fruit is a tree, then by logic, no fruit can be a flower either.II. Some trees are flowers. This also follows.Statement 1 clearly says all flowers are trees, so definitely some trees are flowers.Answer:Both Conclusion I and II follow.

Q177. Statement: There is an unprecedented increase in the migration of villagers to urban areas as repeated crop failure has put them into a precarious financial situation.Courses of Action:I. The villagers should be provided with an alternate source of income in their villages which will make them stay put.II. The migrated villagers should be provided with jobs in the urban areas to help them survive.

• A. Only I follows✓
• B. Only II follows
• C. Either I or II follows
• D. Both I and II follow

Explanation: Action I addresses the root cause of the problem (lack of income in villages) and is a sustainable, long-term solution. Action II is a temporary fix that could encourage more migration and further strain urban resources. Therefore, Action I is the more logical course.

Q178. Statements: All chocolates are sweet. Some sweets are unhealthy. Conclusions: I. Some chocolates are unhealthy. II. All sweets are chocolates. III. Some sweets are not chocolates

• A. None follow✓
• B. Only III
• C. Only I
• D. I and III

Explanation: The correct answer is 'None follow'. Conclusion I suggests a direct relationship between chocolates and unhealthy items, which is not explicitly stated or implied. Conclusion II is incorrect because it suggests all sweets are chocolates, but the statement only says 'some sweets are unhealthy'. Conclusion III is also incorrect because there is no information to determine that some sweets are not chocolates. Thus, none of the conclusions can be logically derived from the given statements.

Q179. Statements: No bottle is glass. All glasses are cups. Conclusions: I. Some bottles are cups. II. No bottle is cup. III. Some cups are not bottles

• A. Only II and III✓
• B. Only I
• C. Only III
• D. None

Explanation: To solve this question, we need to analyze the logical deductions based on the given statements:1. Statement: No bottle is glass. All glasses are cups.2. Conclusion I: Some bottles are cups. This conclusion does not follow because there is no direct relationship between bottles and cups established by the statements.3. Conclusion II: No bottle is cup. This conclusion follows because if no bottle is a glass and all glasses are cups, then no bottle can be a cup.4. Conclusion III: Some cups are not bottles. This conclusion follows because since all glasses are cups and no bottle is a glass, then at least some cups will not be bottles.Therefore, the correct option is II and III, as they logically follow from the given statements.

Q180. Statements: All metals are solids. Some solids are liquids. Conclusions: I. Some metals are liquids. II. All solids are metals. III. Some solids are not liquids.

• A. Only III✓
• B. Only II
• C. Only I and II
• D. None

Explanation: To solve this problem, use logical deduction and Venn diagrams to analyze the relationships between the categories defined in the statements.Statement Analysis:All metals are solids: This means the set of metals is wholly contained within the set of solids.Some solids are liquids: This indicates there is an overlap between solids and liquids, but not all solids are liquids.Conclusion Evaluation:Some metals are liquids: The statements do not establish a direct relationship between metals and liquids, so this conclusion does not logically follow.All solids are metals: This conclusion is incorrect because the statements indicate that some solids are liquids, thus some solids are not metals.Some solids are not liquids: This conclusion is valid because the statement 'Some solids are liquids' implies that there are solids that are not liquids.Therefore, the only correct conclusion is III: 'Some solids are not liquids.'