Free Kpk Mdcat Mock — Solved Past Paper with Answers

Q1. In each of the following questions, four alternative sentences are given.Choose the CORRECT option.

• A. Norma had put the button unit besides her on the couch.
• B. Norma put an button unit besides her on the couch.
• C. Norma put the button unit besides hers on the couch.
• D. Norma put the button unit beside her on the couch.✓

Explanation: The phrase "beside her" is used in the sentence to indicate the location of the button unit in relation to Norma. "Beside" means next to or by the side of someone or something. In this case, it suggests that Norma placed the button unit in close proximity to herself, likely on the couch within reach. The phrase provides spatial information and helps to specify the position of the button unit in relation to Norma.

Q2. Each sentence below has a blank indicating that something has been omitted. Choose the word that best fits in with the meaning of the sentence as a whole.Though the apprentice was a highly proficient IT professional, he had little or no _ in designing educational software.

• A. Creativity✓
• B. Prospect
• C. Configuration
• D. Competition

Explanation: The word that best fits in the blank is "Creativity."So the sentence reads:"Though the apprentice was a highly proficient IT professional, he had little or no creativity designing educational software."

Q3. Each sentence below has a blank indicating that something has been omitted. Choose the word that best fits in with the meaning of the sentence as a whole.The foreman’s clemency, particularly in being candid, had its _, one of which was _workmanship.

• A. Damages ... intolerable
• B. Ambiguities ...superior
• C. Shortcomings ...shoddy✓
• D. Occurrences ...attractive

Explanation: The sentence can be completed with the following words:"Shortcomings" and "shoddy" So the completed sentence is:"The foreman’s clemency, particularly in being candid, had its shortcomings, one of which was shoddy workmanship."

Q4. Select the most suitable synonym for the given word:CAPACIOUS

• A. Ample✓
• B. Cramped
• C. Stagnant
• D. Dormant

Explanation: Capacious is synonymous to roomy, having a generous amount of space.

Q5. In the following question, four alternative sentences are given. Choose the CORRECT one.

• A. In my experience, the awakening of that clear judgemeent as to what the college is for, is not as difficult as is often supposed.
• B. In my experience, the awakening of a clear judgement as for what the college is for,is not as difficult as is often supposed.
• C. In my experience, the awakening of a clear judgement as to what the college is for, is not as difficult as is often supposed.✓
• D. In my experience, the awakening of a clear judgement as to what the college is for, is not as much as difficult as often supposed.

Explanation: The best way to solve this question is to read each option in your head and then decide which one sounds the best and fits in well.In Option A, the spelling of judgement is incorrect.In Option B, there is no space left after the comma which is the wrong punctuation.In Option D, as much as is unnecessary text and can be removed.

Q6. Fill in the blank: I'm now going to give you 10 _ tips for running a really successful website.

• A. Invaluable✓
• B. Valueless
• C. Unworthy
• D. Unprofitable

Explanation: Invaluable is defined as something extremely useful; indispensable. Since these tips will help create a successful website, they are extremely helpful and thus important, making them invaluable.

Q7. His bag was quite _ so I easily carried it to his room.

• A. Cheap
• B. Heavy
• C. Light✓
• D. Short

Explanation: As the sentence says, the bag was easy to carry; therefore, option C (i.e., light) fits here.

Q8. Demonstrate the control of tenses and sentence structure in the following sentence:I _ (eat) an apple.

• A. Eat
• B. Eaten
• C. Eating
• D. Ate✓

Explanation: ‘Eaten’ [Option B] and 'Eating' [Option C] require helping verbs before them [e.g. am/have]. As these assisting verbs to are not given in the sentence these options are eliminated.Option D: ‘ate’ is the past tense of ‘eat’ which is the most appropriate choice.

Q9. Comprehend key vocabulary.This medicine is _ from plants.

• A. Deprived
• B. Derived✓
• C. Derisive
• D. Derivative

Explanation: “Derived” is correct because it means obtained or extracted from a source, which fits the sentence. “Deprived” means denied or robbed of something, which changes the meaning completely.

Q10. In which model do proteins float in membrane like icebergs in sea:

• A. Lock and key model
• B. Induced fit model
• C. Fluid mosaic model✓
• D. Lotka volterra model
• E. All of these

Explanation: As per the fluid mosaic model cell membranes have proteins as icebergs in a sea of lipids (lipid). Hence option C is correct.

Q11. Which of the following are not polymer?

• A. Proteins
• B. Polysaccharides
• C. Lipids✓
• D. Nucleic acids

Explanation: The correct answer is Lipids. Unlike proteins, polysaccharides, and nucleic acids, lipids do not consist of repeating monomer units and therefore are not classified as polymers. Proteins are polymers because they are made up of amino acids linked in chains. Polysaccharides are polymers of monosaccharides, and nucleic acids are polymers of nucleotides, each consisting of a series of repeated structural units.

Q12. What are the components of a nucleotide?

• A. Purine, pyrimidine and phosphate
• B. Purine, sugar and phosphate
• C. Nitrogen base, sugar and phosphate✓
• D. Pyrimidine, sugar and phosphate

Explanation: The correct answer is Nitrogen base, sugar and phosphate.A nucleotide is the basic unit of DNA and RNA. It is made up of three components:A nitrogenous base: This is a ring-shaped molecule that contains nitrogen atoms. There are two types of nitrogenous bases: purines and pyrimidines. Purines are larger and have two rings, while pyrimidines are smaller and have one ring. The nitrogenous bases in DNA are adenine (A), guanine (G), cytosine (C), and thymine (T). In RNA, the base uracil (U) takes the place of thymine.A pentose sugar: This is a five-carbon sugar molecule. In DNA, the sugar is deoxyribose. In RNA, the sugar is ribose.A phosphate group: This is a group of three oxygen atoms bonded to a phosphorus atom.The nitrogenous base is attached to the sugar molecule, and the phosphate group is attached to the other end of the sugar molecule. The nucleotides are linked together by the phosphate groups, forming a long chain.The sequence of nucleotides in DNA and RNA determines the genetic information of an organism.

Q13. A segment of DNA has 120 adenine and 120 cytosine bases. The total number of nucleotides present in the segment is:

• A. 120
• B. 240
• C. 60
• D. 480✓

Explanation: According to Chargaff's rule, in DNA, the amount of adenine (A) is equal to the amount of thymine (T), and the amount of cytosine (C) is equal to the amount of guanine (G). This means that the number of adenine bases is equal to the number of thymine bases, and the number of cytosine bases is equal to the number of guanine bases.In the given segment of DNA, it is mentioned that there are 120 adenine (A) bases and 120 cytosine (C) bases. Since the number of A bases is equal to the number of T bases and the number of C bases is equal to the number of G bases, the total number of nucleotides can be calculated as follows:Number of nucleotides = Number of A bases + Number of T bases + Number of C bases + Number of G basesSince the number of A bases and T bases is equal, and the number of C bases and G bases is equal, we can calculate the total number of nucleotides as:Total number of nucleotides = 120 + 120 + 120 + 120 = 480Therefore, the total number of nucleotides present in the segment is 480.

Q14. _ fats are solid at room temperature.

• A. Saturated✓
• B. Unsaturated
• C. All
• D. None

Explanation: Saturated fats are solid at room temperature because their molecular structure is straight and allows for tight packing, which is typical of animal fats. Unsaturated fats, however, have kinks in their structure that prevent them from packing tightly, causing them to be liquid at room temperature, which is typical of plant oils. Therefore, the correct answer is that saturated fats are solid at room temperature, distinguishing them from unsaturated fats.

Q15. A → B → C → D → EAccumulation of “E” will control the above pathway through:

• A. Feedback mechanism
• B. Feedback activation
• C. Positive feedback
• D. Feedback inhibition✓

Explanation: Feedback inhibition is a crucial regulatory mechanism in metabolic pathways where the accumulation of the end product 'E' inhibits an enzyme earlier in the pathway, usually the first enzyme, to prevent overproduction. This maintains balance and prevents resource wastage. Other options, such as 'feedback mechanism' and 'positive feedback', are either too general or describe different regulatory processes that do not directly apply to the scenario described in the question. 'Feedback activation' is not a recognized term in this context and inaccurately suggests activation rather than inhibition.

Q16. Which of these enzymes works best at a pH of 9.70?

• A. Arginase✓
• B. Pancreatic enzyme
• C. Catalase
• D. Chymotrypsin

Explanation: Arginase operates optimally at a pH of around 9.7. It is involved in the urea cycle, converting arginine into ornithine and urea.

Q17. Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?

• A. Ribosome
• B. Chloroplast
• C. Mitochondrion✓
• D. Lysosome

Explanation: The mitochondrion is called the "powerhouse of the cell" because it breaks down carbohydrates through cellular respiration and produces ATP. Ribosomes make proteins, chloroplasts perform photosynthesis, and lysosomes carry out digestion, but ATP formation mainly occurs in mitochondria.

Q18. Microtubules are the constituents of:

• A. Centrioles, spindle fibres and chromatin
• B. Centrosome, nucleosome and centrioles
• C. Cilia, flagella and peroxisomes
• D. Spindle fibres, centrioles and cilia✓

Explanation: Spindle fibres, centrioles, and cilia all contain microtubules. Microtubules are crucial for their structure and function.

Q19. Sarcolemma is primarily made up of:

• A. Lipoprotein✓
• B. Glycolipids
• C. Glycoprotein
• D. Nucleoproteins

Explanation: The correct answer is lipoprotein. The sarcolemma, as the cell membrane of muscle cells, consists mainly of a lipid bilayer with embedded proteins, together forming lipoproteins. These components are essential for maintaining the integrity and function of the sarcolemma. Options B, C, and D are incorrect because they either refer to minor components or are unrelated to the structural makeup of the sarcolemma.

Q20. Sertoli cells are found in:

• A. Ovaries and secrete progesterone
• B. Adrenal cortex and secrete adrenaline
• C. Seminiferous tubules and provide nutrition to germ cells✓
• D. Pancreas and secrete cholecystokinin

Explanation: Sertoli cells are specialized cells found in the seminiferous tubules of the testes. They play a crucial role in supporting and nourishing the developing germ cells (sperm cells) during spermatogenesis, which is the process of sperm production. Sertoli cells provide physical support, produce essential proteins and growth factors, regulate hormone levels, and create a suitable microenvironment for the maturation of sperm cells.

Q21. When you hold your breath, which of the following gas changes in blood would first lead to the urge to breathe?

• A. Falling CO2 concentration
• B. Rising CO2 and falling O2 concentration
• C. Falling O2 concentration
• D. Rising CO2 concentration✓

Explanation: During breath-holding, the accumulation of CO in the blood is the primary factor that triggers the urge to breathe. This rise in CO concentration stimulates the respiratory center in the brain, prompting the body to resume breathing to expel excess CO and maintain homeostasis.

Q22. Which of the following is derived from the Latin word venom which means poisonous?

• A. Bacteria
• B. Fungi
• C. Virus✓
• D. Malaria

Explanation: Option A: Bacteria are single-celled organisms that can cause disease. They are not derived from the Latin word venemus. The word bacteria comes from the Greek word bakterion, which means "small staff". Otion B: Fungi are multicellular organisms that can cause disease. They are not derived from the Latin word venemus. The word fungi comes from the Latin word fungus, which means "mushroom". Option C: The Latin word venume is the root of the English word venom, which is a poisonous substance produced by some animals, such as snakes, spiders, and scorpions. The Latin word virus also comes from venume, and it refers to a small, infectious agent that can cause disease. The word virus was first used in English in the 16th century. It was borrowed from the Latin word virus, which originally meant "poison" or "venom". The modern meaning of virus as a small, infectious agent was not established until the 20th century. Option D: Malaria is a disease caused by a parasite. The parasite is not derived from the Latin word venemus. The word malaria comes from the Italian word mala aria, which means "bad air".

Q23. Avidin is a protein that:

• A. Binds specifically to biotin in egg white✓
• B. Binds egg white with egg albumin
• C. Binds both biotin and albumin
• D. This protein is not found in egg white

Explanation: Avidin is a glycoprotein present in egg white that binds specifically to biotin (vitamin B7) with high affinity. This binding prevents the absorption of biotin if raw egg whites are consumed excessively, potentially leading to biotin deficiency. Cooking the egg whites denatures avidin, reducing its binding capacity. Other options are incorrect because avidin does not bind to egg albumin, and it is indeed found in egg white.

Q24. Cells that kill cells that display foreign motifs on their surface are:

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

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

Q25. Amphibians are poikilotherm. Therefore, they use to hibernate in:

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

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

Q26. Purkinje fibers are connected with the impulse conducting system of:

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

Explanation: The correct answer is Option A: Heart. Purkinje fibers are a critical component of the heart's impulse conducting system. They rapidly transmit electrical impulses, which orchestrate the nearly simultaneous contraction of the ventricles, facilitating efficient blood circulation. In contrast, the brain relies on neurons for impulse transmission, the skin uses nerve endings for sensory signaling, and the nephron uses epithelial cells for its physiological processes.

Q27. Some marine fishes possess salt-excreting organs known as:

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

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

Q28. Piriformis syndrome is associated with which of the following disorder:

• A. Arthritis
• B. Sciatica✓
• C. Spondylosis
• D. Disc slip

Explanation: Piriformis syndrome is a condition where the piriformis muscle in the buttocks compresses the sciatic nerve, leading to sciatica symptoms such as pain, numbness, and tingling in the lower limbs. While a slipped disc can also cause sciatica, piriformis syndrome is specifically related to muscle compression rather than disc issues. Arthritis and spondylosis, on the other hand, are primarily joint and spinal disorders that do not typically involve the piriformis muscle or cause sciatica directly.

Q29. Diphtheria vaccines are an example of:

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

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

Q30. DNA polymerase III works always in:

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

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

Q31. A heterozygote fruit fly has more florescent pigments in their eyes than a wild homozygote fruit fly, this is an example of:

• A. Co-dominance
• B. Incomplete dominance
• C. Over dominance✓
• D. Complete dominance

Explanation: When the phenotypic expression of heterozygotes becomes more intense than the homozygous state of the dominant allele, is called over-dominance. So, the above example is an example of Over dominance.

Q32. Vomit centre is located in:

• A. Pons
• B. Midbrain
• C. Cerebellum
• D. Medulla✓

Explanation: The medulla oblongata is the posterior part of the brainstem and plays a key role in controlling autonomic functions and reflexes, such as vomiting. This is why the vomit center is located here. In contrast, the pons is more involved with sleep and respiratory rhythms, the midbrain with sensory information processing, and the cerebellum with motor control and balance. None of these has a direct role in the vomiting reflex.

Q33. Which of the following is the suitable vector to be incorporated with the large external DNA fragment?

• A. Small size vector✓
• B. Large size vector
• C. Large size vector with no origin of replication
• D. Small size vector with no origin of replication

Explanation: In molecular cloning, a vector is a DNA molecule used as a vehicle to artificially carry foreign genetic material into another cell, where it can be replicated and/or expressed. A vector containing foreign DNA is termed recombinant DNA. A small size is the suitable vector to be incorporated with the large external DNA fragment because it should be small in size so that it can easily integrate into the host cell. It should be capable of inserting a large segment of DNA. It should possess multiple cloning sites.

Q34. If black and white true breeding mice are mated and they result is all gray offspring, what inheritance pattern would this be indictive of ?

• A. Dominance
• B. Co dominance
• C. Multiple alleies
• D. Incomplete dominance✓

Explanation: Codominance means that no allele can block or mask the expression of the other allele. While incomplete dominance is a condition in which a dominant allele does not completely mask the effects of a recessive allele so the correct option is D as black and white parents produce gray offsprings

Q35. Calcium ions bind with the troponin molecule and cause them to:

• A. Extend
• B. Change shape✓
• C. Contract
• D. Remain in the same position

Explanation: Upon binding with calcium ions, the troponin molecule undergoes a conformational change, which shifts the position of tropomyosin on the actin filament. This movement uncovers the myosin binding sites on the actin, allowing the myosin heads to attach and initiate muscle contraction. Option B is correct because it describes this process accurately.Option A is incorrect as troponin does not extend.Option C is incorrect because troponin does not contract; it facilitates contraction.Option D is incorrect since troponin changes position upon calcium binding.

Q36. 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.

Q37. During muscle relaxation, the calcium ions are:

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

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

Q38. Which of the following is not part of the first line of defense?

• A. Sebum
• B. Perspiration
• C. Interferon✓
• D. Epidermis

Explanation: The first line of defense includes physical and chemical barriers like the skin (including the epidermis), sebum, and perspiration, which help prevent pathogens from entering the body. Sebum and perspiration are secretions that support these barriers. Interferon, however, is part of the second line of defense. It is a protein that aids in the immune response by inhibiting viral replication and activating immune cells. This distinction makes interferon not a component of the first line of defense.

Q39. In which of the following the female workers are sterile?

• A. Ants
• B. Honeybee✓
• C. Baboon
• D. Parrots

Explanation: The worker bee and the queen bee are both female but only the queen bee can reproduce. All drones are male. Worker beans clean the hive, collecting Pollen and nectar to feed the colony and they take care of the offsprings. The drones only job is to mate with the queen with the queen

Q40. The usual position of the two centrioles in relation to each other is at right angle in:

• A. Higher plant cells
• B. Lower plant cells
• C. Animals cells
• D. Both (B) &(C)✓

Explanation: Only lower plant cells and animal cells have centrioles at right angle to each other so correct option is D

Q41. Of 100 ml of arterial blood, oxygen provided to the tissues is:

• A. 2ml
• B. 3ml
• C. 4ml
• D. 5ml✓

Explanation: 6 mL of oxygen is delivered, at tissue level in normal conditions so option D is correct

Q42. Enzymes work by lowering the _ of the reactions they catalyse:

• A. Kinetic energy
• B. Activation energy✓
• C. Heat energy
• D. Potential energy

Explanation: Activation Energy is the minimum amount of energy that the reacting species must possess in order to undergo a specified reaction. Enzymes speed up reactions by lowering this energy and hence the reaction occurs faster by the virtue of more reactants being capable of initiating and taking part in the reaction due to the lowered energy threshold.Kinetic energy (KE) of an object is the energy that it possesses due to its motion. Heat energy is the result of the movement of tiny particles called atoms, molecules or ions in solids, liquids and gases.Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.

Q43. First stable compound during Calvin cycle is:

• A. 3-phosphoglycerate✓
• B. Glyceraldehyde 3 - phosphate
• C. 1, 3 bisphosphoglycerate
• D. Ribulose bisphosphate

Explanation: The first stable compound formed during the Calvin Cycle is 3-phosphoglycerate (3-PGA). The Calvin Cycle, also known as the light-independent reactions of photosynthesis, is a series of enzyme-catalyzed reactions that take place in the stroma of chloroplasts. The cycle begins with the process of carbon fixation, where carbon dioxide from the atmosphere is combined with a five-carbon sugar called ribulose-1,5-bisphosphate (RuBP) using the enzyme ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO).

Q44. Which of the following is a copper containing protein in electron transport chain?

• A. Plastoquinone
• B. Cytochrome - C
• C. Plastocyanin✓
• D. Ferredoxin

Explanation: It serves as an electron transfer agent between the cytochrome complex which follows Photosystem II and the entry point to Photosystem I of the non-cyclic electron transfer process. The electron transport chain is a series of electron transporters embedded in the inner mitochondrial membrane that shuttles electrons from NADH and FADH2 to molecular oxygen. In the process, protons are pumped from the mitochondrial matrix to the intermembrane space, and oxygen is reduced to form water.

Q45. This theory says that “mitochondria and chloroplasts are, in effect, ancient bacteria which now live inside the larger cells”:

• A. Darwin’s theory of evolution
• B. Lamarckism
• C. Neo - darwinism
• D. Endosymbiont theory✓

Explanation: The endosymbiotic theory states that some of the organelles in today's eukaryotic cells were once prokaryotic microbes. They eventually lost their cell wall and much of their DNA because they were not of benefit within the host cell. Mitochondria and chloroplasts cannot grow outside their host cell. The illustration of the prokaryotic microbes penetrating into the now called eukaryotes is presented below:

Q46. The complete, mature, and infectious virus particle is known as:

• A. Capsomere
• B. Genome
• C. Virion✓
• D. Capsid

Explanation: The correct answer is c. Virion. A virion is the complete, mature, and infectious virus particle, consisting of the viral genome (DNA or RNA) enclosed within a protective protein coat called the capsid. Unlike the genome or capsid alone, which are only parts of a virus, the virion represents the fully assembled unit capable of infecting host cells.

Q47. Regarding structure of the human heart, Chordae tendinae are present in:

• A. Atria
• B. Pulmonary valve
• C. Ventricles✓
• D. Aortic valve

Explanation: Chordae tendineae (or simply chordae) are fibrous cords or tendons that are present in the heart to anchor the atrioventricular (AV) valves (specifically, the tricuspid and bicuspid/mitral valves) to the papillary muscles located in the ventricles. This prevents the valves from inverting into the atria when the ventricles contract and helps in maintaining proper valve function. Therefore, the correct location for chordae tendineae is Ventricles

Q48. Each air-sac consists of several microscopic single layered structures called:

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

Explanation: Alveoli are tiny air sacs in your lungs that take up the oxygen you breathe in and keep your body going. Although they're microscopic, alveoli are the workhorses of your respiratory system. People have an average of 480 million alveoli in their lungs, located at the end of bronchial tubes.

Q49. A chemical component that is NOT found in all viruses is:

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

Explanation: The correct answer is B "Lipids". Non-enveloped viruses do not have a "lipid" covering, but their effects on humans can be just as devastating. These “naked” viruses only need their protein-based capsid and host detector proteins to infect host cells. Also viruses have either DNA or RNA as it's genetic material.

Q50. Sperms are produced in:

• A. Urethra
• B. Sperm dust
• C. Pancreas
• D. Testes✓

Explanation: The testes are the male reproductive organs responsible for producing sperm through a process called spermatogenesis.

Q51. It is the length of myofibril from one Z-line to the next:

• A. Plasma membrane
• B. Sarcomere✓
• C. Sarcoplasm
• D. Sarcolemma

Explanation: The sarcomere is the length of a myofibril from one Z-line to the next and is the basic contractile unit within muscle cells. During muscle contraction, the sarcomeres shorten as the actin and myosin filaments slide past one another, bringing Z-lines closer together. The other options, such as the plasma membrane, sarcoplasm, and sarcolemma, refer to different components of the muscle cell that do not define this specific structural unit.

Q52. 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.

Q53. 50 secondary oocytes in females and 50 secondary spermatocytes in males give rise to:

• A. 100 ova and 100 sperms
• B. 50 ova and 200 sperms
• C. 200 ova and 50 sperms
• D. 50 ova and 100 sperms✓

Explanation: Spermatogenesis is the process of formation of spermatid (n) from spermatogonia(2n). The process goes as follows:-Spermatogonia(2n)⟶ Primary Spermatocyte(2n)⟶2 Secondary Spermatocyte(n)⟶4 Spermatid(n)Oogenesis is the production of an ovum from oogonia. The mechanism, in short, can be depicted as below:-Oogonia(2n)⟶ Primary Oocyte(2n)⟶ Secondary Oocyte(n)+ Polar Body(n)This meiosis I is an unequal division leading to the formation of a haploid ovum and a polar body. The first polar body is said to be atretic and does not divide further. However, the secondary oocyte completes its meiosis during fertilization to form an egg(n) and the second polar body(n).So, 50 secondary oocytes produce 50 ovum and 50 secondary spermatocytes produce 100 sperms.

Q54. The genotypes of a husband and wife are IAIB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?

• A. 3 genotypes; 4 phenotypes
• B. 4 genotypes; 3 phenotypes✓
• C. 4 genotypes; 4 phenotypes
• D. 3 genotypes; 3 phenotypes

Explanation: The correct answer is that there are 4 genotypes and 3 phenotypes. The parents' genotypes are IAIB (father) and IAi (mother). The possible genotypes for their children are:IAIA - Blood group AIAi - Blood group AIAIB - Blood group ABIBi - Blood group BThese genotypes result in three phenotypes: Blood group A (from IAIA and IAi), Blood group B (from IBi), and Blood group AB (from IAIB). Options A, C, and D are incorrect because they either miscount the number of genotypes or phenotypes.

Q55. The partial pressure of oxygen in the alveoli of the lungs is:

• A. Equal to that in the blood
• B. More than that in the blood✓
• C. Less than that in the blood
• D. Less than that of carbon dioxide.

Explanation: The partial pressure of oxygen in the alveoli is higher than in the blood, facilitating the diffusion of oxygen from the alveoli into the blood, where it can be transported by hemoglobin to tissues throughout the body.

Q56. Which of the following properties of an amino acid is affected by a change in pH?

• A. Oxidation of amino acid
• B. Atomization of amino acids
• C. Reduction of amino acid
• D. Ionization of amino acids✓

Explanation: The ionization state of amino acids is directly affected by pH. Both acidic and basic groups in amino acids can gain or lose protons based on the pH, altering their charge and affecting interactions with other molecules.

Q57. The Lock and Key Model for enzyme action proposed by Emil Fischer suggests that:

• A. Enzymes are unbiased for the substrate.
• B. Enzymes can modify their active sites.
• C. Enzymes are restricted to one reaction type.✓
• D. An enzyme can catalyze a variety of reactions.

Explanation: Emil Fischer (1890) proposed a Lock and Key model to visualize substrate and enzyme interaction. According to this model, as one specific key can open only a specific lock, in the same manner, a specific enzyme can transform only one substrate into products(s). According to the Lock and Key Model, the active site is a rigid structure. There is no modification or flexibility in the active site before, during, or after the enzyme action, and it is used only as a template. Later studies did not support this model in all reactions.

Q58. The transport of secretory proteins takes place through organelles in which of the following order?

• A. RER-Golgi apparatus-Secretory vesicles
• B. SER-RER-Golgi apparatus-Secretory vesicles
• C. RER-SER-Golgi apparatus-Secretory vesicles✓
• D. RER-Golgi apparatus-Secretory vesicles-SER

Explanation: The correct answer is C: RER-SER-Golgi apparatus-Secretory vesicles. This sequence accurately represents the pathway of secretory proteins. Initially, proteins are synthesized in the rough endoplasmic reticulum (RER), where ribosomes translate mRNA into polypeptide chains. These proteins then move to the smooth endoplasmic reticulum (SER) for further folding and modifications. Next, they are transported to the Golgi apparatus, where they undergo further processing, sorting, and packaging into secretory vesicles for transport out of the cell. Options A, B, and D are incorrect due to their flawed sequence of organelles. Option A misses the SER step, while Option B incorrectly starts with the SER instead of the RER. Option D places the SER at the end of the sequence, which is not where it functions in the transport process.

Q59. Which of the following organelles is responsible for the production of steroid hormones?

• A. Lysosomes
• B. Golgi bodies
• C. Smooth endoplasmic reticulum✓
• D. Rough endoplasmic reticulum

Explanation: The smooth endoplasmic reticulum (SER) is crucial for the synthesis of steroid hormones, as it harbors enzymes necessary for the production of cholesterol and its conversion into various steroid hormones. This function is distinct from that of other organelles: lysosomes are involved in the degradation of cellular debris, the Golgi apparatus modifies and packages proteins and lipids, and the rough endoplasmic reticulum is dedicated to protein synthesis. This specialized role makes the SER essential in steroidogenic cells like those found in the adrenal glands and ovaries.

Q60. Which of the following describes the fluid mosaic model of plasma membrane structure?

• A. Phospholipid monolayer with embedded proteins
• B. Phospholipid bilayer with embedded proteins✓
• C. Triglyceride bilayer with embedded proteins
• D. Triglyceride monolayer with embedded proteins

Explanation: The fluid mosaic model describes the plasma membrane as a phospholipid bilayer with proteins embedded throughout. This model emphasizes the flexibility of the membrane, as well as the diverse proteins that create a 'mosaic' appearance. The bilayer consists of two layers of phospholipids, with hydrophobic tails facing inward and hydrophilic heads facing outward. This arrangement provides the membrane with its fluid characteristic. Option A is incorrect as it suggests a monolayer, which is not the structure of the plasma membrane. Options C and D are incorrect because triglycerides are not the primary components of the plasma membrane; instead, phospholipids are the structural backbone.

Q61. How many nuclear pores are present in typical differentiated cells such as RBCs?

• A. About 3000 per nucleus
• B. About 30,000 per nucleus
• C. About 1-2 per nucleus
• D. About 3-4 per nucleus✓
• E. Zero

Explanation: The correct answer is about 3-4 nuclear pores per nucleus in fully differentiated cells such as red blood cells (RBCs). These pores are essential for the transport of molecules between the nucleus and the cytoplasm. While undifferentiated cells may possess a higher number of nuclear pores to facilitate various cellular processes, differentiated cells like RBCs have adapted to their specialized functions, resulting in a reduced number of nuclear pores. The other options either overestimate or underestimate the typical count found in these cells. For instance, egg cells possess around 30,000 nuclear pores due to their larger size and different functional demands, while stating zero would be incorrect as immature RBCs do contain nuclei.

Q62. What is the correct location of ATP synthase in mitochondria?

• A. Mitochondrial matrix
• B. Inner mitochondrial membrane✓
• C. Outer mitochondrial membrane
• D. Inter-membrane space

Explanation: The correct location of ATP synthase is the inner mitochondrial membrane. ATP synthase is essential for the production of ATP, utilizing the proton gradient established by the electron transport chain across this membrane. The inner membrane's cristae offer a large surface area for ATP synthesis. In contrast, the mitochondrial matrix, while important for other metabolic processes, does not contain ATP synthase. The outer membrane serves mainly as a gateway for molecules entering and exiting the mitochondria, and the inter-membrane space is a site of proton accumulation but not where ATP synthase is located.

Q63. The inner membrane of mitochondria form extensive infoldings called _.

• A. Cristae✓
• B. Cisternae
• C. Lamella
• D. Bifida

Explanation: The inner membrane of mitochondria is highly specialized and forms numerous infoldings known as cristae. These structures are vital as they increase the surface area of the membrane, allowing for more space for the electron transport chain and ATP synthase, which are key components in the production of ATP. The other options are incorrect: cisternae are associated with the endoplasmic reticulum and Golgi apparatus, lamella refers to a general thin layer or membrane, and bifida pertains to a medical condition unrelated to cell structure.

Q64. Below is the diagram of a bacteriophage. In which one of the options are all four parts, A, B, C, and D, correct, respectively?

• A. Tail fibres, Head, Sheath, Collar
• B. Sheath, Collar, Head, Tail fibres
• C. Head, Sheath, Collar, Tail fibres✓
• D. Collar, Tail fibres, Head, Sheath

Explanation: A–Head, B–Sheath, C–Collar, D–Tail fibreThe head contains the genetic material of the bacteriophage, usually in the form of DNA or RNA, enclosed within a protein coat. It plays a vital role in protecting the genetic material and injecting it into the host cell during infection.The sheath is a contractile structure surrounding the tail tube that helps in injecting the phage DNA into the bacterial cell. When the phage attaches to the host, the sheath contracts to push the DNA through the bacterial membrane.The collar is a small structure that connects the head to the tail and helps in stabilizing the phage during attachment. It also assists in the transfer of genetic material from the head to the tail during infection.The tail fibres are long, thin structures that help the phage recognize and attach to specific receptor sites on the bacterial surface. They play a crucial role in determining the host range and initiating the infection process.

Q65. Which of the following statements is wrong for viroids?

• A. They cause infections
• B. Their RNA is of high molecular weight✓
• C. They lack a protein coat
• D. They are smaller than viruses

Explanation: "Their RNA is of high molecular weight" This statement is Incorrect. Viroids have RNA that is of very low molecular weight. In fact, one of the distinguishing features of viroids is their small size, as they consist of a single-stranded RNA molecule with a low molecular weight. This sets them apart from viruses, which typically have larger and mare complex genomes.

Q66. Which of the following statements is not true for retroviruses?

• A. DNA is not present at any stage in the life cycle of retroviruses✓
• B. Retroviruses carry a gene for RNA-dependent DNA polymerase
• C. The genetic material in mature retroviruses is RNA
• D. Retroviruses are causative agents for certain kinds of cancer in humans

Explanation: Retroviruses are unique viruses with a complex life cycle, and their features help explain which statements are true or false. Statement (a) is false because retroviruses indeed have a DNA stage in their replication cycle. After infection, they use the enzyme reverse transcriptase to convert their RNA genome into DNA, which integrates into the host cell’s chromosomes. Statement (b) is true because retroviruses encode RNA-dependent DNA polymerase, better known as reverse transcriptase, which is essential for converting viral RNA into DNA. Statement (c) is also true since retroviruses carry RNA as their genetic material inside mature virions, which is used as a template after infection. Finally, statement (d) is true because some retroviruses, such as HTLV-1, are oncogenic, meaning they can disrupt normal host cell growth regulation and potentially cause cancers like adult T-cell leukaemia. These features distinguish retroviruses from other types of viruses.

Q67. Name the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls:

• A. Pneumonia
• B. Asthma
• C. Pleurisy
• D. Emphysema✓

Explanation: In emphysema, the walls of the alveoli are damaged, leading to fewer and larger air sacs instead of many small ones. This reduces the surface area for gas exchange and causes breathing difficulties.

Q68. Lipid bilayer makes the membrane differently permeable barrier that allows the transport of:

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

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

Q69. _ enzyme needs a primer for the initiation of its function.

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

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

Q70. Male having Down's syndrome have sex chromosomes:

• A. XXY
• B. XY✓
• C. XYY
• D. XYYY

Explanation: In Down’s syndrome, the affected person contains an extra autosomal chromosome number 21 (trisomy 2n+1). Males having this syndrome have XY sex chromosomes. They have 3 21st chromosomes.

Q71. Given the reaction: C3H8 + 5O2 → 3CO2 + 4H2O At STP , how many litres of O2 are needed to completely burn 5.0 litres of C3H8?

• A. 5
• B. 10
• C. 10.5
• D. 15
• E. 25✓

Explanation: Consider that no. of moles of reactants = volume of the reactants.As the molar ratio between C3H6 and O2 is 1:5, the volume ratio is also the same. For 1L of C3H6 we need 5L of O2. Since the volume of C3H6 is 5L now, to burn this, we require 25L of O2.All other options are incorrect because they do not follow the exact mole-volume relationship required in this case.

Q72. The partial pressure of oxygen in the alveoli of the lungs is:

• A. Equal to that in the blood
• B. More than that in the blood✓
• C. Less than that in the blood
• D. Less than that of carbon dioxide.

Explanation: The partial pressure of oxygen in the alveoli is higher than in the blood, facilitating the diffusion of oxygen from the alveoli into the blood, where it can be transported by hemoglobin to tissues throughout the body.

Q73. 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.

Q74. Reduction in pH of blood will directly cause which of the following?

• A. Decrease the affinity of haemoglobin with oxygen✓
• B. Release bicarbonate ions by the liver
• C. Reduce the rate of heart beat
• D. Reduce the blood supply to the brain.

Explanation: A reduction in blood pH (an increase in acidity) is typically associated with a condition called acidosis. In acidosis, the affinity of hemoglobin for oxygen decreases. This means that hemoglobin is less effective at binding with oxygen, which can lead to impaired oxygen transport to tissues.

Q75. Asthma may be attributed to:

• A. Inflammation of the trachea
• B. Accumulation of fluid in the lungs
• C. Bacterial infection of the lungs
• D. Allergic reaction of the mast cells in the lungs✓

Explanation: Asthma is often associated with allergic reactions where allergens trigger the activation of mast cells in the lungs. These cells release chemicals like histamines, leading to airway inflammation and bronchoconstriction, which are hallmarks of asthma.

Q76. The net pressure gradient that causes the fluid to filter out of the glomeruli into the capsule is:

• A. 50 mm Hg
• B. 75 mm Hg
• C. 20 mm Hg✓
• D. 30 mm Hg

Explanation: The net filtration pressure in the glomeruli, which facilitates the movement of fluid into the Bowman’s capsule, is determined by subtracting the opposing pressures from the blood pressure in the glomeruli. The blood pressure in the glomeruli is approximately 70 mm Hg. This is counteracted by the osmotic pressure of plasma proteins (approximately 30 mm Hg) and the hydrostatic pressure in the Bowman’s capsule (approximately 20 mm Hg). Thus, the net filtration pressure is 70 - (30 + 20) = 20 mm Hg. This net pressure is crucial for the filtration process, allowing water and small solutes to pass from the blood into the nephron. Options A, B, and D represent different pressure components but not the net filtration pressure, making option C the correct answer.

Q77. Name the chronic respiratory disorder caused mainly by cigarette smoking.

• A. Respiratory acidosis
• B. Respiratory alkalosis
• C. Emphysema✓
• D. Asthma

Explanation: Emphysema is a chronic respiratory disorder primarily caused by cigarette smoking, leading to damaged lung tissue and breathing difficulties.

Q78. Name the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls:

• A. Pneumonia
• B. Asthma
• C. Pleurisy
• D. Emphysema✓

Explanation: In emphysema, the walls of the alveoli are damaged, leading to fewer and larger air sacs instead of many small ones. This reduces the surface area for gas exchange and causes breathing difficulties.

Q79. When you hold your breath, which of the following gas changes in blood would first lead to the urge to breathe?

• A. Falling CO2 concentration
• B. Rising CO2 and falling O2 concentration
• C. Falling O2 concentration
• D. Rising CO2 concentration✓

Explanation: During breath-holding, the accumulation of CO in the blood is the primary factor that triggers the urge to breathe. This rise in CO concentration stimulates the respiratory center in the brain, prompting the body to resume breathing to expel excess CO and maintain homeostasis.

Q80. Approximately seventy percent of carbon­ dioxide absorbed by the blood will be transported to the lungs

• A. As bicarbonate ions✓
• B. In the form of dissolved gas molecules
• C. By binding to R.B.C
• D. As carbamino ­haemoglobin.

Explanation: The correct option is a) as bicarbonate ions.Explanation of each option:A. As bicarbonate ions (Correct Option): Approximately seventy percent of carbon dioxide (CO2) absorbed by the blood is transported in the form of bicarbonate ions (HCO3-). This is the primary method for CO2 transport in the blood. It involves a chemical reaction where CO2 combines with water (H2O) to form carbonic acid (H2CO3), which rapidly dissociates into bicarbonate ions (HCO3-) and hydrogen ions (H+). This bicarbonate ion is then carried in the bloodstream to the lungs for elimination.B. In the form of dissolved gas molecules: Some CO2 (around 7-10%) is transported in the bloodstream as dissolved gas molecules in plasma. However, this is a relatively small fraction compared to the bicarbonate ion transport method.C.By binding to R.B.C (Red Blood Cells): A smaller portion of CO2 (around 20-23%) binds to hemoglobin within red blood cells, forming carbamino compounds. This is a significant but secondary method compared to bicarbonate ion transport.D. As carbamino-hemoglobin: This option refers to the CO2 that binds to hemoglobin within red blood cells. It is a less dominant method of CO2 transport compared to bicarbonate ions.Brief Summary: Approximately seventy percent of carbon dioxide (CO2) absorbed by the blood is transported as bicarbonate ions (HCO3-). This process involves the conversion of CO2 to bicarbonate ions in a chemical reaction and is the primary method for CO2 transport in the bloodstream. While CO2 can also be transported as dissolved gas molecules, by binding to red blood cells, and as carbamino compounds, these are secondary methods in comparison to the predominant bicarbonate ion transport.

Q81. Photophosphorylation takes place in the _ of the chloroplasts:

• A. Stroma
• B. Granum✓
• C. Inner membrane
• D. Outer membrane

Explanation: Photophosphorylation takes place in the thylakoid membrane of the chloroplasts. This process, part of the light-dependent reactions of photosynthesis, uses light energy to create a proton gradient across the thylakoid membrane, which then drives the synthesis of ATP from ADP and inorganic phosphate.

Q82. A person who is on a long hunger strike and is surviving only on water, will have:

• A. Less amino acids in his urine
• B. More glucose in his blood
• C. Less urea in his urine✓
• D. More sodium in his urine.

Explanation: During starvation, the body must rely on body proteins for the amino acids. On high protein diets the carbon skeletons of the amino acids are oxidized for energy or stored as fat and glycogen, but the amino nitrogen must be excreted. To facilitate this process, enzymes of the urea cycle are controlled at the gene level. When dietary proteins increase significantly, enzyme concentrations rise. On return to a balanced diet, enzyme levels decline. Under conditions of starvation, enzyme levels rise as proteins are degraded and amino acid carbon skeletons are used to provide energy, thus increasing the quantity of nitrogen that must be excreted in the form of urea.

Q83. The experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding that :

• A. Radioactive phage were found in the pellet
• B. Radioactive cells were found in the supernatant
• C. Radioactive sulfur was found inside the cell
• D. Radioactive phosphorus was found in the cell✓

Explanation: The experiments by Hershey and Chase helped confirm that DNA was the hereditary material as radioactive phosphorus was found in the cell.The Hershey-Chase experiments were not the first studies to oppose the prevailing theory in the early 1900s that genetic material was composed of proteins. In 1944, nearly a decade before Hershey and Chase’s work, scientists published sound evidence that genes were made of DNA rather than protein so it was confirmed by presence of phosphorus.

Q84. If the medulla oblongata of a person brain is damaged which of the following process will be disturbed?

• A. Thinking
• B. Sleep
• C. Thirst
• D. Swallowing✓

Explanation: The medulla oblongata is a structure found in the human brain. The major function of the Medulla oblongata is to control automatic functions (that is, not under human control) breathing, respiration, and heartbeat. It also helps in body secretion, reflexes, swallowing, coughing, and sneezing.

Q85. For which purpose are myeloma cells (cancerous B lymphocytes) used in the production of monoclonal antibodies?

• A. Increased rate of cell division
• B. Immunization with antigen✓
• C. To avoid contamination
• D. As nutrient in media

Explanation: Multiple myeloma cells are abnormal plasma cells (a type of white blood cell) that build up in the bone marrow and form tumors in many bones of the body. They are produced for immunization with an antigen, as Vaccines contain weakened or inactive parts of a particular organism (antigen) that triggers an immune response within the body.

Q86. An autoimmune disorders in which stiffness and inflammation of vertebrae occurs is called as:

• A. Lupus
• B. Scleroderma
• C. Ankylosing spondylitis✓
• D. Juvenile dermatomyositis

Explanation: An autoimmune disorder in which stiffness and inflammation of vertebrae occur is called Ankylosis spondylitis.

Q87. Choose reaction that is not correct?

• A. Option A: Acetic acid reacts with sodium to form sodium acetate and hydrogen gas.
• B. Option B: Acetic acid reacts with ethanol in the presence of sulfuric acid to form ethyl acetate and water.
• C. Option C: Acetic acid reacts with phosphorus pentoxide to form acetic anhydride.
• D. Option D: Acetic acid reacts with ammonia to form acetyl chloride and water.✓

Explanation: The correct answer is Option D. Acetic acid does not react with ammonia to form acetyl chloride and water; instead, it usually forms ammonium acetate. Options A, B, and C describe well-established reactions: A is the reaction with sodium forming sodium acetate, B is esterification with ethanol forming ethyl acetate, and C is dehydration with phosphorus pentoxide forming acetic anhydride.

Q88. A person is undergoing prolonged fasting. His urine will be found to contain abnormal quantities of:

• A. Fats
• B. Amino acids
• C. Glucose
• D. Ketones✓

Explanation: During prolonged fasting, the body relies on fat stores for energy, leading to increased fatty acid oxidation in the liver. This process produces ketone bodies such as acetoacetate, b-hydroxybutyrate, and acetone. The presence of excess ketone bodies in urine is termed as ketonuria, indicating a state of increased fat metabolism. While fats, amino acids, and glucose can also be present in urine under certain conditions, the abnormal quantities of ketones in the urine during prolonged fasting specifically point towards increased fatty acid oxidation and the risk of diabetic ketoacidosis.

Q89. Low partial pressure of oxygen in tissues favors _ of oxyheamoglobin.

• A. Dissociation✓
• B. Stability
• C. Formation
• D. Transformation

Explanation: More CO2, less O2, high temperature and less pH facilitate oxygen dissociation from oxyhaemoglobin.When the partial pressure of oxygen (PO2) is low in the tissues, it indicates that the tissues are in need of oxygen supply. In this situation, according to the principle of the oxygen-hemoglobin dissociation curve, hemoglobin's affinity for oxygen decreases. As a result, the oxyhemoglobin complex readily releases its bound oxygen molecules, allowing the oxygen to diffuse from the blood into the tissues where it is needed for cellular respiration and metabolism.This phenomenon is a crucial aspect of the Bohr effect, which refers to the influence of pH, carbon dioxide (CO2) concentration, and partial pressure of oxygen on hemoglobin's ability to bind and release oxygen. In metabolically active tissues with higher CO2 levels and increased hydrogen ion (H+) concentration (lower pH), the partial pressure of oxygen is lower, and this environment favors the release of oxygen from oxyhemoglobin, enabling efficient oxygen delivery to the tissues. Conversely, in the lungs where oxygen levels are high, hemoglobin readily binds to oxygen, forming oxyhemoglobin for efficient oxygen transport from the lungs to the tissues.

Q90. About 70-85% CO2 in blood is carried:

• A. As carboxylase myoglobin
• B. Freely as CO2
• C. With proteins in plasma
• D. As bicarbonate✓

Explanation: About 70% of CO2 is transported as bicarbonate ions.

Q91. Quantum number which describes the orientation of orbitals in three dimensional space is:

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

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

Q92. Substance that has sharp melting point in the following is:

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

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

Q93. Which of the following pairs is an example of completely immiscible liquids?

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

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

Q94. The number of isomers of pentane is:

• A. 2
• B. 4
• C. 5
• D. 3✓

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

Q95. The compound which has the higher boiling point in the following is:

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

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

Q96. Which one of the following is addition polymer:

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

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

Q97. Which one is more reactive?

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

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

Q98. For exothermic reversible reaction, the activation energy for forward direction depends upon:

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

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

Q99. As the polarizing power of cation increases thermal stability of carbonates:

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

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

Q100. Which ions are used as catalysts in the reaction between persulfate ions and iodide ions?

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

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

Q101. Which one is stronger nucleophile?

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

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

Q102. C4H11N gives the type of isomerism:

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

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

Q103. Hydration of hydrocarbon give carbonyl compound, the general formula of that hydrocarbon is:

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

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

Q104. Consider reversibility in free radical substitution reaction of alkane then Kc value is smallest for:

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

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

Q105. SiO2 is the only oxide that reacts with:

• A. HClaq
• B. KOHaq✓
• C. Steam
• D. SO3

Explanation: SiO2 is an acidic oxide and reacts with strong bases. Among the given options, it will react with hot, concentrated KOH(aq) due to the need for strong conditions to overcome its extensive covalent bonding. It does not react with HCl(aq) because it is not a base. It doesn't react with steam as the strong covalent network structure of SiO2 is too stable to break under such conditions. Lastly, SO3 is also an acidic oxide, and thus, does not react with another acidic oxide like SiO2.

Q106. Students were heating CaCO3 in an open container to produce CO2 gas,CaCO3(s) ⟶ CaO(s) + CO2(g)If we increase pressure on this system the:

• A. Equilibrium will shift towards right
• B. Equilibrium will shift towards left
• C. Equilibrium will not be disturbed
• D. System does not obey equilibrium rules✓

Explanation: The correct answer is that the system does not obey equilibrium rules. Chemical equilibrium can only be established in a closed system where reactants and products are contained. In this scenario, the CaCO3 is heated in an open container, allowing the CO2 gas to escape. As a result, the system cannot reach equilibrium, as the continuous loss of CO2 gas prevents the establishment of a dynamic balance between reactants and products.All other options are incorrect because they assume that equilibrium can be achieved or disturbed in an open system, which is not possible. The nature of an open system prevents the maintenance of chemical equilibrium as it allows mass to be lost to the surroundings.

Q107. Initially one mole each N2 and O2 were made to react as,If at equilibrium 0 ∙ 25 moles of O2 is present the equilibrium concentration of NO will be:

• A. 0.50 moles
• B. 0.125 moles
• C. 1.50 moles✓
• D. 1.75 moles

Explanation: N2(g) + O2(g) 2NO(g) As from the balance chemical equation it is clear that 1 mole O2 gives 2 mole NO so as 0.25 mole O2 left so 0.75 mole consumed so it will give 1.50 mole NO.2x0.75-1=1.50

Q108. The type of isomerism present in the compound given, is:

• A. Structural
• B. Optical
• C. Stereo✓
• D. None of the above

Explanation: Stereoisomers are isomers that have the same composition (that is, the same parts) but that differ in the orientation of those parts in space. There are two kinds of stereoisomers: enantiomers and diastereomers. The two conditions must be met for compounds to show Stereoisomerism(geometric isomerism) 1- There should be a C-C double bond to restrict the free rotation of attached groups.2- The two groups attached to the same C must be different.

Q109. The volume of 𝐶𝑂2 produced by heating 33.5𝑔 𝐿𝑖2𝐶𝑂3 at room temperature and pressure is (𝑀𝑟 𝐿𝑖2𝐶𝑂3 = 67𝑔/𝑚𝑜𝑙):

• A. 22.4 𝑑𝑚3
• B. 12.0 𝑑𝑚3✓
• C. 11.2 𝑑𝑚3
• D. 24.0 𝑑𝑚3

Explanation: moles of Li2CO3= mass/Mr =33.5/67 = 0.5 Moles of CO2: moles of Li2CO3 1 : 1Hence moles of CO2 =0.5 0.5 × 24= 12dm3 Molar volume at ROOM temperature and pressure is 12 dm³ while molar volume at STANDARD temperature and pressure is 22.4 dm³.

Q110. The compound which can form hydrogen bond with water is:

• A. CH3–O-CH3
• B. CH3 –CH2 –OH✓
• C. CH3 –CH2–NH2
• D. None of the above.

Explanation: Option B contains OH that can interact with the water to form a hydrogen bond with water. Hydrogen bonding occurs when hydrogen has a covalent bond to an atom that is more electronegative than it. The more electronegative atom pulls the shared electrons unequally towards it, creating a polar bond.

Q111. Choose the molecule that could not be represented by single electronic structure formula:

• A. CH4
• B. H2O
• C. SO2✓
• D. O2

Explanation: C is the correct option as the electronic structure of the SO2 molecule is best represented as a resonance hybrid of two equivalent structures.

Q112. 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.

Q113. Stable electronic configuration of Cu(29) is:

• A. [Ar] 4S2 3d4
• B. [Ar] 4S0 3d10
• C. [Ar] 4S1 3d10✓
• D. [Ar] 4S2 3d7 4p2

Explanation: It is a transition element. Copper (Cu) is a chemical element, a reddish, extremely ductile metal of Group 11 (IB) of the periodic table that is an unusually good conductor of electricity and heat. Due to the interelectronic repulsions one electron present in 4s enters to 3d subshell. We can write a stable electronic configuration of copper is written as 1s22s22p63s23p63d104s1. There is additional stability because of the completely occupied 3d subshell and half-filled 4s subshell.orThe electronic configurations of copper (and chromium) are unique since without filling the 4s orbitals, electrons entered into 3-d orbitals. Copper has an atomic number of 29 hence 29 electrons are fulfilled by option C as well.

Q114. The proton acceptor is:

• A. NH3✓
• B. BF3
• C. HCl
• D. H+

Explanation: According to the Bronsted-Lowry theory, a base is a substance that accepts protons. NH3 is a classic example of a weak base that acts as a proton acceptor. In water, it reacts to form NH4+ and OH- ions, demonstrating its ability to accept a proton. In contrast, BF3 is a Lewis acid that accepts electron pairs, not protons. HCl is a proton donor, releasing H+ ions in water, and is therefore classified as an acid. Lastly, H+ is already a proton, so it cannot act as a proton acceptor.

Q115. Select alkene of the following hydrocarbons:

• A. C5 H22
• B. C5 H10✓
• C. C5 H8
• D. C4 H10

Explanation: The general formula for alkenes is CnH2n, where n=5 is the formula for alkene C5H10

Q116. The oxidation state of carbon in Na2C2 is:

• A. +4
• B. +2
• C. –1✓
• D. –4

Explanation: We know that the charge of Na is always +1, and the overall charge on the Na2C2 molecule is 0. We can equate this to form2(+1) + 2x = 02x = -2x(Charge on C) = -1

Q117. Choose atom that having spin quantum number ½

• A. 12C
• B. 15N✓
• C. 16O
• D. 32S

Explanation: The spin is a quality of particles, it tells us the angular momentum (½) of an electron of a substance. Since momentum is a vector quantity, the spin quantum number can be both positive and negative. This question asks which atom has a ½ spin number, the positive and negative spins of atoms with an even number of electrons will cancel out, thus only an atom with an odd number of electrons has a spin of ½ which is B.

Q118. A tertiary carbon is bonded directly to:

• A. 2 Hydrogens
• B. 2 Carbons
• C. 3 Carbons✓
• D. 4 Carbons

Explanation: A tertiary carbon is a carbon atom bonded directly to three other carbon atoms. This classification helps in understanding the structure and reactivity of hydrocarbons. The correct option is Option C: 3 Carbons. Other options are incorrect as they either involve fewer carbon atoms or incorrectly include hydrogen atoms, which do not change the classification of the carbon atom in this context. Option D is incorrect as it describes a quaternary carbon, which is bonded to four carbon atoms.

Q119. Which of the following compounds will react with methyl magnesium Iodide followed by acid hydrolysis to give ethyl alcohol?

• A. Ethylene
• B. Acetone
• C. Acetaldehyde
• D. Formaldehyde✓

Explanation: When formaldehyde (CH2O) reacts with methyl magnesium iodide (CH3MgI), followed by acid hydrolysis, it gives ethyl alcohol (ethanol). The Grignard reagent reacts with the formaldehyde carbonyl group, forming a compound that can be hydrolyzed to yield ethyl alcohol.The reaction can be represented as follows:CH2O + CH3MgI → CH3CH2OH

Q120. 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

Q121. 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.

Q122. 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.

Q123. 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.

Q124. Rain water becomes acidic, when the pH-value of rain water becomes _

• A. Greater than 6
• B. Greater than 6.5
• C. Less than 5.6
• D. Less than 5✓

Explanation: Normal clean rain has a pH value between 5.0 - 5.5, which is slightly acidic, however when rain combines with nitrogen oxides or SO2, it becomes more acidic has the pH drops below 5.0. Therefore the answer is D.

Q125. How many molecules are present in 0.20 g of Hydrogen gas? Select the correct option.

• A. 0.20 x 3.01 x 1023✓
• B. 0.20 x 2.016
• C. 0.70 x 6.02 x 1023
• D. 1.008 x 6.02 x 1023

Explanation: To find the number of molecules in 0.20 g of hydrogen gas, first recognize that hydrogen gas (H2) is diatomic. This means that 1 mole of hydrogen gas, which weighs 2.016 g, contains 6.02 x 1023 molecules. Therefore, 1 g of hydrogen gas contains 3.01 x 1023 molecules. To find the number of molecules in 0.20 g, multiply 0.20 by 3.01 x 1023, yielding 0.20 x 3.01 x 1023 molecules.Option A correctly uses this method. Option B fails by using the molar mass without Avogadro's number, Option C uses an incorrect proportion, and Option D uses the atomic mass of hydrogen instead of the diatomic gas.

Q126. 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.

Q127. 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.

Q128. Which of the following structure has a bond formed by an overlap of sp2 hybrid orbital with that of sp2 hybrid orbital?

• A. HC = CH
• B. H2C = CH2✓
• C. H2C = C= CH2
• D. CH2= CHCH3

Explanation: The C-C sigma bond in ethylene is formed by the overlap of an sp2 hybrid orbital from each carbon. The overlap of hybrid orbitals or a hybrid orbital and a 1s orbtial from hydrogen creates the sigma bond framework of the ethylene molecule.

Q129. The bond angle between H-C-H bond in ethane is:

• A. 109.5✓
• B. 120
• C. 90
• D. 107.5

Explanation: Ethane, C2H6, has a geometry related to that of methane. The two carbons are bonded together, and each is bonded to three hydrogens. Each H-C-H angle is 109.5° and each H-C-C angle is 109.5°.

Q130. Which is the correct product formed when monohydric alcohol reacts with sodium metal?

• A. Alkene
• B. Sodum alkoxide✓
• C. Alkane
• D. Ether

Explanation: When alcohols react with metals a salt is formed,the reaction when a monohydric alcohol reacts with sodium metal results in the production of sodium alkoxide and hydrogen gas, hence the answer is B.

Q131. Thermal processing of industrial waste material aims at:

• A. Burning of waste material in pits
• B. Converting the solid waste into useful products by thermal treatment✓
• C. Energy recovery from organic matter prior to its final disposal
• D. Size reduction and compaction by thermal process

Explanation: The primary function of thermal treatment is to convert the waste to a stable and usable end product. Therefore the answer will be B.

Q132. Which of the following titrants would most likely be used as ths‘ own indicator in acid medium?

• A. K2Cr2O7 (Potassium dichromate)
• B. Iodine (I2)
• C. KMnO4 (Potassium permanganate)✓
• D. H2O2 (Hydrogen peroxide)

Explanation: Potassium permanganate (KMnO4) is used as its own indicator because it changes color distinctly at the endpoint of a titration in an acidic medium. The intense purple color of KMnO4 fades to colorless (or a very faint pink) as it is reduced, providing a clear visual cue without the need for an additional indicator. In contrast, K2Cr2O7, Iodine, and H2O2 lack such a distinct color change when used in acidic conditions, making them unsuitable as self-indicators.

Q133. Chlorosis is caused by the deficiency of:

• A. Magnesium✓
• B. Iron
• C. Phosphorus
• D. Potassium

Explanation: Chlorosis is typically caused when leaves do not have enough nutrients to synthesise all the chlorophyll they need. It can be brought about by a combination of factors including: a specific mineral deficiency in the soil of Mg2+

Q134. How many different values can m, assume in the electron sub-shell designated by quantum number n=5, l=4?

• A. 4
• B. 5
• C. 6
• D. 9✓

Explanation: 'n' is referred to as the principal quantum number and 'l' is the azimuthal quantum number. l=4 means the azimuthal quantum number is associated with 5g. Since 'g' has 9 subshells 'm' will be -4, -3, -2, -1, 0, +1, +2, +3, +4.

Q135. Which one of the following salts will produce an alkaline solution when dissolved in water?

• A. NH4Cl
• B. NaNO3
• C. Na2CO3✓
• D. Na2SO4

Explanation: The correct answer is Na2CO3. Sodium carbonate forms an alkaline solution because it undergoes hydrolysis in water, releasing hydroxide ions (OH-) which increase the pH. In contrast:NH4Cl forms an acidic solution due to ammonium ions generating hydronium ions.NaNO3 remains neutral as neither ion affects pH significantly.Na2SO4 also results in a neutral solution for similar reasons to NaNO3.

Q136. In an inelastic collision, Momentum is _ and Kinetic Energy is _ respectively.

• A. Not conserved ; not conserved
• B. conserved ; not conserved✓
• C. Conserved ; conserved
• D. Not conserved ; conserved

Explanation: For elastic and inelastic collisions, momentum and total energy are always conserved provided no external force is applied; however, for inelastic collisions, kinetic energy is NOT conserved.

Q137. A bullet of mass 20g leaves the gun with a velocity of 200 m/s. If the mass of gun is 2kg then the speed of recoil of the gun is:

• A. 2000 m/s
• B. 2 m/s✓
• C. 20 m/s
• D. 200 m/s
• E. 100 m/s

Explanation: The principle of conservation of momentum states that the total momentum of an isolated system remains constant. Here, the gun and the bullet together form a system. So, the momentum of the gun and the bullet before firing will be equal to the momentum of the gun and the bullet after firing.Initial momentum = Final momentum(mass of gun) x (velocity of gun) = (mass of bullet) x (velocity of bullet) + (mass of gun) x (velocity of recoil)Substituting the values, we get:(2 kg) x 0 m/s = (0.02 kg) x (200 m/s) + (2 kg) x (velocity of recoil)Solving for the velocity of recoil, we get:velocity of recoil = - (0.02 kg) x (200 m/s) / (2 kg) = -2 m/sThe negative sign indicates that the recoil is in the opposite direction to the direction of the bullet. So, the magnitude of the speed of the recoil of the gun is 2 m/s.

Q138. A car at rest starts moving with linear uniformly increasing velocity. After 20 seconds, it attains the maximum velocity of 80 m/s. What is the distance covered during this time interval?

• A. 200 m
• B. 400 m
• C. 800 m✓
• D. 1600 m

Explanation: The acceleration of the car is constant as it undergoes uniformly increasing velocity. Let the initial velocity of the car be u = 0 m/s, and final velocity be v = 80 m/s. The time taken to achieve this velocity is t = 20 s.Using the formula for uniformly accelerated motion, we have:v = u + atwhere a is the acceleration. Rearranging the equation, we get:a = (v - u)/t = 80 m/s / 20 s = 4 m/s2To find the distance covered by the car during the given time interval, we use the formula:s = ut + (1/2)at2where s is the distance traveled. Substituting the given values, we get:s = (1/2)(4 m/s2)(20 s)2 = 800 mTherefore, the correct answer is C) 800 m.

Q139. An object is thrown vertically upward with a velocity of 20 m/s. How much time it will take to reach the highest point?

• A. 2 sec✓
• B. 4 sec
• C. 1 sec
• D. Insufficient information

Explanation: Given: Initial velocity of object, u = 20 m/s Final velocity of object, v = 0 Acceleration, g = 10 m/s To find, 1. Distance travelled, s = ? 2. Time taken, t = ? Solution: 1. Distance travelled, s We know that, 3rd equation of motion, i.e v² - u² = 2(- g)s ⇒ (0)² - (20)² = 2 × (- 10) × s ⇒ 400 = 20s ⇒ 400/20 = s ⇒ s = 20 m Hence, the distance travelled by an object is 20 m. 2. Time taken, t We know that, 1st equation of motion i.e v = u + (- g)t= 20 + (- 10) × t 20 = 10t 20/10 = t ⇒ t = 2 seconds Hence, the time taken by the object is 2 seconds.

Q140. An object is falling down with a speed of 20 m/s. After 3 secondsits velocity will be _ m/s (g = 10 m/s2).

• A. 05
• B. 50✓
• C. 55
• D. 95

Explanation: To find the final velocity of the object after 3 seconds of free fall, we use the kinematic equation: v = u + at. The initial velocity (u) is 20 m/s, the acceleration due to gravity (a) is 10 m/s2, and the time (t) is 3 seconds. Substituting these values into the equation, we get v = 20 + 10(3) = 20 + 30 = 50 m/s. Thus, the correct answer is 50 m/s. Option A (05) incorrectly assumes deceleration, Option C (55) miscalculates the velocity increase, and Option D (95) is an extreme overestimate.

Q141. A paratrooper is falling down with uniform velocity and also rotating with a constant angular velocity of 0.2 rad/sec. The body satisfies _.

• A. The first condition of equilibrium but not the second
• B. The second condition of equilibrium but not the first
• C. Both the first and second conditions of equilibrium✓
• D. Neither the first nor the second condition of equilibrium

Explanation: The correct answer is Option C: Both the first and second conditions of equilibrium. The first condition of equilibrium states that the sum of all forces acting on an object must be zero for it to move with constant velocity; since the paratrooper is falling with uniform velocity, this condition is met. The second condition of equilibrium states that the sum of all torques acting on an object must also be zero for it to rotate with constant angular velocity; since the paratrooper is rotating with a constant angular velocity of 0.2 rad/sec, this condition is also satisfied. Options A and B are incorrect as they each claim that only one condition of equilibrium is satisfied while in fact, both are. Option D is also incorrect because it suggests that neither condition is satisfied, which contradicts the provided information about uniform velocity and constant angular velocity.

Q142. A car of mass 1200 kg initially at rest has been accelerated to a speed of 8 m/s in 16 meters. Average acceleration of the car is _ m/s2? And force is _ N?

• A. 1.5 and 1500
• B. 2.5 and 2400
• C. 3.5 and 3500
• D. 2 and 2400✓

Explanation: To solve this problem, we start by determining the acceleration using the kinematic equation v² = u² + 2as, where:v is the final velocity (8 m/s),u is the initial velocity (0 m/s, since the car starts from rest),a is the acceleration, ands is the displacement (16 m).Substituting the known values into the equation:(8)² = (0)² + 2(a)(16)64 = 32aa = 2 m/s²Next, apply Newton's second law to find the force: F = ma.F = 1200 kg × 2 m/s² = 2400 NTherefore, the correct values are an acceleration of 2 m/s² and a force of 2400 N. The other options are incorrect because they do not satisfy both the acceleration and force calculations based on the given conditions.

Q143. A car starts from rest and moves with constant acceleration. During the 4th second of its motion, it covers a distance of 21 meters. The acceleration of the car is _ m.

• A. 04
• B. 06✓
• C. 08
• D. 16

Explanation: To determine the acceleration, utilize the formula for distance covered in the nth second of uniformly accelerated motion:Sn = u + ½ a (2n-1)Given:u = initial velocity = 0 m/s (as the car starts from rest)a = acceleration (unknown)n = 4 (since we are considering the 4th second)Sn = 21 metersSubstitute these values into the equation:21 = 0 + ½ × a × (2×4 - 1)21 = ½ × a × 742 = 7aTherefore, a = 6 m/s².The other options are incorrect due to errors in either substituting values into the formula or performing the calculations.

Q144. A wire of resistance 4 is bent into a circle. The resistance between the ends of a diameter of the circle is:

• A. 4Ω
• B. 1 Ω✓
• C. 1/4 Ω
• D. 1/16 Ω

Explanation: When a wire of resistance 4R is bent in the form of circle, then the resistance of each segment across the diameter becomes 2R.Equivalent resistance between A and B: +1/Rb (in parallel combination)∴ 1/RAB = 1/(2R) +1/(2R) 1/RAB = 2/(2R) ⟹ RAB=RHere is the link from where you understand this MCQ:

Q145. The state of thermal equllibrium between two systems is determined by equality of:

• A. Pressure
• B. Volume
• C. Temperature✓
• D. Mass

Explanation: Thermal equilibrium between two systems is determined by the equality of temperature, as stated by the Zeroth Law of Thermodynamics. This law implies that if no heat flows between the two systems, they are at the same temperature. Pressure and volume relate to mechanical and volumetric properties, respectively, but do not define thermal equilibrium. Mass influences heat capacity but is not a determining factor for thermal equilibrium.

Q146. Half cell reaction standard reduction potential, Eo=Fe2+ 2e- → Fe –0.41Cu2+ 2e- → Cu 0.34Ni2+ 2e- → Ni -0.25Zn2+ 2e- → Zn -0.76Referring to the table above which metal could be used to prevent iron from corrosion?

• A. Cu only
• B. Zn only✓
• C. Cu & Ni only
• D. Ni and Zn only

Explanation: The correct answer is Zn only. Zinc has a standard reduction potential of -0.76 V, which is more negative than iron's -0.41 V. This makes zinc an effective sacrificial anode, as it will oxidize before iron, thus protecting it from corrosion. Copper and nickel have higher (less negative) reduction potentials than iron, making them unsuitable for this purpose. Therefore, only zinc can provide the necessary protection.

Q147. If a wave can be polarized, it must be:

• A. An electromagnetic wave
• B. A stationary wave
• C. Transverse wave✓
• D. A longitudinal wave

Explanation: Transverse waves which include all electromagnetic waves such as light, radio waves, and x-rays can be polarized, hence the answer will be C.

Q148. 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.

Q149. A racing car accelerates uniformly through three gear changes with the following average speeds:20 ms-1 for 2.0s ;40ms-1 for 2.0 s and 60 ms-1 for 6.0s. What is the overall average speed of the car?

• A. 12 ms-1
• B. 13.3 ms-1
• C. 40 ms-1
• D. 48 ms-1✓

Explanation: To find the overall average speed, we use the formula: Average speed = Total distance / Total time. First, calculate the individual distances traveled during each gear change:For the first gear change: Distance = 20 ms-1 x 2 s = 40 mFor the second gear change: Distance = 40 ms-1 x 2 s = 80 mFor the third gear change: Distance = 60 ms-1 x 6 s = 360 mThe total distance is 40 m + 80 m + 360 m = 480 m. The total time is 2 s + 2 s + 6 s = 10 s. Thus, the overall average speed is 480 m / 10 s = 48 ms-1, making Option D the correct answer.Options A, B, and C are incorrect because they do not correctly apply the formula for average speed, either due to incorrect calculations or by using incorrect values.

Q150. A generator produces 100 kW of power at a potential difference of 10KV. The power is transmitted through cables of total resistance 5Q. How much power is dissipated in the cables?

• A. 50 W
• B. 750 W
• C. 500 W✓
• D. 1000 W

Explanation: To find the power dissipated in the cables, you need to first calculate the current flowing through the circuit. The power produced by the generator is 100 kW, which is equal to 100,000 W, and the potential difference is 10 kV, equal to 10,000 V. Using the formula P = VI, where P is power, V is voltage, and I is current, you can solve for current:I = P / V = 100,000 W / 10,000 V = 10 A.Now, using the formula for power dissipation in the cables, P = I²R, where I is the current (10 A) and R is the resistance of the cables (5 ohms), you calculate:P = 10² x 5 = 100 x 5 = 500 W.Therefore, 500 W of power is dissipated in the cables.Options A (50 W), B (750 W), and D (1000 W) are incorrect as they do not result from the correct application of the power loss formula I²R with the calculated current.

Q151. Ultraviolet rays differ from the X-rays such that ultraviolet rays _

• A. Cannot be diffracted
• B. Cannot be polarized
• C. Have a low frequency✓
• D. Do not affect a photographic plate

Explanation: Option C is correct

Q152. For transformer, if Ns/Nr = 2 : 1 then Ip/Is =

• A. 1:2
• B. 2:1✓
• C. 4:1
• D. 1:1

Explanation: Explanation is given below:

Q153. Astronomers calculate speed of distant stars and galaxies using which of the following phenomena:

• A. Beats
• B. Interference
• C. Superposition principle
• D. Doppler effect✓

Explanation: The Doppler effect is a phenomenon where the observed frequency of a wave changes depending on the relative motion between the source of the wave and the observer. This effect is commonly used in astronomy to determine the motion of celestial objects, such as stars and galaxies.In the context of astronomy, the most common application of the Doppler effect is the redshift or blueshift of light. When a celestial object is moving away from an observer, its light is shifted towards longer wavelengths (redshift), and when it is moving towards the observer, the light is shifted towards shorter wavelengths (blueshift).Astronomers can measure these shifts in the spectrum of light emitted by stars or galaxies to determine their radial velocity (motion along the line of sight). This information is crucial for understanding the dynamics of galaxies, the expansion of the universe, and other aspects of celestial motion and structure. The redshift of light from distant galaxies played a key role in the discovery of the expansion of the universe.

Q154. In a full-wave bridge rectification, the number of diodes used are:

• A. 1
• B. 2
• C. 3
• D. 4✓

Explanation: A full-wave bridge rectifier consists of 4 diodes arranged in a bridge configuration. This arrangement allows the rectifier to convert both halves of an AC waveform into a pulsating DC output. When the positive half-cycle of the AC input is applied, diodes D1 and D2 conduct, allowing current to flow through the load. During the negative half-cycle, diodes D3 and D4 conduct, again directing current through the load in the same direction. This ensures that the load always receives a unidirectional current. Options A, B, and C are incorrect because they do not provide the necessary number of diodes to achieve full-wave rectification in a bridge configuration.

Q155. The number of diodes in bridge rectifier is:

• A. 4✓
• B. 3
• C. 2
• D. 5

Explanation: A bridge rectifier consists of 4 diodes arranged in a bridge configuration.Working of a Bridge Rectifier:The four diodes are connected in such a way that they convert AC (alternating current) to DC (direct current).During the positive half-cycle of the AC input, two diodes conduct, allowing current to flow in one direction.During the negative half-cycle, the other two diodes conduct, maintaining the same direction of current flow through the load.Thus, a bridge rectifier requires 4 diodes to function properly.

Q156. What is the instantaneous power dissipated through a resistor of resistance 10 ohm if it is connected to an AC source of frequency 15 Hz at 1/180 sec if peak value is 20 V?

• A. 10 W✓
• B. 100 W
• C. 40 W
• D. Cannot be determined

Explanation: UsingVo = V sin(2𝜋ft) = 20 x sin (2𝜋 x 15 x 1/180) = 10 VUsingP = v2/ R = 102 / 10 = 10 W

Q157. Which one of the following is not electromagnetic in nature?

• A. X-rays
• B. Gamma rays
• C. Cathode rays✓
• D. Infrared rays

Explanation: Cathode rays EM waves travel with a constant velocity of 3.00x10⁸ ms-1 in vacuum. They are deflected neither by the electric field, nor by the magnetic field. These waves do not require a medium to travel from one place to another. As cathode rays negate all these properties, they are not considered as EM waves.

Q158. Which of the following has zero average value in a plane electromagnetic wave:

• A. Kinetic energy
• B. Magnetic field
• C. Electric field
• D. Both (b) and (c)✓

Explanation: In an electromagnetic wave, the average value of the electric field or magnetic field is zero, because electromagnetic waves are sinusoids and the average of a sinusoid over 1 period is zero.

Q159. A current carrying solenoid is cut into a ratio 1:3 such that both are independently connected with the same source of current. Now if the magnetic field of the first part is “BA” while the second part respectively is “BB” then what is true?

• A. BA=3BB
• B. BA=(BB)/3
• C. BA=BB✓
• D. BA=4BB

Explanation: Magnetic field in solenoid is independent of length and cross section area;B= 𝛍 nIn=I/Ln= number of turnsI=currentL=lengthB magnetic flux density 𝛍= magnetic Constant Even if solenoid is cut in any ratio; n would still remain the same because magnetic field in solenoid is independent of length and cross section area .If n remains same so will be the magnetic field

Q160. The region that lies between IR and UV regions is called:

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

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

Q161. Magnetic force on a charged particle is:

• A. QvBl
• B. None of these
• C. QvB cosθ
• D. QvB sinθ✓

Explanation: When a charge moves through a magnetic field, a force is applied on it which is given by:F = QvB sinθQvBl is the motional EMF produced when a conductor moves through a static magnetic field.

Q162. An electron is moving along the axis of a solenoid carrying a current. Which of the following is a correct statement about the magnetic force acting on the electron?

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

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

Q163. The motional E.M.F depends upon:

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

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

Q164. The cyclotron frequency of an electron projected with velocity perpendicular to a magnetic field B is given by:

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

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

Q165. An electron is projected horizontally from south to north in a uniform horizontal magnetic field acting from west to east. The direction along which it will be deflected is:

• A. Northwards
• B. Southwards
• C. Vertically upwards✓
• D. Vertically downwards

Explanation: Apply the rule as shown in the following image.By this, we get a downward direction. But keep in mind we use the right-hand rule for positive charge. In case of a negative charge, we will take opposite conclusions so the direction of force (deflection) would be vertically upward in the case of an electron.

Q166. An electron is moving along the axis of a solenoid carrying a current. Which of the following is a correct statement about the electromagnetic force acting on the electron?

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

Explanation: The question states along the axis of a solenoid. This implies that the Magnetic feild is parallel to the direction of current (For magnetic force to act, it is necessary that the direction of current and magentic feild are perpendicular to eachother) Therefore, no force acts and electron moves straight through without getting deflected.

Q167. The magnetic lines of force are directed in a manner that they:

• A. Originate at south pole and terminate at north pole
• B. Pass through the magnet
• C. Originate at north pole and terminate at south pole✓
• D. Go away from both the poles

Explanation: The correct answer is Option C: The magnetic lines of force originate at the north pole and terminate at the south pole. This is because magnetic field lines emerge from the north pole and enter the south pole, completing a loop.Option A is incorrect because it reverses the direction of the magnetic lines of force, which do not originate at the south pole.Option B is incorrect in context because, while magnetic lines do pass through the magnet as continuous loops, it does not address their directional origin and termination.Option D is incorrect because magnetic lines of force do not move away from both poles; they form loops, ensuring the magnetic field is present outside the magnet.

Q168. Which of the following physical phenomena connot be described only by the wave theory of the electromagnetic radiation?

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

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

Q169. Which of the following is the same unit as the farad?

• A. Ωs
• B. Ωs–1
• C. Ω–1s✓
• D. Ω–1 s–1

Explanation: Farad is defined as charges held per unit of volt across the capacitor. Charges are current x time. So farad is the current x time over volt, or time over ohm.

Q170. In a ripple tank, 40 waves pass through a certain point in 1 second. If the wavelength of the wave is 5 cm, then speed of the wave is:

• A. 0.5 ms-1
• B. 1 ms-1
• C. 1.5 ms-1
• D. 2 ms-1✓

Explanation: Use the formula “v=fλ”As it is known that 40 waves pass through the point in 1 second so frequency is 40.Wavelength is 5cm which can be converted to 0.05m as all options are having units in ms-1(metres per second).v=(0.05)(40)v=2ms-1

Q171. If the potential at a point which is 1m from a charge is 1 volt, then the potential at a point which is 2m from the same charge will be:

• A. 2 V
• B. 1 V
• C. 0.5 V✓
• D. 3 V

Explanation: The electric potential V at a distance r from a charge is given by the formula V = kQ/r, where k is Coulomb's constant and Q is the charge. When the distance from the charge is doubled, the potential is halved because V is inversely proportional to the distance. Thus, if the potential is 1 V at 1 m, it will be 0.5 V at 2 m. Option C is correct because it reflects this relationship. Options A and D are incorrect because they suggest the potential increases with distance, which is not the case. Option B is incorrect because it implies the potential remains the same regardless of distance, which contradicts the inverse relationship between potential and distance.

Q172. Find the missing number in the series: 6, 3, 3, 4.5, ?

• A. 5.5
• B. 7
• C. 9✓
• D. 11.5

Explanation: The sequence follows a pattern where each number is multiplied by an incrementally increasing factor. The operations are: 6 × 0.5 = 3; 3 × 1 = 3; 3 × 1.5 = 4.5. The next step is 4.5 × 2 = 9.

Q173. Find the missing number in the series: 4, 6, 10, 14, 22, 26, 34, 38, 46, ?

• A. 62
• B. 58✓
• C. 26
• D. 42

Explanation: This series is generated by multiplying 2 by the sequence of prime numbers. The terms are 2×2, 2×3, 2×5, 2×7, 2×11, and so on. The next prime after 23 is 29, so the missing term is 2 × 29 = 58.

Q174. Find the missing number in the series: 1, 5, 13, 25, ?

• A. 41✓
• B. 44
• C. 23
• D. 21

Explanation: The difference between consecutive terms increases by 4 each time. The differences are +4, +8, +12. The next difference will be +16. Therefore, the missing term is 25 + 16 = 41.

Q175. 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.

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. 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

Q178. 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.

Q179. 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.

Q180. 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.