Kmu Mdcat 2023 — Solved Past Paper with Answers

Q1. I like _ blue T-shirt over there better than _ red one.

• A. No article, the
• B. The, the✓
• C. The, a
• D. A, the

Explanation: The correct answer is Option B: the, the.In the sentence I like the blue T-shirt over there better than the red one., both 'the' articles specify that the speaker is referring to particular items that are already known to both the speaker and listener. The definite article 'the' is used when discussing specific objects. In contrast, the other options are incorrect for the following reasons:Option A fails to provide an article for 'blue T-shirt', making it ambiguous.Option C uses 'the' for 'blue T-shirt' but incorrectly uses 'a' for 'red one', suggesting it's a non-specific item.Option D incorrectly suggests a generic reference for 'blue T-shirt' with 'a', while 'the' for 'red one' is accurate but does not fit the context. Therefore, only Option B correctly conveys the intended meaning.

Q2. Choose the correct preposition. All you think and hanker _ is government service.

• A. An
• B. For✓
• C. At
• D. Upon

Explanation: The correct answer is Option B: for. The verb 'hanker' is typically followed by the preposition 'for,' which means to have a strong desire for something. In the sentence, the phrase 'hanker for' indicates that the person has a strong longing for government service, suggesting an obsession with that career path.Option A: 'an' is incorrect because it does not fit the context of expressing desire. Option C: 'at' does not work with 'hanker' and fails to convey the intended meaning. Option D: 'upon' is also incorrect, as it is not used with 'hanker' and does not express the longing that the sentence implies.

Q3. Choose the correct sentence.

• A. Who isn't here yet asked Parveen.
• B. "Who isn't here yet?" asked Parveen.✓
• C. Who isnt here yet? asked Parveen.
• D. "Who isnt here yet?" asked Parveen.

Explanation: Option B is the correct answer as the punctuation and the format are correct. The basic rules for dialogue include using quotation marks around the spoken words, separating the dialogue from the rest of the text, and including the appropriate punctuation within the question marks.

Q4. The director as well as the manager of the hotel _ attending the meeting.

• A. Is✓
• B. are
• C. were
• D. has

Explanation: Option A is correct. The sentence 'The director as well as the manager of the hotel is attending the meeting' uses 'as well as' to connect the subject, but it does not change the singular nature of the first subject, 'the director'. Therefore, the verb should also be singular, which is why 'is' is the correct choice.Options B, C, and D are incorrect because they do not maintain subject-verb agreement with the singular subject 'the director'. 'Are' suggests a plural subject, 'were' indicates past tense and implies a different number, and 'has' does not fit the present continuous form needed in this sentence.

Q5. Read the passage and answer the question given at the end.For my 25th birthday, my uncle gave me a gift and to go skydiving at a special place near Miami. I was happy because I wanted to do something wild. On the day of my jump, I could not take breakfast because of the nerves. Before I knew the danger involved in it, I was on the plane with a parachute on my back After the fall, I could feel adrenaline running through me. Then the parachute opened, and I was floating freely like a bird- Those wonderful moments helped me to realize that I am the kind of person who can take risks in life.The author learns from the experience of skydiving that:

• A. Skydiving is dangerous
• B. Miami is a good place for skydiving
• C. He is capable of doing risky things in life✓
• D. Skydiving made him so nervous that he could not take breakfast in the morning

Explanation: Option C is correct because it captures the essence of the author's realization from his skydiving experience. The passage illustrates how the experience led him to understand that he can take risks in life, which is a pivotal conclusion.

Q6. The novice teacher was excited about her first teaching job.[The word closest in meaning to the underlined word is]:

• A. Experienced
• B. Beginner✓
• C. Erudite
• D. Stem

Explanation: Option B is correct.The word closest in meaning to the underlined word novice is beginner. A novice is someone who is new to something, especially a skill or job. A beginner is also someone who is new to something, especially a skill or job.Experienced means having a lot of knowledge and skill from doing something for a long time.Erudite means having a lot of knowledge and learning.Stem is an acronym that stands for science, technology, engineering, and mathematics.

Q7. Choose the correct spelling:

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

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

Q8. If i were younger, I _ around the world.

• A. Will travel around the world.
• B. Would have travelled
• C. Would travel✓
• D. Travel

Explanation: Option C is correct.The correct answer is would travel. The second conditional tense is used to talk about imaginary or hypothetical situations in the present or future. It is formed using the following structure:If + past tense verb, would + infinitive verbIn the sentence "If I were younger, I _ around the world," the imaginary situation is that the speaker is younger. The speaker is saying that if they were younger, they would travel around the world. The other options are incorrect:Will travel around the world is the future tense, not the second conditional tense.Would have travelled is the third conditional tense, which is used to talk about imaginary or hypothetical situations in the past.Travel is the infinitive verb, not the past tense verb.

Q9. The flower girl sold him flowers.The underlined word is:

• A. Direct Object
• B. Indirect Object✓
• C. Complement
• D. Predicate

Explanation: Option B is correct.The underlined word in the sentence "The flower girl sold him flowers" is an indirect object. An indirect object is the person or thing that receives the direct object of a sentence. The direct object is the person or thing that is acted upon by the verb. In this sentence, the direct object is "flowers" and the indirect object is "him". The indirect object tells us who or what the direct object is being given to. In this case, the flowers are being given to "him".

Q10. The flower girl sold him flowers.The underlined word is:

• A. Direct object
• B. Indirect object✓
• C. Complement
• D. Predicate

Explanation: “Flowers” is the direct object of the verb “sold,” as it receives the action of selling.The indirect object receives the direct object. “Him” is the indirect object because he receives the flowers.A complement completes the meaning of a verb or adjective. “Flowers” is not a complement here.The predicate contains the verb and tells something about the subject. Here, the predicate is “sold him flowers.”Hence, option B is correct.

Q11. Identify the past simple tense in the following

• A. We have been reading in the classroom since morning
• B. He had broke this window.
• C. They spoke in Ilrdu to the waitress.✓
• D. She is enjoying the weather on the roof.

Explanation: Option C is correct. It uses the past simple tense, which is used to indicate actions that were completed at a specific time in the past. The sentence 'They spoke in Ilrdu to the waitress' clearly demonstrates this tense. Option A is in the present perfect tense, which describes actions that began in the past and continue to the present, thus not fitting the criteria for past simple. Option B contains a grammatical error with 'broke' instead of 'broken', and while it indicates a past action, it is not in the past simple form. Option D is in the present progressive tense, which indicates an action currently in progress rather than a completed action. Therefore, the only sentence that correctly uses the past simple tense is Option C.

Q12. identify the given clause:Something odd was happening all around the world.

• A. Subordinate clause
• B. Independent clause✓
• C. Dependent clause
• D. Relative clause

Explanation: Option B is correct. It is an independent clause because an independent clause can form a complete sentence on its own, containing a subject and a predicate. In the given clause, 'something odd' serves as the subject, while 'was happening all around the world' serves as the predicate, thus making it a complete thought.

Q13. We met rather few people who spoke English.The sentence is:

• A. Complex✓
• B. Simple
• C. Compound
• D. Compound complex

Explanation: Option A is correct. A complex sentence is defined as having one independent clause and at least one dependent clause. In this case, "we met rather few people" is the independent clause, and "who spoke English" is the dependent clause that provides additional detail. Option B is incorrect because a simple sentence contains only an independent clause with no dependent clauses. Option C is incorrect as a compound sentence requires at least two independent clauses. The provided sentence contains only one. Option D is incorrect since a compound-complex sentence needs at least two independent clauses along with one dependent clause, which this sentence does not have.

Q14. Identify the sentence with correct punctuation and capitalization.

• A. The automobile dealer handled three makes of cars: Toyota, Suzuki, and Honda.✓
• B. The automobile dealer handled three makes of cars, Toyotas, Suzuki, and Honda.
• C. The automobile dealer handled three makes of cars Toyota, Suzuki, and Honda.
• D. The automobile dealer handled three makes of cars, Toyota, Suzuki, and Handa.

Explanation: Option A is correct because it appropriately uses a colon to introduce the list of car brands and capitalizes proper nouns accurately. The colon is essential here as it signals that what follows is a list related to the preceding clause.Option B is incorrect because it pluralizes 'Toyota' as 'Toyotas', which is not the proper way to refer to car makes in this context.Option C is incorrect due to the absence of punctuation. A colon is needed to clarify that a list follows, making the sentence grammatically flawed.Option D is incorrect because it contains a factual error: 'Honda' is misspelled as 'Handa', even though the punctuation is correctly placed.

Q15. Which of the following sentences is in correct order?

• A. We up right drove to Karachi in two days.
• B. We to Karachi drove in two days right up.
• C. We right drove to Karachi up in two days.
• D. We drove right up to Karachi in two days.✓

Explanation: Option D is the only correct sentence structure. The rest are grammatically incorrect.The first sentence, "We up right drove to Karachi in two days," is grammatically incorrect because it places the adverb "right" before the verb "drove." Adverbs should typically be placed after the verb they modify.The second sentence, "We to Karachi drove in two days right up," is awkward because it places the prepositional phrase "to Karachi" between the auxiliary verb "drove" and the main verb "up." Prepositional phrases should typically be placed at the end of a sentence.The third sentence, "We right drove to Karachi up in two days," is awkward because it places the adverb "up" after the prepositional phrase "to Karachi." Adverbs should typically be placed before prepositional phrases.The fourth sentence, "We drove right up to Karachi in two days," is the only sentence that is both grammatically correct and natural-sounding. It clearly states that the speaker and their companions drove to Karachi in two days, and that they arrived there quickly and directly

Q16. You can either come with me now nor walk home later.The word which does not fit in the sentence correct order is:

• A. Can
• B. Either come
• C. Now
• D. Nor✓

Explanation: Option D is correct. The sentence uses 'either' but incorrectly pairs it with 'nor' instead of 'or'. The correct conjunctions would be 'either/or' or 'neither/nor'. Therefore, 'nor' does not fit in the sentence.Options A, B, and C are incorrect because they do not disrupt the grammatical integrity of the sentence. 'Can', 'either come', and 'now' can all be part of a grammatically correct sentence, while 'nor' creates a mismatch in conjunction pairing.

Q17. Find one error in the sentence, if any. I just don't want to get out off bed today.

• A. Get out
• B. Just
• C. Off✓
• D. No error

Explanation: In the sentence 'I just don't want to get out off bed today,' the error lies in the use of the word 'off.' The correct preposition should be 'of,' making the correct sentence: 'I just don't want to get out of bed today.' The preposition 'out' is usually followed by 'of' when indicating movement from a container or location, such as a bed. Therefore, the speaker is expressing a desire to remain in bed, which is a type of container. The other options are incorrect because they either point out words that do not contain errors or claim that the sentence has no errors at all.

Q18. Read the passage and answer the questions given at the end;Efforts should be made to control the use of tobacco in the campuses of educational institutions because it is gateway to drug abuse. Most of the youth usually starts with soft drugs like cigarettes, chhaliya, gutka. naswar and pan, and then move to hard drugs like heroin, opium, cocaine, ke and sheesha, etc. People who start-smoking cigarettes of drink alcohol at a young age are much more likely to experiment with illegal drugs than people who do not smoke of drink.According to this paragraph

• A. Educational institutions are gateways to the control of tobacco
• B. Youth who starts with soft drugs are likely to experiment with illegal drugs
• C. People who do not smoke or drink are less-likely, to experiment with illegal drugs✓
• D. People who start smoking at a young age are less likely to experiment with illegal drugs

Explanation: Option C is most appropriate according to the last line of the passage that "people who do not smoke or drink" are less likely to experiment with drugs.

Q19. Smoking cigarettes or drinking alcohol leads to:

• A. Legal drugs
• B. Illegal drugs✓
• C. Soft drugs
• D. The use of tobacco in educational institutions

Explanation: Option B is correct. The passage indicates that individuals who smoke cigarettes and drink alcohol often find themselves using illegal drugs. This connection highlights the concern around the escalation of substance use. Option A (legal drugs) is incorrect because the question implies a transition to more harmful substances, not those that are legally sanctioned. Option C (soft drugs) also fails to capture the severity of the situation described in the passage, as soft drugs are typically less dangerous than illegal drugs. Option D (The use of tobacco in educational institutions) is not relevant to the question's main focus on the relationship between smoking, drinking, and illegal drug use.

Q20. Which of the following statements Is not correct about alveoli:

• A. Alveoli form the gas exchange surface
• B. The wall of each alveolus is 0.1 cm thick✓
• C. Alveoli has a dense network of capillaries
• D. Collagen and elastin present in alveoli allow it to expand and recoil easily during breathing

Explanation: Option B regarding alveoli is not correct.The walls of alveoli are extremely thin, only about 0.0002 cm thick. This allows for the rapid diffusion of gases between the alveoli and the blood capillaries that surround them.while all other statements regarding alveoli are correct.

Q21. Which of the following connective tissues consists of cells known as chondrocytes?

• A. Myocardium
• B. Cartilage✓
• C. Bone
• D. Periosteum

Explanation: Option B is correct. The connective tissue that consists of cells known as chondrocytes is cartilage. Chondrocytes are the only type of cell found in cartilage. They are responsible for producing and maintaining the extracellular matrix of cartilage, which is made up of collagen, elastin, and proteoglycans. Myocardium is the muscle tissue of the heart. It contains muscle cells, also known as cardiomyocytes. Bone is a type of connective tissue that is made up of a hard, mineralized matrix. It contains osteocytes, which are bone cells that are responsible for maintaining the bone matrix. Periosteum is a tough, fibrous membrane that covers the outer surface of bones. It contains fibroblasts, which are cells that produce collagen and other extracellular matrix proteins.

Q22. The joint present between wrist bones is;

• A. Fibrous joint
• B. Cartilaginous joint /sliding joint✓
• C. Hinge joint
• D. Ball and socket joint

Explanation: Option B is correct. The joint between wrist bones is a cartilaginous joint. Cartilaginous joints are slightly movable joints that are connected by cartilage. The cartilage acts as a cushion and helps to reduce friction between the bones. Cartilaginous joints are found throughout the body, including in the spine, pelvis, and ribs. They are also found in the joints between the skull bones. Fibrous joints are immovable joints that are connected by fibrous connective tissue. Fibrous joints are found in the skull and pelvis. Hinge joints are slightly movable joints that allow for movement in one plane. Hinge joints are found in the elbows, knees, and fingers. Ball and socket joints are freely movable joints that allow for movement in all planes. Ball and socket joints are found in the hips, shoulders, and head of the thighbone.

Q23. The sarcomere is the region of a myofibril between two successive:

• A. I-Line
• B. A-Line
• C. Z-Line✓
• D. M-Line

Explanation: Option C is correct. The sarcomere is the region of a myofibril between two successive Z-lines. A myofibril is a long, thin strand of muscle tissue that is made up of sarcomeres. Sarcomeres are the basic contractile units of muscle tissue. The Z-line is a thin, dark line that marks the boundary of a sarcomere. It is made up of a protein called α-actinin, which helps to anchor the actin filaments in place. I-band: The I-band is the light-colored band in the sarcomere. It contains only actin filaments. A-band: The A-band is the dark-colored band in the sarcomere. It contains both actin and myosin filaments. M-line: The M-line is a thin line in the center of the A-band. It is made up of a protein called myomesin, which helps to stabilize the myosin filaments.

Q24. The size of a synaptic cleft (the gap separating nerve cells) in electrical synapse is

• A. 0.2 nm✓
• B. 2 nm
• C. 20 nm
• D. 200 nm

Explanation: Option A is correct. The size of a synaptic cleft in an electrical synapse is typically 0.2 nm, which is much smaller compared to the synaptic cleft of chemical synapses, which ranges from 20 to 40 nm. This small size is crucial because it allows ions and small molecules to pass quickly between the two cells through gap junctions, facilitating rapid electrical signal transmission. Conversely, Option B (2 nm) and Option C (20 nm) are incorrect as they exceed the actual size of the cleft in electrical synapses, which would hinder the speed of signal transmission. Option D (200 nm) is also incorrect, being significantly larger than the typical size, and does not represent the characteristics of electrical synapses.

Q25. The activities like sleeping, walking, and dreaming are controlled by which part of the brain:

• A. Pons✓
• B. Cerebellum
• C. Olfactory bulb
• D. Hippocampus

Explanation: The activities like sleeping, walking, and dreaming are controlled by the pons. The pons is a small but important part of the brain that is located in the brainstem. It is involved in a variety of functions, including: 1. Sleep-wake cycle: The pons helps to regulate the sleep-wake cycle by controlling the production of melatonin, a hormone that is involved in sleep regulation. 2. Motor control: The pons helps to coordinate the movement of the eyes, face, and tongue. It also helps to control the muscles that are involved in breathing and swallowing. 3. Sensory processing: The pons helps to process sensory information from the body, including information from the ears, nose, and tongue. The cerebellum, olfactory bulb, and hippocampus are three important parts of the brain with distinct functions.

Q26. The hormone 'calcitonin' is secreted by:

• A. Thyroid gland✓
• B. Parathyroid gland
• C. Pancreas
• D. Adrenal gland

Explanation: Option A is correct. The hormone calcitonin is secreted by the thyroid gland. The thyroid gland is a small, butterfly-shaped gland located in the neck. It is responsible for producing hormones that regulate metabolism, growth, and development, and body temperature. Calcitonin is a hormone that helps to regulate calcium levels in the blood. It works by inhibiting the activity of osteoclasts, which are cells that break down bone. This helps to prevent calcium from being released into the blood from bone. Calcitonin also helps to increase the excretion of calcium in the urine. This also helps to lower calcium levels in the blood. The other options are not involved in the secretion of calcitonin: Parathyroid gland: The parathyroid gland produces parathyroid hormone, which is a hormone that helps to raise calcium levels in the blood. Pancreas: The pancreas produces insulin and glucagon, which are hormones that regulate blood sugar levels. Adrenal gland: The adrenal gland produces hormones such as cortisol, adrenaline, and aldosterone, which are involved in a variety of functions, including stress response, metabolism, and blood pressure regulation.

Q27. Fallopian tube is also known as:

• A. Oviduct✓
• B. Ovaries
• C. Uterus
• D. Cervix

Explanation: Option A is correct.Fallopian tube is also known as oviduct.Fallopian tubes are muscular tubes that connect the ovaries to the uterus. They transport eggs from the ovaries to the uterus, where they can be fertilized by sperm. If an egg is fertilized, it begins to travel down the fallopian tube to the uterus, where it implants and grows into a fetus. Fallopian tubes are essential for conception. If the fallopian tubes are blocked or damaged, it can make it difficult or impossible for a woman to get pregnant. Ovaries: The ovaries are two small, oval-shaped glands located on either side of the uterus. They are responsible for producing eggs (also called ova) and hormones, such as estrogen and progesterone.Uterus: The uterus is a hollow, pear-shaped organ located in the lower abdomen. It is where a fertilized egg implants and grows into a fetus. The uterus is also responsible for menstruation, which is the monthly shedding of the lining of the uterus if a fertilized egg does not implant. Cervix: The cervix is the narrow opening at the bottom of the uterus that leads to the vagina. It produces mucus, which helps to protect the uterus from infection and sperm. The cervix also dilates and opens during childbirth to allow the baby to pass through.

Q28. The syphillis is a _ disease.

• A. Bacterial✓
• B. Fungal
• C. Protozoic
• D. Viral

Explanation: Option A is correct.Syphilis is a bacterial disease. It is caused by the bacterium Treponema pallidum. Syphilis is a sexually transmitted infection (STI) that can also be transmitted from mother to child during pregnancy or childbirth. Bacterial diseases are caused by bacteria. Bacteria are microscopic organisms that can cause a variety of diseases, including pneumonia, strep throat, and urinary tract infections.Fungal diseases are caused by fungi. Fungi are microscopic organisms that can cause a variety of diseases, including athlete's foot, ringworm, and candidiasis.Protozoan diseases are caused by protozoa. Protozoa are microscopic organisms that can cause a variety of diseases, including malaria, giardiasis, and toxoplasmosis.Viral diseases are caused by viruses. Viruses are microscopic organisms that can cause a variety of diseases, including the common cold, influenza, and HIV/AIDS.

Q29. The most common symptom of gonorrhea in male is:

• A. Skin rash on neck
• B. Joint pain
• C. Painful urination✓
• D. Chancre

Explanation: Option C is correct.The most common symptom of gonorrhea in males is painful urination. Other symptoms of gonorrhea in males can include;White, yellow, or green discharge from the penisPainful or swollen testiclesRedness and inflammation at the opening of the penisGonorrhea is a sexually transmitted infection (STI) caused by the bacterium Neisseria gonorrhoeae. It can infect the urethra, rectum, and throat. Gonorrhea can be treated with antibiotics, but if left untreated, it can lead to serious health problems, such as infertility and pelvic inflammatory disease (PID).The other options are not the most common symptoms of gonorrhea in males:Skin rash on neck: A skin rash on the neck is not a typical symptom of gonorrhea.Joint pain: Joint pain can be a symptom of gonorrhea, but it is not the most common symptom.Chancre: A chancre is a small, round ulcer that can be a sign of syphilis, another STI.

Q30. G.J Mendel chose _ as an experimental plant for his study

• A. Medicago sativa
• B. Piscum Sativum✓
• C. Delonicx regia
• D. Lens Culinaris

Explanation: Option B is coorect.G.J Mendel chose Pisum sativum, the garden pea plant, as an experimental plant for his study.Mendel chose pea plants because they have a number of characteristics that make them ideal for studying inheritance:1.They have a short life cycle, so Mendel could study multiple generations of plants in a relatively short period of time.2.They produce a large number of seeds, which gave Mendel a large sample size to work with.3.They have a variety of easily observable traits, such as seed color, flower color, and plant height.4.They are relatively easy to cross-pollinate, which allowed Mendel to control the inheritance of traits.Mendel's experiments with pea plants led to the discovery of the basic laws of inheritance, which are still taught in biology classrooms today.The other options are not plants that Mendel used in his experiments:Medicago sativa is alfalfa, a legume plant that is used as forage for livestock.Delonix regia is the flamboyant tree, a flowering tree that is native to Madagascar.Lens culinaris is the lentil, a legume plant that is grown for its edible seeds.

Q31. When round and yellow color seeded plant (RrYy) is self-crossed the number of plants with rounds and yellow color seeds obtained is _ in 32 plants:

• A. 18✓
• B. 9
• C. 3
• D. 1

Explanation: Option A is correct.When a round and yellow seeded plant (RrYy) is self-crossed, the expected phenotypic ratio of the offspring is 9:3:3:1. In this ratio, 9 represents the plants with round and yellow seeds. To find the number of these plants in a total of 32, you can calculate:(9/16) * 32 = 18.This indicates that out of 32 plants, we expect 18 to exhibit the round and yellow phenotype. The other options are incorrect because they do not accurately reflect the expected ratios from a dihybrid cross. Option B (9) represents only a fraction of the expected round and yellow plants; Option C (3) and Option D (1) both underestimate the number of plants based on the Mendelian inheritance ratios. Thus, they do not align with the expected outcomes of the genetic cross.

Q32. In crossing over, the exchange of maternal and paternal chromatids occurs while homologous chromosomes are paired during:

• A. Prophase meiosis I✓
• B. Prophase meiosis II
• C. Metaphase of meiosis I
• D. Anaphase meiosis II

Explanation: Option A is correct. Crossing over occurs during prophase I of meiosis I, when homologous chromosomes pair up and exchange segments of their genetic material, leading to genetic variation in the offspring. In contrast, Option B is incorrect because prophase II occurs after the homologous chromosomes have been separated, and no crossing over takes place. Option C is also incorrect; during metaphase I, homologous chromosomes align but do not exchange genetic material. Lastly, Option D is incorrect as anaphase II focuses solely on the separation of sister chromatids, with no involvement of crossing over.

Q33. Drosophila Melanogaster is the scientific name of:

• A. House fly
• B. Fruit fly✓
• C. Flash fly
• D. Sand fly

Explanation: Option B is correct.Drosophila Melanogaster is the scientific name of the fruit fly. It is a small fly that is commonly found in homes and restaurants. Fruit flies are attracted to ripe fruit and decaying organic matter. The other options are not the scientific names of the fruit fly:House fly: The scientific name of the house fly is Musca domestica.Flash fly: The scientific name of the flash fly is Phoridae.Sand fly: The scientific name of the sand fly is Phlebotominae.

Q34. A condition that renders the individual less able to form blood clots is:

• A. Alports syndrome
• B. Coffin-Lowry syndrome
• C. Hemophilia✓
• D. Thalassemia

Explanation: Option C is correct.Hemophilia is a condition that renders the individual less able to form blood clots. It is a genetic disorder that affects the blood's ability to clot. People with hemophilia have low levels of clotting factors, which are proteins that help the blood to clot. This can lead to excessive bleeding, even from minor injuries.Hemophilia is a serious condition, but it can be managed with treatment. There are two types of hemophilia: hemophilia A and hemophilia B. Hemophilia A is the most common type of hemophilia. It is caused by a deficiency in clotting factor VIII. Hemophilia B is less common than hemophilia A. It is caused by a deficiency in clotting factor IX.Treatment for hemophilia typically involves replacing the missing clotting factor with infusions. This can be done as needed to stop bleeding or on a regular basis to prevent bleeding. The other options are not conditions that render the individual less able to form blood clots:Alport syndrome is a genetic disorder that affects the kidneys, eyes, and ears.Coffin-Lowry syndrome is a genetic disorder that causes intellectual disability, short stature, and distinctive facial features.Thalassemia is a genetic disorder that affects the production of hemoglobin, the protein in red blood cells that carries oxygen.

Q35. Intra-specific struggle refers to:

• A. Competition between members of the same species✓
• B. Competition between members of the different species
• C. Environmental struggle
• D. Prey or predation

Explanation: Option A is correct. Intra-specific struggle refers to competition between members of the same species. This can include competition for food, water, mates, and territory. Intra-specific struggle is a very important factor in natural selection, as it helps to ensure that only the fittest individuals survive and reproduce. Intra-specific struggle can be very intense, as the individuals involved are all very similar and have the same needs. This can lead to aggression, fighting, and even death. However, intra-specific struggle can also have positive benefits, such as driving innovation and promoting cooperation. The other options are not correct: Competition between members of different species is called interspecific struggle. Environmental struggle is the competition between organisms of a population against one another in nature to maintain themselves in a given environment to survive and reproduce. Prey or predation is the interaction between two species, where one species captures and kills the other for food.

Q36. Identify fossil bird among the following:

• A. Equus
• B. Python
• C. Archaeopteryx✓
• D. Dawn horse

Explanation: The correct answer is Archaeopteryx, which is a significant fossil in the study of evolution as it exhibits both avian and non-avian characteristics. This makes it a key piece of evidence for the transition from dinosaurs to modern birds.The other options are incorrect for the following reasons: Equus is associated with horses and is a mammal, Python is a genus of snakes, and Dawn horse is a type of ancient horse. None of these represent fossil birds, highlighting the unique position of Archaeopteryx in the fossil record.

Q37. Which of the following organelle is regarded as self-replicable organelle

• A. Endoplasmic reticulum
• B. Golgi bodies
• C. Mitochondria✓
• D. Vacoule

Explanation: Option C is correct.Mitochondria are regarded as self-replicating organelles. They have their own DNA and can replicate independently of the nucleus. This means that they can divide and reproduce independently, without the assistance of other organelles.Mitochondria are essential for energy production in the cell. They convert food into energy in the form of ATP, which is the cell's energy currency. Mitochondria are also involved in other cellular processes, such as cell signaling and apoptosis (programmed cell death).The endoplasmic reticulum (ER) is not regarded as a self-replicating organelle because it does not have its own DNA. DNA is required for self-replication, as it contains the instructions for building new organelles. The ER is a network of membranes that is involved in the production and transport of proteins and lipids. It is made up of two types of membranes: the rough ER and the smooth ER. The rough ER is covered in ribosomes, which are the organelles that synthesize proteins. The smooth ER is involved in the production of lipids and the detoxification of drugs and toxins. Golgi bodies are also not self-replicating organelles. Golgi bodies are involved in the processing and packaging of proteins and lipids for transport to other parts of the cell. Vacuoles are not self-replicating organelles. Vacuoles are storage organelles that can contain water, salts, and other molecules.

Q38. The prokaryotic cell wall is composed of:

• A. Cellulose
• B. Chitin
• C. Murein✓
• D. Pectin

Explanation: Option C is correct.The prokaryotic cell wall is composed of murein, also known as peptidoglycan. Murein is a polymer made up of two sugar molecules and a short peptide chain. It is a very strong and rigid material that helps to protect the cell from osmotic lysis and other environmental stresses.Cellulose is a polymer of glucose molecules that is found in the cell walls of plants. Chitin is a polymer of N-acetylglucosamine molecules that is found in the exoskeletons of insects and other invertebrates. Pectin is a polymer of galacturonic acid molecules that is found in the cell walls of plants.

Q39. _ is found in the exoskeleton of crabs

• A. cellulose
• B. chitin✓
• C. murein
• D. hemi-cellulose

Explanation: Option B is correct. Chitin is found in the exoskeleton of crabs. Chitin is a tough, fibrous material that is made up of long chains of sugar molecules. It is the main component of the exoskeletons of insects, spiders, crabs, and other invertebrates. Chitin is also found in the cell walls of fungi and some algae. The other options are not found in the exoskeleton of crabs: Cellulose is a tough, fibrous material that is made up of long chains of sugar molecules. It is the main component of the cell walls of plants. Murein is a polymer made up of two sugar molecules and a short peptide chain. It is the main component of the cell walls of bacteria. Hemicellulose is a polymer of sugar molecules that is found in the cell walls of plants.

Q40. Choose the unsaturated fatty acid among the following:

• A. Palmitic acid
• B. Archachidic acid
• C. Stearic acid
• D. Oleic acid✓

Explanation: Option D is correct. Oleic acid is the unsaturated fatty acid. Fatty acids are classified into saturated and unsaturated fatty acids. Saturated fatty acids do not have any double bonds between the carbon atoms in their chain, while unsaturated fatty acids have at least one double bond. Unsaturated fatty acids are considered to be healthier than saturated fatty acids because they can help to lower cholesterol levels and reduce the risk of heart disease. Oleic acid is a monounsaturated fatty acid, which means that it has one double bond in its chain. It is the most common monounsaturated fatty acid in the human diet and is found in many plant-based oils, such as olive oil and avocado oil. The other options are saturated fatty acids:Palmitic acid is a long-chain saturated fatty acid that is found in many animal products, such as meat, butter, and cheese. Arachidic acid is a long-chain saturated fatty acid that is found in peanuts, peanut oil, and chocolate. Stearic acid is a long-chain saturated fatty acid that is found in beef, lamb, and cocoa butter.

Q41. The conjugated molecule that is primarily present in egg albumin is

• A. Lipoprotein
• B. Nucleoprotein
• C. Glycolipid
• D. Glycoprotein✓

Explanation: Option D is correct.The conjugated molecule that is primarily present in egg albumin is glycoprotein. Egg albumin is a water-soluble protein that is found in the white of eggs. It is a major source of protein in the human diet and is also used in a variety of food products, such as mayonnaise, pasta, and baked goods. Egg albumin is a glycoprotein, which means that it is conjugated to carbohydrates. The carbohydrates are attached to the protein through glycosidic bonds. Glycosylation is a process that is important for the folding, stability, and function of many proteins. The other options are not conjugated molecules that are primarily present in egg albumin: Lipoproteins are conjugated molecules that are composed of proteins and lipids. They are found in the blood and transport lipids throughout the body.Nucleoproteins are conjugated molecules that are composed of proteins and nucleic acids. They are found in the nucleus of cells and play a role in gene expression. Glycolipids are conjugated molecules that are composed of carbohydrates and lipids. They are found in the cell membrane and play a role in cell communication and adhesion.

Q42. FSH (Follicle stimulating hormones) stimulates spermatogenesis by stimulating _ cells to complete the development of sperms

• A. Leydig
• B. Inhibin
• C. Sertoli✓
• D. TSH

Explanation: Option C is correct.FSH (Follicle stimulating hormones) stimulates spermatogenesis by stimulating Sertoli cells to complete the development of sperms. Sertoli cells are supporting cells that are found in the seminiferous tubules of the testes. They play a vital role in spermatogenesis, the process by which sperm cells are produced. FSH stimulates Sertoli cells to produce a number of factors that are essential for spermatogenesis, including: Androgen binding protein (ABP), which binds to testosterone and makes it available to developing sperm cells Growth factors, which promote the growth and development of sperm cells Inhibitin, which inhibits the production of FSH.Leydig cells are another type of cell that is found in the testes. They are responsible for producing testosterone, the male sex hormone. Testosterone is also essential for spermatogenesis, but it is produced in response to LH (luteinizing hormone), not FSH.Inhibin is a hormone that is produced by Sertoli cells. It inhibits the production of FSH and LH. This helps to ensure that the production of sperm cells is regulated.TSH (thyroid stimulating hormone) is a hormone that is produced by the pituitary gland. It stimulates the thyroid gland to produce thyroid hormones. Thyroid hormones are not involved in spermatogenesis.

Q43. The process of photophosphorylation takes place in:

• A. Nucelolus
• B. Chloroplast✓
• C. Nucleus
• D. Golgi bodies

Explanation: Option B is correct.The process of photophosphorylation takes place in the chloroplast. Photophosphorylation is the process by which light energy is used to produce ATP, the energy currency of the cell. It is the light-dependent reaction of photosynthesis. Photophosphorylation takes place in the thylakoid membranes of the chloroplast. The thylakoid membranes are stacked sacs that contain chlorophyll, the light-absorbing pigment. When sunlight hits the chlorophyll, it excites the electrons in the chlorophyll molecules. These excited electrons are then passed through a series of electron carriers, which eventually leads to the production of ATP. The ATP that is produced by photophosphorylation is used to drive the Calvin cycle, the light-independent reaction of photosynthesis. The Calvin cycle uses ATP to convert carbon dioxide into sugar.The other options are not organelles where photophosphorylation takes place:The nucleolus is a small structure within the nucleus that is involved in the production of ribosomes.The nucleus is the control center of the cell and contains the cell's DNA.The Golgi apparatus is involved in the packaging and processing of proteins and lipid

Q44. Which of the following hormones is released by the posterior lobe of the pituitary gland?

• A. Antidiuretic hormones✓
• B. Growth hormone
• C. Estrogen
• D. Prolactin

Explanation: Option A is correct. The posterior lobe of the pituitary gland releases antidiuretic hormone (ADH). ADH is also known as vasopressin. It is a hormone that helps the body retain water. ADH is released when the body's fluid levels are low. It acts on the kidneys to cause them to reabsorb more water from the urine. This helps to increase the body's fluid levels and prevent dehydration. The other options are not hormones that are released by the posterior lobe of the pituitary gland: Growth hormone is released by the anterior lobe of the pituitary gland. Estrogen is a female sex hormone that is produced by the ovaries and other tissues in the body. Prolactin is a hormone that is produced by the anterior lobe of the pituitary gland. It stimulates milk production in women.

Q45. Which of the following agents can easily pass through filter paper?

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

Explanation: Viruses can easily pass through filter paper. Viruses are the smallest type of biological agent, with most being smaller than 0.1 micrometers in diameter. Filter paper typically has pores that are several micrometers in diameter, so viruses are small enough to pass through the pores. Viruses can easily pass through filter paper, while bacteria, fungi, and yeast cannot.Bacteria are single-celled organisms that are typically 0.5 to 5 micrometers in size. They are too large to pass through the pores of filter paper, which are typically 0.2 to 0.45 micrometers in diameter.Fungi are eukaryotic organisms that can be unicellular or multicellular. They typically range in size from a few micrometers to several meters. Fungi are too large to pass through the pores of filter paper.Yeast is a type of fungus that is unicellular and typically 5 to 10 micrometers in size. Yeast is too large to pass through the pores of filter paper.

Q46. HIV belongs to a special class of virus known as:

• A. Retrovirus✓
• B. Flavivirus
• C. Novavirus
• D. Tetanus

Explanation: HIV belongs to a special class of virus known as retrovirus. Retroviruses are a type of virus that uses RNA as its genetic material. They are also unique in that they use an enzyme called reverse transcriptase to convert their RNA genome into DNA. This DNA is then integrated into the host cell's genome, where it can be transcribed and translated into viral proteins. Retroviruses are responsible for a number of diseases in humans and animals, including HIV/AIDS, leukemia, and lymphoma. HIV is a retrovirus that attacks the immune system, making it difficult for the body to fight off infection.

Q47. The capsule of a bacterial cell is a sticky gelatinous structure made up of protein and:

• A. Lipids
• B. Carbohydrates✓
• C. Nucleotide
• D. Alcohol

Explanation: Option B is correct.The capsule of a bacterial cell is a sticky gelatinous structure made up of carbohydrates. The capsule is a layer of polysaccharide material that surrounds the cell wall of some bacteria. It is composed of repeating units of sugar molecules. The capsule is a dynamic structure that can be produced or shed by the bacteria in response to environmental conditions. The capsule has a number of functions, including: Protecting the bacteria from dehydration Helping the bacteria to adhere to surfaces Evasion of the host immune system Some bacteria, such as Streptococcus pneumoniae, have a thick capsule that is essential for their virulence. The capsule protects the bacteria from phagocytosis, a process by which white blood cells engulf and destroy bacteria. The other options are not components of the bacterial capsule: Lipids are a type of organic molecule that is made up of fatty acids and glycerol. They are found in the cell membrane and other cellular structures. Nucleotides are the building blocks of DNA and RNA. Nucleotides are joined together by phosphodiester bonds to form DNA and RNA molecules. The order of the nucleotides in DNA and RNA determines the genetic code. Alcohols are a type of organic molecule that contains a hydroxyl group (OH). They are found in a variety of foods and beverages.

Q48. The transgenic plants, such as golden rice, are produced primarily with the help of:

• A. Escherichia coli
• B. Proteus mirabilis
• C. Agrobacterium tumerfaciens✓
• D. Ralstonia solanacearum

Explanation: Transgenic plants such as golden rice are produced primarily with the help of Agrobacterium tumefaciens. Agrobacterium tumefaciens is a bacterium that causes crown gall disease in plants. It is also a natural genetic engineer, as it can transfer its own DNA into plant cells. This ability has been exploited to create transgenic plants, which are plants that have had their DNA modified to include genes from other organisms.

Q49. Which of the following chemicals is most commonly used to preserve biological specimen?

• A. Formalin✓
• B. Glutaraldehyde
• C. Ethylene Oxide
• D. Iodine

Explanation: Formalin is a solution of formaldehyde in water. It is the most common preservative used for biological specimens, as it is effective and inexpensive. Formalin preserves specimens by cross-linking proteins, which prevents them from decaying.

Q50. Coral reef are the characteristic feature of phylum:

• A. Porifera
• B. Coelenterata/Cnidaria✓
• C. Arthropoda
• D. Platyhelminthes

Explanation: Coral reefs are the characteristic feature of the phylum Cnidaria. Cnidaria is a phylum of aquatic invertebrates that includes corals, sea anemones, jellyfish, and hydras. Cnidarians are characterized by their radial symmetry, the presence of nematocysts (stinging cells), and a gastrovascular cavity (a single opening that serves as both a mouth and an anus). Coral reefs are built by colonial cnidarians called scleractinian corals. Scleractinian corals secrete a hard calcium carbonate skeleton that supports the reef structure. Coral reefs are found in warm, shallow waters around the world and are home to a wide variety of marine life.

Q51. According to the sliding filament hypothesis, the release of_ from the sarcoplasmic reticulum causes the reorientation of certain components in the thin filament, permitting them to bind with extension from the thick myosin filaments:

• A. Sodium
• B. Calcium✓
• C. Magnesium
• D. Potassium

Explanation: Option B is correct. According to the sliding filament hypothesis, the release of calcium ions from the sarcoplasmic reticulum causes the reorientation of certain components in the thin filament, permitting them to bind with extension from the thick myosin filaments. Calcium ions play a crucial role in muscle contraction by binding to the protein troponin on the thin filament. This binding causes tropomyosin to move out of the way, exposing myosin binding sites on the actin filament. Myosin heads can then bind to these sites, and cross-bridges can be formed. The cross-bridges then power the sliding of the thin filaments past the thick filaments, which results in muscle contraction. The other options are not involved in muscle contraction: Sodium ions are involved in nerve conduction and muscle action potentials. Magnesium ions are cofactors for many enzymes, including those involved in muscle metabolism. Potassium ions are involved in maintaining the resting membrane potential of muscle cells.

Q52. The green glands found in arthropods are concerned with:

• A. Excretion✓
• B. Respiration
• C. Digestion
• D. Circulation

Explanation: Option A is correct.The green glands found in arthropods are concerned with excretion. Green glands are paired excretory organs that are found in the head of crustaceans. They are responsible for filtering waste products from the blood and excreting them from the body. Green glands are also thought to play a role in osmoregulation, the process of maintaining a balance of water and salts in the body. The other options are not functions of the green glands:Respiration is the process of exchanging oxygen and carbon dioxide between the body and the environment. Arthropods respire through a variety of structures, including gills, tracheae, and book lungs. Digestion is the process of breaking down food into smaller molecules that can be absorbed by the body. Arthropods digest food in a variety of ways, including through their mouths, stomachs, and intestines.Circulation is the movement of blood throughout the body. Arthropods have an open circulatory system, in which the blood does not flow in closed vessels.

Q53. The Purkinje fibers are the specialized fibers found in:

• A. Human heart✓
• B. Human lungs
• C. Human stomach
• D. Human brain

Explanation: Option A is correct.The Purkinje fibers are the specialized fibers found in the human heart. Purkinje fibers are a network of specialized cardiac muscle fibers that conduct electrical impulses from the atrioventricular node (AV node) to the ventricles of the heart. They are responsible for coordinating the contraction of the ventricles, which allows them to pump blood efficiently. The Purkinje fibers are located in the subendocardial layer of the heart, which is the layer of muscle tissue that is closest to the endocardium, the inner lining of the heart. They are large, pale fibers that are easily distinguished from the surrounding cardiac muscle cells. The other options are not organs where Purkinje fibers are found: Lungs are the organs of respiration. They are responsible for exchanging oxygen and carbon dioxide between the body and the environment. The stomach is a digestive organ that is responsible for breaking down food into smaller molecules that can be absorbed by the body. The brain is the central nervous system organ that is responsible for controlling all of the body's functions.

Q54. The largest body of lymphoid tissue in the human body is:

• A. Tonsils
• B. Pancreas
• C. Spleen✓
• D. Thymus

Explanation: Option C is correct.The largest body of lymphoid tissue in the human body is the spleen. The spleen is a soft, purplish organ located in the upper left abdomen, beneath the diaphragm and posterior to the stomach. It is about 5 inches long and 2 inches wide. The spleen is a major part of the lymphatic system, which is responsible for fighting infection and disease. The spleen contains two types of tissue: white pulp and red pulp. The white pulp contains lymphocytes, which are white blood cells that play a key role in the immune system. The red pulp contains macrophages, which are scavenger cells that engulf and destroy bacteria and other foreign invaders. The spleen has a number of important functions, including: Filtering blood to remove bacteria, dead cells, and other debris Producing and storing lymphocytes Regulating the production of red blood cells Helping to fight infection and diseaseThe other options are not the largest body of lymphoid tissue in the human body: Tonsils are small, round masses of lymphoid tissue located at the back of the throat. They help to protect the body from infection by trapping bacteria and viruses.The pancreas is a digestive organ that produces enzymes that help to break down food. It also produces hormones, such as insulin and glucagon, which regulate blood sugar levels. The thymus is a lymphoid organ located in the chest, behind the breastbone. It is responsible for the maturation of T lymphocytes, a type of white blood cell that plays a key role in the cell-mediated immune response.

Q55. The immune cells that clean up pus as a part of the healing process is:

• A. Natural killer cell
• B. Macrophages✓
• C. Antibodies
• D. Neutrophils

Explanation: Neutrophils are the first responders — they rush to the infection site, attack bacteria, and die in large numbers.During healing, macrophages arrive after neutrophils and clean up the dead cells, bacteria, and debris — including pus (which is mostly dead neutrophils and tissue).They “eat up” the waste through phagocytosis, helping tissues heal properlyAntibodies are proteins that are produced by B lymphocytes in response to infection. They bind to specific antigens on the surface of pathogens, which helps to neutralize them and mark them for destruction by other immune cells.

Q56. Lipid synthesis occurs in which one of the following cell organelles?

• A. Ribosomes
• B. Golgi bodies
• C. Mitochondria
• D. Endoplasmic reticulum✓

Explanation: Option D is correct.Lipid synthesis occurs in the endoplasmic reticulum (ER) of the cell. The ER is a network of interconnected tubules and sacs that is found in the cytoplasm of all eukaryotic cells. It is involved in a variety of cellular functions, including protein synthesis, lipid synthesis, and detoxification. Lipid synthesis occurs on the smooth ER, which is a type of ER that does not have ribosomes. The smooth ER contains a number of enzymes that are involved in the synthesis of different types of lipids, including triglycerides, phospholipids, and cholesterol. Once lipids have been synthesized in the ER, they are transported to other parts of the cell, such as the Golgi apparatus and the plasma membrane. Lipids are also used to build cell membranes and other cellular organelles. The other options are not organelles where lipid synthesis occurs: Ribosomes are organelles that are responsible for protein synthesis. Golgi bodies are organelles that are involved in the processing and packaging of proteins. Mitochondria are organelles that are responsible for energy production.

Q57. The testes are male gonads which are situated outside the abdomen within a skin pouch called :

• A. Vas deferens
• B. Epididymis
• C. Scrotum✓
• D. Vasa efferentia

Explanation: Option C is correct.The testes are male gonads which are situated outside the abdomen within a skin pouch called scrotum. The scrotum is a pouch of skin that hangs below the penis. It contains the testes, epididymis, and part of the vas deferens.The epididymis is a long, coiled tube that lies along the back of each testis. It stores and matures sperm.The vas deferens is a tube that carries sperm from the epididymis to the ejaculatory duct.The vasa efferentia are a series of thin tubes that carry sperm from the testes to the epididymis.

Q58. The term polydipsia refers to:

• A. A condition in which abnormally large volume of urine is produced
• B. A condition of exessive thirst✓
• C. A condition of excessive hunger
• D. A condition in which the concentration of blood decreases in the body

Explanation: Option B is correct.The term polydipsia refers to a condition of excessive thirst. It is a symptom of many medical conditions, including diabetes, diabetes insipidus, psychogenic polydipsia, and certain medications. Polydipsia can also be caused by dehydration, hot weather, and strenuous exercise. When a person has polydipsia, they may drink more than three liters of fluids per day. This can lead to frequent urination, which can be disruptive to daily life. If you are experiencing polydipsia, it is important to see a doctor to determine the underlying cause and get the appropriate treatment.The other options are not the definition of polydipsia: Polyuria is a condition in which a person produces abnormally large amounts of urine. Polyphagia is a condition of excessive hunger. Hypovolemia is a condition in which the concentration of blood decreases in the body.

Q59. Lactose is composed of:

• A. Glucose + Fructose
• B. Glucose + Galactose✓
• C. Glucose + Glucose
• D. Fructose + Galactose

Explanation: Option B is correct.Lactose is composed of glucose + galactose. Lactose is a disaccharide, which means that it is made up of two monosaccharides (simple sugars). The two monosaccharides that make up lactose are glucose and galactose. Lactose is found in milk and other dairy products. It is also used in a variety of food products, such as bread, candy, and ice cream. Lactose is broken down into glucose and galactose in the small intestine by the enzyme lactase. Glucose and galactose are then absorbed into the bloodstream and used by the body for energy. People who are lactose intolerant are unable to digest lactose properly. This is because they lack the enzyme lactase.

Q60. Saprophytic bacteria are:

• A. Autotrophs
• B. Decomposers✓
• C. Parasites
• D. Photosynthetic

Explanation: Option B is correct.Saprophytic bacteria are decomposers. Decomposers are organisms that break down dead organic matter. Saprophytic bacteria play an important role in the environment by breaking down dead plants and animals, which releases nutrients back into the soil and air. Examples of saprophytic bacteria include: Bacillus subtilis, Clostridium tetani ,Escherichia coliAutotrophs are organisms that can make their own food from sunlight and carbon dioxide. Decomposers are not autotrophs because they cannot make their own food. Parasites are organisms that live on or inside another organism and get their nutrients from that organism. Saprophytic bacteria are not parasites because they do not live on or inside another organism. Photosynthetic organisms are organisms that use sunlight to convert carbon dioxide into oxygen and carbohydrates. Saprophytic bacteria are not photosynthetic because they do not use sunlight to make their own food.

Q61. In human heart's right atrium communicates with the right ventricle through:

• A. Inter-auricular septum
• B. Ventricular septum
• C. Tricuspid valve✓
• D. Bicuspid valve

Explanation: Option C is correct.In the human heart, the right atrium communicates with the right ventricle through the tricuspid valve. The tricuspid valve is a one-way valve that prevents blood from flowing back from the right ventricle to the right atrium. It is made up of three thin cusps (flaps) that are attached to chordae tendineae (tendinous cords). The chordae tendineae are attached to the papillary muscles, which are small muscles located on the inner walls of the right ventricle. When the right atrium contracts, it forces blood through the open tricuspid valve into the right ventricle. When the right ventricle contracts, it closes the tricuspid valve to prevent blood from flowing back into the right atrium. The other options are not structures that connect the right atrium to the right ventricle: The interatrial septum is a wall that separates the right atrium from the left atrium. The ventricular septum is a wall that separates the right ventricle from the left ventricle. The bicuspid valve, also known as the mitral valve, is a valve that connects the left atrium to the left ventricle.

Q62. The first segment of the small intestine is:

• A. Duodenum✓
• B. Jejunum
• C. Ileum
• D. Colon

Explanation: Option A is correct. The first segment of the small intestine is the duodenum. The small intestine is a long, coiled tube that is divided into three parts: the duodenum, the jejunum, and the ileum. The duodenum is the shortest and widest part of the small intestine. It is about 10 inches long and curves around the head of the pancreas. The duodenum is responsible for the initial digestion of food. It receives food from the stomach and mixes it with digestive juices from the pancreas and liver. The digestive juices in the duodenum break down carbohydrates, proteins, and fats into smaller molecules that can be absorbed into the bloodstream. The other options are not the first segment of the small intestine: The jejunum is the second segment of the small intestine. It is about 16 feet long and is located between the duodenum and the ileum. The jejunum is responsible for the absorption of most of the nutrients from food. The ileum is the third and longest segment of the small intestine. It is about 20 feet long and is located between the jejunum and the colon. The ileum is responsible for the absorption of the remaining nutrients from food, as well as the absorption of bile salts and water. The colon is the large intestine. It is about 5 feet long and is located between the ileum and the rectum. The colon is responsible for the absorption of water and the storage of waste products.

Q63. Which one of the following factors does not affect the rate of enzyme action?

• A. Enzyme concentration
• B. Substrate concentration
• C. Light intensity✓
• D. Temperature

Explanation: Light intensity does not directly affect enzyme activity because enzymes are not influenced by light. They are affected by factors like temperature, pH and concentrations of enzyme and substrate.

Q64. Light reaction occurs in the _ of chloroplast:

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

Explanation: Option C is correct.The light reaction occurs in the grana of the chloroplast. The chloroplast is a specialized organelle in plant cells that is responsible for photosynthesis. Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to produce oxygen and sugar. The chloroplast is divided into two parts: the stroma and the thylakoids. The stroma is a gel-like fluid that contains the enzymes needed for the Calvin cycle, the second stage of photosynthesis. The thylakoids are a series of flattened sacs that are stacked together to form grana. The light reaction occurs in the thylakoids of the chloroplast. The thylakoids contain chlorophyll, a green pigment that absorbs sunlight. The sunlight energy is used to split water molecules into hydrogen and oxygen. The hydrogen atoms are used to produce NADPH, a molecule that stores energy. The oxygen atoms are released into the atmosphere. The products of the light reaction, NADPH and ATP, are used in the Calvin cycle to produce sugar. The other options are not parts of the chloroplast where the light reaction occurs: The outer membrane is a thin membrane that surrounds the chloroplast. It protects the chloroplast from the outside environment. The inner membrane is a thicker membrane that surrounds the stroma. It controls the movement of molecules into and out of the chloroplast. The stroma is the gel-like fluid that contains the enzymes needed for the Calvin cycle.

Q65. How many pieces of ribonucleic acid (RNA) make up the genome of influenza virus?

• A. 4
• B. 6
• C. 8✓
• D. 10

Explanation: The influenza virus genome is composed of eight pieces of ribonucleic acid (RNA). This virus is classified as a segmented RNA virus, which means its genetic material is divided into multiple RNA segments. Each of the eight segments encodes one or more proteins that are crucial for the virus's ability to replicate and cause infection. The segmentation of the genome allows the virus to undergo reassortment, which is a process that contributes to its genetic diversity and ability to evade the immune system, leading to new strains and potential pandemics. The other options (A, B, and D) are incorrect as they misrepresent the number of RNA segments in the influenza virus genome.

Q66. In Mendelian cross for the heterozygous flower, what is the probability that the dominant allele will be in sperm and the recessive in the egg:

• A. 0.05
• B. 0.5
• C. 0.25✓
• D. 0.7

Explanation: In a Mendelian cross involving a heterozygous flower (say, Rr, where:R = dominant allele (e.g., red color)r = recessive allele (e.g., white color)We want the probability that:The sperm carries the dominant allele (R)The egg carries the recessive allele (r)Step-by-step Explanation:In a heterozygous (Rr) individual:Probability of sperm carrying R = ½Probability of egg carrying r = ½Since the sperm and egg assort independently (law of independent assortment), we multiply the probabilities:P(sperm = R and egg = r) = ½ × ½ = ¼

Q67. Mastication causes exocrine glands under the tongue and in the back of the mouth to secrete a watery liquid called:

• A. Bolus
• B. Bile
• C. Pancreatic juice
• D. Saliva✓

Explanation: Option D is correct. Mastication causes exocrine glands under the tongue and in the back of the mouth to secrete a watery liquid called saliva. Saliva is a complex fluid that contains water, enzymes, and other molecules. It plays a number of important roles in digestion. Saliva is secreted by three pairs of major salivary glands: the parotid glands, the submandibular glands, and the sublingual glands. These glands are located in the head and neck, and they release saliva into the mouth through ducts. When you chew food, the salivary glands are stimulated to produce more saliva. This helps to moisten the food and break it down into smaller pieces. The saliva also contains enzymes that start to break down carbohydrates, such as starch, into smaller molecules that can be absorbed by the body. Saliva is an essential part of digestion, and it helps to keep the mouth healthy. Without saliva, it would be difficult to swallow food, and digestion would be less efficient. The other options are not liquids secreted by exocrine glands in the mouth: a Bolus is a mixture of chewed food and saliva that is formed in the mouth. Bile is a digestive fluid that is produced by the liver and stored in the gallbladder. It is released into the small intestine to help digest fats. Pancreatic juice is a digestive fluid that is produced by the pancreas. It is released into the small intestine to help digest carbohydrates, proteins, and fats.

Q68. In phylum Aschelminths (Nematoda) nervous system consists of a nerve ring which encircles the _ and sends its branches in different parts of the body

• A. Lips
• B. Teeth
• C. Stomach
• D. Pharynx✓

Explanation: Option D is correct.In phylum Aschelminths (Nematoda), the nervous system consists of a nerve ring which encircles the pharynx. The pharynx is a muscular tube that is located at the anterior (front) end of the nematode body. It is responsible for sucking in food and moving it to the digestive system. The nerve ring is a cluster of ganglia (nerve cell bodies) that encircles the pharynx. The nerve ring serves as a central nervous system for the nematode, and it sends nerve branches to all parts of the body. The other options are not parts of the nematode body that are encircled by the nerve ring: Lips are not present in nematodes.Teeth are not present in nematodes.The stomach is located in the posterior (back) end of the nematode body.

Q69. Salmonella typhi causing typhoid and Clostridium tetani causing tetanus are which type of bacteria?

• A. Bacilli✓
• B. Spirilla
• C. Cocci
• D. Comma

Explanation: Option A is correct.Salmonella typhi and Clostridium tetani are both bacilli. Bacilli are rod-shaped bacteria. They are one of the three main shapes of bacteria, along with cocci (spherical bacteria) and spirilla (spiral bacteria). Bacilli are found in a wide variety of environments, including soil, water, and the human body. Some bacilli are beneficial, such as those that produce lactic acid in yogurt and cheese. Other bacilli are pathogens, which means that they can cause disease. Salmonella typhi and Clostridium tetani are both examples of pathogenic bacilli. Salmonella typhi causes typhoid, a serious disease that can be fatal if it is not treated. Typhoid is transmitted through contaminated food and water. It causes fever, headache, and diarrhea. Clostridium tetani causes tetanus, a disease that affects the nervous system. Tetanus is caused by a toxin that is produced by the bacteria. The toxin causes muscle spasms, which can be severe and life-threatening. Tetanus is transmitted through contact with contaminated soil or other objects. Both Salmonella typhi and Clostridium tetani can be prevented by vaccination. The typhoid vaccine is recommended for people who are traveling to areas where typhoid is common. The tetanus vaccine is recommended for everyone as part of the routine childhood immunization schedule.Spirilla:Spiral bacteria.Example; Campylobacter (food poisoning), Leptospira (leptospirosis), Treponema pallidum (syphilis) Cocci: Spherical bacteria.Examples; Streptococcus (strep throat), Staphylococcus (staph infection), Neisseria (meningitis, gonorrhea) Comma-shaped bacteria are a type of spirilla. Comma-shaped bacteria include: Vibrio cholerae (cholera), Vibrio parahaemolyticus (food poisoning), and Helicobacter pylori (stomach ulcers).

Q70. Who formulated the principle that "ontogeny recapitulates phylogeny"?

• A. Von bear
• B. Hackel✓
• C. Herbert Spencer
• D. Darwin

Explanation: Option B is correct.The principle that "ontogeny recapitulates phylogeny" was formulated by Ernst Hackel. Hackel was a German biologist who lived from 1834 to 1919. He was a major proponent of Darwin's theory of evolution, and he made a number of important contributions to the field of biology. The principle of ontogeny recapitulating phylogeny states that the development of an individual organism (ontogeny) mirrors the evolutionary history of its species (phylogeny). In other words, as an embryo develops, it goes through stages that resemble the adult forms of its evolutionary ancestors. Hackel based his theory on observations of the development of various animals, including humans. He noted that at certain stages of development, human embryos have features that are similar to those of fish, reptiles, and other animals. The other options are not scientists who formulated the principle of ontogeny recapitulating phylogeny: Karl Ernst von Baer was a German biologist who lived from 1792 to 1876. He is known for his work on the development of embryos. Herbert Spencer was an English philosopher and sociologist who lived from 1820 to 1903. He is known for his work on social Darwinism. Charles Darwin was an English naturalist who lived from 1809 to 1882. He is known for his theory of evolution by natural selection.

Q71. Aquatic mammals belong to the order _

• A. Cetacea✓
• B. Pholidote
• C. Chiroptera
• D. Proboscidea

Explanation: Option A is correct.Aquatic mammals belong to the order Cetacea. Cetaceans are a group of marine mammals that includes whales, dolphins, and porpoises. They are characterized by their streamlined bodies, flippers, and lack of hind limbs. Cetaceans are also known for their intelligence and complex social behavior. The other options are orders of mammals that are not aquatic: Pholidota is the order of mammals that includes pangolins. Pangolins are scaly mammals that are found in Africa and Asia. Chiroptera is the order of mammals that includes bats. Bats are the only mammals that can fly. Proboscidea is the order of mammals that includes elephants. Elephants are the largest land mammals on Earth.

Q72. All of the following paired organs/structures have homology except:

• A. Human hand: Bat's wing
• B. Bat's wing: Wings of butterfly✓
• C. Horse front leg: Front flappers of Whale
• D. Forelimb of a cat: Forelimb of a whale.

Explanation: The wings of a bat and the wings of a butterfly are analogous structures, but not homologous. Analogous structures are structures that have similar functions but different evolutionary origins. Homologous structures are structures that have similar evolutionary origins but may have different functions. The wings of a bat and the wings of a butterfly are both used for flight, but they have evolved independently from each other. The wings of a bat are modified forelimbs, while the wings of a butterfly are modified outgrowths of the thorax.

Q73. The head of some phages are icosahedral, which means that the head possesses:

• A. 6 Sides
• B. 10 Sides
• C. 15 Sides
• D. 20 Sides✓

Explanation: The head of some phages are icosahedral, which means that the head possesses 20 sides. An icosahedron is a polyhedron defined by having 20 equilateral triangular faces. It is one of the five Platonic solids, recognized for its symmetry and structural efficiency. The icosahedral shape allows for optimal packing of viral proteins and DNA, contributing to the stability and resilience of the virus against environmental stressors. Examples of phages with icosahedral heads include T4 phage, T7 phage, Lambda phage, and MS2 phage. The other options are incorrect because they do not represent the true structural nature of an icosahedron, which is specifically characterized by its 20 triangular faces.

Q74. In sickle cell, haemoglobin molecule, glutamic acid, is replaced by:

• A. Proline
• B. Glutamine
• C. Valine✓
• D. Glycine

Explanation: Option C is correct. In sickle cell anemia, glutamic acid in the hemoglobin molecule is replaced by valine. This single amino acid substitution leads to the hemoglobin molecules becoming sticky and clumping together when oxygen levels are low. This causes the formation of sickle-shaped red blood cells, which can obstruct blood vessels and lead to severe complications such as pain, stroke, and other health issues. Sickle cell anemia is a genetic condition resulting from a mutation in the hemoglobin gene and is especially prevalent among individuals of African descent. The other options (Proline, Glutamine, Glycine) do not correctly identify the amino acid involved in this specific mutation, and therefore, do not explain the pathology of sickle cell anemia.

Q75. Where do phospholipids arrange themselves in a cell?

• A. Inside the nucleus
• B. Inside the cytoplasm
• C. Inside the plasma membrane✓
• D. inside the mitochondrial matrix

Explanation: Option C is correct.Phospholipids arrange themselves in a bilayer in the plasma membrane of a cell. The plasma membrane is a thin, flexible barrier that surrounds the cell and protects it from the external environment. It also controls the movement of molecules into and out of the cell. Phospholipids are amphipathic molecules, meaning that they have a hydrophilic (water-loving) head and a hydrophobic (water-hating) tail. In a bilayer, the hydrophilic heads face the outside and the inside of the cell, while the hydrophobic tails face each other. This arrangement creates a barrier that prevents water and other polar molecules from passing through the plasma membrane. The plasma membrane is a dynamic structure, meaning that it is constantly changing and moving. This is important for a number of cellular processes, such as cell signaling and cell division. The other options are not places where phospholipids arrange themselves in a cell: The nucleus is the central organelle of the cell and contains the cell's DNA. The cytoplasm is the jelly-like substance that fills the cell outside of the nucleus. The mitochondrial matrix is the fluid inside of mitochondria, which are organelles that produce energy for the cell.

Q76. Which type of nerve impulse occurs in non-myelinated neuron fibres?

• A. Saltatory impulse
• B. Continuous impulse✓
• C. Rapid impulse
• D. Jumping impulse

Explanation: Option B is correct. The type of nerve impulse that occurs in non-myelinated neuron fibers is a continuous impulse. A continuous impulse is a type of nerve impulse that travels down the axon of a neuron without jumping. This is in contrast to a saltatory impulse, which is a type of nerve impulse that jumps from node to node along the axon of a myelinated neuron. Myelinated neurons are neurons that have a myelin sheath, which is a fatty substance that insulates the axon. The myelin sheath helps to speed up the transmission of nerve impulses by allowing them to jump from node to node. Non-myelinated neurons do not have a myelin sheath, so nerve impulses travel down the axon continuously. This means that nerve impulses travel more slowly in non-myelinated neurons than in myelinated neurons. Saltatory impulse: A type of nerve impulse that jumps from node to node along the axon of a myelinated neuron. Rapid impulse: A type of nerve impulse that travels quickly. Jumping impulse: A type of nerve impulse that jumps from node to node along the axon of a myelinated neuron.

Q77. In alcoholic fermentation _ ATP Molecules are produced from one glucose molecule:

• A. 2✓
• B. 18
• C. 32
• D. 36

Explanation: Option A is correct. In alcoholic fermentation, two ATP molecules are produced from one glucose molecule. This process occurs in the absence of oxygen and is utilized by yeast and certain microorganisms to convert glucose into ethanol and carbon dioxide. The steps of alcoholic fermentation include:Glycolysis: Glucose is metabolized into two molecules of pyruvate, generating two ATP and two NADH in the process.Pyruvate decarboxylation: Each pyruvate is converted into acetaldehyde, releasing one molecule of carbon dioxide.Alcohol dehydrogenase: Acetaldehyde is then transformed into ethanol, which regenerates NAD+ from NADH.Overall, the net production from one glucose molecule in alcoholic fermentation is two ATP molecules.Options B, C, and D are incorrect as they suggest ATP yields associated with aerobic respiration, which are not applicable in the anaerobic process of alcoholic fermentation.

Q78. Amino acids mainly differ from each other by the difference in their:

• A. R-group✓
• B. Amino group
• C. Carboxyl group
• D. Hydrogen of alpha carbon

Explanation: Option A is correct.Amino acids mainly differ from each other by the difference in their R-group. The R-group is a side chain that is attached to the alpha carbon of an amino acid. The R-group can be any of 20 different functional groups, which is what gives different amino acids their unique properties. The other options are all common to all amino acids:All amino acids have an amino group (-NH2) attached to their alpha carbon. All amino acids have a carboxyl group (-COOH) attached to their alpha carbon. All amino acids have a hydrogen atom attached to their alpha carbon.

Q79. All of the following are related to enzymes except:

• A. Speed up reaction
• B. Remain unchanged after reaction
• C. Increase the activation energy✓
• D. Possess the active site

Explanation: Option C is correct.Enzymes are biological catalysts that speed up chemical reactions by lowering the activation energy of the reaction. The activation energy is the amount of energy that must be added to the reactants in order to start the reaction. Enzymes do this by binding to the reactants and forming an enzyme-substrate complex. The enzyme-substrate complex lowers the activation energy of the reaction by stabilizing the transition state not increase activation energy , which is a high-energy intermediate state of the reaction. The other options are all related to enzymes: Speed up reaction: Enzymes speed up chemical reactions by lowering the activation energy of the reaction. Remain unchanged after reaction: Enzymes are catalysts, which means that they remain unchanged after the reaction is complete. Possess the active site: Enzymes have an active site, which is a region of the enzyme that binds to the reactants and catalyzes the reaction.

Q80. The cytoplasm contain(s):

• A. Cell organelle only
• B. Insoluble waste only
• C. Storage products only
• D. Cell organelle, insoluble waste and storage products✓

Explanation: The cytoplasm contains cell organelles, insoluble waste, and storage products. The cytoplasm is the jelly-like substance that fills the cell outside of the nucleus. It is made up of water, proteins, carbohydrates, lipids, and other molecules. The cytoplasm also contains a variety of cell organelles, which are specialized structures that carry out different functions within the cell. Some examples of cell organelles that are found in the cytoplasm include:Mitochondria: Produce energy for the cellEndoplasmic Reticulum: Synthesizes and transports proteins and lipidsGolgi apparatus: Modifies and packages proteins and lipids for secretionRibosomes: Synthesize proteinsCytoskeleton: Provides support and structure to the cellThe cytoplasm also contains insoluble waste products, such as dead organelles and pigments. These waste products are usually stored in vacuoles, which are fluid-filled sacs.Storage products, such as glycogen and starch, are also found in the cytoplasm. These storage products are used by the cell for energy or to build new molecules.

Q81. Tay-Sach's disease particularly results from the malfunctioning of:

• A. Cell membrane
• B. Nucleus
• C. Lysosomes✓
• D. Vacuole

Explanation: Option C is correct.Tay-Sach's disease particularly results from the malfunctioning of lysosomes. Lysosomes are cellular organelles that contain enzymes that break down complex molecules, such as carbohydrates, proteins, and lipids. In Tay-Sach's disease, there is a deficiency of the enzyme hexosaminidase A, which is essential for the breakdown of a fatty substance called GM2 ganglioside. As a result, GM2 ganglioside accumulates in lysosomes, causing damage to nerve cells and other tissues. Tay-Sach's disease is a genetic disorder that is inherited from both parents. It is a rare disease, but it is more common in people of Ashkenazi Jewish descent. There is no cure for Tay-Sach's disease, but there are treatments that can help manage the symptoms and improve the quality of life for people with the disease. The other options are not directly involved in the malfunctioning that causes Tay-Sach's disease: The cell membrane is a barrier that surrounds the cell and protects it from the external environment. The nucleus is the central organelle of the cell and contains the cell's DNA. Vacuoles are fluid-filled sacs that store water, nutrients, and waste products.

Q82. Enzymes present in mammals work best at about:

• A. 20°C
• B. 30°C
• C. 40°C✓
• D. 50°C

Explanation: Option C is correct. Enzymes in mammals typically work best at around 40°C, closely correlating with their normal body temperature of approximately 37°C. At these temperatures, enzymes achieve optimal catalytic efficiency, allowing them to facilitate biochemical reactions at the highest rates. While enzymes can operate at lower and higher temperatures, their activity diminishes as temperatures deviate from this optimal range. Options A (20°C) and B (30°C) are incorrect because they are below the optimal temperature range, leading to reduced enzyme activity. Option D (50°C) is also incorrect, as many mammalian enzymes would likely denature at this temperature, resulting in loss of function. Therefore, understanding the relationship between temperature and enzyme activity is crucial for answering the question correctly.

Q83. An example of competitive inhibitor molecule is:

• A. Insecticides
• B. Sulphonamide✓
• C. Cyanide
• D. Metal ions

Explanation: Option B is correct.An example of a competitive inhibitor molecule is sulfonamide. Sulfonamides are a class of antibiotics that work by inhibiting the enzyme dihydropteroate synthase. Dihydropteroate synthase is an enzyme that is essential for the synthesis of folic acid, which is a vitamin that is necessary for bacterial growth. Sulfonamides are competitive inhibitors of dihydropteroate synthase because they bind to the active site of the enzyme and prevent the substrate, para-aminobenzoic acid (PABA), from binding. This prevents the synthesis of folic acid and kills the bacteria. The other options are not examples of competitive inhibitors: Insecticides are chemicals that are used to kill insects. They work by a variety of mechanisms, but they do not typically inhibit enzymes. Cyanide is a toxic gas that kills cells by inhibiting the enzyme cytochrome c oxidase. However, it is not a competitive inhibitor because it does not bind to the active site of the enzyme. Metal ions are often used as cofactors for enzymes, but they can also be inhibitors. However, they do not typically act as competitive inhibitors.

Q84. The ATP formed in the preparatory phase of glycolysis is (are):

• A. 1
• B. 2
• C. 4
• D. 0✓

Explanation: The correct answer is Option D: 0 ATP. In the preparatory phase of glycolysis, two ATP molecules are utilized to phosphorylate glucose and its derivatives, resulting in no ATP being produced during this phase. This phase is essential for priming the glucose molecule for subsequent breakdown in the energy-producing phase, where four ATP molecules are generated. Thus, the net gain of ATP from glycolysis is two ATP per glucose molecule, derived from the four produced in the energy-generating phase minus the two consumed in the preparatory phase.Options A, B, and C are incorrect because they incorrectly imply that ATP is produced in the preparatory phase. Instead, this phase is characterized by an energy investment, where ATP is consumed rather than created.

Q85. Mitochondria are usually absent in:

• A. Muscle cells
• B. Cardiac cells
• C. Mature RBC's✓
• D. WBC's

Explanation: Option C is correct.Mitochondria are usually absent in mature RBC's. Mature RBC's, or erythrocytes, are the most abundant type of blood cell. They are responsible for transporting oxygen from the lungs to the tissues of the body. Mature RBC's are also responsible for removing carbon dioxide from the tissues and transporting it back to the lungs. Mature RBC's do not have mitochondria because they do not need to produce their own energy. They obtain the energy they need from the glucose that is present in the blood. The other options all have mitochondria: Muscle cells: Mitochondria are essential for muscle cells because they provide the energy needed for muscle contraction. Cardiac cells: Cardiac cells, or heart muscle cells, also have mitochondria because they need a lot of energy to pump blood throughout the body. WBC's: WBC's, or white blood cells, have mitochondria because they need energy to fight infection and protect the body from disease

Q86. _ consists of units called dictyosomes

• A. Ribosomes
• B. Lysosomes
• C. Glyoxisomes
• D. Golgi complex✓

Explanation: Option D is correct.he Golgi complex consists of units called dictyosomes. A dictyosome is a stack of flattened sacs called cisternae. The Golgi complex is responsible for processing, modifying, and packaging proteins and lipids for secretion or transport to other parts of the cell. The other options are all cell organelles, but they are not made up of dictyosomes: Ribosomes are the site of protein synthesis. Lysosomes are organelles that contain enzymes that break down complex molecules. Glyoxisomes are organelles that are found in plant cells and are involved in the metabolism of fatty acids.

Q87. The optimal pH of lipase (pancreas) is:

• A. 4.0-5.0
• B. 1.5-1.6
• C. 8.0✓
• D. 6.1-6.8

Explanation: Option C is correct. The optimal pH of lipase (pancreas) is 8.0. This is the pH at which the enzyme is most active and can most efficiently catalyze the breakdown of triglycerides into fatty acids and glycerol. The pancreas secretes lipase into the small intestine, where it helps to digest fats. Lipase is most active in the slightly alkaline environment of the small intestine, which has a pH of around 7.5-8.0. Factors such as the pH of the environment, the temperature, and the presence of bile salts can all affect the activity of lipase. For example, bile salts help to emulsify fats, making them more accessible to lipase. If the pH of the environment is too low (as in Options A and B) or too high (as in Option D), the activity of lipase will be compromised, leading to issues with fat digestion and potential symptoms such as diarrhea and steatorrhea (fatty stools).

Q88. Consider the reaction below, if 5 moles each of hydrogen and oxygen are reacted to form water, the reaction reveals:2H2 + O2 --> 2H2O

• A. H2 is the excess agent
• B. O2 is the limiting agent
• C. H2 is the limiting reagent✓
• D. The reaction has no limiting reagent

Explanation: Option C is the correct answer. In the reaction 2H2 + O2 --> 2H2O, we start with 5 moles of both hydrogen and oxygen. To find the limiting reagent, we need to calculate how many moles of water can be produced from each reactant:Moles of water from hydrogen: 5 moles H2 * (2 moles H2O / 2 moles H2) = 5 moles H2OMoles of water from oxygen: 5 moles O2 * (2 moles H2O / 1 mole O2) = 10 moles H2OSince hydrogen can produce only 5 moles of water while oxygen could produce 10 moles, hydrogen is the limiting reagent. This means that all of the hydrogen will be consumed in the reaction before all the oxygen is used up, leaving some oxygen unreacted. Thus, the maximum amount of product, which is water, that can be formed is 5 moles.The other options are incorrect because:Option A: Hydrogen is not the excess agent; it is the limiting reagent.Option B: Oxygen is not the limiting agent; it is present in excess.Option D: There is indeed a limiting reagent in this reaction.

Q89. Oxygen can be prepared by the decomposition of potassium chlorate (KClO3). How many moles of oxygen (O2(g)) can be formed by taking 12 moles of potassium chlorate (KClO3) according to the following equation?2KClO3(l) + heat --> 2KCl(s) + 3O2(g)

• A. 12 moles of oxygen
• B. 15 moles of oxygen
• C. 18 moles of oxygen✓
• D. 21 moles of oxygen

Explanation: Option C is correct. The balanced chemical equation is: 2 KClO3(l) + heat → 2 KCl(s) + 3 O2(g). This indicates that for every 2 moles of KClO3, 3 moles of O2 are produced. To find out how many moles of O2 are produced from 12 moles of KClO3, we set up the following calculation: 12 moles KClO3 * (3 moles O2 / 2 moles KClO3) = 18 moles O2. Therefore, the answer is 18 moles of oxygen. The other options are incorrect because they do not align with the stoichiometry of the reaction. Specifically, options A, B, and D propose amounts of O2 that exceed or do not reach the correct product ratio based on the given moles of KClO3.

Q90. 117.0g of NaCl have

• A. 12.04 x 1023 Formula unit of NaCl✓
• B. 12.04 x 1022 Formula unit of NaCl
• C. 12.04 x 1023 Molecules of NaCl
• D. 6.023 x 1023 Molecules of NaCl

Explanation: Option A is correct.To calculate the number of formula units of NaCl in 117.0g of NaCl, we can use the following steps: Calculate the number of moles of NaCl in 117.0g of NaCl. Multiply the number of moles of NaCl by Avogadro's number (6.022 x 10^23 atoms/mol) to get the number of formula units of NaCl. Step 1: Calculate the number of moles of NaCl in 117.0g of NaCl. The molar mass of NaCl is 58.5g/mol. This means that 1 mole of NaCl weighs 58.5g. Number of moles of NaCl = mass of NaCl / molar mass of NaCl Number of moles of NaCl = 117.0g / 58.5g/mol Number of moles of NaCl = 2.0 moles Step 2: Multiply the number of moles of NaCl by Avogadro's number to get the number of formula units of NaCl. Number of formula units of NaCl = number of moles of NaCl * Avogadro's number Number of formula units of NaCl = 2.0 moles * 6.022 x 10^23 atoms/mol Number of formula units of NaCl = 1204 x 10^23 formula units Therefore, 117.0g of NaCl have 12.04 x 10^23 formula units of NaCl.

Q91. The number of orientations of a sub – shell for which n= 4 and l= 3 will be:

• A. 5
• B. 3
• C. 7✓
• D. 1

Explanation: Option C is the correct choice. To determine the number of orientations of a sub-shell with quantum numbers n=4 and l=3, we use the formula: Number of orientations = 2l + 1. Here, l=3, so:Number of orientations = 2(3) + 1 = 6 + 1 = 7.Thus, the number of orientations is 7, which corresponds to Option C.Options A, B, and D are incorrect as they fail to apply the formula correctly, leading to lower counts of orientations that do not reflect the quantum mechanical principles governing atomic structure.

Q92. Photon of the lowest wavelength is related to:

• A. Balmer series✓
• B. Pfund series
• C. Bracket series
• D. Paschen series

Explanation: Option A is correct.The photon of the lowest wavelength is related to the Balmer series. The Balmer series is a spectral series in which photons are emitted when an electron in a hydrogen atom transitions from an energy level with a principal quantum number greater than 2 to the energy level with a principal quantum number of 2. The wavelength of the emitted photon is inversely proportional to the difference in energy between the two energy levels. The lowest wavelength in the Balmer series is known as the Balmer alpha line, which has a wavelength of 656.3 nm. This is the wavelength of the photon that is emitted when an electron in a hydrogen atom transitions from the third energy level to the second energy level. The Paschen series corresponds to transitions between the third energy level (n = 3) and higher energy levels (n ≥ 4).The Brackett series corresponds to transitions between the fourth energy level (n = 4) and higher energy levels (n ≥ 5). The Pfund series corresponds to transitions between the fifth energy level (n = 5) and higher energy levels (n ≥ 6).

Q93. The maximum e/m ratio for positive rays is obtained when they discharge tube contains:

• A. He
• B. N2
• C. Ne
• D. H2✓

Explanation: Option D is correct. The maximum e/m ratio for positive rays is achieved with H2 because it is the lightest atom, resulting in the highest charge-to-mass ratio. The e/m ratio is defined as the charge of a particle divided by its mass, and it indicates how much a particle is influenced by electric and magnetic fields. A higher e/m ratio means greater deflection in these fields. The other options (He, N2, and Ne) are heavier than hydrogen, which results in lower e/m ratios due to their greater mass. Understanding the e/m ratio is crucial in identifying elements in gases and was pivotal in early atomic research, such as J.J. Thomson's work on the electron's mass.

Q94. 1 atm of pressure is equal to all of the following except:

• A. 76 cm Hg
• B. 769 mm Hg✓
• C. 760 torr
• D. 101325 Pa

Explanation: Option B is correct.The answer is 769 mm Hg. 1 atm is equal to 760 mm Hg, so 769 mm Hg is slightly greater than 1 atm. All of the other options are equal to 1 atm.

Q95. Volume of a given mass of an ideal gas at certain pressure is x at constant temperature. What will be its volume when the pressure is reduced to half:

• A. 1/2
• B. 2x✓
• C. 4x
• D. 1/4

Explanation: Option B is correct. When the pressure of an ideal gas is reduced to half at a constant temperature, the volume of the gas will double. This relationship is described by Boyle's Law, which states that for a given mass of an ideal gas at constant temperature, the product of pressure (P) and volume (V) is constant (PV = constant). Hence, if the pressure is halved, to maintain the constant product, the volume must double. Therefore, if the initial volume is x, the new volume will be 2x.Option A is incorrect because it suggests the volume reduces to half, which is not in accordance with the principle of gas behavior. Option C is incorrect as it implies that the volume quadruples, which is not true under these conditions. Option D is also incorrect as it indicates a significant reduction in volume, which contradicts the expected increase in volume when pressure decreases.

Q96. The volume occupied by 3.2 g of oxygen gas at S.T.P is:

• A. 1.12 dm3
• B. 2.24 dm3✓
• C. 22.4 dm3
• D. 24 dm3

Explanation: Option B is correct. The volume occupied by 3.2 g of oxygen gas at STP is 2.24 dm3. STP stands for standard temperature and pressure, defined as 0 °C (273.15 K) and 1 atm (101325 Pa). To find the volume of a gas at STP, we use the ideal gas law: PV = nRT. Here, P is the pressure in atm, V is the volume in L, n is the number of moles, R is the ideal gas constant (0.08206 L atm mol-1 K-1), and T is the temperature in K.Given 3.2 g of oxygen, we calculate the moles of oxygen using its molar mass (32 g/mol): n = m / M = 3.2 g / 32 g/mol = 0.1 mol. At STP, we plug the values into the ideal gas law:V = nRT / P = (0.1 mol)(0.08206 L atm mol-1 K-1)(273.15 K) / (1 atm) = 2.24 L. Thus, 2.24 L equals 2.24 dm3.Options A, C, and D are incorrect as they either miscalculate the volume based on the number of moles or present incorrect standard gas volumes.

Q97. All of the following crystal systems have α=β=γ=90° except:

• A. Cubic
• B. Orthrombic
• C. Tetragonal
• D. Rhombohedral✓

Explanation: Option D is correct.The crystal system that does not have β=γ=90° is rhombohedral. The other three crystal systems, cubic, orthorhombic, and tetragonal, all have the following properties: Cubic: a=b=c, α=β=γ=90° Orthorhombic: a≠b≠c, α=β=γ=90° Tetragonal: a=b≠c, α=β=γ=90° The rhombohedral crystal system, on the other hand, has the following properties: a=b=c, α=β=γ≠90° The angle α=β=γ in the rhombohedral crystal system is typically between 60° and 120°, but it can be outside of this range.

Q98. Compund having lowest boiling point is:

• A. Carbon tetrachloride✓
• B. Ethyl alcohol
• C. Benzene
• D. Acetic acid

Explanation: Option A (Carbon tetrachloride) is correct. It has the lowest boiling point because it is a nonpolar molecule with weak intermolecular London dispersion forces. In contrast, Option B (Ethyl alcohol) has a higher boiling point due to its ability to form hydrogen bonds, which are significantly stronger than dispersion forces. Option C (Benzene) has a higher boiling point than carbon tetrachloride primarily due to stronger π-π interactions despite being nonpolar. Lastly, Option D (Acetic acid) has the highest boiling point in this set due to its ability to form hydrogen bonds, which necessitates more energy to break these interactions during boiling. Hence, all other options have higher boiling points than carbon tetrachloride due to their stronger intermolecular forces.

Q99. Which of the following can form Hydrogen bonding with each other?

• A. Methanal and Ethanol
• B. Propane and Ethyl Methyl Ketone
• C. Amine and H2O✓
• D. Acetone and Acetaldehyde

Explanation: Option C is correct. Amines and water can form hydrogen bonds due to the presence of hydrogen atoms attached to the electronegative nitrogen in amines, which can interact with the oxygen in water. This interaction is crucial for many biological and chemical processes.In contrast, Option A (Methanal and Ethanol) cannot bond due to the lack of sufficiently electronegative hydrogen in methanal. Option B (Propane and Ethyl Methyl Ketone) involves nonpolar molecules that do not form hydrogen bonds. Lastly, Option D (Acetone and Acetaldehyde) fails to bond as neither has hydrogen directly bonded to an electronegative atom, preventing hydrogen bonding.

Q100. Identify the molecular formula of Furan:

• A. C4H4O✓
• B. C5H3O
• C. C4H5O
• D. C4H3O

Explanation: Option A is correct. The molecular formula of furan is C4H4O. Furan is a heterocyclic aromatic compound characterized by a five-membered ring comprised of four carbon atoms and one oxygen atom. The oxygen atom is part of the ring structure, contributing to its aromatic properties. Furan has applications in the production of various chemicals, including plastics and pharmaceuticals, and is recognized for its distinctive odor. Option B is incorrect because C5H3O corresponds to furfural, which is a derivative of furan but not the molecular formula for furan itself. Option C is incorrect as C4H5O is the molecular formula for cyclobutanol, a cyclic alcohol, and does not describe furan. Option D is incorrect because C4H3O is not a valid molecular formula for any stable compound, as it suggests that a carbon atom would have fewer than four bonds, which violates the octet rule.

Q101. Haber's process is used for the synthesis of ammonia. The optimum temperature for the Haber process is:

• A. 35-50°C
• B. 130-150°C
• C. 400-450°C✓
• D. 500-600°C

Explanation: Option C is correct. The optimum temperature for the Haber process is 400-450°C. This temperature range strikes a balance between reaction rate and yield. While lower temperatures favor the formation of ammonia, they slow the reaction significantly, making production inefficient. Conversely, higher temperatures increase the reaction rate but diminish the yield due to the equilibrium of the reaction shifting away from ammonia production. At 400-450°C, the reaction is fast enough to be practical, and the yield is high enough to ensure that the process remains economically viable. The Haber process is typically conducted at high pressures (200-400 atm) with an iron catalyst to further enhance ammonia production.

Q102. If ionic product is less than Ksp then:

• A. Solution will be saturated
• B. Precipitation will occur
• C. Solution will be super saturated
• D. No precipitation will occur✓

Explanation: Option D is correct. The ionic product is less than the solubility product, meaning the solution contains fewer ions than it could have at equilibrium. There is room for more ions and the saturation point is not reached. As saturation is not attained, therefore, precipitation does not occur. There are insufficient ions to exceed Ksp and form a solid precipitate. ==> Precipitation only occurs when the ionic product exceeds Ksp.

Q103. For 𝚫n=0

• A. Kp=Kn✓
• B. Kp ≠ Kn
• C. Kp>Kn
• D. Kp<Kn

Explanation: When the change in the number of moles of gas during a reaction (Δn\Delta nΔn) is zero, it means that the total number of moles of gaseous reactants equals the total number of moles of gaseous products. In this case, the relationship between the equilibrium constants KpK_pKp​ (which is based on partial pressures) and KcK_cKc​ (which is based on concentrations) simplifies to Kp=KcK_p = K_cKp​=Kc​ because the equation Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}Kp​=Kc​(RT)Δn results in Kp=KcK_p = K_cKp​=Kc​ when Δn=0\Delta n = 0Δn=0.

Q104. Reaction that follows third order kinetics is:

• A. Decomposition of nitrogen dioxide
• B. Decomposition of hydrogen iodide
• C. Gas phase oxidation of nitric oxide✓
• D. Formation of hydrogen iodide

Explanation: Option C is correct as gas phase oxidation of nitric oxide is a third order reaction. The reaction is as follows: 2NO(g) + O2(g) --> 2NO2(g) This reaction is second order with respect to NO and first order with respect to O2, making the overall order to be 3. Option A and D are second order reaction and option B is a zero order reeaction.

Q105. As compared to exothermic-reaction, formation of activated complex in endothermic reaction requires:

• A. Greater Ea✓
• B. Smaller Ea
• C. Equal Ea
• D. No Ea

Explanation: The correct option is A as the endothermic reactions have a greater activation energy because the products of this reaction have more energy than the reactants. On the otherhand products in exothermic reaction have lesser energy than the reactants, therefore, the activation energy is lesser. The graph is shown below

Q106. SI Unit of specific heat capacity is:

• A. Jg-1K-1✓
• B. JK-1
• C. JmolK-1
• D. KJ/mol

Explanation: The correct answer is Jg-1K-1, which is the SI unit of specific heat capacity. Specific heat capacity (c) is defined as the amount of heat energy (in joules) required to raise the temperature of one gram of a substance by one degree Kelvin (or Celsius). Therefore, it is expressed in joules per gram per Kelvin.Option B (JK-1) is incorrect because it lacks the mass component, making it unsuitable for specific heat capacity. Option C (JmolK-1) refers to molar heat capacity, applicable to moles rather than grams, and is thus incorrect for this question. Finally, Option D (KJ/mol) also measures energy but in relation to moles, not grams, and therefore does not represent specific heat capacity.

Q107. Processes involving solid and liquid have their:

• A. 𝚫H=𝚫E +P𝚫V
• B. 𝚫H=P𝚫V
• C. 𝚫H=𝚫E✓
• D. 𝚫H=0

Explanation: Option C is correct as ΔH=ΔE+PΔV and for a process involving solids and/or liqiuds, ΔV≅0. Therefore, 𝚫H=𝚫E. There is a change in the internal energy of the process with the environment.

Q108. In the redox reaction of SO2 with KMnO4 for an acidic medium:

• A. SO2 oxidizes KMnO4
• B. SO2 reduces KMnO4✓
• C. KMnO4 is inert in acidic medium
• D. SO2 can't undergo redox reaction

Explanation: Option B is correct as SO2 is a strong reducing agent and KMnO4 is a strong oxidising agent. SO2 reduces Mn7+ in KMnO4 to Mn2+ and gets oxidised itself from S4+ to S6+.

Q109. Zinc displaces copper from its solution because:

• A. Atomic number of zinc is higher than that of copper
• B. Zinc has higher reduction potential than copper
• C. Zinc is more soluble
• D. Zinc has smaller reduction potential than copper✓

Explanation: Option D is correct because zinc’s smaller reduction potential (-0.76V) compared to copper’s (+0.34V) signifies that zinc is more reactive. In electrochemical terms, a metal with a lower reduction potential can displace a metal with a higher reduction potential from its compound in solution. Therefore, zinc can easily displace copper from its solution. Option A is incorrect because the atomic number does not dictate the reactivity of metals. Option B misstates the relationship between the reduction potentials, as zinc's is lower than that of copper. Option C, while potentially true in some contexts, does not pertain to the displacement reaction, as reactivity is the determining factor.

Q110. The molecular shape of a molecule with three bonded atoms and one lone pair electron on the central atom will be:

• A. Trigonal planar
• B. Tetrahedral
• C. Trigonal Pyramidal✓
• D. Linear

Explanation: Option C is correct. If there are three bond pairs and one lone pair of electrons then the molecular shape of that molecule is trigonal pyramidal. An example of such a molecule is NH3. Its geometrical structure is given below. Trigonal planar = 3 bond pairs. Tetrahedral = 4 bond pairs. linear = 2 bond pairs.

Q111. Molecule are having the highest bond energy (experimentally) is:

• A. HCl
• B. HF✓
• C. HBr
• D. HI

Explanation: The correct option is B as the decreasing order of bond dissociation energy is HF > HCl > HBr > HI. There is a large difference between electronegativity of H and F, so the bond is stronger. As we go down the group, the difference between electronegativity decreases and therefore, bond dissociation energy also decreases.

Q112. The bond length between double-bonded carbon atom is:

• A. 1.34A°✓
• B. 1.10A°
• C. 1.54A°
• D. 1.82A°

Explanation: The bond length between double-bonded carbon atoms is 1.34A°, which reflects the nature of double bonds being shorter due to increased electron density between the bonded atoms. In contrast, the other options represent lengths that are not characteristic of double bonds: 1.10A° is too short and would suggest a triple bond; 1.54A° is more aligned with a single bond; and 1.82A° is even longer, indicating a weaker single bond. Therefore, 1.34A° is the only correct answer for the bond length of double-bonded carbon.

Q113. Who is the element that forms normal oxide only

• A. Li✓
• B. Na
• C. K
• D. Rb

Explanation: When Lithium is heated in air it only forms Li2O which is a normal oxide as it contains an O2- ion. On the other hand sodium, potassium and rubidium do not readily form normal oxides when heated in air. They tend to form per and superoxides with charges -1 and -1/2 on oxygen respectively.

Q114. Element whose salt do not impart colour to flame test is:

• A. Be✓
• B. Cs
• C. Ca
• D. K

Explanation: Option A is correct as Beryllium is one of the 2 alkaline earth metals that does not impart colour to flame test. Cs gives violet flame, Ca gives brick red flame and K gives lilac flames to flame test.

Q115. Indicate superoxide in the following

• A. CaO
• B. BaO2
• C. Na2O2
• D. RbO2✓

Explanation: Option D is correct. In superoxides oxygen has a charge of -1/2 which is the case in option D. All the rest of the options are normal(-2) or per(-1) oxides.

Q116. Which one of the following is covalent in nature?

• A. Mg3N2
• B. Be3N2✓
• C. Ca3N2
• D. Ba3N2

Explanation: Be3N2 is a covalent compound as Be and N have a small difference in electronegativity atoms and are more likely to share electrons rather then transferring them from one to another. All other compounds have a greater difference in electronegativity between the atoms making them polar and ionic.

Q117. Which one has one unpaired electron in the valance shell?

• A. Zn+2
• B. Cu+
• C. Ti+3✓
• D. Fe+3

Explanation: Titanium has an atomic number of 22, and its electron configuration is [Ar]3d24s2[Ar] 3d^{2} 4s^{2}[Ar]3d24s2. Upon losing three electrons to become Ti³⁺, it loses the two 4s electrons and one 3d electron, resulting in [Ar]3d1[Ar] 3d^{1}[Ar]3d1. This means Ti³⁺ has one unpaired electron in its valence shell.

Q118. The oxidation number of cobalt in the given coordination complex is:[Co(H2NCH2CH2NH2)3]2(SO4)3

• A. III✓
• B. II
• C. IV
• D. VI

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

Q119. Pyrole belongs to which class of compounds?

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

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

Q120. Ethyl alcohol in the presence of H2SO4 at 170°C produces

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

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

Q121. Lindlar's catalyst is used in hydrogenation of alkyne:

• A. To increase activation energy
• B. To enhance hydrogenation of alkene
• C. To produce cis alkene✓
• D. To start polymerisation reaction

Explanation: Lindlar's catalyst is specifically designed for the partial hydrogenation of alkynes to form cis-alkenes. It does this by providing a controlled environment where hydrogenation occurs without over-reduction to alkanes, ensuring the formation of the cis configuration.

Q122. Which one of the following is not meta directing group?

• A. -COR
• B. -NO2
• C. -CHO
• D. -NR2✓

Explanation: Examples of meta– directors include nitriles, carbonyl compounds (such as aldehydes, ketones, and esters), sulfones, electron-deficient alkyl groups, nitro groups, and alkylammoniums. While examples of ortho-, para– directors are hydroxyl groups, ethers, amines, alkyl groups, thiols, and halogens.

Q123. Benzene-1,3-diol is also known as:

• A. Catechol
• B. Resorcinol✓
• C. Hydroquinone
• D. O-cersol

Explanation: Resorcinol is a 1,3-isomer (or meta-isomer) of benzenediol with the formula C6H4(OH)2.

Q124. Identify the electrophile as acylium ion:

• A. RNa+
• B. RCO+✓
• C. RCOO+
• D. RNO+

Explanation: Acylium ions are cations of the formula RCO +. The carbon–oxygen bond length in these cations is near 1.1 Å (110-112 pm), which is shorter than the 112.8 pm of carbon monoxide and indicates triple-bond character.

Q125. Among the following hydrocarbons which one has acidic hydrogen?

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

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

Q126. Greater number of Alkyl groups on substrate favours:

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

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

Q127. SN1 reactions mostly result in partial racemization, in partial racemization there is:

• A. Inversion only
• B. Retention only
• C. Equal inversion and retention
• D. More inversion and retention✓

Explanation: In SN1 reactions, the formation of a planar carbocation allows for nucleophilic attack from either side, resulting in both inversion and retention. However, due to steric factors and the nature of the nucleophile, there tends to be more inversion than retention, leading to a mixture of products with both configurations, thus resulting in partial racemization.

Q128. Which of the following involves the same steps?

• A. E1 and E2
• B. E1 and SN1✓
• C. SN1 and SN2
• D. E2and SN1

Explanation: E1 (unimolecular elimination) and SN1 (unimolecular nucleophilic substitution) both involve the same first step: the formation of a carbocation after the leaving group departs. After that, E1 leads to elimination (forming a double bond), and SN1 involves nucleophilic substitution.

Q129. The correct sequence of electronic configuration is:

• A. 4p 5s 4d 5p 6s 4f 5d✓
• B. 4p 4s 4d 5p 5s 5f 6d
• C. 4p 4s 4d 5p 5s 4f 5d
• D. 4p 3s 4d 5p 6s 4f 5d

Explanation: This sequence follows the correct order of filling orbitals according to the Aufbau principle. The orbitals are filled in the increasing order of their energy levels, and this sequence correctly reflects the way electrons occupy these orbitals (4p → 5s → 4d → 5p → 6s → 4f → 5d).

Q130. When phenol reacts with excess of bromine in aqueous solution it results in the formation of:

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

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

Q131. Propylene glycol and trimethylene glycol are:

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

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

Q132. The reduction of aldehyde and ketone is to alkanes in the presence of Zinc amalgam and HCl is called

• A. Clemmensen reduction✓
• B. Williamson's synthesis
• C. Wolf-Kishner reduction
• D. Dow process

Explanation: Clemmensen reduction is a chemical reaction described as a reduction of ketones or aldehydes to alkanes using zinc amalgam and concentrated hydrochloric acid.

Q133. Aldehyde and ketone on reaction with hydroxylamine form:

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

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

Q134. Fehling's solution works on the principle of redox reaction which results in:

• A. Reduction of aldehydes
• B. Oxidation of Copper(II)
• C. Oxidation of aldehyde✓
• D. Oxidation of Ketone

Explanation: Fehling's solution is specifically designed to test for aldehydes, which are oxidized to carboxylic acids during the reaction. In this process, the Cu²⁺ ions from the Fehling's solution are reduced to Cu₂O, which precipitates as a red solid. Therefore, the correct answer is the oxidation of aldehyde. The other options are incorrect as they either misinterpret the nature of the reaction (like reduction of aldehydes) or focus on the wrong substrate (ketones), which do not react under these conditions. While copper(II) is involved in the reaction, the key reaction occurring is the oxidation of the aldehyde itself.

Q135. In conversion of acid halides to Ester, pyridine is used to:

• A. Stabilize acid halides
• B. Consume HCl formed in the reaction✓
• C. Dehydrate alcohol
• D. Dehydrogenate acid halides

Explanation: This option is correct. During the reaction of acid halides with alcohols to form esters, hydrogen halide (such as HCl) is released as a byproduct. Pyridine, being a weak base, reacts with the HCl to form pyridinium chloride, effectively neutralizing the acid. This helps to shift the reaction equilibrium toward the formation of the ester.

Q136. Number of carbon atoms in valeric acid is:

• A. 4
• B. 5✓
• C. 6
• D. 7

Explanation: Valeric acid is a carboxylic acid with the IUPAC name pentanoic acid. It consists of a straight chain of 5 carbon atoms (C5H10O2), which is why the correct answer is 5. The other options are incorrect because:4 carbon atoms would correspond to butanoic acid, which is different.6 carbon atoms would correspond to hexanoic acid, which also does not fit.7 carbon atoms would correspond to heptanoic acid, which is not valeric acid.Thus, the only valid option is 5, as it accurately reflects the number of carbon atoms in valeric acid.

Q137. The IUPAC name of formyl chloride is:

• A. Methanoic acid
• B. Chloro methane
• C. Methanoyl chloride✓
• D. Chloro methanoate

Explanation: The correct IUPAC name for formyl chloride is Methanoyl chloride. This name indicates that the compound is derived from methane (the simplest alkane with one carbon) with a carbonyl group (C=O) and a chlorine atom attached. The prefix 'formyl' corresponds to the presence of the aldehyde functional group, which is equivalent to the methanoyl structure. The other options are incorrect for the following reasons: - Chloro methanoic acid implies the presence of a carboxylic acid group, which is not present in formyl chloride. - Chloro methane suggests a chlorinated methane without considering the carbonyl group, leading to an insufficient description of the compound. - Chloro methanoate implies an ester structure, which is also incorrect as there is no alcohol component in formyl chloride.

Q138. The inhibition, in which the inhibitor does not combine directly with the enzyme but binds to the enzyme-substrate complex is called:

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

Explanation: The correct answer is Uncompetitive inhibition. This type of inhibition specifically involves the inhibitor binding to the enzyme-substrate complex, preventing the formation of products. In contrast, Reversible inhibition is a broader category that includes various forms of inhibition not limited to the enzyme-substrate complex. Competitive inhibition is characterized by the inhibitor directly competing with the substrate for the active site, while Non-Competitive inhibition refers to the inhibitor binding to a different site on the enzyme, not the enzyme-substrate complex. Therefore, uncompetitive inhibition is the only option that directly addresses the relationship between the inhibitor and the enzyme-substrate complex.

Q139. Trypsinogen can be activated by the action of:

• A. Amylase
• B. HCl
• C. Glucokinase
• D. Entrokinase✓

Explanation: Trypsinogen is a zymogen (inactive enzyme precursor) that requires activation to become trypsin, an active enzyme essential for protein digestion. The activation occurs primarily in the small intestine, where enteropeptidase (also known as entero kinase) converts trypsinogen into trypsin. This process is crucial for initiating the digestive cascade, as trypsin can further activate other digestive enzymes. Amylase, HCl, and glucokinase do not play a role in this specific activation process. Amylase is involved in carbohydrate digestion, HCl aids in protein denaturation and creates an acidic environment for enzymes, while glucokinase is related to glucose metabolism.

Q140. Crystal having orthorhombic crystal system is:

• A. PbCrO4
• B. BaSO4✓
• C. ZnO
• D. NaOH

Explanation: The correct answer is Barium sulfate (BaSO₄), which crystallizes in the orthorhombic crystal system. This system has three mutually perpendicular axes that are of unequal lengths, a characteristic feature of orthorhombic crystals.In contrast, Lead chromate (PbCrO₄) is not correct as it belongs to the tetragonal system, characterized by one axis being longer than the other two, which are equal in length. Zinc oxide (ZnO) is classified under the hexagonal system, which features a different arrangement altogether, while Sodium hydroxide (NaOH) is cubic, with all axes equal. Thus, the only correct option that matches the orthorhombic classification is Barium sulfate.

Q141. Alkoxy carbonyl functional group is present in:

• A. Ether
• B. Aldehyde
• C. Carboxylic acid
• D. Ester✓

Explanation: The alkoxy carbonyl functional group is a defining feature of esters, which are characterized by the presence of a carbonyl group (C=O) bonded to an alkoxy group (–O–R). In this case, option D (Ester) is correct because it contains both components of the alkoxy carbonyl functional group. The other options do not fit the definition:Option A (Ether): Ethers lack the carbonyl group altogether and consist solely of an oxygen atom bonded to two hydrocarbon groups.Option B (Aldehyde): Aldehydes have a carbonyl group bonded to a hydrogen atom and do not have an alkoxy component.Option C (Carboxylic acid): While carboxylic acids contain a carbonyl, they also include a hydroxyl group, which is different from the alkoxy structure.Thus, the correct identification of the alkoxy carbonyl functional group is essential for recognizing that esters (option D) are the correct answer.

Q142. Equal forces F act on isolated bodies A and B. The mass of B is 1/5 times that of A. The magnitude of the acceleration of A is:

• A. 1/5 times that of B✓
• B. 1/3 times that of B
• C. The same as B
• D. nine times that of B

Explanation: According to Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration. Since equal forces F act on both bodies A and B, we can express their accelerations as follows:For body A: a_A = F/m_AFor body B: a_B = F/m_BGiven that the mass of B (m_B) is 1/5 times that of A (m_A), we can substitute:m_B = (1/5)m_ATherefore, the acceleration of B becomes: a_B = F/(1/5)m_A = 5F/m_ANow, we know that:a_A = F/m_AThus, by comparing the two accelerations, we find:a_A = (1/5) * a_BThis means that the magnitude of acceleration of A is indeed 1/5 times that of B.Options B, C, and D are incorrect because they either miscalculate the relationship between the accelerations based on the mass ratio or ignore the fundamental principle that acceleration is inversely proportional to mass when the same force is applied.

Q143. A body is moved through a displacement of 15m towards the North, the total displacement will be zero when the body covered the same displacement towards:

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

Explanation: Displacement refers to the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. In this question, the body initially moves 15m towards the North, resulting in a positive displacement of 15m in that direction. To achieve a total displacement of zero, the body must move back the same distance in the opposite direction. Therefore, moving 15m towards the South will cancel out the initial Northward displacement, resulting in a total displacement of zero.Options North, West, and East do not negate the initial displacement of 15m North, hence they do not result in zero total displacement.

Q144. When the range of a projectile is equal to one fourth of the height then which one is true:

• A. Tanθ=10
• B. Tanθ=12
• C. Tanθ=16✓
• D. Tanθ=1

Explanation: Since Rtan θ = 4H 1/4 H tan θ = 4H tan θ = 16

Q145. The body is moving with with momentum of 100 kg m/s. What is the magnitude of force required to stop this body in 25 seconds?

• A. 4 N✓
• B. 25 N
• C. 100 N
• D. 2500 N

Explanation: The momentum of the body is given as 100 kg m/s, and the time taken to stop it is 25 seconds. To find the force required to stop the body, we can use the formula: Force = Change in Momentum / Time. Here, the change in momentum is equal to the initial momentum since the final momentum will be 0 when the body is stopped. Thus, Force = 100 kg m/s / 25 s = 4 N. This means that a constant force of 4 N is required to bring the body to rest in 25 seconds. The other options are incorrect because they either imply an unrealistic time frame for stopping the body or do not correspond to the given momentum and time constraints.

Q146. Doubling the initial velocity of a projectile while keeping all other parameters the same, the height reached by the projectile increases by:

• A. two times
• B. three times
• C. four times✓
• D. remains the same

Explanation: When a projectile is launched, the maximum height it reaches is directly related to the square of its initial vertical velocity. The formula for maximum height (H) is given by H = (v0y2)/(2g), where v0y is the initial vertical velocity and g is the acceleration due to gravity. When the initial velocity is doubled (v0y becomes 2v0y), the new height becomes H' = (2v0y)2/(2g) = 4(v0y2)/(2g) = 4H. Thus, the height increases by a factor of four. The other options (two times, three times, and remains the same) do not account for the quadratic relationship between velocity and height, which is why they are incorrect.

Q147. The effect of air resistance, how long a stone takes to drop off a 125m high building, landing on the ground, take g=10m/s2:

• A. 3 sec
• B. 4 sec
• C. 18 sec
• D. 5 sec✓

Explanation: To determine the time it takes for a stone to drop from a height of 125 meters, we can use the formula for free fall:t = √(2h/g)Where:t = time in secondsh = height in meters (125m)g = acceleration due to gravity (10m/s²)Substituting the values into the formula:t = √(2 * 125 / 10) = √(25) = 5 secondsThis means the stone will take 5 seconds to hit the ground. The other options are incorrect because:Option A (3 seconds) is too short for the height of 125m.Option B (4 seconds) also underestimates the time required for such a drop.Option C (18 seconds) miscalculates the fall time by a large margin, significantly exceeding the expected duration.Therefore, the correct answer is Option D: 5 sec.

Q148. The weight of a body is 120N and it is lifted to a height of 10m, work done on the body is:

• A. 120 J
• B. 100 J
• C. 1000 J
• D. 1200 J✓

Explanation: The work done on an object when it is lifted can be calculated using the formula: Work = Weight x Height. In this case, the weight of the body is 120N and it is lifted to a height of 10m. Plugging in the values:Work = 120N x 10m = 1200 J.Thus, the correct answer is 1200 J. The other options are incorrect because:120 J: This underestimates the work done as it does not use the correct height.100 J: This is significantly lower than the calculated work.1000 J: While closer, it still fails to multiply the weight and height correctly.Therefore, the correct answer is 1200 J.

Q149. A 200N force acts on 8 kg crate that starts from rest. At the instant the object has gone 2m, the rate at which the force is doing work is:

• A. 2.5 W
• B. 25 W
• C. 75 W
• D. 2000 W✓

Explanation: To find the rate at which the force is doing work (power), we first need to determine the velocity of the crate after it has moved 2 meters. Using Newton's second law, we can calculate the acceleration:F = ma200 N = 8 kg * aa = 25 m/s²Next, we use the kinematic equation to find the velocity after traveling 2m from rest:v² = u² + 2asv² = 0 + 2 * 25 m/s² * 2mv² = 100v = 10 m/sNow we can calculate the power using P = F * v:P = 200 N * 10 m/s = 2000 WThis explains why option D is correct. The other options are incorrect as they either miscalculate the power output or do not take into account the velocity of the crate after it has traveled the specified distance.

Q150. Which of the following bodies has the largest kinetic energy?

• A. Mass 3M and speed v
• B. Mass 2M and speed 3v✓
• C. Mass 3M and speed 2v
• D. Mass M and speed 4V

Explanation: K.E = 1/2Mv2The K.E will be maximum when the product of "M" and "v²" is maximum. In (B) the product has maximum value i.e. 9Mv².

Q151. A 2kg object is released from rest 8m above the surface of the Earth, During the fall work done against air resistance is 60J. Just before it hits the surface its speed is: (Hint-take g=10 m/sec2)

• A. 10 m/sec✓
• B. 36 m/sec
• C. 40 m/sec
• D. 45 m/sec

Explanation: To solve the problem, we first calculate the gravitational potential energy (PE) of the object when it is released from a height of 8 m. The formula for potential energy is PE = mgh, where m = 2 kg, g = 10 m/s², and h = 8 m. Thus, PE = 2 kg * 10 m/s² * 8 m = 160 J.Next, we account for the work done against air resistance, which is given as 60 J. The effective energy available for conversion into kinetic energy (KE) just before hitting the ground is PE - work done against air resistance = 160 J - 60 J = 100 J. Using the kinetic energy formula KE = (1/2)mv², we can rearrange it to find v: v = sqrt((2 * KE) / m) = sqrt((2 * 100 J) / 2 kg) = sqrt(100) = 10 m/s.Thus, the final speed of the object just before it hits the ground is 10 m/sec. The other options (36 m/s, 40 m/s, and 45 m/s) do not correctly account for the work done against air resistance, leading to incorrect calculations.

Q152. The speedometer of a car shows 36 km/hr. The angular speed of the wheels having 0.5m radius is:

• A. 5 rad/sec
• B. 10 rad/sec
• C. 15 rad/sec
• D. 20 rad/sec✓

Explanation: To solve for the angular speed of the wheels, we first need to convert the car's speed from kilometers per hour to meters per second. The car's speed is given as 36 km/hr, which can be converted to m/s by multiplying by a factor of (1000 m / 1 km) * (1 hr / 3600 s) = 10 m/s. Next, we apply the relationship between linear speed (v) and angular speed (ω) using the formula: ω = v / r, where r is the radius of the wheel. Given that the radius (r) is 0.5 m, we substitute the values into the formula: ω = 10 m/s / 0.5 m = 20 rad/sec. Thus, the correct answer is 20 rad/sec. The other options are incorrect because they do not accurately reflect the conversion and calculation based on the given data.

Q153. The angular speed in radian/hour for the daily rotation of our earth is?

• A. π/12 radian/hour✓
• B. 12π radian/hour
• C. 24π radian/hour
• D. π/24 radian/hour

Explanation: The correct answer is π/12 radian/hour. To find the angular speed in radians, we recognize that one full rotation (2π radians) occurs in 24 hours. Thus, the angular speed can be calculated as:Angular speed = Total radians per rotation / Time in hours = 2π / 24 = π/12 radian/hour.

Q154. A fly wheel rotates at constant speed of 600rpm. The angle described by the shaft in radians in one second is:

• A. 100
• B. 20π✓
• C. 0
• D. 20

Explanation: To determine the angle described by the shaft in radians in one second, we first need to convert the rotational speed from revolutions per minute (rpm) to radians per second. Since one revolution equals 2π radians, we can calculate:600 rpm = 600 revolutions/minute * (1 minute/60 seconds) * (2π radians/revolution) = 600 * (2π/60) = 20π radians/second.This calculation shows that the correct answer is 20π radians. Option A (100π) suggests an incorrect larger angle, while Option C (0) states no movement, which contradicts the rotating nature of the flywheel. Option D (20) misses the necessary π factor, making it incorrect as well.

Q155. The angular speed of the second hand of a watch is:

• A. (∏/1800) rad/sec
• B. (∏/60) rad/sec
• C. (∏/30) rad/sec✓
• D. (2n) rad/sec

Explanation: The correct answer is (π/30) rad/sec. The second hand of a watch completes one full revolution (2π radians) in 60 seconds, which means its angular speed is calculated as: Angular Speed = Total Rotation / Time = (2π radians) / (60 seconds) = (π/30) rad/sec. Let's examine the other options: (π/1800) rad/sec suggests an extremely slow speed that is not applicable to a watch's second hand, as it would imply a highly unrealistic time for a full rotation. (π/60) rad/sec is also incorrect because it does not accurately represent the complete rotation in the given time frame of 60 seconds. (2n) rad/sec is completely irrelevant, as 'n' is not defined and does not pertain to the angular motion of the second hand.

Q156. A stretched string vibrates with frequency f, when tension in the string is T, for what value of tension, the frequency of the same string is doubled?

• A. 2T
• B. 4T✓
• C. 8T
• D. 16T

Explanation: The frequency of a stretched string is given by the formula f = (1/2L)√(T/μ), where T is the tension, L is the length of the string, and μ is the linear mass density. To double the frequency (2f), we set up the equation: 2f = (1/2L)√(T'/μ), where T' is the new tension. By squaring both sides and solving for T', we find that (2f)² = (1/2L)²(T'/μ). This leads to the conclusion that T' = 4T. Therefore, to double the frequency, the tension must be quadrupled. The other options are incorrect because:2T: Simply doubling the tension does not suffice to double the frequency, as frequency increases with the square root of tension.8T: Increasing to 8T would increase the frequency by a factor of 2.83, which is more than double.16T: Quadrupling the tension results in a frequency increase of 4 times, not just double.

Q157. The longitudinal waves travel more slowly in_ than in _

• A. solids, gas
• B. solids, liquids
• C. gases, solids✓
• D. liquid, gases

Explanation: Longitudinal waves, such as sound waves, travel slower in gases than in solids. This is because the particles in gases are more spread out, making it harder for the sound waves to transmit energy quickly. In contrast, solids have densely packed particles, allowing for faster energy transfer and wave propagation. Therefore, the correct answer is that longitudinal waves travel more slowly in gases than in solids. Option A is incorrect because it suggests that solids allow for slower wave travel compared to gases. Option B misrepresents the relationship between solids and liquids in terms of wave speed. Option D also incorrectly states that liquids allow for slower wave travel compared to gases. Hence, the only accurate statement is offered by Option C.

Q158. Sinusoidal water waves are generated in a large ripple tank. The waves travel at 20 cm/s and their adjacent crests are 5.0 cm apart. The time required for each new whole cycle to be generated is:

• A. 100 s
• B. 2.0 s
• C. 4.0 s
• D. 0.25 s✓

Explanation: The time period (T) of a wave is calculated using the formula T = λ/v, where λ is the wavelength and v is the wave speed. Given that the wave speed is 20 cm/s and the distance between adjacent crests (wavelength) is 5.0 cm, the time period is calculated as T = 5.0 cm / 20 cm/s = 0.25 s. Therefore, the correct answer is 0.25 s. The other options are incorrect because they do not match the result of this calculation.

Q159. The distance between adjacent node and anti-node is equal to:

• A. 𝛌
• B. 𝛌/2
• C. 2𝛌
• D. 𝛌/4✓

Explanation: The correct answer is 𝛌/4 because in wave motion, a node is a point of destructive interference where the amplitude is zero, and an anti-node is a point of constructive interference where the amplitude is maximum. The distance between a node and its adjacent anti-node is one quarter of the wavelength (𝛌), as the wave oscillates between these two points in its cycle. The other options are incorrect because they either represent the full wavelength (𝛌), half the wavelength (𝛌/2), or double the wavelength (2𝛌), which do not accurately describe the specific distance between a node and its adjacent anti-node.

Q160. A string clamped at its both ends, vibrates in four segments (loops). The length of string is 150cm. The wavelength of the wave is:

• A. 33.3 cm
• B. 66.7 cm
• C. 150 cm
• D. 75 cm✓

Explanation: The length of the string is 150 cm and it vibrates in four segments. Each segment corresponds to half a wavelength, so the total length of the string represents two wavelengths. Therefore, the wavelength can be calculated using the formula:wavelength = (length of string) / (number of segments / 2)Substituting the given values, we get:wavelength = 150 cm / (4 / 2) = 150 cm / 2 = 75 cm

Q161. The square root of the ratio of elasticity to mass density is equal to:

• A. Force
• B. Product of frequency and wavelength✓
• C. Co-efficient of viscosity
• D. Refractive index

Explanation: The correct answer is the product of frequency and wavelength. This relationship arises from the wave equation, where the speed of sound in a medium is given by the square root of the ratio of the elastic modulus (elasticity) to the mass density. Therefore, the square root of this ratio yields the speed of sound, which directly relates to how waves propagate through solids.Option A, Force, is incorrect because force is not a derived quantity from elasticity and mass density in this context. Option C, Co-efficient of viscosity, pertains to fluid dynamics and is unrelated to the elastic properties of solids. Lastly, Option D, Refractive index, is a concept in optics and does not apply to the relationship expressed in the question.

Q162. If the pressure of air enclosed in a long tube is increased 10 times, then the speed of sound will:

• A. Increase
• B. Decrease
• C. Remain constant✓
• D. Becomes zero

Explanation: The speed of sound in a gas is primarily determined by the temperature of the gas rather than the pressure. While increasing the pressure of air might intuitively suggest a change in sound speed, it is important to note that in ideal gas conditions, the speed of sound remains constant if the temperature does not change. Thus, in this scenario, the speed of sound will remain constant. Option A is incorrect because increasing pressure does not necessarily increase sound speed in a constant temperature environment. Option B is also incorrect as sound speed does not decrease with increased pressure. Option D is incorrect since sound cannot become zero simply because of pressure changes.

Q163. The expression, Cp - R is equal to:

• A. Cv✓
• B. R
• C. Cp - Cv
• D. R + Cv

Explanation: The correct answer is Cv, as it represents the specific heat at constant volume. The equation Cp - R = Cv is derived from the relationship between the specific heats of gases. Specifically, for an ideal gas, the specific heat at constant pressure (Cp) is always greater than the specific heat at constant volume (Cv) by the gas constant R. Thus, when we subtract R from Cp, we arrive at Cv. The other options are incorrect because:Option B (R) is not equal to Cp - R; R is a separate constant.Option C (Cp - Cv) is not equal to Cp - R; rather, Cp - Cv equals R, which is a different relationship.Option D (R + Cv) is also incorrect; it does not derive from the relationship we are examining.Understanding these relationships is crucial in thermodynamics, particularly when dealing with ideal gases.

Q164. In a certain process, 200J of heat energy is supplied to a system and at the same time 50J of work is done by the system. The increase in the internal energy of the system is:

• A. 25 J
• B. 100 J
• C. 150 J✓
• D. 250 J

Explanation: The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). In this case:ΔU = Q - WGiven that 200J of heat energy is supplied to the system (Q = 200J) and 50J of work is done by the system (W = 50J), we can substitute these values into the equation:ΔU = 200J - 50J = 150JThus, the increase in internal energy is 150J, making Option C the correct choice.Options A, B, and D are incorrect because they either misapply the first law of thermodynamics or ignore the work done by the system, leading to inaccurate calculations of internal energy change.

Q165. Two charges of magnitude q1= 1𝜇C and q2 = 5𝜇C, are separated at a distance r = 1x10-3m apart, the ratio of the magnitude of the forces acting on them will be:

• A. 1:5
• B. 1:25
• C. 1:3
• D. 1:1✓

Explanation: The magnitude of force on both charges is same. Hence the ratio will be 1:1.

Q166. For 0.5 Siemens of conductance, resistance will be:

• A. 1 Ω
• B. 2 Ω✓
• C. 20 Ω
• D. 10 Ω

Explanation: The correct answer is 2 Ω. This is determined by applying the formula for resistance, which is the inverse of conductance. Given that conductance (G) is 0.5 Siemens, we calculate the resistance (R) as follows: R = 1/G = 1/0.5 = 2 Ω. Options A, C, and D are incorrect because they do not satisfy the relationship between resistance and conductance. Specifically, option A suggests a resistance of 1 Ω, which would indicate a conductance of 1 Siemens, option C indicates a significantly higher resistance that implies very low conductance, and option D suggests a resistance that would indicate a conductance of only 0.1 Siemens.

Q167. 1 volt x 1 ampere is equal to:

• A. 1 coulomb
• B. 1 newton
• C. 1 watt✓
• D. 1 hp

Explanation: According to the formula for electrical power, Power (P) = Voltage (V) x Current (I). Hence, when you multiply 1 volt by 1 ampere, you get 1 watt. That's why option C is the correct answer. Option A (1 coulomb) is incorrect because coulombs measure electric charge, not power. Option B (1 newton) is also incorrect as newtons measure force and do not relate to electrical power. Lastly, Option D (1 hp) is incorrect because while it is a unit of power, it is much larger than 1 watt and does not represent the correct value derived from the given voltage and current.

Q168. A wire of length l has resistivity 𝛒. If the wire Is divided in two halves, then resistivity of each halve is:

• A. 𝛒/2
• B. 𝛒✓
• C. 2𝛒
• D. 𝛒/3

Explanation: When a wire is cut into two equal halves, the resistivity (𝛒) of the material from which the wire is made remains unchanged. Resistivity is an intrinsic property that depends on the material itself and its temperature. Therefore, regardless of the length of the wire, each half retains the same resistivity as the original wire. Thus, the correct answer is 𝛒. Option A (𝛒/2) is incorrect because resistivity does not decrease with the length of the wire. Option C (2𝛒) mistakenly suggests that cutting the wire increases the resistivity, which is not the case. Option D (𝛒/3) similarly misrepresents the concept, as resistivity remains constant for a given material.

Q169. In how many hours a 1000 watt AC will consume one unit of electricity?

• A. 0.5 hr
• B. 1 hr✓
• C. 1.5 hr
• D. 2 hr

Explanation: The power of the AC is given as 1000 watts, which is equivalent to 1 kilowatt. One unit of electricity is defined as 1 kilowatt-hour (kWh). Therefore, a 1000 watt device will consume one unit of electricity in one hour. To understand why the other options are incorrect: 0.5 hours would only consume 0.5 units, 1.5 hours would consume 1.5 units, and 2 hours would consume 2 units of electricity. Thus, the correct answer is 1 hour.

Q170. Of the following, the copper conductor that has the least resistance must be:

• A. Thick, short and cool✓
• B. Thin, long and hot
• C. Thick, long and hot
• D. Thin, short and cool

Explanation: The correct answer is Thick, short and cool because resistance is inversely related to the cross-sectional area and directly related to the length of the conductor. A thicker conductor (larger cross-sectional area) lowers resistance, while a shorter conductor reduces the path for current flow. Cool temperatures decrease the resistivity of copper, contributing to even lower resistance. In contrast, the options Thin, long and hot, Thick, long and hot, and Thin, short and cool do not meet all criteria for minimizing resistance. The first option suffers from both thinness and length, the second has increased length and temperature, and the last option, despite being short and cool, is thin which increases resistance.

Q171. A certain X-ray tube requires a current of 5mA at a voltage of 60kv. The rate of energy dissipation (in watts) is:

• A. 560
• B. 300✓
• C. 200
• D. 800

Explanation: To calculate the rate of energy dissipation (power) in an X-ray tube, we use the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts. In this case, the current is given as 5mA, which is equal to 0.005A (since 1A = 1000mA). The voltage is 60kV, which is equal to 60000V (since 1kV = 1000V).Now, substituting the values into the formula: P = 0.005A * 60000V = 300 Watts. Therefore, the correct answer is Option B: 300 watts.The other options are incorrect because:Option A (560): This value does not correspond to the correct calculation.Option C (200): This is too low for the given current and voltage.Option D (800): This is too high and does not match the calculated power.

Q172. Ampere.second is the unit of:

• A. Power
• B. Charge✓
• C. Potential Difference
• D. Current

Explanation: The correct answer is Charge because the unit ampere.second represents a coulomb, which is the standard unit of electric charge. According to the definition of a coulomb, it is the amount of electric charge transported by a constant current of one ampere in one second. In contrast, the options for Power (watts), Potential Difference (volts), and Current (amperes) are all different physical quantities with their own distinct units. Power relates to energy transfer, potential difference measures energy per charge, and current measures the flow of electric charge per unit time, but none of these relate directly to the unit of charge itself.

Q173. The magnitude of the magnetic flux through area A has exactly the same value at angles:

• A. 0° and 180°✓
• B. 90° and 180°
• C. 180° and 270°
• D. 270° and 360°

Explanation: The magnitude of magnetic flux (Φ) through a surface is given by the formula Φ = B·A·cos(θ), where B is the magnetic field strength, A is the area, and θ is the angle between the magnetic field and the normal to the surface. For angles of 0° and 180°, cos(0°) = 1 and cos(180°) = -1, indicating that the magnitude of the flux remains the same, although the direction is opposite.In contrast, at 90°, the flux becomes zero because the angle leads to cos(90°) = 0. Similarly, at 270°, the flux is also zero due to the same cosine factor. Thus, options B, C, and D are incorrect as they do not yield the same non-zero flux values as found at 0° and 180°.

Q174. A hydrogen atom that has lost its electron is moving east in a region where the magnetic field is directed from south to north. It will be deflected:

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

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

Q175. When the magnetic force acts on a charged particle, change occurs in:

• A. Magnitude of velocity only
• B. Magnitude and direction of velocity
• C. Only the direction of velocity✓
• D. Neither direction nor magnitude of velocity

Explanation: The magnetic force acts on a charged particle moving in a magnetic field, which is given by the equation F = q(v × B), where F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field. The force is always perpendicular to both the velocity of the particle and the magnetic field, which results in a change in the particle's direction of motion but not its speed. Thus, the correct answer is that only the direction of velocity changes. Option A is incorrect because it suggests that only the magnitude changes, which is not the case. Option B incorrectly states that both magnitude and direction change, which also does not happen. Option D is incorrect because it ignores the directional change caused by the magnetic force.

Q176. A constant magnetic field of 5T is passing through a static conducting loop of area 0.8m2, the induced emf is

• A. 4 V
• B. 10 V
• C. Zero✓
• D. 32 V

Explanation: The induced EMF (electromotive force) in a conducting loop is determined by Faraday's law of electromagnetic induction, which states that an EMF is induced when there is a change in magnetic flux through the loop. In this scenario, we have a constant magnetic field of 5T and a static conducting loop with an area of 0.8m². Since the magnetic field strength does not change, the magnetic flux through the loop remains constant, leading to an induced EMF of zero. Therefore, Option C is correct. Options A (4 V), B (10 V), and D (32 V) all propose values for the induced EMF that imply a change in magnetic flux, which does not exist in this instance due to the constancy of the magnetic field.

Q177. Magnetic flux linked with the conducting loop "A" is increased at the rate of 15 weber/second per and another conducting loop "B" is decreased at the rate of 15 weber/second, the ,magnitude of induced emf of:

• A. Loop A will induce an EMF equal to that of Loop B.✓
• B. Loop A is smaller than Loop B.
• C. Loop B is smaller than Loop A.
• D. Both will have zero induced EMF.

Explanation: The induced EMF in a loop is directly proportional to the rate of change of magnetic flux through the loop, as described by Faraday's law of electromagnetic induction. In this case, Loop A is having its magnetic flux increased at a rate of 15 weber/second, while Loop B is having its magnetic flux decreased at the same rate of 15 weber/second. Therefore, both loops will experience an induced EMF of equal magnitude but opposite direction. This leads to the conclusion that Loop A will have an induced EMF equal to that of Loop B. Option B is incorrect because the size of the loops does not affect the induced EMF in this context. Option C is similarly incorrect for the same reason. Option D is incorrect because both loops are experiencing changes in magnetic flux, resulting in non-zero induced EMF.

Q178. For induced emf to be produced in a coil, the magnetic flux linked with the coil must:

• A. Only decrease
• B. Only increase
• C. Be constant
• D. Be Changed✓

Explanation: According to Faraday's law of electromagnetic induction, induced emf is generated when there is a change in magnetic flux through a coil. This change can be an increase or decrease in the magnetic flux. Therefore, the correct answer is that the magnetic flux must be changed to induce emf. Options A and B are incorrect because, while they mention changes in flux, they limit the scope to only decreases or increases, respectively, which does not encompass all scenarios of change. Option C is incorrect as it states that the magnetic flux must be constant, which would result in no induced emf. Hence, only a change in magnetic flux (either increase or decrease) can lead to the production of induced emf.

Q179. For half wave rectification, the number of diode needed in a circuit is:

• A. 1✓
• B. 2
• C. 3
• D. 4

Explanation: Half-wave rectification allows only one half of the AC signal to pass through, which is achieved using a single diode. This diode conducts current only during one half of the AC cycle (either positive or negative), effectively blocking the other half. Therefore, only one diode is necessary to implement this process. The other options are incorrect because:Option B (2 diodes): This configuration is used for full-wave rectification, not half-wave.Option C (3 diodes): Three diodes are not applicable in standard rectification methods for half-wave.Option D (4 diodes): This is typical of a bridge rectifier setup, used for full-wave rectification, which is not relevant to the half-wave process.

Q180. The frequency of a light beam "A" is half of that of light beam "B". The ratio EA/Ea of photons energies is:(hint: where EA & Es are the respective energies of light Beam "A" & "B")

• A. 1/2✓
• B. 1/4
• C. 1
• D. 2

Explanation: The correct answer is 1/2. The energy of a photon can be calculated using the formula E = hf, where E is energy, h is Planck's constant, and f is frequency. If the frequency of light beam 'A' is half that of light beam 'B' (i.e., fA = 1/2 * fB), we can express the energies of the two beams as EA = h * fA and EB = h * fB. Substituting for fA gives us EA = h * (1/2 * fB) = (1/2) * h * fB = (1/2) * EB. Thus, the ratio EA/EB = (1/2) and is equal to 1/2. The other options are incorrect for the following reasons: - 1/4: This implies a larger difference in energy than the frequency relationship allows. - 1: This suggests equal energies, which contradicts the stated frequency relationship. - 2: This incorrectly suggests that beam 'A' has higher energy, which is not possible given its lower frequency.

Q181. The wavelength of light beam "A" is twice the wavelength of light beam "B" The energy of a photon in beam "A" is:

• A. Half the energy of a photon in beam B✓
• B. One-fourth the energy of a photon in beam B
• C. Equal to the energy of a photon in beam B
• D. Twice the energy of a photon in beam B

Explanation: Since energy is inversely proportional to wavelength, doubling the wavelength cuts the energy in half. So a photon in beam A has half the energy of a photon in beam B.

Q182. If frequency of light is greater than threshold frequency, then in photoelectric effect the number of electrons increases with an increase in _ of light :

• A. Frequency
• B. Kinetic energy
• C. Intensity✓
• D. Momentum

Explanation: In the photoelectric effect, when the frequency of light exceeds a certain threshold frequency, photons can eject electrons from a material. The number of electrons emitted is directly related to the intensity of the light. Intensity measures the number of photons hitting the surface per unit area per unit time; thus, a higher intensity means more photons are available to knock electrons loose. Therefore, increasing the intensity results in more electrons being emitted.Option A (Frequency) is incorrect because while frequency determines the energy of the emitted electrons, it does not influence the number of electrons emitted once the threshold is surpassed. Option B (Kinetic energy) is also incorrect as it relates to the energy of the emitted electrons, not their quantity. Option D (Momentum) does not apply in this context, as it does not affect the count of emitted electrons.

Q183. Shortest wavelength associated to Paschen series is:

• A. 𝜆=8/RH
• B. 𝜆=9/RH✓
• C. 𝜆=19/RH
• D. 𝜆=4/RH

Explanation: The Paschen series corresponds to electron transitions from higher energy levels down to the n=3 level. The shortest wavelength associated with this series occurs during the transition from n=4 to n=3, which can be calculated using the Rydberg formula. The correct formula for this transition is 𝜆=9/RH, where RH is the Rydberg constant. The other options are incorrect because they correspond to transitions that are not part of the Paschen series or suggest incorrect energy levels.

Q184. The ratio of longest wavelength to shortest of Lyman series is:

• A. R
• B. 4/3✓
• C. 3/4
• D. 4

Explanation: The explanation of this question is given below;Maximum wavelength/Minimum wavelength = 121.6 nm/91.2 nm = 1.333... So the ratio of the maximum wavelength to the minimum wavelength in the Lyman series is approximately 1.33 = 4/3

Q185. The half-life of a radioactive element is 100 days. How much quantity of the 88 g of such element will remain after 3 half-lives:

• A. 8g
• B. 10g
• C. 11g✓
• D. 44g

Explanation: To find the remaining amount of a radioactive element after a certain number of half-lives, you can use the formula: Remaining Quantity = Initial Quantity × (1/2)^(number of half-lives). Here, the initial quantity is 88g and the number of half-lives is 3. Therefore, Remaining Quantity = 88g × (1/2)^3 = 88g × (1/8) = 11g. Thus, after 3 half-lives, 11g of the element remains.Option A (8g) is incorrect because it suggests an even more significant decay than calculated. Option B (10g) underestimates the remaining amount after 3 half-lives. Option D (44g) incorrectly assumes only one half-life has occurred, which is not the case.

Q186. Radioactive kcSr has a half-life of 30 years. What percent of a sample of kcSr will remain after 90 years?

• A. 0%
• B. 12.5%✓
• C. 50%
• D. 75%

Explanation: To solve the question, we first determine how many half-lives have passed in 90 years. Given the half-life of kcSr is 30 years, we calculate:90 years ÷ 30 years/half-life = 3 half-lives.After each half-life, the remaining percentage of the original sample is halved. Starting with 100%:After 1 half-life (30 years): 100% ÷ 2 = 50%After 2 half-lives (60 years): 50% ÷ 2 = 25%After 3 half-lives (90 years): 25% ÷ 2 = 12.5%Thus, 12.5% of the original kcSr sample remains after 90 years. The other options are incorrect because they represent amounts after fewer half-lives or assume complete decay.

Q187. At the end of 15 min, 1/32 of a sample of radioactive polonium remains. The corresponding half-life is:

• A. 0.469 min
• B. 32 min
• C. 15 min
• D. 3 min✓

Explanation: The correct answer is 3 minutes. To determine this, we first need to understand the relationship between the half-life and the remaining fraction of the radioactive substance. After each half-life, the amount of the sample halves. Therefore, we can calculate how many half-lives have occurred in 15 minutes by examining the fraction that remains:Starting with 1 (the whole sample), after one half-life (3 min), we have 1/2 remaining; after two half-lives (6 min), we have 1/4 remaining; after three half-lives (9 min), we have 1/8 remaining; after four half-lives (12 min), we have 1/16 remaining; and after five half-lives (15 min), we have 1/32 remaining. This shows that 5 half-lives have occurred in 15 minutes, confirming that the half-life is indeed 3 minutes.Other options are incorrect as they do not result in 1/32 remaining after 15 minutes. For example, with a half-life of 15 minutes, only half of the sample would remain after 15 minutes (1/2), and with a half-life of 32 minutes, it would take much longer to reach 1/32 of the original sample.

Q188. Two equal and opposite charges of 10 C are separated at a distance of 10 cm, the electric potential at mid-point between the charges is:

• A. 20 V
• B. 10 V
• C. 5 V
• D. 0 V✓

Explanation: At the midpoint between two equal and opposite charges, the electric potential contributions from each charge are equal in magnitude but opposite in sign. Therefore, they cancel each other out, resulting in a total electric potential of 0 V at that point. This is why the correct answer is 0 V. The other options suggest a non-zero potential, which does not consider the crucial fact that the charges are equal and opposite, leading to a complete cancellation of their effects at the midpoint.

Q189. The force of repulsion between two alike charges is 10 N in vacuum. When a material of Ep= 2 is placed between them. New force will be:

• A. 20 N
• B. 15 N
• C. 10 N
• D. 5 N✓

Explanation: In this problem, we start with a repulsive force of 10 N between two like charges in a vacuum. When a dielectric material with an electric polarization factor of Ep= 2 is introduced, the force between the charges is reduced by this factor. The relationship governing this change is given by:F' = F / Ep,where F' is the new force, F is the original force (10 N), and Ep is the electric polarization factor (which is 2 in this case). Thus:F' = 10 N / 2 = 5 N.This shows that the force is halved due to the presence of the dielectric. The other options are incorrect as they either suggest an increase in force or no change at all, which contradicts the effect of the dielectric material on the electric forces between the charges.

Q190. The slope-of the charge-time graph for a charging capacitor gives:

• A. Current✓
• B. Voltage
• C. Force
• D. Energy

Explanation: The correct answer is Current, as the slope of the charge-time graph indicates how quickly charge is flowing into the capacitor, which is defined as current (I = dq/dt). A steeper slope corresponds to a higher current, meaning more charge is accumulating over time. The other options do not represent this relationship: Voltage is related to the charge stored but not the rate of charge flow, Force is unrelated to charge and time, and Energy pertains to the work done to charge the capacitor rather than the rate of charge accumulation.

Q191. Two small charged objects attract each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes:

• A. F/16
• B. F/8
• C. F/4✓
• D. F/2

Explanation: According to Coulomb's Law, the force F between two point charges is given by the formula: F = k * (q1 * q2) / d^2, where k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and d is the distance between them. In this scenario, the charges are reduced to one-fourth of their original values, so we can write this as:F' = k * ((q1/4) * (q2/4)) / (d/2)^2Now simplifying this gives:F' = k * (q1 * q2) / (16 * (d^2/4)) = k * (q1 * q2) / (4 * d^2)This means that the new force F' is F/4. Therefore, option C is correct. The other options miscalculate the force due to incorrect consideration of both charge reduction and distance reduction.

Q192. If both the plate area and the plate separation of a parallel-plate capacitor doubled, the capacitance is

• A. Doubled
• B. Halved
• C. Unchanged✓
• D. Triple

Explanation: The capacitance (C) of a parallel-plate capacitor is given by the formula C = (ε₀ * A) / d, where A is the area of the plates, d is the separation between the plates, and ε₀ is the permittivity of free space. When both the plate area (A) and the plate separation (d) are doubled, the new capacitance becomes C' = (ε₀ * (2A)) / (2d) = (ε₀ * A) / d = C. Thus, the capacitance remains unchanged.Option A is incorrect as it mistakenly assumes that only the area increase affects capacitance without accounting for the separation change. Option B incorrectly assumes that the increase in separation alone would reduce capacitance. Option D miscalculates the relationship entirely by suggesting an increase by a factor of three.

Q193. A 20𝜇F capacitor is charged 200 V. Its stored energy is:

• A. 4000 J
• B. 4 J
• C. 0.4 J✓
• D. 2000 J

Explanation: The energy stored in a capacitor can be calculated using the formula E = 0.5 * C * V^2. Here, C is the capacitance in farads, and V is the voltage. Given that C = 20μF (which is 20 × 10^-6 F) and V = 200 V, we can substitute these values into the formula:E = 0.5 * (20 × 10^-6) * (200^2) = 0.5 * (20 × 10^-6) * 40000 = 0.5 * 0.8 = 0.4 J.Thus, the correct answer is 0.4 J. The other options are incorrect because they either vastly overestimate or underestimate the energy based on the formula and the given values.

Q194. The primary winding of a transformer has a 120 V AC supply. What is the value of secondary voltage if the turn ratio is 20.(Hints: turn ratio =N1/N0)

• A. 2400V✓
• B. 1200V
• C. 24000V
• D. 200 V

Explanation: The primary winding of the transformer has a voltage of 120 V and a turn ratio of 20. The formula to find the secondary voltage (V2) is V2 = V1 × (N2/N1), where N2/N1 is the turn ratio. Here, the turn ratio is 20, which indicates that for every 1 turn in the primary winding, there are 20 turns in the secondary winding. Thus, V2 = 120 V × 20 = 2400 V, making Option A the correct answer. The other options either miscalculate the voltage based on incorrect turn ratios or misunderstand the relationship between the primary and secondary voltages.

Q195. Statements:Some teachers are students.All students are girls.No student is a boy.Condusions:I. Some teachers are boys.II. Same girls are teachers.IlI. Some girls are students.

• A. Only I follows
• B. Only I, II and III follow
• C. Only II follows
• D. Only II and III follow✓

Explanation: The correct answer is D: 'Only II and III follow.' To analyze the conclusions:- Conclusion I ('Some teachers are boys') cannot be true because it contradicts the third statement that 'No student is a boy.' Since some teachers are students, it implies that those teachers cannot be boys.- Conclusion II ('Some girls are teachers') is valid because we know that some teachers are students, and since all students are girls, it follows that some of those girls must be teachers.- Conclusion III ('Some girls are students') is also valid as it directly follows from the statement that all students are girls.Thus, the only valid conclusions are II and III, making option D the correct choice.

Q196. Complete the following series: KCD, LEF, MGH,_ OKL.

• A. NQM
• B. PJI
• C. NIJ✓
• D. NJI

Explanation: The series can be analyzed by examining the letters in each segment. For the second letter in each term, the pattern follows:K (11) → L (12) → M (13) → N (14) → O (15) indicates a +1 increment.C (3) → E (5) → G (7) → I (9) → K (11) shows a +2 increment.Following this pattern leads us to the next term, which would be NIJ. The other options do not fit these increments:NQM fails to establish the correct sequence.PJI deviates from the progression.NJI, while similar, does not maintain the required increments.Thus, the correct answer is NIJ.

Q197. Read the following statements and decide about the arguments.Statement: Does Pakistan need so many plans for development?Arguments:_ I. Yes. Nothing can be achieved without proper planning.II. No. Too much time, money and energy is wasted on planning.

• A. Only I is true✓
• B. Only II is true
• C. Either I or II is true
• D. Neither I nor II is true

Explanation: Argument I states that proper planning is essential for achieving development objectives, which is a strong argument. It highlights the necessity of having a clear direction and allocation of resources, which is fundamental in any development program. On the other hand, argument II suggests that planning is a waste of resources, which is a flawed perspective. Effective planning not only helps in utilizing resources efficiently but also guides the implementation process. Therefore, the correct answer is that only argument I is valid. The other options are incorrect as they either misinterpret the necessity of planning or disregard the validity of argument I.

Q198. Methyl tetrachloride (MTC) is a chemical found in some pesticides, glues, and sealants. Exposure to MTC can cause people to develop asthma. In order to halve the nation's asthma rate, the government plans to ban all products containing MTC. The government's plan to halve the nation's asthma rate relies on which of the following assumptions?

• A. Exposure to MTC is responsible for no less than half of the nation's asthma cases.✓
• B. Products containing MTC are not necessary to the prosperity of the country's economy.
• C. Asthma has reached epidemic proportions.
• D. MTC products are helpful for asthma.

Explanation: The correct answer is Option A: 'Exposure to MTC is responsible for no less than half of the nation's asthma cases.' This is the underlying assumption of the government's plan, as it directly correlates the ban on MTC products with a significant reduction in asthma cases. Without this assumption, the effectiveness of such a ban in achieving the goal of halving asthma rates would be unfounded.Option B is incorrect because the economic implications of MTC products do not directly affect their relationship with asthma cases. Option C is not pertinent to the assumption needed for the plan, as the epidemic status of asthma does not imply that banning MTC will lead to a decrease in cases. Lastly, Option D is incorrect because it suggests a beneficial role for MTC products in asthma, which contradicts the rationale behind the proposed ban.

Q199. Statements:I. The Importance of yoga and exercise is being realized by all sections of the societyII. There is an increasing awareness about health in the society particularly among middle ages group of people.

• 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 the statements I and II are independent causes.
• D. Both the statements I and II are effects of independent causes.

Explanation: Statement II indicates an increasing awareness about health, particularly among the middle-aged group. This awareness leads to the realization of the importance of yoga and exercise across all sections of society, as stated in Statement I. Thus, Statement II serves as the cause, and Statement I is the effect. The other options incorrectly interpret the relationship between the statements; Option A reverses the cause-effect order, while Options C and D suggest independence, which contradicts the clear connection presented in the statements.

Q200. The rates of interest on Post office recurring deposit accounts have been increased with effect from 1st March. This has been done to attract more deposits.Course of ActionI. Efforts should also be made to make the public aware about this increase in the rate of Interest.II. If the deposits don't increase in next six months, the rate of interest should be further Increased.

• A. Only conclusion I follows
• B. Only conclusion II follows
• C. Neither I nor II follows
• D. Both conclusions I and II follow✓

Explanation: The question presents a scenario where the interest rates on post office recurring deposit accounts have been increased to attract more deposits. Conclusion I advocates for public awareness of this increase, which is essential for informing potential depositors and encouraging them to invest. Conclusion II suggests that if deposits do not rise in the following six months, the interest rate should be increased further, allowing for a responsive strategy to enhance deposit attraction.Both conclusions are valid because they address different aspects of the same objective: to increase deposits. The first focuses on communication, while the second emphasizes evaluation and adjustment based on initial results. Therefore, the correct answer is that both conclusions follow.Option A is incorrect as it only considers the need for public awareness, ignoring the evaluation aspect. Option B wrongly prioritizes only the evaluation without acknowledging the need for awareness. Option C dismisses both actions, which is inaccurate given their relevance to the situation.