Sindh Mcat Nts 2014 Duhs And Jsmu — Solved Past Paper with Answers

Q1. Choose the correct antonym of the given word:Distress

• A. Suffering
• B. Lash out
• C. Upset
• D. Happiness✓

Explanation: The word 'DISTRESS' refers to a state of emotional pain or discomfort. The opposite of this emotional state is 'HAPPINESS', which signifies contentment and joy.

Q2. Choose the lettered word or phrase that is most nearly opposite in meaning to the word in capital letters:SWIFT:

• A. Slow✓
• B. Fast
• C. Brief
• D. At once

Explanation: The word 'swift' means quick or fast. The opposite of this would be 'slow,' which indicates a lack of speed or movement. This makes 'slow' the correct antonym.

Q3. Read the passage to answer the question:What is life? A little scum of no importance on the surface of an unimportant globe circling round a second-rate star? An accidental conglomeration of atoms which have come together by an odd chance, the result of an exceedingly improbable happening? That is what some astronomers would have us think. Looking out into the depth of space, they have discovered a universe of unthinkable dimensions. A billion suns in our own galaxy, beyond it perhaps a billion galaxies, only revealed to us as tiny smudges on a photographic plate. No wonder they are impressed by the enormous disparity between the scaffolding and the result. Life seemed to be, as Jeans said, ‘an utterly unimportant by-product’ in a universe which was clearly not designed for life, and which, to all appearances, is either totally indifferent or definitely hostile to it’. It seemed ‘incredible that the universe can have been designed primarily to produce life like our own; had it been so, surely we have might expected to find a better proportion between the magnitude of the mechanism and the amount of the product.The title of the passage can be:

• A. Gathering of atoms
• B. The life outside the earth
• C. Universe and it's unneeded vastness
• D. Life versus universe✓

Explanation: In this passage, the author can be seen talking about the majesty of the universe with respect to the existence of “life” in various respects. Hence the best title to this will be Life Versus Universe.

Q4. Read the passage to answer the question:What is life? A little scum of no importance on the surface of an unimportant globe circling round a second-rate star? An accidental conglomeration of atoms which have come together by an odd chance, the result of an exceedingly improbable happening? That is what some astronomers would have us think. Looking out into the depth of space, they have discovered a universe of unthinkable dimensions. A billion suns in our own galaxy, beyond it perhaps a billion galaxies, only revealed to us as tiny smudges on a photographic plate. No wonder they are impressed by the enormous disparity between the scaffolding and the result. Life seemed to be, as Jeans said, ‘an utterly unimportant by-product’ in a universe which was clearly not designed for life, and which, to all appearances, is either totally indifferent or definitely hostile to it’. It seemed ‘incredible that the universe can have been designed primarily to produce life like our own; had it been so, surely we have might expected to find a a better proportion between the magnitude of the mechanism and the amount of the product.According to the author if 'the universe had been designed primarily to produce life like our own' then:

• A. There would have been a smaller proportion for us
• B. There would have been a better proportion for us✓
• C. There would have been more luxuries in the universe
• D. There would have be many galaxies containing many creatures

Explanation: The author demeans the idea that the universe is primarily designed to produce life, supporting it by the fact that there is so little life in-universe as mentioned in the lines below:“It seemed ‘incredible that the universe can have been designed primarily to produce life-like our own; had it been so, surely we might be expected to find a better proportion”

Q5. Complete the sentence by choosing the most appropriate option, from the given lettered choices (A to E) below:The tree must _ planted over fifty years ago.

• A. been
• B. be been
• C. would been
• D. have been✓
• E. have had

Explanation: The sentence is in the past tense, and it indicates an action that was completed before the present moment. In such cases, we use the present perfect tense, which consists of the auxiliary verb "have" followed by the past participle of the main verb. Therefore, the correct sentence is: "The tree must have been planted over fifty years ago."

Q6. Complete the sentence by choosing the most appropriate option, from the given lettered choices (A to E) below:It is quite _ for poor people to be happier than rich people.

• A. Possible✓
• B. Risky
• C. Potential
• D. Liable
• E. Different

Explanation: The best option is "possible", as it expresses the idea that it is indeed possible for poor people to be happier than rich people.

Q7. Identify the word or phrase that needs to be changed for the sentence to be correct:The office is so busy that two extra clerks have had to be taken on.

• A. The
• B. Busy
• C. Extra
• D. Have had
• E. No error✓

Explanation: There is no error in the above sentence. The sentence is grammatically correct. "The office" is a proper subject, "is so busy" correctly describes its state, "two extra clerks" is correctly phrased, and "have had to be taken on" is properly structured in the present perfect passive voice.

Q8. Identify the word or phrase that needs to be changed for the sentence to be correct:Mr. Ahmed is to old to work now; he depends upon his son.

• A. Mr.
• B. To✓
• C. Work now
• D. Depends
• E. No error

Explanation: The sentence contains a common error with the use of "to" instead of "too." The word "too" is used to indicate an excessive degree, which fits the context of Mr. Ahmed being "too old" to work. The other parts of the sentence are correct: "Mr." is an appropriate title, "work now" accurately describes the present situation, and "depends" is correctly used to show reliance. Therefore, the sentence should read: "Mr. Ahmed is too old to work now; he depends upon his son."

Q9. Choose the word most similar in meaning to the given word:Meditative

• A. Selfish
• B. Thoughtful✓
• C. Heedless
• D. Opinion

Explanation: The correct answer is Thoughtful as it captures the essence of being deeply engaged in thought, which is a core aspect of being meditative. The other options do not align with this meaning.

Q10. Choose the word most similar in meaning to the capitalized one:AMAZEMENT

• A. Surprise✓
• B. Appreciation
• C. Criticism
• D. Praise

Explanation: 'Amazement' is characterized by a feeling of wonder or astonishment, often due to something unexpected. The word 'surprise' best captures this notion of being unexpectedly astounded. In contrast, 'appreciation' is about valuing something, 'criticism' is disapproval, 'praise' is admiration, and 'objection' is opposition, none of which entail the sensation of astonishment associated with 'amazement'.

Q11. The first cell to contain the diploid number of chromosomes is:

• A. 1
• B. 2
• C. 3✓
• D. 4

Explanation: The zygote, formed by the fusion of a sperm and an ovum, is the first cell to contain the diploid number of chromosomes. The sperm and egg are haploid cells, each contributing one set of chromosomes. Upon fusion, the resulting zygote has two sets of chromosomes—one from each parent—making it diploid. This makes the zygote the first diploid cell in the sequence. The other options refer to later stages or cells that do not initially have diploid chromosomes.

Q12. Haploid and monoploid numbers of chromosomes of hexaploid wheat are:

• A. 21 and 42
• B. 7 and 21
• C. 21 and 7✓
• D. 42 and 21

Explanation: Hexaploid wheat has a total of 42 chromosomes. The haploid number, representing the number of chromosomes in a gamete, is half of the total, which is 21. The monoploid number refers to the basic set of chromosomes, which is 7 in wheat. Thus, the correct answer is Option C: 21 and 7. The other options either reverse these numbers or mistake the relationship between haploid and monoploid numbers.

Q13. An anticodon is the sequence of the nitrogenous bases on the:

• A. Complementary strand of mRNA which codes for one amino acid
• B. tRNA molecule where the amino acid is attached
• C. tRNA molecule which recognizes the appropriate sequence of bases on the mRNA✓
• D. mRNA molecule which instructs the ribosomes to initiate

Explanation: The correct answer is Option D. An anticodon is a sequence of three nucleotides on a tRNA molecule that recognizes and pairs with a complementary codon on an mRNA molecule. This pairing is crucial for the translation process, ensuring that the correct amino acid is added to the growing polypeptide chain. Option A is incorrect because DNA does not contain anticodons. Option B is incorrect as anticodons are not found on mRNA. Option C is misleading because while tRNA is where the amino acid is attached, the anticodon itself is responsible for recognizing codons on mRNA, not for amino acid attachment. Option E is incorrect because the mRNA's start codon, not the anticodon, initiates protein synthesis.

Q14. The diagrams below show chromosomes in a cell undergoing mitosis and in a cell undergoing meiosis. Which of the following names the stages of division correctly?

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

Explanation: Metaphase I is shown in meiosis. As the spindle fibers are released from the centrosome and the pairs of homologous chromosomes are arranged on the equator, which is a hallmark of metaphase II. Prophase is shown in mitosis as condensed homologous chromosomes can be seen, which is a hallmark of prophase.

Q15. Flower color is controlled by a single pair of alleles. The allele for red flowers is dominant to the allele for white flowers. A plant homozygous for red flowers is crossed with a plant homozygous for white flowers. All the resulting plants have red flowers (F1 generation). When the F1 generation crosses with each other, 18 plants are obtained. 12 plants have red flowers and 6 have white flowers (F2 generation). What ratio is expected in the F2 generation and what ratio has been obtained?

• A. 3:1 expected, 2:1 obtained✓
• B. 2:1 expected, 3:1 obtained
• C. 1:1 expected, 3:1 obtained
• D. 1:1 expected, 2:1 obtained

Explanation: The correct answer is that the expected ratio in the F2 generation is 3:1, and the obtained ratio is 2:1. In a typical Mendelian monohybrid cross (Rr x Rr), the expected phenotypic ratio is 3:1, with three plants having the dominant phenotype (red flowers) for every one plant with the recessive phenotype (white flowers). The obtained ratio of 2:1 suggests a deviation from the expected, possibly due to a small sample size or experimental error. The other options incorrectly present the expected and obtained ratios.

Q16. The following observations refer to evolution: I. Inherited variations that are ‘favored’ in particular environments are passed on. II. There is a struggle for existence. III. In time, ‘favored’ inherited variations may accumulate causing gradual changes in the organism. IV. Although populations tend to overproduce, they remain more or less constant in numbers from generation to generation. In what sequence should the statements be placed to support Darwin’s theory of evolution?

• A. I, II, III, IV
• B. II, I, III, IV
• C. III, I, IV, II
• D. IV, I, II, III
• E. IV, II, I, III✓

Explanation: Summarizing Darwins’ theory in correct sequence: 1. Organisms overproduce but remain constant in number. 2. There is a struggle for existence. 3. Only those traits which survive the best are passed on to the next generation. 4. As a result gradually the traits are passed on to form an organism.

Q17. The diagram shows the ultrastructure of a chloroplast as seen in section. What are the functions of P, Q, and R?

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

Explanation: Option D is correct. In the chloroplast, P is the thylakoid membrane, which plays a critical role in absorbing light energy during the light-dependent reactions of photosynthesis. Q refers to starch molecules that serve as storage for the carbohydrates produced. Although the explanation originally states cytoplasm, R is actually the stroma of the chloroplast, where the Calvin cycle takes place, leading to the synthesis of carbohydrates. Other options incorrectly describe these functions or misidentify the structures.

Q18. Consider the following statements about biological communities.I. Their members share a common gene poolII. The community remains stable even though some physical aspect of the environment may undergo a changeIII. It consists of all the populations living in a particular area.IV. A community interacts with a non-living environment and both function together to form an ecosystemWhich two of the above statements are true?

• A. 1 and 2
• B. 1 and 3
• C. 2 and 4
• D. 2 and 3
• E. 3 and 4✓

Explanation: Different living and nonliving things in a single habitat form a single community. Since different species are present they have different characteristics so the gene pool is not the same. Change in the environment does cause a change in the balance of genes in the environment so the community does not remain stable.The last two statements iii and iv are right that as previously mentioned the community is all the population living in a particular area. Secondly, the community involves biotic(living) and abiotic(non-living) factors interacting with one another.

Q19. Four events in the transmission of nerve impulses across synapses are:I. Depolarisation of the presynaptic membraneII. Propagation of postsynaptic action potentialIII. Reabsorption of the transmitter substanceIV. Release of transmitter substance into the synaptic cleftIn which sequence do these events occur?FIRST -> LAST

• A. I, III, II, and IV
• B. I, IV, II and III✓
• C. IV, I, III and II
• D. IV, III, I and II
• E. II, I, IV and III

Explanation: The sequence of transmission of nerve impulses is as follows:1. Depolarisation occurs at the presynaptic terminal.2. This leads to the release of the transmitter substance to the cleft.3. This leads to the depolarization at the postsynaptic membrane.4. Then the transmitter substance is reabsorbed back to the presynaptic membrane.So, from the above summary, it shows that the sequence in the transmission of nerve impulses across synapses is 1, 4, 2, and then 3.

Q20. Joints found at the vertebrae are:

• A. Gliding joints✓
• B. Sliding joints
• C. Partially moveable joints
• D. Fixed joints
• E. Pivot joints

Explanation: Gliding joints are present between vertebrates.

Q21. How many meninges cover the human brain?

• A. 5
• B. 4
• C. 3✓
• D. 2
• E. 1

Explanation: There are three layers of meninges around the brain and spinal cord. The outer layer, the dura mater, is tough, white fibrous connective tissue. The middle layer of meninges is the arachnoid, a thin layer resembling a cobweb with numerous threadlike strands attaching it to the innermost layer.

Q22. The site and principal mechanism for the passage of glucose into the bloodstream in the human kidney is the:

• A. Collecting duct …. By active secretion
• B. Distal convoluted tubule …. By passive diffusion
• C. Glomerulus …. By selective reabsorption
• D. Glomerulus …. By ultrafiltration
• E. Proximal convoluted tubule …. By active reabsorption✓

Explanation: All the nutrients, including glucose, are absorbed back into the proximal convoluted tubule by active reabsorption. The glomerulus is not involved in absorption but just “ultrafiltration”.DCT and Collecting ducts are primarily used to absorb water.

Q23. Booklungs serve the function of respiration in:

• A. Scorpion✓
• B. Earthworm
• C. Frog
• D. Cockroaches

Explanation: A book lung is a type of respiration organ used for atmospheric gas exchange that is present in many arachnids, such as scorpions and spiders. Each of these organs is located inside an open ventral abdominal, air-filled cavity (atrium) and connects with the surroundings through a small opening for the purpose of respiration.

Q24. The scientific name of potato is Solanum tuberosum and that of brinjal is Solanum melongena. This indicates that both potato and brinjal are members of the:

• A. Same species but different genera
• B. Same genus but different classes
• C. Same species but different classes
• D. Same genus but different species✓
• E. Same class but different genera

Explanation: The scientific names you mentioned indeed place both potato (Solanum tuberosum) and brinjal (Solanum melongena) in the same genus, Solanum. Therefore, both potato and brinjal belong to the same plant genus, which is Solanum. The Solanaceae family encompasses a diverse group of plants that includes not only potatoes and brinjals but also tomatoes, peppers, and other species commonly known as nightshades.

Q25. The following are all features of eukaryotic cells:I. chloroplastII. peroxisomeIII. lysosomeIV. mitochondrionv. nucleusWhich of these has a double membrane?

• A. I, II and IV
• B. I, III and V
• C. I, IV and V✓
• D. II, III and V
• E. II, III and IV

Explanation: Chloroplasts, mitochondria, and the nucleus are unique in that they all have double membranes, which are essential for their respective functions in energy conversion and genetic protection. Chloroplasts facilitate photosynthesis, mitochondria are the powerhouses of the cell, and the nucleus houses genetic material. In contrast, peroxisomes and lysosomes are single-membrane organelles involved in detoxification and digestion, respectively, and do not require a double membrane for their functions. Therefore, the correct answer is I, IV, and V.

Q26. Which plant process is responsible for water loss from leaves?

• A. Guttation
• B. Transpiration✓
• C. Photosynthesis
• D. Respiration

Explanation: Water vapor is lost from leaves through stomata, creating a transpiration pull.

Q27. Which of the following is true in angiosperm life cycle?

• A. Gametophyte are photosynthetic and partially independent than sporophyte
• B. Sporophytes are totally dependent on gametophytes
• C. Gametophytes are totally dependent on sporophytes✓
• D. Both gametophytes and sporophytes are totally dependent on each other
• E. Both gametophytes and sporophytes are totally independent of each other

Explanation: The statement "Gametophytes are totally dependent on sporophytes" is generally accurate in the context of the life cycle of plants that exhibit alternation of generations. This alternation involves two distinct phases. Plant sporophytes represent the asexual phase of the cycle and produce spores. In seed-bearing vascular plants, such as angiosperms and gymnosperms, the gametophyte is totally dependent on the sporophyte for development.

Q28. The adaptive feature(s) which help(s) the fish to live in water include(s):

• A. A tail and an air bladder
• B. Unpaired and paired fins
• C. Streamlined body
• D. Gills and strong sense of smell
• E. All of these options✓

Explanation: All of the mentioned features are essential adaptations that enable fish to live successfully in water. Unpaired and paired fins provide stability and maneuverability; a streamlined body reduces drag for efficient swimming; gills are necessary for extracting oxygen from water, and a strong sense of smell aids in finding food and navigating. While a tail is vital for movement, an air bladder helps with buoyancy control but is not present in all fish, making option A less universally applicable. Thus, the comprehensive set of adaptations described in options B, C, and D collectively supports aquatic living, making option E the correct choice.

Q29. The diagram represents the human respiratory system. While inhalation, which of the labelled muscles contract?

• A. X only
• B. X and Y only
• C. X and Z only
• D. Y and Z only✓
• E. X, Y and Z

Explanation: During inhalation, the diaphragm (labelled as Z) and the external intercostal muscles (labelled as Y) contract to allow the lungs to expand and fill with air. The intercostal muscles labelled as X do not contract during this process; they relax. Hence, the correct answer is option D: Y and Z only. Options involving X are incorrect because X does not contract during inhalation.

Q30. What is the correct order of arthropod groups, from those with the most legs to those with the fewest legs?

• A. Arachnids-Crustaceans-Insects-Myriapods
• B. Crustaceans-Myriapods-Insects-Arachnids
• C. Insects-Arachnids-Myriapods-Crustaceans
• D. Myriapods-Arachnids-Crustaceans-Insects
• E. Myriapods-Crustaceans-Arachnids-Insects✓

Explanation: The correct order from most to fewest legs is Myriapods > Crustaceans > Arachnids > Insects. Myriapods, which include centipedes and millipedes, have many pairs of legs, often more than 20. Crustaceans, like crabs and lobsters, typically have 5 pairs of legs (10 legs). Arachnids, such as spiders and scorpions, have 4 pairs of legs (8 legs). Insects, which include ants and butterflies, have 3 pairs of legs (6 legs). Each option was incorrect if it did not follow this descending order based on the typical number of legs.

Q31. When a red stain is added to a culture containing both living and dead cells, only the dead cells take up the stain.Which structure(s) prevent(s) the stain entering the living celis?

• A. cell membrane✓
• B. cell wall
• C. cytoplasm
• D. vacuole
• E. all of these options

Explanation: The correct answer is the cell membrane. The cell membrane is selectively permeable, meaning it controls what enters and exits the cell. This property allows it to prevent substances like stains from penetrating living cells. In contrast, dead cells lose membrane integrity, allowing the stain to enter. The cell wall provides structural support but does not have a role in selective permeability. The cytoplasm is involved in metabolic processes, and vacuoles are for storage, neither of which control the entry of external substances.

Q32. Which of the following statements concerning nucleolus is correct?

• A. It disappears at the time of cell division.✓
• B. There is only one nucleolus in every cell.
• C. It plays an important role in the synthesis of ribonucleic acid and ribosomes in prokaryotic cells.
• D. It helps in destroying worn-out organelles.
• E. It captures energy for the cell.

Explanation: The nucleolus is essential for ribosome synthesis and is prominent during interphase. However, it disappears during prophase as the cell enters mitosis or meiosis, making Option A correct. The nucleolus disappears during prophase because the cell shifts its energy and resources from protein synthesis (ribosome production) to cell division. Chromosomes condense tightly, making the DNA inaccessible for transcription, which leads to the halting of ribosomal RNA (rRNA) synthesis and the breakdown of the nucleolus. This disassembly also allows the nuclear membrane to break down and the chromosomes to become free to move to the metaphase plate. Option B is incorrect because cells can have multiple nucleoli. Option C is incorrect as the nucleolus is absent in prokaryotic cells. Option D is incorrect because lysosomes, not the nucleolus, are responsible for breaking down worn-out organelles. Option E is incorrect because the nucleolus is not involved in capturing energy; this is a function of chloroplasts in plant cells and mitochondria in eukaryotic cells.

Q33. In birds the male is the homogametic sex. A male bird showing the recessive trait was mated with a female showing the dominant trait of a characteristic governed by a pair of alleles which are sex-linked. What is the probability that the male offspring will show the dominant trait?

• A. 0
• B. 0.25
• C. 0.50
• D. 0.75
• E. 1.00✓

Explanation: In this problem, we are dealing with a sex-linked trait where the male parent (XbY) carries the recessive allele (b), and the female parent (XBXB) has two dominant alleles (B). The key is to recognize that in birds, males are homogametic, meaning they have two identical sex chromosomes (XY), while females have two different sex chromosomes (XX).The cross between the male (XbY) and the female (XBXB) yields the following gametes:Xb from the male and XB from the female leads to XbXB (female with dominant trait)Xb from the male and XB from the female leads to XbXB (female with dominant trait)Y from the male and XB from the female leads to XBY (male with dominant trait)Y from the male and XB from the female leads to XBY (male with dominant trait)Thus, all male offspring (XBY) will inherit the dominant allele from the mother, resulting in a probability of 1.00 that they will express the dominant trait.The other options are incorrect because they misinterpret the inheritance pattern and the genotypes of the parents involved in the cross.

Q34. In an experiment, the production of hormone secretin was blocked. As a result, levels of all of the following enzymes were affected EXCEPT:

• A. Trypsin
• B. Pepsin✓
• C. Chymotrypsin
• D. Amylase
• E. Lipase

Explanation: Secretin is a hormone that primarily stimulates the pancreas to release digestive enzymes such as trypsin, chymotrypsin, amylase, and lipase. These enzymes are crucial for the digestion of proteins, carbohydrates, and fats in the small intestine. However, pepsin is an enzyme produced in the stomach and is not under the regulatory influence of secretin. Instead, pepsinogen (the inactive form of pepsin) is activated in the acidic environment of the stomach. Therefore, blocking secretin production affects all listed pancreatic enzymes but does not influence pepsin levels. Hence, the correct answer is Pepsin, as it would not be affected by the blockage of secretin.

Q35. At what point are two populations descending from the same ancestral stock considered separate species?

• A. When they can no longer produce viable, fertile offspring✓
• B. When they look significantly different from each other
• C. When they can interbreed successfully and produce offspring
• D. When their habitats are separated by a significantly large distance so that they cannot meet
• E. Both B and C

Explanation: The correct answer is Option A: When they can no longer produce viable, fertile offspring. This aligns with the biological species concept, which defines species based on their ability to interbreed and produce fertile offspring. Option B is incorrect because morphological differences alone do not define species. Option C contradicts the definition of separate species. Option D is incorrect because geographical separation does not necessarily result in speciation. Option E is incorrect as both B and C are not valid reasons to consider populations as separate species.

Q36. Living things that would be the first to experience adverse effects if large amounts of carbon dioxide were taken out of the biosphere are:

• A. Decomposers (e.g., bacteria and fungi)
• B. Producers (e.g., green plants)✓
• C. Primary consumers (e.g., mice)
• D. Secondary consumers (e.g., snakes)
• E. Tertiary consumers (e.g., hawks)

Explanation: Producers are the only organisms that directly use carbon dioxide from the atmosphere for the process of photosynthesis, converting it into energy and oxygen. If carbon dioxide levels were to drop significantly, producers would be the first to suffer since they require it for their survival. Decomposers, primary consumers, secondary consumers, and tertiary consumers rely on the flow of energy and carbon through the food chain, making producers the foundational element. Without sufficient carbon dioxide, the producers cannot perform photosynthesis, affecting the entire ecosystem above them.

Q37. Which of the following organelles are present in all the bacteria?

• A. Mesosome
• B. Ribosomes✓
• C. Golgi bodies
• D. Mitochondria

Explanation: The correct answer is 'Ribosomes.' In bacteria, which are prokaryotic organisms, ribosomes are present and essential for protein synthesis. These are not membrane-bound and are found throughout the cytoplasm. Other options such as Golgi bodies, mitochondria, and chloroplasts are membrane-bound organelles found in eukaryotic cells, not in bacteria. Mesosomes were once thought to be a common organelle in bacteria, but they are now largely considered artifacts of cell preparation.

Q38. The gland known as the" gland of emergency" is the

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

Explanation: The adrenal glands are often referred to as the "glands of emergency" because they play a crucial role in the body's stress response. These glands produce hormones such as adrenaline and cortisol, which prepare the body to respond to stressful situations by increasing heart rate, raising blood pressure, and mobilizing energy resources.

Q39. The autonomic nervous system controls all of the following activities except:

• A. Digestion of food
• B. Heart beat
• C. Contraction of pupil of eye
• D. Thought✓
• E. Breathing rate

Explanation: The autonomic nervous system does not directly control thought processes. Thought is primarily regulated by the central nervous system, which includes the brain and spinal cord. While the autonomic nervous system can be influenced by certain thought patterns, such as stress or relaxation, it does not directly control the generation or content of thoughts.

Q40. In the northern hemisphere, a tundra type of growth:

• A. is impossible
• B. occurs only in winter
• C. lasts only for two to three months
• D. is in the form of a wide land✓
• E. is in the form of small patches of land

Explanation: Tundra growth occurs in the form of a wide land, covering expansive areas in the northern hemisphere. Tundra is a major zone of treeless level or rolling ground found in cold regions, because it is treeless and flat rolling ground it looks very wide as shown below:

Q41. SN2 reaction at an asymmetric carbon of a compound always gives:

• A. An enantiomer of the substrate.
• B. A product with opposite optical rotation.
• C. A mixture of diastereomers.
• D. A single stereoisomer.✓

Explanation: SN2 reaction proceeds with inversion of configuration. Since the attacking nucleophile is not same as that of leaving group, the product cannot be enantiomer of the substrate so the product will not necessarily have opposite optical rotation. Moreover only one product is obtained, so we cannot obtain diastereomers.

Q42. In the reaction, R-C≡C-R → the reagent used to convert alkyne into cis alkene is:

• A. Ni
• B. Lindlar catalyst✓
• C. B2H6/CH3COOH
• D. Li/NH3
• E. C6H6

Explanation: To convert an alkyne into a cis alkene, the reagent commonly used is Lindlar's catalyst. Lindlar's catalyst consists of palladium (Pd) deposited on calcium carbonate (CaCO3) or barium sulfate (BaSO4) support, with quinoline or lead acetate as a poison to prevent further reduction to an alkane. Lindlar's catalyst selectively reduces the triple bond of an alkyne to a cis-alkene or trans-alkene, depending on the reaction conditions. By using Lindlar's catalyst under appropriate conditions (such as low temperature and low hydrogen pressure), cis-alkenes can be obtained. The reaction mechanism involves the partial reduction of the alkyne, stopping at the alkene stage. The presence of the poison helps to hinder the complete reduction to an alkane and directs the selectivity towards the cis-alkene product. Therefore, Lindlar's catalyst is the reagent commonly employed to convert an alkyne into a cis alkene.

Q43. Ethanol, when reacted with PCI5 , gave A, POCl3, and HCI. A reacts with AgNO2 to form B and AgCl. A and B are respectively:

• A. C2H5Cl and C2H5OC2H5
• B. C2H6 and C2H5OC2H5
• C. C2H5Cl and C2H5NO2✓
• D. C2H6 and C2H5NO2
• E. C2H6 and C2H6NO

Explanation: The sequence of reactions begins with ethanol (C2H5OH) reacting with phosphorus pentachloride (PCl5), resulting in the formation of ethyl chloride (C2H5Cl), phosphorus oxychloride (POCl3), and hydrochloric acid (HCl). The ethyl chloride then reacts with silver nitrate (AgNO2), producing nitroethane (C2H5NO2) and silver chloride (AgCl). The correct products for A and B in the sequence are thus ethyl chloride and nitroethane, respectively. The other options involve incorrect products, such as diethyl ether, ethane, or incorrect formulas, and do not match the reaction steps provided.

Q44. The false statement regarding saline hydrides is:

• A. They conduct electricity in molten state
• B. They are used as reducing agents
• C. They give H2 from H2O
• D. They are covalent in nature✓

Explanation: Saline hydrides (also known as ionic hydrides or pseudohalides) are compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements, thus they are ionic, not covalent.

Q45. Which of the following compounds is formed when sodium burns in excess of air?

• A. Na2O
• B. Na2O3
• C. Na2O2✓
• D. NaO2

Explanation: When sodium burns in excess air, it reacts with more oxygen than is available in a limited supply, leading to the formation of sodium peroxide, Na2O2. This is because the additional oxygen allows for further oxidation of sodium oxide (Na2O), which is initially formed. The reaction is represented by the equation: Na2O + 1/2O2 → Na2O2.Other options such as Na2O and NaO2 are either formed in limited oxygen conditions or are not typically stable products of this reaction. Na2O3 is not a recognized compound in this context.

Q46. H2SO4 has great affinity for water because:

• A. It decomposes the acid
• B. It hydrolyses the acid
• C. Acid decomposes the water
• D. Acid forms hydrates with water✓

Explanation: Sulphuric acid is kept in air-tight bottles because if kept in the open it absorbs water vapors from the surrounding. It is a dehydrating agent. Sulphuric acid forms hydrate with water because the reaction of water and acid is highly exothermic and a great amount of heat in the form of fumes is liberated in that exothermic reaction of acid and water, acid produces stable hydrates-this shows the acid’s affinity for water.

Q47. Which of the following is not an interstitial compound?

• A. Cu-Zn
• B. Cu-Zn-Sn
• C. TiH1.73
• D. V2O5✓

Explanation: V2O5 is not an interstitial compound. Interstitial compounds are those that do not have an exact composition as shown in V2O5. Interstitial compounds are formed when small atoms of H, C, B, or N get trapped inside the crystal lattice of metals/ enter the interstices of transition metals and impart useful features to them. These are non-stoichiometric compounds (their element ratio can’t be represented by a ratio of well-defined numbers.) and are also sometimes termed interstitial alloys. An interstitial alloy is formed when one type of metal atoms are much smaller than the other type. V2O5 is a stoichiometric compound as shown by the formula i.e., one molecule of vanadium pentaoxide has 2 vanadium atoms and 5 oxygen atoms.

Q48. Monosaccharides contain _ carbon atoms.

• A. 2-3
• B. 3-7✓
• C. 5-20
• D. 20-25

Explanation: Monosaccharides are the simplest sugars containing 3 to 7 carbon atoms. Examples include triose (3C), tetrose (4C), pentose (5C), hexose (6C) sugars. They serve as the building blocks for disaccharides and polysaccharides.

Q49. Zinc reacts with dil. H2SO4 to give H2. It also reacts with conc. H2SO4 to form SO2. In these reactions:

• A. Zn reduces H+ to H2
• B. Zn oxidizes H+ to H2
• C. Zn reduces SO42- to SO2
• D. Zn oxidises SO42- to SO2
• E. Both A and C✓

Explanation: Both options A and C are correct.

Q50. The reaction: Cl2 + H2O → HCl + HOCl is an example of:

• A. Oxidation reaction
• B. Reduction reaction
• C. Auto-oxidation and reduction reaction✓
• D. Substitution reaction
• E. Addition reaction

Explanation: Auto-oxidation and reduction reaction is a reaction in which the same element is oxidized as well as reduced. In the reaction given above, the Cl element reduces to Cl-(HCl) and oxidizes to Cl+(in HOCl).

Q51. Which one of the following ions exhibits d-d transition and paramagnetism as well?

• A. MnO4-
• B. CrO42-
• C. Cr2O72-
• D. MnO42-✓

Explanation: The correct answer is MnO42-, where manganese is in the +6 oxidation state. In this state, manganese retains d electrons, making d-d transitions possible. Moreover, the presence of unpaired electrons in the d orbitals results in paramagnetism.The other options, MnO4-, CrO42-, and Cr2O72-, have their respective metal ions in oxidation states where no d electrons are present (Mn⁷⁺ and Cr⁶⁺). Consequently, these ions cannot exhibit d-d transitions and are not paramagnetic.

Q52. Glass is a/an:

• A. Pure solid
• B. Supercooled liquid✓
• C. Mixture of sodium and calcium
• D. The crystalline form of Na2CO3
• E. Alloy

Explanation: Glass is sometimes called a supercooled liquid because it does not form a crystalline structure, but instead forms an amorphous solid that allows molecules in the material to continue to move, while it is also true that it’s a homogeneous mixture of sodium and calcium silicates.

Q53. Which of the following is correct about ascorbic acid?

• A. Soluble in water
• B. Helps in healing wounds
• C. Easily destroyed by oxidation
• D. All of these✓

Explanation: Ascorbic acid is indeed soluble in water, easily destroyed by oxidation, and helps in healing wounds through collagen synthesis.

Q54. Catalyst used in reaction CHCl3 + ½ O2 → COCl2 + HCl is _ and its nature is _.

• A. 5% methyl alcohol ... negative
• B. 2% Ethyl alcohol ... negative✓
• C. V2O5 … positive
• D. Al2O3 ….. negative

Explanation: The correct catalyst for the reaction CHCl3 + ½ O2 → COCl2 + HCl is 2% ethyl alcohol, which acts as a negative catalyst. This means it decreases the rate of the reaction, aiding in the controlled formation of the toxic gas phosgene (COCl2). Other options like methyl alcohol, V2O5, and Al2O3 are incorrect as they do not fulfil the role of a negative catalyst in this specific reaction.

Q55. If products of a reaction act as a catalyst, such process is called:

• A. Positive catalyst
• B. Negative catalyst
• C. Auto catalyst✓
• D. Both A and B

Explanation: An autocatalyst is a catalyst that is also a product of the reaction it catalyzes. This unique type of catalysis, known as autocatalysis, involves the product enhancing the reaction rate as it forms, leading to a potentially faster and self-sustaining reaction. The concept of positive and negative catalysts does not apply here, as they do not involve the products of the reaction acting as catalysts.

Q56. In the following reaction:3Br2 +6CO32- + 3H2O → 5Br-1 + BrO3-1 + 6HCO3-1

• A. Bromine is reduced and water is oxidized.
• B. Bromine is both reduced and oxidized.✓
• C. Bromine is oxidized and carbonate is reduced.
• D. Bromine is neither reduced nor oxidized.

Explanation: The correct answer is that bromine is both reduced and oxidized. This reaction is an example of a disproportionation reaction. In Br2, bromine has an oxidation state of 0. It is reduced to form bromide ions (Br-1), where the oxidation state is -1, and it is oxidized to form bromate ions (BrO3-1), where the oxidation state is +5. Therefore, bromine undergoes both a decrease and an increase in oxidation state. Other elements, such as hydrogen, carbon, and oxygen, do not experience any change in their oxidation states during this reaction.

Q57. The lower part of the “Solvay tower” has been cooled during the manufacture of soda ash because:

• A. this facilitates the production of soda ash
• B. it decreases the solubility of Na2CO3
• C. this controls the flow of brine
• D. it decreases the solubility of NaHCO3✓

Explanation: In the Solvay process, the goal is to produce sodium carbonate (soda ash). One of the key steps involves converting ammonium bicarbonate and sodium chloride into sodium bicarbonate (NaHCO3). This reaction is exothermic, which increases the temperature and the solubility of NaHCO3. By cooling the lower part of the Solvay tower, the solubility of NaHCO3 decreases, allowing it to precipitate and be separated for further processing into Na2CO3. Option D is correct because it directly addresses the need to precipitate NaHCO3 for soda ash production. The other options do not relate to this specific cooling function.

Q58. Which of the following elements has the highest boiling point?

• A. Li
• B. Mg
• C. Sr
• D. Be✓

Explanation: Group II elements have a higher boiling point than group I elements because of a higher charge, thus better attractive forces, so Li can’t be the answer given other group II elements. Down the group, the melting point and boiling point decreases so we can say that Be has the highest boiling point.

Q59. For which one of the following gaseous equilibrium will an increase in pressure increase the yield of the product:

• A. 2HI ⇌ H2 + I2
• B. 2SO2 + O2 ⇌ 2SO3✓
• C. H2O + CO ⇌ H2 + CO2
• D. H2 + Br2 ⇌ 2HBr

Explanation: In A, C and D, the number of moles of reactants and products are same. Therefore, increase in pressure does not affect the rate of reaction. In B, however, number of moles of product are less, therefore increase in pressure increases the rate of reaction.

Q60. Which of the following statements is NOT true for the first law of thermodynamics?

• A. Total energy of the system and surroundings is conserved
• B. Energy can neither be created nor destroyed
• C. The first law is synonymous with the law of conservation of energy
• D. Total energy of the system is increasing✓

Explanation: The first law of thermodynamics, also known as the law of conservation of energy, asserts that the total energy within an isolated system is constant. Energy may be transformed from one form to another or transferred between the system and its surroundings, but it cannot be created or destroyed. Therefore, the statement in Option D is incorrect as it contradicts this fundamental principle by suggesting that the total energy of the system is increasing.Options A, B, and C correctly reflect the principles of the first law of thermodynamics, emphasizing the conservation of energy and its equivalence to the law of conservation of energy.

Q61. Nitrogen and phosphorus have 3 of their valence electrons unpaired because of:

• A. Auf bau principle
• B. Heisenberg's principle
• C. Hund’s rule✓
• D. Planck’s statement
• E. None of these

Explanation: Hund's rule states that every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Q62. The chemical analysis of a compound having a molecular mass 188 gives, C=12.8%, H=2.1%, and Br=85.1%. Its molecular formula is:

• A. CH2Br
• B. C2H4Br2✓
• C. C2H4Br
• D. CH2(Br)2
• E. C2H2(Br)3

Explanation: To determine the molecular formula, first calculate the empirical formula based on the given percentage composition:Convert the percentage of each element to mass: 12.8 g of C, 2.1 g of H, and 85.1 g of Br.Convert mass to moles: C (12.8 g / 12.01 g/mol), H (2.1 g / 1.008 g/mol), Br (85.1 g / 79.904 g/mol).Find the mole ratio by dividing each by the smallest number of moles.The empirical formula is CH2Br.The empirical formula mass is approximately 94.5 g/mol.Divide the molecular mass (188 g/mol) by the empirical formula mass (94.5 g/mol) to find the multiplier, which is 2.Multiply the empirical formula by this factor to get the molecular formula: C2H4Br2.Options A, C, D, and E do not match the molecular mass or fail to maintain the correct elemental ratio.

Q63. The stability of ionic crystal depends principally on:

• A. High electron affinity of anion forming species
• B. Lattice energy of crystal✓
• C. Low ionization energy of cation forming species
• D. Low heat of sublimation of cation forming solid

Explanation: The stability of an ionic crystal is primarily determined by its lattice energy, which is the energy released when ions in the gaseous state form an ionic solid. This energy reflects the strength of the electrostatic forces between oppositely charged ions in the crystal lattice. The higher the lattice energy, the more stable the ionic crystal is. Option B correctly identifies lattice energy as the key factor contributing to the stability of ionic crystals.Option A, while addressing electron affinity, pertains to ion formation rather than crystal stability. Option C discusses ionization energy, relevant to cation formation but not to the stability of the formed crystal. Option D involves heat of sublimation, which is unrelated to ionic crystal stability.

Q64. How many sigma groups are there in benzene?

• A. 3
• B. 6
• C. 9
• D. 12✓

Explanation: Benzene (C6H6) is a cyclic hydrocarbon with alternating double and single bonds. It consists of 6 carbon atoms, each bonded to one hydrogen atom. Each carbon forms a sigma bond with its neighboring carbon, resulting in 6 C—C sigma bonds. Additionally, each carbon is bonded to a hydrogen atom, accounting for 6 C—H sigma bonds. Therefore, the total number of sigma bonds in benzene is 12. The incorrect options (3, 6, and 9) fail to account for all the sigma bonds present in benzene's structure.

Q65. If we balanced the following reaction, what is the net ionic charge on the right side of the equation? H+ + MnO4- + Fe2+ →Mn2+ + Fe3+ + H2O

• A. +5
• B. +7
• C. +10
• D. +17✓

Explanation: To solve this problem, we must balance the redox reaction by ensuring that both mass and charge are conserved. Manganese is reduced from +7 in MnO4- to +2 in Mn2+, gaining 5 electrons. Iron is oxidized from +2 in Fe2+ to +3 in Fe3+, losing 1 electron per iron atom. By balancing the electron transfer, we multiply the iron half-equation by 5, resulting in the balanced equation:8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2OOn the right side, the charges are:Mn2+: +25Fe3+: +15 (5 x +3)H2O: NeutralTotal charge on the right: +2 + 15 = +17Thus, the correct net ionic charge on the right side is +17. Other options are incorrect as they do not reflect the correct balance of charges based on the stoichiometry of the reaction.

Q66. In the reaction, 2SO2 (g) + O2 ⇌ 2SO3 (g) ∆H= -X cal most favorable conditions of temperature and pressure for a greater yield of SO3 are:

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

Explanation: The correct answer is Option D: Low temperature and high pressure. According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer gas moles, which is the product side (2SO3), resulting in a higher yield of SO3. Additionally, since the reaction is exothermic (∆H = -X cal), lowering the temperature will also favor the forward reaction, further increasing the yield of SO3. The combination of low temperature and high pressure is thus the most favorable for maximizing the production of SO3. Other options either incorrectly prioritize temperature and pressure conditions that do not favor the formation of SO3 or apply an incorrect understanding of Le Chatelier's principle.

Q67. The chemical reaction in which reactions require high amount of activation energy are generally:

• A. Slow✓
• B. First fast then slow
• C. First slow then fast
• D. Spontaneous

Explanation: The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. This is because molecules can only complete the reaction once they have reached the top of the activation energy barrier.

Q68. In those reactions where determination of enthalpy value is difficult by experiments, in such cases enthalpy value can be calculated by:

• A. Hess's law✓
• B. Henry's law
• C. Kirchoff's law
• D. Clapeyron equation ‘
• E. Boyle’s law

Explanation: Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function. It is used to find out the enthalpy of the reaction.

Q69. IUPAC name of the given compound is:

• A. 1,4-Dimethyl hexanol
• B. 2,4-Diethyl hexanol
• C. 4,5-Dimethyl hexanol
• D. 4-methyl,5-ethyl hexanol
• E. 2,4-Dimethyl hexanol✓

Explanation: The correct IUPAC name for the compound is 2,4-Dimethyl hexanol. The longest carbon chain that includes the hydroxyl group consists of six carbon atoms (hexanol). The chain should be numbered from the end nearest the hydroxyl group to assign the lowest possible numbers to the substituents. In this case, methyl groups are attached to the 2nd and 4th carbon atoms, resulting in the name 2,4-Dimethyl hexanol. The other options are incorrect due to either improper identification of substituents or incorrect numbering of the carbon chain.

Q70. The only o, p−directing group which is deactivating in nature is:

• A. -NH2
• B. -OH
• C. -X (halogens)✓
• D. -R (alkyl groups)

Explanation: Halides are ortho-para- directing groups but unlike most ortho-para- directors halides tend to deactivate benzene. This unusual behaviour can be explained by two properties:1. Since the halogens are very electronegative they cause inductive withdrawal (withdrawal of electrons from the carbon atom of benzene).2. Since the halogens have non-bonding electrons they can donate electron density through pi bonding (resonance donation).Option C is correct.

Q71. A circuit in which there is a current of 5 amp is changed so that the current falls to zero in 0.1s. If an average e.m.f of 200 volts is induced, what is the self-inductance of the circuit?

• A. 4 henrys✓
• B. 8 henrys
• C. 12 henrys
• D. 16 henrys
• E. 20 henrys

Explanation: Initial currect= I1 = 5.0 A Final current= I2= 0 A change in current , dI= I1 -I2 =5A Time taken for the change= 0.1s Avg emf= 200V e=L di/dt L= e /di/dt 200/5/01 = 4H Hence the self inductance of the coil is 4H.

Q72. The phenomena in which certain metals emit electrons when exposed to high-frequency light is known as?

• A. Photoelectric Effect✓
• B. Compton’s Effect
• C. Henry's Effect
• D. Principle of Relativity
• E. Coulomb's Law

Explanation: When a metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength (or equivalently, above a threshold frequency), the incident radiation is absorbed and the exposed surface emits electrons. This phenomenon is known as the photoelectric effect.

Q73. A galvanometer has a resistance of 20 ohms and a full-scale deflection when a current of 0.001 ampere flows in it. What is the value of the series resistance to convert it into a voltmeter of range 10 volts?

• A. 7780 ohms
• B. 9980 ohms✓
• C. 5580 ohms
• D. 4480 ohms
• E. 3380 ohms

Explanation: As we know when converting a galvanometer into a voltmeter, we use the formula Rh=V/Ig -Rg where Rh is the series resistance V is the voltage Ig is a full-scale deflection Rg is the resistance of the galvanometer so by putting values: Rh=10/0.001 -20 Rh=9980 ohm

Q74. Sodium nucleus consists of 11 protons and 12 neutrons. The conventional symbol of this nucleus is?

• A. 11Na11
• B. 12Na12
• C. 11Na23✓
• D. 12Na23
• E. 12Na11

Explanation: The conventional symbol of the element contains the proton number(number of protons) and the mass number(proton number+number of neutrons), which is 11 and (11+12=)23, in this case. So the correct option is C. The conventional symbol of a sodium nucleus is 11Na23.

Q75. The atomic spectra deals with the measurement of:

• A. Wavelengths
• B. Intensities of electromagnetic radiations emitted by atoms
• C. Intensities of electromagnetic radiations absorbed by atoms
• D. All of these options✓
• E. Both B and C

Explanation: The correct answer is D: All of these options. Atomic spectra encompass the study of both the wavelengths and the intensities of electromagnetic radiation emitted and absorbed by atoms. When electrons transition between energy levels, they emit or absorb photons, whose characteristics—like wavelength and intensity—can be measured. Options A, B, and C all address partial aspects of these measurements, but the full scope of atomic spectra includes all these components.

Q76. The detection and estimation of an element in a mixture is sometimes nearly impossible, if it is present in very minute traces or if it’s chemical properties are very similar to those of other elements in the mixture. An effective technique is developed for these purposes is known as?

• A. Simple Analysis
• B. Spectral Analysis
• C. Activation Analysis✓
• D. Geometric Analysis
• E. Mechanical Analysis

Explanation: Activation analysis is a highly sensitive technique used to detect and measure trace elements in a mixture, especially when they are present in very small amounts or have similar chemical properties to other elements. By irradiating the sample, it becomes radioactive, and the subsequent measurement of emitted radiation allows for precise identification and quantification of the elements. Other methods like simple analysis lack the sensitivity required, while spectral analysis may face challenges with low concentrations or similar spectra. Geometric and mechanical analyses are unrelated to chemical detection.

Q77. How much energy is dissipated as heat in 20 s by a 100 Ω resistor that carries a current of 0.5 A?

• A. 50 J
• B. 100 J
• C. 250 J
• D. 500 J✓
• E. 1,000 J

Explanation: To determine the energy dissipated as heat in a resistor, use the formula: H = I²Rt. For this question, I = 0.5 A, R = 100 Ω, and t = 20 s. Applying these values: H = (0.5)² × 100 × 20 = 500 J, confirming that Option D is correct. Other options result from either incorrect squaring of the current, misapplication of the formula, or arithmetic errors.

Q78. A sphere of charge +Q is fixed in a position. A smaller sphere of +q is placed near the larger sphere and released from the rest. The small sphere will move away from the large sphere with?

• A. Decreasing velocity and decreasing acceleration
• B. Decreasing velocity and increasing acceleration
• C. Decreasing velocity and constant acceleration
• D. Increasing velocity and decreasing acceleration✓
• E. Increasing velocity and increasing acceleration

Explanation: In this scenario, the small sphere (+q) is initially at rest near a larger fixed sphere (+Q). Because both spheres are positively charged, they repel each other. Initially, the repulsive force causes the small sphere to accelerate, increasing its velocity. However, as the sphere moves further away, the force of repulsion decreases in magnitude due to Coulomb's Law (F ∝ 1/r2), which causes the acceleration to decrease. Therefore, the correct answer is that the small sphere moves away with increasing velocity and decreasing acceleration.The other options are incorrect because:Decreasing velocity and decreasing acceleration: The velocity increases due to initial acceleration.Decreasing velocity and increasing acceleration: Acceleration decreases as the distance increases.Decreasing velocity and constant acceleration: Neither velocity decreases nor acceleration remains constant.Increasing velocity and increasing acceleration: Acceleration does not increase; it decreases as the force diminishes.

Q79. A 10 nanofarad (10 x 10-9 F) parallel plate capacitor holds a charge of magnitude 50 µC on each plate.If the plates are separated by a distance of 0.885mm, what is the area of each plate?

• A. 1.0 m2✓
• B. 3.0 m2
• C. 5.5 m2
• D. 7.5 m2

Explanation: To find the area of each plate, we use the formula for the capacitance of a parallel plate capacitor: C = ε0 * (A/d), where C is the capacitance, ε0 is the permittivity of free space (approximately 8.85 x 10-12 F/m), A is the area of the plates, and d is the distance between the plates. Given C = 10 x 10-9 F and d = 0.885 x 10-3 m, we rearrange the formula to solve for A: A = (C * d) / ε0. Plugging in the values gives A = (10 x 10-9 F * 0.885 x 10-3 m) / 8.85 x 10-12 F/m = 1.0 m2. Thus, Option A is correct. The other options provide areas that are either too large or too small when calculated properly with the given values.

Q80. Kelvin, the unit of thermodynamic temperature is _ of the thermodynamic temperature of the triple point of water.

• A. 1/100
• B. 1/212
• C. 1/273.16✓
• D. 1/32
• E. 1/98

Explanation: The correct answer is that the Kelvin is defined as 1/273.16 of the thermodynamic temperature of the triple point of water. This precise definition ensures that the Kelvin scale is based on an absolute scale, where absolute zero is 0 K. The other options suggest incorrect fractions that do not align with the scientific definition of the Kelvin scale. These values, such as 1/100 or 1/32, do not have any relation to the Kelvin scale's definition based on the thermodynamic temperature of the triple point of water.

Q81. The scalar product of (2i - j + 3k).(3i + 2j - k)?

• A. 1✓
• B. 2
• C. 10
• D. 20
• E. 25

Explanation: The scalar product (also known as the dot product) of two vectors is calculated by multiplying their corresponding components and adding the results: (2i-j+3k) • (3i+2j-k) = 2×3 + (-1)×2 + 3×(-1) = 6 - 2 - 3 = 1. Options B, C, D, and E are incorrect as they result from miscalculations in either the component multiplication or the addition of these products.

Q82. A rock is thrown straight upward from the edge of a 30 m cliff, rising 10 m then falling all the way down to the base of the cliff.Find the rock's displacement.

• A. 20 meters downward
• B. 30 meters downward✓
• C. 40 meters upward
• D. 50 meters upward
• E. 60 meters upward

Explanation: The displacement is the distance of the rock from the initial position, the initial position of the rock was at the cliff which was 30m. So the displacement of the rock is 30m downwards, although the distance traveled is 50m in total.

Q83. A stone dropped from a certain height can reach the ground in 5 seconds. It is stopped after 3 seconds of its fall and then allowed to fall again. Find the time taken by the stone to reach the ground for the remaining distance.

• A. 2 s
• B. 4 s✓
• C. 6 s
• D. 8 s
• E. 10 s

Explanation: The stone is initially dropped from a height and falls for 3 seconds before being stopped. During this time, it falls a certain distance which can be calculated using the equation of motion: s = ut + (1/2)gt², where u is the initial velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is 3 seconds.After being stopped, the stone falls the remaining distance, which is the difference between the total height and the distance fallen in the first 3 seconds. It takes 4 seconds to fall this remaining distance, calculated using the same equation of motion. Therefore, the correct answer is 4 seconds.The other options are incorrect because they either underestimate or overestimate the time required for the stone to fall the remaining distance.

Q84. A moon of mass ‘m’ orbits a planet of mass 100 m. Let the strength of the gravitational force exerted by the planet on the moon be denoted by F1, and let the strength of the gravitational force exerted by the moon on the planet be F2. Which of the following is true?

• A. F1 is ten times greater than F2
• B. F1 is ten times smaller than F2
• C. F2 is ten times greater than F1
• D. F2 is ten times smaller than F1
• E. F1 is equal to F2✓

Explanation: The gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Since the mass of the planet is 100 times the mass of the moon, the strength of the gravitational force exerted by the planet on the moon will be much greater than the strength of the gravitational force exerted by the moon on the planet. However, the forces will still be equal in magnitude due to Newton's third law.

Q85. Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45° to the horizontal?

• A. The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.
• B. The speed at the top of the trajectory is zero.
• C. The object's total speed remains constant during the entire flight.
• D. The horizontal speed decreases on the way up and also decreases on the way down.
• E. The vertical speed decreases on the way up and increases on the way down.✓

Explanation: The only external force acting on the object is the gravitational force vertically downwards. Thus the acceleration of the projectile is g downwards always. Thus there is no change in the speed of the object in the horizontal direction. The speed of an object in the vertical direction at the top of the trajectory is zero. However, the net speed at the point is equal to the constant horizontal component. Due to downward constant acceleration, the vertical speed decreases way up, becomes zero, and then increases way down. Thus option E is correct.

Q86. A football, at rest on the ground, is kicked with an initial velocity of 10 m/s at a launch angle of 30°. Calculate its total flight time, assuming that air resistance is negligible.

• A. 0.5 s
• B. 1 s✓
• C. 1.7 s
• D. 2 s
• E. 4 s

Explanation: The total flight time of a projectile is determined by the formula: T = (2u sinθ) / g, where 'u' is the initial velocity, 'θ' is the launch angle, and 'g' is the acceleration due to gravity. For this question, the initial velocity (u) is 10 m/s, the launch angle (θ) is 30°, and g is 10 m/s². Calculating this gives:T = (2 x 10 x sin(30°)) / 10Since sin(30°) is 0.5, the equation simplifies to T = (2 x 10 x 0.5) / 10 = 1 s.Therefore, the correct total flight time is 1 s. The incorrect options arise from miscalculating or misunderstanding the application of the formula.

Q87. Guttation is the result of:

• A. Diffusion
• B. Transpiration
• C. Osmosis
• D. Root pressure✓
• E. Becomes half

Explanation: Various ions from the soil are actively transported into the vascular tissues of roots, water follows its potential gradient and increase the pressure inside the xylem. This positive pressure is called root pressure. Effect of root pressure is observable at night and early morning when evaporation is low and excess water collects in the form of droplets near the tip of leaves of many herbaceous plants. Such water loss in its liquid phase is known as guttation. It is the movement of molecules from high to low concentration down the gradient. Transpiration is the loss of water in vapour form from leaves. Osmosis is the movement of water molecules from higher to lower concentration through semi permeable membrane. So correct option is D.

Q88. A body having translatory motion possesses _ and _. In the same way a body having rotatory motion possesses _ and _.

• A. None of them
• B. Linear velocity... Linear momentum... Angular velocity... Angular momentum✓
• C. Linear momentum... Angular momentum... Linear velocity... Angular velocity
• D. Angular velocity... Angular momentum... Linear momentum... Linear velocity

Explanation: The correct answer is: Linear velocity... Linear momentum... Angular velocity... Angular momentum.In translatory motion, all points of a body move uniformly in a single direction, involving linear properties such as linear velocity and linear momentum. This type of motion does not change the orientation of the object.In contrast, rotatory motion occurs when a body rotates around its own axis, involving angular properties like angular velocity and angular momentum. For example, the Earth's rotation about its axis is a type of rotatory motion.All options except B do not properly align the motion types with their respective properties, which is why they are incorrect.

Q89. When a body moves in the direction of gravitational force i.e. towards the earth, the work is done by the force of gravity on the body and is _, whereas when the body moves against the direction of gravitational force, the corresponding work done is _.

• A. Negative, Positive
• B. Positive, Negative✓
• C. Positive, Positive
• D. Negative, Negative
• E. Insufficient Information

Explanation: When the work is done in the direction of the force then work done is said to be positive. However, if the work is in opposition to the direction of force then the work done is said to be negative.

Q90. A man pushes a box, initially at rest towards another man by exerting a constant horizontal force F of magnitude 5 N through a distance of 1m. Its final kinetic energy is?

• A. 5 J✓
• B. 10 J
• C. 15 J
• D. 20 J
• E. 25 J

Explanation: The correct answer is Option A (5 J) because the final kinetic energy of the box is equal to the work done on it by the applied force. The work done in this case is 5 J, as calculated by force x distance moved. The other options provide incorrect explanations based on flawed assumptions about the relationship between work, energy, and force.

Q91. A sound wave with a frequency of 343 Hz travels through the air. What is its wavelength? (speed of sound through air = 343 m/s)

• A. 1m✓
• B. 2m
• C. 3m
• D. 4m
• E. 5m

Explanation: λ The speed of a wave v is related to its frequency f and wavelength λ by the formula: v = fλIn this case, the speed of sound v through the air is given as 343 m/s, and the frequency f of the sound wave is also 343 Hz. We can rearrange the formula to solve for the wavelength λλ= v/fSubstitute the given values:λ= 343/343λ= 1m So, the wavelength of the sound wave is 1 meter.

Q92. When a force acts at right angles to the displacement (θ=90°) the work done is zero i.e. the force does not produce work. Identify the example/s from the following when work is zero.I. It is considered ‘hard work to hold a heavy stone stationary at stretched handII. A person walks along a level surface while carrying a boxIII. When a body moves in a circular path

• A. I only
• B. II only
• C. III only
• D. II and III only
• E. I, II and III✓

Explanation: I. It is considered 'hard work to hold a heavy stone stationary at stretched hand: In this example, when you hold a heavy stone stationary at stretched hand, you are exerting an upward force against gravity. However, the displacement of the stone is zero because it remains stationary. Since the force and displacement are perpendicular (θ=90°), the work done by the force is zero. II. A person walks along a level surface while carrying a box: In this case, if the person walks along a level surface while carrying a box, the force of gravity acting vertically downwards is perpendicular to the displacement along the horizontal level surface. As a result, the work done against gravity is zero because the force and displacement are at right angles to each other (θ=90°). III. When a body moves in a circular path: In this example, when a body moves in a circular path, a centripetal force is acting towards the center of the circle. However, the displacement of the body is tangential to the circle, perpendicular to the force. Since the force and displacement are at right angles (θ=90°), the work done by the centripetal force is zero. Summary: When a force acts at right angles (θ=90°) to the displacement, the work done is zero. In all the provided examples, the force acts perpendicular to the displacement, resulting in zero work. Holding a heavy stone stationary, walking along a level surface while carrying a box, and moving in a circular path are situations where the force and displacement are perpendicular, leading to no work done by the force.

Q93. A neutron travels a distance of 12 m in a time interval of 3.6 x 10-4 s. Assuming its speed was constant, its kinetic energy is: (take 1.7 x 10-27 kg as the mass of neutron)

• A. 3.1 eV
• B. 4.7 eV
• C. 5.78 eV✓
• D. 6.91 eV
• E. 7.81 eV

Explanation: speed of neutron = distance traveled by neutron/time taken by neutron to cover that distance. speed of neutron = 12/3.6 × 10^-4 m/s speed of neutron ( V)= 3.3 × 10⁴ m/s now, kinetic energy = 1/2mv² K.E = 1/2 × 1.7 × 10^-27 ×( 3.3 × 10⁴)²j = 1/2 × 1.7 × (3.3)² × 10^-19 j =9.2565 × 10^-19 j we know, 1ev = 1.6 × 10^-19 j so, K.E = 9.2565 × 10^-19/1.6 × 10^-19 ev = 9.2565/1.6 ev = 5.785 ev

Q94. A student is performing a lab experiment on simple harmonic motion. He has two different springs (with force constants k1 and k2) and two different blocks (of masses m1 and m2). If k1 =2k2, and m1 = 2m2, which of the following combinations would give the student the spring-block simple harmonic oscillator with the shortest period?

• A. The spring with force constant k1 and the block of mass m1
• B. The spring with force constant k1 and the block of mass m2✓
• C. The spring with force constant k2 and the block of mass m1
• D. The spring with force constant k2 and the block of mass m2
• E. All the combinations above would give the same period

Explanation: The period of the spring-block simple harmonic oscillator is given by the equation T =2π√m/k so, to make T as small as possible, we want m to be as small as possible and k to be as large as possible. Since m2 is the smaller mass and k1 is the larger spring constant, this combination will give the oscillator the shortest period.

Q95. A microscope has an objective of 10 mm focal length and an eyepiece of 25 mm focal length. What is the distance between the lenses, if the object is in sharp focus when it is 10.5 mm from the objective?

• A. 115 mm
• B. 235 mm✓
• C. 417 mm
• D. 716 mm
• E. 617 mm

Explanation: The correct answer is 235 mm. To find the distance between the lenses, use the formula: 1/f = 1/v - 1/u for the objective lens, where f is the focal length of the objective lens, u is the distance of the object from the lens, and v is the image distance. For the given microscope, f = 10 mm, and u = 10.5 mm. Solving for v, we get v = 210 mm. The total length of the microscope, which is the sum of v and the focal length of the eyepiece (25 mm), gives us 235 mm. The other options are incorrect due to errors in applying the lens formula or arithmetic mistakes in the calculations.

Q96. Light can be polarized by which of the following methods?I. scattering of lightII. double refractionIII. reflection

• A. I only
• B. II only
• C. III only
• D. I and III only
• E. I, II, and III✓

Explanation: Polarization can be done using:1. Scattering: When light interacts with particles or molecules in a medium, it can be scattered. Depending on the scattering mechanism, polarization can occur.2. Double refraction: Double refraction, also known as birefringence, is a property exhibited by certain transparent materials like calcite. In such materials, light is split into two beams, each polarized differently. This phenomenon is often used to polarize light.3. Reflection: When light reflects off a surface, the reflected light can be partially polarized, especially if the surface is non-metallic and the angle of incidence is specific (known as Brewster's angle). This polarized reflection is commonly used in polarizing filters.

Q97. A steel rod has a length of 15 m at a temperature of 30°C. If the temperature is raised to 45°C. The increase in its length is:(α= 1.1 x 10-5 K-1)

• A. 537.1 x 10-5 m
• B. 447.5 x 10-5 m
• C. 327.5 x 10-5 m
• D. 247.5 x 10-5 m✓
• E. 127.5 x 10-5 m

Explanation: The increase in length of the steel rod is calculated using the linear expansion formula: ΔL=L0⋅α⋅ΔT, where L0 is the initial length, α is the coefficient of linear expansion, and ΔT is the change in temperature. Here, L0=15 m, α=1.1 x 10-5 K-1, and ΔT=15°C (since 45°C - 30°C = 15°C). Applying these values gives: ΔL = 15 × 1.1 x 10-5 × 15 = 247.5 x 10-5 m. This is the correct result, which matches option D.The other options are incorrect due to either a miscalculation or misunderstanding of the application of the formula.

Q98. The volume occupied by a gram mole of a gas at 0°C and a pressure of 1 atmosphere is?

• A. 44.2 Liters
• B. 34.2 Liters
• C. 32.4 Liters
• D. 35.5 Liters
• E. 22.4 Liters✓

Explanation: The ideal gas equation is given by PV = nRT, where:P is the pressure (1 atm at STP).V is the volume.n is the number of moles (1 mole for this question).R is the ideal gas constant (8.314 J mol–1 K–1).T is the temperature (273 K at STP).Calculating the volume at STP: V = (1 mole) x (8.314 J mol–1 K–1) x (273 K) / (1.013 × 105 N/m2) = 0.0224 m3 = 22.4 liters.Hence, the volume occupied by a gram mole of a gas at STP is 22.4 liters.All other options exceed this standard volume, thus are incorrect.

Q99. If you wish to decrease the resistance of a circuit, you will add the resistors in _.

• A. Series
• B. Parallel✓
• C. Does not make a difference
• D. Cannot be determined
• E. 160

Explanation: As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit decreases, and the total current of the circuit increases. Adding more resistors in parallel is equivalent to providing more branches through which charge can flow.

Q100. Germanium and silicon are semiconductors having crystalline structures. Both these materials have _ valence electrons in their outermost shells.

• A. 2
• B. 4✓
• C. 6
• D. 8
• E. 10

Explanation: Germanium atoms have one more shell than silicon atoms, but what makes for interesting semiconductor properties is the fact that both have four electrons in the valence shell. As a consequence, both materials readily constitute themselves as crystal lattices.