Etea Mdcat 2018 — Solved Past Paper with Answers

Q1. I am afraid we have not got _ sugar for making tea.

• A. Some
• B. No
• C. Any✓
• D. Plenty

Explanation: The correct answer is 'any'. In English, 'any' is used in negative sentences and questions. The sentence 'I am afraid we have not got any sugar for making tea' is a negative sentence, hence 'any' fits perfectly. 'Some' is used for affirmative sentences, 'no' would create a double negative with 'not', and 'plenty' contradicts the negative context implied by 'afraid', thus none of these options are suitable.

Q2. It’s raining cats and dogs. So there are _ cars on the road today.

• A. Few✓
• B. A few
• C. A big number of
• D. A great deal of

Explanation: The phrase 'It's raining cats and dogs' suggests heavy rain, which typically leads to fewer cars on the road. The correct word is 'few', which implies 'not many cars.' The phrase 'a few' would suggest some cars, which doesn't emphasize the reduction in number. 'A big number of' and 'a great deal of' imply a large quantity, which contradicts the expected decrease in car numbers due to the rain.

Q3. I had an unexpected guest today. _ my old classmate.

• A. It was✓
• B. It is
• C. He was
• D. She was

Explanation: The sentence "I had an unexpected guest today. It was my old classmate." uses "It was" to correctly match the past tense context set by "had". The use of "it" is appropriate because the gender of the classmate is not specified, and "it" can refer to an unspecified or gender-neutral subject. "He was" and "She was" incorrectly assume a gender, and "It is" uses the wrong tense.

Q4. Choose the correct indirect speech:She said, “What a lovely dress it is.”

• A. She exclaimed that it is a lovely dress.
• B. She exclaimed that it was a lovely dress.✓
• C. She exclaimed that what a lovely dress it was.
• D. She exclaimed what a lovely dress it is.

Explanation: In indirect speech, the tense of the verb usually shifts back to reflect the fact that the reporting is happening after the original speech. As a result, 'is' becomes 'was'. Additionally, exclamatory sentences are often reported by using verbs like 'exclaimed' and converting the sentence into a statement. The correct answer, Option B, successfully makes these adjustments. Option A fails to change the verb tense. Option C improperly retains the exclamatory form. Option D also fails to change the verb tense and retains the exclamatory structure.

Q5. Choose the correct indirect speech:The teacher said, “Amna, watch your steps.”

• A. The teacher ordered Amna that she should watch her steps.
• B. The teacher ordered Amna to watch your steps.
• C. The teacher ordered Amna to watch her steps.✓
• D. The teacher requested Amna to watch your steps.

Explanation: The correct answer is: 'The teacher ordered Amna to watch her steps.' In indirect speech, the verb 'said' is typically changed to match the speaker's intention—in this case, 'ordered' is appropriate as the teacher is giving an instruction. Additionally, the possessive adjective should match the subject (Amna), so 'her' is used instead of 'your'. Other options either use incorrect pronouns or verbs that do not accurately reflect the speaker's intention.

Q6. Saba was sick on the bus.The underlined prepositional phrase functions as a _ in this sentence.

• A. Adjunct✓
• B. Disjunct
• C. Conjunct
• D. Adverbial

Explanation: The underlined prepositional phrase 'on the bus' functions as an adjunct, which is a type of adverbial that provides additional information about the action or verb in the sentence. It tells us where the action of being 'sick' took place. Although all adjuncts are adverbials, not all adverbials are adjuncts. The other options are incorrect because:Disjunct: This option is incorrect as disjuncts express the writer's or speaker's attitude, which 'on the bus' does not do.Conjunct: This option is incorrect as conjuncts are used to connect clauses or sentences, which 'on the bus' does not do in this context.Adverbial: While technically correct, this option is too broad. 'On the bus' serves a more specific role as an adjunct.

Q7. Which way shall we go?The underlined word is:

• A. Demonstrative adjective
• B. Interrogative pronoun
• C. Interrogative adjective✓
• D. Exclamatory adjective

Explanation: The underlined word 'which' is an interrogative adjective because it modifies the noun 'way' and directly precedes it. Interrogative adjectives are used to ask questions about nouns. The other options are incorrect: a demonstrative adjective would indicate position or distance, which 'which' does not; an interrogative pronoun would stand alone and be followed by a verb, not a noun; and an exclamatory adjective is used to express strong feelings, which is not applicable in this question context.

Q8. Choose the correct indirect speech:He said to me, “traitor”.

• A. He said to me that I was a traitor.
• B. He told me that I have been a traitor.
• C. He called me a traitor.✓
• D. He exclaimed with anger that I was a traitor.

Explanation: The correct answer is: He called me a traitor. In this case, the direct speech 'He said to me, "traitor"' involves the speaker directly addressing the listener with a specific term, 'traitor.' When converting such expressions to indirect speech, the verb 'called' is most appropriate as it conveys the act of naming or labeling someone. The other options either alter the tense, add unnecessary emotional context, or do not accurately reflect the original tone and intent of the direct speech.

Q9. My mother offered me milk. But for my life, I could not drink it.[The underlined expression means:]

• A. However hard I may try✓
• B. Because of my life
• C. For the sake of my life
• D. During my life

Explanation: The phrase But for my life in the given context means However hard I may try, indicating an inability to perform the action regardless of effort. The correct answer reflects this meaning. The other options misinterpret the phrase: 'Because of my life' incorrectly suggests a life-threatening reason, 'For the sake of my life' implies a protective action, and 'During my life' is irrelevant to the context of effort and inability.

Q10. Out of the following indicate the matching item for PUPPIES.

• A. School
• B. Litter✓
• C. Covey
• D. Group

Explanation: Out of the following words that indicate ''puppies' is "Litter".

Q11. Choose the related word for Broom on the analogy of "Water : Splash".

• A. Whisper
• B. Gush
• C. Swish✓
• D. Screech

Explanation: The correct answer is Swish. When water moves rapidly, it creates a splash sound. Similarly, when a broom is used, it makes a swishing sound as it moves through the air.

Q12. Choose the related word for Rat on the analogy of "Elephant : Stride".

• A. Scamper✓
• B. Loiter
• C. Whimper
• D. Gallop

Explanation: Elephants take long steps while walking which is known as a stride. Similarly, a rat scampers which means it runs in quick, light movements.

Q13. Which of the following is correct in all respects?

• A. I have done matric in 2010.
• B. This is an utensil.
• C. The population of the world rises.
• D. This is the best peach-producing valley.✓

Explanation: This sentence is both grammatically correct and contextually appropriate. It uses the correct article and hyphenates the compound adjective "peach-producing."

Q14. Which one of the following is opposite in meaning to the word SYMPATHY?

• A. Apathy✓
• B. Pathos
• C. Empathy
• D. Jealousy

Explanation: Apathy lack of interest, enthusiasm, or concern. While sympathy means feelings of pity and sorrow for someone else's misfortune.

Q15. Choose the correct indirect speech:He said, "What is the matter"?

• A. He said what the matter was.
• B. He asked what the matter was✓
• C. He enquired that what was the matter
• D. He asked that what the matter had been

Explanation: Option (b), "He asked what the matter was," is the correct choice. This option correctly conveys the original question in indirect speech, changing "is" to "was" and properly using "asked" to indicate that it was a question.

Q16. Choose the correct voice:You are called names by him.

• A. He is calling you names.
• B. He calls you names.✓
• C. He called you names.
• D. You are being called names by him.

Explanation: The sentence 'You are called names by him.' is in passive voice and present simple tense. To convert it to active voice, rearrange the components so that 'he' is the subject performing the action: 'He calls you names.' This maintains the present simple tense. The other options either change the tense or maintain the passive structure, which is why they are incorrect.

Q17. Fill in the blank with the most appropriate words.Don’t poke your nose _ my affairs

• A. In
• B. On
• C. Into✓
• D. By

Explanation: This is the most appropriate choice as "poke your nose into my affairs" is the standard expression meaning to interfere or meddle in someone else's business.

Q18. "Enlarge upon" means:

• A. Explain in more detail✓
• B. To make taller
• C. To become large
• D. To measure

Explanation: It means to speak or write about something in greater detail or much more emphasis.

Q19. ‘To the letter’ means:

• A. Cursory
• B. Enveloping a letter
• C. Precisely✓
• D. Reporting a problem

Explanation: "To the letter" means adherence to every detail or precision.

Q20. A person who leaves his country and settles in another country is called:

• A. Emigrant✓
• B. Immigrant
• C. Migrant
• D. Aborigine

Explanation: An emigrant is someone who leaves their country of origin to settle in another country, focusing on the movement away from the home country.In contrast, an immigrant is someone who moves into a new country to live permanently, emphasizing the arrival and settlement in a new location. A migrant can refer to both scenarios but is a broader term that includes temporary moves. An aborigine is unrelated to this context, as it describes native inhabitants of a land.

Q21. Alcoholic fermentation is the sole mean of respiration in:

• A. Saccharomyces✓
• B. Armillaria
• C. Trichonympha
• D. Balantidium

Explanation: Alcoholic fermentation occurs in yeast and plants. Alcoholic fermentation is the sole mean of respiration in Saccharomyces cerevisiae (yeast). Saccharomyces can ferment carbohydrates, breaking down glucose to form ethanol and carbon dioxide.

Q22. The edible part of Morchella esculanta is:

• A. Ascocarp✓
• B. Basidiocarp
• C. Zygocarp
• D. Pseudocarp

Explanation: The fruiting body of ascomycetes is called ascocarp. As Morchella esculenta is an edible fungus from class Ascomycota, its ascocarp part is edible.

Q23. XO in drosophila result in sterile:

• A. Female
• B. Male✓
• C. Both (A) & (B)
• D. No effect

Explanation: XO in drosophila results in a sterile male. As XO means the absence of the second chromosome. The male has only one X chromosome and one Y chromosome, while the female has two X chromosomes. So, when offspring of drosophila carry one X chromosome from a male and one X chromosome from a female it results in a normal female but when offspring of drosophila carry one X chromosome from a female and no chromosome from a male then it results in a sterile male.

Q24. Which of the following is not a fern?

• A. Pteris
• B. Tmesipteris✓
• C. Dryopteris
• D. Pteridium

Explanation: Tmesipteris is actually a genus of ferns belonging to the class Psilotopsida within the sub-division Psilophytina. Tmesipteris is a small genus of ferns that belongs to the class Psilotopsida. Psilotopsida is a class within the division Psilophyta, and this division includes ferns that are often considered primitive vascular plants. The class Psilotopsida includes not only Tmesipteris but also other genera like Psilotum. These ferns are unique because they lack true leaves and roots. Instead, they have small, scale-like appendages on their stems. Tmesipteris, like other members of Psilotopsida, exhibits simple vascular tissues and has a relatively simple overall structure compared to more advanced ferns.

Q25. Umbel of umbels is present in:

• A. Hydrocotyl
• B. Carrot✓
• C. Iberis
• D. Grapes

Explanation: The term "umbel of umbels" refers to an inflorescence arrangement where multiple smaller umbels are arranged in an overall larger, compound umbel. This structure is commonly found in plants belonging to the Apiaceae (or Umbelliferae) family. The Apiaceae family is known for its characteristic inflorescence called an umbel. One of the well-known examples of an umbel of umbels is seen in plants of the genus Daucus. The common carrot, Daucus carota, is a member of this genus and exhibits an umbel of umbels in its inflorescence.

Q26. Male having Down's syndrome have sex chromosomes:

• A. XXY
• B. XY✓
• C. XYY
• D. XYYY

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

Q27. Darwins finches are found in:

• A. New Zealand
• B. New Guinea
• C. Galapagos Islands✓
• D. Australia

Explanation: Darwin finches (13 types) are found in the Galapagos islands, west of Ecuador.

Q28. Which of the following is the most economically important plant family?

• A. Poaceae✓
• B. Asteraceae
• C. Rosaceae
• D. Fabaceae

Explanation: Poaceae, commonly known as the grass family, is one of the largest and most economically important families of flowering plants. The Poaceae family includes a wide variety of grasses, ranging from small, herbaceous plants to large, woody bamboos. Many grasses in the Poaceae family are of great economic importance for humans and animals. They are used as staple food crops, for forage, as ornamental plants, and for the production of materials such as paper and thatch. This family includes major cereal crops that serve as staples worldwide, such as rice (Oryza sativa), wheat (Triticum aestivum), maize (corn, Zea mays), barley (Hordeum vulgare), and oats (Avena sativa).

Q29. The living phloem, cork and cork cambium is collectively called:

• A. Periderm
• B. Protoderm
• C. Periblem
• D. Bark✓

Explanation: The living phloem, cork and cork cambium is collectively called Bark.

Q30. All of the following acts as cloning vector except:

• A. BAC
• B. YAC
• C. Cosmids
• D. EcoRI✓

Explanation: Cloning: It is a technique for developing a large number of genes, identical cells, tissues, or organisms.Vector: Vectors are required to make rDNA molecules for gene cloning. Ecor1 does not act as a cloning vector. As it is a restriction enzyme, which cut double-stranded DNA into single-stranded complementary strands.

Q31. How many sperms are produced from fifty secondary spermatocyte?

• A. 200
• B. 100✓
• C. 50
• D. 150

Explanation: The secondary spermatocyte continues into meiosis II and each spermatocyte forms two sperm. So, 2x50=100.

Q32. Which one of the following is a shrub?

• A. Parmelia
• B. Aster
• C. Rhus✓
• D. Banana

Explanation: Rhus are shrubs.

Q33. In nitrogen-fixing bacteria, the nitrogenase complex is sensitive to:

• A. O2✓
• B. CO2
• C. NO2
• D. NO3

Explanation: Nitrogen fixing bacteria contain an enzyme complex known as Nitrogenase which catalyses the conversion of nitrogen gas to ammonia. The nitrogenase complex is sensitive to oxygen. Oxygen will make the enzyme irreversibly inactive. So, the nitrogen-fixing bacteria needs to protect the nitrogenase enzyme from oxygen.

Q34. Over eating psychological disorder is called:

• A. Dyspepsia
• B. Septicemia
• C. Anorexia
• D. Bulimia✓

Explanation: Bulimia nervosa is an eating disorder in which a person may eat a lot of food at once and then try to get rid of the food by vomiting, using laxatives, or sometimes over-exercising.

Q35. The first successful surgery of heart was performed by Dr. Ludwig by repairing a wound on which part:

• A. Right auricle
• B. Right ventricle✓
• C. Left auricle
• D. Left ventricle

Explanation: The first successful surgery of the heart was performed by Dr. Ludwig by repairing a wound on the Right ventricle.

Q36. Lignin could not be expected in which part of the plant cell wall:

• A. Secondary cell wall
• B. Middle lamella
• C. Primary cell wall
• D. Both "B" and "C"✓

Explanation: The primary cell wall of the plant is composed of cellulose, pectic compounds mostly polysaccharides, and hemicellulose. It lacks lignin. Middle lamella also lacks lignin.

Q37. The following statement is true for the absorption spectra of photosynthesis:

• A. Chlorophyll a and b have the same absorption spectra.
• B. Chlorophyll a and b have different absorption spectra.✓
• C. Chlorophyll a and carotenoids have the same absorption spectra.
• D. Carotenoids and chlorophyll b have the same absorption spectra.

Explanation: Yes, chlorophyll-a and chlorophyll-b, the two main types of chlorophyll found in photosynthetic organisms, have slightly different absorption spectra. Chlorophyll-a and chlorophyll-b absorb light most efficiently in the blue and red regions of the electromagnetic spectrum, but they have different absorption peaks. While chlorophyll-a is directly involved in the conversion of light energy to chemical energy in the photosystems, chlorophyll-b expands the range of light that can be captured and utilized for photosynthesis.

Q38. An autoimmune disorders in which stiffness and inflammation of vertebrae occurs is called as:

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

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

Q39. The study of fishes is:

• A. Ornithology
• B. Ichthyology✓
• C. Herpetology
• D. Serpetology

Explanation: The correct answer is Ichthyology, which is the scientific study of fishes, including their biology, classification, and conservation. Ornithology refers to the study of birds, Herpetology covers amphibians and reptiles, and 'Serpetology' is not an established scientific term.

Q40. Which of the following is absent in C4 Plants:

• A. Calvin Cycle
• B. Bundle Sheath Cells
• C. PEP Carboxylase
• D. CO2 Fixation in Mesophyll Cells✓

Explanation: In C4 plants, CO2 fixation occurs in the mesophyll cells, but it follows a unique pathway compared to C3 plants. C4 plants are adapted to minimize photorespiration by initially fixing CO2 in the mesophyll cells using the enzyme PEP Carboxylase, forming a four-carbon compound (oxaloacetate). This compound is then transported to the bundle sheath cells, where the Calvin cycle occurs. Therefore, the statement that CO2 fixation is absent in mesophyll cells is incorrect.The Calvin cycle is indeed present in C4 plants, but it occurs in the bundle sheath cells, not the mesophyll cells. Bundle sheath cells are essential for the Calvin cycle in C4 plants, and PEP Carboxylase is vital for the initial CO2 fixation. Thus, options A, B, and C are incorrect.

Q41. The stage of plasmodium life cycle not related to human body is:

• A. Merozoite
• B. Ookinetes✓
• C. Trophozoites
• D. Gametozoites

Explanation: Ookinetes are formed in the mosquito gut after fertilization of the female gametocyte by the male gametocyte. This stage is part of the Plasmodium's life cycle within the mosquito and does not occur in the human body.

Q42. In protein synthesis the initiator tRNA carrying amino acid methionine land on which site of ribosome:

• A. E site
• B. P site✓
• C. A site
• D. C site

Explanation: In protein synthesis, the initiator tRNA molecule carrying chemically modified initial amino acid, methionine (known as N-formyl methionine) binds to a small ribosomal subunit at P-site (peptidyl site) where a peptide bond will form. This binding is controlled by an enzyme called initiation factors (IF1 and IF2).

Q43. Goblet cells are:

• A. Unicellular exocrine gland✓
• B. Unicellular endocrine glands
• C. Multicellular exocrine gland
• D. Multicellular endocrine glands

Explanation: The goblet cells are unicellular exocrine gland. Goblet cells are scattered in the epithelial linings of the intestinal and respiratory tracts. They secrete mucin and create a protective mucus layer.

Q44. Which process of cell division is involved in gametes formation in Funaria:

• A. Mitosis✓
• B. Meiosis
• C. Amitosis
• D. Binary Fission

Explanation: In the life cycle of mosses such as Funaria, the process of cell division involved in gamete formation is known as mitosis. The life cycle of mosses involves alternation of generations between a haploid gametophyte phase and a diploid sporophyte phase.

Q45. Closed vascular system is the characteristic of:

• A. Lycopsida
• B. Sphenopsida
• C. Dicot
• D. Monocot✓

Explanation: Monocots (monocotyledons) typically have a scattered vascular bundle and they are collateral and closed. The vascular bundles in monocots are distributed throughout the stem without a clearly defined ring pattern. This is in contrast to dicots, which often have a ring-shaped arrangement of vascular bundles in their stems.

Q46. Common name of the Loligo pealeii is:

• A. Squid✓
• B. Laligo
• C. Slug
• D. Oyster

Explanation: The correct answer is Squid. Loligo pealeii is a species of squid, which are marine cephalopods known for their elongated bodies and tentacles. They are distinct from other mollusks such as slugs and oysters. Slugs are terrestrial and belong to a different class (Gastropoda), while oysters are bivalve mollusks, not cephalopods. 'Laligo' is not a valid common name for any known species.

Q47. Tornaria larva resembles with:

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

Explanation: Tornaria larva resembles with Bipinnaria larva. Because Tornaria larva of hemichordates and Bipinnaria of Echinodermata are very similar in shape and structure. The pattern of cleavage of a fertilised egg, formation of mesoderm, anus, mouth, and coelom in echinoderms and hemichordates is similar. So, most resemblance is with Bipinnaria larva.

Q48. Taxus baccata is the botanical name of:

• A. Fever tree
• B. Deadly nightshade
• C. English Yew✓
• D. Daffodils

Explanation: Taxus baccata is the botanical name for the European yew, a species of coniferous tree. The European yew belongs to the family Taxaceae. The plant is known for its dark green needles, red berries, and its association with various cultural and historical contexts. It contains compounds called taxanes, which have been of interest for their potential medicinal properties, particularly in the development of certain cancer treatments such as paclitaxel (Taxol).

Q49. Venous flower basket belong to which group of organisms:

• A. Angiosperms
• B. Sponges✓
• C. Marine Algae
• D. Fungus-like protists

Explanation: Venous flower basket belong to sponges (Phylum porifera).The Venus' flower basket is a species of glass sponge found in the deep waters of the Pacific Ocean, usually at depths below 500 m. Like other glass sponges, they build their skeletons out of silica, which forms a unique lattice structure consisting of spicules

Q50. The main difference between catalysts and enzymes is:

• A. Enzymes are more specific in action than catalysts
• B. Catalysts are required in larger quantities than enzymes
• C. Catalysts are generally inorganic while enzymes are organic and protein-based✓
• D. Enzymes require specific pH conditions, whereas catalysts do not

Explanation: Enzymes are organic molecules, primarily proteins, whereas most catalysts are inorganic substances.

Q51. The first hormone to be discovered was:

• A. Secretin✓
• B. Testosterone
• C. Insulin
• D. Thyroxin

Explanation: Wm. M. Bayliss and Ernest H. Starling discovered secretin, the first hormone, in 1902. They discovered this hormone when they were studying pancreatic secretion. Secretin hormone stimulates the pancreas to secrete digestive fluid.

Q52. For defense against a virus attack, the body produces:

• A. Antibodies
• B. Histamines
• C. Antigens
• D. Interferons✓

Explanation: The correct answer is Interferons. Interferons are proteins produced by host cells in response to viral infections. They play a crucial role in the immune response by inhibiting viral replication within host cells and activating immune cells. While antibodies are essential for neutralizing viruses outside cells, they are not the primary defense against viruses that have already entered cells. Histamines are related to allergic responses, and antigens are not produced by the body but are targets for the immune system.

Q53. Short life cycle is a plant adaptation to survive in:

• A. High temperature
• B. Low temperature✓
• C. High soil pH
• D. Low soil pH

Explanation: Plants growing in low temperatures may suffer from ill effects. To manage low temperatures, they possess well-developed bark for protection and short life cycles. Such plants bring changes in the composition of solutes in the cell to prevent ice crystal formation. The leaves and stems are hardy to withstand low temperatures. Most of them possess scale leaves, and the rate of transpiration is low to retard cooling.

Q54. Baroceptors are the sensors in body responsible for determination of:

• A. Blood Glucose
• B. Blood Ammonia
• C. Blood 𝑝H
• D. Blood Pressure✓

Explanation: Baroreceptors are mechanoreceptors located in blood vessels near the heart that provide the brain with information about blood volume and pressure by detecting the level of stretch on vascular walls.

Q55. Which one of the following is not a draught animal?

• A. Buffalo
• B. Mule
• C. Elephant✓
• D. Yak

Explanation: Elephants, although used for heavy lifting in industries such as logging, are not traditional draught animals in agriculture. Draught animals like buffaloes, mules, and yaks are specifically bred and trained for agricultural tasks such as plowing and transporting goods across fields. These animals are chosen for their ability to work efficiently in these environments, unlike elephants, which are not typically utilized in such roles.

Q56. Which of the following is not isotonic to sea water?

• A. Myxine (Hagfish)
• B. Skates
• C. Sharks
• D. Teleost (Bony Fishes)✓

Explanation: Marine teleosts are hypoosmotic to seawater, so they face osmotic loss of water and diffusional gain of NaCl across the gill.

Q57. Which of the following is not part of the first line of defense?

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

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

Q58. Which one of the following is not an insect?

• A. Ticks✓
• B. Honey bee
• C. Beetle
• D. Wasp

Explanation: Ticks are not insects because they are arachnids.Ticks are parasitic arachnids of the order Ixodida. They are part of the mite superorder Parasitiformes. Adult ticks are approximately 3 to 5 mm in length depending on age, sex, and species, but can become larger when engorged.

Q59. A condition called Goose pimples, are caused by:

• A. Overcooled body✓
• B. Bacteria
• C. Environmental changes
• D. Pollution

Explanation: Option A is correct.The hairs in mammals act as insulating organs and reduce heat loss. Thus the heat is retained in the body to a certain extent. To increase the effect of insulation, the hairs are erected. This occurs involuntarily when the body is over-cooled. In humans, it produces Goose pimples.

Q60. Piriformis syndrome is associated with which of the following disorder:

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

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

Q61. Which one of the following is not an exclusive trait of arthropoda?

• A. Presence of wings✓
• B. Jointed appendages
• C. Haemocoel
• D. Chitinous exoskeleton

Explanation: Phylum Arthropoda doesn’t possess wings.Yes, some arthropods have wings, but not all of them do; only members of the class Insecta within the subphylum Hexapoda have wings. These wings are attached to the thorax, along with the three pairs of legs that are characteristic of insects. Many other arthropods, such as spiders, scorpions, and crustaceans, lack wings entirely.

Q62. Locomotary organ in leech is called?

• A. Setae
• B. Chatae
• C. Parapodia
• D. None of the above✓

Explanation: Leech has no locomotory organs.

Q63. The number of Hyoid bone in human skull region is:

• A. 1✓
• B. 6
• C. 22
• D. 206

Explanation: There are overall 29 bones in the skull region, but there is only 1 hyoid bone in the human skull region.

Q64. Which factor is not involved in release of Oxytocin in females:

• A. Stretching of uterus
• B. Stretching of Cervix
• C. Low level of testosterone✓
• D. Low level of progesterone

Explanation: The correct answer is Option C: Low level of testosterone. Testosterone is a male sex hormone and does not influence the release of oxytocin in females. Oxytocin is primarily involved in childbirth and breastfeeding, and its release is stimulated by the stretching of the uterus and cervix (Options A and B) as well as hormonal changes such as a decrease in progesterone levels (Option D). Therefore, low testosterone levels do not play a role in this process.

Q65. Chymotrypsin acts upon:

• A. Starch in duodenum
• B. Proteins in stomach
• C. Proteins in duodenum in acidic medium
• D. Proteins in duodenum in alkaline medium✓

Explanation: Chymotrypsin acts in the duodenum, the first part of the small intestine, where it helps break down proteins into smaller peptides. It is a digestive enzyme secreted by the pancreas as an inactive precursor, chymotrypsinogen, which is then activated in the small intestine by another enzyme called trypsin.

Q66. In E.C.G QRS complex represent:

• A. Atrial systole
• B. Atrial diastole
• C. Ventricle systole✓
• D. Ventricle diastole

Explanation: The second part of ECG is the QRS complex which features a small drop in voltage (Q) a large voltage peak ( R ) and another small drop in voltage (S). The QRS complex corresponds to the depolarization of the ventricle during ventricle systole.

Q67. The common name of rubella is:

• A. Whooping cough
• B. German measles✓
• C. African sleeping sickness
• D. Tay Sach’s Disease

Explanation: It is also called German measles, but it is caused by a different virus than measles.

Q68. A heterozygote fruit fly has more florescent pigments in their eyes than a wild homozygote fruit fly, this is an example of:

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

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

Q69. A vein differs from an artery in having:

• A. Strong muscular walls
• B. Narrow lumen
• C. Valves control direction of blood flow opposite to heart
• D. Valves control direction of blood flow towards heart✓

Explanation: All the veins transport blood to the heart.

Q70. Expiratory centre in medulla is:

• A. Dorsal✓
• B. Ventral
• C. Lower part
• D. All of the above

Explanation: The dorsal and lateral regions contain the expiratory centre, they stimulate expiration and inhibit inspiration.

Q71. Polyploidy is more common in:

• A. Animals
• B. Plants✓
• C. Both equally
• D. Viruses

Explanation: Polyploidy (extra sets of chromosomes) is rare in animals but common in plants, where it often leads to larger size and hybrid vigor (e.g, wheat, cotton).

Q72. The first stage in development of Xerosere is appearance of:

• A. Foliose lichens
• B. Crustose lichens✓
• C. Fruticose lichens
• D. Climax stage

Explanation: Due to great exposure to the sun and extreme deficiency of water, the first pioneers in the bare rock area are a few simple organisms. The most successful of such organisms is crustose lichens.

Q73. Mg++ and Ca++ are excreted in fishes through:

• A. Kidney✓
• B. Gills
• C. Skin
• D. All of the above

Explanation: Option A is correct.The kidneys in fishes play a vital role in osmoregulation, helping maintain the balance of water and salts in their bodies despite different aquatic environments. In addition to filtering nitrogenous wastes, they are especially important in the excretion of divalent ions such as magnesium (Mg²⁺) and calcium (Ca²⁺), which, if accumulated, can disturb cellular and physiological processes. Marine fishes actively excrete these excess salts through their kidneys and gills, while freshwater fishes conserve essential ions and excrete large amounts of dilute urine to maintain osmotic balance. Thus, fish kidneys are central to both waste removal and the regulation of ionic composition of body fluids.

Q74. Vomit centre is located in:

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

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

Q75. Which of the following is a summer variety?

• A. Figs✓
• B. Cabbages
• C. Oranges
• D. Pears

Explanation: Figs (Ficus carica) are generally known for their sweet and succulent fruits, and they are indeed associated with the summer season in many regions. The timing of the fig fruiting season can vary based on factors such as the fig variety, climate, and local conditions. However, in many places, figs tend to ripen and are harvested during the summer months.

Q76. Depolarization of neuron is characterized by:

• A. Na+ into the axon and K+ out of the axon
• B. K+ into the axon and Na+ out of the axon
• C. Na+ and K+ within the axon toward the axon terminal
• D. Increased Na+ permeability leading to an influx of Na+✓

Explanation: When a neuron fiber is stimulated by a stimulus of adequate strength (threshold stimulus), the stimulated area of the fiber becomes several times more permeable to Na+ than to K+ due to the opening of voltage-regulating Na+ gates. As a result, Na gates permit the influx of Na+ ions by diffusion. Since more Na+ ions are entering than leaving, the electrical potential of the membrane changes from -70 mV towards zero and then reaches 50 mV. This reversal of polarity across two sides of the membrane is called depolarization. This electropositive inside and electronegative outside last for about one millisecond till the Na+ gates are closed. Note that K+ is not involved in the depolarization of neurons. It is involved in repolarization and hyperpolarization.

Q77. The center of porphyrin in the head region of hemoglobin is occupied by:

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

Explanation: Haemoglobin consists of four polypeptide chains. Each chain is associated with a haem group. Haeme group is an iron-containing group, which consists of porphyrin with a central atom of ferrous (iron) between four pyrrole rings.

Q78. Syphilis is caused by:

• A. Spirochete✓
• B. Nostoc
• C. Water blooms
• D. Cyanobacteria

Explanation: Syphilis is caused by the bacterium Treponema pallidum, which belongs to the spirochete group. Spirochetes are notable for their spiral shapes and unique motility mechanisms. This bacterial group includes pathogens known for causing various diseases, with Treponema pallidum specifically linked to syphilis. In contrast, the other options are related to cyanobacteria, which are photosynthetic bacteria not associated with sexually transmitted infections. Nostoc and cyanobacteria are both non-pathogenic in the context of syphilis, and water blooms are an environmental concern rather than a bacterial classification.

Q79. The organism developed with two heads and one trunk is called:

• A. Identical twins
• B. Dizygotic twins
• C. Fraternal twins
• D. Siamese twins✓

Explanation: Siamese twins or also known as Conjoined twins. It is a very rare condition in which twins are joined side to side at the pelvis and part or all of the belly (abdomen) and chest but with separate heads. Such twins develop when an early embryo only partially separates to form two individuals.

Q80. This is used as a substrate during the formation of alcohol:

• A. Sucrose✓
• B. Glucose
• C. Galactose
• D. Mannose

Explanation: Sucrose or cane sugar is widely distributed among higher plants. Its commercial source is solely sugarcane and beet. It is used as a substrate for the formation of alcohol. Glucose is a widely distributed sugar.

Q81. If the required excitation voltage is given, for which element the x − rays spectrum consists of three spectral lines i.e. Kα Kβ Lα :

• A. Na✓
• B. B
• C. K
• D. Ca

Explanation: A spectral line is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from the emission or absorption of light in a narrow frequency range, compared with the nearby frequencies. Spectral lines are often used to identify atoms and molecules. A spectral line may be observed either as an emission line or an absorption line.Option A: The electronic configuration of Na is = 1s2,2s2,2p6,3s1 There are 2 electrons in 1st shell, 8 in 2nd and 1 in 3rd. The difference between the highest and lowest shells is 3-1=2. The spectral lines are (2,1). The two are K(alpha), and K(beta) and one is K(gamma).

Q82. SiO2 is the only oxide that reacts with:

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

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

Q83. The cation that distort the electron cloud of NO3− ion more and facilitates its decomposition is:

• A. Li+
• B. Mg2+
• C. Cs+
• D. Be2+✓

Explanation: Polarizability refers to the ability of a cation to distort the electron cloud of an anion in an ionic compound and the anionic cloud is pulled towards the cation when it gets distorted. According to Fajan’s rule, the greater charge density(charge/size) on a cation leads to greater distortion. Due to higher charge and smaller size, the charge density of Be++ is and so it will distort the electron cloud of NO3- more facilitating its decomposition.

Q84. The energy of electron in the first excited state of Hydrogen atom in J/atom is:

• A. 2.8 × 10−18 J/atom
• B. 0.545 × 10−18 J/atom✓
• C. −2.18 × 10−18 J/atom
• D. −1312.36 J/atom

Explanation: By using the formula for the energy of an electron in the nth orbitEn=-{me-4Z2/8 ε2 h2 n2}For hydrogen Z=1 and by putting values of constants also first excited state of hydrogen means, n=2 which is the second shell to which the ground state hydrogen electron has excited to. En= (-2.178×10-18)×1/n2

Q85. Unhybrid "p" orbitals on linear overlap:

• A. Always form Pi(π) bond✓
• B. Always form Sigma(σ) bond
• C. Neither form σ nor π bond
• D. Form more reactive and more unstable π bond

Explanation: Unhybridized orbitals can only form π bonds upon sideways or parallel overlap and cannot form sigma bonds.

Q86. Amino acid leucine is coded by how many codons:

• A. 1
• B. 2
• C. 4
• D. 6✓

Explanation: Amino acid leucine is coded by 6 codons UUA, UUG, CUU, CUC, CUA, and CUG.

Q87. Whenever Pb shows inert pair effect it always form:

• A. Ionic bond✓
• B. Covalent bond
• C. Co-ordinate covalent bond
• D. Metallic bond

Explanation: The inert pair effect is the effect of the s-orbitals in which they do not participate to make bonds with other molecules because the electrons are held tightly due to the strong attraction of the nucleus towards the electrons. The reason for this effect is that the elements having the f-orbital show poor shielding effect due to their diffused shapes. As a result, the nuclear charge on their s orbitals increases due to this poor shielding effect. This results in more attraction and binding of the electrons in the s-orbitals which stops them from getting out and prevents the metal to acquire a +4 oxidation state. The electronic configuration of lead isPb (82) = (Xe) 4f14 5d10 6s2 6p2Due to the inert pair effect, the +2 oxidation state of Pb is more stable than its +4 oxidation state. Further on moving down the group, the ionization enthalpy decreases. As a result, the sum of the first two ionization enthalpies of Pb is quite low and hence Pb can easily lose two electrons to form Pb2+ ions. Consequently, Pb forms ionic compounds.

Q88. In the compound CO2 and H2O the hybridization in oxygen is respectively:

• A. sp2 and sp2
• B. sp2 and sp3✓
• C. sp3 and sp3
• D. sp3 and sp2

Explanation: In the carbon dioxide molecule, oxygen hybridizes its orbitals to produce three sp2 hybrid orbitals. The Unhybridized p orbital in the oxygen atom remains unchanged and is primarily used to form a pi bond. However, out of these three sp hybrid orbitals, only one will be used to produce a bond with the carbon atom.The central atom, which is hybridized, is oxygen. In the formation of the water molecule, there are three 2p orbitals and one 2s orbital. They are combined to generate the sp3 hybrid orbitals. Furthermore, each hydrogen atom forms covalent connections with two hybrid orbitals, and two hybrid orbitals are occupied by lone pairs during the process.

Q89. According to M.O theory the number of molecular orbitals in O2 are:

• A. 10✓
• B. 7
• C. 8
• D. 9

Explanation: Oxygen atoms contribute ground-state molecular orbitals from 1s, 2s, and 2p electrons. The 1s orbitals form σ1s and σ*1s filled Mos. The 2s orbitals form σ2s and σ*2s filled MOs. The σ2p and π2p MOs are also filled, while the antibonding π*2p orbitals are partially filled in triplet oxygen (the ground state), or only one is filled in singlet oxygen. But the antibonding σ*the 2p orbital is ordinarily empty in the ground state, and higher orbitals are not filled unless the oxygen molecule absorbs light energy which is capable of exciting the electrons to higher orbitals.

Q90. Specie with dipole moment equal to zero is:

• A. AlCl3
• B. CH4
• C. 1,4 − Dibromobenzene
• D. All of the above✓

Explanation: Specie with dipole moment equal to zero is: 𝐴𝑙𝐶𝑙3 𝐶𝐻4 1,4 − 𝐷𝑖𝑏𝑟𝑜𝑚𝑜𝑏𝑒𝑛𝑧𝑒𝑛𝑒 because their structures allow for their charges to be 'canceled' out.

Q91. Aqueous KOH causes SN-reaction in alkylhalide. On which of the following alkylhalides KOHaq would like to attack easily.

• A. CH3 − CH2 − Cl
• B. CH3 − CH2 − Br
• C. CH3 − CH2 − F
• D. CH3 − CH2 − I✓

Explanation: Iodide ion is a very good leaving group due to low bond energy so the attack of KOH(aq) to proceed SN reaction and hence it can be easily substituted.

Q92. Three reactions are giveni. H2SO4 + 2HF ⟶ F2 + SO2 + 2H2Oii. H2SO4 + 2HBr ⟶ Br2 + SO2 + 2H2Oiii. H2SO4 + 8HI ⟶ 4I2 + H2S + 4H2OThe strongest reducing agent in these reactions is:

• A. HI✓
• B. HF
• C. HBr
• D. All of the above

Explanation: HI is the strongest reducing agent. The compounds having low dissociation energy are a stronger reducing agent than the others having high dissociation energy. The HI has low thermal stability as compared to HF, HCI, and HBr. HI can release H atoms easily as compared to HF, HCI, and HBr to reduce the other compound. The decreasing order of the reducing character is HI>HBr>HCl>HF

Q93. An unknown gas diffuses 5 times slower than that of H2. The molecular mass of the unknown gas is:

• A. 50✓
• B. 10
• C. 15
• D. 25

Explanation: According to Graham’s law of diffusion of gases, the rate of diffusion is inversely proportional to the square root of molar masses of gases. Using the formular1/r2 =√M1/M2(5rH2/r2)2 =(√M2/2)225×2=M2M2=50

Q94. Hydrogen bonding in H − F is stronger than H2O and NH3. The highest boiling point among the three is of:

• A. HF
• B. NH3
• C. H2O✓
• D. All have equal boiling points

Explanation: Each water molecule has two hydrogen atoms. Whereas, each HF molecule has only one hydrogen atom. And hence, water molecule takes part in extensive and more hydrogen bonding than Hydrofluoric acid. Water molecule through their extensive hydrogen bonding forms bulky molecule and it is very difficult to break their bonds. A large amount of energy is required to break all its bonds. Thus H2O has a higher boiling point than HF.

Q95. We have three test tubes having aqueous solutions of Ca(NO3)2, Ba(NO3)2 and K2CO3 respectively. On addition of dil H2SO4. Which test tube solution turns milky?

• A. Ca(NO3)2 solution✓
• B. Ba(NO3)2 solution
• C. K2CO3 solution
• D. All turned milky

Explanation: When sulfuric acid is added to calcium nitrate, calcium sulfate is produced, which is slightly soluble and often forms a fine precipitate that appears milky. This contrasts with barium nitrate, which forms a dense white precipitate of barium sulfate, and potassium carbonate, which releases carbon dioxide gas causing bubbling but no precipitate. Thus, only the calcium nitrate solution turns milky due to the formation of a fine calcium sulfate precipitate.

Q96. The chelating ligand out of the following is:

• A. CH3COO−
• B. (CH2)2(NH2)2✓
• C. SCN−
• D. NO2−

Explanation: A chelating ligand is a ligand that is mostly attached to a central metal ion by bonds that are from two or more donor atoms. In other words, these are a type of ligands where the molecules can form several bonds to a single metal ion or they are ligands with more than one donor site.The chelating ligand here Is (𝐶𝐻2)2(𝑁𝐻2)2

Q97. Evaporation depends upon:

• A. Surface area
• B. Temperature
• C. Both (A) & (B)✓
• D. None of the above

Explanation: The correct answer is that evaporation depends on both surface area and temperature. Evaporation is a surface phenomenon, meaning that more molecules can escape to the air when the surface area is larger. Additionally, increasing the temperature provides more energy to the molecules, allowing them to overcome intermolecular forces more easily and escape into the vapor phase. Therefore, both factors are crucial in determining the rate of evaporation. Options that consider only one factor or neither are incomplete explanations of the process.

Q98. Students were heating CaCO3 in an open container to produce CO2 gas,CaCO3(s) ⟶ CaO(s) + CO2(g)If we increase pressure on this system the:

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

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

Q99. The oxidation state of platinum in [Pt(NH3)4(NO2)Cl]SO4 is:

• A. III
• B. II
• C. 0
• D. IV✓

Explanation: Explained in image below:

Q100. The outer electronic configuration of Cu+ ion is 4S03d10 with this configuration the aqueous solution of copper (I) compound is:

• A. Blue
• B. Greenish blue
• C. Bluish green
• D. Colourless✓

Explanation: Transition elements exhibit color due to d-d transition. d-d transition is possible only when the d subshell has unpaired electrons. Cu+ has filled the orbital: 4s0 3d10. Due to a lack of unpaired d- electrons Cu+ is colorless.

Q101. For the reaction,CO(g) + (1/2)O2(g) ⟶ CO2(g)

• A. Kp > Kc
• B. Kp < Kc✓
• C. Kp = Kc
• D. Kp ≥ Kc

Explanation: For the reaction. CO(g)+1/2O2(g) = CO2(g) Kp<Kc

Q102. Initially one mole each N2 and O2 were made to react as,If at equilibrium 0 ∙ 25 moles of O2 is present the equilibrium concentration of NO will be:

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

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

Q103. The compound of manganese with zero reducing power is:

• A. KMnO4✓
• B. MnO2
• C. MnCl2
• D. Mn2(SO4)3

Explanation: The reducing power is the ability to give electrons and reduce others. The compound having maximum oxidation number will have the zero reducing power as it doesn’t have any remaining electrons to give other compounds.

Q104. Chemical reactions associated with hydro-carbons is/are:

• A. Electrophilic addition
• B. Electrophilic substitutions
• C. Free radical substitutions
• D. All are possible✓

Explanation: Chemical reactions associated with hydrocarbons are:Electrophilic addition, Electrophilic substitutions, Free radical substitutions.

Q105. Henderson-Hasselbalch equation is used to calculate the pH of a buffer solutions. The correct representation of the equation is:

• A. pH = pKa + log([Acid]/[Salt])
• B. pH = pKa - log([Salt]/[Acid])
• C. pH = pKa + log([Salt]/[Acid])✓
• D. pH = pKa - log([Acid]/[Salt])

Explanation: Henderson-Hasselbalch equation to calculate the pH of a buffer solution ispH=pka+log[salt]/[Acid]

Q106. If the pH of the solution is 9 its OH− ions concentration is:

• A. 10−5✓
• B. 5
• C. 10−9
• D. 9

Explanation: As we know that,pH+pOH=14pOH=14-9=5Also, pOH=log[OH-][OH-]=antilog(pOH)[OH]=antilog (5)[OH]=10-5

Q107. During the formation of addition polymers, which smaller molecules you think are eliminated:

• A. H2O
• B. HCl
• C. NH3
• D. No molecules are eliminated✓

Explanation: In polymer chemistry, an additional polymer is a polymer that forms by simple linking of monomers without the co-generation of other products. Addition polymerization differs from condensation polymerization, which does co-generate a product, usually water.In addition, polymerization monomers simply react to form polymers without the formation of by products.

Q108. If the overlap of sp3 hybrid orbitals in carbon atoms is smaller, the bond so formed is:

• A. Weak✓
• B. Strong
• C. Less energetic
• D. More stable

Explanation: According to the quantum mechanical approach, a covalent bond is formed when half-filled orbitals in the outer or valence shells of two atoms overlap, so as a result of this overlap, the electrons with opposite spins become paired to stabilize themselves. Larger the overlap, the stronger the bond is formed.

Q109. The minimum energy below which no reaction occur in reactant molecules is:

• A. Average Kinetic Energy of the molecules
• B. Potential Energy of the molecules
• C. Free Energy of the molecules
• D. Activation Energy of the molecules✓

Explanation: Activation energy is the minimum energy that the reactants must have to be converted into products. It can also be described in terms of energy which activate reactants to start a chemical reaction.

Q110. Reactants in a transition state:

• A. Always change to product
• B. Return back to reactants
• C. May return to reactants or proceed to form products✓
• D. Are of low energy

Explanation: The transition state is a high-energy state, and some amount of energy – the activation energy – must be added for the molecule to reach it. Because the transition state is unstable, reactant molecules don't stay there long but quickly proceed to the next step of the chemical reaction.If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs, and products form. But if the k.E of colliding molecules is not enough and so transition state then the colliding molecules just bounce back unchanged.

Q111. Compound in which addition takes place through Markovnikov’s rule is:

• A. CH3 − CH = CH − CH3
• B. CH3 − C(CH3) = CH − CH3✓
• C. C2H5 − CH = CH − CH3
• D. CH3 − CH = CH − C3H7

Explanation: Markovnikov Rule predicts the regiochemistry of HX addition to unsymmetrically substituted alkenes. The halide component of HX bonds preferentially at the more highly substituted carbon, whereas the hydrogen prefers the carbon which already contains more hydrogens.

Q112. The type of isomerism present in the compound given, is:

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

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

Q113. During the formation of aqueous solution of any electrolyte:

• A. Heat is evolved
• B. Heat is absorbed
• C. Heat may be evolved or absorbed✓
• D. Electrolyte does not dissolve in water

Explanation: A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, or the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution can either be positive (endothermic) or negative (exothermic). The heat may be evolved or absorbed during the formation of aqueous solution of electrolytes and it completely depends upon their behaviour and properties of electrolytes being used.

Q114. The compound with more than 10% solubility in pure water is:

• A. M𝑔𝐶𝑂3
• B. A𝑙2(𝐶𝑂3)3
• C. K2𝐶𝑂3✓
• D. 𝑍𝑛𝐶𝑂3

Explanation: The solubility of ionic compounds in water is largely determined by the balance between lattice energy and hydration energy. Lattice energy is the energy required to separate the ions in a compound, and it increases with greater charge and smaller ionic size. Hydration energy is the energy released when ions are solvated by water molecules. For K2CO3, the K+ ions have a relatively large size and a +1 charge, resulting in lower lattice energy and higher solubility. In contrast, MgCO3, Al2(CO3)3, and ZnCO3 all have higher lattice energies due to their higher charges and smaller ionic sizes, making them less soluble in water.

Q115. The mass of 𝑁𝑎𝑂𝐻 needed to prepare 0.2 molal solution in 500𝑔 pure water at 4℃ is:

• A. 0.4𝑔
• B. 4.0𝑔✓
• C. 1.5𝑔
• D. 1.0𝑔

Explanation: As for 0.2 molal solution 0.2 mole of NaOH is needed to dissolve in 1000 grams of water. For 500 grams0.2÷1000×500=0.1 mole of NaOHThe mass for 0.1 mole of NaOH is 40 g/mol×0.1mol = 4 grams.

Q116. The Fridel crafts catalyst “𝐴𝑙𝐶𝑙3” used in the substitution reactions of Benzene is good:

• A. Electrophile
• B. Lewis acid✓
• C. Electron deficient species
• D. Bear all properties

Explanation: The function of AlCl3(anhydrous), in the Friedel-Craft reaction, is to produce electrophile, which later adds to the benzene nucleus. This electrophilic aromatic substitution allows the synthesis of monoacetylated products from the reaction between arenes and acyl chlorides or anhydrides. The products are deactivated and do not undergo a second substitution. Normally, a stoichiometric amount of the Lewis acid catalyst is required for both the substrate and the product form complexes.

Q117. The most reactive compound out of the following is:

• A. Ortho hydroxy toluene✓
• B. Ortho chloro ethyl benzene
• C. Phenol
• D. Para ethyl benzoic acid

Explanation: Substituents: -OH and -CH₃Effect: Both are activating groups. -OH is strongly activating; -CH₃ is weakly activating. Being ortho to each other increases electron density in the ring even more.This is even more activated than phenol, due to combined activating effects.

Q118. Addition of soluble impurities into a liquid and solid respectively causes:

• A. Increase in boiling point of liquid and decrease in melting point of solid✓
• B. Increase in both boiling and melting points
• C. Decrease in boiling point of liquid and increase in melting point of solid
• D. Decrease in both boiling and melting points

Explanation: Whenever a soluble, non-volatile impurity is added to a liquid, some of the solute particles tend to migrate toward the surface of the liquid, and some will remain in the bulk liquid. Due to this, an obstruction is caused by the volatile nature of the liquid. In other words, it becomes harder for the liquid molecules to vaporize due to the obstruction caused by the molecules of the added impurity. The result is that the net vapor pressure of the mixture becomes lower than that of the pure liquid, as fewer liquid molecules can go into the vapor phase. We know that the boiling point of a substance is the temperature at which the vapor pressure of the substance equals the external pressure. What happens during boiling is that we raise the temperature of the liquid, thereby raising its vapor pressure, until it becomes equal to the atmospheric pressure and starts to boil. So, when the impurity is added, vapor pressure is reduced, and thus, more heat needs to be supplied to raise it to the external pressure and make the mixture boil. Hence, the presence of soluble impurities increases the boiling point of liquids. The higher the concentration of impurity, the higher will be the elevation in boiling point. When we add impurities to the pure substance, the melting point of the solid is decreased because impurities weaken the lattice structure of solids due to which it becomes less stable and melts before its original melting point. This is called melting point depression.

Q119. The elevation in boiling point ∆𝑇𝑏 is equal to ebullioscopic constant 𝐾𝑏 when the Molarity (𝑀) of the solution is:

• A. 0.1m
• B. 1.0m
• C. 10.0m
• D. Statement is wrong✓

Explanation: According to the formula for the elevation of boiling point∆Tb=kb×m Where kb is the ebullioscopic constant and m is the molality of the solution. ∆Tb will be equal to kb only if molality is 1.0m and it has no relation with the molarity of solution so the given statement is wrong.

Q120. 𝑂𝐻− + 𝐶2𝐻5 − 𝐼 ⟶ 𝐶2𝐻5 − 𝑂𝐻 + 𝐼− One of the species in the above reaction is a substrate. It is:

• A. 𝑂𝐻−
• B. 𝐶𝐻3 − 𝐶𝐻2 − 𝑂𝐻
• C. 𝐼−
• D. C𝐻3 − 𝐶𝐻2 − I✓

Explanation: The molecule which contains the electrophile and the leaving functional group is called a substrate. It is the molecule on which nucleophiles attack to form a product.

Q121. Cyclic alkanes with greater angle strain are always:

• A. More stable
• B. Less energetic
• C. More reactive✓
• D. Obey the general formula of normal alkanes

Explanation: Angle strain occurs when bond angles deviate from the ideal bond angles to achieve maximum bond strength in a specific chemical conformation. Angle strain typically affects cyclic molecules, which lack the flexibility of acyclic molecules.It occurs in cycloalkanes because the carbons in cycloalkanes are sp3 hybridized, which means that they do not have the expected ideal bond angle of 109.5°; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5°. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy and hence less stability and more reactivity.

Q122. Students were decomposing 𝐶𝑎𝐶𝑂3 placed in a China dish by heating using burner in the laboratory. The “system” in this experiment is:

• A. China dish
• B. Burner
• C. Laboratory
• D. 𝐶𝑎𝐶𝑂3✓

Explanation: The system is the collective substances in the reaction such as the reactants and products. The surroundings are everything around the reaction such as the reaction flask and the room. During a reaction, energy is transferred between the system and its surroundings. The system and surrounding together make up the universe.

Q123. The process that can be both endothermic and exothermic out of the following is:

• A. Dissolution✓
• B. Crystallization
• C. Bond breaking
• D. Condensation

Explanation: The process of dissolving can be endothermic (temperature goes down) or exothermic (temperature goes up). When water dissolves a substance, the water molecules attract and “bond” to the particles (molecules or ions) of the substance causing the particles to separate from each other.The process of dissolving can be endothermic (temperature goes down) or exothermic (temperature goes up).Ammonium nitrate dissolving in solution is an endothermic reaction. As the ammonium nitrate dissolves, heat energy is absorbed from the environment causing the surrounding environment to feel cold. The dissolution of calcium chloride in water is an example of an exothermic reaction.

Q124. The amine which is more reactive towards 𝐻𝐼 is:

• A. Methyl amine
• B. Dimethyl amine✓
• C. Methyl propyl amine
• D. Butyl amine

Explanation: The greater the basicity of amines the higher its reactivity. Order of basicity if the −R group is −CH3 group:Secondary amine > primary amine > tertiary amine > Ammonia > Aryl compounds. Dimethyl amine, a secondary amine, is the most reactive amine towards HI.

Q125. The alcohol given 𝐶𝐻3 − 𝐶𝐻2 − 𝐶(𝐶𝐻3)2 −𝑂𝐻. If oxidized with a strong oxidizing agent given:

• A. Aldehyde
• B. Ketone
• C. Ether
• D. None of the above✓

Explanation: A tertiary alcohol can't form this carbonyl group: the tertiary custom has three single bonds to other C- atoms and one single bond to the oxygen of the hydroxyl-group. To form a carbonyl group, you need a double bond. Carbon can only form 4 bonds and the C-C- bonds are relatively stable, so it can't form the carbonyl group. With enough energy, it is possible to oxidise even tertiary alcohols, but your product will be only carbon dioxide and water (that's what you get by the complete oxidation of every alcohol).

Q126. The aqueous solution of which of the following electrolyte will conduct electric current to large extent:

• A. 𝑀𝑔(𝑂𝐻)2
• B. 𝐻2𝐶𝑂3
• C. N𝐻4𝐶𝑙✓
• D. N𝐻4𝑂H

Explanation: The correct answer is NH4Cl because it is a strong electrolyte that completely dissociates in water, producing a large number of ions (NH4+ and Cl-), which facilitates high electrical conductivity.Mg(OH)2 is not correct as it is sparingly soluble and does not produce a significant amount of ions in solution.H2CO3 is a weak acid that partially dissociates, offering limited conductivity.NH4OH is a weak base and also partially dissociates, thus offering lower conductivity compared to strong electrolytes.

Q127. Read the statement and choose the option correctly.

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

Explanation: The general form of a rate law is: Rate =k[A]x[B]y1. Find the effect of changing [A] (Order with respect to A):Compare Experiment 1 and Experiment 2.In Experiment 1: [A]=0.1, [B]=0.1, Rate =2×10−5In Experiment 2: [A]=0.2, [B]=0.1, Rate =4×10−5Notice that [B] stays the same (0.1).When [A] doubles (from 0.1 to 0.2), the Rate also doubles (from 2×10−5 to 4×10−5).If doubling the concentration doubles the rate, the reaction is first order with respect to A. So, x=1.2. Find the effect of changing [B] (Order with respect to B):Compare Experiment 1 and Experiment 3.In Experiment 1: [A]=0.1, [B]=0.1, Rate =2×10−5In Experiment 3: [A]=0.1, [B]=0.2, Rate =4×10−5Notice that [A] stays the same (0.1).When [B] doubles (from 0.1 to 0.2), the Rate also doubles (from 2×10−5 to 4×10−5).If doubling the concentration doubles the rate, the reaction is first order with respect to B. So, y=1.3. Combine to find the Rate Law: Since x=1 and y=1, the rate law is: Rate =k[A]1[B]1 Which simplifies to: Rate =k[A][B]

Q128. The compound which you think is not the derivative of acetic acid is:

• A. Acetyl chloride (CH3COCl)
• B. Acetic anhydride ((CH3CO)2O)
• C. Acetamide (CH3CONH2)
• D. None of these✓

Explanation: All the listed compounds are derivatives of acetic acid based on their structural transformations.

Q129. The reducing agent in the reaction given, 𝐾𝑀𝑛𝑂4 + 𝐾𝐼 + 𝐻2𝑆𝑂4 ⟶ 𝑀𝑛𝑆𝑂4 + 𝐾2𝑆𝑂4 +𝐼2 + 𝐻2𝑂 is:

• A. 𝐾𝑀𝑛𝑂4
• B. 𝐾𝐼✓
• C. 𝐻2𝑆𝑂4
• D. 𝐾𝐼 and 𝐻2𝑆𝑂4 both

Explanation: Reducing agent is the one which has the ability to give its unshared pair of electrons to other species to reduce them.

Q130. Students calculated the cell voltage for the reaction, 𝐵𝑟2 + 2𝑁𝑎𝐶𝑙 ⟶ 2𝑁𝑎𝐵𝑟 + 𝐶𝑙2 through the formula 𝐸°𝑐𝑒𝑙𝑙= 𝐸°𝑟𝑒𝑑+ 𝐸°𝑜𝑥𝑑 the answer was negative. It means that:

• A. The reaction is non-spontaneous and feasible
• B. The reaction is non-spontaneous and not feasible✓
• C. The reaction is spontaneous and feasible
• D. The reaction is spontaneous and not feasible

Explanation: The energy associated with the separation/transfer of charge is the property of potential, E. A cell potential's arithmetic sign determines the spontaneity of the cell reaction, with positive values indicating spontaneous and feasible reactions and zero or negative values indicating nonspontaneous and non-feasible reactions (spontaneous in the reverse direction). If the value is positive, the oxidation-reduction reaction is spontaneous. Only spontaneous reactions are feasible. That is, without the assistance of a third party. The cathode undergoes reduction, while the anode undergoes oxidation. If the value is negative, it signifies that only the opposite reaction is spontaneous. It refers to the cathode's oxidation and the anode's reduction. The change in Gibbs free energy in a galvanic cell, where a spontaneous redox reaction drives the cell to produce an electric potential, must be negative. The cell potential, which is positive when electrons flow freely through the electrochemical cell, is the polar opposite of this.

Q131. The non-carbonyl compound out of the following is:

• A. Option A: CH3-CO-CH3
• B. Option B: C2H5-CH(OH)-CH3✓
• C. Option C: CH3-C(NH2)=O
• D. Option D: CH3-C(OR)=O

Explanation: Carbonyl groups have a carbonyl functional group (C=O) which may be either aldehyde or ketone.

Q132. The empirical formula of the compound was found to be 𝐶𝐻2𝑂. If the molar mass of the compound is 150𝑔/𝑚𝑜𝑙. The molecular formula of the compound is:

• A. C6H12O6
• B. C4H8O2
• C. C5H10O4
• D. C5H10O5✓

Explanation: The molecular formula is given byMolecular formula = n x empirical formulaWhere n= molecular formula mass/empirical formula massThe empirical formula mass of CH2O is =12+2+16=30g/mol.n = 150/30=5. So Molecular formula is Molecular formula = 5×CH2O = C5H10O5

Q133. The volume of 𝐶𝑂2 produced by heating 33.5𝑔 𝐿𝑖2𝐶𝑂3 at room temperature and pressure is (𝑀𝑟 𝐿𝑖2𝐶𝑂3 = 67𝑔/𝑚𝑜𝑙):

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

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

Q134. The equation used to describe the behavior of ideal gases under standard conditions is:

• A. P𝑉 = 𝑛𝑅T✓
• B. 𝑃𝑀 = 𝑑𝑅T
• C. 𝑃𝑉𝑀 = 𝑚𝑅T
• D. All of the above

Explanation: The correct equation to describe the behavior of ideal gases under standard conditions is PV = nRT. This equation is fundamental in chemistry and physics, relating the pressure (P), volume (V), temperature (T), and moles of gas (n) with the universal gas constant (R). The other options represent derived forms of the ideal gas equation under specific conditions, such as when considering density (Option B) or involving mass (Option C), but they are not the primary form of the ideal gas law.

Q135. The nuclei you think is invisible in NMR spectroscopy is:

• A. N14✓
• B. P31
• C. Cl35
• D. C13

Explanation: NMR spectroscopy requires nuclei with non-zero spin to be detected. Cl³⁵ has a quadrupolar nucleus with spin 3/2 and often gives broad, weak, or undetectable signals, making it practically "invisible" in routine NMR. Nuclei like C¹³, H¹, N¹⁴, and P³¹ are commonly detectable.

Q136. The amount of methane in biogas is approximately:

• A. 10-30%
• B. 50-90%
• C. 50-75%✓
• D. 60-75%

Explanation: Biogas is predominantly composed of methane (CH4) and carbon dioxide (CO2). The methane content typically ranges from 50% to 75% by volume. Option C accurately reflects this range. Option A suggests a much lower methane content than is typical, Option B overestimates the possible upper limit, and Option D does not encompass the full typical range, omitting the lower end.

Q137. By the absorption of visible light, which of the following compounds gives smogy air and has brown tint?

• A. NO
• B. 𝑆𝑂3
• C. 𝑁𝑂−3
• D. 𝑁𝑂2✓

Explanation: Nitrogen dioxide absorbs radiation in the ultraviolet (UV) and visible (vis) regions due to the electron transitions within the nitrogen dioxide molecule. It is a pungent gas that, along with fine airborne particulate matter, contributes to the reddish-brown haze characteristic of smoggy air.

Q138. Regarding reactivity of the compounds having carbonyl group. The most reactive compound out of the following is:

• A. CH3CH2COCH3
• B. C2H5COOH
• C. CH3COH✓
• D. CH3COOH

Explanation: Compound C is the most reactive because it has a simple carbonyl group with minimal electron donating influence, enhancing its electrophilic nature.

Q139. Whenever nitrile group is hydrolysed (𝑑𝑖𝑙 𝐻𝐶𝑙) with water it always produces:

• A. Alcohol
• B. Carboxylic acid✓
• C. Amines
• D. Amides

Explanation: Nitrile gets hydrolyzed in two steps; amides are formed first. While in the second step, an ammonium salt of a carboxylic acid is formed. For example, Ethanenitrile on getting hydrolyzed gives ethanamide in the first step, while ammonium ethanoate in the second step.

Q140. Elements of which block of the periodic table shows variable oxidation state?

• A. The s-block elements
• B. The p-block elements
• C. The d and f block elements✓
• D. None of the blocks

Explanation: In the case of d-block elements due to the presence of electrons at d orbitals, is closer to the outermost shell of the metal. They show variable oxidation states. With increasing the number of electrons of the d orbitals (up to 5 electrons), the number of oxidation states increases. In the case of d electrons due to a lower effective nuclear charge of attraction, the electrons can be removed to form different oxidation states.Actinides are F-block elements(atomic number 89 to103) with the general electronic configuration of the outermost shell being [Rn]5f1,146d0,17s2. Where the last electron enters the inner 5f-orbital of the actinides. Actinides are also known as rare earth metals. Now according to the Aufbau principle(L+S value), the energy order of the orbitals should be 5f<6d<7s but due to the more diffuse orbitals, their energy becomes more or less the same. As a result, electrons can be excited easily. Due to this reason, actinides show a greater range of oxidation states. But if we consider lanthanides due to their comparatively small size of 4f orbital they have a limited number of oxidation states.

Q141. What is the strain energy per unit volume of a rod made from this metal when the strain of the rod is 0.010?

• A. 10 kJ m−3
• B. 100 kJ m−3
• C. 1.0 MJ m−3
• D. 10 MJ m−3✓

Explanation: To calculate the strain energy per unit volume (also known as strain energy density), you use the formula: strain energy per unit volume = (1/2) × stress × strain.Given that the strain is 0.010, and assuming the stress is calculated using Young's modulus of the material (for example, 2 × 109 Pa), the calculation would be:strain energy per unit volume = (1/2) × (2 × 109 Pa) × 0.010 = 10 × 106 J m−3 = 10 MJ m−3Thus, the correct answer is 10 MJ m−3. All other options are incorrect because they either underestimate or overestimate the strain energy density based on improper calculations or incorrect assumptions.

Q142. A diffraction grating is used to measure the wavelength of monochromatic light, as shown in the diagram. The spacing of the slits in the grating is 1.00 × 10−6m. The angle between the first-order diffraction maxima is 60.0°. What is the wavelength of the light?

• A. 287 nm
• B. 470 nm
• C. 574 nm✓
• D. 940 nm

Explanation: The explanation is given below:

Q143. A filament lamp has a resistance of 180 Ω when the current in it is 500mA. What is the power dissipated in the lamp?

• A. 45 W✓
• B. 90 W
• C. 290 W
• D. 360 W

Explanation: The power dissipated in an electrical component can be calculated using the formula P = I2R, where P is the power, I is the current, and R is the resistance. Given that the current (I) is 500 mA (or 0.5 A) and the resistance (R) is 180 Ω, we can calculate the power as follows: P = (0.5)2 × 180 = 0.25 × 180 = 45 W Therefore, the correct answer is 45 W. Other options are incorrect as they result from either using wrong calculation methods or misinterpretation of the question.

Q144. Orange light in a vacuum has a wavelength of 600 nm. What is the frequency of this light?

• A. 180 Hz
• B. 5.0 × 105 Hz
• C. 1.8 × 1011 Hz
• D. 5.0 × 1014 Hz✓

Explanation: Explanation is given in image.

Q145. A stationary sound wave has a series of nodes. The distance between the first and the sixth node is 30.0 cm. What is the wavelength of the sound wave?

• A. 5.0 cm
• B. 6.0 cm
• C. 10.0 cm
• D. 12.0 cm✓

Explanation: The following is the solution:

Q146. In Simple Harmonic Motion the acceleration of the particle is zero when its:

• A. Velocity is zero
• B. Displacement is zero✓
• C. Both velocity and displacement are zero
• D. Both velocity and displacement are maximum

Explanation: As we know; a ∝-x Where a=acceleration x=displacement As long as the pendulum in harmonic motion is moving, it covers a displacement and thus has an acceleration. But as soon as it stops and comes to rest, the velocity and change in velocity are zero thus the acceleration is also zero and the displacement covered by it is also zero. Acceleration depends upon the change in velocity, if either final or initial velocity is zero it doesn't mean that acceleration is zero.The velocity is maximum at the mean position and displacement is maximum at the extreme so this option can't be justified to be correct.

Q147. A typical mobile phone battery has an e.m.f. of 5.0 V and an internal resistance of 200 mΩ.What is the terminal P.D. of the battery when it supplies a current of 500 mA?

• A. 4.8 V
• B. 4.9 V✓
• C. 5.0 V
• D. 5.1 V

Explanation: Explanation is given in image.

Q148. Which combination of up (u) and down (d) quarks forms a neutron?

• A. u u u
• B. u u d
• C. u d d✓
• D. d d

Explanation: Quarks are fundamental and elementary particles, there are six flavors of quarks =up, down, top, bottom, charm, and strange.Down, bottom, and strange are the antiparticles of up, top, and charm respectively.Three of these are combined to make any particle such as the proton, neutron, etc.A neutron consists of two down quarks and one up quark.A proton consists of two up quarks and one down quark.Both protons and neutrons are baryons. Baryons are heavy subatomic particles made up of three quarks.

Q149. How many cubic nanometres, nm3, are in a cubic micrometer, μm3?

• A. 103
• B. 106
• C. 109✓
• D. 1012

Explanation: The following is the solution:

Q150. A man stands in a lift that is accelerating vertically downwards. Which statement describes the force exerted by the man on the floor?

• A. It is equal to the weight of the man.
• B. It is greater than the force exerted by the floor on the man.
• C. It is less than the force exerted by the floor on the man.
• D. It is less than the weight of the man.✓

Explanation: As the lift is accelerating vertically downward so (g-a). The apparent weight of a man would become less than the actual weight. F=W-Rma=mg-RR=mg-maR= m(g-a)If the lift would be moving upward then (g+a), both g and acceleration would be parallel F=R-Wma=R-mgma+mg=RR=m(a+g)In the case of acceleration, the weight and the exerted force can't be equal. The weight of man and exerted force will be equal if the lift is at rest or moving with constant speed. The floor would apply a force on a man equal to the weight of a man.

Q151. The efficiency of a heat engine working between the freezing point and the boiling point of water is near to:

• A. 50%
• B. 25%✓
• C. 12.5%
• D. 6.25%

Explanation: Boiling point of water= 100° =373K Freezing point of water=0° = 273K The formula for efficiency; =1-TCold/Thotx100 = 1- 273/373x100 =26.80% ≈25% NOTE: make sure the temperature values being used in the formula are in Kelvin. In the numerator, we always take the cold body's temperature. In the denominator, we always take the hot body's temperature.

Q152. Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together. What is the speed of the trucks after the collision?

• A. v/4✓
• B. v/2
• C. v
• D. 5v/4

Explanation: This problem involves an inelastic collision where two railway trucks stick together after colliding. The principle of conservation of momentum is used to solve for the final speed of the combined mass. Initially, the momentum of the system is calculated by considering the direction and magnitude of each truck's momentum:The momentum of the truck with mass 'm' moving at '2v' is 2mv, while the truck with mass '3m' moving at '-v' (opposite direction) has a momentum of -3mv. The total initial momentum therefore is 2mv - 3mv = -mv.After the collision, the trucks stick together, forming a combined mass of 4m. Let the final velocity of this combined mass be 'v_final'. According to conservation of momentum, the total momentum before the collision equals the total momentum after the collision:-mv = 4m × v_finalSolving for v_final gives v_final = -v/4. The negative sign indicates the direction is opposite to the initial direction of the truck with mass 'm', and the magnitude of the speed is v/4.Incorrect options:v/2: This assumes a different distribution of momentum that does not align with the conservation principle.v: This implies no change in velocity, which contradicts the nature of inelastic collisions.5v/4: This results from a miscalculation of momentum distribution, leading to an incorrect final speed.

Q153. A thin horizontal plate of area 0.036 m2 is beneath the surface of a liquid of density 930 kg m−3. The force on one side of the plate due to the pressure of the liquid is 290 N. What is the depth of the plate beneath the surface of the liquid?

• A. 0.88 m✓
• B. 1.1 m
• C. 1.8 m
• D. 8.7 m

Explanation: Pressure P = hρgwhere h is the depth, ρ is the density of the liquid and g is the acceleration of free fall.But,Pressure P = Force / Area = F / ATherefore,Depth h = P / ρg = F / (Aρg) = 290 / [0.036 (930) (9.81)] = 0.88m

Q154. An electron and a proton enter a magnetic field with equal velocities which one of them experiences more force:

• A. Electron
• B. Proton
• C. Both experience same force✓
• D. Cannot be predicted

Explanation: F=q(V×B) F=qvBsin The formula infers that F∝v As both proton and electron have the same velocity they will experience the same value of force. As both have opposite charges, therefore the force being applied would be the same in magnitude but opposite in direction. In simple words both particles are deflected with the same force but in the direction opposite to each other.

Q155. An electromagnetic wave travels in a straight line through a vacuum. The wave has a frequency of 6.0 THz. What is the number of wavelengths in a distance of 1.0 m along the wave?

• A. 5.0 × 10–5
• B. 2.0 × 101
• C. 2.0 × 104✓
• D. 5.0 × 107

Explanation: Explanation is given in image.

Q156. What is the magnitude of a point charge which produces an electric field of 2 N/C at a distance of 60 cm?

• A. 8 × 10−11C✓
• B. 2 × 10−12C
• C. 3 × 10−11C
• D. 6 × 10−10C

Explanation: The electric field (E) due to a point charge (q) at a distance (r) is given by the formula E = k * |q| / r2, where k is the electrostatic constant (approximately 8.99 × 109 N m2/C2). Here, E = 2 N/C and r = 60 cm = 0.6 m. Rearranging the formula to solve for |q|, we have |q| = E * r2 / k. Substituting the values, |q| = (2 N/C) * (0.6 m)2 / (8.99 × 109 N m2/C2) = 8 × 10−11C. Therefore, the correct answer is 8 × 10−11C. The other options fail to provide the correct magnitude as they either miscalculate or misinterpret the formula and given values.

Q157. The speed v of a liquid leaving a tube depends on the change in pressure ΔP and the density ρ of the liquid. The speed is given by the equation. where k is a constant that has no units. What is the value of n?

• A. 1/2✓
• B. 1
• C. 3/2
• D. 2

Explanation: Explanation is given in image.

Q158. A uniform horizontal footbridge is 12 m long and weighs 4000 N. It rests on two supports X and Y as shown. A man of weight 600 N is at a distance of 4 m from support X. What is the upward force on the footbridge from support X?

• A. 2200 N
• B. 2300 N
• C. 2400 N✓
• D. 2600 N

Explanation: As the total length of the bridge is 12m and a man of 600N weight is standing at a distance of 4 m from X. The torque acting at a 4m distance is τ1 =12F Nm. The distance of Y from the center of the bridge is 6m. The weight of the bridge acting from its center is 4000N.So, the torque acting at a 6m distance is 6(4000).=24000Nm Since the weight of the bridge is acting downward and the torque is clockwise and negative so τ2 = -24000NmThe distance of man from Y is 8m.So the torque acting at 8m distance is 8(600)= 4800NmSince the force due to man is acting downward and torque is clockwise and negative so τ3 =-4800 NmApplying 2nd condition of equilibrium;∑τ = 0τ1 + τ2 + τ3 =012F -24000 - 4800 =012F = 24000 + 480012 F = 28800F = 28800/12F = 2400N

Q159. A diameter 'd' and length 'l' wire hangs vertically from a fixed point. The wire is extended by hanging a mass 'M' on its end. The Young modulus of the wire is 'E'. The acceleration of free fall is 'g'. Which equation is used to determine the extension 'x' of the wire?

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

Explanation: Use the formula given:𝑥 =4𝑀𝑔𝑙 / 𝐸𝜋𝑑2

Q160. A sound wave has a frequency of 2500 Hz and a speed of 1500 m s−1. What is the shortest distance from a point of maximum pressure in the wave to a point of minimum pressure?

• A. 0.15 m
• B. 0.30 m✓
• C. 0.60 m
• D. 1.20 m

Explanation: Frequency=2500 hzWave speed=1500m/sNow wave speed = wave length*FrequencySo wavelength=Wave speed/FrequencyWavelength=1500/2500 =3/5 =0.6 mA sound wave is also known as a pressure wave. The wavelength can be said to be the distance between 2 consecutive maximum pressure points or the distance between 2 consecutive minimum pressure points.In the middle of 2 consecutive maximum pressure points, there is a minimum pressure point.So the wavelength between the point of maximum pressure in the wave to the point of minimum pressure is=0.6/2=0.3m. Hence the shortest distance between the point of maximum pressure in the wave to the point of minimum pressure is 0.3m.

Q161. The instantaneous current in a circuit is given by I = √2 sin(ωt + ∅) ampere what is the rms value of the current?

• A. 2 A
• B. √2 A
• C. 1 A✓
• D. (1/√2)A

Explanation: The RMS (Root Mean Square) value of a sinusoidal alternating current is given by dividing the peak current by √2. For the given current I = √2 sin(ωt + ∅) A, the peak current I0 is √2 A. Thus, the RMS current is IRMS = I0/√2 = √2/√2 = 1 A. Therefore, the correct answer is 1 A. The other options are incorrect because they either represent the peak current or a miscalculated RMS value.

Q162. Which list shows electromagnetic waves in order of increasing frequency?

• A. Radio waves → gamma rays → ultraviolet → infra-red
• B. Radio waves → infra-red → ultraviolet → gamma rays✓
• C. Ultraviolet → gamma rays → radio waves → infra-red
• D. Ultraviolet → infra-red → radio waves → gamma rays

Explanation: Electromagnetic waves don't require a medium to travel. They are 7 types, each of them having the same kind of vibrations but different frequencies and wavelengths and hence energies. Cosmic waves are not electromagnetic waves.

Q163. What is the momentum of X −rays having wavelength 0.001nm?

• A. 6.63 × 10−22 kg m s−1✓
• B. 6.63 × 10−27 kg m s−1
• C. 6.63 × 10−31 kg m s−1
• D. 6.63 × 10−34 kg m s−1

Explanation: We know that,λ= h / pp=6.26×10 −22 kgm/s

Q164. The maximum energy of the electrons released in photocell is independent of:

• A. Frequency of incident light
• B. Intensity of incident light✓
• C. Nature of cathode surface
• D. Wavelength of light

Explanation: The working of a photocell can be explained by the photoelectric effect.•The photoelectrons which are released when a beam of light falls on the cathode are independent of the intensity of light. •It does depend upon the frequency of incident light, the effect will only be observed if the incident light has the frequency of a specific value known as threshold frequency.•And should have a specific wavelength known as cutoff wavelength.• As long as the incident wave doesn't have the frequency and wavelength equal or greater than the required values no effect will be observed.• Different metals have different values of threshold frequencies so the nature of the cathode is also taken into consideration. •consider the following equationK.Emax= hv-work functionThis shows that the max kinetic energy of electrons depends upon frequency, wavelength, and work function and work function depends upon the nature of the metal being used as a cathode.So it's independent of the intensity of light.

Q165. A mass m is suspended from a spring of spring constant k. The angular frequency of oscillations of the spring is:

• A. k/m
• B. √[k/m]✓
• C. m/k
• D. √[m/k]

Explanation: Explanation is given in image.

Q166. Which expression using SI base units is equivalent to the volt?

• A. kg m2s−1A−1
• B. kg m s−2 A
• C. kg m2s−1A
• D. kg m2s−3A−1✓

Explanation: Explanation is given in image.

Q167. On a planet, a vertically-launched projectile takes 12.5 s to return to its starting position. The projectile gains a maximum height of 170 m. The planet does not have an atmosphere. What is the acceleration of free fall on this planet?

• A. 2.2 m s−2
• B. 8.7 m s−2✓
• C. 27 m s−2
• D. 54 m s−2

Explanation: Explanation is given below:

Q168. The ionization potential of a hydrogen atom is 13.6V what will be the energy of the electron in the second orbit?

• A. −10.2 eV
• B. −3.40 eV✓
• C. +3.40 eV
• D. −1.51 eV

Explanation: Explanation is given in image.

Q169. A man has a mass of 80 kg. He ties himself to one end of a rope which passes over a single fixed pulley. He pulls on the other end of the rope to lift himself up at an average speed of 50 cm s−1. What is the average useful power at which he is working?

• A. 40 W
• B. 0.39 kW✓
• C. 4.0 kW
• D. 39 kW

Explanation: Explanation is given in image.

Q170. A transverse wave travels along a rope. The graph shows the variation of the displacement of the particles in the rope with distance along it at a particular instant. At which distance along the rope do the particles have maximum upwards velocity?

• A. 0.5 m✓
• B. 1.0 m
• C. 1.5 m
• D. 2.0 m

Explanation: Since the wave is moving to the right, we can draw the position of the wave at the next instant by simply replicating the wave, but shifted to the right.In this way, we can know how each point has moved in the next instant. It can be seen that particles at both 0.5 m and 1.0 m will move upward at the next instant but the largest upward change in displacement occurs at a distance of 0.5 m.

Q171. Which of the following will be a better shield against 𝛾-rays?

• A. Ordinary water
• B. Heavy water
• C. Lead✓
• D. Aluminum

Explanation: Gamma rays are known to have maximum penetration they can penetrate several metals. The best shielding against them is elements with high atomic numbers and high density, such as lead. The atomic number of Lead is 82.Gamma rays travel at the speed of light, and they can travel thousands of meters in the air before spending their energy. They can be shielded by water but a huge amount would be required: 13.8 feet of water, about 6.6 feet of concrete, or about 1.3 feet of Lead is required to shield against gamma rays. Aluminium is less dense and gamma rays can penetrate through them completely.

Q172. In a uniform electric field, which statement is correct?

• A. All charged particles experience the same force
• B. All charged particles move with the same velocity
• C. All electric field lines are directed towards positive charges
• D. All electric field lines are parallel✓

Explanation: In case of the uniform electric field, the two electric field lines are1. straight2. parallel3. equidistantA uniform electric field is a field in which the value of the field strength remains the same

Q173. A metal cube with sides of length “𝑎” has electrical resistance 𝑅 between opposite faces. What is the resistance between the opposite faces of a cube of the same metal with sides of length 3𝑎?

• A. 9𝑅
• B. 3𝑅
• C. R/3✓
• D. 𝑅/9

Explanation: the resistance can be found by the relation R=pL/A p(resistivity) In this case L=a and A=a2 soR=p 1/aWhen L=3a then A=9a2So R=p 1/3a Hence 1/3 times

Q174. Which particle is a fundamental particle?

• A. Electron✓
• B. Hadron
• C. Neutron
• D. Proton

Explanation: •An electron is a kind of lepton which is considered a fundamental or elementary particle because they are not further divisible.•Hadrons are the heaviest particles; they are further composed of 2 quarks.•Protons and neutrons are baryons which are further composed of 3 quarks

Q175. The nuclear equation shown has a term missing. What is represented by the missing term?

• A. An antineutrino✓
• B. An electron
• C. A neutrino
• D. A positron

Explanation: Antineutrinos are produced in nuclear beta decay together with a beta particle (in beta decay a neutron decays into a proton, electron, and antineutrino).

Q176. Two wave sources are oscillating in phase. Each source produces a wave of wavelength 𝜆. The two waves from the sources meet at point 𝑋 with a phase difference of 90°.What is a possible difference in the distances from the two wave sources to point 𝑋?

• A. 𝜆/8
• B. 𝜆/4✓
• C. 𝜆/2
• D. 𝜆

Explanation: The two given sources are in phase. A distance equal to the wavelength corresponds to a phase difference of 360, this means that they are in phase. When two things are in phase it infers that they have no phase difference.This shows us that a phase difference of 90 corresponds to a distance equal to 𝜆/4.

Q177. The activity of a certain nuclide is governed by the relation ∆𝑁/∆𝑡= −𝜆𝑁 where 𝜆 = 2.4 × 10−8𝑠−1. What is the half-life of the nuclide?

• A. 2.9 × 107𝑠✓
• B. 1.3 × 107𝑠
• C. 1.2 × 10−8𝑠
• D. 3.4 × 10−8𝑠

Explanation: Using the following formulaT1/2= 0.693/decay constant

Q178. Four resistors are connected in a square as shown. The resistance may be measured between any two junctions. Between which two junctions is the measured resistance greatest?

• A. P and Q
• B. Q and S✓
• C. R and S
• D. S and P

Explanation: Formula for total resistance in a parallel circuit will be used 1/Rt=1/R1 + 1/R2

Q179. The power loss 𝑃 in a resistor is calculated using the formula: 𝑃 =𝑉2/R The uncertainty in the potential difference 𝑉 is 3% and the uncertainty in the resistance 𝑅 is 2%. What is the uncertainty in 𝑃?

• A. 4%
• B. 7%
• C. 8%✓
• D. 11%

Explanation: •The fractional error in V would be3×2=6•The fractional error in R would be2•So the fractional error of power would be equal to the sum of the fractional errors of both potential difference and resistance;6+3=8•NOTE= The given % error is multiplied by the power as in case of potential difference.

Q180. A quantity 𝑥 is to be determined from the equation, 𝑥 = 𝑃 – 𝑄. 𝑃 is measured as (1.27 ± 0.02) 𝑚 and 𝑄 is measured as (0.83 ± 0.01) 𝑚. What is the percentage uncertainty in 𝑥 to one significant figure?

• A. 0.4%
• B. 2%
• C. 3%
• D. 7%✓

Explanation: The following is the solution:•The first figures are to be subtracted:1.27-0.83=0.44•The second figures (least counts) are to be added:0.02+0.01=0.03•Now the results from both steps are divided:=(0.03/0.44)x100=6.82% ≈ 7%Hence, option D is correct.

Q181. The number of electrons in one coulomb of charge are:

• A. 6.25 × 1021
• B. 1.6 × 1019
• C. 6.25 × 1018✓
• D. 9.1 × 1031

Explanation: 6.25×1018 are the number of electrons in one coulomb of charge.

Q182. Which of the following series lie in the visible region?

• A. Lyman
• B. Paschen
• C. Balmer✓
• D. Pfund

Explanation: Four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths between 400nm−700nm.The Lyman, Balmer, Paschen, and Pfund series are terms used to describe different series of spectral lines in the emission spectrum of hydrogen and other elements. These series correspond to transitions of electrons between different energy levels within the atom.1. **Lyman Series:** The Lyman series is a series of spectral lines in the ultraviolet region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 1 energy level (ground state) of the hydrogen atom. The lines in this series are named after the scientist Theodore Lyman who studied them.2. **Balmer Series:** The Balmer series is a series of spectral lines in the visible region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 2 energy level of the hydrogen atom. The lines in this series were discovered by Johann Balmer.3. **Paschen Series:** The Paschen series is a series of spectral lines in the infrared region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 3 energy level of the hydrogen atom. The lines in this series were studied by Friedrich Paschen.4. **Pfund Series:** The Pfund series is a series of spectral lines in the infrared region of the electromagnetic spectrum. It corresponds to transitions of electrons from higher energy levels to the n = 4 energy level of the hydrogen atom. The lines in this series were observed by August Herman Pfund.These series of spectral lines occur because of the quantized nature of energy levels in atoms. When electrons transition from higher energy levels to lower energy levels, they emit photons of specific energies, which correspond to specific wavelengths or frequencies of light. Each series corresponds to transitions ending at a specific lower energy level, and the lines within each series represent different possible paths for electrons to take while transitioning to that level.These spectral series provided crucial insights into the quantization of energy in atoms and played a significant role in the development of quantum mechanics.

Q183. Two springs of spring constant 𝐾1 and 𝐾2 are arranged in parallel and a body of mass 𝑚 is attached to it then calculate the time period of the system:

• A. T = 2π√(m/(K1 + K2))✓
• B. T = 2π√(2m/(K1 + K2))
• C. T = 2π√(m/(K1 * K2))
• D. T = 2π√((K1 + K2)/m)

Explanation: T=2π(M/K)½T=2π(M/K1+K2)½

Q184. To determine the resistance of a voltmeter by discharging a capacitor through it, the instantaneous voltage is then given by the relation:

• A. V(t) = V0e-t/RC✓
• B. V(t) = V0(1 - e-t/RC)
• C. V(t) = V0et/RC
• D. V(t) = V0t/RC

Explanation: Option A is the correct option since Voe-t/RC gives the correct instantaneous voltage.

Q185. The amount of heat required to raise the temperature of 10 moles of water from 70𝐾 to 80𝐾 (molar heat capacity of water 75.24𝐽) is:

• A. 0.7524𝐽
• B. 7524𝐽✓
• C. 95.24𝐽
• D. 752.4𝐽

Explanation: We use the formula E= nC x (change in temperature) i.e 10 x 75.24 x 10 = 7524 J

Q186. To determine Young’s modulus of a material of a given wire of length 𝐿 we use:

• A. Melda's Apparatus
• B. Young’s Apparatus
• C. Searle’s Apparatus✓
• D. Cavendish Apparatus

Explanation: •Searle's apparatus is used for the measurement of Young's modulus. It consists of two equal-length wires that are attached to a rigid support.•Melde's Apparatus is a simple way to introduce students to the concept of standing waves. The apparatus consists of a string and an oscillator to generate different frequencies. Melde's experiment is ideal to study the behavior of standing waves. Supplied with accessories like pulley and clamp, scale pan, and cord.•Young's modulus apparatus is used to measure Young's modulus of a bar, not a wire.•By measuring the angle of the rod and knowing the twisting force (torque) of the wire for a given angle, Cavendish was able to determine the force between the pairs of masses, by using the Cavendish apparatus.

Q187. An electron is projected horizontally from south to north in a uniform horizontal magnetic field acting from west to east. The direction along which it will be deflected is:

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

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

Q188. When we are measuring the internal resistance of a cell by potentiometer, the emf of the battery must be greater than the :

• A. Emf of the cell✓
• B. P.D in the circuit
• C. Current in the cell
• D. Current in the circuit

Explanation: The emf of the battery in the potentiometer experiment should be greater than the emf of the cell to be determined because The fall of potential across the potentiometer wire should not be less than the emf of the cell to be determined. And the potential across the wire is determined by the emf of the battery. If either the emf of the battery or the potential difference across the whole potentiometer wire is lesser than the emf of the experimental cell, the balance point will not be obtained. The balance point indicates that the emf of the cell to be determined is just balanced by the voltage drop that appears across the wire.

Q189. The Young’s modulus of a given rod of uniform length 𝐿 is given by the relation:

• A. 𝐹𝐿/𝐴
• B. 𝐹𝐴/L
• C. F𝐿/𝜋𝑟2𝑙✓
• D. F𝑙/𝜋𝑟2L

Explanation: Young modulus=stress/strain FL/Al where l is the extension

Q190. The inward and outward electric flux from a closed surface is respectively 8×103units and 4×103 units, then the net charge inside the closed surface is:

• A. A:+4×103 ε0
• B. B:-4×103 ε0✓
• C. C:0 ε0
• D. D:+8×103 ε0

Explanation: Flux2-flux1= q/epsilon notQ= (4×103 - 8×103) epsilon

Q191. A radioactive isotope has a half-life of 3 days. The time after which its activity is reduced to 6.25% of its original activity is:

• A. 6 days
• B. 8 days
• C. 12 days✓
• D. 16 days

Explanation: 50% in 3 days25% in 6 days12.5% in 9 days6.25% in 12 days

Q192. There are two charges +2𝜇𝐶 and -3𝜇𝐶. The ratio of the force acting on them will be:

• A. 3 ∶ 1
• B. 1 ∶ 1✓
• C. 11 ∶ 8
• D. 3 ∶ 8

Explanation: Electrostatic force is a mutual force that will not act on a single charge.

Q193. Two radioactive samples 𝑆1 and 𝑆2 have half-lives of 3 hours and 7 hours respectively. If they have the same activity at certain instant 𝑡, what is the ratio of the number of atoms of 𝑆1 to 𝑆2 at instant 𝑡?

• A. 9: 49
• B. 49: 9
• C. 3: 7✓
• D. 7: 3

Explanation: Explanation is given below:

Q194. The reciprocal of resistivity is called:

• A. Resistance
• B. Inductance
• C. Conductivity✓
• D. Flexibility

Explanation: Conductivity is the measure of a material's ability to conduct electric current. It is the reciprocal of resistivity.The higher the conductivity, the better the material is at allowing the flow of electric current

Q195. Doubly ionized atoms 𝑋 and 𝑌 of two different elements are accelerated through the same P.D. on entering a uniform magnetic field they describe circular paths of radii 𝑅1 and 𝑅2. The masses of 𝑋 and 𝑌 are in the ratio of:

• A. R1: 𝑅2
• B. 𝑅2: 𝑅1
• C. R21 : 𝑅22✓
• D. R22: 𝑅21

Explanation: Refer to the following

Q196. The rest mass of Photon is 𝑚𝑜. Its linear momentum, when it moves with the speed equal to half of the speed of light in space, will be:

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

Explanation: Refer to the following.

Q197. The charge on the electron and proton is reduced to half. If the present value of Rydberg constant is 𝑅, then the new value of Rydberg constant will be:

• A. 𝑅/2
• B. 𝑅/4
• C. 𝑅/8
• D. 𝑅/16✓

Explanation: Explaination is given in the image.

Q198. The paratrooper of mass 80 𝑘𝑔 descends vertically at a constant velocity of 3 𝑚 𝑠−1taking the acceleration of free fall as 10 𝑚 𝑠−2. Find out what is the net force acting on him?

• A. Zero✓
• B. 800 𝑁 upward
• C. 800 𝑁 downward
• D. 240 𝑁 downward

Explanation: The correct answer is zero net force. Since the paratrooper is descending at a constant velocity, there is no acceleration (Newton's first law). This means the forces are balanced: the gravitational force (weight) is exactly balanced by the upward force from the parachute. Thus, the net force is zero. Other options are incorrect because they suggest a net force which would imply acceleration, conflicting with the situation of constant velocity.

Q199. The dimensional formula for change in momentum is same as that for:

• A. Force
• B. Impulse✓
• C. Acceleration
• D. Velocity

Explanation: The following is the solution:Impulse can be defined as “The sudden force acting on an object for a short interval of time”.It is equal to ∆PThe units of both momentum and impulse are the same so their dimensional formulas are also same.

Q200. What is NOT true of 2 forces that give rise to a couple?

• A. They act in opposite directions
• B. They both act at the same point✓
• C. They both act on the same body
• D. They both have the same magnitude

Explanation: Two equal and opposite parallel forces not acting along the same line form a couple.