Etea Mdcat 2019 — Solved Past Paper with Answers

Q1. Sadia wore her rain boots; _ her feet stayed dry during the storm.

• A. However
• B. Therefore✓
• C. On the other hand
• D. Still

Explanation: This is the correct answer. "Therefore" is used to show a logical consequence or conclusion. In this context, wearing rain boots logically leads to the result that Sadia's feet stayed dry during the storm.

Q2. Anum asked me, "Did you see the drama on television last night?" [Choose the correct indirect speech]

• A. Anum asked me whether I saw the drama on television the earlier night.
• B. Anum asked me whether I had seen the drama on television the earlier night.✓
• C. Anum asked me did I see the drama on television the last night.
• D. Anum asked me whether I had seen the drama on television the last night.

Explanation: Interrogative sentences beginning with an auxiliary verb are changed into indirect speech using the connective if or whether. The reporting verb said (or any other word used as the reporting verb) changes to asked, queried, questioned, demanded of, or enquired of in the indirect speech. Note that of is used after enquired and demanded only when the reporting verb has an object. So B will be the correct option.

Q3. Do not make so much noise, Farrah _ to study for her ESL test!

• A. Try
• B. Tries
• C. Tried
• D. Is trying✓

Explanation: This is the correct answer. "Is trying" is the present progressive (continuous) tense form of the verb "try," used to indicate an action that is currently happening. In this context, it shows that Farrah is currently making an effort to study for her ESL test, which is in line with the instruction not to make noise.

Q4. Zara changed the flat tire. [Choose the correct voice]

• A. The flat tire was changed by Zara✓
• B. The flat tire is changed by Zara
• C. The flat tire has been changed by Zara
• D. The flat tire had changed by Zara

Explanation: In the past, the passive voice uses the verbs was and were + past participle of the main verb.

Q5. The simplest form of learning is

• A. Imprinting
• B. Insight learning
• C. Latent learning
• D. Habituation✓

Explanation: The simplest form of learning is nonassociative learning such as habituation and sensitization.

Q6. Sorry she can't come to the phone. She _ a bath!

• A. Is having✓
• B. Having
• C. Have
• D. Has

Explanation: This option is correct. "Is having" is the present progressive (continuous) tense form of the verb "have," used to indicate an action that is currently happening. In this context, it means that the person is currently in the process of taking a bath, which prevents them from coming to the phone.

Q7. Choose the word nearest in meaning to ENIGMA"

• A. Evaluation
• B. Puzzle✓
• C. Answer
• D. Account

Explanation: Enigma means a person or thing that is mysterious or difficult to understand So the answer should be PUZZLE as it is difficult to understand.

Q8. When I went back to my hometown three years ago, I found that a lot of changes _

• A. Had taken place✓
• B. Have taken place
• C. Are taken place
• D. Were taken place

Explanation: This option is incorrect because "have taken place" is the present perfect tense, which indicates actions that started in the past and have relevance to the present. It does not fit the context of the sentence, which is discussing a specific past event (going back to the hometown three years ago).

Q9. Choose the correct sentence indicating facts and habits

• A. He is clever and he lacks experience.✓
• B. He is clever but he is lacking experience
• C. He is clever but he is lacked experience
• D. He is clever but he is lack experience

Explanation: The simple present tense is used to describe facts and habits. Option A is correct as it is a fact that he is clever and he lacks experience.

Q10. Look! A hamster _ by cat

• A. Has been chased
• B. Was being chased
• C. Is being chased✓
• D. Is chased

Explanation: The word LOOK with an exclamation mark tells that the cat is chasing the hamster right now so the present continuous tense will be used.

Q11. Choose the word opposite in meaning to "VOCIFEROUS"

• A. Silent✓
• B. Boisterous
• C. Blatant
• D. Noisy

Explanation: ociferous means expressing or characterized by vehement opinions; loud and forceful. Boisterous means noisy, cheerful and energetic. Blatant means have done openly and unashamedly. So the correct option is A as it is the opposite of Vociferous.

Q12. I'm sorry the house is not available any longer, It _ to timber tycoon

• A. Was being sold
• B. Will be sold
• C. Is sold
• D. Has been sold✓

Explanation: Has been sold” will be used here as the event took place in the past.

Q13. I always liked to lean _ the side of mercy

• A. Over
• B. On✓
• C. Towards
• D. About

Explanation: It should be lean on as it is taken from the quote

Q14. They already _ some of the old ones and _ them more comfortable

• A. Repair, make
• B. Repaired, made✓
• C. Repaired, make
• D. Repair, made

Explanation: The sentence involves two actions that have already been completed. Therefore, both verbs should be in the past tense."Repaired" is the past tense of "repair.""Made" is the past tense of "make."

Q15. I was born in Peshawar but I _ most of my childhood in the Mardan

• A. Spends
• B. Have spent✓
• C. Was spending
• D. Is spending

Explanation: The sentence talks about a completed period in the past (childhood) with relevance to the present, which is best expressed using the present perfect tense "have spent."The present perfect is often used for an action that started at some time in the past and is still continuing now. So the correct option will be B.

Q16. _ you win first place. you will receive a prize

• A. Whenever
• B. If✓
• C. Unless
• D. So forth

Explanation: You use if in conditional sentences to introduce the circumstances in which an event or situation might happen, might be happening, or might have happened. As this sentence is conditional IF is used.

Q17. The train was _

• A. Halt
• B. Halted✓
• C. Had halted
• D. Has halted

Explanation: The past is used to describe things that have already happened. So B is the correct answer

Q18. He confided _ me

• A. About
• B. In✓
• C. On
• D. Of

Explanation: The phrasal verb "confide" usually goes with the preposition "in", and it is actually an idiom

Q19. He said. ''you need not wait". Choose the correct indirect speech

• A. He said that I need not wait.✓
• B. He said you needed no wait.
• C. He said that wait was not need by you
• D. He said that you must not wait

Explanation: In the direct speech, "He said" is the reporting verb indicating that someone is conveying information or a statement. The subject "He" remains the same in indirect speech, but the verb "said" changes according to the context and tense. The direct statement "you need not wait" is directly quoted but needs to be transformed into reported speech. In indirect speech, the pronoun "you" changes according to who is reporting the statement. Here, it changes to "I" because it's now the narrator reporting what was said. The verb "need" remains the same.

Q20. It is natural for us to exert _ our own success

• A. In
• B. At✓
• C. Against
• D. Regarding

Explanation: Exert means to apply force. And in this question, it is natural for us to exert AT our own success.

Q21. The vagus nerve is part of which system?

• A. Sympathetic nervous system
• B. Parasympathetic nervous system✓
• C. Somatic nervous system
• D. None of the above options are correct

Explanation: The vagus nerve,also known as the vagal nerves are the main nerve of the parasympathetic nervous system.This system controls specific body functions such as digestion,heart rate,and immune system.

Q22. The genome of influenza virus is made up of _.

• A. Single strand RNA✓
• B. Double strand DNA
• C. Single strand DNA
• D. Double stranded RNA

Explanation: Influenza viruses belong to the Orthomyxoviridae family and have a segmented, negative-sense, single-stranded RNA genome. The influenza virus genome consists of multiple RNA segments, typically eight in the case of influenza A and B viruses. The negative-sense RNA means that the RNA strand cannot be directly translated by the host cell machinery. Instead, the virus carries its own RNA-dependent RNA polymerase to transcribe the negative-sense RNA into positive-sense RNA, which can then serve as a template for protein synthesis.

Q23. Galantamine hydrobromide is a compound derived from:

• A. Cannabis
• B. Coca
• C. English yew
• D. Daffodil✓

Explanation: Galantamine Hydrobromide is a tertiary alkoid that has been extracted from plant sources and is now synthesized for use in the treatment of mild to moderate Alzeimer’s disease. It is isolated from the bulbs and flowers of Galanthus nivalis (Common snowdrop) and some other members of the family Amaryllidaceae, such as Narcissus (daffodil).It can also be produced synthetically.

Q24. Mark the correct match:

• A. Haemophilia - Blood cancer
• B. SA node - Pacemaker✓
• C. ECG - Brain
• D. Alpha cell - Insulin

Explanation: Haemophilia is a disorder and not blood cancer, EEG is only a test that detects abnormalities in your brain waves, or in the electrical activity of your brain, Insulin is produced by Beta cells not Alpha. Whereas, The SA node is often referred to as a natural pacemaker because it generates a series of electrical pulses at regular intervals. Thus B is the correct option.

Q25. Cells which kills cells that display foreign motifs on their surface are:

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

Explanation: T cells are direct fighters of foreign invaders. Also called cytotoxic T cell and cytotoxic T lymphocyte.

Q26. Chitin is a

• A. Lipoprotein
• B. Polysaccharides✓
• C. Glycoprotein
• D. Phospholipids

Explanation: Chitin is a modified polysaccharide that contains nitrogen

Q27. Organization of photosynthetic pigment into clusters is:

• A. Photosynthesis
• B. Photosystem✓
• C. Photosynthetic cluster arrangement
• D. Calvin system

Explanation: The organization of photosynthetic pigments into clusters is referred to as a "photosystem." A photosystem is a complex of pigments and proteins that work together to capture light energy and initiate the process of photosynthesis. There are two main types of photosystems in photosynthetic organisms: Photosystem I (PSI) and Photosystem II (PSII).

Q28. Amphibians are poikilotherms, therefore they use to hibernate in

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

Explanation: Poikilotherm is an organism with a variable body temperature that is usually slightly higher than the temperature of its environment : a cold-blooded organism—called also heterotherm.Poikilothermic animals include types of vertebrate animals, specifically some fish, amphibians, and reptiles, as well as many invertebrate animals.Poikilothermic,or cold blooded animals face a risk of death due to cold or freezing over the winter so they go into hibernation during winter season.

Q29. All of the following are macronutrient except

• A. Cu ions✓
• B. Ca ions
• C. Mg ions
• D. K ions

Explanation: Sixteen elements have been found essential for plant growth.Nine of these are required in fairly large quantities and are therefore known as macronutrients.These include carbon,hydrogen,oxygen,nitrogen,phosphorus,potassium,sulfur,calcium and magnesium. The remaining seven elements are needed in trace or small amounts for normal plant growth and development that are known as micronutrients.These include iron,boron,manganese,copper,molybdenum,chlorine and zinc.

Q30. Purkinje fibres are connected with the impulse conducting system of:

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

Explanation: The Purkinje fibers are located in the inner ventricular walls of the heart, just beneath the endocardium in a space called the subendocardium. The Purkinje fibers are specialized conducting fibers composed of electrically excitable cells. It is the extension of the autonomic nervous system and first discovered by Jan Avenglesta Purkine in 1839.

Q31. The alveoli represent total surface area of

• A. 10-30m
• B. 30-60m
• C. 70-90m✓
• D. 90-110m

Explanation: The alveoli form the gas exchange surface.There are about 700 million alveoli present in the lungs,representing approximately a total surface area of 70-90m2.

Q32. Some marine fishes possess salt excreting organs known as

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

Explanation: The rectal gland, The salt gland is an organ for excreting excess salts. It is found in the cartilaginous fishes subclass elasmobranchii (sharks, rays, and skates), seabirds, and some reptiles.

Q33. Tetanus is the infection of

• A. Respiratory system
• B. Nervous system✓
• C. Circulatory system
• D. Bones and muscles

Explanation: Tetanus is an infection of the nervous system with the potentially deadly bacteria Clostridium tetani. Spores of the bacteria C.tetani live in the soil and are found around the world.

Q34. _ regulate the body temperature

• A. Hypothalamus✓
• B. Thalamus
• C. Hippocampus
• D. Amygdala

Explanation: Our internal body temperature is regulated by a part of our brain called the hypothalamus.

Q35. A man has to face interview, but during his first five minutes before the interview he experiences sweating, increased heart rate and respiration. Which hormone is responsible for his restlessness

• A. Adrenocorticotropic hormone
• B. Insulin and glucagon
• C. Epinephrine and norepinephrine✓
• D. Aldosterone

Explanation: Epinephrine and norepinephrine are the hormones behind your “fight-or-flight” response which involves increased heart rate and heavy breathing. thus C is the correct option.

Q36. Hypothalamus connected to pituitary gland via:

• A. Nerves
• B. Infundibulum✓
• C. Blood
• D. No connection

Explanation: Hypothalamus and the Pituitary gland is connected by the pituitary stalk, or more technically, the infundibulum

Q37. Second meiotic division in oocyte is completed:

• A. When oocyte is fertilized by sperm✓
• B. When ovum is discharged from the ovary
• C. Just before fertilization
• D. Before the onset of menstruation

Explanation: In human beings, the secondary oocyte is released from the mature Graafian follicle of an ovary (ovulation). The oocyte is received by the nearby Fallopian funnel and sent into the Fallopian tube by movements of fimbriae and their cilia. The secondary oocyte can be fertilized only within 24 hours after its release from the ovary. The secondary oocyte is surrounded by numerous sperms but only one sperm succeeds in fertilizing the oocyte. Since, the second meiotic division is in progress, so the sperm enters the secondary oocyte. Second meiotic division is completed by the entry of the sperm into the secondary oocyte (fallopian tube). After this secondary oocyte is called ovum (egg).

Q38. The enzyme which is found in saliva accelerates the conversion of starch into sugar is:

• A. Pepsin
• B. Thrombin
• C. Ptyalin✓
• D. Fumarase

Explanation: Amylase is the enzyme in saliva that digests starch and ptyalin is an amylase

Q39. A pure breeding tall plant was crossed to a dwarf plant. What would be the probability of "Tt" genotype n F2?

• A. 0
• B. 0.25
• C. 0.5✓
• D. 0.75

Explanation: probability of "Tt" genotype in F2 2/4 = 0.5

Q40. The number of human spinal nerves is

• A. 60
• B. 62✓
• C. 64
• D. 66

Explanation: The Peripheral Nervous System (PNS) consists of 12 cranial nerves, and 31 pairs of spinal nerves.Here the number is asked so 31+31=62

Q41. Diphtheria vaccines is an example of:

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

Explanation: Toxoid vaccines contain a toxin or chemical made by the bacteria or virus. They make you immune to the harmful effects of the infection, instead of to the infection itself. Examples are the diphtheria and tetanus vaccines.

Q42. Which one of the following items gives its correct total number?

• A. Cervical vertebrae -7✓
• B. Floating ribs in human - 3
• C. Auditory ossicles - 8
• D. Cranium bones - 4

Explanation: There are two pairs of floating ribs, The ossicles are three bones, There are eight cranial bones, The cervical spine is comprised of seven cervical vertebrae. Thus, A is the correct option.

Q43. Find mismatch

• A. Thyroid gland-T3 and T4
• B. Parathyroid gland-Calcitonin✓
• C. Pancreas-Insulin
• D. Gonads-Testes and ovaries

Explanation: The gonads are the testes in the male and the ovaries in the female.T3 and T4 are the two major hormones made by the thyroid, The most important hormone that the pancreas produces is insulin. The parathyroid gland lies just behind the thyroid gland in the neck and produces PTH which appears to work against calcitonin. Thus, option B is the correct answer

Q44. To the end of the first trimesters, the embryo can now technically describe as a:

• A. Zygote
• B. Infant
• C. Toddler
• D. Fetus✓

Explanation: At the beginning of the 11th week of pregnancy, or the ninth week after conception, The baby's head still makes up about half of its length. However, The baby's body is about to catch up. The baby is now officially described as a fetus.

Q45. How many pairs of homologous chromosomes are present in Pisum sativum

• A. Seven pairs✓
• B. Eight pairs
• C. Nine pairs
• D. Ten pairs

Explanation: The total number of chromosomes in Pisum sativum (pea plant) is 14.

Q46. Ozone layer in upper atmosphere is being destroyed by:

• A. Chlorofluorocarbon
• B. Freon
• C. Smog
• D. Both a and b✓

Explanation: This option is correct. The ozone layer in the upper atmosphere can be depleted or destroyed by certain substances known as ozone-depleting substances (ODS). The primary culprits responsible for ozone layer depletion are chlorofluorocarbons (CFCs), halons, carbon tetrachloride, methyl chloroform, and bromine-containing compounds such as bromofluorocarbons (BFCs) and hydrochlorofluorocarbons (HCFCs).When these ozone-depleting substances are released into the atmosphere, they can rise to the stratosphere, where the ozone layer is located. There, they undergo photodissociation due to the absorption of high-energy ultraviolet (UV) radiation from the sun. This process releases chlorine and bromine atoms, which act as catalysts in the destruction of ozone molecules.The chlorine and bromine atoms can undergo catalytic reactions with ozone, breaking down ozone molecules and reducing the overall concentration of ozone in the ozone layer. This leads to the thinning or depletion of the ozone layer, particularly in regions like the ozone hole over Antarctica.Freon is a common name used to refer to a group of chlorofluorocarbon (CFC) compounds. CFCs were widely used as refrigerants, propellants in aerosol products, and solvents in various industrial applications.Unfortunately, CFCs like Freon, specifically chlorofluorocarbon-12 (CFC-12) or dichlorodifluoromethane (CCl2F2), have been identified as potent ozone-depleting substances (ODS). When released into the atmosphere, CFCs can eventually reach the stratosphere, where they undergo photodissociation due to the absorption of high-energy ultraviolet (UV) radiation. This process leads to the release of chlorine atoms, which act as catalysts in the destruction of ozone molecules.

Q47. The percentage of fresh water on earth is

• A. 1%✓
• B. 3%
• C. 5%
• D. 7%

Explanation: Water is the most important renewable resource of environment.Water covers about 70% of the surface of earth.It is the major constituent of the protoplasm of organism.Sea water constitute 97% of the total water which is not usable,2% is in frozen form and only 1% of water is available as freshwater (lakes,streams and rivers).

Q48. Recombinants contains DRNA from

• A. 2 different sources✓
• B. Single source
• C. 2 same sources
• D. 3 same sources

Explanation: Recombinant DNA technology is the joining together of DNA molecules from two different species. The recombinant DNA molecule is inserted into a host organism to produce new genetic combinations that are of value to science, medicine, agriculture, and industry.

Q49. The inner surface of kidney has a deep notch called

• A. Renal pelvis
• B. Hilus✓
• C. Medulla
• D. Pyramid

Explanation: The medial concave border of a kidney contains a notch known as hilus through which the renal artery enters and the renal vein and ureter leave the kidney.

Q50. _ is considered as chief structural and functional unit of nervous system

• A. Cell
• B. Neuron✓
• C. Nephron
• D. Brain

Explanation: Neurons are the structural and functional unit of the nervous system.

Q51. The bacteriophage replicates only inside a _.

• A. Animal cell
• B. Bacterial cell✓
• C. Fungal cell
• D. Animal and bacterial cell

Explanation: Bacteriophages replicate only inside a bacterial cell. Bacteriophages, or simply phages, are viruses that infect and replicate within bacteria. They are composed of genetic material (either DNA or RNA) surrounded by a protein coat called a capsid. Some bacteriophages also have an outer lipid envelope.

Q52. _ is stored in animal cells

• A. Starch
• B. Cellulose
• C. Sucrose
• D. Glycogen✓

Explanation: Animal cells store food in the form of glycogen. The excess of glucose in the blood is converted to glycogen in the muscle cells. The process of formation of glycogen is known as glycogenesis. During the muscle exercise when extra energy is needed the glycogen is broken down to glucose and channelized into the bloodstream by glycogenolysis.

Q53. A bacterium which has a group of two or more flagella inserted at one pole of the cell

• A. Monotrichous
• B. Peritrichous
• C. Lophotrichous✓
• D. Amphitrichous

Explanation: There are four types of flagellar arrangements based on the number and placement of flagella on the cell. These are, 1. Monotrichous - A single flagellum extends from one end of the cell. 2. Amphitrichous - A single flagellum or multiple flagella extend from two ends of the cell. 3.Lophotrichous - Tufts of flagella extend from one end or both ends of the cell. 4. Peritrichous - Multiple flagella randomly distributed over the entire bacterial cell. Thus, the correct answer is 'Lophotrichous.'

Q54. The gametophyte of lycopsida is mainly:

• A. Aerial
• B. Partly aerial and partly underground
• C. Underground✓
• D. Photosynthetic

Explanation: In the plant group Lycopsida, which includes clubmosses, spike mosses, and quillworts, the gametophyte generation is predominantly underground. The life cycle of these plants exhibits alternation of generations, with a distinct gametophyte and sporophyte phase. In Lycopsida, the sporophyte is the dominant and more visible phase. It is the familiar, upright plant that produces spores through sporangia. Spores are dispersed, and upon germination, they give rise to the gametophyte generation.

Q55. Opossum and koala bear belongs to sub-class

• A. Prototheria
• B. Eutheria
• C. Metatheria✓
• D. Monotremata

Explanation: Opossums and koala bears belong to subclass metatheria.

Q56. The form of immunity which inherit from mother is:

• A. Active immunity
• B. Passive immunity
• C. Acquired immunity
• D. Innate immunity✓

Explanation: During the last 3 months of pregnancy, antibodies from mothers are passed to their unborn babies through the placenta. This type of immunity is called innate immunity because the baby has been given antibodies of its mother.

Q57. The least toxic excretory product is

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

Explanation: Uric acid is the least toxic excretory product.

Q58. Chemically hormones are

• A. Carbohydrates
• B. PSteroidsroteins
• C. Steroids
• D. Both Proteins and Steroids✓

Explanation: Hormones are chemical substances secreted by cells that act to regulate the activity of other cells in the body.Hormones are synthesized and secreted by endocrine glands.Chemically,hormones are of three basic types i.e. steroids,amino acids or their derivatives,proteins or polypeptides.

Q59. DNA polymerase III works always in

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

Explanation: DNA polymerase can only make DNA in the 5’ to 3’ direction.

Q60. The biogas plant is tank which is

• A. 5-10 ft deep
• B. 10-15 ft deep✓
• C. 15-20 ft deep
• D. 20-25 ft deep

Explanation: The correct answer is b) 10-15 ft deep. This depth provides enough space for the efficient anaerobic digestion of organic waste and the production of biogas. The other options are incorrect because they either have depths that are too shallow or too deep for a typical biogas plant.

Q61. Which wave lengths are mainly absorbed by chlorophyll?

• A. Violet blue and red✓
• B. Green and blue
• C. Violet and orange
• D. Red and indigo

Explanation: Chlorophyll primarily absorbs light in the blue and red regions of the electromagnetic spectrum. The violet-blue and red wavelengths are most efficiently absorbed by chlorophyll, while green wavelengths are less effectively absorbed and are, instead, reflected, which is why chlorophyll-containing plants appear green to our eyes.

Q62. For hepatitis B the incubation period is between _.

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

Explanation: The incubation period of the hepatitis B virus ranges from 30 to 180 days. Which is 4 - 20 weeks. So, option A is correct.

Q63. Sulphur bacteria belongs to which group of bacteria called

• A. Beta-proteo bacteria
• B. Alpha-proteobacteria
• C. Gamma-proteo bacteria✓
• D. Delta proteo bacteria

Explanation: Sulfur bacteria belongs to gamma-proteo bacteria which take part in nitrogen and sulfur cycling processes

Q64. Nuclear mitosis occurs in

• A. Plants
• B. Animals
• C. Fungi✓
• D. Monrea

Explanation: Plants and Animals undergo regular mitosis, Monera reproduce asexually by binary fission while fungi undergo nuclear mitosis

Q65. Excess glucose is converted in the liver to glycogen in response to the hormone

• A. Glucagon
• B. Insulin✓
• C. Bile
• D. Both Glucagon and Insulin

Explanation: Bile is not related to the homeostasis of blood glucose concentration so option C is incorrect. The answer is insulin as it converts extra glucose to glycogen in the liver and muscle cells

Q66. During muscles relaxation, the calcium ions are:

• A. Released from sarcoplasmic reticulum into Sarcoplasm
• B. Forced back from the sarcoplasm to sarcoplasmic reticulum✓
• C. Further Forced from sarcoplasmic reticulum into sarcoplasm
• D. Neither released more nor forced back but remain constant

Explanation: When muscles relax, the calcium ions are forced / pumped back from the sarcoplasm into the sarcoplasmic reticulum so that the globular head can remove itself from the acting binding site

Q67. In male luteinizing hormone also known as

• A. ACTH
• B. ICSH✓
• C. TRF
• D. MSH

Explanation: LH (luteinizing hormone) stimulates testes to make testosterone, another name for this hormone is “ICSH” (interstitial cell stimulating hormone).

Q68. Particular amino acid and RNA molecule binds together by the action of an enzyme named

• A. tRNA synthetase
• B. Amino tRNA synthetase
• C. tRNA ligase
• D. Aminoacyl tRNA synthetase✓

Explanation: The enzyme associated with the binding of amino acids and RNA molecules is called “Aminoacyl tRNA synthetase”.

Q69. Lipid bilayer makes the membrane differently permeable barrier that allows the transport of:

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

Explanation: Lipids,proteins and carbohydrates are responsible for functional diversity of the membranes. Lipid bilayer makes the membrane differentially permeable barrier that allows the transport of non-polar materials across it and prevents ionic materials.

Q70. The following are sexual reproduction methods in bateria except

• A. Transformation
• B. Transduction
• C. Binary fission✓
• D. Conjugation

Explanation: In bacteria,genetic recombination takes place. Genetic recombination is primitive type of sexual reproduction. There are three types of genetic recombination: Transformation,Transduction, Conjugation.

Q71. Lichen is the symbiotic association of fungus with

• A. Bacteria
• B. Algae✓
• C. Other fungus
• D. Animals

Explanation: Symbiosis: Fungi develop symbiotic associations with other organisms e.g. Lichens and Mycorrhizae Lichen: It is a symbiotic association of a fungus with algae.

Q72. The possible reason(s) for cyanosis, one of congenital heart disease is:

• A. Formation of carboxy-hemoglobin✓
• B. The high concentration of oxyhemoglobin
• C. Low level of CO
• D. Low level of hemoglobin

Explanation: Cyanosis indicates there may be decreased oxygen attached to red blood cells in the bloodstream such as due to the formation of carboxy-hemoglobin. Carbon Monoxide has a higher affinity with hemoglobin and it doesn’t allow oxygen to bind to Hb molecules

Q73. The deficiency of which micronutrient causes goiter formation?

• A. Iron
• B. Zinc
• C. lodine✓
• D. Sodium

Explanation: Goiter is a condition in which the thyroid in neck enlarges, this happens mainly due to the lack of iodine in diet.

Q74. Phosphatases belong to which group of the following:

• A. Lyases
• B. Ligases
• C. Hydrolases✓
• D. None

Explanation: Phosphatases belong to the group of enzymes known as hydrolases. Specifically, they are a type of hydrolase called phosphohydrolases. Phosphatases catalyze the removal of phosphate groups from molecules through hydrolysis reactions. The removal of phosphate groups is a crucial regulatory step in various cellular processes, as it can activate or deactivate proteins and other molecules.

Q75. The ribosomes responsible for protein synthesis are present in the cell _.

• A. Floating in the cytosol
• B. Localized in the nucleus
• C. Bound to rough endoplasmic reticulum
• D. Both floating in the cytosol and bound to rough endoplasmic reticulum✓

Explanation: Ribosomes are the organelles that help in protein synthesis. Protein is required for many cell activities such as damage repair and other chemical processes. Ribosomes exist in two forms;either freely dispersed in cytosol or attached with a rough endoplasmic reticulum as tiny granules and are the site of protein synthesis.

Q76. _ enzyme needs a primer for the initiation of its function

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

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

Q77. The following histone proteins form a nucleosome complex except

• A. H1✓
• B. H2A
• C. H2B
• D. H3

Explanation: A nucleosome is the basic structural unit of DNA packaging in eukaryotes. The structure of a nucleosome consists of a segment of DNA wound around eight histone proteins. The nucleosome is the fundamental subunit of chromatin. Each nucleosome is composed of a little less than two turns of DNA wrapped around a set of eight proteins called histones, which are known as a histone octamer. Each histone octamer is composed of two copies each of the histone proteins H2A, H2B, H3, and H4.

Q78. The bond that is formed b/w two monosaccharide units is called

• A. lonic bond
• B. Hydrogen bond
• C. Peptide bond
• D. Glycosidic bond✓

Explanation: A glycosidic bond is a type of covalent bond which is formed between two units of monosaccharides

Q79. Which statement about chlorophyll is not true?

• A. It contains terminal carbonyl group✓
• B. It contains phyto tail
• C. It contains porphyrin ring
• D. It contains magnesium

Explanation: Chlorophyll is a complex molecule involved in photosynthesis, and it does contain a terminal carbonyl group. The carbonyl group is a functional group consisting of a carbon atom double-bonded to an oxygen atom. In chlorophyll, the carbonyl group is typically present in a phytol side chain. Chlorophyll molecules consist of a porphyrin ring system with a magnesium ion at the center. The phytol side chain, which contains the terminal carbonyl group, is hydrophobic and anchors the chlorophyll molecule in the thylakoid membrane of chloroplasts. The terminal carbonyl group is involved in the overall structure and function of chlorophyll, contributing to its interaction with light and its role in capturing energy during the light-dependent reactions of photosynthesis. Chlorophyll absorbs light energy, and the carbonyl group is part of the chromophore responsible for the absorption of specific wavelengths of light.

Q80. In humans the disease symptoms develop during the:

• A. Log phase✓
• B. Lag phase
• C. Growth phase
• D. Decline phase

Explanation: Bacterial cells divide at exponential rate during log phase, hence disease symptoms become prominent during this phase.

Q81. Independent gametophyte and sporophyte are found in:

• A. Selaginella
• B. Polytrichum
• C. Ectocarpus✓
• D. Liverworts

Explanation: Ectocarpus is a genus of brown algae, and it exhibits an alternation of generations life cycle, which involves both gametophytic and sporophytic phases. In the case of Ectocarpus, the life cycle includes an independent gametophyte and a sporophyte phase.

Q82. Tmesipteris is an example of:

• A. Horsetail
• B. Club mosses
• C. Psilopsida✓
• D. Pteropsida

Explanation: Tmesipteris is a genus of plants belonging to the class Psilopsida, which is commonly referred to as whisk ferns. Psilopsida is a class of primitive vascular plants that are characterized by their lack of true leaves and roots. They have simple, dichotomously branching stems and reproduce by spores.

Q83. The larva formed during the life cycle of annelida is

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

Explanation: Trochophore, also called trochosphere, small, translucent, free-swimming larva characteristic of marine annelids and most groups of mollusks and it is commonly formed during life cycle of annelida

Q84. Ebner's gland on the dorsal surface of the tongue secrete an enzyme:

• A. Amylase
• B. Ptyalin
• C. Lingual lipase✓
• D. Both Amylase and Ptyalin

Explanation: The von Ebner glands, also known as serous glands, are minor salivary glands located on both sides of your oral cavity toward the back of your tongue.

Q85. Antibodies consists of _ polypeptide chains:

• A. 2
• B. 4✓
• C. 6
• D. 8

Explanation: Antibodies are protective proteins produced by your immune system. They attach to antigens (foreign substances) — such as bacteria, fungi, viruses and toxins — and remove them from your body. Each antibody has four polypeptides (peptides that consist of two or more amino acids), including two heavy chains and two light chains

Q86. Platyhelminthes are

• A. Bilaterally symmetrical and diploblastic
• B. Bilaterally symmetrical and triploblastic✓
• C. Radially symmetrical and triploblastic
• D. Radially symmetrical and diploblastic

Explanation: Phylum Platyhelminthes belongs to the kingdom Animalia. This phylum includes 13,000 species. The organisms are also known as flatworms. Platyhelminthes are triploblastic, bilaterally symmetrical. They may be free-living or parasites. The body has a soft covering with or without cilia. They respire by simple diffusion through the body surface. They are devoid of the anus and circulatory system but have a mouth etc.

Q87. The scientific name of freshwater mussel is

• A. Mytilus edulis
• B. Loligo peali
• C. Anodonta grandis✓
• D. Anodonta bairdi

Explanation: Freshwater mussels are an essential component of our rivers and streams. By their siphoning actions, mussels filter bacteria, algae, and other small particles, which make them one of the few animals that improve water quality. Mussels also serve as a food source to many species of fish, reptiles, birds, and mammals. Freshwater mussel is a common for ‘Anodonta grandis’.

Q88. Potamogeton is an example of _

• A. Xerophytes
• B. Mesophytes
• C. Hydrophytes✓
• D. Halophytes

Explanation: Potamogeton is a genus of aquatic, mostly freshwater, plants of the family Potamogetonaceae. Most are known by the common name pondweed. They are hydrophytes.

Q89. _ and cytokinins stimulates fruit ripening.

• A. Cytokinins
• B. Abscisic acid
• C. Ethylene✓
• D. Auxin

Explanation: Ethylene stimulates fruit ripening by increasing the fruit’s metabolism.

Q90. The bulbourethral glands produce:

• A. Acidic fluid
• B. Alkaline fluid✓
• C. Semen
• D. Mucus

Explanation: In response to sexual stimulation, the bulbourethral glands secrete an alkaline mucus-like fluid. This fluid neutralizes the acidity of the urine residue in the urethra.

Q91. A condition in which an abnormally large volume of urine is produced is

• A. Polydipsia
• B. Polyuria✓
• C. Polyphagia
• D. Polyanypsida
• E. Both A and B

Explanation: Option B is correct according to KPK Book Line (Page: 128)

Q92. HIV destroys a type of defense cell in the body called _ helper lymphocyte.

• A. TD4
• B. T4
• C. C4
• D. CD4✓

Explanation: HIV (Human Immunodeficiency Virus) primarily targets and infects CD4-positive T cells, which are a type of immune cell. CD4 cells play a crucial role in coordinating the immune response against infections. The HIV virus specifically binds to the CD4 receptor on the surface of these cells, allowing it to enter and infect them. Once inside the CD4 cell, HIV uses its reverse transcriptase enzyme to convert its RNA genome into DNA. The viral DNA then integrates into the host cell's genome, becoming a permanent part of the infected cell. As the virus replicates, new viral particles are produced and released from the infected CD4 cells. This process ultimately leads to a gradual depletion of CD4 cells in the immune system.

Q93. Acetabularia crenulata has a _ shaped cap:

• A. Irregular✓
• B. Umbrella
• C. Regular
• D. Disc

Explanation: Acetabularia is a unicellular alga. It is 2 to 3 inches in length and inhabits European seawater. an It has a stalk with a characteristic cap and base containing nucleus. The two species of this alga are different in the shape of their cap. Acetabularia mediterranea has an umbrella-shaped cap, while acetabularia crenulata has an irregular head.

Q94. The safranin stain is usable for:

• A. Fungal hyphae
• B. Cytoplasm/Cellulose
• C. Blood cells
• D. Lignin✓

Explanation: Safranin stain is usable for lignin in plant cells

Q95. In the human skull the unpaired bones are:

• A. Frontal, occipital, ethmoid and sphenoid✓
• B. Frontal, ethmoid, sphenoid and zygomatic
• C. Ethmoid, sphenoid, zygomatic and frontal
• D. Temporal, sphenoid, frontal and ethmoid

Explanation: Skull consists of 8 bones. Unpaired Bones Paired Bones Frontal Parietal Occipital Temporal Sphenoid Ethmoid

Q96. Functionally _ pairs of cranial nerves are sensory in nature and _ pairs are mixed in nature and _ are motor in nature

• A. 3,4 and 5✓
• B. 4,5 and 3
• C. 3,5 and 4
• D. 4,3 and 5

Explanation: There are 12 pairs of cranial nerves. 3 pairs of cranial nerves are sensory,4 pairs are mixed and 5 pairs are motor in nature.

Q97. DNA finger printing refer to:

• A. Techniques used for the identification of fingerprints of individuals
• B. Molecular analysis of profiles of DNA samples✓
• C. Analysis of DNA samples using imprinting devices
• D. Both Techniques are used for the identification of fingerprints of individuals and Analysis of DNA samples using imprinting devices

Explanation: DNA fingerprinting is also called DNA typing, DNA profiling in genetics, it is the method in which a small amount of DNA material is isolated, and the variables of elements of DNA are identified. DNA fingerprinting refers to the molecular analysis of the profile of DNA samples. Short nucleotide repeats are important for DNA fingerprinting which vary in number from person to person but are inherited.

Q98. Oleic acid is a fatty acid with 18 carbon atoms. It breaks down into 9 acetyl groups. It is estimated that these nine acetyl groups would generate _ ATP molecules.

• A. 81
• B. 98
• C. 101
• D. 108✓

Explanation: To calculate the ATP yield from the breakdown of oleic acid into 9 acetyl groups, we can consider the process of beta-oxidation and the subsequent metabolism of the resulting acetyl-CoA molecules in the citric acid cycle and oxidative phosphorylation. Each molecule of acetyl-CoA yields 12 ATP so 9 acetyl groups would generate 12 x 9 = 108 ATP.

Q99. Horsetails are included in class:

• A. Pteropsida
• B. Lycopsida
• C. Psilopsida
• D. Sphenopsida✓

Explanation: In Sphenopsida (Horsetails), the sporophyte is differentiated into root, stem and leaves. The leaves may be expanded or scale-like and are always arranged in whorls. Plants belonging to this group are also called arthrophytes because the whole plant body is composed of large number of joints. Main stem is not smooth, it has large number of ridges and furrows. Each node has whorl of branches. The sporangia are born on structures called sporangiophores, aggregated to form strobili.

Q100. Which one of the following bone is the only movable portion of the skull?

• A. Maxilla
• B. Frontal bone
• C. Mandible✓
• D. Zygomatic

Explanation: Mandible is the only movable portion of the skull

Q101. Which one of the following is the correct oxidation state of Mn in KMnO4?

• A. -6
• B. 7✓
• C. 9
• D. 10

Explanation: Since the oxidation state of potassium and oxygen are +1 & -2 respectively. So, KMnO4 1 + Mn + 4(-2) =0 1 + Mn - 8 = 0 Mn - 7 = 0 And Mn = 7 Thus, oxidation state of Mn is +7 in KMnO4.

Q102. What will be the geometrical shape of a molecule that contains two lone pairs and two bond pairs of electron in valence shell of central atom?

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

Explanation: If there are two bond pairs and two lone pairs of electrons the molecular geometry is angular or bent (e.g. H2O).

Q103. Quantum number which describes the orientation of orbitals in three dimensional space is

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

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

Q104. Which one of the following gas has the highest rate of diffusion at the same temperature and pressure?

• A. HCL
• B. CO2
• C. C2H2✓
• D. C2H6

Explanation: Graham Law states Rate of Diffusion ∝ 1/√(MolecularMass) Option A: HCl has molecular mass 36g Option B: CO2 has molecular mass 44g Option C: C2H2 has molecular mass 26g Option D: C2H6 has molecular mass 30g As C2H2 has lowest molecular mass, it has highest rate of diffusion.

Q105. At higher altitude, the boiling point of water is less than 100°C. This is because of:

• A. Higher atmospheric pressure
• B. Weak hydrogen bonding
• C. No change in atmospheric pressure
• D. Lower atmospheric pressure✓

Explanation: Boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure

Q106. Substance that has sharp melting point in the following is:

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

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

Q107. Which of the following pairs is an example of completely immiscible liquids?

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

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

Q108. How many elements are there in the third period of the periodic table?

• A. 18
• B. 8✓
• C. 32
• D. 10

Explanation: Second and third period are called short periods and contain eight elements.The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulphur, chlorine, and argon

Q109. The number of isomers of pentane is:

• A. 2
• B. 4
• C. 5
• D. 3✓

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

Q110. When ammonium cyanide (NH4CN) salt is dissolved in water the solution will be:

• A. Neutral
• B. Acidic
• C. Basic✓
• D. Both acidic and basic

Explanation: A solution of NH4CN is alkaline because CN− is more strong (conjugate base) than NH4+ (which is conjugate acid).

Q111. Consider the reversible reaction, N2+3H2 = 2NH3 + Heat. The yield of NH3 will be maximum at:

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

Explanation: Exothermic reaction is a reaction in which heat is given out. For exothermic reaction, by increasing temperature reaction move backwards and value of Kc decreases and by decreasing temperature reaction moves forward and value of Kc increases.Reactions involving only gases and having unequal number of moles,by increasing pressure reaction move to that side where number of moles are less. Since the reaction is exothermic, low temp will favour the product’s (NH3) side. High pressure favours the side with lower no of moles which in this case is the product’s side. So lower temperature and higher pressure will ensure maximum lead of NH3

Q112. In the complex, potassium hexacyanoferrate (III) K3 [Fe(CN)6] the coordination of the Fe is

• A. 9
• B. 3
• C. 6✓
• D. 5

Explanation: Coordination number, also known as ligancy, is the number of atoms, ions, or molecules that a central atom or ion carries in a complex or coordination compound or a crystal as its closest neighbours coordination number of Fe in K3[Fe(CN)6] is 6. Since the Fe atom is surrounded by six cyanide ligands which is mono-dentate, the coordination number is 6.

Q113. The compound which has the higher boiling point in the following is

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

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

Q114. Which one of the following is addition polymer:

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

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

Q115. Photochemical smog is primarily caused by:

• A. O3
• B. NO2✓
• C. SO3
• D. CO2

Explanation: Photochemical smog appears to be initiated by nitrogen oxides that are emitted into the air as pollutants mainly from internal combustion engines. Absorbing the visible or ultraviolet energy of sunlight, it forms nitric oxide (NO) to free atoms of oxygen (O), which then combine with molecular oxygen (O2) to form ozone (O3). In the presence of hydrocarbons (other than methane), certain other organic compounds, and sunlight, various chemical reactions take place to form photochemical smog.

Q116. Which of the following is not the major source of organic compound?

• A. Natural gas
• B. Petroleum
• C. Coal
• D. Ammoniacal Liquor✓

Explanation: Ammoniacal liquor is an organic compound. It is a concentrated solution of ammonia, ammonium compounds, and sulfur compounds, obtained as a by-product in the destructive distillation of bituminous coal.

Q117. Which of the following concentration unit is temperature dependent:

• A. Molality
• B. Mole Fraction
• C. Molarity✓
• D. Both Molality and Molarity

Explanation: Molarity of a solution depends upon temperature because volume of a solution is temperature dependent. What happens is that when the temperature is increased the distance between the molecules in a liquid increase leading to volume expansion. In turn leading to decrease in molarity. Molarity = Moles of solute / volume of solution

Q118. Which one is more reactive?

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

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

Q119. Tertiary alcohols are not oxidized into carbonyl compounds because

• A. They contain more alkyl group
• B. They have no alpha-hydrogen✓
• C. Suitable oxidizing agent is not available
• D. None of the above

Explanation: Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. During oxidation,oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom is attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.

Q120. Which compound shows the highest boiling point

• A. CH3COOH
• B. C2 H5 OH
• C. C2 H2-0 C2 H5
• D. (CH3CH2)3N✓

Explanation: By increasing the molecular mass of a compound, boiling point increases.

Q121. Which of the following pollutant decolorizes the skin?

• A. Mercury
• B. Arsenic✓
• C. Lead
• D. Cadmium

Explanation: Air pollutants damage the skin by inducing oxidative stress. Although human skin acts as a biological shield against pro-oxidative chemicals and physical air pollutants, prolonged or repetitive exposure to high levels of these pollutants may have profound negative effects on the skin.

Q122. Which contains more atoms?

• A. 7 gram Mg
• B. 8 gram Na✓
• C. 9 gram Al
• D. All same

Explanation: As we know, to calculate the number of atoms, [Mass / Molar Mass] x [6.022 x 1023] Hence, Na has the most atoms by putting the values.

Q123. Which contains the highest percentage of nitrogen?

• A. NO
• B. NO2
• C. N2O✓
• D. N2O5

Explanation: N2O contains the highest percentage of nitrogen i.e, 28/44 × 100 = 63.6 %

Q124. Fe+2 will form the most ionic bonds with:

• A. N-3
• B. S-2✓
• C. P-3
• D. F-1

Explanation: Fe is the symbol for iron (a transition metal), and S is the symbol for sulfur (a nonmetal). When the iron atoms bond with the sulfur atoms, they create a sulfide anion with a charge of 2-, and an iron cation, with a charge of 2+, hence the (II). This compound is used in the stainless steel and iron industries. Fe + S → FeS (heat)

Q125. For exothermic reversible reaction activation energy for forward direction depends upon

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

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

Q126. As the polarizing power of cation increases thermal stability of carbonates

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

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

Q127. Which of the following elements has lower first ionization energy?

• A. N
• B. O
• C. C
• D. B✓

Explanation: The first ionization decreases from beryllium to boron, and from magnesium to aluminum, as electrons from the p-block start to come into play. In the case of boron, which has an electron configuration of 1s2 2s2 2p1, the 2s electrons shield the higher-energy 2p electron from the nucleus, making it slightly easier to remove. A similar effect occurs in aluminum, which has an electron configuration of 1s2 2s2 2p6 3s2 3p1. Even though oxygen is to the right of nitrogen in period 2, its first ionization energy is slightly lower than that of nitrogen. Nitrogen has an electron configuration of 1s2 2s2 2p3, which puts one electron in each p orbital, making it a half-filled set of orbitals.Half-filled sets of p orbitals are slightly more stable than those with 2 or 4 electrons, which makes it slightly harder to ionize a nitrogen atom. Oxygen has an electron configuration of 1s2 2s2 2p4, which puts another electron in one p orbital; since this is one electron away from being half-filled, it is slightly easier to remove this additional electron.

Q128. The anhydride of HClO4 is

• A. ClO3
• B. ClO2
• C. Cl2 O7✓
• D. CI2 O7

Explanation: Anhydride literally means 'without water. It can be defined as the chemical compound formed by eliminating water from another compound 2HClO4 → Cl2O7 + H2O ​Thus, Anhydride of 2HClO4 is Cl2O7

Q129. A gas diffuses 1/2 times as fast as hydrogen, its molecular mass is:

• A. 50 amu
• B. 25 amu
• C. 16 amu
• D. 8 amu✓

Explanation: Graham found experimentally that the rate of diffusion of a gas is inversely proportional to the square root of the mass of its particles. v1/v2=√(M2/M1) where M1=H2 v1/(1/2)v1 =√(M2/2) 2²=M2/2 => M2 = 4×2 = 8

Q130. Which one of the following ions has more electrons than protons and more protons than neutrons?

• A. D
• B. d-
• C. H-✓
• D. He

Explanation: H is Hydrogen, it’s proton number is 1 so it has 1 proton, 1 electron and no neutrons. The ion of H gains an electron to attain the negative charge so now it has 2 electrons, 1 proton and 0 neutrons

Q131. Ice and water are in equilibrium with each other. By increasing the pressure, the equilibrium will shift in:

• A. Forward direction✓
• B. Reverse direction
• C. To all system at equilibrium
• D. None of the above

Explanation: By increasing the pressure, the equilibrium between ice and water will shift in the direction of the phase with the lower volume, which is the forward direction.

Q132. Steam causes severe bums than boiling water. It is due to

• A. Absence of hydrogen bonding
• B. High latent heat of vaporization✓
• C. Freely moving molecules
• D. Statement is incorrect

Explanation: Steam has more warmth and vitality than water. Due to its dormant warmth of vaporization, it will produce more burn than boiling water. Steam contains the heat energy of boiling water as well as the latent heat of vaporization

Q133. Which oxides of "K" contain more oxygen than its normal oxide?

• A. Peroxide
• B. Super oxide✓
• C. Both contain equal quantity
• D. None of the above

Explanation: Potassium form normal oxide ( in which oxygen has oxidation state -2) K2O. KO is peroxide (oxygen has oxidation state -1) has same no.of oxygen as normal oxide. KO2 is super oxide (oxygen has oxidation state -½) has more no. of oxygen than normal oxide.

Q134. A gas decolorized alkaline KMnO4 solution but does not give any PPT with ammoniacal AgNO3

• A. Methane
• B. Ethylene✓
• C. Ethane
• D. None of the above

Explanation: Option d) "None of the above" is incorrect because ethylene (option b) decolorizes KMnO4 but does not give a precipitate with ammoniacal AgNO3. Methane (option a) and ethane (option c) are both saturated hydrocarbons with only single bonds and are not reducing agents, so they would not decolorize KMnO4 or react with ammoniacal AgNO3 to form a precipitate. Therefore, the correct answer is option b) Ethylene.

Q135. Why ethanoic acid is a stronger acid in the liquid ammonia than in water

• A. Ammonia is strong base than water✓
• B. Ethanoic acid molecules form H-bonding with water
• C. Ethanoic acid is more soluble in liquid ammonia than in water
• D. None of the above

Explanation: Both water and ammonia can form hydrogen bond, and this is a very strong intermolecular force. In order for ethanoic acid to dissociate, the attraction between the protons and the surrounding solvent must be more favorable then the strength of the hydrogen bond to the ethanoic acid which ammonia provides because it is more basic than water.

Q136. Which ions are used as catalysts in the reaction between persulfate ions and iodide ions?

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

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

Q137. Which one is stronger nucleophile?

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

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

Q138. Which one of the following elements has the largest second ionization energy:

• A. O
• B. F
• C. Na✓
• D. N

Explanation: Na has the largest second ionization energy as the second electron is removed from an inner shell, thus the nuclear charge acting on it is stronger and more energy is needed to remove an electron. After losing 1st electron, sodium gains electronic configuration of Neon (noble gas) which has completely filled orbital so it has the highest second ionization energy.

Q139. Which of the following species has the maximum number of unpaired electrons?

• A. O2✓
• B. O+2
• C. O-2
• D. O2-2

Explanation: O2 has 2 unpaired electrons, O+2 has no unpaired electrons, O-2 has paired electrons while O2-2 also has no unpaired electrons hence the correct option is A The number of unpaired electrons in an atom or ion can be determined by examining its electron configuration. Let's analyze the given species: 1. **O2**: Molecular oxygen (O2) has a total of 16 electrons (8 electrons per oxygen atom). Its electron configuration is 1s² 2s² 2p⁴. There are two unpaired electrons in the 2p orbital. 2. **O+2**: Oxygen cation (O+2) is formed by losing two electrons from O2. Its electron configuration becomes 1s² 2s² 2p³. There is one unpaired electron in the 2p orbital. 3. **O-2**: Oxygen anion (O-2) is formed by gaining two electrons by O2. Its electron configuration becomes 1s² 2s² 2p⁶. There are no unpaired electrons. 4. **O2-2**: This is a hypothetical species, as it would mean adding two more electrons to O-2, which is not energetically favorable. If you meant O-2, please refer to the explanation for O-2 above. Based on the analysis, **O2** has the maximum number of unpaired electrons with **two unpaired electrons**.

Q140. A mixture of 10cm3 of oxygen and 50cm3 of hydrogen is sparked continuously. Calculate volume of non limiting reagent left after reaction completion:

• A. 10cm3
• B. 15cm3
• C. 20cm3
• D. 30cm3✓

Explanation: We'll begin by writing the balanced equation for the reaction. This is given below: O2 + 2H2 —> 2H2O From the balanced equation above, we can say that: 1 cm³ of O2 reacted with 2 cm³ of H2 to produce 2 cm³ of H2O. Next, we shall determine the excess reactant. This can be obtained as follow: From the balanced equation above, 1 cm³ of O2 reacted with 2 cm³ of H2. Therefore, 10 cm³ of O2 will react with = (10 × 2)/1 = 10 × 2 = 20 cm³ of H2. From the calculations made above, we can see that only 20 cm³ out of 50 cm³ of H2 given is needed to react completely with 10 cm³ of O2. Therefore, O2 is the limiting reactant and H2 is the excess reactant (non limiting reactant). Finally, we shall determine the volume of the non limiting reactant (excess react) that is remaining after the reaction. This can be obtained as follow: Volume of non limiting reactant (H2) = 50 cm³ Volume of non limiting reactant (H2) that reacted = 20 cm³ Volume of non limiting reactant (H2) remaining = (Volume of non limiting reactant) – (Volume of non limiting reactant that reacted) Volume of non limiting reactant (H2) remaining = 50 – 20 = 30 cm³

Q141. The oxidation states of Nitrogen in NH4NO3 are:

• A. -3 and +5✓
• B. +5 and-3
• C. -3 & -3
• D. Zero

Explanation: NH₄⁺ N + 4 = 1 N = 1-4 N = 3 NO3- N + 3(-2)= -1 N - 6 = -1 N = -1 + 6 N= 5

Q142. Which equation relates to the first ionization energy of bromine?

• A. Br(g) → Br-(g) + 1e
• B. Br(g) → Br+(g) + 1e-✓
• C. 1/2 Br2(g) → Br-(g) + 1e-
• D. 1/2 Br2(g) → Br+(g) + 1e-

Explanation: Refer to folllowing explanation

Q143. Co-ordination number of [Co(en)2Cl2]is

• A. -2
• B. 6✓
• C. 4
• D. None

Explanation: In [Co(en)2​Cl2] No. of monodentate ligand (Cl) =2 No. of bidentate ligand,(Ethylene diamine) =2 ∴ Co-ordination no. of the metal =2+2×2=6

Q144. An olefin "X" on ozonolysis gives CH3CH2COCH3 and CH3COCH3. The IUPAC name of X is:

• A. 2 - Butene
• B. 2,3 di Methyl-2 Pentene✓
• C. 2 - Pentene
• D. 1 - Hexene

Explanation: The longest carbon chain is of 5 carbons so the name ends with pentene and since there are two methyl groups as branches so option B is correct

Q145. Which one is more soluble in water?

• A. Secondary amines✓
• B. Tertiary amines
• C. Quatemary amines
• D. All are insoluble

Explanation: Lower amines can form hydrogen bonds with water thus making them more soluble.

Q146. The number of peaks given by ethane thiol in the H-NMR spectrum is:

• A. 2
• B. 3✓
• C. 4
• D. None

Explanation: Ethane thiol has the formula CH3- CH2- SH so the number of peaks of proton NMR are going to be 3

Q147. C4H11N gives the type of isomerism:

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

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

Q148. From the following, identify the incorrect statement regarding gas having high value of coefficient of attraction.

• A. Easy to be liquified✓
• B. Having higher critical temperature
• C. Less soluble in water
• D. None

Explanation: A gas having high value of coefficient of attraction is tough to be liquified. It is because, the repulsive forces become more significant and the attractive forces become less dominant. Hence these gases are difficult to be condensed.

Q149. Which one can form more acidic oxide?

• A. Sc
• B. Mn✓
• C. V
• D. Ti

Explanation: The acidic strength of oxides of different elements will depend on the tendency of oxygen atoms to accept electrons. When the oxygen atom is attached to some electronegative element its acidic character increases as the electronegative element pulls the electrons towards its side. In the periodic table, the electropositive character decreases from right to left across a period and increases down the group. Since Mn appears after Sc, V, and Ti in the periodic table it is comparatively more electronegative, and hence Mn is the most acidic oxide among these transition elements.

Q150. Hydration of hydrocarbon give carbonyl compound, the general formula of that hydrocarbon is

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

Explanation: Alkynes on hydration in presence of H2SO4 and HgSO4 yield carbonyl compound through enol All alkynes except ethyne give ketones while ethyne gives aldehyde.

Q151. Consider reversibility in free radical substitution reaction of alkane then Kc value is smallest for:

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

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

Q152. Ethylenediamine diacetate is

• A. Didentate
• B. Tridentate
• C. Tetradentate✓
• D. Hexadentate

Explanation: Ethylenediamine diacetate is a chemical compound with tetradentate coordination properties, not bidentate. A tetradentate ligand can form four coordination bonds with a central metal ion, while a bidentate ligand can form two coordination bonds. In the case of ethylenediamine diacetate, it has two ethylenediamine groups, each of which can bind to a metal ion through two coordination bonds, totaling four bonds and making it tetradentate.

Q153. Epoxide obtained from isobutylene is further hydrolyzed in the presence of acid The final product is

• A. 2,3 - Butanediol
• B. 1,2 - Butanediol
• C. 2-Methyl-1,2-Propandiol✓
• D. All of them

Explanation: Epoxide obtained from isobutylene is further hydrolyzed in the presence of acid The final product is 2-Methyl-1,2-Propandiol

Q154. In the detection of nitrogen in an organic compound The appearance of Prussian blue colouration is due to the formation of:

• A. Fe4[Fe(CN6)]3✓
• B. Na3[Fe(CN)6]
• C. K3[Fe(CN)6]
• D. None

Explanation: In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour.

Q155. The bond angle in H2S is less than H2O. it is due to

• A. Small size of oxygen atom
• B. Greater E.N of oxygen atom✓
• C. Oxygen contain two bone pairs of electrons
• D. All of the above

Explanation: Bond angle of H2S(92° )<H2O(104°). As the electronegativity of the central atom decreases, bond angle decreases. In the present case, S is less electronegative than oxygen. Thus bond pairs in H2S are more away from the central atom than in H2O and thus repulsive forces between bond pairs are smaller producing smaller bond angle

Q156. The auxochrome not concern with Metanil yellow dye:

• A. -SO3H✓
• B. -OH
• C. -NH2
• D. Both-SO3H and -NH2

Explanation: Auxochrome increases the color of any organic compound and SO3H is not associated with Metanil yellow dye

Q157. The molecular theory for metals is also called as:

• A. Electron pool theory
• B. Valence bond theory
• C. Band theory✓
• D. None of these options are correct

Explanation: Metals conduct electricity with the help of valence electrons present in them. The atomic orbitals of the metals having the same energy combine to form molecular orbitals which are close in energy to each other to form a band. Molecular orbital theory is called band theory for metals because in the metals, which are periodic structures.

Q158. The general trend of ionic radii from left to right is:

• A. Increase
• B. Decrease✓
• C. Increase then decrease
• D. Decrease then increase

Explanation: Ionic radius increases as you move from top to bottom on the periodic table. Ionic radius decreases as you move across the periodic table, from left to right.

Q159. Actual yield is a/an:

• A. Theoretical term
• B. Experimental term✓
• C. Stoichiometric term
• D. Supposed term

Explanation: The actual yield is the quantity of a product that is obtained from a chemical reaction. In contrast, the calculated or theoretical yield is the amount of product that could be obtained from a reaction if all of the reactant converted to product. Theoretical yield is based on the limiting reactant.

Q160. The angular acceleration of second hand of watch is

• A. π rad/sec2
• B. 2π rad/sec2
• C. π/2 rad/sec2
• D. None of the above✓

Explanation: Angular acceleration (α) is defined as the change in angular velocity (ꞷ) per unit time. The second-hand watch as a constant angular velocity of rad/s. Therefore, change in (ꞷ) will be zero and α will be zero as well. Option A, B and C are all hence wrong.

Q161. Which one of the following is not a state function?

• A. Work✓
• B. Enthalpy
• C. Internal energy
• D. Pressure

Explanation: The thermodynamic state of a system refers to the temperature, pressure and quantity of substance present. Examples of state functions include density, internal energy, enthalpy, entropy. State functions only depend on these parameters and not on how they were reached, So Work is NOT a state function and hence is the correct option.

Q162. The viscous drag on small spherical body (moving with slow speed v) is propotional to

• A. v✓
• B. Square root of v
• C. 1/square root of v
• D. v2

Explanation: Formula for viscous drag force is Fs = 6πrηv. Form the given formula it can be seen that keeping r and η constant for a specific object in a fluid, F is proportional to v. Option A is hence the only correct relation. Option B, C & D are wrong relations.

Q163. The transverse nature of light is shown by

• A. Interference of light
• B. Refraction of light
• C. Polarization of light✓
• D. Dispersion of light

Explanation: A light wave is an electromagnetic wave which travels through the vacuum of outer space and transparent materials like glass. Electromagnetic waves are transverse waves in the sense that the plane of the wave vibrations is different from the plane of wave propagation. The electromagnetic waves in the true sense have the direction of propagation perpendicular to the direction of its electric and magnetic fields. In the case of polarized waves, the electric and magnetic fields are also at right angles to each other. Polarized light waves are electromagnetic waves in which the vibrations occur in a single plane. Light can be polarized by transmission, reflection, refraction or scattering. Using a polaroid filter, different planes of vibration of polaroid filters can be blocked due to the special material used there. Due to the polarization of light, it is evident that light is a transverse wave.

Q164. An electron is moving along the axis of a solenoid carrying a current. Which of the following is a correct statement about the magnetic force acting on the electron?

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

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

Q165. The motional E.M.F depends upon:

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

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

Q166. When zinc electrode is coupled with copper electrode in a galvanic cell:

• A. Reduction takes place at zinc electrode
• B. Oxidation takes place at copper electrode
• C. Reduction takes place at copper electrode✓
• D. Both Reduction takes place at zinc electrode and Oxidation takes place at copper electrode

Explanation: A galvanic cell is made by joining two half cells. Cu2+ has a higher reduction potential so it is reduced and reduction talks place at Cu electrode. Consequently, oxidation takes place at Zn electrode as it has lower reduction potential (higher oxidation potential).

Q167. Which one of the following physical quantity does not the dimension of force per unit area:

• A. Stress
• B. Strain✓
• C. Young modulus
• D. Pressure

Explanation: The strain unit is dimensionless. It is the ratio between the length shift and the initial length, so it is unit-less. Thus B, strain, is the correct option

Q168. In the case of germanium, the value of potential barrier develops across the depletion region is

• A. 0v
• B. 0.3V✓
• C. 0.7V
• D. 0.9V

Explanation: 0.3 V for Germanium and 0.7 V for silicon.

Q169. Electron microscope makes practical use of:

• A. Particle nature of electron
• B. Wave nature of electron✓
• C. Dual nature of electron
• D. None of the above

Explanation: The electron microscope is a powerful scientific instrument that employs a beam of electrons to illuminate an object and magnify its image. By harnessing the wave-like properties of electrons, this technology allows researchers to capture high-resolution images of tiny samples and gain unprecedented insights into their structure and composition.

Q170. The projectile is thrown in such a way that its maximum height equals its range, and the angle of projection is

• A. Tan-1 (45)
• B. Tan-1 (60)
• C. Tan-1 (30)
• D. None✓

Explanation: We will make use of the formulas: Options A, B, and C are all wrong as they give wrong numerical values.

Q171. Car "X" is travelling at half the speed of car "Y" and mass of car X is twice as compared to the mass of car 'Y". Which of the following statements is correct?

• A. Car "X' has half of the kinetic energy of car "Y'✓
• B. Car "X" has one quarter of the K.E of car "Y"
• C. Car 'X" has twice the K.E of car "Y'
• D. The two cars have the same KE

Explanation: Car X speed = v/2 amd mass = 2m Car Y speed = v and mass = m

Q172. If the wavelength of a transverse is 2cm and the period is 2 seconds, then the wave speed in CGS is:

• A. 0.1 cms-1
• B. 0.2 cms-1
• C. 11 cms-1
• D. 1 cms-1✓

Explanation: Option A, B & C are all numerically wrong.

Q173. A car battery has an EMF of 12 volts and an internal resistance of 5 x 10-2 ohms. If it draws 60 ampere current, then the terminal voltage of the battery will be:

• A. 5 volts
• B. 3 volts
• C. 15 volts
• D. 9 volts✓

Explanation: Here is the terminal potential:V = E - IR =12-(60×5×10−2) =12-3 =9V

Q174. The cyclotron frequency of an electron projected with velocity perpendicular to a magnetic field B is given by:

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

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

Q175. If A, B = 1/2 AB then angle between A and B is

• A. Zero
• B. 30
• C. 60✓
• D. 90

Explanation: In this case dot product is equal to scalar product of two vector A.B= A x B x cos θ In this case on right hand side of equation, ½ will be Hence θ will be 60. Option A, B and D give wrong value of angle

Q176. A train is 200m long and is moving with uniform velocity of 36 km/h the time it will take to cross a bridge of 1 km is

• A. 100 sec
• B. 120 sec✓
• C. 60 sec
• D. 50 sec

Explanation: Consider one point on the train as your frame of reference , so at any given point the train needs to travel 1200m ( 1km +200 m) to reach the other side of the bridge.And converting 36kmph into m/s gives us 10m/s. So by using time = displacement/velocity Time = 1200/10= 120 seconds

Q177. The escape velocity of body from a planet does not depend upon;

• A. The mass of a body✓
• B. The mass of the planet
• C. The average radius of the planet
• D. The density of the planet

Explanation: The escape velocity 'v' of a body depends upon the acceleration due to gravity of the planet and the radius of the planet but not on mass of planet.

Q178. In order to increase the stopping potential, there should be increase in

• A. Intensity of radiation
• B. Wavelength
• C. Frequency of radiation✓
• D. Both wavelength and intensity

Explanation: The energy of incident light is directly proportional to its frequency. Hence, the stopping potential increases when the frequency of the incident light is increased.

Q179. Two meter high tank is full of water. A hole is made in the middle of the tank. The speed of efflux is

• A. 4.9 m/s
• B. 9.8 m/s
• C. 4.42 m/s✓
• D. 3.75 m/s

Explanation: The speed of efflux is equal to the velocity gained by the fluid in falling through the distance (h1-h2) under the action of gravity. Using Bernoulli's equation

Q180. A hail and a rain drop of the same radius are released from the same height, the rain drop will reach

• A. Before hail✓
• B. After hail
• C. At the same time
• D. None of the above

Explanation: Hail drop will be less dense as ice is less dense. Lesser dense object falls slower. Hence rain drop will reach before hail drop. B, C and D cannot be correct.

Q181. Two springs A and B (kA​=2kB​) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

• A. E/2
• B. 2E✓
• C. E
• D. E/4

Explanation: EA = ½ kAxA2 Keeping force constant and using F= kx As kA = 2kB Then 2xA=xB Energy stored in B will be: EB= ½ (kA/2) (2xA)2 EB= ½ (2xA2 kA) Replace EA = ½ kAxA2 EB = 2 EA Hence only B is correct (A, C, D give wrong answer)

Q182. The general form of path difference in young's double slit experiment is "mλ" Its corresponding phase shift (in radians) is

• A. mπ
• B. 2 mπ✓
• C. mπ/2
• D. None

Explanation: Factual. General form for path difference is given as =2𝞹 Option A, C and D are not correct according to this.

Q183. An α-particle is accelerated through a potential difference of 106 volts, K.E is:

• A. 1 MeV
• B. 2 MeV✓
• C. 4 MeV
• D. 8 MeV

Explanation: For a particle that is accelerated, its KE equals VqApplying the formula,KE= Vq AND M=106KE= (106) (2e)KE= 2MeVOption A, C and D give wrong numerical values.

Q184. If there are "n" capacitors each of capacity "c" connected in parallel to "V" volts source then the energy stored is equal to:

• A. CV
• B. ½ nCV2✓
• C. CV2
• D. CV2/2n

Explanation: The energy of a capacitor is calculated using the formula = ½ C V2 multiplied by n i.e number of capacitors

Q185. The electric field strength between a pair of plates is "E". If the separation of the plates is doubled and the potential difference between the plates is increased by a factor of four, the new field strength is:

• A. E
• B. 2E✓
• C. 4E
• D. 8E

Explanation: Electric field is calculated by the formula = Voltage / Distance so doubling the separation of plates decreases the electric field by 2 times and increasing pd between plates by factor of four, increases electric field 4 times so the over all increase is 2E

Q186. Two satellites of masses "3M" and "M" orbit the earth in a circular orbit of radius Y and "3r" respectively, the ratio of their speed is:

• A. 1:1
• B. Square root of 3 :1✓
• C. 3:1
• D. 9:1

Explanation: To move in a circular orbit, gravitational force provides centripetal force. Gravitational force between the two is the same as it is a mutual force. For the same reason, distance between the satellites is the same so speed does not depend on radius.

Q187. The optimum pH of enzyme urease is

• A. 7.8-8.7
• B. 7.0✓
• C. 4.5
• D. 8.0

Explanation: The optimum pH of enzyme urease is 7.0 i.e neutral

Q188. Two wires A and B are made of the same material. Wire A has a length of I, and a diameter of R while wire B has a length of 2L and a diameter of R/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is:

• A. ⅛✓
• B. ¼
• C. 4
• D. 8

Explanation: Wires made of the same material have the same Young Modulus.Hence, option A is the correct choice.

Q189. A wire can sustain the weight of 20kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of

• A. 10kg
• B. 20kg✓
• C. 40kg
• D. 80kg

Explanation: The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same each part can sustain a weight of 20 kg.

Q190. Which of the following is not E.M wave?

• A. Radio waves
• B. X-rays
• C. Light waves
• D. Sound waves✓

Explanation: Factual. Electromagnetic waves include:Gamma rays, X rays, Ultra violet, Visible light, Infra-Red, Microwaves, RADIO WAVES.So, D is not a EM wave.

Q191. A shell of mass m moving with velocity v suddenly breaks into two pieces. The part having mass m/4 remains stationary. The velocity of the other shell will be:

• A. v
• B. 2v
• C. 3v/4
• D. 4v/3✓

Explanation: Using the law conservation of mass, Total momentum Before = Total momentum After Let x be the new speed of remaining mass (3m/4) mv = m1v1 + m2v2 mv = (m/4) (0) + (3m/4) (x) mv = (3m/4) x 4v/3 = x Options A, B and C cannot be derived from the given situation as they donot follow law of conservation of momentum.

Q192. Two blocks "A" and "B" having masses 3kg and 4kg are raised to the same height from earth surface. The ratio of gravitational potential of "A" to that of "B" is

• A. 3:4
• B. 4:3
• C. 1:1✓
• D. None

Explanation: Factual. Gravitational potential for an equipotential surface (one that is at same height) is same for all masses. Hence ratio will be 1:1

Q193. Heat and work are equivalent. This means

• A. When we supply heat to a body we do work on it
• B. When we do work on a body we supply heat to it
• C. The temperature of a body can be increased by doing work on it✓
• D. Heat and work are not inter convertible

Explanation: From the given statement it can be derived Q= W and hence Internal energy change must me zero. (Following from first law of thermodynamics Q= W + U) the temperature of a body can be increased by doing work on it. When work is done on a body, it can lead to an increase in the body's internal energy, which, in turn, can result in a rise in its temperature.Work done on a body involves transferring energy to the body through mechanical means. This energy is converted into the body's internal energy, which is the sum of the kinetic energy and potential energy of its particles. As the internal energy increases, the average kinetic energy of the particles also increases, leading to a rise in temperature.One common example is compressing a gas in a container. When you compress a gas, you are doing work on it by reducing its volume against an external force. The work done on the gas increases its internal energy, and since temperature is a measure of the average kinetic energy of the gas particles, the temperature of the gas increases as well.It's important to note that the increase in temperature due to work done on the body is a result of the transfer of energy and is not a direct heating process like supplying heat through a temperature difference. Nonetheless, work done on a body can indeed raise its temperature by increasing its internal energy.

Q194. The velocity time plot for a particular moving on a straight line is shown in the figure:

• A. The particle has a constant acceleration.✓
• B. The particle has never turned around.
• C. The particle has zero displacement.
• D. The data is insufficient.

Explanation: The particle has a constant acceleration because the slope of the curve is constant.The particle turns around at t=10 seconds.The particle has non-zero displacement because the area above x-axis is not equal to area below the x-axis.Hence, A is correct.

Q195. Mark out the correct option.

• A. The energy of any small part of a string remains constant in a travelling wave
• B. The energy of any small part of a string remains constant in standing wave✓
• C. The energies of all small parts of equal length are equal in a travelling wave
• D. The energies of all small parts of equal length are equal in a travelling wave

Explanation: Option A, C and D are wrong as in a travelling wave energy does not remain constant and is transferred from one point to another.However, in a standing wave as mentioned in option B, Energy of small part remains constant or STANDING in that part.

Q196. A system can be taken from the initial state P1V1 to the final state P2V2 by two different methods. Let Q and W represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods?

• A. Q
• B. W
• C. Q + W
• D. Q - W✓

Explanation: Internal energy would be the same irrespective of the path.U (internal energy) = Q - W

Q197. At what angle two forces 2F and square root of 2F must act sc that their resultant is F * square root of 10

• A. π/4✓
• B. π/2c
• C. 2π
• D. None of them

Explanation: Explaination for this question will be added shortly.

Q198. When 20J of work was done on a gas, 40J of heat energy was released If the initial internal energy of the gas was 70J. what is the final internal energy?

• A. 50J✓
• B. 60J
• C. 90J
• D. 110J

Explanation: The change in internal energy is 20-40 = -20 J so the final internal energy is x where x-70= -20X=50

Q199. The time required by the projectile to reach the summit point is:

• A. T = square root of (2H/g)✓
• B. T = square root of (3H/g)
• C. T= square root of (4H/g)
• D. T = square root of (H/g)

Explanation: T = square root of (2H/g) is the time required to reach maximum height, this formula has to be memorized

Q200. If energy of activated complex is close to energy of reactants, it means that the reaction is:

• A. Fast✓
• B. Moderate
• C. Slow
• D. Very slow

Explanation: If energy of activated complex is close to energy of reactant then reaction is fast, spontaneous and endothermic. Activated state is the state when reactants gain sufficient energy to be converted into product hence, if energy of reactants is already close to the energy of activated complex, so they will not take time and energy to get converted to activated state.Thus they will react spontaneously and fast.