Uhs Mdcat 2017 — Solved Past Paper with Answers

Q1. She is intelligent enough to _ things to serve her own purpose.

• A. Pick
• B. Manoeuvre✓
• C. Give
• D. Away

Explanation: “Manoeuver” is a verb, which means to perform a movement in order to secure an advantage.This is the actual context in which this vocabulary word is used here.

Q2. She _ about the excitement of hearing the news of her sister’s wedding.

• A. Ran
• B. Jigged✓
• C. Talked
• D. Wept

Explanation: Dance performed out of excitement is called a “jig”, which is the proper context for using this word in this case. Weep means to cry which happens in cases of sadness. Talking and running also do not suit the sense of the sentence.

Q3. Everyone should be _ duties and assignments according to his/her abilities.

• A. Prevented
• B. Advised
• C. Delegated✓
• D. Suggested

Explanation: “To delegate a task” means “to assign a task” which fits most appropriately in this context “Everyone should be assigned duties...according to his/her abilities.

Q4. Spot the error. We were ten miles up the highway when I happened to saw this classified advertisement in the newspaper.

• A. We were
• B. Up the
• C. To saw this✓
• D. In the

Explanation: ‘To saw this’ is the part of the sentence that contains a grammatical error. Always remember that ‘TO’ is always followed by the 1st form of the verb [the present form of the verb]. It can be corrected as ‘to see’. ("to" + base form of the verb)Example: I want to speak to the manager.

Q5. Identify the incorrect part of the given sentence. “All is well what ends well”, said the father when he had finished the story.

• A. Is well what✓
• B. Said the
• C. When he had
• D. The story

Explanation: ‘ALL IS WELL THAT ENDS WELL’ is the right saying. It is used to say that a person can forget about how unpleasant or difficult something was because everything ended in a good way. Historically, It is a play by William Shakespeare, published in the First Folio in 1623, where it is listed among the comedies.

Q6. Identify the incorrect part of sentence.Rubber tubes upon which children had swing in backyards hung suspended like stopped clock pendulums in the blazing air.

• A. Upon which
• B. Had swing in the backyards✓
• C. Suspended like
• D. In the

Explanation: “Had swing” is wrong. We always use the third form of the verb with has/have/had.

Q7. Identify the incorrect part of the sentence.The child was fully dressed and sitting in her father’s lap near the kitchen table.

• A. Fully
• B. Sitting in✓
• C. Kitchen table
• D. No error

Explanation: ‘Sitting in’ is not correct. The word ‘father’ suggests the use of ‘sitting on’ instead.If there had been the word ‘mother’ in the sentence, then we would have chosen ‘sitting in’.

Q8. Identify the incorrect part of the sentence.The three Abdul Rahman, like his illustrious predecessor, was a young man of twenty-three when he took office.

• A. The three✓
• B. Like his
• C. Was a
• D. When he

Explanation: “The three” is a wrong use as “the” is not used to describe a proper noun like it is done in this case. The correct use would have been “Abdul Rahman the third”.

Q9. Identify the incorrect part of the sentenceEnlarged and beautified by later Caliphs, Al-Zahra become the nucleus of a royal suburb whose remains partly evacuated in and after 1910, can still be seen.

• A. By later
• B. Become the✓
• C. Whose remains
• D. In and after

Explanation: The sentence is in past tense so the verb should also be in past tense hence it is wrong to use “become” instead we should use “became”.

Q10. Choose the correct sentence from the following options:

• A. I thought it over very carefully before broaching the subject to Asma.✓
• B. I thought it by very carefully before broaching the subject to Asma.
• C. I thought through very carefully before broaching the subject to Asma.
• D. I thought over very carefully before broaching the subject to Asma.

Explanation: Correct sentence structure is present in option A, having proper use of subject-verb and object.Options B and C are wrong because of the use of “by”, “through” and “over”, which goes against the basic rules of subject-verb-object.

Q11. Choose the correct sentence from the following options:

• A. He left into a blaze of anger.
• B. He left with a blaze of anger.
• C. He left in a blaze of anger.✓
• D. He left back in a blaze of anger.

Explanation: “Blaze of anger” is a mood so the person has to be “in” it, not “into”, with, or “back in”, which are all inappropriate use of verbs. ‘Blaze of anger’ is reflecting an angry emotion. A person can be ‘in’ option C in an angry mode not into option A, with option B, or back in option D.

Q12. Choose the correct sentence from the following options:

• A. I remember going to British Museum one day.
• B. I remember to go to British Museum one day.
• C. I remember to go to the British Museum one day.
• D. I remember going to the British Museum one day.✓

Explanation: "I remember going to the British Museum one day": This sentence is the correct sentence. It is grammatically correct and idiomatic. The word "to" is used as a preposition to show the direction of movement, and the verb "go" is in the past tense to show that the speaker is remembering an event that happened in the past.

Q13. Select the NEAREST CORRECT MEANING of the given word:DISDAIN

• A. Vice
• B. Dislike
• C. Contempt✓
• D. Ignorance

Explanation: “The feeling that someone or something is unworthy of one's consideration or respect.” is called disdain, the same as what is meant by contempt.

Q14. Select the NEAREST CORRECT MEANING of the given word:SAGACITY

• A. Suspicious
• B. Cruelty
• C. Wisdom✓
• D. Foolishness

Explanation: SAGACITY means the quality of being intelligent which is synonymous with “WISDOM”.

Q15. Select the NEAREST CORRECT MEANING of the given word:FLAUNT

• A. Snipe
• B. Dance
• C. Show off✓
• D. Preserve

Explanation: FLAUNT means to “display (something) ostentatiously, especially in order to provoke envyor admiration or to show defiance” which is what is meant by “show off”.

Q16. Select the NEAREST CORRECT MEANING of the given word:URBANE

• A. Suave✓
• B. Rough
• C. Bad
• D. Dishonest

Explanation: URBANE means a person who is courteous and refined in a manner which is exactly what is meant by suave.

Q17. Select the NEAREST CORRECT MEANING of the given wordDIASPORA

• A. Gathering
• B. Dispersion✓
• C. Alliance
• D. Animosity

Explanation: A diaspora is a scattered population whose origin lies in a separate geographic locale.

Q18. Select the NEAREST CORRECT MEANING of the given word:IMPETUOUS

• A. Honest
• B. Impulsive✓
• C. Lazy
• D. Liar

Explanation: Impetuous suggests eagerness, violence, and rashness. Impulsivity is the tendency to act without thinking. Being impulsive leads to being impetuous. These two words are most closely related to the given options, hence impulsive is our answer.

Q19. Select the NEAREST CORRECT MEANING of the given word:VOCIFEROUS

• A. Hidden
• B. Loud✓
• C. Strong
• D. Weak

Explanation: VOCIFEROUS is characterized by being “loud and forceful.”

Q20. Select the NEAREST CORRECT MEANING of the given word:TRANSIENT

• A. Permanent
• B. Temporary✓
• C. Long
• D. Good

Explanation: Transient means something that ends after a certain time/fades away, which is synonymous with “temporary”.

Q21. Select the NEAREST CORRECT MEANING of the given word:PROWESS

• A. Hindrance
• B. Skill✓
• C. Reservation
• D. Bad name

Explanation: PROWESS means “a skill or expertise in a particular activity or field”, hence the answer is B.

Q22. Select the NEAREST CORRECT MEANING of the given word:BEQUEATH

• A. Grant✓
• B. Imbibe
• C. Irrigate
• D. Hope

Explanation: To BEQUEATH means “to leave (property) to a person or other beneficiary by a will". Exactly what is meant by grant.

Q23. Identify the word or phrase that needs to be changed for the sentence to be correct:We are expecting to go abroad this summer.

• A. Are
• B. Expecting
• C. To go
• D. This
• E. No error✓

Explanation: The above sentence has no error.

Q24. The low partial pressure of oxygen in tissues favours _ of oxyhaemoglobin.

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

Explanation: When oxygen pressure falls below 60 mm mercury, as in many cells and tissues, the oxygen saturation of haemoglobin decreases very sharply. This results in the liberation of large quantities of oxygen from haemoglobin. In this way in the tissue where oxygen tension is low oxyhaemoglobin dissociates rapidly.

Q25. Respiratory tubules are termed as bronchioles when they attain the diameter _ or lesser:

• A. 1.2 cm
• B. 1 cm
• C. 1 mm✓
• D. 1.2 mm

Explanation: When the smaller bronchi attain a diameter of one mm or less, then they are called bronchioles. Bronchi have the same cartilage rings as the trachea, but the rings are progressively replaced by irregularly distributed cartilage plates, and the bronchioles totally lack cartilage. Bronchioles are made up mainly of circular smooth muscles.

Q26. Elastic fibres are absent in the walls of _.

• A. Aorta
• B. Arteries
• C. Veins✓
• D. Both options A and B

Explanation: The elastic fibres are essential in the recoil and the stretch of the blood vessel in order to survive high blood pressure. In veins, however, due to the lack of significant blood pressure, elastic fibres are not needed.

Q27. A type of blood cell that produces heparin is _.

• A. Basophil✓
• B. Neutrophil
• C. Eosinophil
• D. Monocyte

Explanation: Basophils release heparin to prevent blood clots and histamine which causes inflammation.

Q28. The thoracic lymph duct of the lymphatic system opens into _.

• A. Superior vena cava
• B. Subclavian Vein✓
• C. Inferior vena cava
• D. Renal vein

Explanation: Lymph capillaries join to form larger and larger lymph vessels; and ultimately form the thoracic lymph duct, which opens into the subclavian vein. The low of lymph is always towards the thoracic duct.

Q29. Select the part of nephron which is NOT permeable to water and stops its outflow:

• A. Glomerulus
• B. Proximal Tubule
• C. Ascending loop✓
• D. Descending loop

Explanation: The ascending loop is only permeable to salt ions such as Cl- and Na+. The descending loop, on the other hand, is permeable to water. The idea behind the impermeability of the ascending loop is to allow salts to move out of the loop of Henle via active transport so they can be reabsorbed.

Q30. Vessels which carry blood to the glomerulus are called:

• A. Efferent arterioles
• B. Renal vein
• C. Vasa recta
• D. Afferent arterioles✓

Explanation: The renal artery divides to give afferent arterioles which further divide into the glomerulus network at the bowman’s capsule, forming the renal corpuscle. After the glomerulus, the capillaries reconnect to give the efferent arterioles.

Q31. In ECG, QRS wave represents:

• A. Ventricular systole✓
• B. Atrial systole
• C. Diastole
• D. Recovery systole

Explanation: The QRS complex represents depolarization of the ventricles and is followed by ventricular contraction which is collectively known as the ventricular systole.

Q32. When water content in body becomes high, what will happen:

• A. ADH release will be inhibited✓
• B. ADH will be released in large amount
• C. Aldosterone will be released
• D. Anterior pituitary will produce ADH

Explanation: The anterior pituitary gland would stop releasing ADH which would consequently make the collecting duct impermeable to water so that excess water is not reabsorbed from the filtrate making the urine more dilute.

Q33. The major factor in producing hypertonic urine is:

• A. Glomerulus.
• B. Influence of aldosterone.
• C. ADH influencing on collecting duct.
• D. Gradual increase in osmolarity from cortex to medullary portion.✓

Explanation: The major factor in producing hypertonic urine in mammals is gradually increasing osmotic concentration from the cortex to the medulla. Juxtamedullary nephrons are specifically instrumental in the production of concentrated urine via a counter current mechanism which uses high medullary osmolarity.

Q34. What is the least selective process during urine formation:

• A. Reabsorption
• B. Pressure filtration✓
• C. Secretion
• D. Differential permeability

Explanation: Pressure filtration at the renal corpuscle is the least selective since all molecules with a molecular mass below 69000 pass through the basement membrane and into the bowman’s capsule as the filtrate.

Q35. The nerve impulse which jumps from node to node in myelinated neurons is called as:

• A. Resting membrane potential
• B. Saltatory nerve impulse✓
• C. Threshold stimulus
• D. Initial nerve impulse

Explanation: Saltatory conduction describes how an electrical impulse skips from node to node down the full length of an axon, speeding the arrival of the impulse at the nerve terminal compared to the slower continuous progression of depolarization spreading down an unmyelinated axon.

Q36. The CNS is protected by:

• A. Three layers of meninges✓
• B. One layer of moninx
• C. 4 layers of meninges
• D. 2 layers of meninges

Explanation: The CNS is covered with three layers of protective coverings called meninges. The outermost layer is the dura mater having its primary function being the protection of the brain and the spinal cord. The dura mater also contains vein-like structures that carry blood from the brain back to the heart. The middle layer is the web-like arachnoid mater. The last layer is the pia mater, which directly contacts and covers the brain and spinal cord like plastic wrap. The space between the arachnoid and pia mater is filled with cerebrospinal fluid

Q37. White matter of spinal cord is made up of:

• A. Sensory nerve fibres
• B. Myelinated nerve fibres✓
• C. Motor nerve fibres
• D. Mixed nerve fibres

Explanation: White matter is found in the deeper tissues of the brain (subcortical). It contains nerve fibres (axons), which are extensions of nerve cells (neurons). Many of these nerve fibres are surrounded by a type of sheath or covering called myelin. Myelin gives the white matter its colour.

Q38. There are evidences that high levels of aluminium can lead to the onset of:

• A. Parkinson's disease
• B. Alzheimer's disease✓
• C. Lesch-Nyhan syndrome
• D. Fragile X-syndrome

Explanation: There is evidence that high levels of aluminium may contribute to the onset of Alzheimer's disease.

Q39. _ is the structure in female reproductive system in which fertilization takes place:

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

Explanation: The zygote is formed in the oviduct and moves towards the uterus where it could be embedded.

Q40. Which of the following directly develops into sperms?

• A. Primary spermatocytes
• B. Spermatids✓
• C. Secondary spermatocytes
• D. Spermatogonia

Explanation: Spermatids are the end result of the process of meiosis. They, then, undergo specialization which includes growing flagella, and acrosomes and having a greater quantity of mitochondria to form the spermatozoa.Each testis consists of a highly complex duct system called seminiferous tubules, in which repeated division by the cells of the germinal epithelium produce spermatogonia[option D].These increase in size and differentiate into primary spermatocytes[option A] which undergo meiotic division to form secondary spermatocytes[option C] and spermatids. Eventually, the spermatids differentiate into mature sperms.[option B]

Q41. FSH stimulates the production of oestrogen hormone which has two targets _; and _:

• A. Uterus, posterior pituitary
• B. Ovaries, uterus
• C. Uterus, anterior pituitary✓
• D. Ovaries, hypothalamus

Explanation: Uterus: Estrogen enhances and maintains the mucous membrane that lines the uterus. It also regulates the flow and thickness of uterine mucus secretions.Anterior pituitary: At high concentrations near the end of the follicular phase, estrogen becomes a positive inducer of the anterior pituitary. Positive feedback triggers the anterior pituitary to release more FSH and LH. more FSH and LH causing the ovary to produce more estrogen. The ensuing LH surge is responsible for ovulation.

Q42. Select the organelle which is only present in animal cells:

• A. Centrioles✓
• B. Rough Endoplasmic Reticulum
• C. Microtubules
• D. Ribosomes

Explanation: Centrioles are present in (1) animal cells and (2) the basal region of cilia and flagella in animals and lower plants (e.g. Chlamydomonas). In cilia and flagella, centrioles are called 'basal bodies' but the two can be considered inter-convertible. Centrioles are absent from the cells of higher plants.

Q43. Syphilis is a sexually transmitted disease and can also damage:

• A. Hair
• B. Heart✓
• C. Peripheral Nervous System
• D. Birth canal

Explanation: Syphilis, at its late stages, is known to cause cardiovascular diseases which involve the enlargement of the heart such that the aorta can not keep up with excess blood pressure through dilation and recoilation. It is caused by a spirochaete, Treponema pallidum. It damages the reproductive organs, eyes bones joints, central nervous system, heart, and skin. Sexual contact is the major source of its dissemination.

Q44. Spongy bone is always surrounded by:

• A. Compact bone✓
• B. Cartilage
• C. Osteoblast cells
• D. Osteoclast cells

Explanation: The spongy bone is usually surrounded by a shell of the compact bone which provides added strength and rigidity. Spongy bone is usually located at the ends of the long bones (the epiphyses), with the harder compact bone surrounding it. It is also found inside the vertebrae, in the ribs, in the skull and the bones of the joints. Spongy bone is softer and weaker than compact bone but is also more flexible.

Q45. Bone matrix is hardened by the:

• A. Haversian canals
• B. Canaliculus
• C. Bone marrow tissues
• D. Calcium phosphate✓

Explanation: Calcium phosphate is responsible for the mineralization of the bones that essentially gives them their rigidity.

Q46. The number of bones forming skull in man is:

• A. 8
• B. 14
• C. 20
• D. 22✓

Explanation: The skull is a complex of 22 bones which can be divided into two categories: the cranium and the facial bones. The cranium bones are responsible for surrounding and protecting the brain whereas the facial bones provide support to soft facial tissues.

Q47. The spine consists of linear series of:

• A. 33 bones✓
• B. 24 bones
• C. 12 bones
• D. 7 bones

Explanation: The spine or the vertebrae consists of 33 bones interlocked with each other to give the spinal column.

Q48. Which of the following changes occurs when skeletal muscles contract:

• A. I-band shortens only
• B. A-band shortens and Z-lines move apart
• C. I-band shortens and Z-lines come close to each other✓
• D. Actin filament contracts

Explanation: The I- band is the region near the Z - disc where we only have the actin filament. As the sarcomere contracts, the actin overlaps with the myosin filament, shortening the I band as well as bringing the Z discs closer to each other.

Q49. The thyroxine hormones of thyroid glands act directly on:

• A. Iodine metabolism
• B. Protein metabolism
• C. Glucose metabolism
• D. Basal metabolic rate✓

Explanation: Thyroxine hormone stimulates various metabolic activities such as the metabolism of glucose, cholesterol metabolism etc leading to an increase in basal metabolic rate.

Q50. All the hormones released by the anterior pituitary are tropic hormones except:

• A. TSH
• B. STH✓
• C. ACTH
• D. Gonadotropin hormone

Explanation: Somatotrophic hormone (STH), also known as growth hormone (GH), is the only hormone released by the anterior pituitary that is not primarily a tropic hormone. It directly influences growth and metabolism in tissues rather than causing another gland to release hormones. Other hormones like TSH, ACTH, and gonadotropins are considered tropic because they stimulate other endocrine glands, such as the thyroid, adrenal cortex, and gonads, to produce and release hormones.

Q51. Which of the following is endocrine as well as exocrine?

• A. Liver
• B. Adrenals
• C. Thyroid
• D. Pancreas✓

Explanation: The pancreas has the dual function of secreting glucagon and insulin into blood (endocrine) and secreting enzymes in the form of the pancreatic juice through ducts (exocrine).

Q52. Ovulation is suppressed by progesterone via:

• A. Only by inhibition of Luteinizing Hormone.✓
• B. Inhibition of FSH & stimulation of LH.
• C. Inhibition of LH & stimulation of FSH.
• D. Inhibition of both FSH & LH.

Explanation: Luteinizing Hormone is the one which stimulates the ovulation of the follicle. The surge in the amount of LH, before the ovulation, is prevented which in turn prevents ovulation and the release of the egg.

Q53. The antibody molecule consists of _ polypeptide chains.

• A. Eight
• B. Four✓
• C. Six
• D. Two

Explanation: An Antibody molecule is made up of 4 polypeptide chains i.e. two heavy chains and two light chains. (PTB page:326)

Q54. _ cells survive for a few days and secrete a huge no. of antibodies in blood, tissue fluids or lymph.

• A. Memory cells
• B. B-lymphocytes
• C. T-lymphocytes
• D. Plasma cells✓

Explanation: Plasma B cells are cells produced or specialized in the bone marrow. They tend to release significant amounts of antibodies in response to the presence of foreign antigens.

Q55. The intermediate protection from infection of snake bite can be obtained by:

• A. Active Immunity
• B. Natural active immunity
• C. Passive immunity✓
• D. Vaccination

Explanation: The method of passive immunization is used to combat active infections such as tetanus, rabies, snake venom, etc. (PTB page:327)

Q56. Chlorophyll molecule contains:

• A. Mg2+✓
• B. Ca2+
• C. K+
• D. Na+

Explanation: Mg2+ ions are essential in making chlorophyll similar to how iron ions are substantial in making haemoglobin. The Mg ions are found in the centre of the chlorophyll molecule.

Q57. The tail of chlorophyll molecule is embedded in:

• A. Membrane of mitochondria
• B. Thylakoid membrane✓
• C. Membrane of Smooth Endoplasmic Reticulum
• D. Membrane of Rough Endoplasmic Reticulum

Explanation: The thylakoid membrane of the chloroplasts has these light-harvesting clusters called photosystems which contain various pigments needed to absorb light. These pigments, including chlorophyll, have their tails embedded in the thylakoid membrane as both are hydrophobic. The head of the molecule acts as a solar panel, facing and absorbing light.

Q58. Carotenoids absorb light of:

• A. Yellow-orange range
• B. Yellow-red range
• C. Orange-red range
• D. Blue-violet range✓

Explanation: Carotenoids are one of the accessory pigments in the chloroplasts which absorb light in the Blue-Green and Violet range.

Q59. Chlorophyll 'a' and chlorophyll 'b' differ in one of the functional group. Chlorophyll 'a' has:

• A. -CHO
• B. -OH
• C. -CH3✓
• D. -NH2

Explanation: The only difference between chlorophyll-a and b is that chlorophyll-a has methyl (CH3) on the 2nd pyrrole ring and whereas chlorophyll-b has an aldehyde (-CHO) at this point.

Q60. Glycerate-3-phosphate in the presence of ATP and reduced NADP from light dependent stage is reduced to:

• A. 3-carbon compound✓
• B. Ribulose bisphosphate
• C. 5-carbon compound
• D. 6-carbon compound

Explanation: Glycerate-3-phosphate, being a carboxylic acid, gets reduced to glyceraldehyde phosphate which is a 3-carbon compound.

Q61. The Calvin cycle occurs in:

• A. Chloroplast✓
• B. Mitochondria
• C. Glyoxysomes
• D. Cytoplasm

Explanation: In the stroma of the chloroplast, carbon dioxide is fixed into organic molecules using ATP and NADPH produced during the light reactions. Prokaryotes do not have chloroplasts, although some photosynthetic bacteria carry out similar processes in their cytoplasm. Mitochondria are the sites of cellular respiration, not photosynthesis, and are involved in energy release rather than carbon fixation. The cytoplasm is where other metabolic processes like glycolysis occur, but it does not carry out the Calvin cycle. Glyoxysomes are specialised organelles involved in fat metabolism in germinating seeds, not in photosynthesis. Therefore, the Calvin cycle occurs only in the chloroplasts.

Q62. Restriction enzyme EcoR1 cuts DNA to produce:

• A. Blunt ends
• B. Non-palindromic ends
• C. Sticky ends✓
• D. Split ends

Explanation: EcoRI is a restriction enzyme. Restriction enzymes either make staggered cuts or blunt cuts. A staggered cut is one in which the duplex fragment shows single-stranded projected ends called sticky ends. Blunt cuts do not produce sticky ends.

Q63. Restriction endonucleases are produced by:

• A. Fungi
• B. Algae
• C. Bacteria✓
• D. Viruses

Explanation: Restriction endonucleases are also called restriction enzymes. They are naturally found in bacteria where they serve a host-defense role. (NBF page:254)Thus option C is the correct option.

Q64. DNA segments of different lengths can be separated by a process of:

• A. Western blotting
• B. Northern blotting
• C. Autoradiography
• D. Gel electrophoresis✓

Explanation: Gel electrophoresis is a technique used in molecular biology to separate different-sized fragments of charge-bearing polymers (proteins, RNA or DNA) under the influence of an electric field in a semi-solid gel medium of agarose or polyacrylamide.

Q65. This is the 1st heat-stable component used in PCR:

• A. Taq-isomerase
• B. Taq-helicase
• C. Taq-polymerase✓
• D. Taq SSBp

Explanation: Taq-polymerase is a polymerase enzyme extracted from the heat-stable bacteria called Thermus aquaticus which lives in hot springs. The bacteria has its enzymes adapted to work in extreme heat making Taq polymerase a heat-stable compound.

Q66. Patients of cystic fibrosis (CF) produces thick mucus because of faulty:

• A. Trans-membrane carrier✓
• B. Cl- ions
• C. Na+ ions
• D. Mucus membrane

Explanation: The trans-membrane carrier is responsible for allowing Chlorine ions to enter into the mucus so that water may follow due to osmosis, making the mucus less viscous. In CF, this carrier protein is defective so it doesn’t provide a path for the chlorine ions to enter into the mucus, rendering the mucus thick.

Q67. Chemicals used for destroying agricultural competitors are known as:

• A. Antibiotics
• B. Pesticides✓
• C. Disinfectants
• D. Chemotherapeutic agents

Explanation: Pesticides are a class of chemicals that are used to kill crop-eating insects, unwanted weeds etc. Insecticides, herbicides, nematicides; they all come under the umbrella of pesticides.

Q68. How denitrification does occur in soils?

• A. Bacterial reduction of NO3- ions to N2 gas.✓
• B. Active uptake of Nitrate ions by plant roots
• C. Drainage of manure from fields
• D. Leaching of nitrate ions

Explanation: Denitrification is a natural soil microbial process where nitrate ions are converted to nitrogen gas which is then lost to the atmosphere. Denitrification occurs when soil bacteria use nitrate for their respiration in place of oxygen in the air.

Q69. Process by which unrelated species evolve to functionally resemble each other is called:

• A. Convergent evolution✓
• B. Divergent evolution
• C. Coevolution
• D. Parallel evolution

Explanation: The pattern of evolution in which different species have evolved from different ancestors in a common habitat is known as convergent evolution. Such organisms have similar functions but differ in structures.

Q70. Which of the following shows evidence from evolution through molecular biology?

• A. Development of branchial arches invertebrate embryo.
• B. Distribution of species
• C. Comparison of genes and proteins in different species✓
• D. Study of vestigial organs

Explanation: To compare species to see how they have evolved differently, we look at their gene sequences or the amino acid sequence of essential proteins such as the sequence of the cytochrome C complex. The comparison gives us an idea of how closely related the species are to each other and how they have evolved to be different.

Q71. Large population size, random mating, no mutation and no emigration or immigration are the postulates of:

• A. Hardy-Weinberg Theorem✓
• B. Mendel's law of independent assortment
• C. Mendel's law of segregation
• D. Theory presented by Schleiden and Schwann

Explanation: Hardy-Weinberg Theorem allows us to calculate the allele frequency in populations where the phenotype that we are examining is the same for a homozygous dominant and heterozygous condition but different for the homozygous recessive condition. The postulates for this theorem to be applicable to any population are; no mutation, random mating, no gene flow (no migration from or into the population), infinite population size (large population), and no natural selection or selective advantage.

Q72. Pure breeding lines of pea were taken regarding seed shape —Round and wrinkled and were crossed with no intermediate between parents. All offspring were found to be round. These results show:

• A. Codominance
• B. Dominance-recessive relationship✓
• C. Incomplete dominance
• D. Over dominance relationship

Explanation: The round phenotype is due to the presence of the dominant allele which expresses itself regardless of the presence of the recessive allele. All the offsprings are, apparently, heterozygous and inherited the dominant allele coding for the round shape hence all of them turned out to be round.

Q73. Base substitution, deletion and insertion are examples of:

• A. Chromosomal aberration
• B. Aneuploidy
• C. Point mutation✓
• D. Euploidy

Explanation: All these mutations revolve around a single nucleotide being replaced, inserted or deleted hence the term ‘point’.

Q74. The condition in which the heterozygote has a phenotype intermediate between contrasting homozygous parents is called as:

• A. Dominance
• B. Incomplete dominance✓
• C. Codominance
• D. Over-dominance

Explanation: Incomplete dominance is when a dominant allele, or form of a gene, does not completely mask the effects of a recessive allele, and the organism’s resulting physical appearance shows a blending of both alleles.

Q75. The interaction between different genes occupying different loci is:

• A. Dominance
• B. Codominance
• C. Pleiotropy
• D. Epistasis✓

Explanation: Epistasis is an example of nonallelic interactions. It is defined as the phenomenon in which the effect caused by genes on one locus interferes or hides the effect caused by another gene at another locus.

Q76. Locus stands for:

• A. Position of gene on homologous chromosomes
• B. Regions of chromosomes
• C. Position of an allele within a DNA molecule.
• D. Close regions of the same chromosomes
• E. A and C✓

Explanation: The correct answers are Options A and C. A locus refers to the specific position of a gene on a chromosome, and homologous chromosomes have the same genes at the same loci, but may carry different alleles, which is covered by both options. Option B is incorrect as it does not specify the precise nature of a locus. Option D is incorrect as it confuses the concept of a locus with genomic proximity unrelated to specific gene positions.

Q77. Self fertilization of F-1 dihybrids, following independent assortment of alleles result in:

• A. 3/16 Tall-round ; 3/16 dwarf-wrinkled
• B. 9/16 Tall-wrinkled ; 3/16 dwarf-round
• C. 9/16 Tall-round ; 3/17 Dwarf-round
• D. 3/16 Tall-wrinkled ; 3/16 Dwarf-round✓

Explanation: As we talk about the F1 generation, we can assume them to be heterozygous for both characteristics. We can consider T as the allele coding for Tall, t for dwarf, R for round and r for wrinkled. By drawing a Punnett square, we can find the ratio of the different phenotypes.

Q78. As a result of cross-fertilization of a pure breeding pea plant having purple coloured flowers; with that of white coloured flowers, the offsprings will have flowers with:

• A. ¼ purple ; ¾ white
• B. ¼ white ; ¾ purple
• C. All white
• D. All purple✓

Explanation: As it is mentioned in the question that it is pure breeding, we can assume the pea plants involved in the breeding to be homozygous dominant and recessive, accordingly. As the allele coding for purple flowers is dominant over white, we would get all offsprings having purple flowers since all the offsprings would be heterozygous.

Q79. The gene for red-green color blindness is present on:

• A. Y-chromosome
• B. X-chromosome✓
• C. Autosome 7
• D. Autosome 9

Explanation: Red- Green Color blindness is a sex-linked recessive disease which is passed down through the X chromosomes.

Q80. Which of the following structures is present in both plant and animal cells but is absent in prokaryotic cells?

• A. Centrioles
• B. X-chromosome✓
• C. Plastids
• D. Sieve-tubes

Explanation: Since prokaryotic cells do not indulge in sexual reproduction, they don’t have either the X chromosomes or the Y chromosomes.

Q81. Cilia and flagella are absent in:

• A. Viruses
• B. Bacteria
• C. Higher plants✓
• D. Lower animals

Explanation: Both prokaryotic and eukaryotic cells contain structures known as cilia and flagella. They are motile cellular appendages found in most microorganisms and animals, but not in higher plants since they don't need to move. The explanation is given below.

Q82. DNA molecule in prokaryotes is:

• A. Single, circular, double-stranded molecule not bound by a membrane✓
• B. Double, circular molecule
• C. Linear, double-stranded molecule
• D. Single, circular, double-stranded and membrane bound

Explanation: Prokaryotic cells, such as bacteria, have a unique DNA structure that is essential for their function and reproduction. The genomic DNA is a single, circular, double-stranded molecule that is not contained within a membrane-bound nucleus, allowing it to be freely dispersed in the cytoplasm. This structure is fundamental for the process of transcription and replication in prokaryotes. In contrast, the other options incorrectly describe prokaryotic DNA. Option B suggests it is a double molecule, which is false as prokaryotic DNA is single. Option C describes it as linear, which is characteristic of eukaryotic DNA. Option D incorrectly states that it is membrane-bound, which contradicts the nature of prokaryotic cells.

Q83. Nucleoid is a structure not found in:

• A. Campylobacter
• B. Cyanobacteria
• C. Spirochete
• D. Goblet cells✓

Explanation: Nucleoid is exclusive to prokaryotic cells only where the circular DNA exists without a membrane. Goblet cells are eukaryotic.

Q84. Cell wall structure of a cell of unknown origin was studied and was found to contain a polysaccharide chain linked with short chains of amino acid.What do you think it can be?

• A. Bacteria✓
• B. Fungi Cell
• C. Algae
• D. Cortex cells

Explanation: Bacterias have their cell walls made from peptidoglycan which is known to have sugars and amino acids as its constituents.

Q85. Ribosomes present in prokaryotes are:

• A. 80S
• B. 50S
• C. 60S
• D. 70S✓

Explanation: The size of the ribosomes present in prokaryotic and eukaryotic cells is 70s and 80s, respectively. 70S ribosomes are, however, present in some eukaryotic organelles like mitochondria and chloroplasts.

Q86. Functionally mesosomes can be compared with:

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

Explanation: Mesosomes are found in prokaryotes and have functions similar to mitochondria as they both assist in cellular respiration. It also has functions in DNA replication and cell division or excretion of exoenzymes.

Q87. Students were asked to give a guess about a unicellular organism with darkly stained nucleus.Which of the following can be straight away excluded from the list:

• A. Paramecium
• B. Amoeba
• C. Plasmodium
• D. Lactobacillus✓

Explanation: Eukaryotes have a definite nucleus and prokaryotes do not. Instead, prokaryotes have a nuclear region called the nucleoid. Among the given options, lactobacillus is only prokaryotic thus it does not have a nucleus. The explanation is given below.

Q88. Binary fission is a characteristic cell division NOT found in:

• A. Pseudomonas
• B. Campylobacter
• C. Euglena✓
• D. E.coli

Explanation: Euglena is a genus of single-cell flagellate eukaryotes. All other mentioned organisms are prokaryotes. Binary fission is a characteristic of prokaryotes only.Thus option C is the correct answer as Euglena does not undergo binary fission.

Q89. _ are the specific structures related to monosaccharides:

• A. Glycosidic Bond
• B. Keto group✓
• C. Maltose
• D. Fructose

Explanation: Monosaccharides are polyhydroxy aldehydes or polyhydroxy ketones. This implies that monosaccharides contain either an aldehyde group or a ketone group.

Q90. _ are the major site for storage of glycogen in animal's body.

• A. Muscle and liver✓
• B. Around thighs and belly
• C. Around belly and hips
• D. Liver and kidneys

Explanation: Our body mainly stores glycogen in the liver and muscles. When the body doesn’t immediately need glucose from the food we eat for energy, it stores glucose primarily in our muscles and liver as glycogen for later use.

Q91. The number of amino acids that have been found to occur in cells and tissues are:

• A. 170✓
• B. 25
• C. 20
• D. 45

Explanation: The correct answer is 170. This number represents the total variety of amino acids that have been discovered in cells and tissues. Out of these, about 25 are used in protein synthesis, known as the standard or proteinogenic amino acids. The other amino acids may have various roles in metabolism or structural functions but are not typically involved in protein construction. Options B, C, and D are incorrect because they either only account for proteinogenic amino acids or underestimate the total number of amino acids present in cells and tissues.

Q92. Most proteins are made up of _ type of amino acids.

• A. 20✓
• B. 170
• C. 25
• D. 200

Explanation: Many naturally occurring amino acids have been found however, only 20 of those types are known to be found in the proteins making up the body.

Q93. If in lipids there is a higher proportion of unsaturated fatty acid, then it will be:

• A. Oils✓
• B. Phenols
• C. Waxes
• D. Fats

Explanation: Option A is correct. When lipids have a higher proportion of unsaturated fatty acids, they tend to be liquid at room temperature, thus classified as oils. This is due to the presence of double bonds in unsaturated fatty acids, which prevent the fatty acid chains from packing closely together, increasing fluidity. In contrast, phenols are unrelated aromatic compounds, waxes are solid lipids with saturated or monounsaturated fatty acids, and fats are typically solid at room temperature due to a higher proportion of saturated fatty acids.

Q94. When X-rays are passed through crystalline DNA, it shows helix making one twist every:

• A. 2nm
• B. 3.4nm✓
• C. 34nm
• D. 4nm

Explanation: The two strands of DNA are wound around each other so that there are 10 base pairs in each turn of about 3.4nm.

Q95. Following is the structure of:

• A. Uracil
• B. Thymine
• C. Guanine
• D. Cytosine✓

Explanation: The given structure is of a pyrimidine base which bears resemblance to the structure of Cytosine.

Q96. All enzymes are _:

• A. Fibrous proteins
• B. Low molecular weight proteins
• C. Lipoproteins
• D. Globular proteins✓

Explanation: Enzymes being globular proteins have multiple polypeptide chains interacting with each other to give a 3D structure. This 3D structure is unique and in enzymes, its importance is essential with regard to the active site of the enzymes.

Q97. The reactants on which enzyme works are:

• A. Products
• B. Metabolites
• C. Substrates✓
• D. Catabolites

Explanation: Enzymes work by acting on substrates and converting them into products. The substrate (reactant which is to be converted into the product) molecule is attached to the active site of the enzyme by non-covalent interactions.

Q98. Which of the following comprises inorganic ions?

• A. Coenzymes
• B. Activators✓
• C. Prosthetic group
• D. Apoenzyme

Explanation: Activators are inorganic ions that bind to enzymes and play a crucial role in enzyme activity. They are often metal ions like Mg²⁺, Fe²⁺, Cu²⁺, and Zn²⁺. These ions stabilize the enzyme structure or assist in the conversion of substrates to products. In contrast, coenzymes and prosthetic groups are organic and serve different roles in enzymatic reactions. An apoenzyme is the inactive protein form of an enzyme that requires a cofactor, either inorganic or organic, to become active.

Q99. Which of the following is a non-cellular infectious entity?

• A. Mycoplasma
• B. Escherichia coli
• C. Herpes virus✓
• D. Diplococcus

Explanation: Herpes virus is essentially a virus and since viruses do not fall under the category of cellular beings, herpes virus is a non-cellular infectious entity.

Q100. The viruses can reproduce:

• A. Without invading any cell.
• B. In bacterial cell.✓
• C. By Mitosis
• D. By Meiosis

Explanation: Viruses are incapable of reproducing on their own, so they hunt for a host cell for hijacking. Once the host cell has been hijacked, the cell machinery is used to make copies of the genetic material of the virus.

Q101. The life cycle in which the phage kills the bacteria is known as:

• A. Transduction
• B. Temperate phage cycle
• C. Lytic cycle✓
• D. Lysogenic phage cycle

Explanation: In the lytic cycle the phage infects a bacterium, hijacks the bacterium to make lots of phages, and then kills the cell by making it explode.

Q102. In which of the following shapes, gut living symbiont Escherichia coli is found?

• A. Round
• B. Oval
• C. Spiral
• D. Rod✓

Explanation: Escherichia coli is an example of rod shaped bacteria, also called Bacilli.

Q103. Chitin, a chemical found in exoskeleton of arthropods is also found in cell wall of:

• A. Bacteria
• B. Fungi✓
• C. Cyanobacteria
• D. Algae

Explanation: Chitin is a substance made from fibrous polysaccharides. They are famously known for being used to construct the cell walls of eukaryotic Fungi.

Q104. Snails are the intermediate hosts in:

• A. Fasciola hepatica✓
• B. Schistosoma
• C. Taenia solium
• D. Ancylostoma duodenale

Explanation: Snails are the intermediate hosts of Fasciola hepatica.It is an endoparasite in sheep and occasionally in human beings. It has suckers used for attachment to host tissue. It completes its life cycle in two hosts, a snail, sheep, or man. It lives in the bile duct of its hosts.

Q105. _ is an intestinal parasite of man belonging to phylum Nematoda.

• A. Taenia solium
• B. Wuchereria bancrofti
• C. Ascaris lumbricoides✓
• D. Schistosoma

Explanation: Ascaris Lumbricoides is a large, parasitic roundworm known for surviving in the human intestine. Ascaris lumbricoides is a roundworm which belongs to Phylum Nematoda.

Q106. Food is diverted in the esophagus by:

• A. Glottis
• B. Tongue
• C. Cheeks
• D. Epiglottis✓

Explanation: The epiglottis is a leaf-shaped flap of cartilage located behind the tongue, at the top of the larynx. The main function of the epiglottis is to seal off the windpipe during eating so that food is not accidentally inhaled and goes into the oesophagus.

Q107. Label 'a' in the following diagram:

• A. Cardiac sphincter
• B. Sinoatrial valve
• C. Stomach valve
• D. Pyloric sphincter✓

Explanation: Refer to the following diagram

Q108. Enzyme pepsin acts on:

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

Explanation: Pepsin is a proteolytic enzyme, meaning it breaks down proteins. It functions in the acidic environment of the stomach, where the stomach lining secretes hydrochloric acid (HCl) to create an acidic environment (pH around 1-2). In this acidic environment, pepsinogen, the inactive precursor of pepsin, is secreted by the chief cells of the gastric glands. The acidic conditions in the stomach convert pepsinogen into its active form, pepsin. Pepsin then acts on dietary proteins, cleaving the peptide bonds between amino acids and breaking the proteins down into smaller peptides.

Q109. Following is the structure of gastric glands in stomach wall where 'x' is:

• A. Mucosa
• B. Visceral fat cells
• C. Mucus cells
• D. Oxyntic cells✓

Explanation: Since the other labeled cells are Zymogen cells (also known as acinar cells, which secrete digestive enzymes), the unlabeled cell type marked "X" is the Oxyntic cells (also called parietal cells) are found in the gastric glands of the stomach.They secrete hydrochloric acid (HCl) and intrinsic factor, which are involved in digestion and vitamin B12 absorption.

Q110. Label the part 'Y' in the following diagram:

• A. Pleura
• B. Chest cavity
• C. Diaphragm✓
• D. Inter-coastal muscle

Explanation: The diaphragm is a thin skeletal muscle that sits at the base of the chest and separates the abdomen from the chest. The pleura, chest cavity, and intercostal muscle are not present below the lungs. So, the correct answer is diaphragm.

Q111. Which of the following is a respiratory disorder related to malnutrition:

• A. Cancer
• B. Asthma
• C. Emphysema
• D. Tuberculosis✓

Explanation: Tuberculosis, in addition to causing symptoms such as coughing and fever, often results in significant weight loss and lack of appetite. The association between TB and malnutrition is bi-directional: TB leads the patient to malnutrition, and malnutrition increases the risk of developing active TB by 6 to 10 times.

Q112. In NO3- the oxidation number of N is:

• A. 5+✓
• B. 2+
• C. 3+
• D. 3-

Explanation: NO3N = xO = - 2x + (-2) × 3 = - 1x + (- 6) = - 1x – 6 = - 1x = - 1 + 6x = 5

Q113. The E0 value of the standard copper half cell is 0.34, measured when it is connected withthe SHE i.e. Standard Hydrogen Electrode. In this case the half cell reaction taking place atSHE is:

• A. 2H+(aq) + 2e- → H2(g)
• B. H2(g) → 2H+(aq) + 2e-✓
• C. 2H+(aq) + 2e- → 2H(g)
• D. 2H2- → 2H(g) + 2e-

Explanation: The Standard Hydrogen Electrode (SHE) serves as a benchmark for measuring standard electrode potentials, with a potential of 0 V. In the given context, the copper electrode is positive, making the SHE the anode. Thus, oxidation occurs at the SHE, involving the conversion of hydrogen gas (H2) into hydrogen ions (2H+) and electrons (2e-), as represented by the correct option B. Option A represents reduction, not oxidation. Option C describes an atom formation not typical at the SHE. Option D depicts an erroneous reaction involving a non-existent species.

Q114. Consider the following reversible reaction:Initial concentrations:CH3-CH2-OH) = 1 mol.dm-3(CH3-COOH)= 1 mol.dm-3(CH3-CH2-O-CO-CH3)= 0 mol.dm-3(H2O)= 0 mol.dm-3Equilibrium concentrations:(CH3-CH2-OH) = 0.33mol.dm-3(CH3-COOH)= 0.33 mol.dm-3(CH3-CH2-O-CO-CH3)= 0.66 mol.dm-3(H2O)= 0.66 mol.dm-3Kc= 4 at temperature 100 CWhat are the new equilibrium concentrations of all species if 1 mol.dm-3 of CH3-CH2-OH and CH3-COOH are added to this equilibrium mixture? (Apply Le-Chatelier’s principle)(Temperature and Kc remain constant)

• A. (CH3-COOH)= 0.333 mol.dm-3(CH3-CH2-OH) = 1.333mol.dm-3(CH3-CH2-O-CO-CH3)= 1.666 mol.dm-3(H2O)= 1.666 mol.dm-3
• B. (CH3-COOH)= 1.333 mol.dm-3(CH3-CH2-OH) = 0.333mol.dm-3(CH3-CH2-O-CO-CH3)= 0.666 mol.dm-3(H2O)= 0.666 mol.dm-3
• C. (CH3-COOH)= 0.666 mol.dm-3(CH3-CH2-OH) = 0.666 mol.dm-3(CH3-CH2-O-CO-CH3)= 1.333 mol.dm-3(H2O)= 1.333 mol.dm-3✓
• D. (CH3-COOH)= 0.333 mol.dm-3(CH3-CH2-OH) = 0.333mol.dm-3(CH3-CH2-O-CO-CH3)= 1.333 mol.dm-3(H2O)= 1.333 mol.dm-3

Explanation: Change in concentration of of CH3COOH is 1-0.33 = 0.66 mol.dm-3Change in concentration of of CH3CH2OH is also 1-0.33 = 0.66 mol.dm-3Where as change in conc of CH3CH2OCOCH3 is 1 + 0.333 = 1.333 moldm-3And change in conc for H2O is also 1 + 0.333 = 1.333 mol.dm-3

Q115. For which of the following equilibrium reaction, Kc has no units:

• A. N2 + 3H2--------> 2NH3
• B. SO2 + 2O2--------> 2SO3
• C. CO + H2O ----------> CO2 + H2✓
• D. 2NO2 + O2---------> 2NO3

Explanation: There are equal moles of reactants and products therefore Kc will have no units as they get cancelled out.

Q116. Choose the type of catalysis in the following reaction:

• A. Homogeneous catalysis✓
• B. Heterogeneous catalysis
• C. Biological catalysis
• D. Gas catalysis

Explanation: The given reaction 2SO2(g) + O2(g) -> 2SO3(g) with NO2(g) as a catalyst represents Homogeneous catalysis. In this type of catalysis, the catalyst and reactants are in the same phase, specifically gaseous in this case. NO2(g) temporarily forms intermediates, facilitating the conversion without being consumed.In contrast, Heterogeneous catalysis would involve different phases, Biological catalysis requires enzymes, and Gas catalysis is not a standard type of catalysis.

Q117. Which of the following graphs is the representation for more rapidly catalysed reaction?

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

Explanation: The activation energy is the lowest in Option C, hence it would be the most rapid reaction among the other options.

Q118. The following graph shows a physical property along the period 3 elements. Which physical property is shown in the graph?

• A. Electron Affinity
• B. Non-metallic character
• C. Atomic Radius✓
• D. Melting point upto group IV

Explanation: The atomic radius decreases as we move along the period from left to right. Because, the shell remains the same among all the elements (since they're all in the same period) and, since the proton number increases the force on attraction increases on the valence shell, hence the radius decreases along the period.

Q119. The following sketch shows the melting point of eight elements with consecutive atomic numbers. Which element is Silicon?

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

Explanation: Silicon has a giant covalent structure and a lot of heat is required to break the strong covalent bonds hence it has the highest melting point in the graph above.

Q120. 6NaOH + 3Cl2 ————> 5NaCl + NaClO3+3H2OIn the above disproportion reaction, the oxidation state of chlorine is changed from zero to _ and _.

• A. -1, +1
• B. -1, +3
• C. -1, +5✓
• D. +1, +5

Explanation: In NaCl, Na has an oxidation state of +1 so Cl is -1 and in NaClO3, Na is +1 and O is -6 while Cl is an unknown x. According to the calculation1+x-6 = 0x= +5 so Cl in NaClO3 is +5.

Q121. Which noble gas is alpha emitter?

• A. Xenon
• B. Radon✓
• C. Krypton
• D. Argon

Explanation: Radon emits alpha particles to become Polonium-128.

Q122. Scandium has the atomic number 21, which one will be its electronic configuration?

• A. 1s2,2s2,2p6,3s2,3p6,3d3
• B. 1s2,2s2,2p6,3s2,3p6,4s2,3d1✓
• C. 1s2,2s2,2p6,3s2,3p6,4s2,4p1
• D. 1s2,2s2,2p6,3s2,3p6,4s1,4p2

Explanation: The electronic configuration for Sc is 1s2, 2s2, 2p6, 3s2, 3p6 and since 4s is filled before 3d, we get 4s2 and 3d1

Q123. Violet color of [Ti(H2O)4]+is due to:

• A. Central metal ion✓
• B. Water molecule
• C. Complex ion
• D. Outer anion

Explanation: The central metal ion has the coordination number 4. To bring the electrons from lower to higher energy levels, the photon is absorbed and the colour seen is complementary to the wavelength absorbed.

Q124. Nitrogen gas reacts under _ conditions.

• A. Standard
• B. Normal
• C. Cool
• D. Harsh✓

Explanation: Nitrogen gas is a molecule with a triple bond between the atoms, which makes Nitrogen unreactive under normal conditions as it takes a lot of energy to break the N-N bond.

Q125. Liquid Ammonia has become an important fertilizer for direct application to soil. It contains _Nitrogen.

• A. 46%
• B. 82%✓
• C. 14%
• D. 17%

Explanation: The Mr of Ammonia (NH3) is 14+3=17It contains ammonia in the percentage of :14/17 x 100 = 82%

Q126. SO3 formed in the contact process is absorbed in _% H2SO2

• A. 90
• B. 80
• C. 98✓
• D. 89

Explanation: In contact process, SO3 is formed in 98% of H2SO2

Q127. The balanced chemical equation to manufacture ammonia by Haber’s Process is:

• A. 3N2 (g) + 3H2 (g) <=====> 2NH3 (g)
• B. N2 (g) + H2 (g) <=====> NH3 (g)
• C. 3N2 (g) + H2 (g) <=====> 2NH3 (g)
• D. N2 (g) + 3H2 (g) <=====> 2NH3 (g)✓

Explanation: Ammonia is synthesized in a process called Haber Process by Nitrogen (N2) and Hydrogen (H2) First of all, write the chemical equation for the formation of Ammonia N2 + H2 —--> NH3 Now balance the number of Nitrogen and Hydrogen on both sides N2 + 3H2 —--> 2NH3

Q128. Which one of the following is used as a typical catalyst for catalytic cracking?

• A. Mixture of SiO2 and Ni
• B. Mixture of Pt and Cu
• C. Mixture of Fe and MgO
• D. Mixture of SiO2 and Al2O3✓

Explanation: The catalysts required in catalytic cracking are usually solid acidic catalysts that are a mixture of SiO2 and Al2O3

Q129. The type of structural isomerism which arises due to the differences in nature of carbon chains or carbon skeletons is:

• A. Chain isomerism✓
• B. Position isomerism
• C. Cis-Trans isomerism
• D. Optical isomerism

Explanation: In chain isomerism, the molecular formula of the isomers is same but they differ in the carbon chain or carbon skeleton.

Q130. Which one of the following is the best name according to IUPAC system for the compound with the formula shown below?

• A. 4-methyl-6-chloro heptane
• B. 2-chloro-4-methyl heptane✓
• C. 2-chloro-4-n propyl hexane
• D. 2-chloro-4-n propyl pentane

Explanation: The longest carbon chain is 7 carbons long and has only single carbon-carbon bonds hence heptane.The numbering would favour chloro group over the methyl group, hence 2-chloro-4-methyl heptane.

Q131. Immediate product formed when propanoyl chloride reacts with benzene is:

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

Explanation: This set of reaction is known as Friedal Craft Acylation and in this, CH3CH3COCl i.e propanoyl chloride reacts with benzene and the product obtained is phenylpropane.

Q132. Which of the following are 3,5 (meta) directing groups when second group is induced in them:i) -NH3ii) -CHOiii) -COOHiv) -CH3

• A. ii, iii, iv
• B. ii, iii✓
• C. i, iv
• D. i, ii, iv

Explanation: The 3 5 directing groups or Meta directing groups are as follows:Aldehyde (-CHO), Ketone (-CO-), Ester (-RCOOR’), Carboxyl Group (-COOH) etc.Because the carbon in these groups is partially positive hence it withdraws the electrons.Other than them Nitro groups, Nitriles, Sulphones etc are also called meta-directing groups.

Q133. When benzene reacts with acetyl chloride (CH3COCl) in the presence of AlCl3, acetophenone is formed. The electrophile in this reaction will be:

• A. CH3CO+✓
• B. AlCl3
• C. CH3+
• D. CH3COCl

Explanation: CH3COCl reacts with AlCl3 to form —---> AlCl4- and CH3CO+CH3CO+ has a positive charge thus it is an electrophile and attracted to nucleophiles i.e. negative charges.

Q134. The reaction of bromine with benzene in the presence of FeBr3 follows the mechanism of:

• A. Electrophilic addition
• B. Electrophilic substitution✓
• C. Nucleophilic substitution
• D. Nucleophilic addition

Explanation: Benzene has an extended delocalised pi system above and below the ring which makes it highly attracted to electron-deficient species, which is why it readily undergoes electrophilic substitution.electrophilic substitution because an electrophile (H+) is being substituted from the reactant (benzene ring).

Q135. Which of the following is Halothane?

• A. Cl-CH2-CH2-Cl
• B. Cl-CH2-CH2-Br
• C. CF3-CHCl-Br✓
• D. Br-CH2-CH2-Br

Explanation: The following is the structure of halothane.

Q136. The non-stick lining of pans is:

• A. Difluoroethane
• B. Chloroethane
• C. Chlorofluororyhane
• D. Tetrafluoroethane✓

Explanation: The non-stick lining of pans is made of Teflon/PTFE which is a polymer. Its monomer is tetrafluoroethane.

Q137. In the elimination reaction, alcoholic KOH is used. OH- in this case, will act as:

• A. Electrophile
• B. Base✓
• C. Leaving group
• D. Acid

Explanation: Bases are proton acceptors and OH- is a base as it eliminates H+ (proton) from the reactant and forms a water molecule.

Q138. During the SN1 reactions, the fast reaction involves:

• A. Breakage of covalent bond
• B. Formation of carbocation
• C. Transition state
• D. Attack of nucleophile✓

Explanation: The nucleophile has a negative / partially negative charge and it readily attacks the carbocation in SN1 reactions (which has a positive charge) This is the fast step of the reaction.

Q139. Alcohol reacts slowly with Na metal compared to water because it has low concentration of H+ ion which suggests that it is:

• A. Less acidic than water✓
• B. Less basic than phenol
• C. More acidic than phenol
• D. More acidic than water

Explanation: The concentration of H+ ions determines the acidity. More H+ ions means the solution is more acidic as in the case of water while alcohol has a low concentration of H+ ions so it is less acidic than water.

Q140. CH3—CH2—OH + PCl5————> CH3—CH2—Cl + POCl3 + HClFormation of HCl is the test for the presence of _ in a compound.

• A. Alkyl group
• B. Hydroxyl group✓
• C. Saturated alkyl group
• D. Acidic H+ ion

Explanation: Hydroxyl groups react with chlorinating agents such as PCl5, PCl3 or SOCl2 and they give off misty fumes of HCl.

Q141. What will be the exact product?

• A. Diethyl ether
• B. Methyl propyl ether
• C. Ethyl acetate✓
• D. Butyl alcohol

Explanation: Ethyl Acetate is an ester that is made when Ethanoic acid reacts with Ethanol.

Q142. Choose the correct type for this reaction from the following:

• A. Reduction
• B. Oxidation
• C. Hydroxylation
• D. Hydration✓

Explanation: In a hydration reaction, a water molecule is added to the unsaturated reactants to form an alcohol. This is the last step of the hydration of alkenes in the presence of sulfuric acid.

Q143. Ethanal reacts with HCN to form cyanohydrin. This is an example of:

• A. Nucleophilic addition✓
• B. Electrophilic addition
• C. Electrophilic substitution
• D. Nucleophilic substitution

Explanation: It is an example of nucleophilic addition. The CN- ion has a negative charge and it is attracted to a positive charge so CN- is a nucleophile. CN- forms a sigma bond with the electron-deficient species i.e. C of ethanol.

Q144. The reactions of aldehydes and ketones with ammonia derivative G-NH3 to form compounds containing >C=N-C and water is known as _ reaction.

• A. Nucleophilic addition
• B. Nucleophilic substitution
• C. Electrophilic addition
• D. Addition elimination✓

Explanation: This reaction is called an addition-elimination reaction since a water molecule is eliminated from the reaction whereas the lone pair of electrons on Nitrogen form a dative bond with the electron-deficient carbon of aldehydes and ketones, thus adding it across the carbon.

Q145. Which one of the following compounds will give iodoform test on treatment with aqueous iodine?

• A. 3-pentanone
• B. Propanone✓
• C. Propanal
• D. Butanal

Explanation: Compounds that give a positive iodoform test are primary alcohol, methyl alcohol and methyl ketones. Propanone gives a positive iodoform test since C=O is bonded to CH3

Q146. What will be the product of the reaction given below?

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

Explanation: Ketones react with HCN in nucleophilic addition reaction and the product is a cyanohydrin i.e a compound which has OH and CN both bonded to the same carbon.

Q147. In the reaction below “?” represents which one of the following products:

• A. Ketone
• B. Aldehyde✓
• C. Formic Acid
• D. Ether

Explanation: Primary alcohols are oxidised to the corresponding aldehydes which are then oxidised to carboxylic acids.

Q148. Compounds having the –CN group are called:

• A. Cyano compounds
• B. Nitro compounds
• C. Carbon nitrogen compounds
• D. Nitriles✓

Explanation: In organic chemistry compounds which contain CN are called nitriles such as CH3CH2CN is called ethanenitrile.

Q149. Select the correct acidic strength order of chloro substituted acids:

• A. CH3COOH > ClCH2COOH > Cl2CHCOOH > Cl3CCOOH
• B. CH3COOH > Cl2CHCOOH > Cl3CCOOH > ClCH2COOH
• C. Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH✓
• D. Cl3CCOOH > CH3COOH > Cl2CHCOOH > ClCH2COOH

Explanation: Chlorine is an electronegative atom and it pulls the electron density towards itself thus spreading the charge and making the conjugate base stable. This leads to the carboxylic acid being more acidic. As the number of chlorines increases, the acidity increases too.

Q150. The phenoxide ion is more stable than ethoxide ion as:

• A. The lone pair on the oxygen atom overlaps with the delocalized π- bonding system in benzene.✓
• B. Oxygen atom is directly bonded with the benzene ring in the phenoxide ion.
• C. The negative charge is localized on the oxygen atom of phenoxide ion.
• D. The negative charge is delocalized on oxygen atom of ethoxide ion.

Explanation: In phenoxide ion, the lone pair on the oxygen atom forms a delocalised pi system with the benzene ring by involving the electrons of the ring. Moreover, oxygen is highly electronegative and it is stabilized by resonance. That’s why phenoxide ion is more stable than ethoxide ion.

Q151. Acidic character of amino acid is due to:

• A. –NH2
• B. –NH+3
• C. –COOH✓
• D. –COO-

Explanation: The correct answer is –COOH, which is the carboxylic acid functional group present in amino acids. This group can donate a proton (H+), which is the defining characteristic of acids, thus imparting acidic properties to amino acids.Options –NH2 and –NH+3 are incorrect because they represent amine groups that can accept protons, making them basic rather than acidic. The presence of the basic amino group does not contribute to the acidity of the amino acid.Option –COO- is also incorrect as it represents the deprotonated form of the carboxylic acid. While it is a fundamental part of the amino acid structure, it does not exhibit acidic properties.

Q152. The IUPAC name of alanine is:

• A. 2-aminopropanoic acid✓
• B. 2-aminoethanoice acid
• C. 2-aminobutane-1,4-dioic acid
• D. 2-aminobutanoic acid

Explanation: The IUPAC name of Alanine is 2-aminopropanoic acid since the amino group is bonded to carbon number 2 and the longest chain is of 5 carbons.

Q153. The amide linkage in Nylon-6,6 has the structure:

• A. –NH2–CO–
• B. –CO-O-
• C. –NH–CO–✓
• D. –NH–O–CO–

Explanation: Amide linkage is a covalent bond between COOH and NH2 of the amino acid. The linkage is CONH and a water molecule is eliminated.

Q154. The monomers needed to make ‘Terylene’, i.e. a polyester are:

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

Explanation: The monomers for the compound “Terylene” or Polyester are Ethylene glycol and Terephthalic acid. The reaction is as follows:

Q155. Which one of the following are the main functions of DNA?

• A. Making proteins.
• B. Making amino acids.
• C. Breakdown of ribose sugar
• D. Carries genetic material.✓

Explanation: DNA is a hereditary material which carries the genetic information generation after generation.

Q156. _is the major source of acid deposition in the atmosphere.

• A. SiO2
• B. CO2
• C. SO3✓
• D. Al2O3

Explanation: SO3 reacts with rain water to form H2SO4 (sulfuric acid) which is a major source of acid rain.

Q157. The energy from ultraviolet light is sufficient to break the _ bonds in CCl2F2.

• A. Cl-Cl
• B. Cl-F✓
• C. C-Cl
• D. C-F

Explanation: The Cl-F bond is less stronger than the C-Halide bond as it can be broken down by sunlight or UV radiation.

Q158. There are almost 200 million people alive in Pakistan. If you were to distribute 100 to each Pakistani in the form of 5 Rupee coins, how many moles of coins are you required to have?

• A. 6.67 x 10-14✓
• B. 1.5 x 10-14
• C. 6.67 x 1014
• D. 1.5 x 1014

Explanation: As we have to distribute 5 rupee coins to 100 people so,100/5 =20 coinsNow multiply these coins by the total population alive in Pakistan200000000 or 200 million x 20 = 4000 x 106As we know 1 mole = 6.02 x1023 molecules so,4000 x 106 / 6.02 x1023 = 6.67 x10-14.

Q159. A researcher has prepared a sample of 1-bromopropane from 10g of 1-propanol. After purification he had made 12g of product. Which of the following is percentage yield?

• A. 60%
• B. 58%✓
• C. 90%
• D. 50%

Explanation: The above question states that 1-bromopropane i.e CH3-CH2-CH2-Br has a molar mass of 123g is being prepared from 10g of 1-propanol i.e CH3-CH2-CH2-OH whose molar mass is 60g, we need to figure out the percentage yield as the given mass of product formed after purification is 12g, the ACTUAL YIELD. Our first step is to find out the theoretical yield which is the quantity of product calculated from a balanced chemical equation The balanced chemical equation is given below, the following reaction is an example of SN2 reactionCH3CH2CH2OH + HBr ============> CH3CH2CH2Br Since we know, 60g of propanol reacts to form 123g of bromopropane, then how many grams of bromopropane will be formed if 10g of propanol reacts?Using conversion factor:20.5g is the THEORETICAL YIELD. The next step is to find the percentage yield The formula for percentage yield is given as under %yield= actual yield/ theoretical yield x 100Plugging in the values, we get = 12g/ 20.5 x 100 = 58.5% Hence the correct answer is B, as the value 58.5 is closest to option B.

Q160. Which of the following has the same number of molecules as there are in 11g of CO2?

• A. 4g of O2
• B. 4.5g of H2O✓
• C. 4g of Cl2
• D. 1 moles of NaCl

Explanation: First, we need to calculate the number of moles of CO2 in 11g of CO2 So, 44g of CO2 is 1 mole, then 11g of CO2 is how many moles? Solution: using conversion factor: 11g x 1/ 44g = ¼ moles or 0.25 moles.

Q161. A sample of an organic compound consisting of carbon, hydrogen and oxygen was subjected to combustion analysis. 0.5439g of this compound gave 1.039g of carbon dioxide and 0.63469g of water vapours. The empirical formula of this compound is:

• A. CH2O
• B. C4H12O2
• C. C2H6O✓
• D. CH4O

Explanation: Convert g CO2 to g C.Convert g H2O to g H.g O = total g - g C - g Hg C = g CO2 x (C/CO2) = 1.039 x 12.01/44.01 = approx 0.2833g g H = g H2O x (2H/H2O) = 0.63469 x (2/18) = 0.07052g g O = .5439-0.07052-0.2833 = about 0.1900Now covert g C, g O, g H to mols.mols C = 0.2833/12 = about 0.0237molmols H = 0.0705/1 = about 0.0705molmols H = 0.0705/1 = about 0.0705molNow find the ratio of these elements to each other with the lowest number no less than 1.00 and round to whole numbers. The easy way to do that is to divide the smallest number by itself (making sure that is 1.00), then divide the other two numbers by the same small number. Here is what you have:C = 0.0237H = 0.0705O = 0.0119. Divide each by 0.0119.C = 0.0237/0.0119 = 1.99 = 2.0H = 0.0705/0.0119 = 5.92 = 6.0O = 0.0119/0.0119 = 1.0So the empirical formula is C2H6O.

Q162. 28g of N¬2 at STP will occupy a volume of:

• A. 22.41 dm3✓
• B. 44.82 dm3
• C. 64.82 dm3
• D. 2.241 dm3

Explanation: To find the volume of a gas, we use the formula: Moles = Volume of gas / 22. 41First, we find the moles of Nitrogen which are (mass / mr) so 28/28 = 1The volume of gas = Moles x 22.41Volume = 1 x 22.41 = 22.41 dm3

Q163. Study the following graphs for the boiling points of some substances:Which of the above graph shows that some members of the graph have hydrogen bonding?

• A. I + IV
• B. II + IV
• C. III + IV + V✓
• D. I + II + III

Explanation: Hydrogen bonding is found between the hydrogen atom of one molecule (which is covalently bonded to an electronegative atom) and the fluorine, oxygen and nitrogen of another molecule. It is represented with dotted lines.

Q164. What is the number of electrons in the following ion?

• A. 28✓
• B. 29
• C. 30
• D. 34

Explanation: No of electrons = no of protons = atomic number in a neutral atom. Since Ga-69 has atomic number 31 so its number of electrons is also 31 however by to question which is concerned with the number of electrons in Ga+3 ion, meaning this atom has lost 3 electrons so we subtract 3 from 31, we get 28. So the number of electrons in Ga+3 ion is 28. Hence option A is correct.

Q165. Isotonic symbol of the ion of Sulphur-33 is shown below. How many protons and neutrons are present if the number of electrons are 18?

• A. p= 18, n= 15
• B. p=16, n=17✓
• C. p=16, n=16
• D. p=17, n=16

Explanation: The proton number is 16 and the neutrons are 33-16= 17Note: the questions that follow after this point were not found in the pdf for the paper but were selected from different past papers

Q166. The yellowish colour of photochemical smog is due to the presence of:

• A. Nitrogen dioxide✓
• B. Dinitrogen trioxide
• C. Nitrous oxide
• D. Nitric oxide

Explanation: Explanation:The yellowish color of photochemical smog is primarily due to the presence of nitrogen dioxide (NO2), which is a brownish-yellow gas. Dinitrogen trioxide (N2O3) and nitric oxide (NO) are also nitrogen oxides, but they donot contribute significantly to the color of photochemical smog. Nitrous oxide (N2O) is a different compound that is not typically present in photochemical smog.

Q167. In Modern Periodic Table, the elements in Group II-B are:

• A. Zn, Cd, Pb
• B. Zn, Cd, Hg✓
• C. Zn, Cd, Ba
• D. Zn, Cd, Bi

Explanation: Explanation:In modern periodic table, the elements in group ll-B are Zn, Cd, and Hg. This option is not correct. Lead (Pb) belongs to Group IV-B of the modern periodic table. This option is also incorrect because Barium (Ba) belongs to Group ll-A of the modern periodic table. This option is also incorrect because Bismuth (Bi) belongs to Group V-A of the modern periodic table.

Q168. Chromyl Chloride test is performed to confirm:

• A. Chloride ions✓
• B. Sulfate ions
• C. Phosphate ions
• D. Cr3+ ions

Explanation: The chromyl chloride test is used to detect chloride ions in the qualitative analysis. If any chloride salt like sodium chloride is heated with acidified potassium dichromate it produces red colour fumes of chromyl chloride. It confirms the presence of chloride ions in that salt.

Q169. The charge on one gram of electrons is:

• A. 1.7588x10-11
• B. 1.7588x1011
• C. 1.602x10-19
• D. 1.7588x108✓

Explanation: We know 1 electron has a mass of 9.1x10²⁸g So 1g of electrons have how many number of electrons in it ? No of e= 1g x 1 /9.1x10²⁸g = 1.098x10²⁷ electrons Charge on an electron is 1.6x10-¹⁹ C So charge on 1.098x10²⁷ electrons (weighing one gram) is? 1.098x10²⁷ X 1.6x10-¹⁹ =1.756x10⁸C

Q170. The quantities which can be measured accurately are:

• A. Base Quantities
• B. Physical Quantities✓
• C. Derived Quantities
• D. Supplementary Quantities

Explanation: Physical Quantities, by definition, are quantities that can be measured with the help of an instrument to give a numerical value with a unit representing the quantity itself.

Q171. An observer notes the reading of scale from different angles (parallax) while measuring the length of the wire, what type of error is possible:

• A. Systematic error
• B. Zero error
• C. Precise error
• D. Random Error✓

Explanation: Parallax error is due to the positioning of the eye while recording a value. The positioning and the error in the reading birthed from this positioning is completely random since the inaccurate reading could be either greater or lesser than the true value.

Q172. The ratio of displacement along the diameter of the circle and the total distance along the circle is:

• A. 1:π✓
• B. π:1
• C. 2:π
• D. π:2

Explanation: Since the diameter of the circle is unknown, we can assume it to be 2r, where ‘r’ is the radius of the circle. The total distance along the circle refers to the circumference of the circle, which, mathematically, is 2πr. The ratio of the two values would be 2r:2πr, which can be simplified to 1:π.

Q173. Arshad is driving down 7th street, he drives 150m in 18s.Assume he does not speed up or slow down, what is his speed?

• A. 0.38m/s
• B. 126m/s
• C. 8.33m/s✓
• D. 58.33m/s

Explanation: We need to calculate the constant speed, the formula for it is given: speed= distance/time = 150/18 = 8.33m/s Hence the right answer is option C.

Q174. The distance travelled by a moving car with velocity 15m/s in 2s, decelerates at 2m/s2 is equal to:

• A. 30m
• B. 16m
• C. 34m
• D. 26m✓

Explanation: This is a question from a body moving at constant acceleration for that we have 3 equations of motion vf=vi + at S=vit +1/2at2 2aS=vf2 – vi2 Here S refers to a displacement Vi is the initial velocity Vf is the final velocity t is for time a is for acceleration According to the data given, the most suitable equation to be used here is the 2nd equation of motion, SOL: S=vit+1/2at2 =15x2+½x(-2)22 -ve sign indicates that the car is decelerating =30–4 =26m Hence the correct answer is option D.

Q175. In the shown figure, calculate the total work done on the gas when it is transformed from state a to state c, along the path indicated?

• A. 1500 J
• B. 3000 J
• C. 4500 J✓
• D. 5000 J
• E. 9, 500 J

Explanation: The thermodynamic work done is W = PdVAs dV = 0 for path b to c it will not contribute any work.The work done will come from path a to b and the required work done,W = ( 3 x 105) x (25 - 10) 10-3 = 4500 J

Q176. Work done will be zero if angle between force and displacement is:

• A. 0°
• B. 60°
• C. 270°✓
• D. 360°

Explanation: The formula for work done is displacement multiplied by the force in the direction of the displacement thus we incorporate the angle between the force and the displacement given the directions of the two vectors don’t align. The formula transforms to ‘Displacement times Force Cos (X)’ where X is the angle between the force and the displacement. If the force is perpendicular to the displacement, it won’t have any influence on the displacement making the work done zero because cos 90 or cos 270 equals zero.

Q177. If mass 'm' is dropped from height 'h' vertically, 'f' is the force of friction during downward motion and 'v' is the velocity at bottom, following will hold:

• A. ½mv²=mgh+fh
• B. mgh=½mv²–fh
• C. fh=mgh+½mv²
• D. mgh=½mv²+fh✓

Explanation: Throughout the course of this fall, we would be having the gravitational potential energy converted to kinetic energy and the energy needed to overcome friction. Thus mgh (being the maximum gravitational potential energy) equals the maximum kinetic energy (at the bottom of the fall) and the energy utilized to overcome friction.

Q178. A body moves in a circle with increasing angular velocity, at time 't'=6s the angular velocity is 27 rad/s… What is the radius of circle where linear velocity is 81cm/s:

• A. 6cm
• B. 9cm
• C. 7cm
• D. 3cm✓

Explanation: So here we use the relation between linear and angular quantity i.e. Vt=𝝎r Rearranging the expression, we get r=Vt/𝝎 Here Vt is linear velocity r is radius of the circle And 𝜔 is angular velocity = 81/27 = 3cm Therefore the radius of the circle is 3cm. Note: we do not need to convert the unit of linear velocity i.e cm/s into m/s as all the answers are already given in cm.

Q179. A moon rotates about its axis. In the future, scientists may wish to put a satellite into an orbit around the moon such that the satellite remains stationary above one point on moon surface, the period of rotation of moon about its axis is 27.4 days,what is the radius of required orbit? Mass of the moon =7.35x10²²kg

• A. 3.59x107 m
• B. 4.23x107 m
• C. 8.86x107 m✓
• D. 6.96x106 m

Explanation: In this case, once the satellite has been launched, the gravitational force between the satellite and the moon would be providing the centripetal force hence the centripetal force, mrw2 is equal to the gravitational force, GMm/r2 where r is the radius of the orbit, w being the angular velocity that can be simplified to 2π/T(time period) and M being the mass of the moon. Equating the two expressions would give us r= (GMT2/4π2)1/3 Plugging in all the values, performing mathematical operations and then finally taking cube root of the expression gives you the answer i.e 8.8x10⁷m

Q180. In mass spring system mass 'm' is attached with spring of spring constant 'k' with time period 'T1'..Then the mass is replaced by '2m' with same spring, what is the time period 'T2'

• A. T2=T1
• B. T2=2T1
• C. T2=√2T1✓
• D. T2=T1/(√2)

Explanation: Time period T for a mass attached to a spring can be calculated by using the following formula: T=2𝝅√m/k T1=2𝝅√m/k ==============eq(1)Since 2m is attached to the same spring to which m was attached, spring constant k remains same. T2=2𝝅√2m/k ==============eq(2)Divide eq(2) by eq(1)T2/T1=2𝝅√2m/k / 2𝝅√m/kT2=√2xT1So increasing the mass by 2 would increase the time period by √2Correct answer here is option C

Q181. A body performing SHM with displacement x=x 0sin(wt+Ⲫ),when t=0, x=x0..Then what is the phase angle of Ⲫ?

• A. π
• B. π/2✓
• C. π/4
• D. –π

Explanation: Inside the sin bracket we need π/2 to get us a 1 in order to satisfy the condition which is x=x0. Since t=0, we would have wt as zero as well so the value of fi must be π/2.

Q182. Angular displacement of a point moving in a circle of 10 cm when displacement of projection of this point a long vertical diameter of circle is 8.66cm will be:

• A. 30°
• B. 45°
• C. 60°✓
• D. 75°

Explanation: Explanation is given below.

Q183. A wave travelling with speed of 130m/s having a wavelength of 5m. What is its frequency:

• A. 650Hz
• B. 20Hz
• C. 26Hz✓
• D. 3.8x10² Hz

Explanation: We know, The product of wavelength and frequency is speed c of waveMathematically:fx𝞴=cf=c/𝞴f=130/5f=26Hz.

Q184. A metallic wire of length 2m hooked between two points has tension 10N. If mass per unit length is 0.004kg/m, their fundamental frequency emitted by wire on vibration is:

• A. 48 Hz
• B. 24 Hz
• C. 12.5 Hz✓
• D. 6.25 Hz

Explanation: Explanation is given below.

Q185. Coherent lines emerge from two fine parallel slits 'A' and 'B' as shown in figure:If 'P' is the position of nth dark fringe from centre of interference, then phase difference between wave train 'A' and 'B' is:

• A. nπ radian
• B. 2π n radian
• C. (n+½) π radian✓
• D. (2n+1) π radian

Explanation: Dark fringes are formed when we have destructive interference and since destructive interferences happen at phase differences being odd multiples of π, we have the answer as C.

Q186. The wavelength of light which produces second order spectrum on diffraction grating on which 5000 lines/cm are ruled at an angle of 30° will be:

• A. 6 x 10-7 m
• B. 4 x 10-6 m
• C. 5 x 10-7 m✓
• D. 3 x 10-6 m

Explanation: We know, for diffraction grating: dsin𝜭=m𝞴 Where d is the reciprocal of lines per centimeter. Also convert cm into m For second order spectrum m=2 𝞴=dsin𝜭/m =0.01/5000xsin30÷2 =5x10-⁷m Therefore option C is right.

Q187. Estimate pressure of air molecules at 273 K, if mean square speed is 500m²/s² and density of air is 6kg/m³:

• A. 1 x 10³ Pa✓
• B. 2.5 x 10² Pa
• C. 1 x 10² Pa
• D. 2.7 x 10³ Pa

Explanation: We are going to be using the formula; Pressure= ⅓ x density x mean square speed.

Q188. 1 mole of a gas occupies volume 1.00 x 10-2 m³ in a gas cylinder whose pressure is equal to 2.50 x 105 Pa.The temperature of cylinder is:

• A. 227K
• B. 300K✓
• C. 370K
• D. 390K

Explanation: Using general gas equation aka ideal gas equation PV=nRT Plugging in the values, we get: 2.5x10⁵ x 1x10-²=1x8.3xT Note: make sure that all the values are in SI units, and in physics particularly take R value that has SI units. T= 300K

Q189. The value of pressure and volume of fixed mass of gas in thermometer at triple point of water Pf=1.00x105 Pa and Vf=1x10-3 m³. When P=1.1x105 Pa and V=1.2x10-3 m³. Then temperature of gas is:

• A. 361K✓
• B. 298K
• C. 273K
• D. 250K

Explanation: Use the formula given below: P1V1/T1=P2V2/T2 T1 is triple point of water that is the temperature at which water co exist in all its phases its value is 273.16K Plugging in the values 1x10⁵ x 1x10-³/ 273.16=1.1x10⁵ x1.2x10-³/T2 We get T2= 360.5K which is closest to option A.

Q190. Isobaric processes occur at:

• A. Constant temperature
• B. Constant pressure✓
• C. Constant volume
• D. Constant temperature and constant pressure

Explanation: Isobaric process is a process in which the change in pressure is 0 i.e. pressure remains constant.Processes at constant volume are called ‘isochoric’.Processes at constant temperature are called ‘isothermal’.The word “iso” means same/constant and bar means pressure since we know that unit of pressure is bar.

Q191. The Coulomb's force between two point charges q1=1C and q2 is 2N. Where the distance between them is 3m, the charge q2 is:

• A. 1 x 10-9 C
• B. 1 x 109 C
• C. 2 x 10-9C✓
• D. 4 x 10-9 C

Explanation: From coulomb’s law we knowF=Kq1q2/r²Rearranging the equation, we get:q2=Fxr²/Kxq1 =2x9/9x10⁹x1 =2x10-⁹C Ans

Q192. Electric field strength at position vector r=(4i+3j)m caused by point charge q=5uC placed at origin is:

• A. 1440i+1080jV/m
• B. 1240i+1280jN/C
• C. 1440i+1080jN/C✓
• D. 1240i+1080jN/C

Explanation: The formula for the electric field strength is E=(K x Charge, Q)/distance2. The distance is the magnitude of the position vector which is d=(32 + 42)1/2. We multiply the electrostatic force by the unit vector which is the position vector, 4i + 3j, divided by the magnitude of the vector, 5 to get 1438.4i + 1078.8j. This rounds off to give C as the answer.

Q193. 2.00x106 e passing through a conductor in 1 millisecond. Electric current through conductor is:

• A. 3.2x10-10 A✓
• B. 32.0x10-9 A
• C. 320x10-10 A
• D. 0.320x10-10

Explanation: Charge on an electron is 1.6x10-¹⁹CWhat will be the total charge on n number of electrons?In this case, n=2x10⁶ electrons E=neWhere e is charge on one electron And E is total charge =2x10⁶x1.6x10^-¹⁹ =3.2x10-¹⁹CNow we know:I=Q/t =3.2x10-¹³/1x10-³note:Convert milli sec into sec =3.2x10-¹⁰AA is the correct answer Charge on an electron is 1.6x10-¹⁹CWhat will be the total charge on n number of electrons?In this case, n=2x10⁶ electrons E=neWhere e is charge on one electron And E is total charge =2x10⁶x1.6x10^-¹⁹ =3.2x10-¹⁹CNow we know:I=Q/t =3.2x10-¹³/1x10-³note:Convert milli sec into sec =3.2x10-¹⁰AA is the correct answer Charge on an electron is 1.6x10-¹⁹CWhat will be the total charge on n number of electrons?In this case, n=2x10⁶ electrons E=neWhere e is charge on one electron And E is total charge =2x10⁶x1.6x10^-¹⁹ =3.2x10-¹⁹CNow we know:I=Q/t =3.2x10-¹³/1x10-³note:Convert milli sec into sec =3.2x10-¹⁰AA is the correct answer on an electron is =2x10^6x1.6x10^-19 milli sec into sec

Q194. A carbon resistor connected to a battery of 50V and 2A current is passing through it. If voltage is increased to 75V then current will be:

• A. 1.5A
• B. 3A✓
• C. 4.5A
• D. 6A

Explanation: Firstly, we’ll calculate the resistance of the carbon resistor by using the formula Voltage= Current x Resistance. Once we have the resistance, we use it to calculate the current when voltage increases to 75V using the same formula. SInce we have a simple resistor and not a variable one, its resistance remains constant. First we calculate resistance of the resistor by using ohm’s law : V=IR R=V/I =50/2 = 25ohm Since its a simple resistor and not a variable one hence its resistance remains constant Now we calculate the current through it when voltage changes from 50V to 75V I=V/R =75/25 = 3A ANS

Q195. The effective resistance between the points A and B in the figure is

• A. 2 Ohms✓
• B. 6 Ohms
• C. 3 Ohms
• D. 4 Ohms

Explanation: Resistor AD and DC are in series For series combination: Re=R1+R2 =AD+DC =3ohm+3ohm =6ohm, this is in parallel combination with resistor AC For parallel combination: Re=R1R2/R1+R2 =6X6/6+6 =3ohm. In the 2nd figure resistor AC and CB are in series, so we add them, Re=3+3 =6ohm, now this resistor is in parallel combination with the last resistor of value 3ohm Re=6x3/6+3 =2ohm, here is the answer.

Q196. A metallic conductor is held vertically upright, such that electric current in it flows in the upward direction.What is the direction of the magnetic lines of force around the conductor?

• A. Parallel to the conductor
• B. In clockwise direction
• C. Anticlockwise direction✓
• D. None of these

Explanation: The direction of magnetic field in the conductor can be found out by right hand thumb rule. As the direction of the current is in upward and if we keep our thumb along the direction of the current then the curl fingers will give the direction of the magnetic field which is anticlockwise direction.

Q197. The magnetic flux linked with a solenoid of area 'A', having 'N' turns at right angle to magnetic field is:

• A. NBA✓
• B. BA
• C. ½ NBA
• D. BAcos(θ)

Explanation: In this scenario we have to find the magnetic flux linkage. Since we already have the solenoid’s wire being perpendicular to the field lines, we don’t need cos Theta in the equation. Moreover, BA gives us the flux whereas multiplying BA by N, which is the number of turns, gives us the total flux that is the flux linkage.

Q198. A charge projected with velocity of 10 m/s in a magnetic field of 10 T at an angle of 60°, if force exerted on charge is 2.78 x 10-17 N, then value of charge is:

• A. 1.6 x 10-19C
• B. 2.7 x 10-19C
• C. 3.2 x 10-19C✓
• D. 4.8 x 10-19C

Explanation: The equation for this scenario is; Magnetic Force= Bqv sin(X) where B is the magnetic field strength, q is the charge and v is the velocity of the charge. As we have the angle, the velocity, the magnetic field strength as well as the magnetic force, we can make q the subject to find the magnitude of the charge.

Q199. The value of magnetic flux is 10Wb, when magnetic lines of force containing magnetic field strength of 1T passing through unit area of 10m², then angle between magnetic field and unit area is:

• A. 360°✓
• B. 180°
• C. 90°
• D. 45°

Explanation: The magnetic flux is equal to the magnetic field strength times the area through which the field lines pass times the angle between the field lines and the area; Ⲫ= BAcos(X). By making X the subject, we can find the angle which would be either 0 or 360.

Q200. A loop of 5 turns of wire is placed in uniform magnetic field of 0.5T, then area of loop shrinks at a constant rate of 10m²/s, the emf induced is:

• A. 2.5V
• B. 25V✓
• C. 250V
• D. 0.25V

Explanation: According to faraday’s law of electromagnetic induction, whenever there is change in magnetic flux, in this particular case there’s change in area of loop, an emf e is induced in the coil. Mathematically e = -NdⲪ/dt -sign is due to lenz’s law Solution of the problem is shown in figure below:

Q201. The phase at negative peak of AC voltage is:

• A. π/2
• B. π
• C. 3π/2✓
• D. 2π/3

Explanation: Given the AC voltage is in the form of a sinusoidal wave starting from 0, the positive peak is at π/2 and thus the negative peak is at 3π/2. The AC voltage is in the form of a sinusoidal wave shown below. 270°/ 3π/2 rad is the negative peak as indicated. Option C is correct.

Q202. A 1.25cm diameter cylinder is subjected to load of 2500kg, stress on bar is:

• A. 200Pa
• B. 2x105Pa
• C. 2x106Pa✓
• D. 2x109Pa

Explanation: Explanation is given below.

Q203. Output voltage of rectifier is not smooth, it can be made smooth by a circuit known as:

• A. Wheatstone Circuit
• B. Bridge circuit
• C. Filter circuit✓
• D. Ripple circuit

Explanation: The filter circuit incorporates a capacitor which charges and discharges in order to smooth the circuit. A rectifier is a device that converts AC current into DC current.

Q204. A wire of length 2m is attached with mass of 5kg vertically, tensile strain of wire is 0.3x10-3, the extension in wire is:

• A. 1.5mm
• B. 2mm
• C. 0.15mm
• D. 0.6mm✓

Explanation: Strain is nothing but ratio of change in length to original length i.e Δl/L. We are asked to calculate extension meaning increase in length i.e Δl. strain= Δl/L Δl =strain x L =0.3x10-³ x 2 =0.6x10-³ =0.6mm ANS

Q205. The relation between an alternative current voltage source and time in international system units is:V = 120 sin (100𝜋t) cos (100𝜋t) volt value of peak velocity and frequency will be respectively:-

• A. 120 volt and 100 Hz
• B. 120/√2 volt and 100 Hz
• C. 60 volt and 200 Hz
• D. 60 volt and 100 Hz✓

Explanation: Explanation is given below.

Q206. If signal is applied to input of non-inverting amplifier through resistance of 100kohm, and the value of feedback resistance is 10kohm, the gain is:

• A. 11
• B. 10
• C. 1.1✓
• D. 0.11

Explanation: The formula used for the gain in non inverting amplifier is given as: 1+(Feedback resistance/resistance of other resistor) 1+(10kohm/100kohm) 1+(0.1) 1.1 answer, this is gain.

Q207. The frequency of photon having momentum 4.42x10-26 Ns is:

• A. 2.00 x 1016Hz✓
• B. 2.00 x 1014 Hz
• C. 5.00 x 1016Hz
• D. 2.00 x 1018 Hz

Explanation: Where h is planck’s constant whose value is 6.6x10^-34J.s = 6.6x10^-34/4.42x10^-26 =1.5x10^-8m Now finding frequency of photon: 𝞴xf=c c , speed of photon is speed of light f=c/𝞴 Using de broglie’s wavelength 𝞴=h/mvWhere h is planck’s constant whose value is 6.6x10-³⁴J.s = 6.6x10-³⁴/4.42x10-²⁶ =1.5x10-⁸mNow finding frequency of photon:𝞴xf=cc , speed of photon is speed of lightf=c/𝞴 =3x10⁸/1.5x10-⁸ =2x10¹⁶Hz The answer is A.

Q208. 1 Gauss is equivalent to:

• A. 1 tesla
• B. 10-3tesla
• C. 10 tesla
• D. 10-4tesla✓

Explanation: We know that:1 Tesla = 104Gauss=> 1 gauss = 10-4 TeslaBoth Tesla and Gauss are units of magnetic induction.Electromagnetic or magnetic induction is the production of an electromotive force (i.e.voltage) across an electrical conductor in a changing magnetic field.

Q209. The momentum of wave where wavelength is 1.32 x 10-9 m:

• A. 5.00 x 10-25 Ns✓
• B. 5.00 x 10-26 Ns
• C. 5.00 x 10-43 Ns
• D. 5.00x10–44Ns

Explanation: Using de broglie’s wavelength mv=h/𝞴 Where h is planck’s constant whose value is 6.6x10-³⁴J.s =6.6x10-³⁴/1.32x10-⁹ =5x10-²⁵ N.s Ans

Q210. Ionization energy of hydrogen atom is:

• A. 0.54eV
• B. 0.85eV
• C. 3.39eV
• D. 13.6eV✓

Explanation: The ionization energy for hydrogen is 2.18 x 10-18 Joules which when divided by 1.60 x 10-19 Joules/eV gives us 13.6 eV.

Q211. What will be 'Y':??Z → ??Y + β

• A. ab+1Z✓
• B. a+1b-1Z
• C. abZ
• D. a+1b+1Z

Explanation: Explanation is given below.

Q212. The quantity of uranium is 400g, the amount of uranium left after 3 half lives is:

• A. 25g
• B. 50g✓
• C. 100g
• D. 200g

Explanation: After every half life, the amount of uranium halves. Following this idea, after the first half life, the amount of uranium drops from 400g to 200g. After the second half of life, it drops to 100g. And finally after the third half life, what remains is 50 grams of uranium.

Q213. The mass of Radium atom decreases by 8.6 x 10-3 kg, mass defect equivalent to energy is:

• A. 4.48 MeV
• B. 4.84 MeV✓
• C. 3 x 102 MeV
• D. 4.84 eV

Explanation: Mass defect multiplied by square of speed is equal to the energy (E). You are given themass defect and you know the speed of light so just give it a shot ! And you’ll get your answer as 4.84 MeV.E=mc2; where m = 8.6 x 10-3 kgE= 7.74 x 1014 J or 4.84 MeV

Q214. Spot the error. My brother used to give me an advice but he has stopped doing so.

• A. Give
• B. An✓
• C. Has
• D. Doing

Explanation: The word "an" is incorrect because "advice" is an uncountable noun and should not have an article "an" before it. Instead, it should simply be "advice."

Q215. Find the error of Article.It is hoped to have everything in order for first performance at the end of November.

• A. To have
• B. In
• C. First✓
• D. At

Explanation: The phrase "first performance" is missing an article. It should be "the first performance." The sentence should read: "It is hoped to have everything in order for the first performance at the end of November."

Q216. Spot the error. The faster you run to reduce your belly-fat, more likely it is to achieve the desired results quickly.

• A. The faster
• B. More likely✓
• C. The
• D. Quickly

Explanation: This phrase is correct in meaning but needs the article "the" before it to maintain the parallel structure.

Q217. Spot the error. Our university building is now under construction on a hill side in north of Fukuoka.

• A. Our university
• B. Under construction
• C. A hill side
• D. North✓

Explanation: "In the north of Fukuoka" is more complete than just "north of Fukuoka."

Q218. Choose the correct one.

• A. They apparently walked to the south in the direction of India and at around 6 am were approached by soldiers from a nearby army post.✓
• B. They apparently walked to south in the direction of India and at around 6 am were approached by soldiers from a nearby army post.
• C. They apparently walked to the south in the direction of the India and at around 6 am were approached by soldiers from a nearby army post.
• D. They apparently walked to the south in direction of India and at around 6 am were approached by soldiers from a nearby army post.

Explanation: The correct answer is A. They apparently walked to the south in the direction of India and at around 6 am were approached by soldiers from a nearby army post. The sentence describes a sequence of events, and the correct form is to use the preposition "to" followed by the direction "the south", and the preposition "in" followed by the phrase "the direction of India".

Q219. Choose the segment that contains an error.With the reunification of Germany, the Berlin became once again the capital of the country.

• A. The reunification
• B. Germany
• C. The Berlin✓
• D. The capital

Explanation: The error is the use of the definite article "the" before "Berlin". Since "Berlin" is a city, it should not be preceded by "the" unless it is part of a specific name or title.

Q220. Spot the error. Smoking is a major contributor to many serious diseases, such as heart disease and the lung cancer.

• A. Smoking
• B. A major contributor
• C. Diseases
• D. The lung cancer✓

Explanation: "The lung cancer" should be "lung cancer" (without the definite article "the"), as it's one example of the many serious diseases caused by smoking.