Pk Mdcat Mock 1 — Solved Past Paper with Answers

Q1. Select the nearest correct meaning of the word given.Aghast

• A. Critical
• B. Reluctant
• C. Happy
• D. Horrified✓

Explanation: "Horrified" means to be filled with horror; extremely shocked which is the same meaning as "aghast".

Q2. Shakespeare, a (an) _ writer, entertained audiences by writing many tragic and comic plays.

• A. Numeric
• B. Obstinate
• C. Dutiful
• D. Prolific✓
• E. Generic

Explanation: Prolific means producing a large quantity or number of works. Shakespeare is widely recognized as one of the most prolific writers in the English language.

Q3. Choose the correctly punctuated sentence:

• A. What a fall was there, my countrymen! Long live the king!✓
• B. What a fall was there! My countrymen. Long live the king!
• C. What a fall was there, my countrymen. Long live the king.
• D. What a fall was there, my countrymen, long live the King.

Explanation: There are no conjunctions in the entire statement. Hence, any independent clauses must be separated into different sentences. “Long live the King!” stands as an independent sentence,while “What a fall was there,my countrymen!” can stand as an independent sentence. Moreover since both statements are exclamatory, an exclamation mark should be used to determine their ends. Therefore, Option A is correct.

Q4. Select the nearest correct meaning of the word given.BOURGEOIS

• A. Belonging to the bureaucratic class
• B. Belonging to the middle class✓
• C. Belonging to the upper class
• D. Belonging to the lower class

Explanation: BOURGEOIS means belonging to or characteristic of the middle class, typically with reference to its perceived materialistic values or conventional attitudes.Bourgeois is often mistakenly used to refer to people of considerable wealth or status, possibly because the French pronunciation causes us to associate it with opulence, yet the word is of decidedly middle-class origins.

Q5. Demonstrate correct use of articles and prepositions, subject-verb agreement, writing conventions of spelling, capitalization and punctuation to clarify meaning.You must boil those vegetables before _ in the stew.

• A. Using✓
• B. Use
• C. Their used
• D. The use

Explanation: Suffix ‘ing’ will be added to the verb because the verb is used after a time order word “before”.

Q6. Identify the mistake that needs to be corrected.Stevens was very delighted to see that he was declared more popular than any talk show just in the USA.

• A. Very
• B. To see that
• C. Than any✓
• D. The USA

Explanation: The trick to solving these types of questions is to read the question a few times in your head. It is not appropriate to compare a person with a non tangible thing such as a talk show. Also, it should be:“..than any OTHER talk show…” this sounds more grammatically accurate.

Q7. Read the passage and answer the questions given at the end of the passage:Recent advances in science and technology have made it possible for geneticists to find out abnormalities in the unborn fetus and take remedial action to rectify some defects which would otherwise prove to be fatal to the child. Though genetic engineering is still at its infancy, scientists can now predict with greater accuracy a genetic disorder. It is not yet an exact science since they are not in a position to predict when exactly a genetic disorder will set in. While they have not yet been able to change the genetic order of the gene in germs, they are optimistic and are holding out that in the near future they might be successful in achieving this feat. However, genetic misinformation can sometimes be damaging for it may adversely affect people psychologically. Genetic information may lead to a tendency to brand some people as inferiors. Genetic information can therefore be abused and its application in deciding the sex of the fetus and its subsequent abortion is now hotly debated on ethical lines. But on this issue geneticists cannot be squarely blamed though this charge has often been levelled at them. It is mainly a societal problem. At present genetic engineering is a costly process of detecting disorders but scientists hope to reduce the costs when technology becomes more advanced. This is why much progress in this area is seen in scientifically advanced and rich countries like the U.S.A, UK and Japan. It remains to be seen if in the future this science will lead to the development of a race of supermen or will be able to obliterate disease from this world.Which of the following is the same in meaning as the word "obliterate" as used in Passage?

• A. Wipe off
• B. Eradicate✓
• C. Give birth to
• D. Wipeout

Explanation: Even though “obliterate” literally means to wipe out or to utterly destroy, it’s not used in the given context that way. Such questions require you to use context to construct meaning from words as they are read. Replace obliterate with eradicate in the passage and you’ll realize it makes total sense.

Q8. In the following question, demonstrate the control of tenses and sentence structure:To walk, biking and driving are Pat's favourite ways of getting around.

• A. To walk, biking, and driving
• B. Walking, biking, and driving✓
• C. To walk, biking, and to drive
• D. Walking, biking, and to drive

Explanation: All verbs must be in the same forms. This format is known as " parallelism ". In the given question, the first verb is to † infinitive while others are not like this. So, we have to change V1 i.e walk to walking.

Q9. Choose the right option to complete the sentence:Shaken by two decades of virtual anarchy, the majority of people were ready to buy _ at any price.

• A. Order✓
• B. Hope
• C. Liberty
• D. Emancipation

Explanation: Anarchy refers to disorder due to absence of authority. The people covet something that is opposite in meaning to anarchy, hence, order must be the right option here.

Q10. Choose the correct option. He placed the book _ the vase on the table.

• A. Beside✓
• B. After
• C. On
• D. In

Explanation: The correct preposition in this sentence is beside, which means next to something.

Q11. Choose the CORRECT sentence from the following options:

• A. The best places to eat are casual, fun and you can get a meal for cheap.
• B. The best places to eat are casual, fun, can be termed with inexpensive.
• C. The best places to eat are casual, funny, and cheapest.
• D. The best places to eat are casual, fun, and inexpensive.✓

Explanation: When read aloud, A is correct, unlike B and C. Between A and D, D is correct because a comma appears before "and".

Q12. Fill in the blank with appropriate option:The cattle _ away the crops.

• A. Has eaten
• B. Is eating
• C. Have eaten✓
• D. Have been eating

Explanation: 'Cattle' represents a plural noun. Hence, plural verb is used. Cattle 'are' grazing; 'cattle is grazing' is incorrect. So, similarly, we will use a plural verb. Like 'They have', we use 'Cattle have'.

Q13. A successful businessman is _ of the risks.

• A. Audacious✓
• B. Agoraphobic
• C. Dismay
• D. Coerce

Explanation: Audacious means invulnerable to fear.Agoraphobic means afraid of open or public places.Dismay means discouragement.Coerce means to pressurise or force.Therefore, Audacious is the most appropriate word to fill in the blank.

Q14. Choose the right option to complete the following sentences.Irony can, after a fashion, become a mode of escape: to laugh at the terrors of life is in some sense to _ them.

• A. Overstate
• B. Revitalize
• C. Corroborate
• D. Evade✓

Explanation: Evade means to run away, flee from something. This is the best word which fits in here well.

Q15. Choose the correct sentence from the following:

• A. We should pay maximum accolade for our national heroes
• B. We should pay maximum accolade in our national heroes
• C. We should pay maximum accolade to our national heroes✓
• D. We should pay maximum accolade from our national heroes

Explanation: The sentence 'We should pay maximum accolade to our national heroes' is correct because the preposition 'to' is commonly used to indicate the direction of an action, in this case, directing the accolade towards the national heroes. The other options misuse prepositions, altering the intended direction or source of the accolade.

Q16. Identify the word or phrase that needs to be changed for the sentence to be correct.The bus stopped too take up three or four people who were waiting by Post office.

• A. The bus
• B. Too✓
• C. People
• D. Post office
• E. No error

Explanation: It should be “The bus stopped to take…”. The synonym of “to” is used here wrongly. The preposition “to” should be used here as we need a word that should be indicating position. “Too” is an adverb used to indicate a higher degree or an addition.

Q17. Select the NEAREST CORRECT MEANING of the given word:IMPETUOUS

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

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

Q18. Find out the pair that has the same connection as;Altimeter: height

• A. Speedometer: speed✓
• B. Racetrack : furlong
• C. Observatory : constellation
• D. Vessel : knots

Explanation: A. Speedometer: speed - This analogy is correct because a speedometer is an instrument that measures and displays the speed of a vehicle. Similarly, an altimeter is an instrument that measures and displays the height or altitude of an object or location. In summary, option A is the correct analogy.

Q19. The diagram shows the inheritance of haemophilia in a family. What is the genotype of person 7?

• A. XH XH
• B. XH Y
• C. XH Xh✓
• D. Xh Xh
• E. Xh Y

Explanation: Haemophilia is a sex-linked recessive disorder. This means that if the father has the recessive allele (Xh), then he will have the disease. Since the brother of person 7 had haemophilia, this shows that the mother was a carrier of haemophlia and hence the daughter (person 7) can also be suspected to be a carrier of heamophilia, since one X chromosome is recieved by the daughter from the mother. The indication that one of the sons had haemophlia confirms that the mother was a carrier as they go the recessive allele from her.

Q20. Cyanide for enzymes act as:

• A. Substrate
• B. Prosthetic group
• C. Cofactor
• D. Inhibitor✓

Explanation: Option A: A substrate is a molecule that an enzyme works on. Enzymes are proteins that help to speed up chemical reactions in the body. They do this by binding to the substrate and forming an enzyme-substrate complex. The enzyme then changes the shape of the substrate, which allows it to react with other molecules. Option B: A prosthetic group is a non-protein molecule that is attached to an enzyme. Prosthetic groups are often metal ions, such as iron or copper. They are required for the enzyme to work properly.Option C: A cofactor is a metal ion or other small molecule that is required for an enzyme to work. Cofactors are not permanently attached to the enzyme, but they do bind to the enzyme during the reaction. Option D: Cyanide acts as an inhibitor for enzymes. Enzymes are proteins that help to speed up chemical reactions in the body. Cyanide binds to an enzyme called cytochrome c oxidase, which is responsible for carrying oxygen to the cells. When cyanide binds to cytochrome c oxidase, it prevents the enzyme from working, which can lead to cell death.Cyanide is a very potent poison, and even small amounts can be fatal. It is important to know the signs and symptoms of cyanide poisoning, so that you can seek medical attention if you are exposed.

Q21. Read the passage below to answer the question:Hemophilia is a disorder in which blood fails to clot. Saad, a male hemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Hemophilia. Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmeds or Ali’s children, male or female, display symptoms of hemophilia, but the sons of Alia and Ayesha display symptoms of hemophilia while the daughters of Alia and Ayesha do not. Which of the following individuals are heterozygous for hemophilia:

• A. Saad, Ahmed, and Ali
• B. Ahmed, Ali, Alia, and Ayesha
• C. Saad and Sara
• D. Alia and Ayesha✓
• E. Ahmed and Ali

Explanation: Hemophilia is an X-linked recessive disorder, meaning that males (XY) need only one affected X chromosome to express the disorder, while females (XX) need two affected X chromosomes to have the disease.Saad (hemophiliac male) has only one X chromosome, which carries the hemophilia mutation, and a Y chromosome.Sara (normal female) has two normal X chromosomes.Ahmed and Ali (sons of Saad and Sara) inherit their Y chromosome from Saad and a normal X chromosome from Sara, so they are unaffected.Alia and Ayesha (daughters of Saad and Sara) inherit one affected X chromosome from Saad and one normal X chromosome from Sara, making them carriers (heterozygous) for hemophilia.Since Alia and Ayesha are carriers, they do not show symptoms themselves but can pass the affected X chromosome to their sons. Their sons receive their Y chromosome from their fathers and have a 50% chance of inheriting the affected X chromosome from their mothers, leading to hemophilia.Thus, Alia and Ayesha are heterozygous for hemophilia.

Q22. In cats, the genes controlling coat color are co-dominant and are carried on the X chromosomes. When a black female was mated with a ginger male the resulting litter consisted of black male and tortoise-shell female kittens. What phenotypic ratio would be expected in the F2 generation?

• A. 1 black male : 1 ginger male : 2 tortoise - shell females
• B. 1 black male : 1 ginger male : 1 tortoise - shell female : 1 black female✓
• C. 2 black males : 1 tortoise - shell female : 1 ginger female
• D. 1 black male : 1 tortoise - shell female : 1 ginger female : 1 black female
• E. 2 black males : 1 tortoise - shell female : 1 black female

Explanation: A black male has a genotype of XbY Tortoise-shell female kittens will have: XbXg XbY and XbXg XbXb; XbXg; XbY; XgY Hence: 1 Black-female; 1 tortoiseshell female; 1 black male; 1 ginger male.

Q23. Choose the correct combinations of labeling the carbohydrate molecule involved in the Calvin cycle.

• A. A - RuBP, B - Triose phosphate, C - PGA
• B. A - PGA, B - RuBP, C - Triose Phsophate
• C. A - PGA, B - Triose Phosphate, C - RuBP
• D. A - RuBP, B - PGA, C - Triose Phosphate✓
• E. A - Triose Phosphate, B - PGA, C - RuBP

Explanation: The Calvin cycle is a metabolic pathway found in the stroma of the chloroplast in which carbon enters in the form of CO2 and leaves in the form of sugar. The cycle spends ATP as an energy source and consumes NADPH2 as reducing power for adding high energy electrons to make the sugar. In the first phase, carboxylation of RUBP takes place, which leads to the formation of 1,3-bis phosphoglycerate (PGA). PGA then gets reduced to triose phosphate (glyceraldehyde phosphate). Thus, the labelled molecules in the diagram are A- RUBP, B- PGA, C- triose phosphate. Thus, the correct answer is option D.

Q24. The menstrual cycle can be divided into:

• A. Single cycle
• B. Two cycles✓
• C. Three cycles
• D. Four cycles

Explanation: The menstrual cycle is governed by hormonal changes. These changes can be altered by using hormonal birth control to prevent pregnancy. Each cycle can be divided into two phases based on events in the ovary (ovarian cycle) or in the uterus (uterine cycle).

Q25. Enzyme-catalyzed modifications are?

• A. Reversible
• B. Irreversible
• C. Both a and b✓
• D. None of these

Explanation: Enzyme-catalyzed modifications can be both reversible and irreversible, depending on the specific reaction and the enzyme involved. Some enzyme-catalyzed modifications are reversible, meaning that the reaction can proceed in both the forward and reverse directions. For example, enzymes that catalyze the synthesis and breakdown of glycogen can operate in both directions, depending on the needs of the cell. Other enzyme-catalyzed modifications are irreversible, meaning that the reaction can only proceed in one direction. For example, enzymes involved in the synthesis of DNA and RNA are typically irreversible, since the reaction is energetically favourable in one direction only. The reversibility of enzyme-catalyzed modifications depends on the specific reaction, the enzyme involved, and the conditions under which the reaction takes place.

Q26. Carbohydrates are commonly found as starch in plants storage organs. Which of the following five properties of starch make it useful as a storage material?(1) Easily translocated(2) Chemically non­reactive(3) Easily digested by animals(4) Osmotically inactive(5) Synthesized during photosynthesisThe useful properties are?

• A. (1), (3) and (5)
• B. (1) and (5)
• C. (2) and (3)
• D. (2) and (4)✓

Explanation: Starch is the major storage carbohydrate of plants. In most plant species it is accumulated in the chloroplast of leaves, whereas in storage organs it accumulates in the amyloplast as reserve starch. It is the osmotically inactive form of photosynthetic product and is a hexosan polysaccharide made of a large number of glucose units so it is chemically non-reactive.

Q27. Pharynx leads air into _ through glottis.

• A. Trachea
• B. Bronchus
• C. Alveoli
• D. Nasal sac
• E. Larynx✓

Explanation: Pharynx leads to the air into the larynx through glottis, as shown in the diagram below:

Q28. Three of the following pairs of the human skeletal parts are correctly matched with their respective inclusive skeletal category and one pair is not matched. Identify the nonmatching pair.

• A. Sternum and ribs - Axial skeleton
• B. Clavicle and glenoid cavity - Pelvic girdle✓
• C. Humerus and ulna - Appendicular skeleton
• D. Malleus and stapes - Ear ossicles

Explanation: The correct answer is Option B: Clavicle and glenoid cavity - Pelvic girdle. The clavicle and glenoid cavity are actually components of the pectoral girdle, not the pelvic girdle. The pelvic girdle includes the hip bones, not the clavicle or glenoid cavity.Option A is incorrect because the sternum and ribs are accurately categorized under the axial skeleton. Option C is incorrect as the humerus and ulna do belong to the appendicular skeleton. Option D is correct in stating that the malleus and stapes are part of the ear ossicles.

Q29. Bacteria plays an important role in:

• A. Phosphorus Cycle
• B. Nitrogen Cycle
• C. Carbon Cycle
• D. All of the above✓

Explanation: In nitrogen cycle :Nitrogen-fixing bacteria convert N2 to ammonia Nitrifying bacteria convert ammonia to nitrates and nitrites Denitrifying bacteria release the N2 back to the atmosphere by converting nitrates back to atmospheric nitrogen. In the carbon cycle, bacteria decompose dead organisms and release carbon back to the environment. In the phosphorus cycle, bacteria convert phosphate into the organic form so that it can be utilized by the plants.

Q30. When phenotype of a heterozygote is in between the phenotypes of both the homozygous parents, it is called:

• A. Incomplete dominance✓
• B. Epistasis
• C. Pleiotropy
• D. Codominance

Explanation: This type of relationship between alleles, with a heterozygote phenotype intermediate between the two homozygous phenotypes, is called incomplete dominance. In the question, "in between" is closest in meaning to "intermediate", thus incomplete dominance is the correct choice. Thus, the correct answer is 'Diphtheria, Leprosy, Plague.'

Q31. Which of the following organisms have the greatest problem with photorespiration?

• A. C4 plants
• B. Heterotrophs
• C. C3 plants✓
• D. CAM plants

Explanation: C3 plants have the disadvantage that in hot dry conditions their photosynthetic efficiency suffers because of a process called photorespiration. When the CO2 concentration in the chloroplasts drops below about 50 ppm, the catalyst rubisco that helps to fix carbon begins to fix oxygen instead. Thus the rate of photosynthesis begins to decrease, posing a problemC4 plants—including maize, sugarcane, and sorghum—avoid photorespiration by using another enzyme called PEP during the first step of carbon fixation. This step takes place in the mesophyll cells that are located close to the stomata where carbon dioxide and oxygen enter the plant.

Q32. A 25 years old female with chronic fatigue was diagnosed with iron deficiency anemia and low blood count. What is the cause of her fatigue?

• A. Reduction in amount of Fe-S centers
• B. Lowered production of water from the electron transport chain that cause dehydration
• C. Iron is important for electron transport chain✓
• D. Iron is important for NADH production

Explanation: Option C is the correct option because Fe-S clusters mediate electron transfer within and between the respiratory complexes of the electron transport chain so iron is important for the electron transport chain to continue and for ATP synthesis.Option A is not correct because the reduction in the amount of Fe-S centers doesn't lead to low blood count.Option B is not correct because lower production of water does not lead to anemia.Option D is not correct because NADH is a crucial coenzyme in making ATP; it acts as a hydrogen acceptor in an oxidation-reduction reaction and does not require Iron.

Q33. What happens to the volume of the thorax and the air pressure in the lungs during breathing in?

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

Explanation: When breathing in, the intercostal muscles contract, moving the ribs up and out, increasing the volume of the thorax. When breathing out, the intercostal muscles relax, moving the ribs down and in, decreasing the volume of the thorax. When the volume of the thoracic cavity increases – the volume of the lungs increases and the pressure within the lungs decreases. When the volume of the thoracic cavity decreases – the volume of the lungs decreases and the pressure within the lungs increases.

Q34. Acyl-glycerols like fats and oils are esters formed by condensation reaction between:

• A. Fatty acids and water
• B. Fatty acids and alcohols✓
• C. Fatty acids and glucose
• D. Fatty acids and phosphates

Explanation: Acyl-glycerols, such as fats and oils, are esters formed by a condensation reaction between fatty acids and alcohols. This process involves a glycerol molecule, which is a type of alcohol, reacting with three fatty acid molecules to form a triglyceride. Option A is incorrect because water is a byproduct, not a reactant, in this reaction. Option C is incorrect because glucose, a carbohydrate, is not involved in esterification. Option D is incorrect as phosphates are related to phospholipids formation, not acyl-glycerols.

Q35. What are labelled phases A, B, and C in the below given sigmoidal growth curve?

• A. A: Stationary, B: Log, C: Lag
• B. A: Lag, B: Stationary, C: Log
• C. A: Log, B: Lag, C: Stationary
• D. A: Lag, B: Log, C: Stationary✓

Explanation: When a graph is plotted taking into consideration time on one hand and growth rate, on the other hand, an S-shaped curve is obtained. It is called the "grand period of growth". This curve mainly shows four phases of growth: 1. Initial slow growth (lag phase). 2. The rapid period of growth (log phase), where maximum growth is seen in a short period. 3. The diminishing phase, where growth will be slow. 4. Stationary phase, where finally growth stops.

Q36. Growth movement caused in response to gravitational stimulus is called:

• A. Nutation
• B. Geotropism✓
• C. Nastic movement
• D. Tropic movement
• E. Turgor movement

Explanation: Growth movements induced by the stimulus of gravity are said to be geotropism.Nutation is a rocking, swaying, or nodding motion in the axis of rotation of a largely axially symmetric object. Nastic movements are non-directional responses to stimuli (e.g. temperature, humidity, light irradiance), and are usually associated with plants. The movement can be due to changes in turgor or changes in growth. One can define tropic movement or tropism as the directional movement of particular parts of a plant in response to environmental stimuli.Turgor movement is due to the difference of turgidity of the cells in the lower half and upper half of pulvinus (petiole of leaf).

Q37. Read the passage below to answer the question:Haemophilia is a disorder in which blood fails to clot. Saad, a male haemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Haemophilia. Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmed's or Ali’s children, male or female, display symptoms of haemophilia, but the sons of Alia and Ayesha display symptoms of haemophilia while the daughters of Alia and Ayesha do not.Which of the following best explains the reason that Ahmed, Ali, Ayesha and Alia do not display symptoms of haemophilia, even though their father, Saad, is a haemophiliac.

• A. Haemophilia is an X-linked disorder, and Saad can only pass on his Y chromosome
• B. Haemophilia is an X-linked disorder, and even though Alia and Ayesha receive a hemophiliac X-chromosome from Saad, Sara gave them a normal X-chromosome.✓
• C. Haemophilia is a Y -linked disorder, and therefore cannot be displayed in females
• D. Haemophilia is a Y-linked disorder ,and Ahmed and Ali must have received an X Chromosome from Saad
• E. Haemophilia is an X-linked disorder, and even though Ahmed and All received a Hemophilia X-chromosome from Saad, Sara gave them a normal X- chromosome.

Explanation: Haemophilia is an X-linked recessive disorder. Males (XY) only need one affected X chromosome to express the disorder, but females (XX) need two affected X chromosomes. Since Sara is normal, she provides a normal X chromosome to Alia and Ayesha, making them carriers but not affected.

Q38. When a nerve impulse jumps from one node of Ranvier to the next in a myelinated neuron, it's called _.

• A. Synapses
• B. Saltatory conduction✓
• C. Resting potential
• D. Membrane potential

Explanation: Saltatory conduction is the propagation of action potentials along myelinated axons from one node of Ranvier to the next node, increasing the conduction velocity of action potentials.In the central nervous system, a synapse is a small gap at the end of a neuron that allows a signal to pass from one neuron to the next.Resting potential is the imbalance of electrical charge that exists between the interior of electrically excitable neurons (nerve cells) and their surroundings.Membrane potential (also transmembrane potential or membrane voltage) is the difference in electric potential between the interior and the exterior of a biological cell.Thus, Option B is correct.

Q39. In human females, oocytes are held in the second metaphase until _.

• A. Death
• B. Birth
• C. Ovulation
• D. Fertilization✓

Explanation: The secondary oocyte commences meiosis 2 which arrests at metaphase and will not continue without fertilization. Oocytes, also known as egg cells or ovum, are the female reproductive cells. They are produced within the ovaries through a process called oogenesis. Oocytes are non-motile and larger in size compared to sperm cells. They contain half the number of chromosomes (23) as compared to other body cells, as they undergo meiosis to prepare for fertilization. During a woman's reproductive years, several oocytes begin to mature each month. However, only one oocyte is typically released during each menstrual cycle in a process known as ovulation. Oocytes are surrounded by supportive cells called follicular cells within the ovarian follicles. Once an oocyte is released during ovulation, it travels through the fallopian tube towards the uterus. It remains in a state of arrested development in the second metaphase of meiosis until fertilization occurs. If the oocyte is fertilized by a sperm cell, it completes meiosis, forming the female pronucleus, and becomes a zygote, which develops into an embryo. If fertilization does not occur, the oocyte eventually disintegrates and is shed along with the uterine lining during menstruation. The process of oocyte development, release, and potential fertilization is an essential part of the female reproductive system and plays a central role in human reproduction.

Q40. Shape of the maturing phase of the golgi apparatus is?

• A. Biconcave
• B. Convex
• C. Spherical
• D. Concave✓

Explanation: The correct answer is Option D: Concave. The maturing face of the Golgi apparatus is the trans face, which is characterized by its concave shape. This concave structure is crucial for receiving vesicles from the cis face and dispatching them to their final destinations. Options A, Biconcave, and C, Spherical, are incorrect as they do not describe the typical structure of the Golgi apparatus. Option B, Convex, is incorrect as the maturing face is specifically concave, not convex.

Q41. Which type of bonds are never formed when a substrate fits into the active site of an enzyme?

• A. Hydrogen bonds
• B. Ionic interactions
• C. Hydrophobic interactions
• D. Covalent linkages✓

Explanation: The correct answer is Covalent linkages because these bonds involve a strong and permanent connection between atoms, which is not typical during the transient interaction of a substrate with an enzyme's active site. Enzymes typically bind substrates through weaker interactions like hydrogen bonds, ionic interactions, and hydrophobic interactions, which facilitate the temporary nature of enzyme-substrate complexes. The other options—hydrogen bonds, ionic interactions, and hydrophobic interactions—are all types of interactions that can indeed occur as substrates fit into the active site, making them incorrect choices for this question.

Q42. Which statement correctly outlines some of the main events in photosynthesis?

• A. A 5-Carbon carbohydrate accepts carbon dioxide and is then reduced by NADPH derived from photophosphorylation✓
• B. A 3-Carbon carbohydrate is regenerated and reduced by hydrogen molecules derived from photophosphorylation
• C. Photolysis uses light to produce reduced NADP and oxygen which are used to reduce a 3-Carbon carbohydrate
• D. Photolysis produces NADPH and ATP which are used to reduce a 5-Carbon carbohydrate

Explanation: During the light-independent reactions, a 5-carbon carbohydrate called RuBP accepts carbon dioxide to form 3-carbon phosphoglycerate, which is reduced by NADPH formed in the light reaction.

Q43. Which bones meet at the elbow joint, and what kind of movement do they allow?

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

Explanation: At the elbow joint, the ulna and humerus meet, and they cause back-and-forth movement of the forearm. At the shoulder joint, the humerus and scapula meet, and there they cause full axial movement.

Q44. Pancreas produces secretory granules that help in digestion. These granules, after passing through Endoplasmic reticulum, are pinched off from _ surface of Golgi apparatus?

• A. Forming phase
• B. Maturing phase✓
• C. Cis-end
• D. None of these

Explanation: The maturing phase of the Golgi apparatus is when the secretory granules become fully formed and ready for secretion. During this phase, further processing, sorting, and modification of proteins and other molecules occur within the Golgi apparatus. The maturing phase is characterized by the final packaging and sorting of these molecules before they are transported to their target destinations.

Q45. Glucose → 2 pyruvic acid + 2H2O2ADP + 2Pi → 2 ATP2NAD + 4H → 2NADH + 2H+These three reactions collectively constitute _.

• A. Krebs cycle
• B. Calvin cycle
• C. Electron Transport Chain
• D. Light reaction
• E. Glycolysis✓

Explanation: The reaction in which glucose converts to pyruvate, forming two ATP molecules and two NADH molecules is called glycolysis.

Q46. The types of gametes produced by two pairs of chromosomes can be?

• A. 2
• B. 4✓
• C. 6
• D. 8

Explanation: Humans have 2 pairs of chromosomes. So for each pair, there are two options available. 2C1 x 2C1= 2 x 2 = 4 When two pairs of chromosomes undergo meiosis, the different combinations of maternal and paternal chromosomes lead to four different types of gametes. This is due to the independent assortment of homologous chromosomes during meiosis I. Each homologous pair aligns independently during metaphase I, leading to a random distribution of maternal and paternal chromosomes into the resulting gametes. As a result, four different combinations of alleles are possible for each pair of chromosomes.

Q47. Four plants are present in different environmental conditions. Plant A is present in a warm climate with continuous rainfall, plant B is present in a cool forest, plant C is present in a warm climate with little breeze, while plant D is present in a warm climate with high wind speed. Which one of the above plants will have the highest rate of transpiration?

• A. Plant B
• B. Plant D✓
• C. Plant A
• D. Plant C

Explanation: When it comes to questions like these, do not confuse the option letter with the letter of the plant itself. The factors which affect the rate of transpiration are: temperature, as at a higher temperature, more water vapour is given up by the leaves of the plant; humidity, as more humid conditions decrease the concentration gradient of water vapour between the air spaces of the leaves and the atmosphere so, the rate of transpiration decreases; wind speed, as a greater wind speed moves away more of the water vapour collected around the leaves and stomata, thereby increasing the concentration gradient between the air spaces of the leaves and the atmosphere.Plant A has a warm atmosphere but has continuous rainfall. Rainfall would imply a humid atmosphere so, the rate of transpiration is not very high.Plant B has a cool atmosphere so, the rate is low.Plant C is present in a warm climate but the breeze is slow so, when wind speed is low, more water vapour accumulates around the leaves of the plant thereby reducing the concentration gradient between the air spaces and the atmosphere. Hence, the rate of transpiration is not very high.Plant D is present in a warm climate in addition to having a high wind speed so, both factors contribute to a high rate of transpiration. Hence, the option which mentions plant D is the correct answer, which is option B.

Q48. Which of the statements correctly describes why ions are unable to cross the plasma membrane without channel proteins?

• A. They are unable to cross the hydrophilic phosphate heads of the lipid bilayer
• B. They are unable to cross the hydrophobic tails of the lipid bilayer✓
• C. They are unable to cross both the phosphate heads and fatty acid chains of the lipid bilayer
• D. They are too big to cross the plasma membrane

Explanation: The plasma membrane is made up of a phospholipid bilayer, which is a double layer of phospholipids. Phospholipids have a hydrophilic head and a hydrophobic tail. The hydrophilic heads face the outside and inside of the cell, while the hydrophobic tails face each other. The hydrophobic tails of the lipid bilayer create a barrier that prevents ions from crossing the plasma membrane. Ions are charged particles, and they are attracted to water. The hydrophobic tails of the lipid bilayer are repelled by water, so ions cannot cross through them. Channel proteins are proteins that span the plasma membrane and create pores through which ions can pass. Channel proteins are selective, so they only allow certain types of ions to pass through.

Q49. According to the fluid mosaic model of cell membrane, which zone is embedded inside?

• A. Hydrophobic✓
• B. Hydrophilic
• C. Globular
• D. Filamentous

Explanation: The fluid mosaic model describes the cell membrane as a tapestry of several types of molecules (phospholipids, cholesterols, and proteins) that are constantly moving. This movement helps the cell membrane maintain its role as a barrier between the inside and outside of the cell environment. Phospholipids, arranged in a bilayer, make up the basic fabric of the plasma membrane. They are well-suited for this role because they are amphipathic, meaning that they have both hydrophilic and hydrophobic regions. Since the outer region of the cell membrane comes in contact with the fluid so the hydrophilic heads are present on the surface of the membrane while the hydrophobic tails are embedded inside the membrane. The diagram below represents this model.

Q50. Starch is present in tubers, fruits, and grains but absent in animal cells. Instead, animals have a substance stored in liver and muscles known as:

• A. Galactose
• B. Glucagon
• C. Glycogen✓
• D. Glucose

Explanation: Glycogen is the storage form of carbohydrate found in animals. Glycogen is the body's stored form of glucose, which is a type of sugar that serves as the main source of energy. It is a large, branched molecule made of many glucose units linked together, and it is stored primarily in the liver and muscles. When the body needs energy, glycogen is broken down to release glucose into the bloodstream for the cells to use.Galactose (A) is a monosaccharide, Glucagon (B) is a hormone, Glucose (D) is a monosaccharide and the monomer of glycogen and starch.

Q51. Which combination describes a triglyceride?

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

Explanation: Triglyceride being nonpolar do not dissolve in water hence are insoluble in water; they do store a lot of energy in them; they do hydrolyze to release water, and being non-polar they can dissolve in organic solvents.Restriction enzyme, also called restriction endonuclease, a protein produced by bacteria that cleaves DNA at specific sites along the molecule. In the bacterial cell, restriction enzymes cleave foreign DNA, thus eliminating infecting organisms.

Q52. Which of the following cell organelles involved in the synthesis of cell wall and middle lamella?

• A. Endoplasmic reticulum
• B. Golgi apparatus✓
• C. Lysosomes
• D. Peroxisomes

Explanation: The Golgi and plasma membranes are the two main sites where constituents of cell wall are synthesized. The Golgi body is also known as Golgi apparatus. It synthesizes vesicles which then fuse to form the middle lamella as shown below.

Q53. Tropical plants such as sugarcane show high efficiency of CO2 fixation because of:

• A. Calvin cycle
• B. Hatch-slack✓
• C. TCA-cycle
• D. All of these

Explanation: The high efficiency of carbon dioxide (CO2) fixation in C4 plants, including tropical plants like sugarcane, is indeed associated with the Hatch-Slack pathway. The Hatch-Slack pathway, also known as C4 photosynthesis, is a mechanism that allows these plants to efficiently fix carbon dioxide even under high temperatures and intense sunlight.

Q54. During photosynthesis, in the electron transport chain, ATP synthesis takes place when electrons move from:

• A. Primary Electron Acceptor (PEA) to Plastoquinone (Pq)
• B. Plastoquinone (Pq) to cytochromes
• C. Cytochrome to Plastocyanin (Pc)✓
• D. Plastocyanin (Pc) to Photosystem I (PS I)

Explanation: When an electron leaves PSII, it is transferred first to a small organic molecule (plastoquinone, Pq), then to a cytochrome complex (Cyt), and finally to a copper-containing protein called plastocyanin (Pc). As the electron moves through this electron transport chain, it goes from a higher to a lower energy level, releasing energy. Some of the energy is used to pump protons H+ from the stroma (outside of the thylakoid) into the thylakoid interior. This transfer of H+, along with the release of H+ from the splitting of water, forms a proton gradient that will be used to make ATP. It occurs when electrons move from Cyt b6-f complex to Pc.

Q55. Cytochromes are electron transport intermediates containing?

• A. Myoglobin
• B. Haem✓
• C. Globulin
• D. Fibrin

Explanation: Cytochromes are proteins that contain haem as their prosthetic group and whose principal biological function, in the cells of animals, plants, and microorganisms, is electron transport.

Q56. People who migrated from plains to hills six months back _.

• A. Possess more RBCs with haemoglobin of low O2 binding affinity✓
• B. Possess same RBCs with haemoglobin of high O2 binding affinity
• C. Lose physical fitness to play games like football
• D. Suffer from altitude sickness with nausea and fatigue

Explanation: When one is at a high altitude, he/she has to acclimatize to it. Breathing becomes difficult at high altitudes, because of the low density of the air. There is not sufficient oxygen in the atmosphere to oxygenate the blood adequately. Oxygenation of blood depends upon the partial pressure of oxygen in the atmosphere. As atmospheric pressure decreases, so does the partial pressure of the oxygen. For short durations, humans compensate for high altitudes by increasing the respiratory rate (hyperventilation). During a longer period of acclimatization, physiological changes occur including increased levels of hemoglobin. Other changes in the body's acid-base balance can also happen.

Q57. A student of chemical engineering mistakenly engulfed the toxic compound "A" which was a potent inhibitor of a certain enzyme. He was immediately brought to a hospital where the doctor injected, intravenously, substrate "B" to minimize the toxic effect of Compound A. His life was saved from serious damage. The treatment method shows that compound A was a _ inhibitor.

• A. Non competitive
• B. Irreversible
• C. Temperature sensitive
• D. Competitive reversible✓

Explanation: 'Compound A' caused a toxic effect by inhibiting the action of a certain enzyme. 'Substrate B' was injected, which minimized the toxic effect of compound A, and this toxic effect was brought about by inhibiting the enzyme. The addition of a larger quantity of substrate allowed the enzyme to work to its normal capacity once again, as the inhibitory or toxic effect of compound A was minimized. Competitive reversible inhibitors compete with the substrate for the active site and if there is a greater amount of the substrate, the active site is more likely to bind to the substrate, rather than the inhibitor so, the activity of the enzyme is restored. This is the exact scenario described in the question, as the inhibitory effect was reversed upon addition of more substrate. Hence, the correct option is D.Option A, non-competitive inhibitors, decrease the Vmax of the enzyme so, addition of more substrate will not change or restore the enzyme's effectiveness.Option B, irreversible inhibitors, would have an effect that cannot be reversed under any circumstances so, addition of substrate would have no effect.Option C, temperature sensitive inhibitor has no relevance to the given scenario as temperature has not been mentioned as a factor in the question.

Q58. Bacteriochlorophylls does not include which of the following?

• A. Chlorophyll e
• B. Chlorophyll b
• C. Chlorophyll c
• D. All of these options✓

Explanation: Bacteria use bacteriochlorophylls in photosynthesis without the production of energy. Bacteria only use bacteriochlorophylls for capturing light. Bacteriochlorophylls don't include chlorophylls e, b, or c.

Q59. The difference in photosynthesis spectrum and action spectrum occurs due to _.

• A. Carbon dioxide
• B. Oxygen
• C. Carotenoids✓
• D. Wavelength

Explanation: Photosynthesis is a complex process that converts light energy into chemical energy, producing glucose and oxygen. The difference between the photosynthesis spectrum and the action spectrum is due to the presence of carotenoids, which extend the range of light wavelengths that plants can utilize for photosynthesis. Additionally, carbon dioxide is a necessary raw material, and oxygen is a byproduct of this vital biological process. The wavelength of light affects the efficiency of photosynthesis, and only certain wavelengths are effectively used by the plant's pigments to drive the process.

Q60. Which one of the following statements is incorrect?

• A. The competitive inhibitor does not affect the rate of breakdown of the enzyme-substrate complex
• B. The presence of the competitive inhibitor decreases the Km of the enzyme for the substrate✓
• C. A competitive inhibitor reacts reversibly with the enzyme to form an enzyme-inhibitor complex
• D. In competitive inhibition, the inhibitor molecule is not chemically changed by the enzyme

Explanation: In competitive inhibition, the inhibitor competes with the substrate for the active site of the enzyme. This results in an increase in the apparent Km, as it requires a higher concentration of substrate to achieve the same reaction velocity. Option B is incorrect because the presence of a competitive inhibitor does not decrease the Km; instead, it increases it. Option A is true because the inhibitor does not affect the breakdown of the enzyme-substrate complex. Option C is true as competitive inhibitors bind reversibly, and Option D is true because the inhibitor is not chemically changed by the enzyme.

Q61. What contributes to genetic variation during human reproduction?

• A. Random mating
• B. Non random mating
• C. Independent assortment
• D. All of these✓

Explanation: Non random mating can also be outbreeding, wherein there is an increased probability that individuals with a particular genotype will mate with individuals of another particular genotype. Random mating increases the chances of genetic variation. Independent assortment during meiosis is the random assortment of chromosomes and so new combinations are made. All three options cause genetic variation hence the answer is D, all of them.

Q62. PS I have a chlorophyll a molecule that absorbs maximum light of _ nm.

• A. 600
• B. 650
• C. 680
• D. 700✓

Explanation: Photosystem I (PS I) is one of the two photosystems involved in the light-dependent reactions of photosynthesis. The reaction center of PS I contains a chlorophyll-a molecule that absorbs light most efficiently at a wavelength of 700 nm, hence it is often referred to as P700. This absorption peak distinguishes it from Photosystem II, which absorbs light best at 680 nm, known as P680.Options A (600 nm) and B (650 nm) do not match the peak absorption characteristics of PS I. Option C (680 nm) is specifically characteristic of Photosystem II, not Photosystem I. Therefore, the correct answer is Option D (700 nm).

Q63. Oxaloacetate combines with which molecule to enter the Krebs cycle again?

• A. ATP
• B. NADPH
• C. FAD
• D. Acetyl coA✓

Explanation: Acetyl CoA combines with oxaloacetate to form citrate, which marks the beginning of the Krebs cycle. This step is also known as the condensation reaction, and it allows oxaloacetate to re-enter the Krebs cycle for further energy production.

Q64. The most abundant intracellular free nucleotide is _.

• A. UTP
• B. FAD
• C. NAD
• D. ATP✓

Explanation: The most abundant free nucleotide in mammalian cells is ATP.

Q65. The following results of a cross between two individuals shown in the picture is:

• A. One that is homozygous dominant and the other has a dominant phenotype, but has a mother with recessive phenotype
• B. One that is homozygous recessive and other has a dominant phenotype, but has a mother with recessive phenotype.✓
• C. One that is homozygous recessive and other has a dominant phenotype, but has a brother with recessive phenotype.
• D. One that is homozygous recessive and another has a recessive phenotype. but has a father with a dominant phenotype.

Explanation: All statements describing the children’s genetic combination and phenotype are wrong except B and C. One child is homozygous recessive with recessive phenotype and the other is heterozygous with a dominant phenotype.Option A is wrong because it says one child is homozygous dominant whereas according to the box diagram there is no possibility that a child will be homozygous dominant. Option D is wrong because it says one child, apart from one being homozygous recessive, has a recessive phenotype, which is wrong. Option C is wrong because it assumes the gender by saying “brother” but the question does not provide any evidence for what the gender of another offspring is.

Q66. Read the code mentioned in the following picture and arrange the sequence of all five codons in which leucine is at 3rd position, while isoleucine is in 4th position. Keep in mind that CUU is the code for leucine, AUU for isoleucine and CAA or glutamine.

• A. UAA-AUG-CAA-CUU-AUU
• B. AUG-CAA-CUU-AUU-UAA✓
• C. AUU-AUG-CAA-CUU-UAA
• D. AUG-UAA -CUU-AUU-CAA

Explanation: Option B is most appropriate because it starts with a start codon and ends with a stop codon while having CUU and AUU at third and fourth positions respectively.

Q67. Enzymes are produced in inactive form by the cell and enclosed in which of the following?

• A. Mesosomes
• B. Lysosomes✓
• C. Peroxisomes
• D. All of these

Explanation: Lysosomes are membrane-bound organelles that contain digestive enzymes produced by the cell. These enzymes are initially produced in an inactive form to prevent them from digesting the cell itself. When activated, lysosomal enzymes help to break down and recycle cellular waste and foreign material and also play a role in processes such as apoptosis (programmed cell death) and autophagy (self-eating of cellular components).

Q68. They serve as a scaffold for formation of higher order chromatin structure:

• A. Nucleosome✓
• B. Nucleus
• C. Chromosomes
• D. Histones

Explanation: Nucleosomes are the basic units of DNA packaging and serve as a scaffold for the formation of higher-order chromatin structures. They consist of DNA wrapped around histone proteins, allowing the DNA to be compacted efficiently within the nucleus. While histones are essential for nucleosome formation, they do not act as a scaffold themselves. The nucleus is the organelle that contains the DNA but does not directly influence the chromatin structure. Chromosomes are the end result of further condensation of chromatin, representing a more complex level of organization, but they do not serve as the starting scaffold for chromatin formation.

Q69. What do the two peaks in the action spectrum represent?

• A. Absorption of light✓
• B. Carbon dioxide consumption
• C. Emission of light
• D. Both Options A and B are correct

Explanation: The action spectrum represents the efficiency with which different wavelengths of light are absorbed by a photosynthetic pigment to drive the process of photosynthesis. The two peaks in the action spectrum indicate the wavelengths of light that are most efficiently absorbed by the photosynthetic pigment(s) in question. These absorbed photons provide the energy needed to initiate the light-dependent reactions of photosynthesis.

Q70. Which one of the following is a storage plastid?

• A. Chromoplast
• B. Leucoplast✓
• C. None of these
• D. Chloroplast

Explanation: Leucoplasts are colorless plastids that store starch or lipids. Chloroplasts contain a green-colored pigment called chlorophyll and are the site for photosynthesis. Chromoplasts contain colored pigments and are found in fruits and petals of flowers.

Q71. The abundant inhibitory neurotransmitter found in the CNS is called?

• A. Gamma-glutamyltransferase
• B. Gamma-linolenic acid
• C. Gamma-Aminobutyric acid✓
• D. None of these

Explanation: The abundant inhibitory neurotransmitter found in the central nervous system (CNS) is "c) Gamma-Aminobutyric acid (GABA)." GABA is a crucial neurotransmitter that plays a fundamental role in regulating neuronal excitability and maintaining a balance between inhibitory and excitatory signals in the brain.GABA is Gamma-aminobutyric acid, an amino acid that functions as the primary inhibitory neurotransmitter in the brain and spinal cord, reducing neuronal excitability to promote a calming effect. It counteracts the excitatory neurotransmitter glutamate and is vital for proper neurological function. GABA is associated with regulating anxiety, stress, fear, sleep, and even conditions like high blood pressure, diabetes, and epilepsy.

Q72. For respiratory metabolism, a bacterial cell membrane contains:

• A. Proteins
• B. Lipids
• C. Enzymes✓
• D. Chemicals

Explanation: Bacterial cells lack membrane-bound organelles such as mitochondria. Therefore, the enzymes needed for respiration are present in invaginations of the cell membrane, called 'Mesosomes'.

Q73. The type of gene interaction in which the effect caused by a gene at one locus interferes with the effect caused by another gene at another locus is known as:

• A. Pleiotropy
• B. Epistasis✓
• C. Polygenic inheritance
• D. Gene linkage
• E. Crossing over

Explanation: Option A: Pleiotropy is a gene interaction in which a single gene affects multiple traits. Option B: The type of gene interaction in which the effect caused by a gene at one locus interferes with the effect caused by another gene at another locus is known as epistasis. Option C: Polygenic inheritance is a gene interaction in which multiple genes contribute to a single trait. Option D: Gene linkage is a phenomenon in which genes that are located close together on a chromosome are inherited together more often than would be expected by chance. Option E: Crossing over is a process that occurs during meiosis in which homologous chromosomes exchange genetic material.

Q74. In 'Complete Dominance':

• A. Different alleles of a gene are both expressed in heterozygous condition
• B. One allele (R) is completely dominant over the other (r)✓
• C. The phenotype of the heterozygote is intermediate between phenotypes of the two homozygotes
• D. Gene mutations may produce many different alleles of a gene

Explanation: Complete dominance is a form of dominance in a heterozygous condition wherein the allele that is regarded as dominant completely masks the effect of the allele that is recessive. Option C refers to 'Incomplete Dominance' where 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. Which of the following is true about both bacterial conjugation and meiosis?

• A. Both processes produce four haploid cells
• B. Both processes are a form of asexual reproduction
• C. Both processes involve genetic recombination✓
• D. None of these options are correct

Explanation: Both bacterial conjugation and meiosis involve genetic recombination, which is the process of exchanging genetic material to create diversity. In meiosis, genetic recombination occurs during crossing over in prophase I, resulting in four genetically unique haploid cells. In bacterial conjugation, DNA is transferred from a donor to a recipient cell, introducing new genetic material, but it does not result in the production of new cells. Thus, option C is correct. Option A is incorrect because only meiosis produces four haploid cells. Option B is incorrect because both processes are linked to sexual reproduction due to their involvement in genetic recombination. Option D is incorrect because option C is a true statement.

Q76. FADH2 is produced during which step of the Krebs cycle?

• A. When Isocitrate is converted into α - ketoglutarate
• B. When α - ketoglutarate is converted into Succinate
• C. During oxidation of succinate to fumarate✓
• D. When Fumarate is converted into Maltose

Explanation: Krebs cycle is a series of enzyme catalysed reactions occurring in the mitochondrial matrix, where acetyl-CoA is oxidised to form carbon dioxide and coenzymes are reduced, which generate ATP in the electron transport chainFADH2 is produced during the Krebs cycle, also known as the citric acid cycle or the tricarboxylic acid (TCA) cycle. The Krebs cycle is a series of enzymatic reactions that occur in the mitochondria of eukaryotic cells and is an essential part of cellular respiration.During the Krebs cycle, acetyl-CoA, derived from the breakdown of carbohydrates, fats, or proteins, enters the cycle by combining with a four-carbon molecule called oxaloacetate. This reaction forms a six-carbon molecule called citrate.Throughout the subsequent reactions of the Krebs cycle, citrate is converted back into oxaloacetate while releasing energy in the form of ATP, NADH, and FADH2. The energy-rich molecules NADH and FADH2 carry high-energy electrons that will be used in the electron transport chain to generate ATP.FADH2 is specifically produced in one of the reactions of the Krebs cycle. During the conversion of succinate to fumarate, an enzyme called succinate dehydrogenase catalyzes the reaction and transfers electrons to FAD (flavin adenine dinucleotide), reducing it to FADH2. FADH2 then carries these high-energy electrons to the electron transport chain for ATP production during oxidative phosphorylation.Overall, FADH2 plays a crucial role in the production of ATP by delivering electrons derived from the Krebs cycle to the electron transport chain, where they participate in the generation of a proton gradient

Q77. Which of the following is true regarding the bacteria?

• A. Bacteria contain plasma membranes that are made up of triglycerides
• B. Gram staining reveals that bacteria contain cell walls, but no cell membranes
• C. Bacteria do not contain a membrane-bound nucleus, but they contain an non membrane-bound region called the nucleoid✓
• D. All of these options are false

Explanation: Option A: All cells including bacteria contain cell membrane but it is not made up of triglycerides.Option B: Gram staining tells us about the nature of cell wall gram-positive and gram-negative bacteria. It does not tell about cell membranes.Option C: Bacteria do not contain membrane-bound organelles such as mitochondria or chloroplasts, as eukaryotes do. However, photosynthetic bacteria, such as cyanobacteria, may be filled with tightly packed folds of their outer.

Q78. Which of the following statement is incorrect about skeletal muscle fibres?

• A. They are multinucleated
• B. They are 1-1 nm in diameter and surrounded by sarcolemma✓
• C. They contain O2 storing molecules called myoglobin
• D. Their cytoplasm contains a high concentration of carbohydrates

Explanation: Skeletal muscle fibres are not 1-1 nm in diameter. They are 10-100 micrometers in diameter and are surrounded by sarcolemma.

Q79. Which of the following hormones acts on the uterus wall for thickening?

• A. Progesterone✓
• B. Zona pellucida
• C. Follicle stimulating hormone
• D. Oxytocin

Explanation: Progesterone prepares the endometrium for the potential of pregnancy after ovulation. It triggers the lining to thicken to accept a fertilized egg. It also prohibits the muscle contractions in the uterus that would cause the body to reject an egg.In women, FSH helps control the menstrual cycle and stimulates the growth of eggs in the ovaries. FSH levels in women change throughout the menstrual cycle, with the highest levels happening just before an egg is released by the ovary.The two main actions of oxytocin in the body are contraction of the womb (uterus) during childbirth and lactation.Zona pellucida is the thick transparent membrane surrounding a mammalian ovum before implantation.

Q80. Viroids lack:

• A. RNA
• B. Enzyme
• C. Protective protein coat✓
• D. All of these

Explanation: The correct option is C. Viroids lack a protein coat. Viroids are the smallest known infectious agents. They are composed of a single-stranded RNA molecule that is circular in shape. Viroids do not have a protein coat, which is a protective layer that surrounds the genetic material of other viruses.

Q81. The diagram shows the ultrastructure of a chloroplast as seen in section. What are the functions of P, Q, and R?

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

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

Q82. In humans, ovaries after birth contain:

• A. Oogonia
• B. Primary oocyte✓
• C. Secondary oocyte
• D. Ova

Explanation: Option A: Oogonia are the immature egg cells that are present in the ovaries before birth. They start to develop into primary oocytes during fetal development, but they do not mature until puberty. Option B: Primary oocytes are surrounded by a layer of cells called the granulosa cells. The granulosa cells provide nutrients and support for the primary oocyte. When a girl reaches puberty, one or two primary oocytes mature each month and are released from the ovaries during ovulation. The released primary oocyte is then called a secondary oocyte. So, the ovaries after birth contains primary oocytes. Option C: Secondary oocytes can be fertilized by sperm. If fertilization occurs, the secondary oocyte will develop into an embryo. Option D: Ova are fertilized eggs. They are not produced in the ovaries. Ova are produced when a secondary oocyte is fertilized by sperm.

Q83. The process by which various components of cells including its organelle can be isolated is called:

• A. Homogenization
• B. Cell Fractionation✓
• C. Cell Fixation
• D. Gel Electrophoresis
• E. Ultracentrifugation

Explanation: Option A: Homogenization is a process used in cell fractionation where cells are broken open to release their contents. It involves methods like mechanical disruption, sonication, or enzymatic digestion. Homogenization helps in breaking down the cell structure to release cellular components for further separation and analysis. Option B: Cell fractionation describes the process which is used to to separate individual cell components while maintaining the individual functions of each component. This enables each of the components or organelles to be isolated separately and examined. It involves homogenization first so that the organelles are released from the cells, and the organelles are separated via centrifugation. Hence, this term covers the entire process involved in isolating organelles, so it is the correct answer.Option C: Cell fixation is the preservation of cells or tissues using chemical agents. It involves treating cells with fixatives to immobilize their structures and preserve their components. Fixation helps maintain cellular morphology and prevents degradation or changes in cellular components during subsequent processing steps. Option D: Gel electrophoresis is a technique used to separate and analyze biomolecules, such as proteins or nucleic acids, based on their size and charge. The molecules are placed in a gel matrix and subjected to an electric field. They migrate through the gel at different rates based on their size and charge, allowing for their separation.Option E: Ultracentrifugation is a high-speed centrifugation method used to separate particles based on their density or size. Samples are spun at very high speeds using ultracentrifuges, generating high centrifugal forces. This causes particles to sediment based on their density or size gradients. Ultracentrifugation is commonly used in cell fractionation to separate cellular components, like organelles, based on their density. This is a step involved in the whole cell fractionation process, so option B is the preferred answer over this.

Q84. Eukaryotes can share which of the following structures with prokaryotes:

• A. Cell wall✓
• B. Nucleoid
• C. Golgi
• D. Mitochondria

Explanation: The answer is 'Cell wall'. The Cell wall of eukaryotes is made up of Chitin (Fungi) or Cellulose (Plants) while the prokaryotic cell wall is made up of peptidoglycan. Prokaryotes lack membrane-bound structures like Golgi apparatus and Mitochondria. They also don't have a definite nucleus.

Q85. First step of viral replication is:

• A. Entry
• B. Assembly
• C. Uncoating
• D. Adsorption or Attachment✓

Explanation: Viral infections follow similar steps in the virus replication cycle: attachment or adsorption, penetration, uncoating, replication, assembly, and release.

Q86. Goblet cells are:

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

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

Q87. In an adiabatic process:

• A. Pressure is maintained constant
• B. The gas is isothermally expanded
• C. There is perfect heat insulation✓
• D. System exchanges heat with the surroundings

Explanation: During an adiabatic process, the system is completely insulated from its surroundings. Thus, it takes place without heat entering or leaving the system, i.e.,q=0. During an isobaric process, the pressure of the system is kept constant (ΔP=0). During the isochoric process, the volume of the system is kept constant. (ΔV=0). During an isothermal process. the temperature of the system is kept constant (ΔT=0 but q≠0).

Q88. What is the volume in cm3 of 3.01 x 10^23 molecules of O2 gas at S.T.P.?

• A. 1000 cm3
• B. 11000 cm3
• C. 1120 cm3
• D. 11200 cm3✓

Explanation: 3.01*10^23 /6.02*10^23 = 0.5 mole at STP,1 mole of gas –––> 22,4 L 0,5 mole –––––––> x x= 11,2 L 11,2 L = 11200 cm3 As this is a numerical, it can only have one correct answer which is option D

Q89. When two moles of H2 and one mole of O2 react to form H2O, -484 KJ heat is evolved. What is Δ Hf for one mole of H2O?

• A. -484 KJmol-1
• B. -242 KJmol-1✓
• C. -121 KJmol-1
• D. +242 KJmol-1

Explanation: To determine the enthalpy change of formation (ΔHf) for one mole of H2O, we can use the given information about the heat evolved in the reaction.The balanced equation for the reaction is:2H2(g) + O2(g) → 2H2O(g)According to the given information, when two moles of H2 and one mole of O2 react, 484 kJ of heat is evolved. This heat change represents the enthalpy change of the reaction.Since two moles of H2O are formed in the reaction, we need to find the enthalpy change for one mole of H2O. To do that, we divide the heat evolved by the stoichiometric coefficient of H2O:ΔHf (H2O) = (484 kJ) / 2ΔHf (H2O) = 242 kJTherefore, the enthalpy change of formation for one mole of H2O is 242 kJ.Firstly, since heat is evolved, the formation of water is an exothermic reaction and so the sign of enthalpy change will be negative. Secondly, enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its constituent elements under standard conditions.When 2 moles of H2 and 1 mole of O2 react, 2 moles of H2O will be formed:2H2 + O2 —------------------------> 2H2O ΔH=-484 kJ/mol (given)Now, for the formation of one mole of H2O, we can simply divide the entire equation by 2:H2 + ½ O2 —-----------------------> H20 ΔH= -242 kJ/molSo, the enthalpy change of formation is –242 kJ/mol, which corresponds to option B, hence that is the correct answer.

Q90. The table shown below gives the bond dissociation energies of single covalent bonds of carbon atoms with elements A, B, C, and D. Which of the following has the smallest atom?

• A. A
• B. B
• C. C
• D. D✓

Explanation: The smaller the atom the closer the orbital electrons are to the nucleus, making the forces and thus the bond energy highest. Our answer to this question is hence C-D.

Q91. A sample of 0.7 moles of metal 'M' reacts completely with an excess of fluorine to form 45 g of MF2. How many moles of F are present in it?

• A. 1.4 moles✓
• B. 2.4 moles
• C. 2 moles
• D. 1.2 moles

Explanation: 1 mole of metal formed one mole of MF2 molecule ∴ 0.7 mole of metal formed = 0.7 mole MF2 molecule ∵ Each molecule of MF2 contains = 2 F atoms ∴ 0.7 mole of MF2 molecule contains = 2 x 0.7 moles of F atom = 1.4 mole of F atom Hence, 45 g of MF2 molecule contains 1.4 moles 'F' atom.

Q92. When does a gas deviate the most from its ideal behavior ?

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

Explanation: At high pressure and low temperature, a gas deviates the most from its ideal behaviour.Under these conditions, the gas molecules are close to each other and they exert significant intermolecular forces on each other. Also, the volume of gas molecules becomes a significant portion of total volume and cannot be neglected.Hence, the correct option is D.

Q93. If we move down the electrochemical series:

• A. Reduction potential will increase✓
• B. Reduction potential will decrease
• C. Oxidizing ability increase
• D. None of them

Explanation: Moving down the electrochemical series, the reduction potential will increase.The electrochemical series, also known as the activity series, is a list of elements or compounds arranged in order of their reduction potentials. Reduction potential is a measure of the tendency of a species to gain electrons and undergo reduction in a redox reaction.In the electrochemical series, elements or compounds higher up in the series have more negative reduction potentials, indicating a lower tendency to gain electrons and be reduced. Conversely, elements or compounds lower down in the series have more positive reduction potentials, indicating a greater tendency to gain electrons and be reduced.As you move down the electrochemical series, the reduction potential increase because the species are mire likely to accept electrons and be reduced. This is due to factors such as increasing atomic size, decreasing electronegativity, and decreasing ionization energy as you move down the series.In summary, as you move down the electrochemical series, the reduction potential increase, indicating an increasing tendency to undergo reduction.

Q94. Red lead is:

• A. PbO
• B. Pb3O4✓
• C. Pb2O4
• D. Pb2O5

Explanation: Lead(II, IV) oxide, also called minium, red lead or triplumbic tetroxide, is a bright red or orange crystalline pigment. Chemically, red lead is Pb3O4, or 2PbO, PbO2. Lead(II, IV) oxide is used in the manufacture of batteries, glass and rustproof primer paints.

Q95. The equation shows the reaction between element X and dilute hydrochloric acid. X(s) + 2HCl(aq) -> XCl2(aq) + H2(g) What types of bonding is present in element X and in compound XCl2?

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

Explanation: Elements, in this case, X, can not possess a covalent bond alone so it should be a metallic bond, this rules out options A and B. Since now it is established that X is metal and chlorine is non-metal, now the type of bonding possessed between a metal ion and a non-metal ion is ionic so that rules out option C while we are left with correct option D.

Q96. The value K for H2(g) + CO2(g) <-> H2O(g) + CO(g) is 1.80 at 1000 C. If 1.0 mole of each H2 and CO2 are placed in 1 litre flask, the final equilibrium concentration of CO at 1000 C will be:

• A. 0.295 M
• B. 0.385 M
• C. 0.531 M
• D. 0.573 M✓

Explanation: The given reaction is :- H2(g)+CO2(g)⇌H2O(g)+CO(g) Initial moles : 1 1 0 0 At eqm : 1−x 1−x x x Now, KC=[H2O][CO]/[H2][CO2] 1.8=x.x/(1-x)(1-x) ⇒x2=1.8(1−x)2 Taking square root on both sides ⇒x=1.34(1−x)⇒2.34x=1.34 ⇒x=0.573 So, concentration of CO at eqb=0.573M

Q97. Name the orbital n = 5; l = 2; m = 0.

• A. 5p
• B. 5d✓
• C. 5s
• D. 5f

Explanation: Quantum numbers are the basis of the sequence, arrangement, shape, and orientation of the elements in periodic table. N = 5 indicates the 5th energy level (5th period of the periodic table or the 5th principal quantum number) L = 2 indicates the shape of the orbital (2 corresponds d-orbital-shaped quantum number) M = 0 indicates the orientation of the orbital (z - axis magnetic quantum number)

Q98. If 9.8 g of sulfuric acid is dissolved in excess quantity of water, it will yield _ moles of hydrogen ion (H+) and _ moles of sulphate ions (SO42-).

• A. 0.1, 0.2
• B. 0.1, 0.3
• C. 0.2, 0.4
• D. 0.2, 0.1✓

Explanation: Mr (H2SO4) = 2(1) + 32 + 4 (16) = 98g.mol-1n(H2SO4)= 9.8/98= 0.1n(H+ ) = 2 x0.1 = 0.2n(SO4) = 0.1 x 1= 0.1

Q99. An intermolecular force of attraction X is relatively stronger than the other intermolecular forces. It stabilizes alpha-helix and Beta-pleated sheets of proteins. The double helical structure of DNA is also stabilized by this force of attraction. Identify X.

• A. Van Der Waals Forces
• B. Hydrogen bonding✓
• C. Ionic interactions
• D. Dipole dipole attraction

Explanation: Option A, Van der Waals forces are the weakest intermolecular forces and are not found in the examples, mentioned in the question given.Option B is correct, as H bonding stabilizes DNA structure as it is between bases in a complementary base pair as well as it stabilizes the secondary structures, such as alpha-helix and beta-pleated sheet. It is also stronger than the other intermolecular attraction, which is Van der Waals forces.Option C is present between ions and not molecules, and is not found within DNA or in secondary protein structures.Option D is an example of a Van der Waals force.

Q100. Which of the following is an insecticide?

• A. DDT✓
• B. Teflon
• C. Iodoform
• D. Dynamite

Explanation: DDT or dichlorodiphenyltrichloroethane is a colorless crystalline compound commonly used as an insecticide in the agriculture sector. It works by ceasing the nervous function in insects.

Q101. Consider the following reaction:R—CHO + 2[Ag(NH3)2]OH + R—COONH4 + 2Ag + 2NH3 + H2OThis reaction represents:

• A. Fehling test
• B. Benedict test
• C. Ninhydrin test
• D. Tollen's test✓

Explanation: The reaction you mentioned is actually the Tollens test. It's a chemical test that helps us detect the presence of aldehydes. In this test, an aldehyde (represented by R-CHO) reacts with Tollens' reagent, which is a mixture of silver nitrate (AgNO3) and ammonia (NH3). The reaction produces a silver mirror on the inner surface of the reaction vessel. This silver mirror formation is a positive result for the presence of an aldehyde.

Q102. The only thing which can predict that if the reaction will be spontaneous or nonspontaneous is:

• A. Enthalpy change
• B. Energy Change
• C. Free energy of the system✓
• D. Temperature Change

Explanation: Free energy of the system also known as the Gibbs free energy determines if the reaction is spontaneous or not. The formula for Gibss free reaction = enthalpy change- T(entropy change). If Gibbs free reaction is 0 or less, then the reaction is spontaneous.

Q103. The amount of energy released by absorbing an electron in the valence shell is:

• A. Ionization Energy
• B. Electron Affinity✓
• C. Electronegativity
• D. Atomic Radius
• E. Atomization Energy

Explanation: Electron Affinity is the correct answer because it is defined as the amount of energy released when an electron is added to a neutral atom to form an anion. This process is exothermic for most elements, meaning energy is released.Ionization Energy is incorrect because it involves removing an electron, not adding one, and thus requires energy. Electronegativity is a measure of an atom's ability to attract electrons in a bond and does not directly involve energy release upon electron addition. Atomic Radius is a measurement of size and has no direct correlation with energy release. Atomization Energy pertains to breaking molecules into atoms, not adding electrons to individual atoms.

Q104. Which type of isomerism depends on the distribution of carbon atoms on each side of the functional group?

• A. Structural isomerism
• B. Functional isomerism
• C. Chain isomerism
• D. Metamerism✓

Explanation: This type of isomerism arises due to the presence of different alkyl chains on each side of the functional group.Structural isomer is a type of isomerism that uses the same formula but different structures. Functional isomerism is a type of isomerism in which two structures have the same formula but different functional groups. Chain isomers have the same molecular formula, but the way their carbon atoms are joined together differs from isomer to isomer.

Q105. The influence of temperature on reaction rate is predicted by:

• A. Free energy change
• B. Van Der Waals equation
• C. Arrhenius equation✓
• D. Kinetic equation

Explanation: The effect of temperature on reaction rate is described by the Arrhenius equation.It shows that increasing temperature increases the rate constant exponentially

Q106. The Unit cell with crystallographic dimensions a=b≠c , α=β=γ=90 degrees is:

• A. Cubic
• B. Tetragonal✓
• C. Monoclinic
• D. Hexagonal

Explanation: A tetragonal unit cell has two equal dimensions (a = b ≠ c) and all angles equal to 90 degrees (α = β = γ = 90°).

Q107. Identify the common name of the following compound:CH₃CH₂COCH₂OH

• A. Α-hydroxy-ethyl methyl aldehyde
• B. Α-hydroxy-ethyl methyl ketone✓
• C. Β-hydroxy-ethyl methyl ketone
• D. None of the above

Explanation: For the common naming of a ketone, the two alkyl groups on both the sides of the carbonyl group are first identified, and then Greek letters are used for specifying the position of the carbon atoms near the carbonyl group. Let us now understand the concept of α, β, and γ carbons. α carbon= the one attached to carbonyl group β carbon= the one next to α carbon γ carbon= the one next to β carbon For the compound in question, the hydroxy group is attached to the α carbon and the alkyl groups are ethyl (2 carbons on the left side) and methyl (1 carbon on the right side). Additionally, the alphabetical order is followed while writing alkyl groups. Hence, the common name of the compound becomes α-hydroxy-ethyl methyl ketone.

Q108. The given diagram shows the enthalpy changes during a chemical reaction.

• A. An endothermic reaction
• B. An exothermic reaction✓
• C. An isothermic process
• D. A non-spontaneous process

Explanation: This diagram shows a reaction in which the reactants are at a greater energy level than the products, indicating that energy is released during the reaction, i.e. change in enthalpy or (H2-H1) would be negative, indicating that it is an exothermic reaction so, option B is correct.Option A describes an endothermic reaction which would have a diagram depicting products, at a higher energy level than the reactants.Option C indicates a process in which the heat energy of the system does not change, but the diagram itself shows a change in enthalpy.Option D cannot be determined from a mere energy diagram as shown.

Q109. The thermal energy at constant pressure is called:

• A. Enthalpy✓
• B. Internal energy
• C. Heat capacity
• D. work done

Explanation: Enthalpy is defined as the sum of the internal energy of a system and the product of its pressure and volume. It is specifically used to describe the thermal energy of a system at constant pressure. Enthalpy change is related to the heat transfer occurring at constant pressure, making it the appropriate term for thermal energy under these conditions.

Q110. Which of the following alcohol does not get oxidized when it reacts with K₂Cr₂O₇/H₂SO₄?

• A. CH₃C(CH₃)₂OH✓
• B. CH₃CH(CH₃)OH
• C. CH₃OH
• D. C₂H₅OH

Explanation: Option A is a tertiary alcohol which cannot be oxidized with K2Cr2O7 as the central carbon atom has no free hydrogens.Option B is a secondary alcohol which oxidizes to a ketone. Option C is methanol which oxidizes to carbondioxide. Option D is ethanol which oxidizes to ethanal first then ethanoic acid.

Q111. When CH3 is attached to the benzene ring, it makes the ring a:

• A. Good electrophile
• B. Good nucleophile✓
• C. Resonance hybrid
• D. Extraordinary stable

Explanation: Benzene is a nucleophile because of its delocalized electrons thus the molecule has electron-rich areas which are frequently attacked by electrophiles. As alkyl groups have an electron pushing effect, they tend to push the delocalized electrons inside the ring, making it relatively more concentrated in terms of having more electrons gathered in a particular area.

Q112. Precipitation occurs when the product of ionic concentrations is:

• A. Greater than Ksp✓
• B. Less than Ksp
• C. Equal to Ksp
• D. Equal to unity

Explanation: Precipitation occurs when the product of ionic concentrations exceeds the solubility product constant (Ksp) for the particular ionic compound in question. The solubility product constant is a measure of the equilibrium solubility of a compound in a solution. Mathematically, the condition for precipitation is expressed as: [Product of ion concentrations] > Ksp The "product of ion concentrations" refers to the product of the molar concentrations (in mol/L) of the ions involved in the compound. It is important to note that only the ionic species that are involved in the compound and are present in the solution are considered for this product. When the product of ion concentrations exceeds the Ksp value, the ionic compound becomes supersaturated and precipitation occurs, leading to the formation of a solid precipitate. Conversely, if the product of ion concentrations is below the Ksp value, the system remains unsaturated, and no precipitation occurs. Ksp (solubility product constant), is an equilibrium constant for the quantitative measure of solubility of solids in solution. It is the product of concentrations of ions of a sparingly soluble salt, at equilibrium, raised to the power of their coefficients in the balanced chemical equation of dissociation. The higher the Ksp, the more soluble the salt is. If Q, the reaction quotient (ionic product) is greater than the Ksp, the excess will be precipitated. Therefore, option A is the correct answer.

Q113. According to valence shell electron pair repulsion theory, the repulsive forces between the electron pair of central atom of molecule are in the order:

• A. Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair✓
• B. Lone Pair - Bond Pair > Lone Pair - Lone Pair > Bond Pair - Bond Pair
• C. Bond Pair - Bond Pair > Lone Pair - Lone Pair > Lone Pair - Bond Pair
• D. Lone Pair - Bond Pair > Bond Pair - Bond Pair > Lone Pair - Lone Pair

Explanation: According to the VSEPR theory, lone pairs of electrons occupy more space than bonding pairs. As a consequence, this is an order of the magnitude of repulsion.Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair The greatest repulsion is between 2 lone pairs.Followed by repulsion between one lone pair and one bonding pair.The weakest repulsion is between 2 bonding pairs, the magnitude of repulsion between bonding pairs of electrons depends on the electronegativity difference between a central atom and the other atoms.

Q114. If 4.6 gm of ethyl alcohol and 6.0 gm of acetic acid is kept at a constant temperature until equilibrium was established, 2.0 gm of unused acetic acid were present. What is the Kc?

• A. 2.0
• B. 3.0
• C. 4.0✓
• D. 5.0

Explanation: Initially the moles of ethyl alcohol are:4.6/46=0.1 mol and acetic acid are:6/60=0.1 mol.In the end, we have 2gm of acetic acid that means 2/60=0.0333 mol of acetic acid is left or conversely speaking 0.0667 mol of acetic acid reacted to produce the same amount of ethyl acetate and water.So now Kc can be calculated as we now at the final stage of reaction we have 0.0333 mol of acetic acid and thus 0.0333 mol ethyl alcohol present as reactants, as molar ratio is same in the equation. We also know that since 0.0667 mol reacted so ethyl acetate is present in the same amount as water.Kc=[0.0667]2/[0.0333]2=4.0

Q115. In chemistry the work is generally _.

• A. Done by Temperature change
• B. Pressure - Temperature Work
• C. Pressure Volume Work✓
• D. None of these

Explanation: In chemistry, work is generally PV work.Pressure–volume work (or PV work) occurs when the volume V of a system changes. PV work is often measured in units of liter-atmospheres where 1L·atm = 101.325J.

Q116. 20 cm3 of a gaseous hydrocarbon was completely burnt in an excess of oxygen 60 cm3 of carbon dioxide and 40 cm3 of water vapours were formed, all volumes being measured at the same temperature and pressure. What is the formula of the hydrocarbon?

• A. C2H6
• B. C3H4✓
• C. C3H6
• D. C3H8
• E. C4H10

Explanation: The combustion of a hydrocarbon can be represented by the equation CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O. Given that 20 cm³ of the hydrocarbon produces 60 cm³ of CO₂ and 40 cm³ of H₂O, we can use the volume ratios to determine the stoichiometry. For each 20 cm³ of hydrocarbon, we obtain 60 cm³ of CO₂, indicating x = 3, and 40 cm³ of H₂O, indicating y = 4. Therefore, the formula is C₃H₄. Other options like C₂H₆, C₃H₆, C₃H₈, and C₄H₁₀ would result in different product volume ratios and do not satisfy the conditions given in the problem.

Q117. What does the symbol, || , means ?

• A. Indicator of electrochemical cell
• B. Indicator of salt bridge✓
• C. Indicator of SHE
• D. Indicator of electrode potential

Explanation: In the representation of redox reactions that occur in an electrochemical cell, the symbol, || , represents the salt bridge.

Q118. If the reaction, P + Q → R + S, is described as being of zero-order with respect to P, it means that:

• A. P is a catalyst in this reaction
• B. No P molecules possess sufficient energy to react
• C. The concentration of P does not change during the reaction
• D. The rate of reaction is independent of the concentration of P✓
• E. The rate of reaction is proportional to the concentration of Q

Explanation: If the order of reaction with respect to P is 0 (zero), this means that the concentration of P doesn't affect the rate of reaction. Mathematically, any number raised to the power of zero (x0) is equal to 1. That means that that particular term disappears from the rate equation.

Q119. Which of the following statement about Avogadro's hypothesis is correct?

• A. Under similar conditions of temperature and pressure, gases react with each other in simple ratio.
• B. Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules.✓
• C. At NTP all gases contain same number of molecules.
• D. Gases always react with gases only at given temperature and pressure.

Explanation: Avogadro's hypothesis states that equal volumes of gases at the same temperature and pressure contain the same number of particles, regardless of their chemical nature and physical properties. This hypothesis is the basis for the concept of the mole, which is a unit of measurement used in chemistry to express the amount of a substance. One mole of a substance is defined as the amount of that substance that contains the same number of particles as there are atoms in exactly 12 grams of carbon-12.

Q120. In which of the following reagents, cupric citrate complex is formed?

• A. Fehling's solution test
• B. Benedict's solution test✓
• C. Silver mirror test
• D. Sodium nitroprusside test

Explanation: The correct option for the reagent in which the cupric citrate complex is formed is B) Benedict's solution test. Both Fehling's solution and Benedict's solution contain cupric ions complexed with citrate ions, and they are used to test for reducing sugars (particularly aldehydes). The presence of reducing sugars causes the formation of a cupric citrate complex, which is then reduced to cuprous ions, leading to the characteristic color change in the solution.

Q121. The product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K, raised to the power of their relative concentrations, is the:

• A. Ksp✓
• B. Ka
• C. Kw
• D. Kb

Explanation: This is the definition of the solubility product constant (Ksp ). It represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound into its constituent ions in a saturated solution.

Q122. Linear combination of atomic orbitals (LCAO) results in the formation of:

• A. Sigma bond
• B. Pi bond
• C. Bonding molecular orbitals only
• D. Bonding and antibonding molecular orbitals✓
• E. All of these options

Explanation: According to Molecular Orbital Theory (MOT), two atomic orbitals overlap resulting in the formation of molecular orbitals. Number of atomic orbitals overlapping together is equal to the molecular orbital formed. The two atomic orbital thus formed by LCAO (linear combination of atomic orbital) in the same phase or in the different phase are known as Bonding Molecular Orbital (BMO) and Antibonding Molecular Orbitals (ABMO) respectively.

Q123. 1 mole of N2O4 (g) was placed in an empty 1 dm^3 container and allowed to reach equilibrium according to the following equation:N2O4(g) <-> 2NO2(g)At equilibrium, x mole Of N2O4 (g) had dissociated. What is the value of the equilibrium constant, Kc, at the temperature of the Experiment?

• A. 2x / (1 - x)
• B. 4x2 / (1 - x)2
• C. 2x / (1 - x)2
• D. 4x2 / (1 - x)✓

Explanation: This is the following solution:

Q124. In a zero-order reaction, the rate is independent of:

• A. The temperature of the reaction
• B. The concentration of reactants✓
• C. The concentration of products
• D. The presence of a catalyst

Explanation: In a zero-order reaction, the rate of reaction is independent of the concentration of reactants. This means that even if the concentration of reactants changes, the rate remains constant. This is distinct from other reaction orders, where the rate depends on the concentration of one or more reactants. Temperature affects the rate of reaction as it influences kinetic energy and collision frequency. Catalysts accelerate the reaction rate by providing a lower-energy pathway, and product concentration can shift equilibrium but does not directly affect the rate of a zero-order reaction.

Q125. In contact process, to which substance adequate quantities of water is added to convert it to sulphuric acid?

• A. H2S2O7✓
• B. HSO4-
• C. SO3
• D. S

Explanation: In the contact process, SO₃ is not added directly to water because it forms a mist that is hard to absorb. Instead, SO₃ is first absorbed in concentrated H₂SO₄ to form oleum (H₂S₂O₇). Adding adequate water to oleum safely produces sulphuric acid (H₂SO₄).

Q126. Which of the following compounds are added to ethanol to make it unfit for drinking?

• A. Pyridine
• B. Methanol
• C. Acetone
• D. All of these options are correct✓

Explanation: Although Pyridine is used in some medicines and flavorings as a solvent, it is also a main constituent for toxic substances like dyes, therefore the ingestion of pyridine could cause irritation, inflammation, dizziness, and even coma leading to death.Methanol is common rubbing alcohol which is poisonous for the body and can cause death if ingested. Acetone is the main constituent of nail paint removers, its fumes being so toxic that mass inhalation can cause a fatal coma.

Q127. CO2 and SO2 both are tri-atomic molecules but the heat of vaporization of SO2 is greater than that of CO2 due to:

• A. High electronegativity of S
• B. Greater size of SO2
• C. SO2 is polar and CO2 is non-polar✓
• D. SO2 is more acidic than CO2

Explanation: More polar molecules tend to form strong intermolecular forces which require a significant amount of heat to be overcome hence SO2 has a greater latent heat of vaporization. The polarity difference between the two molecules is explained below:CO2 is a linear molecule so the resultant dipole moment is zero as the individual dipole moments of the two CO bonds cancel each other out. SO2, on the other hand, is an angular molecule due to which it has a net dipole moment hence SO2 is polar and CO2 is non-polar.

Q128. The unit of the rate constant is the same as that of the rate of reaction in:

• A. Zero order reaction✓
• B. First order reaction
• C. Second order reaction
• D. Third order reaction

Explanation: The answer is the Zero order reaction. The rate law equation is given as: Rate=k([A]x)([B]y) for a reaction A+B ---> C. Here x and y are the order of the reaction with respect to A and B. Now the L.H.S is rate, which is defined as the change in an entity with respect to time. In this case, we are talking about concentration hence the rate is the change in concentration over time so, the units of the rate should be: (concentration)/time i.e. molarity/second or (moles/litre) x second. Now consider R.H.S, the units of A and B would be (concentration)x+y. For the rate constant to have units the same as the rate, the value of x+y should be zero. Therefore a zero-order reaction is the one where you have the same units for both rate and rate constant.

Q129. Which of following is an example of electrophilic attack on alcohols?

• A. C2H5OH + CH3COOH → CH3COOC2H5 + H2O
• B. 2C2H5OH + 2Na → 2CH5ONa + H2
• C. C2H5OH + HCl → C2H5Cl + H2O✓
• D. Both A and B

Explanation: The correct answer is Option C: C2H5OH + HCl → C2H5Cl + H2O. In this reaction, ethanol acts as a nucleophile and attacks the electrophile, HCl, resulting in the formation of ethyl chloride and water. This is a classic example of electrophilic substitution where the alcohol is converted to an alkyl halide.Option A represents an esterification reaction, which is not a direct electrophilic attack on the alcohol. Option B is a reaction of alcohol with sodium, which is a redox reaction and does not involve electrophilic attack. Finally, Option D is incorrect because it suggests both A and B are correct, but they do not represent electrophilic attacks.

Q130. A solid melts sharply just above 100-degree centigrade. It does not conduct electricity even when molten. It has weak Vander Waal's forces. What is the structure of the solid most likely to be?

• A. An atomic crystal
• B. Ionic crystal
• C. A molecular crystal✓
• D. Metallic

Explanation: The correct answer is a molecular crystal. Molecular crystals have low melting points and weak intermolecular forces, such as Vander Waals forces, which align with the description of the solid melting sharply just above 100-degree centigrade. They do not conduct electricity even when molten because they do not have any free-moving ions or electrons. In contrast, atomic crystals have strong covalent bonds with high melting points, ionic crystals have high melting points and conduct electricity when molten, and metallic crystals have high melting points and conduct electricity due to delocalized electrons.

Q131. Primary Alcohol is produced by reactions of Grignard's reagent with _ followed by hydrolysis in an acidic medium.

• A. Carbon dioxide
• B. Formaldehyde✓
• C. Acetaldehyde
• D. Ketone
• E. Methyl chloride

Explanation: The Grignard Reaction involves the addition of an organomagnesium halide (Grignard reagent) to carbonyl compounds. When Grignard reagents react with formaldehyde, a primary alcohol is produced upon hydrolysis in an acidic medium. Formaldehyde is unique as it is the simplest aldehyde, thus allowing the formation of a primary alcohol. Other aldehydes and ketones lead to secondary and tertiary alcohols, respectively. Carbon dioxide reacts with Grignard reagents to form carboxylic acids, not alcohols, and methyl chloride does not react to form an alcohol.

Q132. What is the value of molecularity and order of SN1 reactions?

• A. 2, 1
• B. 1, 1✓
• C. 0, 1
• D. 0, 2

Explanation: Option A) 2, 1 is incorrect because the molecularity of an SN1 reaction is 1, not 2.Option B) We define the order of reaction as the number of molecules of the reactant whose concentration changes during the chemical change. Molecularity is the number of ions or molecules that take part in the rate-determining step. It is meaningful only for simple reactions or individual steps of a complex reaction. Option C) 0, 1 is incorrect because the molecularity of a reaction cannot be zero, as it implies that no reactants are involved in the rate-determining step.Option D) 0, 2 is incorrect because the order of an SN1 reaction is 2, not 0. The reaction is second order concerning the substrate, but the nucleophile does not appear in the rate law equation, so it is considered to be in zero order.

Q133. When methylbenzene is treated with bromine in the presence of a catalyst, a mixture of two monobromo isomers is formed. What are the structures of these two isomers?

• A. Ortho-bromotoluene and meta-bromotoluene
• B. Meta-bromotoluene and para-bromotoluene
• C. Ortho-bromotoluene and para-bromotoluene✓
• D. Ortho-bromotoluene and ortho-bromotoluene
• E. Para-bromotoluene and para-bromotoluene

Explanation: The correct answer is Ortho-bromotoluene and para-bromotoluene. The methyl group (-CH3) on the benzene ring acts as an ortho-para directing group, meaning it directs incoming substituents (like bromine in this case) to the ortho (positions 2 and 6) and para (position 4) positions on the ring. Thus, when methylbenzene (toluene) is treated with bromine in the presence of a catalyst, bromine substitutes at these positions, forming ortho-bromotoluene and para-bromotoluene as the two monobromo isomers.Other options are incorrect because they either suggest meta substitution, which is not favored by the -CH3 group, or imply duplication of the same isomer, which does not align with the formation of two distinct isomers.

Q134. The structure of xenon trioxide is shown below:With reference to the valence shell electron pair repulsion theory (VSEPR), the shape of XeO₃ is:

• A. Trigonal planar
• B. Tetrahedral
• C. Bent (or angular)
• D. Trigonal pyramidal✓

Explanation: To determine the shape of the molecule, you must find the number of bonding pairs of electrons and lone pairs of electrons. Pi bonds are to be excluded. The molecule has three single bonds, which correspond to three bonding pairs of electrons, and one lone pair of electrons, resulting in a structure similar to that of ammonia, which is trigonal pyramidal so, the answer is D.Option A would occur in a molecule with only three bonding pairs of electrons.Option B would occur in a molecule with four bonding pairs of electrons.Option C would occur in a molecule with either two bonding pairs with one lone pair or with two bonding pairs and two lone pairs of electrons.

Q135. Consider the following reactionN2 (g) + O2 (g) < - > 2NO (g) Kc - 0.1 at 2000 CIf the original concentration of N2 and O2 were 0.1 M each. Calculate the concentrations of NO at equilibrium.

• A. 0.028 M✓
• B. 0.0012 M
• C. 0.18 M
• D. 0.0018 M
• E. 0.002 M

Explanation: Explanation is given below.

Q136. At equilibrium the concentration of reactants and product become:

• A. Zero
• B. Equal
• C. Constant✓
• D. Infinite

Explanation: Equilibrium occurs when the rate of forward reaction becomes equal to the rate of backward reaction, thus making the concentration of reactants and products constant hence 'option C' is correct. 'Option A' is very wrong because concentrations of reactants and products do not become zero at equilibrium. It can be understood by common sense that if the concentrations were to become zero, then the reaction will not occur. 'Option B' is the most commonly chosen incorrect answer. Equilibrium is frequently misunderstood as equal concentrations of reactants and products, which is extremely wrong! 'Option D' is wrong because concentrations of reactants and products are limited (not infinite).

Q137. In the reaction; H2 + CO2 ⇌ H2O + CO. The decrease in the concentration of reactants and products cause the equilibrium to shift:

• A. Towards left
• B. Towards right
• C. Nothing happens to the equilibrium✓
• D. Equilibrium will shift towards both the directions

Explanation: At equilibrium, the system depends on the ratio of concentrations (the reaction quotient), not the absolute amounts. A proportional decrease in both reactants and products does not change the ratio, so the system remains at equilibrium. Thus, the equilibrium does not shift.

Q138. The total number of neutrons in 5g of H2O are:1 NA = 6.023 x 1023

• A. 25 NA
• B. 1.1 NA
• C. 1.34 x 1024✓
• D. 0.5 NA

Explanation: To find the total number of neutrons in 5g of H2O, we start by calculating the number of moles of water:1. The molar mass of H2O is approximately 18 g/mol (2 for H and 16 for O).2. The number of moles in 5g of H2O is calculated as follows: Number of moles = mass (g) / molar mass (g/mol) = 5g / 18 g/mol ≈ 0.278 moles.3. Each molecule of H2O contains 2 hydrogen atoms (each with 0 neutrons) and 1 oxygen atom (with 8 neutrons), totaling 8 neutrons per molecule.4. The total number of molecules in 0.278 moles is: 0.278 moles x 6.023 x 1023 molecules/mole ≈ 1.67 x 1023 molecules.5. Therefore, the total number of neutrons in 5g of H2O is: 1.67 x 1023 molecules x 8 neutrons/molecule = 1.34 x 1024 neutrons, which corresponds to Option C. The other options are incorrect due to miscalculations or misunderstandings of stoichiometric principles.

Q139. How many molecules are there in 2.1 moles of CO₂:

• A. 3.49 x 10⁻²⁴
• B. 2.53 x 10²⁴
• C. 1.26 x 10²⁴✓
• D. 3.79 x 10²⁴

Explanation: Since there are 2.1 moles of CO₂, we can easily find out the number of molecules present in it by multiplying 2.1 with the Avagadro's constant, which is 6.02 × 10²³. Hence, the answer would be approximately 1.2 x 10²⁴. The answer choice that is closest to the obtained value is 1.26 x 10²⁴. Please do not calculate the total number of atoms in this question; you have only been asked to calculate the total number of molecules.

Q140. Which of the following have the same number of molecules?

• A. 200 cm2 of Carbon dioxide and 500 cm2 of oxygen
• B. 300 cm2 of Carbon dioxide and 150 cm2 of oxygen
• C. 300 cm2 of Carbon dioxide and 150 cm2 of oxygen
• D. 300 cm2 of Carbon dioxide and 300 cm2 of oxygen✓

Explanation: 300 cm2 of Carbon dioxide and 300 cm2 of oxygen would have the same number of molecules since the number of moles in question is equal, thus the number of molecules would also be the same.

Q141. If magnetic field is given by B= (2i + 3j -8k) and a loop of area 10 sq.m is placed in field in y-z plane, the maximum flux will be:

• A. -20 wb
• B. 30 wb
• C. 20 wb✓
• D. 80 wb

Explanation: The maximum flux through a loop of area A placed in a magnetic field B is given by the product of the area and the component of the magnetic field perpendicular to the plane of the loop. In this case, the loop is placed in the y-z plane, which is perpendicular to the x-axis. Therefore, the component of the magnetic field perpendicular to the plane of the loop is given by the x-component of the magnetic field, which is 2 Tesla. So, the maximum flux through the loop is given by the product of the area of the loop and the x-component of the magnetic field, which is: Φ = BAcosθ Cos (0°) = 1 Maximum flux ( Φ)= B×A = 2× 10 = 20 Weber. Therefore, the maximum flux through the loop is 20 Weber.

Q142. A potential difference of 10V is applied across a conductor whose resistance is 2.5 Ohms. What is the value of current flowing through it?

• A. 4 A✓
• B. 2 A
• C. 6 A
• D. 10 A

Explanation: R = V/I ;2.5 = 10/I 10/2.5 = 4 A

Q143. Two positive point charges Q1 = 16 microCoulombs and Q2 = 4 micro Coulombs are seperated in a line by a distance of 3 meters. Find the spot on the line between the charges where the electric field is zero.

• A. 3 m
• B. 5 m
• C. 4 m
• D. 2 m✓

Explanation: We have to search for the null point where the resultant field is zero. Since the two point charges are of similar nature (both positive), there cannot be a null point on the line lying either left of point A or right of point B as in these regions the individual fields produced by the charges are parallel and hence cannot nullify each other. However there is a possibility of a null point between the points A and B. Let P be that point (lying at a distance x from point A). Therefore, the resultant electrostatic field is zero at a point (on the line joining the two charges) lying 2m right of the charge 16 μC or 1m left of the charge 4μC.

Q144. When the output power equals to one-half of the input power, efficiency of the transformer becomes:

• A. 0%
• B. 100%
• C. 50%✓
• D. 200%

Explanation: Following is the formula for efficiency. As we can plug in values of input and output, we will get 50%.

Q145. One complete circle is equal to:

• A. 2π radian✓
• B. 3 radian
• C. 5 radian
• D. 9 radian

Explanation: There are 2π radians in a full circle. So 2π radians should equal 360°.

Q146. If a wheel of radius r turns through an angle of 30°, then the distance through which any point on its rim moves is?

• A. r(π/3)
• B. r(π/6)✓
• C. r(π/30)
• D. r(π/180)

Explanation: The dispance through which any point on the rim moves is equal to the the distance the rim has rolled, S. S = rθ where θ is angle in radians Convert 30° to radians 30 x π/180 = π/6 S = r(π/6)

Q147. A uniform horizontal footbridge is 12 m long and weighs 4000 N. It rests on two supports X and Y as shown. A man of weight 600 N is at a distance of 4 m from support X. What is the upward force on the footbridge from support X?

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

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

Q148. If alpha, beta, and gamma rays carry the same momentum, which has the longest wavelength?

• A. Alpha rays
• B. Beta rays
• C. Gamma rays
• D. All have same wavelength✓

Explanation: Since alpha, beta and gamma rays carry the same momentum. So all these have the same wavelength.

Q149. The ratio of angular speed of moon around the Earth to its angular speed about its own axis is:

• A. 2:1
• B. 1:6
• C. 1:30
• D. 1:1✓

Explanation: The moon orbits the Earth once every 27.322 days. In the same amount of time, the moon rotates once on its axis. As a result, the moon does not seem to be spinning but appears to observers from Earth to be keeping almost perfectly still (this is called synchronous rotation). So its angular speed (W) around its own axis isW =2π /TW(moon) = (2π)/(27.322 x 24 x 3600) = 2.67*10(-6) rad/sFor orbital motion around the earth, the angular speed isW(earth) = (2π)/(27.322 x 24 x 3600) = 2.67*10(-6) rad/sSo the ratio is 1:1

Q150. Ali punches a mattress and then punches a wall with the same force. Why does Ali experience more pain while striking the wall?

• A. The rate of change of momentum while striking the wall is lower
• B. The rate of change of momentum while striking the wall is higher✓
• C. The wall does not move
• D. The mattress applies an equal and opposite force

Explanation: The rate of change in momentum when you hit the wall is more than when you hit a piece of sponge with an equal force.The rate of change in momentum is directly proportional to the force applied.Hitting a wall will hurt more than hitting the mattress. This can be explained as follows-Wall is hard, When you hit a wall, an equal reaction force acts on the hand and hurts it.When you hit the mattress, a part of the momentum of hand is transferred to the sponge. As a result, it gets deformed and the reaction is much less than the action.

Q151. The flux through a surface will be zero when angle between E and △A is:

• A. 90o✓
• B. 60o
• C. 30o
• D. 0o

Explanation: The electric flux (Φ) through a surface is given by the equation:Φ=E⋅A=EAcos⁡θWhere:E\ is the electric field vector,A is the area vector of the surface (whose direction is perpendicular to the surface),θ is the angle between the electric field E and the area vector A.When the angle θ is 90°, the electric field is perpendicular to the surface, meaning the component of the field that contributes to the flux is zero. Therefore, the flux will be zero when the angle between E and A is 90°.

Q152. If the engine power is 3.3kW and it is 60% efficient, how much water will it pump in 5s from a height of 10m?

• A. 60kg
• B. 100kg✓
• C. 75kg
• D. 80kg

Explanation: To calculate the amount of water pumped in 5 seconds from a height of 10 meters, we can use the principle of work and energy. First, let's determine the potential energy of the water at a height of 10 meters.Potential energy (P.E) = mass (m) * gravitational acceleration (g) * height (h)where:g ≈ 9.81 m/s² (gravitational acceleration)Given that the potential energy is transferred into the water by the engine, and the engine is 60% efficient, we can calculate the work done by the engine( Wengine) as follows:Wengine = P.E/efficiency , work done by the engine is also equal to the power (P) of the engine multiplied by time (t):Wengine = P * tWe are given that the engine power (P) is 3.3 kW, and the time (t) is 5 seconds.Now we can set up the equation:3.3 kW * 5 s = PE / 0.60Solving for PE:P.E = 3.3 kW * 5 s * 0.60 ≈ 9.9 kJNow, we can find the mass of the water (m) using the potential energy (P.E):PE = m * g * h9.9 kJ = m * 9.81 m/s² * 10 mSolving for m:m = 9.9 kJ / (9.81 m/s² * 10 m) ≈ 0.101 kg ≈ 100 g

Q153. Five words are shown below:Farthest Universe Spontaneously Photon InfiniteThese words can be used in the spaces P, Q, R, S, and T to complete the sentences below.The _P_ is a stable particle and therefore it does not decay _Q_ into any other particle. Its lifetime is, therefore, _R_ so long it does not undergo interaction with other particles, and is why photons are supposed to be reaching our earth from _S_ distances of the universe. Thus most of our information regarding the _T_ is carried by photons.

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

Explanation: The answer to this question can be reached just by finding what P is. P should be a ‘‘particle’’, in options having only a photon is the particle. · P: Photon · Q: Spontaneously · R: Infinitely · S: Farthest · T: Universe This is the correct order.

Q154. Four wires of same material, the same cross-sectional area and the same length when connected in parallel give a resistance of 0.25 ohms. If the same four wires are connected is series the effective resistance will be?

• A. 1 ohm
• B. 2 ohm
• C. 3 ohm
• D. 4 ohm✓

Explanation: For wires in Parallel1/Rtotal = 1/R + 1/R + 1/R + 1/R1/0.25 = 4/R 4 = 4/R R = 4/4R = 1ΩSo, the resistance of each resistor is 1ΩFor 4 resistors in SeriesRtotal = R+R+R+R = 1+1+1+1 = 4Ω

Q155. Angle between radius vector and centripetal acceleration is

• A. 0°
• B. π✓
• C. 2π
• D. none of these

Explanation: When a particle moves in a circular path, the radius vector points from the center of the circle to the particle, while the centripetal acceleration points towards the center of the circle. Therefore, they are directed opposite each other, making the angle between them 180°. In radians, this angle is represented as π, making Option B the correct choice.Option A (0°) is incorrect because that would imply the vectors point in the same direction. Option C (2π) is incorrect because it suggests a full circle, which is not applicable here. Option D is incorrect because the correct answer is indeed included in the given options.

Q156. Projectile, when launched at 90 degrees to horizontal, its trajectory is:

• A. Parabolic
• B. Periodic
• C. Hyperbolic trajectory
• D. Linear✓

Explanation: When a projectile is launched at 90 degrees with respect to the horizontal, it moves vertically upward and then falls back down under the influence of gravity. Its trajectory is a simple vertical line, not a parabolic, periodic, or hyperbolic curve.

Q157. If electron passes through axis of solenoid the movement will be:

• A. Towards the outward
• B. Towards the inward
• C. Parallel to its motion
• D. No force acts on it✓

Explanation: The electron is moving along the axis with velocity V Force = q(V x B) Velocity and magnetic field are in the same direction or at an angle of 180 V x B = VBsin(0 or 180) = 0 Force = 0 The image shows the direction of the magnetic field inside a solenoid.

Q158. There are three bulbs of 60W, 100W, and 200W. Which bulb has the thickest filament?

• A. 100W
• B. 200W✓
• C. 60W
• D. None of these options are correct

Explanation: 200W bulb has the thickest filament is correct. Higher the power rated, lower will be the resistance. More thickness will lead to low resistance.

Q159. When a train whistling passes near you, a considerable change in the pitch of the sound is heard. When the train is moving away, the pitch of the sound _ whereas the pitch of the sound _ when the train is approaching.

• A. Increases … decreases
• B. Increases … remains same
• C. Decreases … increases✓
• D. Decreases … remains same

Explanation: The correct answer is that the pitch of the sound decreases when the train is moving away and increases when the train is approaching. This phenomenon is explained by the Doppler effect, which describes how the frequency of a wave changes relative to an observer moving in relation to the source of the wave. When the train approaches, the sound waves are compressed, resulting in a higher frequency and thus a higher pitch. Conversely, when the train moves away, the sound waves are stretched out, leading to a lower frequency and lower pitch. The other options misunderstand this effect, suggesting incorrect changes in pitch or that the pitch remains unchanged, which is not consistent with the principles of wave motion and the Doppler effect.

Q160. For a certain organ pipe, three successive resonance frequencies are observed at 425, 595, and 765 Hz. The speed of the sound in air is 340 m/s. The pipe is:

• A. Closed pipe of length 1 m✓
• B. Closed pipe of length 2 m
• C. Open pipe of length 1 m
• D. Open pipe of length 2 m

Explanation: To find out which organ pipe is used, first, we will take the ratio of 3 observed harmonics: 425:595:765 5x 85 : 7x 85 : 9x 85 Since each successive harmonic is an odd number; the organ pipe is most likely a closed one and 85Hz is common for all so it is the fundamental frequency or f=425/5 = 85Hz f=595/7 =85Hz f=765/9= 85Hz Now: For fundamental frequency: L =v/4f = 340/85x4 = 1m answer!

Q161. A bullet of mass 20g leaves the gun with a velocity of 200 m/s. If the mass of gun is 2kg then the speed of recoil of the gun is:

• A. 2000 m/s
• B. 2 m/s✓
• C. 20 m/s
• D. 200 m/s
• E. 100 m/s

Explanation: Let the direction in which the bullet is fired be positive. Therefore, Mass of Bullet(m)=20 grams=0.02 kg Mass of Pistol(M)=2 kg Initial Velocity of Bullet(u)=0 m/s Initial Velocity of Pistol(U)=0 m/s Final Velocity of Bullet(v)=200 m/s Final Velocity of Pistol(V)= Recoil Velocity of Bullet = ? According to the Law of Conservation of Momentum, mu + MU = mv + MV (0.02)(0) + (2)(0) = (0.02)(200) + (2)(V) 0 + 0 = 4 + 2V --4/2 = V V = -2.0 m/s

Q162. What is the speed of 2.0 kg metallic bob at the mean position of a simple pendulum, when released from its extreme position 0.5m high? (g = 10 ms-2)

• A. 3.16 ms-1✓
• B. 10 ms-1
• C. 100 ms-1
• D. 50 ms -1

Explanation: At the extreme displacement, the potential energy, (PE= mgh = 2x10x0.5= 10J) is equal to the total energy of the system.At the mean position, we have all the potential energy being converted to kinetic energy= KE =10 J = ½ mv210J = 1/2mv2mv2=20 -> 2v2 =20 -> v2= 10, hence v =3.16 ms-1

Q163. The dimensions of momentum are the same as that of:

• A. All of these
• B. Power
• C. Impulse✓
• D. Planck's constant

Explanation: The dimensions of momentum are [M1L1T-1] corresponding to its formula mv. The dimensions of impulse are the same as momentum as it is actually the change in momentum. The dimensions of impulse are also [M1L1T-1] corresponding to its formula I = F x t.The dimensions of power are [M1L2 T-3]. The dimensions of Planck's constant are [M1L2T−1], the same as that of angular momentum.

Q164. A sample of a radioactive element with initial mass of 24 gm decayed to 3 gm in 36 minutes. How much of the original sample remained after the first 12 minutes?

• A. 12 g✓
• B. 6 g
• C. 2 g
• D. 8 g

Explanation: Mi = 24 gmf= 3 gTotal time t = 36 min24g→12g→6g→3gIt takes 3 half lives so n= 3As, n = total time/ half life 1 Half life T = t/nT =36/ 3 1 half life T= 12 min So remaining after one half life (means after 12 minutes) is 12 g

Q165. In an A.C. generator, increase in number of turns in the coil does which of the following?

• A. Increases emf✓
• B. Decreases emf
• C. Makes the emf zero
• D. Maintains the emf at a constant value

Explanation: Option A is correct since in an A.C. generator, the emf increases as the number of turns in the coil increases because the emf is directly proportional to the number of turns. A.C. generators work based on the principle of Faraday’s law. Note: Other factors that influence the induced emf are the strength of the magnetic field, the area of the rotating loop, and the angular velocity of the rotating loop.

Q166. The sum of kinetic energy and the potential energy is always constant provided that _ .

• A. There is a greater force of friction involved during motion
• B. The body is in simple harmonic motion✓
• C. There is less force of friction involved during motion
• D. No force of friction involved during motion

Explanation: The law of conservation of energy states that energy can neither be created nor destroyed. According to this, the correct option is B because the total energy in simple harmonic motion will always be constant consisting of kinetic and potential energies. However, kinetic energy and potential energy are interchangeable. Given below is the graph of kinetic and potential energy vs instantaneous displacement.The graph for a body is SHM. Option D although seems to be correct, is incorrect because the type of motion is not specified and hence a particular motion can also include energies other than kinetic and potential which are not interchangeable.

Q167. A car of mass 1200 kg initially at rest has been accelerated to a speed of 8 m/s in 16 meters. Average acceleration of the car is _ m/s2? And force is _ N?

• A. 1.5 and 1500
• B. 2.5 and 2400
• C. 3.5 and 3500
• D. 2 and 2400✓

Explanation: m=1200kg v=8m/s d=16m v2=u2+2as (8)2=(0)2+2(a)(16) 64=32a a=2m/s2 F = ma = 1200×2 =2400N

Q168. The current measuring part of the ammeter consists of a number of low resistors connected _.

• A. At an angle of 180 degrees with the galvanometer
• B. Parallel with galvanometer✓
• C. At an angle of 45 degrees With the galvanometer
• D. Perpendicular with the galvanometer

Explanation: Since Galvanometer is a very sensitive instrument therefore it can't measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low resistance known as "shunt" resistance is connected in parallel to the Galvanometer.

Q169. The minimum indivisible unit of charge is:

• A. 1 coloumb
• B. charge on on alpha particle
• C. charge on one proton✓
• D. 1 micro coloumb

Explanation: The smallest indivisible unit of charge is one electron, which is equal to charge of one proton. So the option is C.The minimum charge is that of an electron: 1.6 x 10-19 C.

Q170. When the anti nodes are all at their extreme displacements:

• A. The energy stored is completely kinetic
• B. The energy stored is completely potential✓
• C. The energy stored is kinetic and potential
• D. The energy stored is wholly chemical

Explanation: In the case of Simple Harmonic Motion, displacement is inversely proportional to velocity. When displacement is maximum, velocity will be zero hence the "energy held will be potential" as it is associated with a change in position.

Q171. A galvanometer acting as a voltmeter will have a coil with:

• A. A high resistance in parallel
• B. A high resistance in series✓
• C. A low resistance in parallel
• D. A low resistance in series

Explanation: The resistance of the coil of the galvanometer on its own is low; hence, to convert a galvanometer into a voltmeter, its resistance must be increased. For this, an appropriate high resistance is joined in series with the galvanometer.

Q172. Two radioactive samples 𝑆1 and 𝑆2 have half-lives 3 hours and 7 hours respectively. If they have the same activity at certain instant 𝑡, what is the ratio of the number of atoms of 𝑆1 to 𝑆2 at instant 𝑡?

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

Explanation: Refer to the following.

Q173. Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together. What is the speed of the trucks after the collision?

• A. v/4✓
• B. v/2
• C. v
• D. 5v/4

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

Q174. A battery whose emf is 40 V has an internal resistance of 5 Ohms. If this battery is connected to a 15 Ohms resistor ‘R’, what will be the voltage drop across ‘R’:

• A. 10 V
• B. 30 V✓
• C. 40 V
• D. 50 V
• E. 70 V

Explanation: 40 = I x (15+ 5) 40= I x 20 I = 40/20 I= 2 AVoltage drop across resistor R V= IR V= 2 x 15 V= 30 V (hence B is the correct option)

Q175. The reciprocal of resistivity is called:

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

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

Q176. A shock wave is produced due to an earthquake which makes the buildings move in the direction of the shock wave. Which progressive wave would this be?

• A. Longitudinal wave✓
• B. Transverse wave
• C. Material wave
• D. Particle wave

Explanation: The correct answer is Longitudinal wave because shock waves, such as those produced during an earthquake, cause the particles in the medium (like the ground) to oscillate in the same direction as the wave travels. This parallel movement of particles is characteristic of longitudinal waves. In contrast, Transverse waves have particle motion perpendicular to the wave direction, which does not apply to shock waves in this context. The term Material wave is too vague and does not specifically describe seismic waves, while Particle wave is not a recognized category in wave physics and does not provide a meaningful description of any wave type.

Q177. A 100 kg golf ball is moving to the right with a velocity of 20 m/s. It makes a head on collision with an 8 kg steel ball, initially at rest. The two balls get stuck together. What will the velocity of these two balls after collision?

• A. 14.3 m/s
• B. 17.5 m/s
• C. 18.5 m/s✓
• D. 19.7 m/s
• E. 20.0 m/s

Explanation: To solve this problem, apply the conservation of momentum principle which states that total initial momentum equals total final momentum in the absence of external forces. Before the collision, the golf ball (100 kg) moves with 20 m/s, and the steel ball (8 kg) is at rest. The initial momentum is calculated as:Initial Momentum = m1v1 + m2v2 = 100 kg × 20 m/s + 8 kg × 0 m/s = 2000 kg·m/s.After the collision, the two balls stick together, so their combined mass is 108 kg. Let the final velocity be u2. The equation becomes:2000 kg·m/s = (100 kg + 8 kg) × u2Solving for u2 gives:u2 = 2000 kg·m/s / 108 kg = 18.5 m/s.Thus, option C is correct. The other options incorrectly calculate the velocity by either misunderstanding the mass involved or misapplying the momentum conservation equation.

Q178. A body having translatory motion possesses _ and _. In the same way a body having rotatory motion possesses _ and _.

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

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

Q179. The path difference for the constructive interference is:

• A. (n-1)λ
• B. (n+1)λ
• C. n λ / 2
• D. 2n λ
• E. n λ✓

Explanation: Option A: (n-1)λ: This is the path difference for destructive interference. When the path difference is one wavelength less than a whole number of wavelengths, the waves will be out of phase and will cancel each other out. Option B: (n+1)λ: This is also the path difference for destructive interference. When the path difference is one wavelength more than a whole number of wavelengths, the waves will be out of phase and will cancel each other out. Option C: nλ/2: This is the path difference for a node. A node is a point where the waves cancel each other out completely. Option D: 2nλ: This is not possible, as the path difference cannot be greater than the wavelength of the wave. Option E: The path difference for constructive interference is nλ, where n is any integer. This means that the difference in the distance traveled by two waves must be a whole number of wavelengths. When this happens, the waves will be in phase and will add together to create a larger wave.

Q180. If there are "n" capacitors each of capacity "c" connected in parallel to "V" volts source then the energy stored is equal to:

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

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

Q181. Electric field lines:

• A. Never cross each other✓
• B. Can cross each other
• C. Depends on the shape of charge
• D. Not enough information is available

Explanation: Electric lines of force never intersect each other because at the point of intersection, two tangents can be drawn to the two lines of force. This means two direction of electric field at the point of intersection, which is not possible

Q182. A mass of 20 kg is lifted from the floor to a height of 2m in 4.9 sec. Calculate the power in Watts.

• A. 60 W
• B. 80 W✓
• C. 100 W
• D. 120 W

Explanation: Firstly, we’ll find the increase in the gravitational potential energy as we have the object traveling vertically and we are unaware of the force by which the object is being lifted. It could be equivalent to the weight of the object or greater. The height is increased by 2 meters and by using the formula 'mgh', where g is the gravitational field strength, we can find the increase in GPE. The GPE increased by 392.4 J in 4.9 seconds and the formula for power is 'energy over time' so, we divide 392.4 J by 4.9 seconds giving us 80W.

Q183. Asimple pendulum has mass M, length L and time period T. What is the period of oscillation of the pendulum with mass 4M and length 0.49L?

• A. 0.7T✓
• B. T
• C. 2T
• D. 3T

Explanation: Option A is correct since the formula for a time period of a simple pendulum is T=2pi * root of l/g. Therefore substituting the values: 2*pi* the root of L/9.81 is T. If 0.49 is placed with l, the time-period becomes 0.7T. Option B is incorrect since it states that the period remains the same, which cannot happen since length changes Option C is incorrect since time-period will double if the length becomes 4 times which is not the case Option D is incorrect since time-period will triple if the length becomes 9 times which is not the case

Q184. The half-life of 14C is approximately 5,730 years, while the half-life of 12C is essentially infinite. If the ratio of 14Cto 12C in a certain sample is 25% less than the normal ratio in nature, how old is the sample?

• A. Less than 5,730 years✓
• B. Approximately 5,730 years
• C. Significantly greater than 5,730 years, but less than 11,460 years
• D. Approximately 11,460 years
• E. Approximately 15,730 years

Explanation: Because the half-life of carbon-12 is essentially infinite, a 25 percent decrease in the ratio of carbon-14 to carbon-12 means the same as a 25 percent decrease in the amount of carbon-14. If less than half of the carbon-14 has deteriorated, then less than one half-life has elapsed. Therefore, the sample is less than 5730 years old. Be careful with the wording here—the question states that the ratio is 25% less than the ratio in nature, not 25% of the ratio in nature, which would correspond to choice (D).

Q185. When brakes of a car are applied, angular velocity of a flywheel reduces from 900 cycle / min to 720 cycle / min in 6 sec. Angular retardation is:

• A. 𝝅 rad/s2✓
• B. 9 𝝅 rad/s2
• C. 8 𝝅 rad/s2
• D. ⅔ 𝝅rad / s2
• E. Insufficient data

Explanation: This is the following solution.First, we need to convert the given angular velocities from cycles per minute to radians per second. We know that 1 cycle is equal to 2π radians. Therefore,Angular velocity at the start = 900 cycles/min = (900 x 2π) / 60 rad/s = 30π rad/sAngular velocity at the end = 720 cycles/min = (720 x 2π) / 60 rad/s = 24π rad/sThe time taken for the flywheel to reduce its angular velocity from 30π rad/s to 24π rad/s is 6 seconds. We can use the formula for angular retardation:Angular retardation (α) = (ω2 - ω1) / tWhere ω2 is the final angular velocity, ω1 is the initial angular velocity, and t is the time taken.Substituting the given values, we get:Angular retardation (α) = (24π - 30π) / 6 = -π rad/s^2Therefore, the angular retardation is -π rad/s^2.

Q186. An example of a non-ohmic resistor is:

• A. Diode✓
• B. Tungsten wire
• C. Carbon resistor
• D. Copper wire

Explanation: Diodes are examples of non-ohmic resistors because they do not have a constant resistance and their current-voltage relationship is nonlinear, meaning they do not adhere to Ohm's Law. The resistance changes depending on the voltage across them. On the other hand, tungsten wire, carbon resistors, and copper wire are all examples of ohmic materials. They maintain a constant resistance over a range of voltages and currents, thus complying with Ohm's Law.

Q187. What is the wavelength of the wave if the phase angle between two points of the medium is 3 Pi/4, and are separated through a distance of 3 cm?

• A. 8 cm✓
• B. 9 cm
• C. 1 cm
• D. 12 cm

Explanation: The phase difference between two points on a wave is given by: Δφ = 2πΔd/λ where Δd is the distance between the two points and λ is the wavelength. In this case, we are given that the phase angle is 3π/4 and the distance between the points is 3 cm. So, we can plug these values into the equation above and solve for the wavelength: 3π/4 = 2π(3 cm)/λ λ = 8 cm Therefore, the wavelength of the wave is 8 cm.

Q188. Three 6 Ω are connected as shown in the diagram What is the resistance between points ‘A’ and ‘B’

• A. 6 Ω
• B. 16 Ω
• C. 4 Ω
• D. 2 Ω✓

Explanation: If looked into closely, these resistors are actually parallel to each other.The total resistance is equal to:1/R=1/R1+1/R2+1/R31/R=⅙+⅙+⅙=3/61/R=½R=2 ohms.

Q189. The force between two charges Q and q, separated by a distance is F. What will be the force between them when the distance between them is d/2?

• A. 4 F✓
• B. 2 F
• C. F
• D. F/2

Explanation: According to Coulomb's law of electrostatic force, the force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. F = k q1q2/d²Where k is constant, q1 and q2 are two point charges and d the distance between them. If the distance between the charges is d/2 and there is no change in chargesThe new force becomesF’=kq1q2/(d/2)²F’=k q1q2/(d²/4)F’=4FWhich means, force will be 4 times of the initial value.

Q190. A particle carrying charge of 2e falls through a potential difference of 3.0 V. Calculate energy acquired by it.

• A. 9.6 × 10-19 J✓
• B. 12.1 × 10-19 J
• C. 14.5 × 10-19 J
• D. 16.7 × 10-19 J
• E. 18.5 × 10-19 J

Explanation: To find the energy acquired by the particle, use the formula: Energy = qV, where q is the charge and V is the potential difference. The charge of the particle is 2e, where e is the elementary charge, approximately 1.6 × 10-19 C. The potential difference is given as 3.0 V. Thus, Energy = 2e × 3 V = 6eV. Converting eV to joules gives: Energy = 6 × 1.6 × 10-19 J = 9.6 × 10-19 J.Options B, C, D, and E are incorrect because they do not follow the correct calculation using the formula and given values.

Q191. E/m of an electron is given by the relationship.

• A. e/m2(VB2r2)
• B. e/m=(V/Br)2
• C. e/m=V.r/B
• D. e/m=2V/B2r2✓

Explanation: The E/m (charge-to-mass ratio) of an electron is given by the formula:em=2V/B2r2​where:e = charge of electronm = mass of electronV = accelerating potential (voltage)B = magnetic field strengthr = radius of the circular path of the electron👉 This relationship comes from balancing the centripetal force and magnetic force on an electron moving in a circular path.

Q192. The sensitivity of a potentiometer can be Increased by _.

• A. Increasing the e.m.f. of the primary coil
• B. Increasing the Potential gradient
• C. increasing the length of the potentiometer wire✓
• D. Decreasing the length of the potentiometer wire

Explanation: The sensitivity of a potentiometer can be increased by increasing the length of the potentiometer wire.The sensitivity of a potentiometer refers to the change in output voltage per unit length of the potentiometer wire. The increased length of the potentiometer wire allows for finer adjustment of the output voltage. It increases the resolution or sensitivity of the potentiometer, enabling more precise control or measurement of the voltage.

Q193. When a force acts at right angles to the displacement (θ=90°) the work done is zero i.e. the force does not produce work. Identify the example/s from the following when work is zero.I. It is considered ‘hard work to hold a heavy stone stationary at stretched handII. A person walks along a level surface while carrying a boxIII. When a body moves in a circular path

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

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

Q194. A longitudinal wave is moving through a medium. Which of the following statements about the direction of the propagation of the wave and displacement of the medium is true?

• A. Displacement of medium = parallel to the energy transfer Direction of the propagation of wave = parallel to the energy transfer✓
• B. Displacement of medium = parallel to the energy transfer Direction of the propagation of wave = perpendicular to the energy transfer
• C. Displacement of medium = perpendicular to the energy transfer Direction of the propagation of wave = parallel to the energy transfer
• D. Displacement of medium= perpendicular to the energy transfer Direction of the propagation of wave = perpendicular to the energy transfer

Explanation: Option A) Longitudinal waves are the waves in which the displacement of the particles/medium is parallel to the direction of propagation of wave/direction of energy transfer. We know that waves transfer energy and a wave will always transfer energy in the direction of its propagation. Hence Option A is the correct option.Option B) Displacement of medium = parallel to the energy transfer and Direction of the propagation of wave = perpendicular to the energy transfer is incorrect because this describes a transverse wave, not a longitudinal wave. In a transverse wave, the displacement of the medium is perpendicular to the direction of propagation of the wave.Option C) Displacement of medium = perpendicular to the energy transfer and Direction of the propagation of wave = parallel to the energy transfer is also incorrect. This does not describe a longitudinal wave.Option D) Displacement of medium = perpendicular to the energy transfer and Direction of the propagation of wave = perpendicular to the energy transfer is also incorrect because this describes a different type of wave called a surface wave.

Q195. Statements:Pakistan is a multilingual country. Urdu is the national language of Pakistan.Conclusions:(l) AII Pakistanis should learn many languages. (ll) To be a Pakistani one needs to learn Urdu.

• A. Only conclusion (I) follows
• B. Only conclusion (II) follows
• C. Both conclusions follow
• D. Both of the conclusions do not follow✓

Explanation: The statement nowhere tells us that Pakistanis should learn several languages, it just states that many languages are spoken in Pakistan. Furthermore, just because Urdu is the national language, it doesn’t mean that a person who doesn’t know it isn’t a Pakistani. Hence both statements do not follow.

Q196. Complete the sequence:2, 15, 41, 80, _

• A. 111
• B. 120
• C. 121
• D. 132✓

Explanation: Multiples of 13 are being added + 13 +26 +39 so now you'll add 52

Q197. Which word does not belong to the group in each of the following questions?

• A. Chest✓
• B. Ear
• C. Lip
• D. Nose

Explanation: Options B, C, and D are body parts found on the head. Chest is the only outlier here.

Q198. Which one of the following has four sides:

• A. Triangle
• B. Square✓
• C. Circle
• D. Right triangle

Explanation: Correct option is B. Square has 4 sides.

Q199. Statement: A large number of students are reported to be dropping out of school in villages as their parents want their children to help them in farms.Courses of Action:I. The government should immediately launch a programme to create awareness among the farmers about the value of education.II. The government should offer incentives to those farmers whose children remain in schools.III. Education should be made compulsory for all children up to the age of 14 and their employment banned.

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

Explanation: Literacy at basic level is the utmost need to prepare good future citizens. So, all children need to be educated. This can be achieved by creating awareness, providing incentives, enforcing education and banning employment of children. Thus, all the three courses follow.

Q200. Statements (I) The university graduate claims that it is hard to find a job in this era. (II) There is a ratio of unemployment on a large scale

• A. Statement I is the cause and Statement II is the effect
• B. Both statements I and II are effects of some common cause
• C. Both statements I and II are effects of independent causes
• D. Statement II is the cause and Statement I is the effect✓

Explanation: The correct answer is that Statement II is the cause and Statement I is the effect. This is because a large scale of unemployment (Statement II) naturally results in fewer job opportunities, making it harder for individuals, such as university graduates, to find jobs (Statement I).Option A is incorrect because the statement about difficulty in finding a job is a consequence, not a cause.Option B is incorrect because there is no indication of a common cause affecting both statements; instead, Statement II directly results in Statement I.Option C is incorrect because the statements are linked, with Statement II directly influencing Statement I, rather than being independent of each other.