Sindh Mdcat Mock Paper 3 — Solved Past Paper with Answers

Q1. Choose the word most similar in meaning to the capitalized ones. UNAMBIGUOUS

• A. Stagnant
• B. Hidden
• C. Clear✓
• D. Muddy
• E. Grubby

Explanation: The word 'unambiguous' means 'without any doubt' or 'clearly expressed.' The term 'clear' is synonymous with unambiguous as it indicates something easy to understand or interpret.

Q2. Read the passage to answer the question:To inquire about the meaning or object of one's own existence or of creation generally has always seemed to me absurd from an objective point of view. And yet everybody has certain ideals that determine the direction of his endeavors and his judgments. In this sense, I have never looked upon ease and happiness as ends in themselves-such an ethical basis I call more proper for a herd of swine. The ideals which have lighted me on my way and time after time given me new courage to face life cheerfully, have been Truth, Goodness, and Beauty. Without the sense of fellowship with men of like mind, of preoccupation with the objective, the eternally unattainable in the field of art and scientific research, life would have seemed to me empty. The ordinary objects of human endeavor property, outward success, luxury-have always seemed to me contemptible.The author of the passage followed which of the following objectives?I.TruthII. GoodnessIII. BeautyIV. Saints

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

Explanation: I, II, AND III are clearly mentioned as the objectives in line 6. This question can be answered by directly inferring from the passage.

Q3. Identify the word or phrase that needs to be changed for the sentence to be correct:The office is so busy that two extra clerks have had to be taken on.

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

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

Q4. Identify the word or phrase that needs to be changed for the sentence to be correct.The guests broke a dozen glass at the party.

• A. The guests
• B. Broke
• C. Glass✓
• D. At
• E. No Error

Explanation: The given sentence is an example of a simple sentence. A simple sentence contains a subject and a verb, and it may also have an object and modifiers. However, it contains only one independent clause.The word "glass" requires a change to the plural form "glasses." The word or phrase that needs to be changed for the sentence to be correct is "glass"" (should be changed to "glasses").

Q5. Choose the correct spelling:

• A. Exantuated
• B. Axantuated
• C. Accenchuated
• D. Accentuated✓

Explanation: To "accentuate" means to “make more noticeable and prominent” and its correct spelling is given in Option D.

Q6. Complete the sentences by choosing the most appropriate option, from the given lettered choices below each. I was having a cup _ tea when he knocked on the door.

• A. Off
• B. At
• C. An
• D. Of✓

Explanation: The correct preposition to use is 'of'. In the phrase 'a cup of tea', 'of' indicates that the tea is contained within the cup, showing a relationship of content. The other options do not correctly express this relationship: 'off' suggests separation, 'at' implies location, and 'an' is not a preposition but an article.

Q7. Choose the correct option.

• A. Samar bought an apple, an orange and an pear.
• B. Samar bought an apple a orange and a pear.
• C. Samar bought an apple, an orange and a pear.✓
• D. Samar bought a apple, an orange and a pear.

Explanation: This option is correct. It correctly uses commas to separate the items in the list ("an apple", "an orange", and "a pear"), and the articles ("an" and "a") are used correctly.

Q8. Complete the sentences by choosing the most appropriate option, from the given lettered choices (A to D) below each. The recent discoveries of medical science have_ life and health to millions of people.

• A. Brought✓
• B. Bring
• C. Had bought
• D. Bringing

Explanation: The question is talking about a thing that has already happened, so bringing and bring would not be correct options. Option C changes the word to “bought” which has a totally different meaning so is incorrect. So option A is correct.

Q9. The diagram shows the four types of human teeth. Which teeth are used for cutting rather than grinding food?

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

Explanation: Most adults have 32 permanent teeth, including eight incisors, four canines, eight premolars, and 12 molars. Incisors and canines are used to cut and chew. Premolars and molars are used for grinding the food.

Q10. The inhibitor which closely resembles the substrate in its molecular structure and inhibits the enzyme activity by binding to the active site of the enzyme is called

• A. Feedback inhibitor
• B. Non-competitive inhibitor
• C. Allosteric modulator
• D. Competitive inhibitor✓

Explanation: In competitive inhibition, the inhibitor (I) closely resembles the real substrate (S) and is regarded as a substrate analog. The inhibitor competes with the substrate and binds at the active site of the enzyme but does not undergo any catalysis. As long as the competitive inhibitor holds the active site, the enzyme is not available for the substrate to bind. A competitive inhibitor diminishes the rate of catalysis by reducing the proportion of enzyme molecules bound to a substrate. At any given inhibitor concentration, competitive inhibition can be relieved by increasing the substrate concentration. So, the correct answer is 'Competitive inhibitor'.

Q11. The gland known as the" gland of emergency" is the

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

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

Q12. The autonomic nervous system controls all of the following activities except:

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

Explanation: Thought is a conscious action, so it is not controlled by the autonomic nervous system rather by the somatic nervous system. The autonomic nervous system regulates certain body processes, such as blood pressure and the rate of breathing. This system works automatically (autonomously), without a person's conscious effort.

Q13. Which of the following is the simplest form of pathogens causing diseases?

• A. Viruses
• B. Prions✓
• C. Fungus
• D. Amoeba

Explanation: Prions are the simplest disease-causing pathogens because they are merely misfolded proteins with with no nucleic acid (no DNA or RNA), making them the most minimal infectious agents compared to viruses, fungi, or protozoa like amoeba.

Q14. The enzyme found in saliva responsible for the digestion of carbohydrates is:

• A. Lysozyme
• B. Amylase✓
• C. Pepsin
• D. Trypsinogen
• E. Lipase

Explanation: The enzyme found in saliva responsible for the digestion of carbohydrates is amylase. Amylase breaks down complex carbohydrates, such as starches, into simpler sugars like maltose and glucose. Salivary amylase is produced by the salivary glands and is an important component of the initial stages of the digestive process, helping to initiate the breakdown of carbohydrates in the mouth before food travels to the stomach and further along the digestive tract.

Q15. The process by which one molecule of Glucose splits up into molecules of Pyruvate is called:

• A. Glycolysis✓
• B. Oxidative Phospholyration
• C. Electron Transport Chain
• D. Kreb's Cycle
• E. Calvin Cycle

Explanation: The process by which one molecule of glucose splits up into molecules of pyruvate is called glycolysis. Glycolysis is the first stage of cellular respiration, which occurs in the cytoplasm of the cell. During glycolysis, a molecule of glucose, a six-carbon sugar, is broken down into two molecules of pyruvate, each containing three carbon atoms.

Q16. Movement of ions and large molecules with the help of protein molecules in and out of the cell is called _

• A. Diffusion
• B. Facilitated diffusion✓
• C. Passive transport
• D. Osmosis

Explanation: Diffusion is the passive movement of molecules down the concentration gradient whereas osmosis is the passive movement of water down the water potential gradient across a semi-permeable membrane. On the contrary, facilitated diffusion is the passive movement of molecules down the concentration gradient with the help of transport proteins (as these molecules/ions are incapable of moving through cell membranes directly).

Q17. Match the parts of the human brain listed under Column I with the functions given under Column II. Choose the answer which gives the correct combination of alphabets of the two columns.

• A. A=r, B=q, C=p, D=s
• B. A=r, B s, C=q, D=t,
• C. A=t, B=p, C=q, D=r✓
• D. A=t, B=q, C=p, D=s

Explanation: Your cerebral hemisphere plays a key role in memory, intelligence, thinking, learning, reasoning, problem-solving, emotions, consciousness, and functions related to your senses. (so A=T, hence options A and B are incorrect). The thalamus is your body's information relay station. All information from your body's senses (except smell) must be processed through your thalamus before being sent to your brain's cerebral cortex for interpretation (so B=P, hence options A, B, and D are incorrect). The cerebellum controls voluntary movements such as walking, posture, balance, coordination, eye movements, and speech (C=Q). Neurons in the medulla also control arousal and sleep (D=R).

Q18. The following statements are about enzymes: 1. They are globular proteins except ribo-enzyme.2. They can be inhibited3. They are formed in the smooth endoplasmic reticulum 4. There are only found attached to Plasma membranes in the cell Which statements are correct for all enzymes?

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

Explanation: Enzymes are a type of protein, and they have a three-dimensional, globular shape. This shape is crucial for their function, as it allows them to interact specifically with their substrates. Competitive inhibitors are molecules that closely resemble the substrate of an enzyme. They compete with the substrate for binding to the active site of the enzyme. When a competitive inhibitor binds to the active site, it prevents the substrate from binding and, as a result, inhibits the enzyme's normal catalytic activity.

Q19. About 55% of the volume of the human blood is?

• A. Plasma✓
• B. Blood proteins
• C. Blood cells
• D. Both B and C

Explanation: The correct answer is plasma, which makes up about 55% of the total blood volume. Plasma is a clear fluid consisting of 90% water and contains essential substances such as electrolytes, proteins, hormones, and nutrients. Blood proteins are an important part of plasma but do not constitute 55% of the blood volume on their own. Blood cells form the remaining 45% of the blood volume, encompassing red blood cells, white blood cells, and platelets, which are essential for various bodily functions but do not make up the majority of the blood volume.

Q20. Human cells maintain concentration gradients across their plasma membranes, such that there is a high sodium concentration outside the cell and a high potassium concentration inside the cell. Suppose that within the cell membrane are sodium leak channels or carrier proteins. These channels would allow sodium to:

• A. Move out of the cell by simple diffusion
• B. Move into the cell by simple diffusion
• C. Move out of the cell by facilitated diffusion
• D. Move into the cell by facilitated diffusion✓

Explanation: As the question mentions, there is a high concentration of sodium ions outside the cell, which means that through sodium leaky channels, the sodium ions should move inside the cell, not outside. The type of movement that occurs through channels or carrier proteins down the concentration gradient is called facilitated diffusion. That is why the answer is D.

Q21. Viruses resemble living things because they:

• A. Circulate
• B. Move
• C. Reproduce✓
• D. Are crystalline
• E. Are able to respond to stimuli in the environment

Explanation: Only thing that makes a group of scientists want the virus to be classified as a living thing is its ability to reproduce. All other options are incorrect as non-living things can also exhibit these characteristics

Q22. A/An _ is mostly a non-protein chemical compound that is required for the protein’s biological activity.

• A. Active site
• B. Substrate
• C. Cofactor✓
• D. Enzyme

Explanation: The correct answer is 'Cofactor'. A cofactor is a non-protein chemical compound required for the biological activity of some proteins, particularly enzymes. Cofactors can be metal ions or organic molecules and are essential for the catalytic activity of many enzymes.The other options are incorrect because:'Active site' refers to the region of the enzyme where the substrate binds, not a separate chemical compound.'Substrate' is the molecule upon which the enzyme acts, not a non-protein compound required for enzyme activity.'Enzyme' itself is a protein that facilitates biochemical reactions, whereas a cofactor is a non-protein component needed for some enzyme activities.

Q23. A group of immune cells that mediate the cellular immune response by processing and presenting antigens for recognition of certain other cells or immune systems is called _.

• A. Natural killer cells
• B. Interferons
• C. Antigen-presenting cells✓
• D. Vaccines

Explanation: Antigen-presenting cells (APCs) like macrophages and dendritic cells display antigens to T-cells, activating immune response.

Q24. A psychological condition usually seen in girls and young women, with loss of appetite, is:

• A. Obesity
• B. Malnutrition
• C. Anorexia Nervosa✓
• D. Dyspepsia
• E. Peptic ulcer

Explanation: Anorexia nervosa is a psychological disorder primarily affecting girls and young women, characterized by an intense fear of gaining weight and a distorted body image, leading to severe food intake restriction and loss of appetite. Unlike the other conditions listed, anorexia is specifically psychological and affects the demographic in question. Obesity is more about excessive weight gain, malnutrition is a broad term for nutrient deficiencies, dyspepsia is a digestive issue, and peptic ulcers are physical conditions of the digestive system.

Q25. The diagram represents the human respiratory system. While inhalation, which of the labelled muscles contract?

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

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

Q26. Which is the incorrectly paired one?

• A. Robert Hooke ... cell wall
• B. Schlelden and Schwann ... cell theory
• C. Robert Brown... nucleus
• D. Watson and Crick ... DNA model
• E. Virchow... Mosaic model of plasma membrane✓

Explanation: The Fluid Mosaic hypothesis was formulated by Singer and Nicolson in the early 1970s, not by Virchow. All others are correctly paired.

Q27. The disease caused by nutritional deficiency is:

• A. Hepatitis
• B. Cholera
• C. Malaria
• D. Beriberi✓
• E. Alzheimer's

Explanation: Beriberi is caused by the deficiency of Vitamin-B1, which is a thiamine.All other diseases mentioned are caused by pathogens.

Q28. Which diagram illustrates the process of active transport?

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

Explanation: Option D is correct because it illustrates active transport, where molecules move against their concentration gradient, requiring energy input, typically from ATP. The diagram likely shows molecules moving into a cell despite a higher concentration inside, indicative of active transport. Option A is incorrect as it shows passive movement, Option B is incorrect as it shows facilitated diffusion, and Option C shows osmosis, all of which do not require energy for the movement of molecules.

Q29. Consider the following statements regarding blood pressure:1. It is the pressure exerted by the blood on the walls of any vessel.2. It decreases in the arteries as the distance from the heart increases.3. It is lower in the capillaries than in the arteries.4. It is usually lower in women than in men.Choose the correct statements:

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

Explanation: All statements are correct: blood pressure varies by vessel type, location, and gender, and is highest near the heart.

Q30. It is the non-living component of the cell, it is secreted and maintained by the living portion of the cell and is found only in plants. What is this structure?

• A. Golgi apparatus
• B. Lysosome
• C. Mitochondrion
• D. Cell wall✓
• E. Ribosomes

Explanation: The correct answer is the cell wall, which is a non-living component unique to plant cells among the options provided. It is secreted and maintained by the living parts of the cell, such as the plasma membrane and the Golgi apparatus. The cell wall provides structure, protection, and supports the plant cell against mechanical stress and osmotic pressure. It is primarily composed of cellulose in plants.Other options such as the Golgi apparatus, lysosomes, mitochondria, and ribosomes are all living organelles involved in various cell functions and are found in both plant and animal cells. They do not serve the structural role that the cell wall does.

Q31. Which of the following are found in both arteries and capillaries?

• A. Connective tissues
• B. Elastic fibres
• C. Endothelial cells✓
• D. Smooth muscle cells

Explanation: Endothelial cells line the inner surface of all blood vessels, including arteries and capillaries, allowing smooth blood flow and exchange of substances.

Q32. A part of the digestive system that is not in contact with food is the:

• A. Small intestine
• B. Stomach
• C. Liver✓
• D. Large intestine
• E. Oesophagus

Explanation: The liver aids in digestion by producing bile, a fluid that emulsifies fats in the small intestine, breaking them down into smaller, more manageable droplets for absorption. While the liver doesn't "digest" food directly, it creates the bile necessary for fat digestion, which is then stored in the gallbladder before being released into the small intestine when food is present. All others are those parts of the digestive tract that come in contact with the food.

Q33. The cell A, in the given figure is:

• A. Gametophyte
• B. Spermatogonium✓
• C. Sperm
• D. Spermatocyte
• E. Spermatid

Explanation: Spermatogenesis is the process by which haploid spermatozoa develop from germ cells in the seminiferous tubules of the testis. During spermatocytogenesis primitive cells called spermatogonia proliferate by mitosis. The primary spermatocyte divides meiotically (Meiosis I) into two secondary spermatocytes; each secondary spermatocyte divides into two equal haploid spermatids by Meiosis II.

Q34. Formation of _ will be greater with the faster breakdown of glucose and glycogen to compensate for energy requirements in anaerobic respiration.

• A. Enzymes
• B. Hormones
• C. Lactic acid✓
• D. Fat

Explanation: In anaerobic respiration, when oxygen is scarce, glucose and glycogen are broken down to meet energy needs, resulting in lactic acid as a byproduct. This process does not produce enzymes or hormones as byproducts. Fat metabolism mainly occurs in the presence of oxygen and is not involved in this type of respiration.

Q35. What does 'a' in the following picture show?

• A. Cellular pathway
• B. Symplast pathway✓
• C. Apoplast pathway
• D. Water pathway

Explanation: Movement from cell to cell through plasmodesmata is a symplast pathway.Movement through the cell walls to the cell walls of other cells is called an apoplast pathway.

Q36. The function of a cell wall in prokaryotes is:

• A. To give cells rigidity
• B. To give specific shape
• C. To protect from osmotic lysis
• D. All of Above✓

Explanation: All prokaryotic cells have a stiff cell wall located underneath the capsule (if there is one). This structure maintains the 'cell's shape', protects the cell interior, and prevents the cell from bursting when it takes up water. Since the bacterial cell wall is 'thick and rigid', it will maintain the shape of the cell and also 'protect it from osmotic lysis' just like the cell wall of plants. Hence, the correct option is 'D'.

Q37. In 1959, Koshland proposed which of the following?

• A. Unit membrane model
• B. Fluid mosaic model
• C. Reflective index model
• D. Induced fit model✓

Explanation: In 1959, Daniel E. Koshland, an American biochemist, proposed the "induced fit" model of enzyme-substrate interaction. This model is an extension of the earlier "lock and key" model proposed by Emil Fischer. The induced fit model suggests that the active site of an enzyme is flexible and can undergo conformational changes upon binding to a substrate. In other words, the enzyme is not a rigid structure with a pre-formed active site (as implied by the lock and key model), but rather, the binding of the substrate induces changes in the enzyme's structure to achieve optimal interaction.

Q38. The diagram shows some organs of the digestive system. Where is amylase present?

• A. 1,4 and 5✓
• B. 1,2 and 3
• C. 2,6 and 4
• D. 3,5 and 4

Explanation: Amylase is mostly present in the mouth, in pancreatic juice, and ileum; hence, option A is correct. Amylase is not present in the liver, no enzyme is present in the large intestine and only pepsin (protease) is present in the stomach. That is why, all other options B, C and D are wrong.

Q39. Which of the following statement(s) is/are correct regarding a nucleus? I.Stores wastes and other SubstancesII.Contains genetic materialIII.Helps in cellular transport systemIV.Control centre of the cell

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

Explanation: Nucleus is the control centre of the cell and stores the genetic material of the cell,so we say that statements II and IV are correct, making option D correct. Nucleus does not store wastes and other substances plus it also does not help in cellular transport system, that eliminate option A,B,C and E.

Q40. Which of the following nutrients is incorrectly paired with its function in a plant?

• A. Potassium - Formation of cell wall.✓
• B. Magnesium - Constituent of chlorophyll.
• C. Nitrogen - Constituent of protein.
• D. Phosphorus - Component of nucleic acid.

Explanation: The plant cell wall is made of Cellulose.Potassium creates the cell membrane potential which allows the cell to generate an action potential.Magnesium in plants is located in the enzymes, in the heart of the chlorophyll molecule.A phosphate backbone is the portion of the DNA double helix that provides structural support to the molecule.

Q41. The disease, which is caused by a defect in a single gene or pair of genes is referred to as:

• A. Unifactorial✓
• B. Multifactorial
• C. Down syndrome
• D. Turner syndrome

Explanation: The disease, which is caused by a defect in a single gene or pair of genes is referred to as a unifactorial gene.

Q42. Which organ(s) in the male reproductive system produce gametes?

• A. Vas deferens
• B. Malpighian tubule system
• C. Seminiferous tubules✓
• D. Ejaculatory duct
• E. Testicular Duct System

Explanation: In mammals production of testes occurs in the seminiferous tubules of the male testes.The male genitals include the testes, the ductus deferens, the seminal vesicles, the ejaculatory ducts, and the penis, together with the following accessory structures: the prostate and the bulbourethral glands.The testes are two glandular organs, which secrete the semen; they are suspended in the scrotum by the spermatic cords.Testosterone is the principal hormone in a group of hormones called androgens. In men, testosterone is produced by the testes, in women, testosterone is primarily secreted by the ovaries.It is derived from cholesterol (like all the sex hormones) and its immediate precursor is DHEA.

Q43. Menstruation, the shedding of the lining of the uterus (endometrium), is accompanied by:

• A. Shivering
• B. Fever
• C. Bleeding✓
• D. Vomiting
• E. Headache

Explanation: Menstruation is the shedding of the endometrium, which results in bleeding from the uterus. It is a normal physiological process that occurs in females of reproductive age. It is not usually accompanied by shivering, fever, vomiting or headache.

Q44. In addition to smaller hind limb muscle mass, the mutant "mini Muscle” gene exhibits lower heart rates during physical activity, and larger kidneys and livers in mice. This is a very good example of _.

• A. Epistasis
• B. Multiple alleles
• C. Codominance
• D. Pleiotropy✓

Explanation: Pleiotropy is a genetic situation where one gene impacts multiple, seemingly unrelated phenotypic traits. The example of the 'mini Muscle' gene impacting muscle mass, heart rate, and organ sizes in mice illustrates this principle. The other options are incorrect because they describe different genetic phenomena: Epistasis involves gene interactions, Multiple alleles involve more than two gene variants, and Codominance involves equal expression of two alleles rather than a single gene affecting multiple traits.

Q45. What does the following picture show?

• A. Motor Neuron
• B. Sensory Neuron✓
• C. Interneuron
• D. Nerve

Explanation: The diagram illustrates a sensory neuron. The defining feature is the cell body, which connects to the dendrites via dendron processes. This structure is specific to sensory neurons and is not found in motor neurons, where the cell body directly attaches to dendrites. Interneurons are characterized by processes that are similar in size to the cell body, which is not evident here. Lastly, the term 'nerve' refers to a collection of neurons, whereas the image shows only one neuron.

Q46. The synapse formed at the sites where the terminal branches of the axon of a motor neuron conduct to a target muscle cell is called:

• A. Sensory end plate
• B. Synapse end plate
• C. Motor end plate✓
• D. Postsynaptic membrane

Explanation: The motor end plate is the specialized area at the neuromuscular junction where the axon terminal of a motor neuron forms a synapse with a muscle fiber. It is responsible for transmitting neural signals to muscle cells, resulting in muscle contraction. The term sensory end plate is associated with sensory neurons, not motor neurons. Synapse end plate is not a recognized term in neuroscience. The postsynaptic membrane refers generically to the membrane receiving neurotransmitters at any synapse, not specifically at neuromuscular junctions.

Q47. Which of the following is a uni-nucleated cell?

• A. Skeletal muscle
• B. Cardiac muscle
• C. Smooth muscle✓
• D. Both B and C

Explanation: Smooth muscles are uni-nucleated and non-striated. It is composed of small spindle-shaped cells with a single, central nucleus. They have unbranched cells that are controlled by the autonomic nervous system. The function of smooth muscle is to contract like any muscle tissue.

Q48. Why are skeletal muscles called striated muscles?

• A. Appear darker than smooth muscles by naked eye
• B. Alternating dark and light bands appear on their surface when visualized by naked eye
• C. Alternating dark and light bands appear on their surface when visualized via a microscope✓
• D. All of these

Explanation: The striated appearance of skeletal muscle tissue is a result of repeating bands of the proteins actin and myosin that are present along the length of myofibrils. Dark A bands and light I bands repeat along myofibrils, and the alignment of myofibrils in the cell causes the entire cell to appear striated or banded.

Q49. The smallest contractile unit of muscle contraction, called a sarcomere, is the area between two?

• A. H zone
• B. M line
• C. Z line✓
• D. Z zone

Explanation: The correct answer is the Z line. The sarcomere is the basic contractile unit of muscle fibers and is defined by the area between two Z lines. These lines serve as anchor points for actin filaments and help maintain the structure during contraction. The H zone and M line are internal components of the sarcomere that do not define its boundaries. The term 'Z zone' is not used in muscle anatomy; thus, it is not a correct option.

Q50. In the sliding filament theory of muscle contraction, which option correctly describes function of ATP?

• A. ATP does all of these things during muscle contraction
• B. It allows the myosin head to detach from the actin filament✓
• C. It moves tropomyosin off of actin binding sites
• D. both A and B

Explanation: What is the role of ATP in sliding filament theory?Energy from ATP is required for the myosin head to release from the actin filament—otherwise the myosin heads would remain in the same place, and the muscle would not contract. Even though the filaments are moving, the filaments themselves never actually get shorter or longer.Sliding filament theory:• During muscle contraction, thin filaments show sliding inward towards the H-zone.• Sarcomere shortens, without changing the length of thin and thick myofilaments.• The cross bridge of the thick myofilaments connects with the portions of actin of the thin myofilaments. These cross bridge move on the surface of the thin myofilaments resulting to sliding of thin and thick myofilaments over each other.• A muscle fibre maintains a resting potential under resting conditions just like a nerve fibre as soon as nerve impulses reach the terminal end of the axon, small sacs called synaptic vesicles to fuse with the axon membrane and release a chemical transmitter, acetylcholine. • Acetylcholine diffuses across the synaptic cleft and binds to the receptor sites of the motor and plate.• As soon as depolarization of the motor end plate reaches and plate reaches a certain level, it creates an action potential. After this, an enzyme cholinesterase present along with the receptor sites for acetylcholine breaks down acetylcholine into acetate and choline. • The area of contact between a nerve and muscle fibre is called motor end plate or muscular junction.• Calcium plays a key regulatory role in muscle contraction. These ions bind to troponin causing a change in its shape and position. This, in turn, alters and the position of tropomyosin. This shift exposes the active sites on the F-actin and myosin cross-bridges are then able to bind these active sites. • The head of each myosin molecule contains an enzyme myosin ATPase. In the presence of myosin ATPase, calcium ions and Magnesium ions, ATP breaks down into ADP and inorganic phosphate and some amount of energy.• Energy from ATP causes energized myosin, cross bridges to bind to actin. The energized cross bridge move causing thin myofilaments to slide along the thick myofilaments.

Q51. Cross bridges are found on:

• A. Actin
• B. Myosin✓
• C. Troponin
• D. Tropomyosin

Explanation: Cross-bridges are formed by the myosin heads, which extend from the thick filaments (myosin) and interact with the thin filaments (actin) during muscle contraction. These cross-bridges are crucial for the sliding filament mechanism, where the myosin heads attach to binding sites on actin, undergo a power stroke, and pull the actin filaments closer together.

Q52. A man who has type AB blood cannot father a child with type _ blood, because he would pass on either the _ or the B allele to all of his offspring.

• A. A,O
• B. O,A✓
• C. B, O
• D. B, A

Explanation: The father can pass on either the A or B allele to each of his offspring. Even if the mother has blood type O, the child cannot have blood type O because of the A or B allele received from the father. A and B alleles are dominant over O. For blood type O (the recessive phenotype) to occur, there should be two copies of the O allele, one from each parent. This is impossible in the given situation.

Q53. A cross between a black cat and a tan cat produces a tabby pattern (black and tan fur together). What percent of kittens would have tan fur if a tabby cat is crossed with a black cat?

• A. 100%
• B. 50%
• C. 25%
• D. 0%✓

Explanation: The correct answer is 0%. The black cat is homozygous for the black fur gene (BB) and does not carry an allele for tan fur. The tabby cat, being a hybrid, has one black fur allele (B) and one tan fur allele (T). When crossed with a black cat, the potential offspring genotypes are BB (black) and BT (tabby), with no possibility of a TT (tan) genotype. Thus, none of the kittens will have tan fur. The other options are incorrect because they suggest the presence of a tan allele in the black parent, which is not possible given the genetic information provided.

Q54. Which of the following is not an important function of bone?

• A. Regulation of ion concentration
• B. Muscular contraction
• C. Organ and nerve protection
• D. Regulation of pH through hydration✓

Explanation: Bones help buffer blood pH by absorbing or releasing alkaline salts, but hydration is not the mechanism involved in this regulation. This phrase is misleading and not considered a valid function of bone.

Q55. Haemophilia is a sex-linked _ trait.

• A. Dominant
• B. Co-dominant
• C. Pleiotropic
• D. Recessive✓

Explanation: Haemophilia is a sex-linked recessive disorder. The abnormal gene responsible for haemophilia is carried on the X chromosome. Recessiveness is proved by the fact that when a woman has this gene on one X and a normal gene on the other X (heterozygous for hemophilia), she does not show haemophilia in phenotype.

Q56. The "d" and "D" alleles are used for lighter and darker skin colour in humans, respectively. By keeping in view the inheritance pattern of skin colour in human beings(controlled by 3 genes), choose which combination shows medium skin colour from the following picture:

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

Explanation: The correct answer is Column C because it presents a balanced number of dominant (D) and recessive (d) alleles across all three genes, which results in a medium skin color. In contrast, Column A is incorrect as it shows homozygous recessive alleles (ddd) for all genes, leading to the lightest possible skin color. Column D is incorrect because it fails to show a balance between dominant and recessive alleles, skewing towards either lighter or darker skin. Column E is incorrect as it does not reflect the polygenic nature of skin color inheritance, which involves multiple genes working together, not just a single gene or row.

Q57. Monosynaptic reflex arc consists of:

• A. One sensory neuron only
• B. One motor neuron only
• C. Two neurons, One sensory neuron, and one motor neuron✓
• D. None of these

Explanation: The correct option is C. When a reflex arc consists of only two neurons, one sensory neuron and one motor neuron, it is defined as monosynaptic. It allows for direct communication between sensory and motor neurons, resulting in muscle stimulation. A common example of a monosynaptic reflex is the knee-jerk reflex.

Q58. The type of plastids found in roots of plants _ .

• A. Chloroplasts
• B. Chromoplasts
• C. Leucoplasts✓
• D. All of them

Explanation: The correct answer is Leucoplasts. These are non-pigmented plastids found in non-photosynthetic tissues such as roots, where their primary role is to store substances like starch, oils, and proteins. Chloroplasts are found in green tissues and are involved in photosynthesis, making them unsuitable for roots. Chromoplasts are responsible for the colors in flowers and fruits due to their pigments. Therefore, neither chloroplasts nor chromoplasts are typically found in roots. The option 'All of them' is incorrect because not all plastids are present in roots; only leucoplasts are typically found there.

Q59. Outbreeding increases which of the following?

• A. Homozygosity
• B. Heterozygosity✓
• C. Gene linkage
• D. Gene pool

Explanation: The correct answer is option 'B', 'heterozygosity'. Outbreeding increases heterozygosity by encouraging the mixing of different alleles, which enhances genetic variation and can improve the fitness of a population. This increased variation allows populations to better adapt to environmental changes and reduce the likelihood of expressing deleterious recessive alleles. Option 'A', 'homozygosity', is incorrect because outbreeding typically reduces homozygosity. Option 'C', 'gene linkage', is unaffected by outbreeding as it is related to the proximity of genes on a chromosome. Option 'D', 'gene pool', refers to the collection of genes within a population, and while outbreeding can influence it, the direct effect of outbreeding is an increase in heterozygosity.

Q60. Homology means?

• A. Similarity in characteristics due to shared ancestry✓
• B. Similarity in function due to acquired characteristics
• C. Study of similar structures regardless of origin
• D. Structures with different functions but similar appearances

Explanation: Homology is a fundamental concept in evolutionary biology that refers to similarities in characteristics due to shared ancestry. For instance, the forelimbs of humans, bats, and whales are homologous because they evolved from a common ancestor, despite having different current functions.Option A is correct because it accurately describes homology as stemming from shared ancestry. Option B is incorrect; it confuses acquired traits with inherited ones, which are central to homology. Option C is incorrect because it disregards the crucial element of common ancestry. Option D is incorrect because it describes analogy, where similar functions evolve independently, not from a shared ancestor.

Q61. Biological molecules which catalyze a biochemical reaction and remain unchanged after completion of reaction are called?

• A. Cofactor
• B. Coenzymes
• C. Activator
• D. Enzymes✓

Explanation: Option D is correct as "enzymes" catalyse reactions and don't take part in the response; therefore, they do not change.

Q62. Out of 31 pairs of spinal nerves, how many pairs of thoracic nerves are there?

• A. 8
• B. 10
• C. 12✓
• D. 15

Explanation: In total, there are 31 pairs of spinal nerves grouped regionally by spinal region. More specifically, there are eight cervical nerve pairs (C1-C8), twelve thoracic nerve pairs (T1-T12), five lumbar nerve pairs (L1-L5), 5 sacral (S1-S5), and a single coccygeal nerve pair.

Q63. What affect do enzymes have on the activation energy of a reaction?

• A. Increases
• B. Decreases✓
• C. No affect
• D. Increases or decreases depending upon individual enzyme

Explanation: b) Decreases:Option (b) "Decreases" is the correct answer. Enzymes significantly decrease the activation energy of a reaction. By providing an alternative reaction pathway with a lower energy barrier, enzymes allow reactant molecules to achieve the transition state more easily and, as a result, the reaction rate is enhanced.

Q64. Virus attachment requires energy?

• A. Yes
• B. No✓
• C. Provided by cell usually
• D. Provided by virus machinery

Explanation: The attachment of a virus to a host cell does not require energy. The process of virus attachment is typically mediated by specific viral surface proteins, which bind to specific receptor molecules on the surface of the host cell. This binding process is typically a highly specific and spontaneous interaction, and it is often the first step in the process of viral infection. Therefore, the attachment of a virus to a host cell does not generally require energy.

Q65. 'Biomolecules' such as starch, cellulose and glycogen yield which of the following upon hydrolysis?

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

Explanation: Option C is correct since the three most abundant polysaccharides, starch, glycogen, and cellulose ( referred to as homopolymers) yield only one type of monosaccharide i.e. 'glucose' after complete hydrolysis.

Q66. Haemoglobin in men increases the oxygen carrying capacity of the blood to about _.

• A. 50 times
• B. 65 times
• C. 70 times
• D. 75 times✓

Explanation: Option D is correct since 'Haemoglobin' is the most important protein present in many animals including man as it increases the oxygen-carrying capacity of blood to about 75 times.Myoglobin has naturally more afiinity for oxygen and is found in muscles.

Q67. Which one is not a monosaccharide?

• A. Galactose
• B. Fructose
• C. Mannose
• D. Lactose✓

Explanation: Lactose is not a monosaccharide but a disaccharide,as shown below:

Q68. The first human hormone that was produced by recombinant DNA technology was:

• A. Insulin✓
• B. Estrogen
• C. Thyroxin
• D. Progesterone

Explanation: The recombinant DNA technological processes have made a great impact in the area of health care by mass production of safe and more effective therapeutic drugs. In 1983, Eli Lily, an American company, first prepared two DNA sequences corresponding to A and B chains of human insulin and introduced them in plasmids of Escherichia coli to produce insulin chains. Chains A and B were produced separately, extracted, and combined by creating disulfide bonds to form human insulin (humulin).

Q69. Pierrie Curie and Marie Curie isolated which radioactive element?

• A. Lithium
• B. Radium✓
• C. Boron
• D. Uranium

Explanation: The correct answer is Radium. Marie and Pierre Curie successfully isolated radium from pitchblende in 1902, contributing greatly to the understanding of radioactivity. Lithium and boron are non-radioactive elements, and uranium, while radioactive, was discovered earlier by Martin Klaproth, not by the Curies.

Q70. Removal of proximal convoluted tubule from the nephron will result in:

• A. No change in quality and quantity of urine
• B. No urine formation
• C. More diluted urine✓
• D. More concentrated urine

Explanation: This is correct because the PCT is responsible for reabsorbing a large amount of water and solutes like glucose, amino acids, and sodium. If it is removed, less water and solutes will be reabsorbed, leading to increased water content in the urine, making it more diluted.

Q71. Which of the following molecules have zero Dipole Moment?

• A. CCl4
• B. CO2
• C. Cl2
• D. C6H6
• E. All of the above✓

Explanation: All the molecules listed have a zero dipole moment due to their symmetrical structures:CCl4: Although it has polar C-Cl bonds, the tetrahedral symmetry results in a net dipole moment of zero.CO2: The linear shape with two opposing C=O bonds results in the cancellation of dipole moments.Cl2: As it consists of identical atoms, no dipole moment arises.C6H6 (Benzene): The planar symmetrical structure leads to the cancellation of dipole moments of C-H bonds.Therefore, the correct answer is 'All of the above'.

Q72. Plants absorb it in the form of soluble phosphates. It is present abundantly in the growing and storage organs of plants. What is it?

• A. H2O
• B. CO2
• C. K
• D. P✓
• E. N

Explanation: Plants absorb phosphorus in the form of soluble phosphates, and it is stored abundantly in growing and storage organs like seeds and fruits, making it the correct answer.

Q73. The Taq polymerase enzyme is obtained from:

• A. Bacillus subtilis
• B. Pseudomonas putida
• C. Thermus aquaticus✓
• D. Thiobacillus ferroxidase.

Explanation: Taq polymerase, generally used in PCR, is isolated from thermophilic bacterium Thermus aquaticus.

Q74. What will be the energy of an X-ray quantum of wave length 1.0 x 10-10 m?

• A. 0.6 x 10^-15 J
• B. 1.98 x 10^-15 J✓
• C. 3.7 x 10^-15 J
• D. 4.7 x 10^-15 J
• E. 7.7 x 10^-15 J

Explanation: So, the energy of x rays is 1.98 x 10 -15 J. Hence, this is the required solution.

Q75. If 92 U 235, decays by emitting two α one β and two γ -rays the new daughter element is = _.

• A. 88 Y 227
• B. 89 Y 227✓
• C. 90 Y 227
• D. 89 Y 231

Explanation: When 92U235 undergoes two alpha decays, it loses 4 protons and 8 neutrons, reducing the atomic number to 88 and the mass number to 227. The emission of a beta particle converts a neutron to a proton, increasing the atomic number by 1 to 89 without affecting the mass number. Gamma rays have no effect on the atomic or mass numbers as they are energy emissions without mass or charge. Thus, the resulting daughter element is 89Y227.Option A is incorrect as it miscalculates the decay process. Option C assumes no beta decay effect. Option D incorrectly calculates the mass number post-decay.

Q76. Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because:

• A. There is a negative pressure in the lungs
• B. There is a negative intarpleural pressure pulling at the lung walls✓
• C. There is a positive intrapleural pressure
• D. Pressure in the lungs is higher than the atmospheric pressure.

Explanation: Negative intrapleural pressure helps keep the lungs expanded even after forceful expiration.

Q77. What forms the basis of DNA fingerprinting?

• A. The relative proportions of purines and pyrimidines in DNA
• B. The relative difference in the DNA occurrence in blood, skin, and saliva
• C. The relative amount of DNA in the ridges and grooves of the fingerprints
• D. Satellite DNA occurring as highly repeated, short DNA segments✓

Explanation: DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. The difference of about 0.1% or 3 × 106 base pairs (out of 3 × 10 9 bp) provides individuality to each human being. The human genome possesses numerous small noncoding but inheritable sequences of bases that are repeated many times. These sequences occur near telomere, centromeres, Y chromosome, and heterochromatic area. The area with the same sequence of bases repeated several times is called repetitive DNA. It is separated as a satellite from the bulk DNA during density gradient centrifugation and hence called satellite DNA where repetition of the bases is in tandem. Satellite DNAs show polymorphism (the occurrence of mutations in a population at high frequency), which is the basis of genetic mapping of human genome as well as DNA fingerprinting. While mutations in genes produce alleles with different expressions, mutations in noncoding repetitive DNA have no immediate impact. These mutations which have piled up with time form the basis of polymorphism.

Q78. The torque will be greater if:

• A. Both magnitude of force and moment arm are smaller
• B. Both magnitude of force and moment arm are greater✓
• C. Only magnitude of force is greater
• D. Only moment arm is greater
• E. None of the above

Explanation: The formula for torque is T=Fr. Hence Torque will be greater if the magnitude of the force and moment arm are greater.

Q79. The translocation of organic solutes in sieve-tube members is supported by:

• A. Cytoplasmic streaming
• B. Root pressure and transpiration pull
• C. P-proteins
• D. Mass flow involving a carrier and ATP✓

Explanation: According to the mass flow hypothesis, the transport of organic solutes takes place from source to sink, this transport also depends on metabolic energy. According to the cytoplasmic streaming hypothesis (which was given by de Vries, in 1885) the transport of organic solutes takes place by a combination of diffusion and cytoplasmic streaming. Cytoplasmic streaming carries organic solutes from one end to the other end of the sieve tube. P-proteins have a role as a defense against phloem-feeding insects and the sealing of damaged sieve tubes. So, the correct answer is 'Mass flow involving a carrier and ATP'. So correct option is C.

Q80. Guttation is the result of:

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

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

Q81. The first clinical gene therapy was given for treating:

• A. Diabetes mellitus
• B. Chicken pox
• C. Rheumatoid arthritis
• D. Adenosine deaminase deficiency.✓

Explanation: Gene therapy is a collection of methods that allows the correction of a gene defect that has been diagnosed in a child/embryo. Here genes are inserted into a person’s cells and tissues to treat a disease. Correction of a genetic defect involves the delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non­functional gene. The first clinical gene therapy was given in 1990 to a 4- year old girl with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function. SCID is caused due to defect in the gene for the enzyme adenosine deaminase. In some children, ADA deficiency can be cured by bone marrow transplantation. Here, the isolated gene from bone marrow cells producing ADA is introduced into cells at early embryonic stages; it can be a permanent cure.

Q82. The basic functional unit of human kidney is

• A. Nephridia
• B. Loop of Henle
• C. Pyramid
• D. Nephron✓

Explanation: A nephron is a unit of structure and function in a kidney. A kidney contains about a million nephrons, each approximately 3 cm long. A nephron is a long tubule differentiated into four regions having different anatomical features and physiological role : Bowman’s capsule, proximal convoluted tubule (PCT), loop of Henle, and distal convoluted tubule (DCT). The latter opens into one of the collecting ducts. Nephridia are the excretory organs of annelids.

Q83. If loop of Henle were absent from mammalian nephron which one of the following is to be expected?

• A. There will be no urine formation.
• B. There will be hardly any change in the quality and quantity of urine formed
• C. The urine will be more concentrated
• D. The urine will be more dilute✓

Explanation: Reabsorption is a process by which useful constituents of glomerular filtrate are returned into the blood streams. It occurs in convoluted tubules (proximal convoluted tubule) as well as loop of Henle. Basically loop of Henle, in association with vasa rectae, plays an important role in the counter current mechanism (the process which makes urine hypertonic, i.e., more concentrated).Therefore, if Henle’s loop was absent from mammalian nephron the urine will be more dilute.

Q84. The haemoglobin of a human foetus

• A. The hemoglobin of a human fetus has only 2 protein subunits instead of 4
• B. The hemoglobin of a human fetus has a higher affinity for oxygen than that of an adult✓
• C. The hemoglobin of a human fetus has a lower affinity for oxygen than that of an adult
• D. Its affinity for oxygen is the same as that of an adult

Explanation: A. The hemoglobin of a human fetus has only 2 protein subunits instead of 4: This statement is not accurate. Hemoglobin, in both fetuses and adults, consists of four protein subunits. The structure of hemoglobin remains consistent in this regard.B. The hemoglobin of a human fetus has a higher affinity for oxygen than that of an adult:This statement is accurate. Fetal hemoglobin (HbF) has a higher affinity for oxygen than adult hemoglobin (HbA). This higher affinity allows fetal hemoglobin to effectively extract oxygen from the mother's bloodstream, where the oxygen concentration is lower, and deliver it to the developing fetus.C. The hemoglobin of a human fetus has a lower affinity for oxygen than that of an adult:This statement is not accurate. As mentioned earlier, fetal hemoglobin actually has a higher affinity for oxygen compared to adult hemoglobin.D. Its affinity for oxygen is the same as that of an adult:This statement is not accurate. Fetal hemoglobin (HbF) and adult hemoglobin (HbA) have different oxygen affinities, with HbF having a higher affinity for oxygen.Correct Option: B. The hemoglobin of a human fetus has a higher affinity for oxygen than that of an adult.Summary: The correct option is B, which accurately states that fetal hemoglobin (HbF) has a higher affinity for oxygen compared to adult hemoglobin (HbA). This difference in oxygen affinity helps facilitate the efficient transfer of oxygen from the mother's bloodstream to the developing fetus during pregnancy. The other options either present inaccurate information or claim that the affinity for oxygen is the same in fetal and adult hemoglobin.

Q85. Given the reaction: C3H8 + 5O2 → 3CO2 + 4H2O At STP , how many litres of O2 are needed to completely burn 5.0 litres of C3H8?

• A. 5
• B. 10
• C. 10.5
• D. 15
• E. 25✓

Explanation: Consider that no. of moles of reactants = volume of the reactants.As the molar ratio between C3H6 and O2 is 1:5, the volume ratio is also the same. For 1L of C3H6 we need 5L of O2. Since the volume of C3H6 is 5L now, to burn this, we require 25L of O2.All other options are incorrect because they do not follow the exact mole-volume relationship required in this case.

Q86. What is the product of both fermentation reactions and fractional distillation?

• A. An ester
• B. An acid
• C. An alcohol✓
• D. A soap
• E. As base

Explanation: Ethyl alcohol can be obtained by fermentation of starch, fermentation of molasses and fractional distillation of pyroligneous acid (obtained through destructive distillation of wood) gives Methyl alcohol.

Q87. For the reaction: N2 + 3H2 ⇌ 2NH3 .The production of NH3 will be favored at:

• A. High pressure and catalyst✓
• B. Low pressure only
• C. Low pressure and catalyst
• D. High pressure only
• E. Catalyst only

Explanation: The forward reaction forms fewer moles of gas (4 → 2), so high pressure shifts equilibrium toward NH₃. A catalyst increases the reaction rate but does not change equilibrium position. Therefore, maximum ammonia production is achieved using high pressure along with a catalyst.

Q88. What volume in dm3 of KCl is obtained in the following equation if the volume of KCIO3 at the beginning of the reaction is 20 dm3? 2KClO3 → 2KCl + 3O2

• A. 2 dm3
• B. 74.5 dm3
• C. 50 dm3
• D. 40 dm3
• E. 20 dm3✓

Explanation: To determine the volume of KCl obtained in the given equation, we need to know the stoichiometric ratio between KCIO3 and KCl. From the balanced equation: 2KClO3 → 2KCl + 3O2 We can see that for every 2 moles of KCIO3, we obtain 2 moles of KCl. Therefore, the ratio is 1:1. Given that the volume of KCIO3 at the beginning of the reaction is 20 dm3, and the stoichiometric ratio is 1:1, the volume of KCl obtained will also be 20 dm3. Therefore, the volume of KCl obtained in the given equation is 20 dm3.

Q89. Amylopectin has _ linkage.

• A. α 1 - 4 glycosidic
• B. β 1 - 4 and 1 - 6 glycosidic
• C. α 1-4 and α 1 - 6 glycosidic✓
• D. β 1 - 4 glycosidic

Explanation: Amylopectin has both α 1-4 and 1 - 6 glycosidic bond linkages.

Q90. The formula for phitkari is?

• A. K2SO4.Al2(SO4)3.24H2O✓
• B. K2SO4.Cr2(SO4)3.24H2O
• C. (NH4)2SO4.Al2(SO4)3.24H2O
• D. (NH4)2SO4.Cr2(SO4)3.24H2O

Explanation: The correct formula for phitkari, also known as Alum, is K2SO4.Al2(SO4)3.24H2O. This form of alum is often used in water purification, as a mordant in dyeing, and in various cosmetic applications. The other options represent different types of alum or similar double salts, but they are not typically referred to as phitkari.

Q91. IUPAC name of the compound (CH3)2-CH-CH(C2H5)-C(CH3)3 is :

• A. 3-ethyl-2,2,4-trimethylpentane✓
• B. 3-ethyl-2,2,4-trimethylpentene
• C. 3-ethyl-2,4-dimethylpentane
• D. 4-ethyl-3-methylhexane
• E. 3-ethyl-2,2-dimethylpentane

Explanation: The correct IUPAC name for the compound is 3-ethyl-2,2,4-trimethylpentane. The numbering of the carbon chain begins from the end closest to the first substituent, ensuring the longest chain includes five carbon atoms (pentane). The substituents are three methyl groups (at carbons 2, 2, and 4) and one ethyl group (at carbon 3). This configuration is described by '3-ethyl-2,2,4-trimethylpentane'.Other options are incorrect because they either suggest the presence of a double bond ('pentene'), a miscount of the longest carbon chain ('hexane'), or an incorrect number of substituents at specified positions, failing to match the given molecular structure.

Q92. (HCHO)nH20 is the chemical formula for?

• A. Formic acid
• B. Paraformaldehyde✓
• C. α-methyl propionaldehyde
• D. Propionic acid

Explanation: The compound (HCHO)nH₂O is known as paraformaldehyde. It is a polymer of formaldehyde, formed when multiple formaldehyde molecules undergo polymerization. This results in a series of repeating units, represented by (HCHO)n, with water (H₂O) being part of its stabilization in the polymer form. Options A, C, and D are incorrect because their chemical formulas do not match the structure of paraformaldehyde. Formic acid (Option A) is a simple carboxylic acid, α-methyl propionaldehyde (Option C) is an aldehyde with a different structure, and propionic acid (Option D) is another carboxylic acid.

Q93. A dipolar charged but overall electrically neutral ion is called:

• A. Zwitterion✓
• B. Unit electron
• C. Cation
• D. Anion

Explanation: A zwitterion is a special type of ion that has both positive and negative charges in different parts of the molecule. This allows it to maintain an overall neutral charge, which is why it fits the description of a dipolar charged but overall electrically neutral ion. Amino acids often exist as zwitterions in biological systems. On the other hand, a unit electron is simply a single elementary charge with a negative value, a cation is positively charged, and an anion is negatively charged. None of these options have both charges simultaneously, so they cannot be considered zwitterions.

Q94. Reaction of Aldehydes with the Tollen's reagent results in the formation of _?

• A. Red precipitate
• B. Yellow precipitate
• C. Silver mirror✓
• D. Brick red precipitate

Explanation: Tollen’s reagent is an alkaline solution of ammoniacal silver nitrate; the reduction of silver ions in the Tollen’s reagent results in a metallic silver.

Q95. Oxidation Number of all the elements in the free state is?

• A. Unity
• B. Positive
• C. Zero✓
• D. Negative

Explanation: The oxidation number of an element in its free state is always zero.

Q96. The strength of London forces depends on the size of:

• A. Electrons.
• B. Electronic cloud.✓
• C. Lone pair on an atom
• D. Poles of atoms

Explanation: The strength of London forces depends upon the size of the electronic cloud of the atom or molecules. When the size of the atom or molecule is big then the dispersion becomes easy and these forces end up being more prominent.

Q97. The number of electrons present in n = 2, l = 1 and m = -1, 0, +1 are:

• A. 6✓
• B. 2
• C. 8
• D. 10
• E. 18

Explanation: n is the principal quantum number and indicates the number of shells. n = 2 indicates that the second shell is being discussed. l is the azimuthal quantum number and indicates the number of subshells. l = 0 means s subshell, l = 1 means p subshell, and so on. m is the magnetic quantum number and indicates the orientation of orbitals in the subshells. m ranges from -1 to +1 through 0. Since m = -1, 0, +1, this means that all three sub-orbitals; 2px, 2py, 2pz are taken into consideration. Since all of these have 2 electrons each, the total number of electrons in them is 6.

Q98. If the difference between electronegativity values between two atoms is less than 1.7 , the bond is necessarily:

• A. Ionic
• B. Covalent✓
• C. Electrovalent
• D. Polar
• E. Non-polar

Explanation: If the electronegative difference is >1.7 then its nature is ionic and <1.7 then its nature is covalent and if its equal to 1.7 then nature of bond will be 50% ionic and 50% covalent. As here, it is less than 1.7, nature is covalent.

Q99. Which one of the following has the lowest % ionic character?

• A. CCl4✓
• B. BCl3
• C. BeCl2
• D. LiCl
• E. HCl

Explanation: Electronegativity increases as we go from left to right in a period. So, the electronegativity difference between elements and chlorine decreases. Ionic character decreases with a decrease in the value of electronegativity difference between elements. thus the correct order is: CCl4<BCl3<BeCl2<LiCl.

Q100. 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.

Q101. Matter changes from one state to another state on change in _.

• A. Density
• B. Pressure✓
• C. Viscosity
• D. All

Explanation: The state of matter changes primarily due to variations in temperature and pressure. For instance, when solids are heated, they can melt into liquids, and liquids can vaporize into gases with further heating. Conversely, increasing pressure can compress gases into liquids and solids by reducing the space between particles. Therefore, pressure is a key factor in changing states of matter. Density and viscosity are properties of matter but they do not directly change its state.

Q102. The bonding pair of electrons are equally shared between the atoms in?

• A. HF
• B. HCl
• C. H2O
• D. H2✓

Explanation: H2 forms a covalent bond with identical atoms, so the electrons in the bond must be shared equally.

Q103. The total of all the possible kind of energies in a system is called:

• A. Total energy
• B. Kinetic Energy
• C. Potential Energy
• D. Internal Energy✓

Explanation: Internal energy is defined as the total energy of a closed system.

Q104. Energy is absorbed?

• A. In bond breaking
• B. In endothermic reactions
• C. In bond formation
• D. Both Options A and B are correct✓

Explanation: Breaking bonds requires adding energy, the opposite process of forming new bonds always releases energy. Bond breaking is an endothermic process while bond formation is an exothermic process.

Q105. Which of the following has the maximum number of unpaired d electrons?

• A. Mg 2+
• B. Ti 3+
• C. V 3+
• D. Fe 2+✓

Explanation: The electron configuration for Fe2+ will be 1s2 2s2 2p6 3s2 3p6 4s2 3d4 because it has lost two electrons.All four electrons in d-orbital are unpaired.The Magnesium ion electron configuration will be 1s22s22p6.For the V3+, called the Vanadium ion, we remove one electron from 3d3 and two from the 3d2 leaving us with: 1s2 2s2 2p6 3s2 3p6 3d2.Electronic configuration of Ti3+ is [Ar] 3d1,which means it only has 1 unpaired electron.

Q106. CH3CH2CH2COCH3 is the functional isomer of:

• A. CH3CH2CH2CH2CHO✓
• B. CH3CH2CH2CH2OH
• C. CH3CH2COCH2CH3
• D. CH3 - (CH)2 COCH2CH3

Explanation: CH3CH2CH2COCH3 and CH3CH2CH2CH2CHO have the same number of atoms for the constituent elements, Carbon, Oxygen, and Hydrogen. However, the second requirement of a 'Functional Isomer' is that the functional groups, of the two isomers, must be different and since, in this case, we have a ketone functional group and an aldehyde functional, both compounds can be positively termed as 'Functional Isomers' of each other.Thus, option A is correct.

Q107. 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.

Q108. The colour of methyl orange in base is?

• A. Red
• B. Yellow✓
• C. Orange
• D. Pink

Explanation: Methyl orange is an indicator that changes color based on the pH of the solution. In an acidic environment, it appears red, while in a basic environment, it turns yellow. Option B is correct because the color change to yellow indicates the presence of a base. Option A is incorrect as red is the color in an acidic solution. Option C is incorrect because methyl orange does not transition through an orange phase; it moves from red to yellow. Option D is incorrect as pink is not a color change associated with methyl orange, but rather with phenolphthalein in basic conditions.

Q109. GN Lewis described which one of the following bonds?

• A. Ionic bond
• B. Covalent bond
• C. Coordinate covalent bond
• D. All of the above✓

Explanation: G.N. Lewis is renowned for his work on the nature of chemical bonding, particularly the sharing of electron pairs between atoms, which forms the basis of covalent bonds. His theories extend to both ordinary covalent bonds and coordinate covalent bonds. While ionic bonds involve electron transfer and are not the primary focus of Lewis's theories, his work laid the groundwork for understanding chemical bonding as a whole. Therefore, 'All of the above' is the correct answer, as it encompasses the types of bonds Lewis described.

Q110. Which of the following compounds shows maximum Hydrogen Bonding?

• A. CH3OH
• B. CH3CH2OH
• C. C6H5OH✓
• D. C6H11OH

Explanation: Phenol has a benzene ring that can participate in pi-bonding (π-bonding) with other electron-rich aromatic systems in addition to forming hydrogen bonds. This makes it more likely to show stronger hydrogen bonding compared to the other options.

Q111. Which of the following solvents favor SN2 reactions?

• A. Water
• B. Ammonia
• C. Carbon tetrachloride✓
• D. Acetic acid

Explanation: The correct answer is Carbon tetrachloride, which is a non-polar solvent that does not stabilize ions and thus allows for a more favorable environment for SN2 reactions. In SN2 mechanisms, the nucleophile attacks the substrate in a single concerted step, and polar protic solvents like water, ammonia, and acetic acid can hinder this process by stabilizing the transition state or the leaving group, making them less favorable for such reactions.In contrast, polar aprotic solvents (like carbon tetrachloride) do not solvate the nucleophile as strongly, allowing it to remain reactive and effectively participate in the nucleophilic substitution process.

Q112. The range of pH below _ and above _ of soil represent its sterilIty.

• A. 5....10
• B. 10....15
• C. 3....10✓
• D. 10....3
• E. 5....3

Explanation: pH below 3 and pH above 10 results in soil infertility. Hence the answer is C.

Q113. Consider NaCl produced in the equation: FeCl3+ NaOH → Fe(OH)3 + NaCl 175.5g of NaCl was dissolved in water to make a liter of solution, the molarity would be:

• A. 0.1 M
• B. 3 M✓
• C. 8 M
• D. 12 M
• E. 16 M

Explanation: Molarity = number of moles/volume in dm3 or liters 175.5 g make up 3 moles of NaCl 175.5/58.5 = 3 mol That means 3 mole of NaCl is dissolved in 1 liter of solution that makes the NaCl solution 3M.

Q114. Harmful and undesirable reaction of a metal when exposed to the atmosphere or any chemical agent is known as:

• A. Allotropy
• B. Electroplating
• C. Collision
• D. Cracking
• E. Corrosion✓

Explanation: The harmful and desirable reaction which occurs due to oxidation of the metal exposed to the surrounding air is called corrosion.

Q115. Catenation is a process in which carbon shows the properties of:

• A. Making single bond
• B. Hybridization
• C. Making chains or rings of carbon atoms✓
• D. Isomerism
• E. Breaking of bonds

Explanation: Catenation is the process by which an atom is able to form a bond with another atom of its kind. Carbon possesses this ability hence is able to form ringed structures.

Q116. Which of the following statements is true for Amorphous solids?

• A. They possess symmetry
• B. They are isotropic✓
• C. They are anisotropic
• D. They cleave along a particular direction
• E. They have a definite shape

Explanation: Amorphous solids are isotropic because their lack of long-range order results in uniform physical properties in all directions. This is in contrast to crystalline solids, which are anisotropic due to their orderly atomic structure. Amorphous solids do not possess symmetry and cannot cleave along particular planes because their particles are randomly arranged. Additionally, they do not exhibit a definite shape as their molecular structure does not form a regular lattice.

Q117. Only two elements are present in:

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

Explanation: Period 1 of the periodic table consists of only two elements: hydrogen and helium. This is because the first electron shell (the 1s orbital) can hold a maximum of two electrons. In contrast, Periods 2 and 3 each have 8 elements, as they include the filling of s and p orbitals. Periods 4 and 5 have even more elements, 18 each, due to the inclusion of the d orbitals. Hence, the correct answer is Period 1, where only two elements are present.

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. The oxidation number of all the elements in the free state is:

• A. 1
• B. 3
• C. 0✓
• D. -1
• E. -3

Explanation: All the elements have an oxidation number of 0 in the free state.

Q120. According to which phenomenon, "the properties of elements are the periodic function of their atomic number."

• A. Mendeleev's periodic law
• B. Newland's law of octaves
• C. Dobereiner's triads
• D. Lothar Meyer’s classification
• E. Modern periodic law✓

Explanation: The modern Periodic law can be stated as: “The physical and chemical properties of the elements are periodic functions of their atomic numbers”. The atomic number is equal to the number of electrons or protons in a neutral atom.

Q121. Comparative rates of diffusion of He and SO2 will be _.

• A. 8
• B. 2
• C. 4✓
• D. 16
• E. 64

Explanation: According to Graham's law of effusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Therefore, the comparative rate of diffusion can be calculated using the formula: Rate of diffusion = √(M1/M2) Here, M1 is the molar mass of SO2 (64 g/mol), and M2 is the molar mass of He (4 g/mol). Plugging in these values gives: √(64/4) = √16 = 4. Therefore, the comparative rate of diffusion of helium to sulfur dioxide is 4. The incorrect options arise from not correctly applying the square root to the ratio of molar masses.

Q122. The maximum possible number of electrons a shell 'n' can accommodate is given by:

• A. n
• B. n2
• C. 2n2✓
• D. n3
• E. 3n2

Explanation: The maximum number of electrons that can be in a shell is 2n2.

Q123. IUPAC name of this is (CH3)2 - CH - CH (C2H5) - C(CH3)3 :

• A. 3-ethyl-2,2,4-trimethylpentane✓
• B. 4-ethyl-2,2,4-trimethylpentane
• C. 5-ethyl-2,2,4-trimethylpentane
• D. 2-ethyl-2,2,4-trimethylpentane
• E. 1-ethyl-2,2,4-trimethylpentane

Explanation: The IUPAC name for the given compound (CH3)2-CH-CH(C2H5)-C(CH3)3 is "3-ethyl-2,2,4-trimethylpentane." Here's how the name is derived:1. Identify the longest continuous carbon chain, which in this case is a five-carbon chain.2. Number the carbon chain to give the substituents the lowest possible locants. In this case, we start numbering from the end nearest to the ethyl group, so the ethyl group is located on carbon 3.3. Assign substituent names and locants. The methyl groups attached to carbon 2 and carbon 4 are called "2,2-dimethyl" and "2,4-dimethyl," respectively. The ethyl group attached to carbon 3 is named "ethyl."4. Combine all the parts to form the complete IUPAC name: "3-ethyl-2,2,4-trimethylpentane."Therefore, the compound (CH3)2-CH-CH(C2H5)-C(CH3)3 is named 3-ethyl-2,2,4-trimethylpentane.

Q124. Metaboric acid when heated produces:

• A. B2O3
• B. HBO2
• C. H2B4O7✓
• D. H3BO3

Explanation: This is the following solution:

Q125. Caster kellners cell evaporated has replaced _ because the Caster Kellner process evaporates_?

• A. Gibbs diaphragm… mercury✓
• B. Iron box… aluminium
• C. Pyrites burners… oleum
• D. Nelson's cell… chlorine

Explanation: The Caster-Kellner cell, also known as the mercury cell, was introduced as a replacement for the Gibbs diaphragm cell. This change was primarily due to the Caster-Kellner process's efficiency in using mercury to produce chlorine and caustic soda through the electrolysis of saltwater. The process involves mercury, which acts as a cathode. Other options like the Iron box and Pyrites burners are not related to this process, and while Nelson's cell also produces chlorine, it does not utilize mercury in the same manner.

Q126. Nitrogen dioxide decomposes on heating according to the following equation: 2NO2 (g) ⇌ 2NO (g)+ O2 (g) When 4 moles of nitrogen dioxide were put into a 1 dm3 container and heated to a constant temperature, the equilibrium mixture contained 0.8 moles of oxygen. What is the numerical value of the equilibrium constant, Kc, at the temperature of the experiment?

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

Explanation: To find the numerical value of the equilibrium constant, Kc, we need to use the equilibrium concentrations of the reactants and products.According to the given information, we start with 4 moles of nitrogen dioxide (NO2) in a 1 dm^3 container. This means the initial concentration of NO2 is 2.4 moles/1 dm^3 = 2.4 M.At equilibrium, we have 0.8 moles of oxygen (O2) in the 1 dm^3 container. The equilibrium concentration of O2 is therefore 0.8 M.Since the stoichiometry of the balanced equation is 2NO2 ⇌ 2NO + O2, the equilibrium concentration of nitrogen monoxide (NO) will also be 2 times the concentration of O2, which is 2 x 0.8 M = 1.6 M.Now we can write the expression for the equilibrium constant, Kc:Kc = [NO]2 [O2] / [NO2]2Kc = (1.6 M)2 * (0.8 M) / (2.4 M)2This is numerical so it can only have one correct option.

Q127. CH ≡ CH + 2AgNO3 -> AgC ≡ CAg + 2HNO3 represents _ property of acetylene.

• A. Basic
• B. Acidic✓
• C. Dehydrating
• D. Physical
• E. None of the above

Explanation: Acetylene is a known weak acid as it can be seen when dissolved in silver nitrate it can give off its two hydrogen ions - thus acting as a acid.

Q128. The stability of carbonium ions following the order:

• A. A. Methyl > Ethyl > Isopropyl (secondary) > Tertiarybutyl
• B. B. Tertiarybutyl > Isopropyl (secondary) > Ethyl > Methyl✓
• C. C. Isopropyl (secondary) > Tertiarybutyl > Ethyl > Methyl
• D. D. Ethyl > Methyl > Isopropyl (secondary) > Tertiarybutyl
• E. E. Methyl > Isopropyl (secondary) > Tertiarybutyl > Ethyl

Explanation: The stability of carbonium ions increases with the number of alkyl groups attached to the positively charged carbon. This is because alkyl groups provide electron-donating effects that help stabilize the positive charge. Therefore, tertiary carbonium ions, with three alkyl groups, are more stable than secondary and primary ions. Thus, the correct order of stability is: Tertiarybutyl > Isopropyl (secondary) > Ethyl > Methyl. Options A, C, D, and E incorrectly represent the stability order by either misplacing the tertiary ion or suggesting a misleading trend.

Q129. _ is used as a preservative for life specimens.

• A. H2SO4
• B. Ammonia
• C. Methanol
• D. NaOH
• E. Formalin✓

Explanation: Formalin is widely used for the preservation of biological specimens due to its ability to fix tissues, making them resistant to decay. It works by crosslinking proteins, which helps maintain the structure of cells and tissues. Formalin also has antiseptic properties, preventing microbial growth. The other options, such as H2SO4, ammonia, methanol, and NaOH, are not suitable for preservation as they can damage tissues or fail to effectively preserve cellular structures.

Q130. EDTA ion is a _ ligand.

• A. Monodentate
• B. Bidentate
• C. Tridentate
• D. Polydentate✓
• E. None of the above

Explanation: EDTA ion is polydentate. Ethylenediaminetetraacetic acid (EDTA) is an amino polycarboxylic acid with the formula [CH2N(CH2CO2H)2]2. This white, water-soluble solid is widely used to bind to iron and calcium ions. It binds these ions as a hexadentate ("six-toothed") chelating agent.

Q131. Sulphur-16 gets its stabilization by gaining 2 electron to become equal to:

• A. Neon
• B. Argon✓
• C. Helium
• D. Krypton

Explanation: Option B is correct since sulfur-16 after gaining two electrons possesses 18 electrons and attains the stable electronic configuration of argon that also contains 18 electrons. Stability is gained by attaining the stable electronic configuration of a noble gas.Option A is incorrect since neon has 10 electrons whereas sulfur ion has 18.Option C is incorrect since helium has two electrons and sulfur ion has 18.Option D is incorrect since krypton has 36 electrons whereas, sulfur ion will have 18 electrons.

Q132. The smaller the distance of the object from the eye, the visual angle will be:

• A. Smaller
• B. Greater✓
• C. Constant
• D. Negligible
• E. None of the above

Explanation: The visual angle of an object is determined by the size of the object's image on the retina. This angle increases as the object comes closer to the observer because the image projected on the retina becomes larger. Therefore, when the distance of the object from the eye decreases, the visual angle becomes greater. Option B is correct because it accurately describes this relationship. Other options fail to capture this dynamic change in visual angle with distance.

Q133. The efficiency of the Carnot's Engine working between 150°C and 50°C is:

• A. 22.3%
• B. 20.0%
• C. 23.6%✓
• D. 30.6%
• E. 33.6%

Explanation: The efficiency of a Carnot engine is calculated using the formula η% = (1 - T2/T1) × 100%, where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir in Kelvin.Convert the given temperatures from Celsius to Kelvin: T1 = 150°C + 273 = 423 K and T2 = 50°C + 273 = 323 K.Substituting these values into the formula, η% = (1 - 323/423) × 100% = 23.6%.Options A, B, D, and E either apply the formula incorrectly or suggest efficiencies that are not supported by the calculated values.

Q134. The principle of a capacitor is based on which of the following facts?

• A. The potential of a conductor increases when the charge on it decreases, but this affects the charge stored in it.
• B. The potential of a conductor is greatly reduced when the charge on it increases.
• C. The potential of a conductor increases without affecting the charge stored in it.
• D. The potential of a conductor is greatly reduced without affecting the charge stored in it.✓
• E. The potential of a conductor increases as the charge on it increases.

Explanation: The correct answer is that the potential of a conductor is greatly reduced without affecting the charge in it. This is the fundamental principle behind capacitors, which are used to store electric charge by reducing the electric potential of a conductor while maintaining its charge. This is achieved by introducing another conductor, either earth-connected or oppositely charged, in proximity to the first. Other options are incorrect because they either involve increasing potential or affecting the charge, which do not describe how capacitors function.

Q135. If in a circuit, 2 Ampere current is drawn from the battery in 10 minutes. How much charge will flow through the circuit in this time?

• A. 1200 Coulombs✓
• B. 600 Coulombs
• C. 500 Coulombs
• D. 20 Coulombs

Explanation: DATA: Current I = 2 amp Time T = 10 minutes = 600 secCharge Q = ??SOLUTION : I= Q/TQ = ITQ = 600×2 Q = 1200 coloumb ans

Q136. 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.

Q137. The SI unit for intensity of sound is:

• A. Joule per second
• B. Watt per metre square✓
• C. Watt metre square
• D. Joule per metre square
• E. Joule metre

Explanation: Intensity is the amount of sound power passing through a unit area, and is measured in watts per square meter (W/m²). The correct answer is Option B: Watt per metre square. Other options do not represent intensity: Option A, Joule per second, measures power; Option C, Watt metre square, is not a recognized unit; Option D, Joule per metre square, measures energy density; and Option E, Joule metre, is not a standard physical unit.

Q138. A 50 watt bulb is operated by 120 volts. What is the current through the bulb?

• A. 0.416 A✓
• B. 1.5 A
• C. 2.5 A
• D. 3.5 A
• E. 5.5 A

Explanation: P=IV(50)=(120)(I)50/120=II=0.416A

Q139. There is a current of 25 A in a long straight wire. What is the flux density at a point 3 cm from the wire?

• A. 1.67 x 10-4 T✓
• B. 3.67 x 10-4 T
• C. 5.67 x 10-4 T
• D. 7.67 x 10-4 T
• E. 9.67 x 10-4 T

Explanation: Given, a=3cm=9×10−2m, I=25AB=μoI/2πa=4π×10−7×25/2π×3×10−2=1.67 x 10-4 T

Q140. Beta particles are the particles emitted from the _ of an atom.

• A. Inner atomic shells
• B. Outer atomic shells
• C. Innermost atomic shell
• D. Nucleus✓

Explanation: The beta particle, which may be either negatively charged (negatrons) or positively charged (positrons), originates from the nucleus of an atom. A beta particle is emitted from the nucleus of an atom during radioactive decay. The electron, however, occupies regions outside the nucleus of an atom.

Q141. Name the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls.

• A. Pneumonia
• B. Asthma
• C. Pleurisy
• D. Emphysema✓

Explanation: Note:The correct option is d) Emphysema.Explanation of each option:a)Pneumonia: Pneumonia is an inflammatory lung condition primarily caused by infection, not by damage to the alveolar walls. It typically results in the air sacs (alveoli) filling with fluid or pus due to infection, rather than damage to alveolar walls.b)cAsthma: Asthma is a chronic respiratory condition characterized by airway inflammation, bronchoconstriction, and excess mucus production. It does not primarily involve damage to the alveolar walls but rather affects the airways.c)Pleurisy: Pleurisy, or pleuritis, is inflammation of the pleura, the lining around the lungs. It does not directly involve damage to the alveolar walls or impact the alveolar surface area responsible for gas exchange.d)Emphysema: This is the correct answer. Emphysema is a pulmonary disease characterized by the gradual destruction of the alveolar walls. This damage reduces the surface area available for gas exchange, leading to impaired lung function, shortness of breath, and other respiratory symptoms. Smoking is a common cause of emphysema. The destruction of the alveolar walls leads to the enlargement of the air sacs and loss of elasticity in the lungs. As a result, the surface area available for gas exchange is drastically reduced, impairing the lung's ability to efficiently transfer oxygen to the bloodstream and remove carbon dioxide. Summary: The correct answer is "Emphysema." Emphysema is a pulmonary disease characterized by damage to the alveolar walls, which drastically reduces the alveolar surface area involved in gas exchange. The other options, pneumonia, asthma, and pleurisy, do not primarily involve damage to the alveolar walls and are distinct conditions with different characteristics.

Q142. How many types of modulations are possible with sinusoidal carrier?

• A. Only one
• B. Two
• C. Three✓
• D. Four

Explanation: The carrier signal is a sine wave at the carrier frequency. The sine wave has three characteristics that can be altered. The term that may be varied is the carrier voltage Ec, the carrier frequency FC, and the carrier phase angle θ. So three forms of modulations are possible.

Q143. Which of the following is not a base unit in the SI system?

• A. Ampere
• B. Second
• C. Meter per second✓
• D. Candela

Explanation: In summary, the correct answer is option c) Meter per second, as it is not a base unit in the SI system. The other options (a, b, and d) represent base units: ampere for electric current, second for time, and candela for luminous intensity.

Q144. In the Compton Scattering experiment the X-ray wavelength change Δλ is _. Here h is Planck constant, m rest mass of electron and θ is angle after scattering.

• A. Δλ = h/m0c (1 + cosθ)
• B. Δλ = h/m0c (1 + cos2θ)
• C. Δλ = h/m0c (1 - cosθ)✓
• D. Δλ = h/m0c2 (1 + cosθ)

Explanation: The correct answer is Option C: Δλ = h/m0c (1 - cosθ). This formula represents the Compton scattering equation, where Δλ is the change in wavelength of the X-ray, h is Planck's constant, m is the rest mass of the electron, c is the speed of light, and θ is the scattering angle.Option A is incorrect because it incorrectly adds the cosine term instead of subtracting it. Option B is incorrect because it squares the cosine term, which is not part of the Compton formula. Option D is incorrect as it erroneously includes \( c^2 \) in the denominator, which does not belong in the equation.

Q145. 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: To solve this problem, we start by determining the acceleration using the kinematic equation v² = u² + 2as, where:v is the final velocity (8 m/s),u is the initial velocity (0 m/s, since the car starts from rest),a is the acceleration, ands is the displacement (16 m).Substituting the known values into the equation:(8)² = (0)² + 2(a)(16)64 = 32aa = 2 m/s²Next, apply Newton's second law to find the force: F = ma.F = 1200 kg × 2 m/s² = 2400 NTherefore, the correct values are an acceleration of 2 m/s² and a force of 2400 N. The other options are incorrect because they do not satisfy both the acceleration and force calculations based on the given conditions.

Q146. Projectile must be launched at which angle with the horizontal to attain maximum range?

• A. 90 degrees
• B. 45 degrees✓
• C. 75 degrees
• D. 105 degrees
• E. 145 degrees

Explanation: To achieve maximum range, a projectile should be launched at an angle of 45 degrees with the horizontal. This angle provides the ideal balance between vertical and horizontal components of velocity, maximizing the distance traveled horizontally. The formula R = (v^2 sin2θ)/g confirms that maximum range occurs when sin2θ equals 1, which happens at θ = 45 degrees. Other angles, such as 90 degrees, result in no horizontal range, while angles greater than or less than 45 degrees reduce the range due to an imbalance in velocity components.

Q147. A block with a mass of 0.1 kg is attached to a spring and placed on a horizontal frictionless table. The spring is stretched 20 cm when a force of 5 N is applied, the spring constant is:

• A. 50 Nm-1
• B. 25 Nm-1✓
• C. 75 Nm-1
• D. 100 Nm-1
• E. 125 Nm-1

Explanation: To find the spring constant k, use Hooke's Law: F = kx. Given that F = 5 N and the extension x = 0.2 m, the equation becomes:5 = k(0.2)Solving for k gives k = 5 / 0.2 = 25 Nm-1.Option B is correct because it accurately reflects this calculation.Options A, C, D, and E are incorrect because they do not match the calculated value of the spring constant. They either overestimate or underestimate the spring constant due to incorrect application of the formula.

Q148. Two blocks, X and Y, of masses m and 2m respectively, are accelerated along a smooth horizontal surface by a force F applied to block X, as shown in the diagram.What is the magnitude of the force exerted by block Y on block X during this acceleration?

• A. 0
• B. F/3
• C. F/2
• D. 2F/3✓
• E. F

Explanation: Both blocks will be accelerated when the block X is applied by a force of magnitude F. Since, the total mass will be m+2m which is 3m.So, the acceleration of the system will be given as F=ma or a=F/3m. So, from this we can find the force acting on the block Y. So, to calculate the force on the block Y we have that F(y) =2m*a or 2m*F/3m which on solving we will get the force on the block y as F(y)= 2F/3Magnitude of the force exerted by block Y on block X during this acceleration is = 2F/3

Q149. The amount of heat at constant volume is called as:

• A. Internal energy✓
• B. Enthalpy
• C. Entropy
• D. Temperature
• E. Pressure

Explanation: At constant volume, the heat of the reaction is equal to the change in the internal energy of the system. Meanwhile, the heat of reaction at constant PRESSURE is known as enthalpy

Q150. A body is hanging from a rigid support by an extensible string of length L. It is struck inelastically by an identical body of mass m with horizontal velocity v =√2gl , the tension in the string increases just after striking by:

• A. mg
• B. 3mg
• C. 2mg✓
• D. None of these

Explanation: Initially, the tension in the string is T = mg, which equals the weight of the hanging body. When struck by an identical body inelastically, the system's mass becomes 2m. Conservation of momentum gives us:mv = 2m(v'), thus v' = v/2.Using the inelastic collision principles, the new tension T' is given by T' - 2mg = 2m(v')^2/L. By substituting v' = v/2 and v = √(2gL), we find T' = 3mg.Therefore, the tension increase is T' - T = 3mg - mg = 2mg.Option A is incorrect as it ignores the momentum change. Option B miscalculates the velocity and resultant tension. Option D is incorrect since the calculated increase in tension is indeed 2mg, making Option C the right choice.

Q151. Which statement describes the electric potential difference between two points in the electric field of charge Q?

• A. The difference of electric field between the points per unit charge.
• B. The ratio of the power dissipated between the points to the mass of charge
• C. The work done in moving a test charge between points divided by the magnitude of the test charge.✓
• D. The force required to move a unit positive charge between the points per unit charge.

Explanation: This question is simply revising how the electric potential is defined as it is work done per unit charge in moving that charge from one place to the other in the electric field.All others are not the definitions of electric potential.

Q152. An electron moving with velocity v has momentum 3 x 10 -26 Kg.m/s. The de Broglie wavelength associated with it is _. Value of h = 6.63 x 10-34 is.

• A. 24.1 nm
• B. 22.14 m
• C. 22.1 nm✓
• D. 22.1 mm

Explanation: The de Broglie wavelength formula is λ = h/p, where h is Planck's constant and p is the momentum of the electron. Given the momentum p = 3 × 10-26 Kg.m/s and h = 6.63 × 10-34 J.s, substitute these values into the formula to find λ:λ = 6.63 × 10-34 / 3 × 10-26 = 22.1 nm.Option A is incorrect due to a calculation error. Option B is incorrect because it suggests a much larger and incorrect unit (meters). Option D is incorrect due to the wrong unit (millimeters). Thus, Option C is the correct answer with the correct units and calculation.

Q153. The vector product of two vectors A and B is _ vectors A and B.

• A. Not equal to the product of magnitudes of
• B. In the plane parallel to
• C. Perpendicular to the plane containing✓
• D. Less in magnitude than the product of magnitudes of

Explanation: The vector product, or cross product, of two vectors A and B results in a vector perpendicular to the plane containing A and B. This is determined using the right-hand rule, which helps visualize the direction of the resulting vector. The magnitude of the cross product is |A × B| = |A||B|sin(θ), where θ is the angle between A and B. Option C correctly identifies the perpendicular nature of the resulting vector. Option A is incorrect because it misrepresents the nature of the cross product as a scalar. Option B is incorrect because it misinterprets the orientation as parallel rather than perpendicular. Option D is incorrect because it inaccurately describes the magnitude relationship without considering the sine component.

Q154. A car starts from rest and moves with a constant acceleration. During the 5th second of its motion, it covers a distance of 36 meters. What is the acceleration of the car?

• A. 12.5 m/s2
• B. 8 m/s2✓
• C. 15 m/s2
• D. 16 m/s2
• E. 14 m/s2
• F. 2.88 m/s2

Explanation: You will find the distance traveled in 5 and 4 seconds individually. After subtracting these two terms and equating it to 36 meters, all variables are fulfilled other than a which we can solve for.In the below equations, s is distance, a is acceleration, and t is time.In 4 seconds, the truck covers:s = 1/2 at2s = 1/2 (a) 42s = 1/2 (a) (16)s = 8 (a) metersIn 5 seconds, the truck coverss = 1/2 a t2s = 1/2 (a) 52s = 1/2 (a) 25s = 12.5 (a) metersThe distance covered in 5 seconds minus the distance covered in 4 seconds is the distance (36 meters) covered in the 5th second.36 = 12.5 a - 8 a36 meters = 4.5 seconds2 (a)a = acceleration = 8 m/s2

Q155. The sinusoidal wave form can be varied by using which of the following parameters?I. Frequency of the carrier wave II. Amplitude of the carrier wave IIl. Phase angle

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

Explanation: To modulate a sinusoidal wave, you can vary its amplitude, frequency, or phase angle. These parameters affect different characteristics of the wave:Amplitude: Changes the wave's intensity or loudness.Frequency: Alters the pitch or tone of the wave.Phase Angle: Shifts the wave's position in time, introducing phase shifts or time delays.Thus, all three parameters can be used to modulate a sinusoidal wave, making option E the correct choice. Other options are incorrect because they exclude one or more of these critical parameters.

Q156. The speed of light is very nearly equal to;

• A. 5x108 m/sec
• B. 3x1016 m/sec
• C. 4x108 m/sec
• D. 3x108 m/sec✓
• E. 7x108 m/sec

Explanation: The speed of light in a vacuum is a fundamental constant of physics, known to be approximately 3x108 meters per second. This value is critical for many areas of physics and technology, including the theory of relativity. Option D correctly represents this speed. The other options are either too high or too low, and do not match the universally accepted value.

Q157. The dimensions of acceleration are:

• A. LT-1
• B. LT-2✓
• C. L3
• D. L2
• E. LT2

Explanation: To find the dimensions of acceleration, start by understanding that acceleration is the change in velocity over time.Velocity is defined as distance over time, which has dimensions of [L][T-1].Thus, acceleration, being velocity per time, is [L][T-1]/[T] = [L][T-2].Option A ([LT-1]) is the dimension of velocity, not acceleration. Option C ([L3]) corresponds to volume, Option D ([L2]) to area, and Option E ([LT2]) is a miscalculation of time's exponent. Only Option B accurately reflects the dimensions of acceleration: [LT-2].

Q158. R1 and R2 are two-position vectors making angles θ1 and θ2 with positive X-axis respectively. Their vector product is:[R1 = 4 cm, R2 = 3 cm , θ1 = 30 degrees , θ2 = 90 ]

• A. 12√3
• B. 6√3✓
• C. 6√12
• D. 12√6
• E. 3√6

Explanation: The vector product of two vectors, R1 and R2, is given by the formula |R1 × R2| = |R1| |R2| sin(θ2 - θ1), where θ is the angle between the vectors. Here, |R1| = 4 cm, |R2| = 3 cm, θ1 = 30°, and θ2 = 90°. Thus, |R1 × R2| = 4 × 3 × sin(90° - 30°) = 4 × 3 × 1/2√3 = 6√3 cm.This calculation confirms that the correct answer is 6√3 cm, directed along the Z-axis.The incorrect options result from errors in either the trigonometric calculations or angle substitutions, leading to incorrect values.

Q159. Coal and petrol are stores of chemical energy when are burnt, chemical energy is converted into heat energy i.e chemical energy = heat energy + _:

• A. Electrical energy
• B. Losses✓
• C. Gains
• D. Wind energy
• E. Nothing

Explanation: When coal or petrol is burned, the chemical energy stored in the fuel is primarily converted into heat energy. However, not all of this energy conversion is perfectly efficient; some energy is inevitably lost to the surroundings as waste heat, which is why 'Losses' is the correct answer. Electrical energy is not directly produced from combustion without additional equipment. Gains are incorrect because energy is not gained but redistributed with losses. Wind energy is unrelated to burning fuels. 'Nothing' is incorrect because energy transformations usually accompany some degree of energy loss.

Q160. A 60 kg boy runs up a hill through a height of 6 m in 3 seconds. How much work does he do against gravitational force?

• A. 3000 J
• B. 2528 J
• C. 3500 J
• D. 3528 J✓

Explanation: W=mgh=(60)(6)(9.8)=3528J

Q161. The velocity of a particle is related to time according to the equationV= ct3. The dimensions of constant are _.

• A. L-1 T-1
• B. L T-3
• C. LT -4✓
• D. L-2 T-2

Explanation: For the principle of homogeneity; V=ct3 (LT-1)=c(T3) c=LT-1/T3 c=LT -1-3 c=LT -4.

Q162. 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.

Q163. The formula for the Paschen series for the Hydrogen spectrum is _.

• A. 1/λ=Rh(1/22 - 1/n2); n=3,4,5
• B. 1/λ=Rh(1/42 - 1/n2); n=2,3,4,5
• C. 1/λ=Rh(1/42 - 1/n2); n=5,6,7
• D. 1/λ=Rh(1/32 - 1/n2); n=4,5,6✓

Explanation: The correct formula for the Paschen series is Option D: 1/λ=Rh(1/32 - 1/n2); n=4,5,6. The Paschen series refers to electron transitions ending at the n=3 energy level, resulting in spectral lines in the infrared region. The other options represent different series or incorrect configurations: Option A represents the Balmer series, transitioning to n=2. Option B does not correspond to any recognised series due to its incorrect configuration. Option C represents the Brackett series, transitioning to n=4.

Q164. A travelling wave, in which the particles of the distributed medium, move parallel to the direction of propagation of the wave is called:

• A. Transverse wave
• B. Circular wave
• C. Longitudinal wave✓
• D. Stationary wave

Explanation: Longitudinal waves are characterized by particle motion that is parallel to the direction of wave propagation. This creates alternating regions of compression and rarefaction. Sound waves in air are a classic example, where air particles oscillate back and forth in the same direction as the wave is moving.Transverse waves, on the other hand, involve particle movement perpendicular to wave direction, as seen in waves on strings or electromagnetic waves. Circular waves are typically seen on water surfaces, where the motion is vertical and not parallel. Stationary waves involve a pattern of nodes and antinodes due to interference, with particles oscillating in place.

Q165. A body starts sliding on a rough horizontal surface with a speed of 10 m/s. If the coefficient of friction is 0.2, find the distance traveled by the body before coming to rest. (g = 10m/s)

• A. 15 m
• B. 25 m✓
• C. 35 m
• D. 40 m
• E. 55 m

Explanation: Given that: v = 10 and u = 0.2 and g = 10 Solution = Using F = μN then, = μmg (here N = mg) So from above (ma = μmg); we got a = μg Now a = 0.2×10 = 2 m/s2Now using equation of motion V2=u2+2as 0 = 10×10 +2×2×s In solving it, we got s = 100/4 = 25 m Hence we get the correct answer.

Q166. A coil of 600 turns is threaded by a flux of 8x10-5 webers if this flux is reduced to 3x10-5 webers in 0.015 seconds. The average induced e.m.f. is:

• A. -2.0 volts
• B. -3.0 volts
• C. +2.0 volts✓
• D. +2.5 volts
• E. +3.0 volts

Explanation: Emf = - N (dФ/dt)where number of turns are denoted by N, dФ is the change in flux and dt is the time interval.Change in flux = Final Flux - Initial FluxChange in flux = (3 × 10⁻⁵) - (8 × 10⁻⁵) WbChange in flux = - (5 × 10⁻⁵) Wbtime interval = 0.015 sectime interval = 15 × 10⁻³ secNumber of turns in coil = 600Putting all the values in formula.Emf = - 600 × (-(5 × 10⁻⁵)/15 × 10⁻³) VoltsEmf = 600 × (5 × 10⁻²/15 ) VoltsEmf = 600 × (10⁻²/3 ) VoltsEmf = 200 × 10⁻² VoltsEmf = 2 Volts

Q167. A rotating wheel of radius 0.5 m has an angular velocity of 5 rad/s at some instant and 10 rad / s after 5 s. Find the angular acceleration of a point on its rim.

• A. 1 rad/s2✓
• B. 3 rad/s2
• C. 5 rad/s2
• D. 7 rad/s2
• E. 9 rad/s2

Explanation: The angular acceleration of a rotating object can be found using the formula α = (ωf - ωi) / t, where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time interval. In this case, the initial angular velocity (ωi) is 5 rad/s, the final angular velocity (ωf) is 10 rad/s, and the time interval (t) is 5 seconds. Substituting these values into the formula gives: α = (10 rad/s - 5 rad/s) / 5 s = 1 rad/s2. Therefore, Option A is correct. All other options are incorrect as they do not result from applying this formula to the given values.

Q168. A shot leaves a gun at the rate of 160 m/s. Calculate the greatest distance to which it could be projected. (Take g = 10 m/s2).

• A. 2460 m
• B. 2560 m✓
• C. 2680 m
• D. 2760 m
• E. 2860 m

Explanation: For maximum distance(range) we assume that it is projected at 45 degrees. So, Rmax = u^2/g. (Following is the correct solution):

Q169. When the aircraft Concorde is moving in a horizontal plane at a constant speed of 650 ms-1, its turning circle has a radius of 80 km. What is the ratio of the centripetal force to the weight of the aircraft? (g = 9.8 m/s2)

• A. 8.3 x 10^4
• B. 0.54✓
• C. 1.9
• D. 52
• E. 540

Explanation: The ratio of Centripetal force to the weight of the aircraft=mv^2/r divided by mgm*(650)^2/80*1000 / m*9.81=0.54.

Q170. If the frequency of a pendulum is four times greater on an unknown planet than it is on Earth, then the gravitational constant on that planet is

• A. 16 times greater✓
• B. 4 times greater
• C. 4 times lower
• D. 16 times lower
• E. 24 times lower

Explanation: f=1 /2π√g/LIf the frequency is increased by a factor of 4 the impact on the value of g will be by a factor of 16.

Q171. A 40 kg block is resting at a height of 5m off the ground. If the block is released and falls to the ground. What is its total energy at a height of 2m? (g= 10 m/s2).

• A. 0 J
• B. 400 J
• C. 2 kJ✓
• D. 6 kJ
• E. It cannot be determined from the information given.

Explanation: The total energy the body possesses is the same at all times: We can calculate the energy of the body when it is 5m off the ground and that’ll be energy when is 2m above the ground too (though in KE form too) PE=mgh PE=(40)(10)(5)=2000J=2kJ

Q172. In an astronomical telescope, the distance between the objective and the eyepiece is called :

• A. Magnifying Power of the telescope
• B. Width of the telescope
• C. Length of the telescope✓
• D. Height of the telescope
• E. Diameter of the lens of the telescope

Explanation: The distance between the eyepiece lens and the objective lens is called the length of the telescope as shown below :

Q173. Look at this series: 12, 11, 13, 12, 14, 13, … What number should come next?

• A. 10
• B. 12
• C. 13
• D. 15✓

Explanation: The sequence alternates between subtracting 1 and adding 2:Start with 12Subtract 1 to get 11Add 2 to get 13Subtract 1 to get 12Add 2 to get 14Subtract 1 to get 13Add 2 to get 15The next number is 15, following the established pattern of alternating operations. The incorrect options fail to maintain this pattern, either by miscalculating the operations or by failing to consider the alternation entirely.

Q174. Statements:I. There is an unprecedented increase in the number of young unemployed in comparison to the previous year.II. A large number of candidates submitted applications against an advertisement for the post of manager issued by a bank.

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

Explanation: An increase in the number of unemployed youth is bound to draw in huge crowds for a single vacancy.

Q175. What should come next to this word, "Recipe"

• A. Desserts
• B. Spoon
• C. Utensils
• D. Cookbook✓

Explanation: A recipe is a list of directions to make something. Recipes may be used to prepare desserts, among other things. Utensils may or may not be used to make a recipe however a cookbook is needed to follow directions for a specific recipe.

Q176. A book cannot exist without?

• A. Education
• B. Pictures
• C. Pages✓
• D. Qualification

Explanation: A book can still exist if it doesn’t have pictures; it can be a reading book without pictures; it might be a book to read for leisure and not for education; and it might not have any qualifications, but it will still be a book. However, without pages, a book cannot physically exist.

Q177. Statement:I. Majority of the citizens in the locality belongs to higher income group.II. The sales in the local supermarket are comparatively much higher than in other localities.

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

Explanation: The reason why sales are comparatively high in a particular locality is because the residents of the locality are capable of paying high prices. So option A is correct.

Q178. Statement:I. Large number of Primary Schools in rural areas are run by only one teacher.II. There has been a huge dropout from primary schools in rural areas.

• A. Statement I is the cause then 2 is its effect✓
• B. Statement 2 is the cause then I is its effect
• C. Both statements are independent causes
• D. Both of the statements are effect of independent causes

Explanation: One teacher running a lot of schools leads to mismanagement and it lowers the educational standard which causes students to drop out from school in rural areas. So, option A is the most accurate answer.

Q179. In an astronomical telescope, the lens called eyepiece forms an image of a distant object that is:

• A. Real and diminished
• B. Real and magnified
• C. Virtual and diminished
• D. Virtual and magnified✓

Explanation: An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies. The final image formed by an astronomical telescope is always virtual, inverted and magnified.

Q180. Statements: 1. During the past year, Josh saw more movies than Stephen. 2. Stephen saw fewer movies than Darren. 3.Darren saw more movies than Josh. If the first two statements are true, the third statement is:

• A. True
• B. False
• C. Uncertain✓
• D. None of these

Explanation: The statements tell us that Josh > Stephen and Darren > Stephen. However, there is no information to directly compare the number of movies seen by Josh and Darren. Darren could have seen more, fewer, or the same number of movies as Josh.