Punjab Mdcat Mock Paper 5 — Solved Past Paper with Answers

Q1. Choose the word most similar in meaning to the capitalized one: DISRUPTION:

• A. Comfort
• B. Luxury
• C. Trouble✓
• D. Freedom
• E. Calm

Explanation: Disruption means “any/ disturbance or problem that interrupts an event, activity or a process”All other words don’t relate to the mentioned word.

Q2. We prefer fruits _ sweets.

• A. Then
• B. On
• C. Over✓
• D. From

Explanation: The correct preposition to use in this case will be “over”. You prefer one thing “over” others.

Q3. Ahmed _ me for a long time.

• A. Know
• B. Has known✓
• C. Knows
• D. Knew

Explanation: The most appropriate word to fill in the blank would be "has known." The complete sentence would be: "Ahmed has known me for a long time." Using the present perfect tense "has known" indicates that Ahmed developed knowledge or familiarity with the speaker in the past and that this knowledge or familiarity has continued up until the present time.

Q4. Choose the correctly punctuated sentence:

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

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

Q5. Choose the correct option:

• A. “Well no, perhaps not sir”
• B. “Well, no, perhaps not sir”✓
• C. “Well, no perhaps not sir”
• D. “Well no perhaps, not sir”

Explanation: Commas should be used at these places to lay emphasis. The key to solving this question is to read the sentence in your head and then decide at which places a pause should be incorporated.

Q6. Not only the parents but also their son _ for interview.

• A. Has called
• B. Have called
• C. Have been called
• D. Has been called✓

Explanation: Parents are plural but the blank follows the noun 'son' which is singular. Hence "has" instead of "have" is the right option. The sentence refers to an act being done to the noun instead of the son doing something. Hence it will be: ‘been called’ instead of just ‘called’

Q7. Select the option which fits appropriately in the blank: Annie and her brothers _ at school.

• A. Is
• B. Are✓
• C. Are being
• D. Have

Explanation: When deciding whether to use 'is' or 'are', look at whether the noun is plural or singular. If the noun is singular, use 'is'. If it is plural or there is more than one noun, use 'are'. Here the noun is plural (brothers) hence, the correct option is 'are'.

Q8. Select the option which fits appropriately in the blank: Dan _ a lot of friends.

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

Explanation: 'Has' is the third person singular present tense of 'is' and it will be used as the term friends is plural, so the verb must be singular.

Q9. Select the option which fits appropriately in the blank: The gardner will have _ the flowers.

• A. Plucked✓
• B. Pluck
• C. Plucks
• D. None of these options are correct

Explanation: "Plucked" is the correct verb form. The future perfect tense refers to a completed activity in the future.

Q10. Choose the correct sentence:

• 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: Option A: This sentence is missing a comma before "and." It should be "an orange, and a pear." The absence of the comma makes the sentence grammatically incorrect.Option B: This sentence is also missing a comma before "and." Similar to option A, it should be "an orange, and a pear." The lack of a comma makes this sentence grammatically incorrect as well.Option C: This sentence is correctly punctuated with commas. It uses the comma before "and" when listing three items. Each item in the list is separated by a comma, and the final comma (known as the Oxford comma) is used before "and." This option is grammatically correct.Option D: Similar to options A and B, this sentence is missing a comma before "and." It should be "an orange, and a pear." The absence of a comma makes this sentence grammatically incorrect.

Q11. Find the error in the following sentence:

• A. They have been✓
• B. Very close friends
• C. Until they quarelled
• D. Over a petty issue.

Explanation: 'They used to be close friends but not anymore' indicates that the sentence is in the past tense hence 'have been' will not be used. 'Had been' would be the correct term here. "Very close friends" is a correct description of the relationship between the two people. "Until they quarreled/quarrelled" is a correct way to indicate that their friendship ended due to an argument. "Over a petty issue" correctly identifies the cause of the quarrel.

Q12. Some of the books on the table _ to me.

• A. Belong✓
• B. Belonging
• C. Belongs
• D. Was belong

Explanation: With pronouns like you, me, and we, 'belong' is used. "Belonging" is a gerund, which is a verb form that functions as a noun. It does not work in this context because the sentence requires a verb. "Belongs" is the singular form of the verb "to belong." Since the subject is plural ("books"), this form of the verb is not appropriate. "Was belong" is not a grammatically correct construction. The correct past tense form of "to belong" is "belonged," but the sentence is in the present tense, so this option is also not appropriate.

Q13. We went to the President as a last _.

• A. Recourse
• B. Resort✓
• C. Resource
• D. Resound

Explanation: "recourse" is a similar word that means seeking help or protection from someone or something when in difficulty or trouble, but it doesn't fit well in the given sentence. A last resort is a final course of action, used only when all else has failed. "resource" refers to a source of aid or support that can be drawn upon when needed. While it may be used in the same context as "last resort", it does not fit the given sentence. "resound" means to echo or reverberate with sound and has no connection to the given sentence.

Q14. Did I say anything to make you angry?

• A. Declarative
• B. Imperative
• C. lnterrogative✓
• D. Exclamatory

Explanation: The sentence "Did I say anything to make you angry?" is a C. interrogative sentence. Interrogative sentences are used to ask questions and typically end with a question mark. In this case, the sentence is asking a question to inquire whether the speaker said something to provoke anger in the listener.

Q15. There was _ article about pollution in _ paper.

• A. A..an
• B. An...the✓
• C. A...the
• D. A...a

Explanation: A/an and the are articles. They are types of determiners and they go before a noun. Use “a” before words that start with a consonant sound and “an” before words that start with a vowel sound. The definite article is used before singular and plural nouns when the noun is specific or particular. The signals that the noun is definite, that it refers to a particular member of a group. For example: 'The dog that bit me ran away.' Here, we're talking about a specific dog, the dog that bit me.

Q16. Which verb is NOT in the present tense?

• A. She listened✓
• B. She talks
• C. She waits
• D. She eats

Explanation: The presence of -ed creates a past tense in the sentence. talks" is in the present tense, describing an action that is currently happening. "waits" is in the present tense, describing an ongoing action or state. "eats" is also in the present tense, describing a habitual or repeated action.

Q17. I borrowed _ pencil from your pile of pencils and pens.

• A. A✓
• B. An
• C. The
• D. No article is required

Explanation: ‘A’ is used before a common noun that starts with a consonant sound. ‘An’ is used before a common noun that starts with a vowel sound. ‘The’ is used before a proper noun. Here, pencil is a common noun, since it refers to any pencil from the pile.

Q18. He was _ of theft in the court.

• A. Charged✓
• B. Blamed
• C. Reported
• D. Accused

Explanation: Charged - criminals in court are charged or sentenced.

Q19. Which of the following is a freshwater sponge?

• A. Sycon
• B. Leucosolenia
• C. Euplectella
• D. Spongilla✓

Explanation: Spongilla is a genus of freshwater sponges in the family Spongillidae found in lakes and slow streams. Sycon is a genus of calcareous sponges belonging to the family Sycettidae. These sponges are small and are tube-shaped and often white to cream in color. Euplectella is a genus of glass sponges that includes the well-known Venus flower basket. So, the correct answer is option D.

Q20. Pseudocoelom is actually derived from:

• A. Blastocoel✓
• B. Gastrocoel
• C. Neurocoel
• D. Haemocoel

Explanation: A pseudocoelom is a fluid-filled space between the body wall and the digestive tract found in roundworms. Pseudocoelom develops from the embryo's blastocoel rather than a secondary cavity within the embryonic mesoderm (which leads to a real body cavity or coelom).

Q21. Species of Phylum Platyhelminthes are:

• A. Roundworms
• B. Flatworms✓
• C. Hook worms
• D. Threadworms
• E. Pinworms

Explanation: Phylum Platyhelminthes are called flatworms. (Fact)

Q22. A characteristics features of Echinoderm is:

• A. Canal system
• B. Water vascular system✓
• C. Tracheal system
• D. Blood vascular system
• E. None of these

Explanation: The characteristics of adult echinoderms are the possession of a water vascular system.There is no true heart and blood and lacks any tracheal or canal system.

Q23. Which one of the following is a fish?

• A. Flying fish
• B. Salmon
• C. Sea horse
• D. All of the above✓

Explanation: Salmon, flying fish, and sea horse (Hippocampus) are all fishes and come under Pisces. They are cold-blooded animals and they possess two-chambered hearts. So, the correct answer is option D.

Q24. The thickest chamber of the human heart is:

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

Explanation: The left ventricle of your heart is larger and thicker than the other chambers. This is because it has to pump the blood further around the body, and against a higher pressure. The right ventricle pumps the oxygen-poor blood to the lungs through the pulmonary valve The right atrium receives oxygen-poor blood from the body and pumps it to the right ventricle through the tricuspid valve and hence don't need very thick walls. The left atrium receives oxygen-rich blood from the lungs and pumps it to the left ventricle through the mitral valve, hence it also doesn’t need thick walls.

Q25. The enzymes required for the Kreb cycle are found in _.

• A. F1 particles
• B. Lysosomes
• C. Cytoplasm
• D. Matrix✓

Explanation: The Krebs cycle enzymes are membrane proteins found within the matrix of the mitochondria except for succinate dehydrogenase which is an integral membrane protein locked to the inner mitochondrial membrane.F1 particle is found in the matrix of mitochondria and is found attached to the cristae. It plays a very important role in the production of ATP molecules as it contains an enzyme called ATPaseA lysosome is a membrane-bound cell organelle that contains digestive enzymes.The cytoplasm is a thick solution that fills each cell and is enclosed by the cell membrane.

Q26. Cell-mediated immune response is given by:

• A. T-lymphocytes✓
• B. B lymphocytes
• C. Neutrophils
• D. Macrophages

Explanation: Cell-mediated immunity is an immune response that does not involve antibodies. Rather, cell-mediated immunity is the activation of phagocytes, antigen-specific cytotoxic T-lymphocytes, and the release of various cytokines in response to an antigen. Humoral immunity is also called antibody-mediated immunity. With assistance from helper T cells, B cells will differentiate into plasma B cells that can produce antibodies against a specific antigen. The humoral immune system deals with antigens from pathogens that are freely circulating, or outside the infected cells. Antibodies produced by the B cells will bind to antigens, neutralizing them, or causing lysis (dissolution or destruction of cells by a lysin) or phagocytosis.

Q27. During breathing air from Pharynx enters to:

• A. Trachea✓
• B. Bronchioles
• C. Alveoli
• D. Bronchi

Explanation: When you inhale through your nose or mouth, air travels down your pharynx (back of your throat), passes through your larynx (voice box) and into your trachea (windpipe). Your trachea is divided into two air passages called bronchial tubes. One bronchial tube leads to your left lung, the other to your right lung.

Q28. The type of carbohydrate which has high molecular weight and is sparingly soluble in water is:

• A. Monosaccharides
• B. Disaccharides
• C. Polysaccharides✓
• D. Maltose

Explanation: Polysaccharides are polymeric carbohydrate molecules composed of long chains of monosaccharide units bound together by glycosidic bonds. They occur as components of high molecular weight compounds significantly less soluble in water due to the absence of hydrogen bonds.

Q29. Among the following, which acid is an unsaturated acid:

• A. Butyric Acid
• B. Oleic Acid✓
• C. Palmitic Acid
• D. Both Oleic Acid and Palmitic Acid

Explanation: •A saturated fatty acid is one that lacks double bonds. Animal fat is commonly composed of saturated fatty acids. They are solid at room temperature. •Unsaturated fatty acid is one that has one or more double bonds in them. They are commonly found in plants and exist in liquid form at room temperature. Oleic acid has a double bond between C9 and C10, making it unsaturated.

Q30. Backbone of amino acids comprises of

• A. R group
• B. Amino group
• C. Peptide bond
• D. Carboxylic group
• E. A,B and D✓

Explanation: Amino acids are molecules that combine to form proteins. Each amino acid contains an amino group,a carboxylic group,an r group and a hydrogen atom which is attached to the central alpha carbon. All of them form the basic skeleton of amino acids.

Q31. Which of the following holds the alpha helix of protein in its place:

• A. R group
• B. Disulfide bond
• C. Amino group
• D. Hydrogen bond✓

Explanation: The α-helix is a right-handed helix with the peptide bonds located on the inside and the side chains extending outward. It is stabilized by the regular formation of hydrogen bonds parallel to the axis of the helix; they are formed between the amino and carbonyl groups of every fourth peptide bond

Q32. The simplest independent unit of life is known as:

• A. Bacterial colony
• B. Cell✓
• C. Chloroplast
• D. DNA

Explanation: The cell is the simplest entity that can exist as an independent unit of life. Every known living organism is either a single cell or an ensemble of a few to many cells.

Q33. The process by which unwanted structures within the cell are engulfed and digested within the lysosome is known as:

• A. Endocytosis
• B. Exocytosis
• C. Hydrolysis
• D. Autophagy✓

Explanation: Autophagy is a type of endocytosis that involves the engulfment and digestion of damaged or unnecessary cell parts by the cell itself. This process is essential for maintaining cell health and survival

Q34. The plants having foreign DNA incorporated into their cells are called:

• A. Colonal plants
• B. Transgenic plants✓
• C. Bio tech plants
• D. Tissue cultured plants

Explanation: The term transgenic indicates an organism that contains genetic material from a foreign source. Transgenic plants are plants that have had foreign DNA incorporated into their cells through genetic engineering techniques.

Q35. Pasteurization technique is widely used for preservation of:

• A. Water
• B. Heat
• C. Milk products✓
• D. Vaccines

Explanation: Pasteurization is a technique to remove pathogenic organisms in certain foods and liquids. In water, bacteria is killed through the process of chlorination. Hence, the only remaining viable option is option C : Milk products.

Q36. The production of genetically identical copies of organisms by asexual reproduction is called:

• A. Genetic engineering
• B. Integrated disease management
• C. Hydroponic culture technique
• D. Cloning✓

Explanation: The production of genetically identical copies of organisms by asexual reproduction is called cloning.

Q37. The _ model of plasma membrane suggests that proteins are embedded in lipid bilayer:

• A. Unit membrane
• B. Fluid mosaic✓
• C. Permeable
• D. Ultracentrifuge

Explanation: According to the Fluid Mosaic Model, cell membranes are quasi-fluidic in nature with a viscous lipid bilayer having proteins at places like mosaic both on the outer surface and embedded inside. Singer and Nicolson have described the cell membrane as a 'protein iceberg' in the sea of lipids.

Q38. Lipid metabolism is the function of:

• A. Mitochondria
• B. Sarcoplasmic reticulum
• C. RER
• D. SER✓

Explanation: Fact: Smooth endoplasmic reticulum(SER) is involved in lipid metabolism.

Q39. In _ response, β-cells produce plasma cells that synthesize antibodies and release in blood plasma and tissue fluid.

• A. Cell-Mediated
• B. Hormonal
• C. Humoral✓
• D. Phototactic

Explanation: This option is the correct response in the given context. Humoral response refers to the immune response involving the production of antibodies by plasma cells. In this context, β-cells are producing plasma cells that synthesize antibodies and release them into the blood plasma and tissue fluid. This accurately describes the process mentioned in the sentence.

Q40. Passive immunity is used against:

• A. Malaria
• B. Dengue
• C. Thypoid
• D. Tetanus✓

Explanation: Tetanus is a bactenal infection which can be treated by anti-tetanus serum. Passive immunity can occur naturally, such as when an infant receives a mother's antibodies through the placenta or breast milk, or artificially, such as when a person receives antibodies in the form of an injection (gamma globulin injection).

Q41. What is the subunit of DNA

• A. Phosphoric Acid
• B. Base
• C. Nucleotide✓
• D. Sugar

Explanation: A nucleotide is the basic building block of nucleic acids(DNA and RNA). •It is composed of pentose sugar,phosphate and a nitrogenous base. •The four nitrogenous bases in DNA are Adenine,Guanine,Cytosine and Thymine. Options A,B and D are incorrect because all of them are subunits of the nucleotide itself,not DNA.

Q42. Enzymes have what effect on the end-product?

• A. They alter the product
• B. The nature of the product is completely changed
• C. No effect✓
• D. The end product is not obtained

Explanation: Enzymes are biocatalysts that speed up chemical reactions by lowering activation energy. Due to this, the products are formed at a faster rate.But enzymes only affect the rate of reaction without being used up, they can't alter the structure of the final product.Options A, B, and C are incorrect because enzymes function as a catalyst only in biochemical reactions.

Q43. Book lungs are present in arthropods for exchange of gases in class:

• A. Crustacea
• B. Insecta
• C. Myriapoda
• D. Arachnida✓

Explanation: Respiration in class arachnida is by gills or special structures called book lungs.So option [D] The remaining options are eliminated as follows:Option [A] crustacea are aquatic arthropods and have gills for respiration.Option [B] insecta are the largest group of animal kingdom and have trachea for respiration.Option [C] myriapoda breathe through .

Q44. Type of respiration which involves step by step breakdown of carbon chain molecules in the cell is called:

• A. External respiration
• B. Pulmonary respiration
• C. Cellular respiration✓
• D. Cutaneous respiration

Explanation: Cellular respiration is a set of metabolic reactions and processes that take place in the cells of organisms to convert biochemical energy from nutrients into adenosine triphosphate (ATP), which is used to power various cellular activities. ATP is often referred to as the "energy currency" of the cell. Cellular respiration occurs in three main stages: glycolysis, the citric acid cycle (also known as the Krebs cycle), and oxidative phosphorylation (including the electron transport chain and chemiosmosis).

Q45. End products of yeast fermentation, bacterial fermentation and anaerobic respiration are:

• A. Citric acid, lactic acid, carbon dioxide and water
• B. Ethyl alcohol, citric acid and carbon dioxide
• C. Ethyl alcohol, lactic acid, carbon dioxide and water✓
• D. Methanol, lactic acid and citric acid

Explanation: Option[A] Citric acid is the product of Aspergillus Niger fermentation, lactic acid is the product of bacterial fermentation,carbon dioxide and water are products of anaerobic respiration Option[B] Ethyl alcohol is a product of yeast fermentation but citric acid is the product of Aspergillus fermentation.Option [C] Ethyl alcohol is produced during yeast fermentation,also known as alcoholic fermentation. Lactic acid is produced as a result of bacterial fermentation and water,carbon dioxide as the products of anaerobic respiration Option[D] Methanol is a product of bacterial fermentation as well as lactic acid also but citric acid is the product of Aspergillus Niger fermentation.

Q46. In human beings, what is the function of amylase in digestion?

• A. Digestion of triglycerides
• B. Digestion of lipids
• C. Digestion of all types of food
• D. Digestion of carbohydrates✓

Explanation: Option [A] Gastric Lipase breaks down Option [b] Lipase breaks down Lipids into fatty acids and glycerolOption[c] Enzymes are specific for each substrate. Different enzymes required for the breakdown of different types of foodOption [D] Amylase is the enzyme which breaks down starch[type of carbohydrate]into sugar

Q47. Where is the ileocolic sphincter located in your body?

• A. At the junction of esophagus and stomach
• B. At the junction of stomach and small intestine
• C. At the junction of ileum and large intestine✓
• D. At the junction of small intestine and large intestine

Explanation: Option[A] cardiac sphincter is present at the junction of oesophagus and stomachOption[b] pyloric sphincter is present at the junction of stomach and small intestineOption [C] ileocolic sphincter is located at the junction of the ileum and first portion of the colon of the large intestine.Option[d] specific positions of intestines are not described in this option.

Q48. The term which is employed to the loss of appetite due to fear of becoming obese is:

• A. Obesity
• B. Anorexia nervosa✓
• C. Dyspepsia
• D. Bulimia nervosa

Explanation: Option[a] Obesity is defined as accumulation of abnormal or excessive fat that may impair healthOption [B] Anorexia nervosa is an eating disorder characterised by an abnormally low body weight and intense fear of gaining weight.Option [c] Dyspepsia means indigestion,refers to discomfort or pain that occurs in upper abdomen often after eating or drinkingOption [d] Bulimia is an eating disorder characterised by uncontrolled episodes of overeating(bingeing)

Q49. Centrioles are made up of _ microtubules:

• A. 9
• B. 27✓
• C. 3
• D. 12

Explanation: Centrioles are cylindrical structures found in animal cells and are involved in cell division. Each centriole is composed of nine microtubule triplets, arranged in a cylindrical shape. Each triplet consists of three microtubules, resulting in a total of 27 microtubules forming the structure of a centriole.

Q50. The use of living organisms in industry for the production of useful products is known as

• A. Parasitology
• B. Biochemistry
• C. Biotechnology✓
• D. Molecular biology

Explanation: The exploitation of biological process for industrial and other purposes, especially the genetic manipulation of microorganisms for the production of useful products.

Q51. Treatment by using attenuated culture of bacteria is called

• A. chemotherapy
• B. sterilization
• C. antisepsis
• D. vaccination✓

Explanation: Vaccination is the administration of a vaccine to help the immune system develop immunity from a disease. For Bacterial Vaccines, a pathogen strain is cultured and inactivated to produce a “whole-cell” vaccine, an attenuated bacterium is used (BCG), or pathogen bacterial strains is cultured to produce inactivated and purified toxins or virulence factors.

Q52. The major cause of hepatitis B is _.

• A. Blood transfusion✓
• B. Blood clotting
• C. Absence of fibrinogen
• D. Contaminated soil

Explanation: Hepatitis B is a potentially life-threatening liver infection caused by the hepatitis B virus (HBV). It is a major global health problem. Hepatitis B is spread when blood, semen, or other body fluids from a person infected with the virus enters the body of someone who is not infected. Hepatitis B is primarily transmitted through contact with infected blood or other body fluids through blood transfusion

Q53. During animal cell division, the spindle fibres are formed from

• A. Mitochondria
• B. Centrioles✓
• C. Ribosomes
• D. Lysosomes

Explanation: At the beginning of the nuclear division, two wheel-shaped protein structures called centrioles position themselves at opposite ends of the cell forming cell poles. Long protein fibers called microtubules extend from the centrioles in all possible directions, forming what is called a spindle.

Q54. Which component of the cell is concerned with cell secretions?

• A. Plasma membrane
• B. Golgi complex✓
• C. Cytoskeleton
• D. Mitochondria

Explanation: The subtances that need to be secreted out of cell are processed and closed in vesicles by Golgi complex. These vesicles are then budded off from plasma membrane.

Q55. During which period of interphase (cell cycle) DNA is synthesized?

• A. G1
• B. G2
• C. S✓
• D. G0

Explanation: The S phase of a cell cycle occurs during interphase, before mitosis or meiosis, and is responsible for the synthesis or replication of DNA. In this way, the genetic material of a cell is doubled before it enters mitosis or meiosis, allowing there to be enough DNA to be split into daughter cells

Q56. The soluble part of cytoplasm or fluid that remains when all organelles are removed is known as:

• A. Solution
• B. Gelatin material
• C. Cytoskeleton
• D. Cytosol✓

Explanation: Cytosol refers to the soluble, fluid portion of the cytoplasm that remains when all organelles are removed. It contains various molecules, ions, and enzymes necessary for cellular metabolism.

Q57. The outer membrane of the nuclear envelope is at places continuous with the:

• A. Golgi Apparatus
• B. Endoplasmic Reticulum✓
• C. Lysozomes
• D. Peroxisomes

Explanation: The outer nuclear membrane is continuous with the endoplasmic reticulum, so the space between the inner and outer nuclear membranes is directly connected with the lumen of the endoplasmic reticulum.

Q58. _ is most abundant carbohydrate in nature.

• A. Waxes
• B. Glycerol
• C. Starch
• D. Cellulose✓

Explanation: Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom.

Q59. Amino acid in which the R-group is hydrogen is:

• A. Glycine✓
• B. Alanine
• C. Leucine
• D. Valine

Explanation: Glycine is the simplest stable amino acid with the molecular formula C2H5NO2.

Q60. Peptidoglycan or murein is a special or distinctive feature of cell wall in

• A. algae
• B. fungi
• C. bacteria✓
• D. plants

Explanation: The peptidoglycan (murein) sacculus is a unique and essential structural element in the cell wall of most bacteria. Peptidoglycan or murein is a unique large macromolecule, a polysaccharide, consisting of sugars and amino acids that forms a mesh-like peptidoglycan layer outside the plasma membrane, the rigid cell wall characteristic of most bacteria

Q61. The most critical phase of mitosis which ensures equal distribution of chromatids in the daughter cells is:

• A. Prophase
• B. Metaphase
• C. Anaphase✓
• D. Telophase

Explanation: The third stage of cell division, between metaphase and telophase, during which the chromosomes move away from one another to opposite poles of the spindle.

Q62. The intake of liquid materials across the cell membrane is

• A. Phagocytosis
• B. Endocytosis
• C. Pinocytosis✓
• D. Exocytosis

Explanation: Pinocytosis (“pino” means “to drink”) is a process by which the cell takes in the fluids along with dissolved small molecules. In this process, the cell membrane folds and creates small pockets and captures the cellular fluid and dissolved substances.

Q63. Which one of the following is the site of oxidative phosphorylation in mitochondria?

• A. Cristae✓
• B. Matrix
• C. Outer membrane
• D. Ribosomes

Explanation: The site of oxidative phosphorylation in mitochondria is the inner mitochondrial membrane. While the inner mitochondrial membrane is the primary site of oxidative phosphorylation, it is essential to mention that the specific structures within the inner membrane called cristae play a crucial role in this process. Cristae are the infoldings or protrusions of the inner mitochondrial membrane, and they significantly increase the surface area available for the various proteins involved in oxidative phosphorylation. Within the inner mitochondrial membrane, there is a series of protein complexes collectively known as the electron transport chain (ETC). This is where the transfer of electrons takes place, and the energy released during this process is used to pump protons across the inner mitochondrial membrane, creating a proton gradient. The enzyme ATP synthase, also located in the inner mitochondrial membrane, utilizes the energy from the proton gradient to synthesize ATP from ADP and inorganic phosphate. This entire process of electron transport and ATP synthesis is referred to as oxidative phosphorylation.

Q64. The most common respiratory substrate as a source of energy is

• A. Glucose✓
• B. Sucrose
• C. Fructose
• D. Insulin

Explanation: Glucose is the most common respiratory substrate. One molecule of glucose produces 38 molecules of ATP, so it's an instant energy source.

Q65. If the genetic code is made up of three nucleotides, then total possible genetic codes will be

• A. 4
• B. 20
• C. 64✓
• D. 61

Explanation: Each group of three nucleotides encodes an amino acid. The total number of genetic codes can be found by 4^3 = 64, where 4 is the total number of nucleotide bases and 3 is the number of bases that can be in one genetic code.

Q66. In translation the terminating codon is

• A. GUA
• B. UAA✓
• C. UUG
• D. AGU

Explanation: The presence of a stop codon—UAA, UAG, or UGA—in the A site of the ribosome is generally a signal to terminate protein synthesis. This process constitutes the last essential stage of translation, as it ensures the formation of full-sized proteins.

Q67. The competitive inhibitors have structural similarity with

• A. Active site
• B. Binding site
• C. Substrate✓
• D. Coenzyme

Explanation: The competitive inhibitor has structural similarity with the substrate, hence it interferes with the substrate binding to the enzyme molecule.

Q68. A cofactor tightly bound to the enzyme on the permanent basis is called

• A. Activator
• B. Coenzyme
• C. Prosthetic group✓
• D. Apo-enzyme

Explanation: Prosthetic groups are cofactors very tightly bound to the inactive enzyme (called apoenzyme), even covalently in some cases. Examples of them are the biotin and heme groups, cytochrome C, flavin mononucleotide, and FAD.

Q69. The ability to distinguish between two separate points/objects is:

• A. Magnification
• B. Fractionation
• C. Centrifugation
• D. Resolution✓

Explanation: Scientists think of resolution as the ability to tell that two objects that are very close together are distinct objects rather than just one.Fractionation is a separation process in which a certain quantity of a mixture (gas, solid, liquid, enzymes, suspension, or isotope) is divided during a phase transition, into a number of smaller quantities (fractions) in which the composition varies according to a gradient.Centrifugation is a technique used for the separation of particles from a solution according to their size, shape, density, viscosity of the medium and rotor speed. The particles are suspended in a liquid medium and placed in a centrifuge tube. The tube is then placed in a rotor and spun at a defined speed.Magnification is the process of enlarging the apparent size, not physical size, of something

Q70. They serve as a scaffold for formation of higher order chromatin structure:

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

Explanation: Nucleosomes are the basic packing unit of DNA built from histone proteins around which DNA is coiled. Nucleosomes are connected by short DNA segments (termed 'linker DNA') into nucleosomal arrays, which undergo short-range interactions with neighbouring nucleosomes to form chromatin fibres. Subsequent fibre–fibre interactions contribute to the high degree of compaction observed in the condensed chromosome. They serve as a scaffold for formation of higher order chromatin structure as well as for a layer of regulatory control of gene expressio.

Q71. Which one of the following acts as functional unit of lungs in man?

• A. Air sac✓
• B. Trachea
• C. Larynx
• D. Bronchioles

Explanation: Option [A] Air sacs,alveoli are present at the end of bronchioles in lung where lungs and blood exchange oxygen and carbon dioxide during inhalation and exhalationOption[b] Larynx(voice box) it contains vocal cords not the part of lungsOption[c] Trachea that connects the larynx to lungs. Acts as a passage of air into lungs.( not the part of lungs)Option[d] Bronchioles are the smaller branches of the bronchi that enter each lung. Bronchioles give rise to alveoli (air sacs)

Q72. Expiration in human beings is carried out by:

• A. Contraction of lungs
• B. Contraction of intercostal membrane
• C. Relaxation of intercostal and diaphragm muscles✓
• D. Contraction of diaphragm muscles

Explanation: Option [a] Lungs do not contractOption[b] Internal intercostal muscles contractOption [C] During expiration External intercostal muscles relax and internal intercostal muscles contract bringing rib cage down to its normal position, muscles of diaphragm Also relax with chest cavity also being reduced Option [d] Diaphragm muscles relaxes during expiration,contraction of diaphragm does not occur

Q73. Granulocytes or white blood cells are produced in:

• A. Lymph nodes
• B. Red bone marrow✓
• C. Tonsils
• D. Spleen

Explanation: Option[a] Lymph nodes don’t have a role in production of WBC,they filter substances that travel through lymphatic fluid.Option [B] Bone marrow makes stem cells, which then produces RBC,WBC and plateletsOption[c] The main function of tonsils is fighting infection, they contain a lot of WBC but do not produce them.Option[d] Spleen controls the levels of blood cells including WBC,RBC and platelets, it removes old or damaged RBC whereas WBC are destroyed in the lymphatic system.

Q74. The flow of lymph in lymphatic vessels is maintained by:

• A. Heart, activity of smooth muscles and valves
• B. Activity of skeletal muscles, heart and breathing movements
• C. Breathing movements, activity of skeletal muscles and valves✓
• D. Exercise, breathing movements and heart

Explanation: Option[a] heart has no role in lymph flowOption[b] Heart has no role in lymphOption [C] All mentioned factors affect the flow of lymph in lymphatic vessels i.e deep breathing helps the flow of the lymph fluid through the body,movement of skeletal muscle increases the lymph flow and valves function to bias lymph flow back towards the heart.Option[d] Heart and exercise play no role in lymph flow

Q75. The central metabolic station and clearing house of a body is:

• A. Liver✓
• B. Kidney
• C. Nephron
• D. Glomerulus

Explanation: Option [A] Liver plays a central role in all metabolic processes in the body. Responsible for metabolism of fats,proteins and carbohydrates. It is also involved in detoxification [as it filters all of the blood and breaks down poisonous substances such as alcohol and drugsOption[b] Kidney is the main excretory organ and is involved in formation of urine to excrete out all the waste products out of the body via urineOption[c] Nephron is the functional unit of kidney involved in formation of urineOption[d] Glomerulus is a network of small blood vessels in each nephron enclosed in a sac called Bowman’s

Q76. Which of the following is purine:

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

Explanation: Guanine is a purine base. It is one of the four nitrogenous bases found in DNA and RNA.

Q77. Optimum pH value for the working of pancreatic lipase is:

• A. 4.50
• B. 7.60
• C. 2.00
• D. 9.00✓

Explanation: Optimum pH values for the catalytic activity of pepsin, sucrose, catalase and pancreatic lipase are 2.00, 4.50, 7.00 and 9.00, respectively.

Q78. In the light independent stage of photosynthesis, the CO2 combines with _ to form an unstable 6-carbon intermediate.

• A. Ribulose bisphosphate✓
• B. Hexose sugar
• C. Glycerate-3-phosphate
• D. Glyceraldehyde-9-phosphate

Explanation: In the light-independent stage of photosynthesis, also known as the Calvin cycle or dark reactions, carbon dioxide (CO2) combines with a five-carbon sugar called ribulose-1,5-bisphosphate (RuBP) to form an unstable 6-carbon intermediate. The enzyme responsible for this initial reaction is ribulose-1,5-bisphosphate carboxylase/oxygenase, commonly abbreviated as Rubisco.

Q79. Malate is oxidized by _ to oxaloacetate in Krebs’s Cycle.

• A. ATP
• B. NADP
• C. NAD✓
• D. FAD

Explanation: NAD (nicotinamide adenine dinucleotide) is involved in the oxidation of malate to oxaloacetate in the Krebs's Cycle. NAD acts as a coenzyme that accepts hydrogen atoms (H+) and electrons (e-) during the redox reactions in cellular respiration, including the Krebs's Cycle.

Q80. What is the percentage of lipids in the chemical composition of plasma membrane?

• A. 60-80%
• B. 20-40%✓
• C. 40-80%
• D. 20-60%

Explanation: The plasma membrane is the outermost covering of an animal cell. Cell membrane is composed of lipids and proteins;60-80% are proteins, while 20-40% are lipids.

Q81. A complex network of channels extending from plasma membrane to nuclear membrane:

• A. Mitochondria
• B. Lysosomes
• C. Endoplasmic reticulum✓
• D. Golgi Apparatus

Explanation: The endoplasmic reticulum is a continuous membrane system that forms a series of flattened sacs within the cytoplasm of eukaryotic cells and serves multiple functions including the synthesis, folding, modification, and transport of lipids and proteins.

Q82. The group of ribosomes joined by mRNA is called as:

• A. Monosome
• B. Polysomes✓
• C. Polypeptide
• D. Lysosome

Explanation: Polysome is defined as the cluster of ribosomes held together by a strand of mRNA which each of them is translating. •Polysomes/Polyribosomes can be found in three forms:free,cytoskeletal bound and membrane bound.

Q83. Which organelle has two Subunits:

• A. Endoplasmic reticulum
• B. Ribosomes✓
• C. Nucleus
• D. Plastids

Explanation: Ribosomes are made of proteins and ribonucleic acid. Ribosomes consist of two major components: the small and large ribosomal subunits. The tinier subunit is the place the mRNA binds and it decodes, whereas the bigger subunit is the place the amino acids are included.

Q84. A condition called Goose pimples, are caused by:

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

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

Q85. Piriformis syndrome is associated with which of the following disorder:

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

Explanation: Sciatica refers to pain, weakness, numbness, or tingling in the leg. It is caused by injury to or pressure on the sciatic nerve. The sciatic nerve starts in the lower spine and runs down the back of each leg.Common causes of sciatica include; slipped disc, piriformis syndrome (a pain disorder involving the narrow muscle in the buttocks), pelvic injury or fracture, and tumors.

Q86. It is an endoparasite of humans, cattle and pig that completes its life cycle in two hosts:

• A. Tapeworm✓
• B. Liver fluke
• C. Aurelia
• D. Aurelia

Explanation: Explanation: The correct answer is A because tapeworm or taenoia is a endoparasite of humans, cats, and pigs that complete its life cycle in two hosts. This option is not correct because liverfluke is an endoparasite of sheep and occasionally humans. Aurelia is a genus of jellyfish, also known as moon jellies, that are found in oceans worldwide. They are not parasitic and donot infect humans or other animals.

Q87. A compound was found to contain nitrogen and oxygen in the ratio 28g:80g. The formula of the compound is:

• A. NO
• B. N2O3
• C. N2O4
• D. N2O5✓

Explanation: Mass of nitrogen is 14 g and mass of oxygen is 16 g. Dividing nitrogen’s mass in compound (28g) by 14g gives us 2. Dividing oxygen’s mass in compound (80g) by 16g gives us 5. So we can now say that the molar ratio of N to O is 2:5 that means the formula is N2O5

Q88. 78 g of a hydrocarbon occupies 22.414 dm3 of volume at S.T.P. The empirical formula of the hydrocarbon is CH. The molecular formula of this hydrocarbon is:

• A. C2H4
• B. C6H6✓
• C. C8H8
• D. C4H4

Explanation: Molar mass of CH =12+1 = 13gMolar mass of the hydrocarbon = 78g as 22.414dm3 is equal to 1 mol.Molecular formula = n * Empirical formula78 = n * 13n = 6So the molecular formula is 6(CH) = C6H6 i.e. option B.

Q89. C12H22O11 + conc. H2SO4 ---> 12 C + ? + H2SO4

• A. 11 H2O✓
• B. 11 O2
• C. 11 CO2
• D. 11 H2

Explanation: Comparing reactant and product side we can see that there is a difference of 22 H and 11O.That means the compound that we can make of it is 11H2O. Hence the answer is A.

Q90. 22.4 dm3 of volume of each H2 and O2 are sparked to produce water vapours (considered as ideal gas) on completion of reaction. What is the decrease in the volume of the vessel?

• A. 44 dm3
• B. 22.4 dm3
• C. 33.6 dm3
• D. 11.2 dm3✓

Explanation: Here we see that 2 moles of H2 react with one mole of O2. We know, volume of one mole of ideal gas is 22.4dm3. 22.4dm3 of H2 reacts with 11.2dm3 of O2. So, volume of remaining O2 = 22.4 dm3- 11.2dm3 = 11.2dm3.So, final volume in the vessel is of O2 which is 11.2 dm3 and of water vapour, which is 22.4 dm3 is 33.6dm3 while initial volume was 22.4 dm3 + 22.4dm3 = 44.8dm3. Thus, decrease in volume = 44.8 - 33.6 = 11.2dm3

Q91. Which of the following has six isotopes?

• A. Manganese
• B. Tin
• C. Calcium✓
• D. Carbon

Explanation: Calcium has six stable isotopes: 40, 42, 43, 44, 46, 48 Manganese has four stable isotopes: 52, 53, 54, 55 Tin has ten isotopes: 112, 114, 115, 116, 117, 118, 119, 120, 122, 124 Carbon has three isotopes: 12, 13, 14. So option C is correct.

Q92. How many molecules are there in 2.1 moles of CO₂:

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

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

Q93. Avogadro's constant is the number of:

• A. Electrons present in 2g of H
• B. Atoms in 24g of Mg✓
• C. Atoms in 1g of He
• D. Molecules in 35.5g of Chlorine

Explanation: Avogadro's Number (NA) is the number of atoms, molecules, ions or anything in one mole of the respective substance. As 6.02 x 1023 atoms, molecules, ions or electrons are present in one mole; therefore, this number is known as Avogadro's number. Only 24g of Mg equals one mole of Mg. Thus, only 24g of Mg has Avogadro's number of atoms. 1g of He equals 1/2 mole of He. Thus it has half the number of atoms than Avogadro's number. 35.5g of chlorine equals 1/2 mole of chlorine gas. Thus, it has half the number of molecules than Avogadro's number. 2g of H equals two moles. Each mole of H atoms has 1 mole of electrons only, so two moles of H have twice the number of electrons than Avogadro's number.

Q94. One mole of substance at STP has:

• A. 2.24 dm³
• B. 24.4 dm³
• C. 2.44 dm³
• D. 22.4 dm³✓

Explanation: Under Standard Temperature and Pressure (STP), one mole of an ideal gas occupies 22.4 dm3. Under room conditions/ Normal Temperature and Pressure (NTP), one mole of an ideal gas occupies 24 dm3.

Q95. The mass number of an atom is:

• A. The number of electrons
• B. The number of neutrons
• C. The number of protons
• D. The number of neutrons and protons✓

Explanation: The mass number of an atom is defined as the sum of protons and neutrons that are present inside the nucleus of an atom. For example, the mass number of carbon is 12.01 amu as it has 6 protons and 6 neutrons in its nucleus.

Q96. Solid NaCl is a bad conductor of electricity because:

• A. Solid NaCl is covalent
• B. In the solid state, there are no ions
• C. In solid NaCl. there is no migration of ions✓
• D. In solid NaCl, there are not electron

Explanation: Conduction of electricity occurs when there is a free-moving charge (electrons or ions), so the answer is C.

Q97. Which of the following solids is an example of a substance with a macromolecular structure?

• A. Aluminium chloride
• B. Ice
• C. Magnesium oxide
• D. Silicon (IV) oxide✓
• E. Sodium chloride

Explanation: SiO2 has a macromolecular structure only among the compounds above. While other compounds possess a simple molecular structure, hence D is the correct option.

Q98. When does a gas deviate the most from its ideal behavior ?

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

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

Q99. When an element exists in more than one crystalline form, the phenomenon is termed as:

• A. Isomorphism
• B. Allotropy✓
• C. Isomerism
• D. Anisotropy
• E. Enthalpy

Explanation: Property of some chemical elements to exist in two or more different forms, in the same physical state is known as allotropy.

Q100. 950 torr corresponds to:

• A. 3.5 atm
• B. 1 atm
• C. 3 atm
• D. 1.25 atm✓
• E. 2.25 atm

Explanation: 1 atm -> 760 torrX atm ->950 torrX=1.25 atm

Q101. Which of the following statements is true of 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 they show thermal and optical properties, same in all directions.They do not possess symmetry,can cleave randomly and do not have a definite shape.

Q102. A child's balloon has a volume 3.80 dm3, when the temperature is 35°C. If the balloon is put in a refrigerator and cooled to 5°c, the approximate volume of the balloon is (assume pressure inside the balloon is equal to atmospheric pressure):

• A. 3.00 dm3
• B. 3.43 dm3✓
• C. 3.08 dm3
• D. 3.25 dm3
• E. 0.54 dm3

Explanation: V1=3.8 dm3 ; T1= 35 +273=308K V2=?; T2=5+273=278K V1/T1=V2/T2 3.8/308=V2/278 V2=3.43 dm3

Q103. If the matter in a given system at a given condition is divided into two equal parts, then the value of the extensive properties will become:

• A. Double of the original value
• B. Half of the original value✓
• C. Remain the same as the original value
• D. One-fourth of the original value
• E. Heat of formation

Explanation: Thermodynamic properties can be divided into two general classes, intensive and extensive properties. Thus, if a quantity of matter in a given state is divided into two equal parts, each part will have the same value of the intensive property as the original and half the value of the extensive property.

Q104. Comparative rates of diffusion of He and SO2 will be _.

• A. 8
• B. 2
• C. 4✓
• D. 16
• E. 64

Explanation: Lets M1 be the mass of SO2 and M2 the mass of He. √M1/M2 = √64/4 = √16 = 4

Q105. The unit of viscosity is:

• A. Joule
• B. N/m2
• C. Dynes/cm
• D. Poise✓
• E. Ergs

Explanation: The cgs unit of viscosity is poise (P) whereas the SI unit of viscosity is newton-second per square meter, which is usually expressed as pascal-second.

Q106. Diameter of an atom is in the range of?

• A. 0.2m
• B. 0.2 nm✓
• C. 2 x 10^-19nm
• D. 0.2 pm

Explanation: The diameter of an atom ranges from about 0.1 to 0.5 nanometer."

Q107. Which of the following is the chemical formula of a compound indicating the relative number of atoms in the simplest ratio?

• A. Empirical formula✓
• B. Empirical formula mass
• C. Molecular formula
• D. Molecular mass

Explanation: Empirical formula shows relative numbers of atoms in the formula in the simplest ratio.

Q108. The equation shows the reaction between element X and dilute hydrochloric acid. X(s) + 2HCl(aq) -> XCl2(aq) + H2(g) What types of bonding is present in element X and in compound XCl2?

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

Explanation: Metallic bonding is a type of chemical bonding that arises from the electrostatic attractive force between conduction electrons (in the form of an electron cloud of delocalized electrons) and positively charged metal ions. The element X is metal as Metals react with halogens to form salts. The compound XCl2 is a salt and has ionic bonds. As two chlorine atoms are attached, this shows the central metal has two valence electrons in free states, so it is a metal of group 2A[Be, Mg, Ca, Sr, Ba].

Q109. Linear combination of atomic orbitals (LCAO) results in the formation of:

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

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

Q110. The raindrop acquires a spherical shape and ink spread over blotting paper due to:

• A. Surface tension✓
• B. Adhesive forces
• C. Viscosity
• D. Polarity
• E. Latent heat of vaporization

Explanation: A raindrop falling through the atmosphere forms a roughly spherical structure due to the surface tension of water. This surface tension is the “skin” of a body of water that makes the molecules stick together. The cause is the weak hydrogen bonds that occur between water molecules.

Q111. Sigma bond is formed by:

• A. Transferring the electrons
• B. Head-on overlapping of atomic orbitals✓
• C. Mutual but unequal sharing of electrons
• D. Parallel overlapping of atomic orbitals
• E. All of these options

Explanation: Sigma bond is formed by head-on overlapping of the two orbitals, as shown below:

Q112. Choose the value of the Rydberg constant among the following values:

• A. 1.09678 x 107 m-1✓
• B. 1.602 x 10-19 C
• C. 1.7588 x 1011 C
• D. 1.007 x 107 m-1

Explanation: You’ll have to rote learn this. Memorise the power and units.

Q113. A 4s orbital has:

• A. One node
• B. Two nodes
• C. Three nodes
• D. Zero nodes✓

Explanation: A node is a region in an orbital where the probability of finding an electron is zero. In a 4s orbital, there are no nodes present. The electron density is maximum at the nucleus and gradually decreases as you move away from it, but there is no specific region where the probability of finding the electron becomes zero.

Q114. Electronic configuration of K is:

• A. [Ar] 4s2
• B. [Ar] 4s1✓
• C. [Kr] 4s1
• D. [He] 4s1

Explanation: The Potassium electron configuration will be 1s22s22p63s23p64s1. The electronic configuration of Argon is 1s22s22p63s23p6

Q115. The spectrum of He is expected to be similar to that of:

• A. Hydrogen
• B. Sodium
• C. Oxygen
• D. Li+✓

Explanation: The spectrum of an atom depends on the number of electrons present in it. Here, helium has two electrons, so the spectrum of 'Li+(Z=3)' is similar to that of helium because both He and 'Li+' have two electrons.

Q116. An element has configuration 2, 8, 1. It belongs to, _.

• A. IA group and 3rd period✓
• B. IIIA group and 1st period
• C. IA group and 8th period
• D. VIIA group and 3rd period

Explanation: Elements’ position in the periodic table can be determined by its electronic configuration. Period can be determined by counting the number of shells its electrons occupy. Since the element shown occupies three shells so we can say that it is in Period III. Group can be determined number of electrons in the last shell. Since the element has 1 electron has 1 electron in its last shell so the group number of this element is 1.Hence the answer is A.

Q117. The maximum number of electrons in an atom with n=3 and l=2 is:

• A. 10✓
• B. 6
• C. 18
• D. 0

Explanation: In this case l = 2then m = −2,−1,0,+1,+2so there are 5 orbitals and each orbital contain 2 electronsSo for value n = 3 and l= 2 there are 10 electrons.The maximum number of electrons in a subshell with l=2 and n=3 is 10.

Q118. Which is the configuration of Cr?

• A. [Ar] 3d4 4s2
• B. [Ar] 3d5 4s1✓
• C. [Ar] 3d6 4s1
• D. [Ar] 3d1 4s2

Explanation: It is [Ar] 3d⁵ 4s¹This happens in Cr, as one 4s electron moves to the 3d sublevel. But why? There are two main reasons:The 3d orbital is slightly lower in energy, and minimizing repulsions in the 4s orbital by moving one of the 4s electrons to a close-lying 3d orbital minimizes the ground-state energy of chromium.Hund’s Rule: It is energetically favorable to maximize the spin state in a sublevel. Since two opposite spins result in a total spin of 0, maximizing this tends to require as many electrons of the same spin in different orbitals as possible. So, in this case, an electron moves to 3d and is unpaired, therefore maximizing the spin state.

Q119. Which of the following does not show variable valency?

• A. Mn
• B. Fe
• C. Zn✓
• D. Cr

Explanation: The electronic configuration of zinc is [Ar]3d10 4s2 It can only lose two electrons from the 4s orbital and show +2 oxidation state. It does not show variable valency. However, all other transition metals like Mn, Fe, Co have partially filled d orbitals and hence, show variable valency.

Q120. Which of the following has the maximum number of unpaired electrons?

• A. Mg2+
• B. V3+
• C. Ti3+
• D. Fe2+✓

Explanation: Practice electronic configurations of as many elements as you can and memorize the periodic table, because they won't provide you with one.

Q121. The common features, among the species CN-, CO, and NO+ are:

• A. Bond order 3 and Isoelectronic✓
• B. Bond order three and weak field ligands
• C. Bond order two and π acceptors
• D. Isoelectronic and weak field ligands

Explanation: Each of these have same number of electrons; Number of electrons in CO = 6+8 = 14 Number of electrons in CN- = 6+7+1 = 14 Number of electrons in NO+ = 7+8-1 = 14 The species having same number of electrons are called iso-electronic. Iso-electronic species have same bond order. Bond order is number of chemical bonds present between a pair of atoms. There are 3 chemical bonds between C and O in CO, C and N in CN- and N and O in NO+ so all of these have same bond order of 3. Hence, all the given species are iso- electronic and have same bond order. So, option A is correct. CO and CN- are strong field ligands, while NO+ is weak field ligand. Weak field ligands are ones that produce small splittings between the d orbitals and form high spin complexes. Hence option B and D are incorrect. Weak field ligands are all π - donors. Strong field ligands are all π - acceptors. Hence option C is also incorrect.

Q122. In an adiabatic process:

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

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

Q123. Enthalpy of a compound is equal to its:

• A. Heat of combustion
• B. Heat of formation✓
• C. Heat of solution
• D. Heat of dilution

Explanation: The enthalpy of reaction for any chemical reaction is equal to the difference between the sum of the enthalpies of the formation of the products and the sum of the enthalpies of the reactants. In an exothermic reaction, the heat content or enthalpy of the product H2 is less than that of the reactants H1. Since the system has lost heat, we can say the enthalpy change for the reaction ∆H is negative, In an endothermic reaction, the enthalpy of product H2 is greater than that of the reactants H1, and the enthalpy change, ∆H is positive. The standard enthalpy of a reaction ∆Ho is the enthalpy change that occurs when a certain number of moles of reactants as indicated by the balanced chemical equation, react together completely to give the products under standard conditions, i.e 25 °C (298K) and one atmosphere pressure. All the reactants and products must be in their standard physical states. Its units are kJ mol-1 .

Q124. 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: Increasing the pressure favors the side where the volume of gases is less. Since NH3 lies at the product side where total moles (volumes) of gases is 2 as compared to volume on the left side (reactant side) which 4. Increasing the pressure favors the NH3 production. Catalyst does not favor any side it increases the forward as well as the backward rate of reaction. Hence the answer is D.

Q125. According to the law of mass action,“The rate of chemical reaction” is proportional to the:

• A. The amount of product formed
• B. Product of the molar concentration of reactants✓
• C. The initial concentration of reactants
• D. Catalyst
• E. Pressure

Explanation: The speed at which the reactants are converted to products is called the rate of reaction. There are many factors that affect the speed of conversion of reactants into products: Nature of reaction: The state and number of reactants and complexity of reaction can affect the rate of reaction. Effect of concentration: From the law of mass action, the rate of reaction is directly proportional to the concentration of reactants, that is, the rate of reaction increases with the increase in the concentration of reactants. Pressure: The concentration of reactants can be increased by pressure. Hence, when pressure is increased the speed of reaction also increases. Catalyst: It can increase speed in both reverse and forward directions. This gives another path having low activation energy. Temperature: The reaction which takes place at a high temperature has more energy than a reaction at a low temperature. Now, we have to know what the law of mass action states which will give us a clearer answer to this question. The law of mass action states that, at a given temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants with their powers raised to their stoichiometric coefficient. The molar concentration of the reactant is also known as the: Q. Hence the correct option is B.

Q126. When 18 x 10-3 moles/dm3 of acetic acid react with 22 x 10-3 moles/dm3 of ethyl alcohol to form 40 x 10-3 moles/dm3 of ethyl acetate and 40 x 10-3 moles/dm-3 of water. Find the value of equilibrium constant (kc):

• A. 4.04✓
• B. 3.14
• C. 3.04
• D. 2.02
• E. 1.04

Explanation: Kc=[40x10-3][40x10-3]/[18x10-3][22x10-3]Kc=4.04

Q127. The enthalpy change accompanying the gain of an electron by a neutral gaseous atom to form a negative ion is called:

• A. Ionization potential
• B. Electronegativity
• C. Electron affinity✓
• D. Lattice energy
• E. Potential energy

Explanation: When an electron is added to a neutral gaseous atom to convert it into a negative ion, the enthalpy change accompanying the process is called Electron affinity enthalpy. Ionization potential, in chemistry, is the amount of energy required to remove an electron from an isolated atom or molecule. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Lattice energy is defined as the energy required to separate a mole of an ionic solid into gaseous ions. Potential energy is energy that is stored – or conserved - in an object or substance.

Q128. The heat of a reaction can be calculated by using:

• A. Joule's Law
• B. Ohm's Law
• C. Hess's Law✓
• D. Faraday's Law
• E. Boyle's Law

Explanation: Hess' law states that the change of enthalpy in a chemical reaction (i.e. the heat of reaction at constant pressure) is independent of the pathway between the initial and final states. In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final conditions are the same). Hess’s law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly.

Q129. Which of the following molecules have zero Dipole moments?

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

Explanation: Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.

Q130. Bond energy: I. is the energy required to break a bond between two atoms in a diatomic molecule II. Is taken as the energy released in forming a bond form free atoms III. Is the measure of the strength of bond

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

Explanation: All of the above statements are correct, it is because bond energy is the amount of energy required to break a bond as well when this bond is formed the same amount of energy is released. It thus provides us with an estimate of the strength of the bond.

Q131. The measurement of heat absorbed or given out in a chemical reaction is referred to as:

• A. Endothermic reaction
• B. Exothermic reaction
• C. Thermochemistry✓
• D. Heat of formation

Explanation: Thermochemistry is the branch of chemistry concerned with the quantities of heat evolved or absorbed during chemical reactions.Endothermic and Exothermic reactions are the type of reactions that are characterized by whether the heat is evolved or absorbed during these processes. The heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements.

Q132. Cl + e ----> Cl- ΔH = -348kJ/mol The value -348KJ/mol in this case will be:

• A. Ionization energy
• B. Electronegativity
• C. Electron affinity✓
• D. Entropy
• E. Free energy

Explanation: As it can be seen in the equation, electrons are gained by the element chlorine to form a negative ion (anion). The process in which electrons are gained is called electron affinity.

Q133. H2S is an example of _ hydride.

• A. Ionic
• B. Covalent✓
• C. Complex
• D. Metallic
• E. Border - line hydride

Explanation: Covalent hydrides are formed when hydrogen reacts with other similar electronegative elements like Si, C,etc. H2S is a covalent hydride because the two elements used in the formation of this hydride are non-metals.

Q134. Sp3 hybridization in CH4 gives it _ geometry.

• A. Linear
• B. Coplanar
• C. Tetrahedral✓
• D. Octahedral
• E. Trigonal pyramid

Explanation: The image below explains the question above.

Q135. Dissolution of ammonium nitrate in water is an endothermic process. Which of the following graphs shows how the temperature alters as the ammonium nitrate is added to water and then the solution is left at room temperature?

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

Explanation: Since it is an endothermic process it takes heat from the surrounding making the temperature fall below room temperature. The temperature then gradually then goes back to room temperature.

Q136. Which of the following statements is NOT true for the first law of thermodynamics?

• A. Total energy of the system and surrounding is conserved
• B. Energy can neither be created nor destroyed
• C. It is the same as law of conservation of energy
• D. Total energy of the system is increasing✓

Explanation: The total energy of the system remains constant. Hence, statement D is not correct.

Q137. The boiling point of water is higher than other hydrides because water molecules can form:

• A. 4 hydrogen bonds✓
• B. 3 hydrogen bonds
• C. 2 hydrogen bonds
• D. 1 hydrogen bonds

Explanation: Intermolecular hydrogen bonding is responsible for the high boiling point of water (100 °C) compared to the other group 16 hydrides that have no hydrogen bonds. For example if we take comparison of fluorine and H20,Fluorine is more electronegative than Oxygen, so technically, HF should engage in more hydrogen bonding than water. But the situation is different. Each HF molecule possesses three lone pairs of fluorine and one hydrogen. HF molecules on average make two hydrogen bonds. Each H2O molecule possesses two lone pairs of oxygen. H2O molecules are thus able to form an average of four hydrogen bonds. Hence, water molecules take part in more hydrogen bonding than Hydrofluoric acid. Water molecules form a bulky molecule and it is very difficult to break its bonds. To break all its bonds, a large amount of the energy is required. Thus, H2O has a higher boiling point than HF. Thus h20 can form four hydrogen bonds.

Q138. In a reaction: A+B < - > 2C When equilibrium was attained, the concentration was [A] = [B] = 4 moles/dm3 [C]= 6 moles/dm3The equilibrium constant Kc of this reaction is:

• A. 1.25
• B. 2.25✓
• C. 3.25
• D. 2.75
• E. 3.75

Explanation: We know Kc is the product of concentration of products raised to the power in a balanced chemical equation divided by the product of concentration of reactants raised to the power in a balanced chemical equation.For given reaction,Kc = [C]² / [A][B]Kc = 6² / 4x4Kc = 36 / 16Kc = 2.25

Q139. If the ratio of initial concentration of the reagents is greater than the Kc then:

• A. The reaction will shift towards the reverse direction✓
• B. More quantity of products is obtained
• C. The ratio increases to the value of Kc
• D. Equilibrium has been attained
• E. There is no shifting of reaction

Explanation: When the ratio of the initial concentration of the reagents is greater than the Kc the reaction shifts towards the reverse direction. The opposite is true for forwarding reactions.

Q140. A bond is not formed:

• A. When both forces become equal to each other
• B. When attraction forces dominate repulsive forces
• C. When repulsive forces become equal to zero
• D. When repulsive forces dominate attraction forces✓

Explanation: Explanation: Bond is formed when either the attraction and repulsion are equal or the force of attraction is greater than repulsion. As a result, the bond is not formed when the repulsion is greater than attraction. Bond is formed when both forces become equal to each other. Bond is also formed when attraction forces dominate repulsive forces. Bond is also formed when repulsive forces becomes equal to zero.

Q141. A player throws a ball at an initial velocity of 36 m/second. The maximum distance the ball can reach (assume ball is caught at the same height at which it was released) is:

• A. 146 m
• B. 130 m
• C. 132 m✓
• D. 129 m
• E. 145 m

Explanation: A football thrown at 45 degree has max range:Range = Vo2sin(2A)/g=362*sin(90)/9.8 = 132.2 m.

Q142. A particle is projected at an angle of 45 degrees with a velocity of 9.8 m/s. The horizontal range will be: (g=9.8 m/s2)

• A. 9.8 / √2 m
• B. 9.8 √3 m
• C. 9.8 m✓
• D. 4.9 m
• E. 3.1 m

Explanation: Attached below is the solution to the question above:

Q143. 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: Explanation is as follows.

Q144. A constant force acting on a body of mass 5 kg changes its speed from 2 /s to 7 m/s in 10 s the direction of motion of the body remains unchanged. Find the magnitude of the force.

• A. 0.5 N
• B. 1.5 N
• C. 2.5 N✓
• D. 3.5 N
• E. 4.5 N

Explanation: F= m(v-u)/t= 5(7-2)/10= 2.5N

Q145. What force should be applied on a 10 kg body so that it moves down in a vacuum with an acceleration of 3 m/s2 ? (Take g=9.8 m/s2)

• A. 42 N
• B. 46 N
• C. 68 N✓
• D. 53 N
• E. 58 N

Explanation: The question wants the object to only move down at 3 m/s2, we need to apply an upward force that would accelerate the object downwards at 9.8–3=6.8 m/s2. Using F=ma, where m = 10 kg and a = 6.8 m/s2, we find F to be 68 N upwards.

Q146. .A rock is thrown straight upward from the edge of a 30 m cliff, rising 10 m then falling all the way down to the base of the cliff. Find the rock's displacement.

• A. 20 meters downward
• B. 30 meters downward✓
• C. 40 meters upward
• D. 50 meters upward
• E. 60 meters upward

Explanation: The displacement is the distance of the rock from the initial position, the initial position of the rock was at the cliff which was 30m. So the displacement of the rock is 30m downwards, although the distance traveled is 50m in total.

Q147. .A stone dropped from a certain height can reach the ground in 5 s. It is stopped after 3 seconds of its fall and then allowed to fall again. Find the time taken by the stone to reach the ground for the remaining distance.

• A. 2 s
• B. 4 s✓
• C. 6 s
• D. 8 s
• E. 10 s

Explanation: The explanation is given below.

Q148. A moon of mass ‘m’ orbits a planet of mass 100 m. Let the strength of the gravitational force exerted by the planet on the moon be denoted by F1 , and let the strength of the gravitational force exerted by the moon on the planet be F2 . Which of the following is true?

• A. F1 is ten times greater than F2
• B. F1 is ten times smaller than F2
• C. F2 is ten times greater than F1
• D. F2 is ten times smaller than F1
• E. F1 is equal to F2✓

Explanation: The gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Since the mass of the planet is 100 times the mass of the moon, the strength of the gravitational force exerted by the planet on the moon will be much greater than the strength of the gravitational force exerted by the moon on the planet. However, the forces will still be equal in magnitude due to Newton's third law.

Q149. Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45° to the horizontal?

• A. The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.
• B. The speed at the top of the trajectory is zero.
• C. The object's total speed remains constant during the entire flight.
• D. The horizontal speed decreases on the way up and also decreases on the way down.
• E. The vertical speed decreases on the way up and increases on the way down.✓

Explanation: The only external force acting on the object is the gravitational force vertically downwards. Thus the acceleration of the projectile is g downwards always. Thus there is no change in the speed of the object in the horizontal direction. The speed of an object in the vertical direction at the top of the trajectory is zero. However, the net speed at the point is equal to the constant horizontal component. Due to downward constant acceleration, the vertical speed decreases way up, becomes zero, and then increases way down. Thus option E is correct.

Q150. If a car collides with a housefly, what will be the magnitude of the force experienced by the housefly?

• A. Much greater than the force experienced by the car
• B. Much lesser than than the force experienced by the car
• C. Same as the force experienced by the car✓
• D. 10 times less than the force experienced by the car

Explanation: Yes, the same amount of force is imparted on each of them at the moment of the collision. However, one is affected much more than the other.Instead of thinking in terms of accelerating cars and flies, assume each has a constant velocity. At the moment of impact, each of them undergoes an acceleration (in the negative direction, so we usually call it deceleration) proportional to the force F of the impact. F is the same for both objects.Now, according to Newton's third law, at the moment of impact, the force that the car imparts on the fly is exactly equal to the force that the fly imparts on the car. Also, according to Newton's second law, the force is equal to mass times acceleration. As such, the train hits the fly with a massive amount of force, equal to the acceleration times the mass of the train.Results:The magnitude of force experienced by the housefly is the same as that experienced by the car neither less nor more because Newton's 3rd law is applicable here.

Q151. 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

Q152. A battery of 6 volts is connected to two resistors of 3 ohm and 2 ohm joined together in series. What will be the current through the circuit?

• A. 6.2 A
• B. 5.2 A
• C. 4.2 A
• D. 2.2 A
• E. 1.2 A✓

Explanation: Resistance of resistors placed in series is calculated by adding the resistance,3+2=5 ohms.Formula connecting voltage,resistance and current is V=IR.V=IR(6)=(5)(I)6/5=I=1.2 Ohms.

Q153. 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

Q154. A circuit in which there is a current of 5 amp is changed so that the current falls to zero in 0.1 s. If an average e.m.f of 200 volts is induced, what is the self inductance of the circuit?

• A. 1 Henrys
• B. 2 Henrys
• C. 3 Henrys
• D. 4 Henrys✓
• E. 6 Henrys

Explanation: Initial current, I1=5.0AFinal current, I2=0AChange in current, dl=I1−I2=5ATime taken for the charge, t=0.1sAverage emf, e=200VFor self inductance(L) of the coil, we have the relation for average emf as:e=L x rate of change of currentL=e/rate of change of currentL=200/ 5/0.1L=4HHence the self inductance of the coil is 4H.

Q155. 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.

Q156. If the body is rotating with uniform angular velocity, then its torque is:

• A. Zero✓
• B. 90
• C. 1
• D. -1

Explanation: A body moving with uniform angular velocity is having zero Angular acceleration so since torque = force * moment arm (where F=ma and a here is zero as object s not accelerating. ) so F=0 So T=0Torque is the moment of force about a body. T = F . rAngular velocity = w = d z /dt Linear velocity = v = rwlinear acceleration=a = r .[ d w/ dt ]F = ma = mr [ dw/dt]T = F.r = mr^2 [dw/dt] , mr^2 = I , the moment of inertiaT = I . [dw/dt]Now if w = angular velocity is a constant , then dw/dt = 0So that T = I[0] =0

Q157. A rescue helicopter drops a package of emergency ration to a stranded party on the ground. If the helicopter is traveling at 40 m/s at a height of 100 m above the ground, where does the package strike the ground relative to the point at which it was released? (g= 9.8 m/s2)

• A. 120 m
• B. 130 m
• C. 140 m
• D. 180.7 m✓
• E. 200.3 m

Explanation: H = 1/2gt² 100 = 1/2 (9.8) t² t² = 100×2/9.8 t² = 100/4.9 t² = 20.4 t = 4.5 X = Vt X = 40×4.5 X = 180.7

Q158. A shot leaves a gun at the rate of 160m/s. Calculate the greatest distance to which it could be projected and the height to which it would rise: (g=10 m/s2)

• A. 1560m, 540m
• B. 2560m, 640m✓
• C. 3560m, 740m
• D. 4560m, 840m
• E. 9595m, 348m

Explanation: Following is the solution to this question:

Q159. Law of conservation of momentum states that: I. there is no external force applied to a system, then the total momentum of that system remains constant II. if there is an external force applied to a system, then the total momentum of that system remains constant III. if there is no external force applied to a system, then the total momentum of that system keeps changing

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

Explanation: If there is no external force applied to the system, then the total momentum of that system remains constant.If an external force is applied then the total momentum will change.

Q160. 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: The cannon ball launched at a 45° -degree angle had the greatest range.The cannonball launched at a 60° -degree angle had the highest peak height before falling.The cannonball launched at the 30-degree angle reached the ground first.

Q161. Example(s) of spin motion is/are:

• A. The daily rotation of the earth about its own axis
• B. Jumping of a paratrooper from an helicopter
• C. Flow of a viscous liquid
• D. Rotation of flywheel about its axle
• E. Both the daily rotation of the earth and flywheel about its own axis✓

Explanation: Spin motion is the movement of an object about its own axis, which is depicted by the daily motion of the earth and flywheel rotation about its own axis.

Q162. 1 radian = _ degrees.

• A. 360
• B. 180
• C. 100
• D. 57.3✓
• E. 1.01745

Explanation: Radian is the SI unit of measuring angles. It is the measure of the central angle whose arc length is the same as the radius of the circle.In one circle,2π radians =3601 radian is equal to 180/π degrees, which is approximately 57.3°

Q163. The following question belongs to the topic of momentum:

• A. Current
• B. Length
• C. Angular momentum✓
• D. Torque
• E. Displacement

Explanation: In three dimensions, the angular momentum for a point article is a pseudovector r × p, the cross product of the particle's position vector r (relative to some origin), and its momentum vector.

Q164. To double the total energy for a mass-spring system oscillating in SHM, by what factor must the amplitude increase?

• A. 4
• B. 2
• C. √2 = 1.414✓
• D. ∜2 = 1.189

Explanation: This is the following solution.

Q165. Which graph in SHM shows the K.E of a body:

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

Explanation: The K.E of an oscillating body is maximum at the equilibirum position because speed is maximum at that point, and is 0 at the extreme positions because speed is 0 at those points. Graph B depcits the P.E, while Graph C depicts the TOTAL energy.

Q166. The time period of a seconds pendulum on the surface of moon becomes:

• A. 2√ 6 seconds✓
• B. 1 seconds
• C. 2 seconds
• D. 6√ 2 seconds

Explanation: Refer to the solution presented below;

Q167. For a simple harmonic motion the acceleration of the body is _ to displacement and always directed towards _.

• A. Directly proportional ; maximum position
• B. Inversely proportional ; maximum position
• C. Directly proportional ; mean position✓
• D. Directly proportional ; minimum position

Explanation: The two conditions that have to be met for an oscillatory motion to be simple harmonic are:1- The acceleration is directly proportional to displacement.2- The displacement is always directed opposite to the direction of acceleration, i.e towards the mean position.

Q168. At which of the following places, motion of simple pendulum becomes slowest:

• A. Murree
• B. Karachi
• C. K-2✓
• D. Peshawar

Explanation: The period of a simple pendulum is T = 2π √L/g , where L is the length of the string and g is the acceleration due to gravity. At mountain regions the value of g falls, which causes the time period to increase due to the inverse proportion, hence slowest motion will be recorded at the place of highest altitude, which in this case is Mountain K-2.

Q169. Equation of SHM, with amplitude ‘a’ is given by:

• A. X = a(sin2 ω t + cos2 ω t)
• B. X = a(sin ω t cos2 ω t)
• C. X = a sin ω t✓
• D. X = a2 sin (sin ω t)

Explanation: X = a sin ω t is the standard equation of SHM, in which a corresponds to amplitude.

Q170. The six strings of a guitar are the same length under nearly the same tension, but they have different thicknesses. On which string do waves travel the fastest?

• A. The thickest string
• B. The thinnest string✓
• C. The wave speed is the same on all the strings
• D. None of the above

Explanation: The velocity of a transverse wave in a stretched string is given by: v = √(T/n). Where T is the Tension and n is the mass per unit lenth n = m/L = ρπr^2/L where r is the radius of the wire r = d/2 where d is the diameter So, v = √ (4T/ρπd^2) v= 2/d √(T/ρπ) As it can be seen, v is inversely proportional to d Thus, the velocity of wave is highest for thinnest string.

Q171. Two point charges of +5 C and -12 C attract each other with a force of 1.48 N. charge of -5 C is added to each of these charges. Now the force will be:

• A. 1.48 N (attractive)
• B. 1.48 N (repulsive)
• C. 2.96 (repulsive)
• D. Zero✓

Explanation: This question does not require any calculation.Just know that if we add -5C to both, it will leave the +5C point charge to become a charge of 0C, in other words without any charge, hence there will neither be an attractive or repulsive force in between.

Q172. A ship approaching a seaport at 10 km/h sends a sonic signal of frequency of 1000 Hz. What is the apparent frequency of the signal as received by a receiver on the seaport?

• A. 984 Hz
• B. 1000 Hz
• C. 1008 Hz✓
• D. 1210 Hz

Explanation: Frequency = 1000 HzSpeed of ship = 10 km/h = 2.78 m/sSonic signal speed = 340 m/sApparent frequency is found as:f’ = vsound / (vsound - vship) x ff’ = 340 / (340 - 2.78) x 1000f’ = 1008 Hz

Q173. In a ripple tank, 40 waves pass through a certain point in one second. If the wavelength of the waves is 5 cm, then find the speed of the wave:

• A. 2.7 m/s
• B. 3 m/s
• C. 200 m/s
• D. 2 m/s✓

Explanation: The formula for the speed of a wave is v=fλ Since 40 waves pass through a certain point in one second, the frequency of the wave is 40 Hz. The wavelength of the wave is 5cm, which can also be written as 0.05m. v=0.05 x 40=2m/s

Q174. The product of frequency and time period is equal to:

• A. 2
• B. 3
• C. 0
• D. 1✓

Explanation: The product of frequency and time is equal to 1 according to the formula f=1/T where f represents the frequency having the units, seconds-1, and T represents the Time period having the units, seconds.

Q175. Trough of a wave acts as:

• A. Concave lens✓
• B. Convex lens
• C. Convex mirror
• D. Plane mirror

Explanation: We can see the waves on the screen. The crest acts as a convex lens while the trough acts as a concave lens.

Q176. In Doppler effect, if listener moves towards a stationary source then:

• A. Observed frequency is greater than original frequency✓
• B. Observed frequency is less than original frequency
• C. Observed frequency is equal to original frequency
• D. Observed frequency is independent of original frequency

Explanation: When a listener moves towards the stationary source, the observed frequency is greater than the actual frequency.

Q177. Two positive point charges are placed 2m apart. The electric potential at midpoint due to these two charges will be _:

• A. Added to double✓
• B. Reduced to half
• C. Remains same (no effect)
• D. Cancel each other effect

Explanation: Electric potential is a scalar quantity. At the midpoint of two positively charged point charges, the electric potentials of both charges will be added so, the potential is doubled. Solution 1m 1m +q--------V=?----------+q V= k q/r + k q/r V=k q/r + k q/r V=2 k q/r.

Q178. When the antinodes, in stationary waves, are passing through their equilibrium position, the energy is:

• A. Potential only
• B. Kinetic only✓
• C. A combination of both forms of energy
• D. There is no energy in stationary waves

Explanation: There is no energy transfer in stationary waves, but the energy of the antinodes keeps on changing from potential to kinetic energy. At the complete stretched position, the energy is wholly potential but as it passes through the equilibrium position, it is kinetic only.

Q179. A pipe open at both ends has a length of 5m. The second harmonic frequency is _. Velocity of sound in air = 340 m/s.

• A. 17 Hz
• B. 34 Hz
• C. 68 Hz✓
• D. 306 Hz

Explanation: In the second harmonic frequency of the sound in an open tube, the wavelength is 1 lambda (as shown in the figure below; the distance of between two antinodes). Thus the length of tube= length of the wave (wavelength), i.e 5 m in our case.Applying the formula for frequency:v=fλ340=f(5)f=340/5f=68 Hz

Q180. A wave, which transfers _ by moving away from the source of disturbance is called _ wave.

• A. Momentum ; standing
• B. Energy ; standing
• C. Energy ; progressive✓
• D. Momentum ; progressive

Explanation: The answer is C as the waves are energy carriers and they must move away from the source to transfer energy. Waves are disturbances that transfer energy without the transfer of the medium. Progressive waves are those waves that transfer energy from a point source to the surrounding areas.

Q181. How much heat is absorbed by 100 g of water when its temperature decreases from 25. C to 5. C? (heat capacity is 4.2 j/gk)

• A. 84,000 J
• B. -2000/4.2 J
• C. 2000/4.2 J
• D. -84000✓

Explanation: To calculate the heat absorbed by a substance, we can use the formula: q = m * C * ΔT Where: q is the heat absorbed or released, m is the mass of the substance, C is the specific heat capacity of the substance, and ΔT is the change in temperature. In this case, we are calculating the heat absorbed by 100 g of water when its temperature decreases from 25°C to 5°C, and the specific heat capacity of water is given as 4.2 J/g·K. ΔT = (Final temperature - Initial temperature) = (5°C - 25°C) = -20°C Note: The change in temperature is negative because the water is decreasing in temperature. Now we can plug in the values into the formula: q = 100 g * 4.2 J/g·K * (-20°C) q = -84,000 J The negative sign indicates that heat is being released or lost by the water. Therefore, 100 g of water releases 84,000 J of heat when its temperature decreases from 25°C to 5°C.

Q182. If a wire having current 10A has a 3T magnetic field, what will be the magnetic field at a point from the wire at double the distance?

• A. Reduces by factor 2✓
• B. Becomes double
• C. Reduces by factor 4
• D. Becomes triple

Explanation: Any current-carrying wire is surrounded by the magnetic field which is B = µ0I/2πr. It can be seen from the above equation that the magnetic field is inversely proportional to the distance and directly proportional to the current. When the distance from the wire becomes double, the magnetic field will reduce by a factor of two, which is 1.5 T.

Q183. What is true regarding magnetic force and magnetic intensity?

• A. If an electron's movement is parallel to the magnetic field. it will rotate
• B. If electron's movement is parallel to magnetic field, it Will rotate anticlockwise
• C. If an electron enters perpendicular to the field force it would be parallel to the plane
• D. If an electron enters perpendicular to the field force will be maximum✓

Explanation: If an electron enters perpendicular to the field, the force will be maximum, because according to theformula of force on a moving chargeF=Bqv Sinθ Where F is the force on moving charge, B is the magnetic field strength, q is the magnitude of chargeand θ is the angle between direction of moving charge and the magnetic field.Sin (θ) is maximum at 90° because Sin (90) = 1. Therefore, when an electron enters perpendicular to themagnetic field, the force will be maximum.There will be no force on the electron if it is moving parallel to the magnetic field.SO OPTION A B AND C ARE INCORRECT To find the direction of magnetic force on the electron, we will also need the direction of the magnetic field.

Q184. A battery whose emf is 40 V has an internal resistance of 5 Ohms. If this battery is connected to a 15 Ohms resistor ‘R’, what will be the voltage drop across ‘R’:

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

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

Q185. Due to the magnetic force, a positively charged particle executes uniform circular motion within a uniform magnetic field ‘B’. If the charge is ‘q’ and radius of its path is r, which of the expressions gives the magnitude of the particle’s linear momentum:

• A. qBr✓
• B. qB/r
• C. q/(Br)
• D. B

Explanation: Here the centripetal force will balance the magnetic force. So, mv2/r=qvB or mv=qBr Thus, linear momentum, p=mv=qBr

Q186. What is the flux density at a point 3 cm from the long straight wire, when there is a current of 25 A in a wire: (u0= 4𝜋 x10-7)

• A. 0.23x10-1 T
• B. 1.67x 10-4 T✓
• C. 2.99x10-6 T
• D. 3.63x10-8 T
• E. 999x10-7 T

Explanation: B= µ0 I / 2 π r = 4 π x 10-7 x 25 / 2 π (3 x 10-2) = 1.666666 x 10-4 = 1.67 x 10-4

Q187. When a conductor of a cross-sectional area 5x10-6 m2 carries a current of 6 A, the drift velocity of the conduction electrons is 1.2 x 10-4 ms1-. What is the number density (number per unit volume) of the conduction electrons?

• A. 4x10-28 m-3
• B. 1.6x10-27 m-3
• C. 2.5x10-27 m-3
• D. 6.3x1028 m-3✓
• E. 1.3x1034 m-3

Explanation: The formula for electric current relating the drift velocity is: I = neAV Where; I = Electric current n = Number of charges per unit volume e = Electron charge A = Cross sectional area V = Drift speed Solving for 'n' : n = I/eAV Here I = 6.0 Amp e = 1.6 x 10-19 C A = 5.0 x 10-6 m2 V = 1.2 x 10^-4 m/s Put all these values inabove equation and do the calculations: n = (6.0)/ (1.6 x 10-19) (5.0 x 10-6) (1.2 x 10-4) n = 6.3 × 1028 m-3 is the answer

Q188. A thermocouple is immersed in water at 373 K and the other in ice at 273K. The emf of the thermocouple Is 90 (μ)V for each 1 K difference in temperature between junctions, and the thermocouple resistance is 6 Ω. What current will flow in the galvanometer connected in series with an internal resistance of 30ohms?

• A. 1.8 μA
• B. 250 μA✓
• C. 300 μA
• D. 1.5 mA
• E. 1.8 mA

Explanation: Given details: Galvanometer resistance = 30 ohm Thermocouple couple resistance = 6 ohm So, Total resistance = 30+6 = 36 ohm Temperature difference = 373K – 273K =100K Emf of thermocouple for 1K temperature difference = 90 microvolt = 90 × 10^-6 V So, for 100K, emf = 90 × 10^-6 x 100 = 90 × 10^-4 V Solution: Current flow in galvanometer = Voltage / Resistance I = 90 × 10^-4 / 36 =0.00025 A = 250 micro Ampere

Q189. According to the first law of thermodynamics, ΔU = Q + W, where ΔU Is the increase in internal energy of the system, Q is the heat transferred to the system and W is the external work done on the system.Which of the following is NOT a correct expression?

• A. At constant temperature: Q = -W
• B. When no work is done: ΔU = Q
• C. In gaseous system: ΔU = Q + P Δ V
• D. When work is done by the system: ΔU = Q - W✓

Explanation: There are two sign conventions: 1. ∆U= Q+W: Here, work done BYthe gas/system is taken POSITIVE. 2. ∆U= Q-W: Here, work done ON the gas/system is taken NEGATIVE. The equation provided in the question follows the first convention, according to which D is incorrect.

Q190. The quantity of heat required to raise the temperature of one mole of a substance through 1 K is called:

• A. Carnot engine
• B. Molar specific heat✓
• C. Kinetic specific heat
• D. General gas law
• E. Boyle's law

Explanation: Specific heat is the amount of heat needed to raise the temperature of one kilogram of mass by 1 degree. The molar heat capacity is the heat capacity per unit amount (SI unit: mole) of a pure substance, and the specific heat capacity often called simply specific heat, is the heat capacity per unit mass of material. Option B is correct.

Q191. The total outward flux over any closed hypothetical surface called _ is equal to the total charge enclosed divided by ε irrespective of how the charge is distributed.

• A. Coulomb surface
• B. Gaussian surface✓
• C. Ohm surface
• D. Faraday surface
• E. Newton surface

Explanation: Gaussian surface:The Gaussian surface is known as a closed surface in three-dimensional space such that the flux of a vector field is calculated. These vector fields can either be the gravitational field or the electric field or the magnetic field.

Q192. Two capacitors C1 (6μF) and C2(12μF) are in series across a 180 volts d.c. supply. Calculate the potential difference across each capacitor (C1 and C2).

• A. 98 volts ... 32 volts
• B. 120 volts ... 60 volts✓
• C. 68 volts ... 96 volts
• D. 30 volts ... 65 volts
• E. 25 volts ... 25 volts

Explanation: Since the voltage is INVERSELY proportional to the capacitance, according to the formula: Q=CV, the voltage division is also in the same way. Since the capacitance ratio is 1:2, the voltage division is 2:1. Hence the correct division of 180V voltage will be 120 V and 60V.

Q193. Charges of +2 μC and -2 μC are situated at points P and Q respectively, as shown below. X is midway between P and Q. Which of the following correctly describes the electric field and the electric potential at point X?

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

Explanation: As the electric field moves away from positive charges and towards negative charges, the direction of the electric field should be towards Q. Electric potential is a scalar quantity that means it should be added without considering the directions. At pSince the charges are equal and opposite with the same distance from point X, the potential at point X will be zero.

Q194. A body is moving with an initial velocity of 2 kms-1. After a time of 50 secs its velocity becomes 1.5 kms-1, Its acceleration will be:

• A. 30 ms-1
• B. 40 ms-1
• C. 20 ms-1
• D. 10 ms-1✓

Explanation: Explanation: The acceleration of the body can be calculated using the formula: a = vf - vi / t Substituting the given values, we get: a = (1.5 km/s - 2 km/s) / 50s = -0.01 km/s² Multiplying by 1000 to convert km/s² to m/s², we get: a = -10 m/s² Therefore, the acceleration of the body is 10 m/s². The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.

Q195. If 7(x-2) = 3x - 2. What is the value of x?

• A. -4
• B. -3
• C. 2
• D. 3✓
• E. 4

Explanation: You can sovle this problem as shown below:

Q196. Suppose in a certain language MADRAS is coded as NBESBT. Then BOMBAY is coded in that language as:

• A. CPNCBZ✓
• B. CPNCBX
• C. CPOCBZ
• D. CQOCBZ

Explanation: As we can see the code exchanges each letter with the letter that follows it, so A becomes B, B becomes C, C becomes D, and so on. We will do this for each letter in Bombay and will arrive to: CPNCBZ.

Q197. STATEMENTS: I. The dialogue between Jinnah and Gandhi failed. Il. Indian National Congress and Muslim League tried to join hands, so they will be united.

• 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 the effects of independent cause.✓

Explanation: Gandhi and Jinnah were unable to reach a decision regarding Pakistan hence the dialogue failed. In order to show a united front to the British as they believe that would help them achieve independence INC and ML tried to join hands(work together). Thus both these statements are effects of independent causes

Q198. The ‘A’ state government has chalked out a plan for the underdeveloped ‘B’ district where66% of the funds will be placed in the hands of a committee of local representatives.Courses of action:I. The ‘A’ state government should decide guidelines and norms for the functioning of thecommittee.II. Other state government may follow similar plan if directed by the Central government.

• A. If only I follows✓
• B. If only II follows
• C. If either I or II follows
• D. If neither I nor II follows

Explanation: Once it is decided to place funds in the hands of the committee of local representatives, it is necessary to decide guidelines and norms for the functioning of the committee. Hence I follows. II is not related to the statement.

Q199. The Emporium Mall has more stores than the Pakages Mall.The Amanah Mall has fewer stores than the Pakages Mall.The Emporium Mall has more stores than the Amanah Mall.If the first two statements are true, the third statement is

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

Explanation: Explanation From the first two statements, you know that the Emporium mall has the most stores, so the Emporium Mall would have more stores than the packages Mall.

Q200. Four people were witnessed as assaulted at an incident. Each of them have their own description of the scene. Which one will be probably right?

• A. He was average height, thin, and middle aged
• B. He was tall, thin, and middle aged✓
• C. He was tall, thin and young
• D. He was tall, of average weight, and middle aged

Explanation: Choice B gives accurate descriptions of all the features of the assailant/attacker. All other options have at least one vague description of the person such as, the average height in average weight in b, and young age in c.