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Sindh Mcat Nts 2014 Lumhs — Solved Past Paper with Answers

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Q1. Complete the sentences by choosing the most appropriate option:The children are going _ visit the museum this afternoon.

A. for
B. to✓
C. off
D. at
E. too

Explanation: "Visit" is utilized as an activity action word here connoting an activity that the youngsters will perform. Before this activity action word, the relational word that ought to be utilized is "to" for the sentence to sound syntactically right.“For” or “at” could have been right if the word ”visit” were being used as a noun! “Too” is incorrect because it shows the intensity, not the preposition.

Why the other options are wrong

A. "Visit" is utilized as an activity action word here connoting an activity that the youngsters will perform. Before this activity action word, the relational word that ought to be utilized is "to" for the sentence to sound syntactically right.“For” or “at” could have been right if the word ”visit” were being used as a noun! “Too” is incorrect because it shows the intensity, not the preposition.
C. "Visit" is utilized as an activity action word here connoting an activity that the youngsters will perform. Before this activity action word, the relational word that ought to be utilized is "to" for the sentence to sound syntactically right.“For” or “at” could have been right if the word ”visit” were being used as a noun! “Too” is incorrect because it shows the intensity, not the preposition.
D. "Visit" is utilized as an activity action word here connoting an activity that the youngsters will perform. Before this activity action word, the relational word that ought to be utilized is "to" for the sentence to sound syntactically right.“For” or “at” could have been right if the word ”visit” were being used as a noun! “Too” is incorrect because it shows the intensity, not the preposition.
E. "Visit" is utilized as an activity action word here connoting an activity that the youngsters will perform. Before this activity action word, the relational word that ought to be utilized is "to" for the sentence to sound syntactically right.“For” or “at” could have been right if the word ”visit” were being used as a noun! “Too” is incorrect because it shows the intensity, not the preposition.

Q2. Complete the sentences by choosing the most appropriate option:She is grieving _ for the loss of her child.

A. bitterly✓
B. gladly
C. happily
D. boldly
E. mockingly

Explanation: Grieving is a state of sorrow,sadness, or regret. Most appropriate adverb to use here will be “bitterly” - referring to how bad the situation is. No one grieves happily, gladly, mockingly, or boldly!

Why the other options are wrong

B. Grieving is a state of sorrow,sadness, or regret. Most appropriate adverb to use here will be “bitterly” - referring to how bad the situation is.No one grieves happily, gladly, mockingly, or boldly!
C. Grieving is a state of sorrow,sadness, or regret. Most appropriate adverb to use here will be “bitterly” - referring to how bad the situation is.No one grieves happily, gladly, mockingly, or boldly!
D. Grieving is a state of sorrow,sadness, or regret. Most appropriate adverb to use here will be “bitterly” - referring to how bad the situation is.No one grieves happily, gladly, mockingly, or boldly!
E. Grieving is a state of sorrow,sadness, or regret. Most appropriate adverb to use here will be “bitterly” - referring to how bad the situation is.No one grieves happily, gladly, mockingly, or boldly!

Q3. Identify the word or phrase that needs to be changed for the sentence to be correct:It is fascinating to watch a potter making a lump of clay into a beautiful-vase.

A. It
B. fascinating
C. lump
D. beautiful-vase✓
E. No error

Explanation: The use of the hyphen is incorrect. The adjective "beautiful" describes the vase so these two words should be separated by a space, and not a hyphen.

Why the other options are wrong

A. It: "It" is correctly used as an impersonal pronoun to introduce the sentence. No error here.
B. fascinating: "Fascinating" is an adjective describing the act of watching the potter, which is grammatically correct. No error here.
C. lump: "Lump" is a correct noun to describe a mass of clay. No error here.
E. No error: Since there is an error in option D, this option is incorrect.

Q4. Identify the word or phrase that needs to be changed for the sentence to be correct:He is among the better novelists in this country today.

A. He
B. among
C. better✓
D. country
E. No error

Explanation: The superlative adjective "best" should be used. Superlative adjectives are used to describe an object which is at the upper or lower limit of a quality (the tallest, the smallest,or the fastest).

Why the other options are wrong

A. A) He: "He" is a correct subject pronoun for the sentence.
B. B) among: "Among" is correctly used to show a relationship within a group of more than two.
D. D) country: "Country" is correctly used as a noun to refer to a nation.
E. E) No error: As explained above, there is an error.

Q5. Choose the word most similar in meaning to the following word:ASTONISH

A. punish
B. surprise✓
C. please
D. warn
E. control

Explanation: Astonish means to surprise or impress (someone) greatly. The only similar word is “surprise”.

Why the other options are wrong

A. * A) punish: To inflict a penalty for an offense. This is not similar to astonish.
C. C) please: To make (someone) feel happy or satisfied. While a pleasant surprise can please someone, the core meaning of "astonish" is not "please."
D. D) warn: To inform someone in advance of a possible danger, problem, or unpleasant situation. This is unrelated to astonish.
E. E) control: To exercise restraint or direction over something. This is not similar to astonish.

Q6. Choose the word most similar in meaning to the following word:RIVAL

A. player
B. friend
C. opponent✓
D. colleague
E. fellow

Explanation: Rival refers to the “arch enemy”- someone against you.The correct word is opponent.

Why the other options are wrong

A. A) player: While a rival can be a player, not all players are rivals. Players can also be teammates.
B. B) friend: A friend is someone with whom you have a bond of mutual affection, which is generally the opposite of a rival.
D. D) colleague: A colleague is someone you work with. While there might be competition among colleagues, the primary meaning of "colleague" is about working together, not competing.
E. E) fellow: "Fellow" implies being in the same group, position, or having shared characteristics, not necessarily being in competition.

Q7. Choose the lettered word or phrase that is most nearly opposite in meaning to the following word:RELIABLE

A. changeable✓
B. stable
C. trustworthy
D. honest
E. balanced

Explanation: Reliable refers to the something/someone on which/whom you can rely on and such a person should be honest and trustworthy.The opposite of which will be “changeable” something that does not remain constant rather keeps on changing/varying.

Why the other options are wrong

B. B) stable: This is a synonym of reliable. A stable entity is steady and not easily moved or changed, which aligns with dependability.
C. C) trustworthy: This is a direct synonym of reliable. To be trustworthy means to be dependable and deserving of trust.
D. D) honest: While an honest person is often reliable, "honest" is not the opposite of "reliable." A person could be honest but still unreliable (e.g., consistently late or forgetful).
E. E) balanced: This implies a state of equilibrium and consistency, which is closer to the meaning of reliable rather than its opposite.

Q8. Choose the lettered word or phrase that is most nearly opposite in meaning to the following word:REGRETTABLE

A. tragic
B. disappointing
C. historic
D. necessary
E. fortunate✓

Explanation: REGRETTABLE refers to a feeling which occurs due to something that should not have happened - is shameful or sorrowful.The opposite of which will be something that is very “fortunate” and welcome.

Why the other options are wrong

A. A) tragic: Is a synonym, meaning very sad or deplorable.
B. B) disappointing: Is a synonym, meaning not meeting expectations
C. C) historic: Refers to something famous or important in history, which is unrelated to the feeling of regret or misfortune.
D. D) necessary: Refers to something essential or unavoidable, which is unrelated to the feeling of regret or misfortune.

Q9. Why did the English language never get planted in Wales and Scotland?

A. Because people in these areas never wanted to learn English.
B. Because the Anglo-Saxons never succeeded in these areas.✓
C. Because the Anglo-Saxon who reached in these areas learnt the native languages there.
D. Because the natives of these areas never came to England.

Explanation: The answer can be found in the third sentence of the text: “The Anglo-Saxons after conquering England never succeeded in penetrating to these remote parts of Britain, so that the English language was never implanted there.”

Why the other options are wrong

A. A) Because people in these areas never wanted to learn English. This is inaccurate. While there was strong cultural and linguistic resistance, English did gain influence over time due to various factors like political union, economic opportunities, and educational policies. Many people in Wales and Scotland became bilingual, learning English for practical reasons, especially for social and economic advancement.
C. C) Because the Anglo-Saxon who reached in these areas learnt the native languages there. There is no historical evidence to support this claim. The general pattern of Anglo-Saxon expansion led to the displacement of Celtic languages, not the widespread adoption of those languages by the Anglo-Saxons.
D. D) Because the natives of these areas never came to England. This statement is irrelevant to the question. The migration patterns of people between these regions do not explain why English didn't become the dominant language within Wales and Scotland. The key factor was the extent of Anglo-Saxon settlement and conquest in those specific territories.

Q10. Which one of the following statements correctly describes the transcription of DNA?

A. It produces amino acids
B. It results in an increases DNA synthesis
C. It produces messenger RNA✓
D. It is a semi conservative process
E. It occurs at the surface of the ribosome

Explanation: DNA transcription is the process by which the genetic information contained in DNA is rewritten by RNA polymerase into messenger RNA (mRNA)

Why the other options are wrong

A. A) It produces amino acids: This is incorrect. The process of producing amino acids (and subsequently proteins) from messenger RNA is called translation, not transcription. Transcription is the step that comes before translation.
B. B) It results in an increases DNA synthesis: This is incorrect. Transcription is the process of synthesizing RNA from a DNA template. DNA synthesis, also known as DNA replication, is a separate process where DNA makes copies of itself. Transcription does not increase DNA synthesis.
D. D) It is a semi conservative process: This is incorrect. DNA replication is a semi-conservative process, meaning that when DNA is replicated, each new DNA molecule consists of one original strand and one newly synthesized strand. Transcription does not involve this semi-conservative mechanism.
E. E) It occurs at the surface of the ribosome: This is incorrect. In eukaryotic cells, transcription primarily occurs in the nucleus. Ribosomes are the sites where translation (protein synthesis) takes place.

Q11. A woman with normal colour vision, whose father was red green colour blind, married a red green colour blind man. What is the probability of her born child being red green colour blind?

A. 1.0
B. 0.75
C. 0.50✓
D. 0.25

Explanation: Males have 1 X chromosome and 1 Y chromosome, and females have 2 X chromosomes. The genes that can give you red-green colour blindness are passed down on the X chromosome.The woman’s father is red-green colour blind. Since the woman got one of her X chromosomes from her father this means that she is a carrier. As it can be seen from the table the probability of having a colourblind child is 0.5%

Why the other options are wrong

A. a) 1.0 (100%):This means that there is a 100% probability of the child being red-green color blind. However, this is not necessarily true. The inheritance of color blindness is not guaranteed, and other factors come into play.
B. b) 0.75 (75%):This means that there is a 75% probability of the child being red-green color blind. This option suggests a high likelihood of inheritance since both parents have a form of color blindness.
D. d) 0.25 (25%):This means that there is a 25% probability of the child being red-green color blind. This option suggests a lower likelihood of inheritance.

Q12. Which of the following ideas was not part of Charles Darwin’s theory of evolution by natural selection?

A. Organisms produce more off springs than the environment can support
B. Variation between individuals arises by gene mutation✓
C. Only those individuals best adapted by environment survive and reproduce
D. Individuals compete for scarce resources

Explanation: The idea that was not part of Charles Darwin's theory of evolution by natural selection is Option B: Variation between individuals arises by gene mutation. While Darwin acknowledged the existence of variation, he did not have a detailed understanding of the genetic mechanisms behind it, such as gene mutations.

Why the other options are wrong

A. This idea is a fundamental aspect of Darwin's theory. Darwin observed that populations have the potential for exponential growth, but resources in the environment are limited. As a result, there is competition for resources, and not all offspring can survive and reproduce.
C. This idea is a central concept in Darwin's theory of natural selection. Darwin proposed that individuals with traits that provide them with advantages in their environment are more likely to survive and reproduce, passing on their advantageous traits to future generations.
D. This idea is also an important part of Darwin's theory. Darwin recognized that resources necessary for survival and reproduction, such as food, water, and shelter, are limited in nature. As a result, individuals within a population must compete with each other for access to these resources.

Q13. What happens during the light phase of photosynthesis?

A. ADP is hydrolyzed and NADPH oxidized
B. ATP is synthesized by photophosphorylation and NADP reduced✓
C. ATP is hydrolyzed and NADP is oxidized
D. ATP is phosphorylated and NADP is oxidized
E. ATP is hydrolyzed and NADP is reduced

Explanation: During the light reaction, ATP is synthesised by photophosphorylation and NADP is reduced to NADPH. ATP is used as the energy source for dark reaction while the reducing power is provided by NADPH.

Why the other options are wrong

A. A) ADP is hydrolyzed and NADPH oxidized: * ADP is hydrolyzed: This is incorrect. ADP is phosphorylated to form ATP, not hydrolyzed. Hydrolysis of ATP (breaking it down into ADP and phosphate) releases energy, which occurs in the light-independent reactions (Calvin cycle) to power carbon fixation. * NADPH oxidized: This is incorrect. NADP+ is reduced to NADPH in the light reactions. NADPH is later oxidized in the Calvin cycle when it donates its electrons.
C. C) ATP is hydrolyzed and NADP is oxidized: * ATP is hydrolyzed: Incorrect, as explained above. * NADP is oxidized: Incorrect. NADP+ is reduced.
D. D) ATP is phosphorylated and NADP is oxidized: * ATP is phosphorylated: This phrasing is imprecise. ADP is phosphorylated to form ATP. ATP itself is the product, not what gets phosphorylated. * NADP is oxidized: Incorrect. NADP+ is reduced.
E. E) ATP is hydrolyzed and NADP is reduced: * ATP is hydrolyzed: Incorrect, as explained above. * NADP is reduced: This part is correct, but because the first part of the statement is wrong, the entire option is incorrect.The primary goal of the light-dependent reactions is to produce ATP and NADPH, which then provide the energy and reducing power for the synthesis of sugars in the light-independent reactions.

Q14. Identify the incorrectly matched pair:

A. Cardiac output…. volume of blood pumped per minute by the left ventricle
B. Arterial systole…. contraction of the two atria
C. Heart murmur…. defect of heat valves
D. Diastole…. phase of contraction of cardiac muscle✓
E. Right atrium in the diastolic phase…. receives deoxygenated blood from the vena cava.

Explanation: Diastole refers to the relaxation of cardiac muscle.

Why the other options are wrong

A. The given statement is correct. Cardiac output is volume of blood pumped per minute by the left ventricle.
B. Systole refers to contraction of heart valves. Arterial systole means contraction of two atria.
C. Calcium deposits can cause a heart valve such as the mitral valve or the aortic valve to become stiff and narrow (valve stenosis). A calcified valve also may not close completely, letting blood flow backward. The changes in blood flow create a murmur
E. The given statement is correct. During diastole, right atrium receives deoxygenated blood form body via superior and inferior vena cava.

Q15. Diptera refers to a(n):

A. Genus
B. Species
C. Order✓
D. Family
E. Class

Explanation: Factual recall: Sindh Board Textbook for Biology, Grade 11 (old edition), Chapter 5: The Variety of Life

Why the other options are wrong

A. Genus: A genus is a taxonomic rank below a family and above a species. Diptera is a much broader group, encompassing numerous genera (e.g., Drosophila is a genus within the order Diptera).
B. Species: A species is the most specific taxonomic rank. Diptera contains hundreds of thousands of individual species, not a single species itself.
D. Family: A family is a taxonomic rank below an order and above a genus. The order Diptera contains many families, such as Muscidae (house flies) and Culicidae (mosquitoes).
E. Class: A class is a broader taxonomic rank than an order, typically above an order and below a phylum. Diptera is an order within the class Insecta.

Q16. he diagram below represents the structure of a chloroplast as seen in section. Which label A,B,C,D, or E indicates the site of light reaction of photosynthesis?

A. A
B. B
C. C
D. D
E. E✓

Explanation: A—> chloroplast envelop B—> storage granule C—> stroma D—> stromal lamellae E—> thylakoid The light reaction takes place in the thylakoid discs.

Why the other options are wrong

A. A—> chloroplast envelop
B. B—> storage granule
C. C—> stroma
D. D—> stromal lamellae

Q17. The scientific name of gumtree is Acacia nilotica and that of the katha plant is Acacia catechu. This indicates that both gumtree and katha plants are members of the:

A. Same species but different genera
B. Same genus but different classes
C. Same species but different classes
D. Same genus but different species✓
E. Same class but different genera

Explanation: According to the principles of binomial nomenclature, the first part of the scientific name indicates the genus and the second part the species.

Why the other options are wrong

A. According to the principles of binomial nomenclature, the first part of the scientific name indicates the genus and the second part the species.
B. According to the principles of binomial nomenclature, the first part of the scientific name indicates the genus and the second part the species.
C. According to the principles of binomial nomenclature, the first part of the scientific name indicates the genus and the second part the species.
E. According to the principles of binomial nomenclature, the first part of the scientific name indicates the genus and the second part the species.

Q18. Which level of protein structure maintains the helix shape of enzymes?

A. Primary
B. Secondary✓
C. Tertiary
D. Quaternary
E. Both A and B

Explanation: The most common types of secondary structures are the α helix and the β pleated sheet. Both structures are held in shape by hydrogen bonds, which form between the carbonyl O of one amino acid and the amino H of another.

Why the other options are wrong

A. Primary Structure (Option A): This is the linear sequence of amino acids in a polypeptide chain, linked by peptide bonds. While the primary structure determines the higher-order structures (including the secondary structure), it does not itself maintain the helix shape. It's like the alphabet that forms words, but not the structure of the paragraphs.
C. Tertiary Structure (Option C): This refers to the overall three-dimensional folding of a single polypeptide chain, including its secondary structures (like helices and sheets) and irregular loops. It's stabilized by interactions between the side chains (R-groups) of amino acids, such as hydrophobic interactions, ionic bonds, hydrogen bonds, and disulfide bridges. While helices are part of the tertiary structure, the tertiary structure describes the global folding rather than maintaining the local helical shape itself.
D. Quaternary Structure (Option D): This level of structure applies to proteins composed of two or more polypeptide chains (subunits). It describes the arrangement of these multiple subunits in a functional protein complex. The interactions are similar to those in tertiary structure but occur between different polypeptide chains. It is not involved in maintaining the internal helical shape of individual polypeptide chains.
E. Both A and B (Option E): As explained, primary structure does not maintain the helix shape, only secondary structure does. So, this option is not correct.

Q19. Why do some root cells have root hairs?

A. For the maintenance of the temperature of the cell sap
B. To increase the surface area of the cells✓
C. To increase the volume of the cell sap
D. To provide a place for cell nuclei
E. To allow movement of soil particles

Explanation: Root hair cells improve plant water absorption by increasing root surface area to volume ratio which allows the root hair cell to take in more water.

Why the other options are wrong

A. Root hair cells improve plant water absorption by increasing root surface area to volume ratio which allows the root hair cell to take in more water.
C. Root hair cells improve plant water absorption by increasing root surface area to volume ratio which allows the root hair cell to take in more water.
D. Root hair cells improve plant water absorption by increasing root surface area to volume ratio which allows the root hair cell to take in more water.
E. Root hair cells improve plant water absorption by increasing root surface area to volume ratio which allows the root hair cell to take in more water.

Q20. The diagram shows a flower in the longitudinal section. Before they had developed fully, a plant breeder removed the structures labelled X, as shown. What is the effect of removing these structures?

A. It prevents asexual reproduction.
B. It prevents the flower from being pollinated.
C. It prevents the flower from producing seeds.
D. It prevents the development of fruit from ovary.
E. It prevents the flower from pollinating itself.✓

Explanation: X represents the anthers, which produce pollen grains. Removing the anthers means that self-pollination can no longer occur because no pollen is being produced! However, cross pollination can still occur because pollen grains from a different flower can still land on the stigma of the carpel.

Why the other options are wrong

A. X represents the anthers, which produce pollen grains. Removing the anthers means that self-pollination can no longer occur because no pollen is being produced! However, cross pollination can still occur because pollen grains from a different flower can still land on the stigma of the carpel.
B. X represents the anthers, which produce pollen grains. Removing the anthers means that self-pollination can no longer occur because no pollen is being produced! However, cross pollination can still occur because pollen grains from a different flower can still land on the stigma of the carpel.
C. X represents the anthers, which produce pollen grains. Removing the anthers means that self-pollination can no longer occur because no pollen is being produced! However, cross pollination can still occur because pollen grains from a different flower can still land on the stigma of the carpel.
D. X represents the anthers, which produce pollen grains. Removing the anthers means that self-pollination can no longer occur because no pollen is being produced! However, cross pollination can still occur because pollen grains from a different flower can still land on the stigma of the carpel.

Q21. Which of the following is/are adaptation(s) for parasitic mode of life in platyhelminthes?

A. Thick body covers
B. High fertility rate
C. Adhesive organs like suckers
D. Simplified digestive system
E. All of the given options✓

Explanation: A) Thick body covers: Parasitic flatworms, such as tapeworms and flukes, have a specialized outer covering called a tegument. This thick, resistant layer protects them from the host's digestive enzymes and immune responses. * B) High fertility rate: To ensure the survival of their species, parasitic platyhelminthes produce a vast number of eggs. This high reproductive capacity increases the chances of their offspring finding a suitable host and completing their complex life cycles, even with significant mortality rates. * C) Adhesive organs like suckers: Parasitic flatworms possess various attachment structures, including suckers and hooks. These organs allow them to firmly anchor themselves to the host's tissues, preventing them from being dislodged by peristalsis or fluid flow. * D) Simplified digestive system: Many parasitic platyhelminthes, especially tapeworms, have a reduced or even absent digestive system. Instead, they absorb digested nutrients directly from their host's body through their tegument. This simplifies their internal structure and conserves energy that would otherwise be spent on digestion.Since all options A, B, C, and D are valid adaptations for a parasitic life in platyhelminthes, option E is the correct choice.

Why the other options are wrong

A. A) Thick body covers: Parasitic flatworms, such as tapeworms and flukes, have a specialized outer covering called a tegument. This thick, resistant layer protects them from the host's digestive enzymes and immune responses. * B) High fertility rate: To ensure the survival of their species, parasitic platyhelminthes produce a vast number of eggs. This high reproductive capacity increases the chances of their offspring finding a suitable host and completing their complex life cycles, even with significant mortality rates. * C) Adhesive organs like suckers: Parasitic flatworms possess various attachment structures, including suckers and hooks. These organs allow them to firmly anchor themselves to the host's tissues, preventing them from being dislodged by peristalsis or fluid flow. * D) Simplified digestive system: Many parasitic platyhelminthes, especially tapeworms, have a reduced or even absent digestive system. Instead, they absorb digested nutrients directly from their host's body through their tegument. This simplifies their internal structure and conserves energy that would otherwise be spent on digestion.Since all options A, B, C, and D are valid adaptations for a parasitic life in platyhelminthes, option E is the correct choice.
B. A) Thick body covers: Parasitic flatworms, such as tapeworms and flukes, have a specialized outer covering called a tegument. This thick, resistant layer protects them from the host's digestive enzymes and immune responses. * B) High fertility rate: To ensure the survival of their species, parasitic platyhelminthes produce a vast number of eggs. This high reproductive capacity increases the chances of their offspring finding a suitable host and completing their complex life cycles, even with significant mortality rates. * C) Adhesive organs like suckers: Parasitic flatworms possess various attachment structures, including suckers and hooks. These organs allow them to firmly anchor themselves to the host's tissues, preventing them from being dislodged by peristalsis or fluid flow. * D) Simplified digestive system: Many parasitic platyhelminthes, especially tapeworms, have a reduced or even absent digestive system. Instead, they absorb digested nutrients directly from their host's body through their tegument. This simplifies their internal structure and conserves energy that would otherwise be spent on digestion.Since all options A, B, C, and D are valid adaptations for a parasitic life in platyhelminthes, option E is the correct choice.
C. A) Thick body covers: Parasitic flatworms, such as tapeworms and flukes, have a specialized outer covering called a tegument. This thick, resistant layer protects them from the host's digestive enzymes and immune responses. * B) High fertility rate: To ensure the survival of their species, parasitic platyhelminthes produce a vast number of eggs. This high reproductive capacity increases the chances of their offspring finding a suitable host and completing their complex life cycles, even with significant mortality rates. * C) Adhesive organs like suckers: Parasitic flatworms possess various attachment structures, including suckers and hooks. These organs allow them to firmly anchor themselves to the host's tissues, preventing them from being dislodged by peristalsis or fluid flow. * D) Simplified digestive system: Many parasitic platyhelminthes, especially tapeworms, have a reduced or even absent digestive system. Instead, they absorb digested nutrients directly from their host's body through their tegument. This simplifies their internal structure and conserves energy that would otherwise be spent on digestion.Since all options A, B, C, and D are valid adaptations for a parasitic life in platyhelminthes, option E is the correct choice.
D. A) Thick body covers: Parasitic flatworms, such as tapeworms and flukes, have a specialized outer covering called a tegument. This thick, resistant layer protects them from the host's digestive enzymes and immune responses. * B) High fertility rate: To ensure the survival of their species, parasitic platyhelminthes produce a vast number of eggs. This high reproductive capacity increases the chances of their offspring finding a suitable host and completing their complex life cycles, even with significant mortality rates. * C) Adhesive organs like suckers: Parasitic flatworms possess various attachment structures, including suckers and hooks. These organs allow them to firmly anchor themselves to the host's tissues, preventing them from being dislodged by peristalsis or fluid flow. * D) Simplified digestive system: Many parasitic platyhelminthes, especially tapeworms, have a reduced or even absent digestive system. Instead, they absorb digested nutrients directly from their host's body through their tegument. This simplifies their internal structure and conserves energy that would otherwise be spent on digestion.Since all options A, B, C, and D are valid adaptations for a parasitic life in platyhelminthes, option E is the correct choice.

Q22. Which of the following deserts is an example of a cold desert?

A. Sahara desert
B. That desert
C. Thal desert
D. Cholistan desert
E. Gobi desert✓

Explanation: E) Gobi desert: The Gobi Desert is a large, cold desert and grassland region in northern China and southern Mongolia. It experiences frost and occasional snow, and its high elevation contributes to its low temperatures.

Why the other options are wrong

A. A) Sahara desert: The Sahara is known as the world's largest hot desert, characterized by extremely high temperatures.
B. B) Thar desert: The Thar Desert, also known as the Great Indian Desert, is classified as a hot subtropical desert.
C. C) Thal desert: The Thal Desert is a subtropical sandy desert in Pakistan, geographically similar to the hot deserts of Cholistan and Thar. It experiences high diurnal temperature variations and is prone to temperature extremes, indicating it is a hot desert.
D. D) Cholistan desert: The Cholistan Desert in Southern Punjab, Pakistan, is explicitly described as a hot sandy saline desert with arid and semi-arid tropical desert climate conditions.

Q23. Chemosynthesis is a kind of nutrition requiring:

A. Inorganic compounds✓
B. Organic compounds
C. Any kind of compounds
D. Light
E. None of the given options

Explanation: The process of carbohydrate formation in which organisms use chemical reactions to obtain energy from inorganic compounds is called chemosynthesis.

Why the other options are wrong

B. The process of carbohydrate formation in which organisms use chemical reactions to obtain energy from inorganic compounds is called chemosynthesis.
C. The process of carbohydrate formation in which organisms use chemical reactions to obtain energy from inorganic compounds is called chemosynthesis.
D. The process of carbohydrate formation in which organisms use chemical reactions to obtain energy from inorganic compounds is called chemosynthesis.
E. The process of carbohydrate formation in which organisms use chemical reactions to obtain energy from inorganic compounds is called chemosynthesis.

Q24. Which of the following is NOT involved in cell mediated immunity?

A. Helper T cells
B. Cytotoxic T cells
C. Suppressor T cells
D. Memory T cells
E. None of the given options✓

Explanation: Cell mediated immunity is contributed by the second family of lymphocytes called T cells, which do not secrete antibodies. They mediate immunity by killing infected cells, and aiding in inflammation. This is particularly important in the defense against viruses as well as some parasites that hide within the host cells, tumor cells and fungi. Several types of T cells contribute to cell mediated immunity: Helper T cells, Cytotoxic T cells, and Suppressor T cells. During CMI response, some T cells turn into memory T cells to protect the body in case of re-infection in future.

Why the other options are wrong

A. Cell mediated immunity is contributed by the second family of lymphocytes called T cells, which do not secrete antibodies. They mediate immunity by killing infected cells, and aiding in inflammation. This is particularly important in the defense against viruses as well as some parasites that hide within the host cells, tumor cells and fungi. Several types of T cells contribute to cell mediated immunity: Helper T cells, Cytotoxic T cells, and Suppressor T cells. During CMI response, some T cells turn into memory T cells to protect the body in case of re-infection in future.
B. Cell mediated immunity is contributed by the second family of lymphocytes called T cells, which do not secrete antibodies. They mediate immunity by killing infected cells, and aiding in inflammation. This is particularly important in the defense against viruses as well as some parasites that hide within the host cells, tumor cells and fungi. Several types of T cells contribute to cell mediated immunity: Helper T cells, Cytotoxic T cells, and Suppressor T cells. During CMI response, some T cells turn into memory T cells to protect the body in case of re-infection in future.
C. Cell mediated immunity is contributed by the second family of lymphocytes called T cells, which do not secrete antibodies. They mediate immunity by killing infected cells, and aiding in inflammation. This is particularly important in the defense against viruses as well as some parasites that hide within the host cells, tumor cells and fungi. Several types of T cells contribute to cell mediated immunity: Helper T cells, Cytotoxic T cells, and Suppressor T cells. During CMI response, some T cells turn into memory T cells to protect the body in case of re-infection in future.
D. Cell mediated immunity is contributed by the second family of lymphocytes called T cells, which do not secrete antibodies. They mediate immunity by killing infected cells, and aiding in inflammation. This is particularly important in the defense against viruses as well as some parasites that hide within the host cells, tumor cells and fungi. Several types of T cells contribute to cell mediated immunity: Helper T cells, Cytotoxic T cells, and Suppressor T cells. During CMI response, some T cells turn into memory T cells to protect the body in case of re-infection in future.

Q25. Blood pressure in the glomerulus is _ because the diameter of efferent arteriole is _ that of afferent arteriole.

A. Low…. much less than
B. High…. much less than✓
C. Low…. much greater than
D. High…. much greater than
E. Normal…. is equal to

Explanation: Blood pressure in the glomerulus is high because the diameter of the efferent arteriole is much less than the diameter of the afferent arteriole.

Why the other options are wrong

A. A) Low.... much less than: This is incorrect because the blood pressure in the glomerulus is high, not low.
C. C) Low.... much greater than: This is incorrect because the blood pressure in the glomerulus is high, not low, and the efferent arteriole's diameter is smaller, not greater, than the afferent arteriole's.
D. D) High.... much greater than: While the blood pressure is high, the statement regarding the diameter of the efferent arteriole is incorrect. The efferent arteriole has a smaller diameter than the afferent arteriole.
E. E) Normal.... is equal to: This is incorrect because the blood pressure in the glomerulus is higher than normal capillary pressure, and the diameters of the afferent and efferent arterioles are not equal; they differ to maintain high glomerular pressure

Q26. Destruction of all beta cells in the pancreas would cause:

A. Glucagon secretion to stop and a decrease in blood glucose
B. Glucagon secretion to stop and an increase in blood
C. Insulin secretion to stop and an increase in blood glucose✓
D. Insulin secretion to stop and a decrease in blood glucose
E. Insulin secretion to increase and a decrease in blood glucose

Explanation: The beta cells of the pancreas secrete insulin, which is responsible for decreasing the concentration of glucose in the blood when it is too high.

Why the other options are wrong

A. A) Glucagon secretion to stop and a decrease in blood glucose: This is incorrect because glucagon is produced by alpha cells in the pancreas, not beta cells (Source: Glucagon: What It Is, Function & Related Conditions). Therefore, its secretion would not stop due to beta cell destruction. Also, a decrease in blood glucose would not occur; rather, an increase would be expected.
B. B) Glucagon secretion to stop and an increase in blood: This is incorrect for the same reason as option A regarding glucagon secretion. Additionally, "an increase in blood" is vague and incomplete.
D. D) Insulin secretion to stop and a decrease in blood glucose: While insulin secretion would stop, this would lead to an increase in blood glucose, not a decrease.
E. E) Insulin secretion to increase and a decrease in blood glucose: This is incorrect. Insulin secretion would stop, not increase, and this would lead to an increase in blood glucose, not a decrease.

Q27. During the action potential:

A. The polarity of neurilemma first changes to -65 mV and then restores to +40 mV again
B. The polarity of neurilemma first changes to +40 mV and then restores to -65 mV again✓
C. The polarity of neurilemma first changes to -85 mV and then restores to +30 mV again
D. The polarity of neurilemma first changes to +30 mV and then restores to -85 mV again
E. The polarity of neurilemma first changes to 0 mV and then restores to +40 mV again

Explanation: An action potential is a temporary localized reversal of the polarity of the neurolemma, which occurs when neurolemma is stimulated (e.g. electric shock, touch, sound, vibrations etc). During the action potential, the polarity of neurolemma first changes to +40mv (i.e. depolarization) and then restores to -65mv again (i.e. repolarization). This action potential is extremly rapid as it occurs only in a few milliseconds. to happen.

Why the other options are wrong

A. An action potential is a temporary localized reversal of the polarity of the neurolemma, which occurs when neurolemma is stimulated (e.g. electric shock, touch, sound, vibrations etc). During the action potential, the polarity of neurolemma first changes to +40mv (i.e. depolarization) and then restores to -65mv again (i.e. repolarization). This action potential is extremly rapid as it occurs only in a few milliseconds. to happen.
C. An action potential is a temporary localized reversal of the polarity of the neurolemma, which occurs when neurolemma is stimulated (e.g. electric shock, touch, sound, vibrations etc). During the action potential, the polarity of neurolemma first changes to +40mv (i.e. depolarization) and then restores to -65mv again (i.e. repolarization). This action potential is extremly rapid as it occurs only in a few milliseconds. to happen.
D. An action potential is a temporary localized reversal of the polarity of the neurolemma, which occurs when neurolemma is stimulated (e.g. electric shock, touch, sound, vibrations etc). During the action potential, the polarity of neurolemma first changes to +40mv (i.e. depolarization) and then restores to -65mv again (i.e. repolarization). This action potential is extremly rapid as it occurs only in a few milliseconds. to happen.
E. An action potential is a temporary localized reversal of the polarity of the neurolemma, which occurs when neurolemma is stimulated (e.g. electric shock, touch, sound, vibrations etc). During the action potential, the polarity of neurolemma first changes to +40mv (i.e. depolarization) and then restores to -65mv again (i.e. repolarization). This action potential is extremly rapid as it occurs only in a few milliseconds. to happen.

Q28. When a sensory receptor receives a threshold stimulus, it will do all of the following EXCEPT:

A. Become depolarized
B. Transduce stimulus to an action potential
C. Inhibit the spread of the action potential to sensory neurons✓
D. C ause the sensory neurons to send action potential to the central nervous system
E. None of the given options

Explanation: The function of sensory receptors is to receive stimuli such as touch, pressure, heat, etc.,trandsuce them to electrical impulses and transmit those impulses to sensory neurons.

Why the other options are wrong

A. The function of sensory receptors is to receive stimuli such as touch, pressure, heat, etc.,trandsuce them to electrical impulses and transmit those impulses to sensory neurons.
B. The function of sensory receptors is to receive stimuli such as touch, pressure, heat, etc.,trandsuce them to electrical impulses and transmit those impulses to sensory neurons.
D. The function of sensory receptors is to receive stimuli such as touch, pressure, heat, etc.,trandsuce them to electrical impulses and transmit those impulses to sensory neurons.
E. Option C is correct

Q29. What role does pepsin play in digestion?

A. It kills the bacteria of stomach
B. It converts soluble milk proteins into insoluble proteins
C. It converts starch and glycogen into methane
D. It activates the gastric glands to produce gastric juices
E. It converts proteins into short chain polypeptides✓

Explanation: Pepsin is responsible for the breakdown of proteins into peptides.

Why the other options are wrong

A. A) It kills the bacteria of stomach: While the highly acidic environment of the stomach (created by hydrochloric acid) is effective in killing most bacteria, this is a function of the stomach's overall environment, not the primary role of pepsin itself.
B. B) It converts soluble milk proteins into insoluble proteins: This is incorrect. Pepsin breaks down proteins, it does not convert soluble proteins into insoluble ones. The curdling of milk proteins (casein) is primarily facilitated by the enzyme chymosin (also known as rennin), especially in infants, not pepsin.
C. C) It converts starch and glycogen into methane: This is incorrect. Pepsin is a protease, meaning it acts on proteins. The digestion of carbohydrates like starch and glycogen begins in the mouth with amylase and continues in the small intestine. Methane is a gas produced by certain microorganisms, not a direct product of enzymatic digestion by pepsin.
D. D) It activates the gastric glands to produce gastric juices: This is incorrect. The production and release of gastric juices from gastric glands are primarily regulated by hormones like gastrin and neural signals from the vagus nerve. Pepsin is an enzyme within the gastric juice, not an activator of the glands themselves.

Q30. Living things that converted nitrogen-containing molecules into nitrates are:

A. Decomposers (e.g. bacteria and fungi)✓
B. Producers (e.g. green plants)
C. Primary consumers (e.g. mice)
D. Secondary consumers (e.g.snakes)
E. Tertiary consumers (e.g. hawks)

Explanation: Decomposer bacteria convert organic nitrogencontaining compounds into ammonia (NH3) . Nitrifying bacteria can convert ammonia (NH3) into nitrite-containing (NO2) compounds , which in turn can be converted into nitrates.

Why the other options are wrong

B. Since decomposers is the correct answer, hence this option is incorrect.
C. Since decomposers is the correct answer, hence this option is incorrect.
D. Since decomposers is the correct answer, hence this option is incorrect.
E. Since decomposers is the correct answer, hence this option is incorrect.

Q31. Vascular cambium produces:

A. Primary xylem and primary phloem
B. Secondary xylem and secondary phloem✓
C. Primary xylem and secondary phloem
D. Secondary xylem and primary phloem

Explanation: The vascular cambium is a lateral meristem that plays a key role in the secondary growth of dicot stems and roots. It forms a thin, actively dividing layer between the xylem and phloem. The cambium produces secondary xylem towards the inside and secondary phloem towards the outside, which together increase the thickness (girth) of the stem. Hence, vascular cambium is responsible for the formation of secondary xylem and secondary phloem.

Why the other options are wrong

A. The primary xylem and primary phloem are produced by the procambium, which is a primary meristem.
C. The option is incorrect because vascular cambium does not produce primary xylem. Vascular cambium is a secondary meristem that arises during secondary growth in plants and gives rise to secondary xylem toward the inside and secondary phloem toward the outside. Primary xylem and primary phloem, on the other hand, are formed by the procambium during primary growth. Therefore, vascular cambium contributes only to the formation of secondary vascular tissues, not primary xylem.
D. This option is incorrect because vascular cambium does not produce primary phloem. It only produces secondary xylem toward the inside and secondary phloem toward the outside. Primary phloem is formed earlier by the procambium during primary growth, not by the cambium. So pairing secondary xylem with primary phloem is wrong.

Q32. The action for the roots of a tree growing into a drain pipe is an example of:

A. Positive hydrotropism✓
B. Negative hydrotropism
C. Positive geotropism
D. Negative geotropism
E. Negative phototropism

Explanation: This is a case of positive hydrotropism because the roots are growing towards the drain pipe, which is a source of water.

Why the other options are wrong

B. B) Negative hydrotropism: This would mean growth away from water, which is contrary to how roots behave when seeking water.
C. C) Positive geotropism: This refers to growth towards the center of the Earth due to gravity (downward growth). While roots do exhibit positive geotropism, the specific action of growing into a drain pipe is primarily driven by the search for water (hydrotropism), not solely by gravity.
D. D) Negative geotropism: This would mean growth against the force of gravity (upward growth), which is characteristic of shoots, not roots.
E. E) Negative phototropism: This refers to growth away from light. Roots generally grow away from light, but this phenomenon doesn't explain their tendency to grow into a drain pipe. The primary stimulus for entering the pipe is the presence of water.

Q33. The diagram below is of a cell from a gametophyte undergoing division. What is the type of division and the stage?

A. Melosis; Prophase I; 2n = 8
B. Melosis; metaphase II; n = 8
C. Mitosis; Prophase; n = 8
D. Mitosis; early metaphase; 2n = 8
E. Mitosis; early anaphase; 2n = 8✓

Explanation: The gametophyte is haploid; a haploid cell cannot undergo meiosis. The cell is in anaphase because the diagram shows sister chromatids being pulled apart by the contraction of spindle fibres.

Why the other options are wrong

A. The gametophyte is haploid; a haploid cell cannot undergo meiosis. The cell is in anaphase because the diagram shows sister chromatids being pulled apart by the contraction of spindle fibres.
B. The gametophyte is haploid; a haploid cell cannot undergo meiosis. The cell is in anaphase because the diagram shows sister chromatids being pulled apart by the contraction of spindle fibres.
C. The gametophyte is haploid; a haploid cell cannot undergo meiosis. The cell is in anaphase because the diagram shows sister chromatids being pulled apart by the contraction of spindle fibres.
D. The gametophyte is haploid; a haploid cell cannot undergo meiosis. The cell is in anaphase because the diagram shows sister chromatids being pulled apart by the contraction of spindle fibres.

Q34. The female gamete of a flower is formed in the:

A. Pollen grain
B. Pollen tube
C. Stigma
D. Style
E. Embryonic sac✓

Explanation: In flowering plants, male and female gametes are produced in the anther and ovule, respectively. Within the ovule is the embryonic sac which contains the female gamete or ovum.

Why the other options are wrong

A. A) Pollen grain: Pollen grains contain the male gametes (sperm cells), not the female gamete.
B. B) Pollen tube: The pollen tube is a structure that grows from the germinating pollen grain down through the style to the ovule. Its function is to transport the male gametes to the embryo sac for fertilization, not to form the female gamete itself.
C. C) Stigma: The stigma is the receptive tip of the pistil (female reproductive organ) where pollen grains land during pollination. It plays a role in receiving pollen but does not form the female gamete.
D. D) Style: The style is the stalk-like part of the pistil that connects the stigma to the ovary. The pollen tube grows through the style, but the style itself is not involved in the formation of the female gamete.

Q35. If the sperm cell of a fruit fly has 4 chromosomes, then the number of chromosomes in each body cell is:

A. 2
B. 4
C. 6
D. 8✓
E. 10

Explanation: Sperm cells are haploid (n) while body cells or somatic cells are diploid (2n). If the haploid chromosomal number is 4, then a somatic cell should have 8 chromosomes.

Why the other options are wrong

A. Sperm cells are haploid (n) while body cells or somatic cells are diploid (2n). If the haploid chromosomal number is 4, then a somatic cell should have 8 chromosomes.
B. Sperm cells are haploid (n) while body cells or somatic cells are diploid (2n). If the haploid chromosomal number is 4, then a somatic cell should have 8 chromosomes.
C. Sperm cells are haploid (n) while body cells or somatic cells are diploid (2n). If the haploid chromosomal number is 4, then a somatic cell should have 8 chromosomes.
E. Sperm cells are haploid (n) while body cells or somatic cells are diploid (2n). If the haploid chromosomal number is 4, then a somatic cell should have 8 chromosomes.

Q36. In the given diagram, which is the first cell to undergo mitosis?

A. 3✓
B. 4
C. 5
D. 6
E. 8

Explanation: Cells 1 and 2 are the sperm and ovum respectively. Their fusion (fertilisation) results in a diploid zygote (cell 3). Further mitotic divisions of the zygote result in different embryonic structures. These initial divisions of the zygote are termed “cleavage”.

Why the other options are wrong

B. Cells 1 and 2 are the sperm and ovum respectively. Their fusion (fertilisation) results in a diploid zygote (cell 3). Further mitotic divisions of the zygote result in different embryonic structures. These initial divisions of the zygote are termed “cleavage”.
C. Cells 1 and 2 are the sperm and ovum respectively. Their fusion (fertilisation) results in a diploid zygote (cell 3). Further mitotic divisions of the zygote result in different embryonic structures. These initial divisions of the zygote are termed “cleavage”.
D. Cells 1 and 2 are the sperm and ovum respectively. Their fusion (fertilisation) results in a diploid zygote (cell 3). Further mitotic divisions of the zygote result in different embryonic structures. These initial divisions of the zygote are termed “cleavage”.
E. Cells 1 and 2 are the sperm and ovum respectively. Their fusion (fertilisation) results in a diploid zygote (cell 3). Further mitotic divisions of the zygote result in different embryonic structures. These initial divisions of the zygote are termed “cleavage”.

Q37. The diagram shows a generalized animal cell, which structure would be involved in the final secretion of digestive enzymes from this cell? (Question canceled in original)

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

Explanation: C is the Golgi apparatus. B is the endoplasmic reticulum. All types of substances that are secreted and excreted are packed in vesicles by Golgi bodies for passage to the outside.

Why the other options are wrong

A. C is the Golgi apparatus. B is the endoplasmic reticulum. All types of substances that are secreted and excreted are packed in vesicles by Golgi bodies for passage to the outside.
B. C is the Golgi apparatus. B is the endoplasmic reticulum. All types of substances that are secreted and excreted are packed in vesicles by Golgi bodies for passage to the outside.
D. C is the Golgi apparatus. B is the endoplasmic reticulum. All types of substances that are secreted and excreted are packed in vesicles by Golgi bodies for passage to the outside.
E. C is the Golgi apparatus. B is the endoplasmic reticulum. All types of substances that are secreted and excreted are packed in vesicles by Golgi bodies for passage to the outside.

Q38. In sickle cell animals, the abnormal hemoglobin, differ from normal hemoglobin in:

A. A single amino acid✓
B. The amount of coenzymes
C. The number or iron atoms
D. The number of magnesium atoms
E. The number of genes

Explanation: The mutation causing sickle cell anemia is a single nucleotide substitution in the codon for amino acid 6. The change converts a glutamic acid codon to a valine codon.

Why the other options are wrong

B. The mutation causing sickle cell anemia is a single nucleotide substitution in the codon for amino acid 6. The change converts a glutamic acid codon to a valine codon.
C. The mutation causing sickle cell anemia is a single nucleotide substitution in the codon for amino acid 6. The change converts a glutamic acid codon to a valine codon.
D. The mutation causing sickle cell anemia is a single nucleotide substitution in the codon for amino acid 6. The change converts a glutamic acid codon to a valine codon.
E. The mutation causing sickle cell anemia is a single nucleotide substitution in the codon for amino acid 6. The change converts a glutamic acid codon to a valine codon.

Q39. Organisms that are included in the protoctista are:

A. Slime molds✓
B. Conidia
C. Mucor
D. Spongocoel
E. Euplectella

Explanation: Slime Molds are single-celled organisms that belong to the kingdom Protoctista.

Why the other options are wrong

B. B) Conidia: Conidia are asexual spores produced by fungi. They are not organisms themselves, but rather reproductive structures of fungi, which belong to the Kingdom Fungi, not Protista.
C. C) Mucor: Mucor is a genus of mold that belongs to the Kingdom Fungi.
D. D) Spongocoel: A spongocoel is the central cavity found within the body of sponges. Sponges belong to the Phylum Porifera, which is part of the Kingdom Animalia. Therefore, spongocoel is an anatomical structure, not an organism, and it's associated with animals, not protists.
E. E) Euplectella: Euplectella is a genus of glass sponges, commonly known as Venus' Flower Baskets. Sponges are classified under the Phylum Porifera, which belongs to the Kingdom Animalia.

Q40. Melting point of alkali metals decreases with:

A. Decrease in size
B. Decrease in density
C. Decrease in atomic mass
D. Increase in size✓
E. No fixed pattern

Explanation: Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.

Why the other options are wrong

A. Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.
B. Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.
C. Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.
E. Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.

Q41. In which of the following reaction Kp and Kc will have the same numerical value?

A. 2SO2 + O2 ⇌ 2SO3
B. N2 + 3H2 ⇌ 2NH3
C. PCI5 ⇌ PCI3 + CI2
D. N2 + O2 ⇌ 2NO✓

Explanation: Kp and Kc have the ame numerical value when the number of moles of reactants and products is equal.

Why the other options are wrong

A. A) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) * Moles of gaseous products = 2 * Moles of gaseous reactants = 2 + 1 = 3 * Δn = 2 - 3 = -1 * Kp = Kc(RT)^(-1) ≠ Kc
B. B) N₂(g) + 3H₂(g) ⇌ 2NH₃(g) * Moles of gaseous products = 2 * Moles of gaseous reactants = 1 + 3 = 4 * Δn = 2 - 4 = -2 * Kp = Kc(RT)^(-2) ≠ Kc
C. Down the group atomic size increases with increase in number of atomic shells and therefore, the strength of metallic bond decreases. So melting point decreases.

Q42. On heating, gypsum gives:

A. CaSO3
B. CaSO4
C. CaSO4 . ½ H2O✓
D. CaO + SO3
E. NaHCO3

Explanation: Plaster of Paris is manufactured by heating gypsum at 120° and 150°K. On heating gypsum at 150°C, it loses water molecules and becomes calcium sulfate hemihydrate. This product is known as the plaster of Paris. Hence option (c) is correct. The chemical formula for the plaster of Paris is (CaSO4).½ H2O and is better known as calcium sulfate hemihydrate.

Why the other options are wrong

A. Plaster of Paris is manufactured by heating gypsum at 120° and 150°K. On heating gypsum at 150°C, it loses water molecules and becomes calcium sulfate hemihydrate. This product is known as the plaster of Paris. Hence option (c) is correct. The chemical formula for the plaster of Paris is (CaSO4).½ H2O and is better known as calcium sulfate hemihydrate.
B. Plaster of Paris is manufactured by heating gypsum at 120° and 150°K. On heating gypsum at 150°C, it loses water molecules and becomes calcium sulfate hemihydrate. This product is known as the plaster of Paris. Hence option (c) is correct. The chemical formula for the plaster of Paris is (CaSO4).½ H2O and is better known as calcium sulfate hemihydrate.
D. Plaster of Paris is manufactured by heating gypsum at 120° and 150°K. On heating gypsum at 150°C, it loses water molecules and becomes calcium sulfate hemihydrate. This product is known as the plaster of Paris. Hence option (c) is correct. The chemical formula for the plaster of Paris is (CaSO4).½ H2O and is better known as calcium sulfate hemihydrate.
E. Plaster of Paris is manufactured by heating gypsum at 120° and 150°K. On heating gypsum at 150°C, it loses water molecules and becomes calcium sulfate hemihydrate. This product is known as the plaster of Paris. Hence option (c) is correct. The chemical formula for the plaster of Paris is (CaSO4).½ H2O and is better known as calcium sulfate hemihydrate.

Q43. Polyvinyl acetate is the polymer of:

A. Vinyl chloride
B. Vinyl acetate✓
C. Formaldehyde
D. Methyl methacrylate

Explanation: Polyvinyl acetate is an addition polymer of vinyl acetate.

Why the other options are wrong

A. Polyvinyl acetate is an addition polymer of vinyl acetate.
C. Polyvinyl acetate is an addition polymer of vinyl acetate.
D. Polyvinyl acetate is an addition polymer of vinyl acetate.

Q44. C25H31-COO-C16H33 is the formula of:

A. Spermaceti
B. Bee wax✓
C. Glycerol
D. Choline acid
E. Pepsinogen

Explanation: The correct answer is Bee wax, which is primarily made up of myricyl palmitate, an ester formed from a long-chain fatty acid and a long-chain alcohol. The formula C25H31-COO-C16H33 suggests a structure similar to those found in waxes, as it includes a long hydrocarbon chain and an ester functional group. The other options, while they may share some structural similarities, do not match the ester composition or the specific carbon counts indicated in the question. Spermaceti is close but does not fit due to the discrepancy in the carbon chain lengths. Glycerol is a simple molecule with no ester bond, while Choline acid and Pepsinogen belong to entirely different classes of compounds (organic cations and proteins, respectively), making them irrelevant for this question.

Why the other options are wrong

A. Spermaceti is primarily made up of cetyl palmitate, a wax ester. While the formula provided differs in carbon count, spermaceti's structure closely resembles that of the formula in question, making it a relevant choice in terms of chemical class.
C. Glycerol is a simple polyol compound with a distinct and much simpler structure compared to the complex ester represented by the formula. It does not match in terms of functional groups present.
D. Choline acid refers to a different class of compounds, specifically organic cations. Its chemical structure and formula differ significantly from the ester structure indicated in the question.
E. Pepsinogen is a large, complex protein that functions as a zymogen. Its chemical structure is vastly different from the ester structure indicated in the question, making it an irrelevant option.

Q45. Which one of the following bond is the longest?

A. C≡C
B. C=C
C. C-C
D. H-bond✓
E. All of the above are equal

Explanation: The official answer key is flawed and marks C as the correct answer. Hydrogen bonds are secondary bonds, and are therefore weaker and longer than all of the aforementioned covalent bonds.

Why the other options are wrong

A. The official answer key is flawed and marks C as the correct answer. Hydrogen bonds are secondary bonds, and are therefore weaker and longer than all of the aforementioned covalent bonds.
B. The official answer key is flawed and marks C as the correct answer. Hydrogen bonds are secondary bonds, and are therefore weaker and longer than all of the aforementioned covalent bonds.
C. The official answer key is flawed and marks C as the correct answer. Hydrogen bonds are secondary bonds, and are therefore weaker and longer than all of the aforementioned covalent bonds.
E. Option D is correct

Q46. Consider the following system, PCl5 ⇌ PCl3 + Cl2. The unit of Kc for this reaction is:

A. mole2/L2
B. L2/mole2
C. L/mole
D. mole/L✓
E. mole/L2

Explanation: The units for Kc change depending on the concentrations of each reactant and product in the equation. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel each other out. However, in other cases, this is how you would work it out: The equation for Kc is [PRODUCTS]/[REACTANTS]. When simplified, the unit of Kc for the equation given in the question is mol/L.

Why the other options are wrong

A. The units for Kc change depending on the concentrations of each reactant and product in the equation. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel each other out. However, in other cases, this is how you would work it out:The equation for Kc is [PRODUCTS]/[REACTANTS].When simplified, the unit of Kc for the equation given in the question is mol/L.
B. The units for Kc change depending on the concentrations of each reactant and product in the equation. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel each other out. However, in other cases, this is how you would work it out:The equation for Kc is [PRODUCTS]/[REACTANTS].When simplified, the unit of Kc for the equation given in the question is mol/L.
C. The units for Kc change depending on the concentrations of each reactant and product in the equation. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel each other out. However, in other cases, this is how you would work it out:The equation for Kc is [PRODUCTS]/[REACTANTS].When simplified, the unit of Kc for the equation given in the question is mol/L.
E. The units for Kc change depending on the concentrations of each reactant and product in the equation. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel each other out. However, in other cases, this is how you would work it out:The equation for Kc is [PRODUCTS]/[REACTANTS].When simplified, the unit of Kc for the equation given in the question is mol/L.

Q47. IUPAC name of Butyric acid is:

A. Methanol acid
B. Butanoic acid✓
C. Ethanoic acid
D. Octadecanoic acid
E. Butyric acid is an IUPAC name itself

Explanation: The common and IUPAC names of some carboxylic acids are given below:

Why the other options are wrong

A. A) Methanoic acid: This is the IUPAC name for Formic acid.
C. C) Ethanoic acid: This is the IUPAC name for Acetic acid, which is the main component of vinegar.
D. D) Octadecanoic acid: This is the IUPAC name for Stearic acid, a common saturated fatty acid.
E. E) Butyric acid is an IUPAC name itself: Butyric acid is a common or trivial name, not an IUPAC (International Union of Pure and Applied Chemistry) systematic name.

Q48. In the reaction: H2 + I2 → 2HI Δ=+12.40 kJ/mole represents:

A. Heat of formation✓
B. Heat of neutralization
C. Heat of oxidation
D. Heat of sublimation
E. Heat of reduction

Explanation: Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid).

Why the other options are wrong

B. Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid).
C. Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid).
D. Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid).
E. Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid).

Q49. A reactant in a chemical reaction is non conductor to electricity colorless and non-responsive towards plane-polarized light but can absorb ultraviolet rays. Then the best technique for the determination of initial rate of such a reaction is:

A. Polarimetry
B. Refractometry
C. Spectroscopy✓
D. Colorimetry

Explanation: We can start by eliminating options A (“non-responsive towards plane-polarized light”) and D (“colorless”). Refractometry is a technique that measures how light is refracted when it passes through a given substance. This does not seem to be relevant to the question at all, hence we can eliminate B. Spectroscopy is the study of electromagnetic radiation and its production from, or its interaction with, matter. This makes sense because the question tells us that the sample can absorb ultraviolet rays.

Why the other options are wrong

A. We can start by eliminating options A (“non-responsive towards plane-polarized light”) and D (“colorless”). Refractometry is a technique that measures how light is refracted when it passes through a given substance. This does not seem to be relevant to the question at all, hence we can eliminate B. Spectroscopy is the study of electromagnetic radiation and its production from, or its interaction with, matter. This makes sense because the question tells us that the sample can absorb ultraviolet rays.
B. We can start by eliminating options A (“non-responsive towards plane-polarized light”) and D (“colorless”). Refractometry is a technique that measures how light is refracted when it passes through a given substance. This does not seem to be relevant to the question at all, hence we can eliminate B. Spectroscopy is the study of electromagnetic radiation and its production from, or its interaction with, matter. This makes sense because the question tells us that the sample can absorb ultraviolet rays.
D. We can start by eliminating options A (“non-responsive towards plane-polarized light”) and D (“colorless”). Refractometry is a technique that measures how light is refracted when it passes through a given substance. This does not seem to be relevant to the question at all, hence we can eliminate B. Spectroscopy is the study of electromagnetic radiation and its production from, or its interaction with, matter. This makes sense because the question tells us that the sample can absorb ultraviolet rays.

Q50. A 5% NaCl solution by weight means:

A. 5g of NaCl/ 95g of water
B. 5g of NaCl/ 95g of solution
C. 5g of NaCl/ 100g of water
D. 5g of NaCl/ 100g of solution✓
E. 95g of NaCl/ 100ml of water

Explanation: The %w/w formula is expressed as follows: It can be argued that option A also satisfies this condition, but, strictly speaking, in percent solutions, the weight of a solute is expressed as a percentage of the total solution in weight. Thus the most suitable answer is D.

Why the other options are wrong

A. The %w/w formula is expressed as follows:It can be argued that option A also satisfies this condition, but, strictly speaking, in percent solutions, the weight of a solute is expressed as a percentage of the total solution in weight. Thus the most suitable answer is D.
B. The %w/w formula is expressed as follows:It can be argued that option A also satisfies this condition, but, strictly speaking, in percent solutions, the weight of a solute is expressed as a percentage of the total solution in weight. Thus the most suitable answer is D.
C. The %w/w formula is expressed as follows:It can be argued that option A also satisfies this condition, but, strictly speaking, in percent solutions, the weight of a solute is expressed as a percentage of the total solution in weight. Thus the most suitable answer is D.
E. The %w/w formula is expressed as follows:It can be argued that option A also satisfies this condition, but, strictly speaking, in percent solutions, the weight of a solute is expressed as a percentage of the total solution in weight. Thus the most suitable answer is D.

Q51. Which of the following is required as the catalyst in the decomposition of laughing gas?

A. Traces of Cl2
B. 2% ethyl alcohol
C. Glycerine
D. MnO2✓
E. H2O

Explanation: MnO2 (Manganese Dioxide): Research indicates that various metal oxides, including Manganese Dioxide (MnO2), act as catalysts for the decomposition of nitrous oxide (N2O), commonly known as laughing gas. Nanocrystalline β–MnO2 and cryptomelane type MnO2 have been specifically cited as effective catalysts for this reaction.

Why the other options are wrong

A. A) Traces of Cl2 (Chlorine): While halogens like chlorine, bromine, and iodine have been shown to have a catalytic effect on the thermal decomposition of nitrous oxide, MnO2 is a more commonly recognized and broadly effective catalyst in this context, especially in typical chemistry curricula.
B. B) 2% ethyl alcohol: There is no scientific evidence to suggest that ethyl alcohol acts as a catalyst in the decomposition of laughing gas.
C. C) Glycerine: There is no scientific evidence to suggest that glycerine acts as a catalyst in the decomposition of laughing gas.
E. E) H2O (Water): Water is often found to inhibit the catalytic decomposition of N2O, rather than promote it.

Q52. The addition reactions shown by alkenes are examples of:

A. Electrophilic addition reactions✓
B. Nucleophilic addition reactions
C. Free radical addition reactions
D. The formation of carbanion as the intermediate
E. None of the given options

Explanation: A) Electrophilic addition reactions (Correct): Alkenes contain a carbon-carbon double bond, which is a region of high electron density due to the presence of pi electrons. This makes them susceptible to attack by electrophiles (electron-loving species). In these reactions, the pi bond breaks, and new single bonds are formed with the electrophile and a nucleophile.

Why the other options are wrong

B. B) Nucleophilic addition reactions (Incorrect): Nucleophilic addition reactions typically occur with compounds containing highly polarized multiple bonds, such as aldehydes and ketones (carbonyl compounds), where a carbon atom is electron-deficient and can be attacked by a nucleophile (electron-rich species). Alkenes, with their electron-rich double bond, do not readily undergo nucleophilic addition.
C. C) Free radical addition reactions (Partially correct, but not the primary answer): While some addition reactions of alkenes can proceed via a free radical mechanism (e.g., addition of HBr in the presence of peroxides), the most common and characteristic type of addition reaction for alkenes, in general, is electrophilic addition. When the question refers to "the addition reactions shown by alkenes," it typically implies their characteristic reactivity.
D. D) The formation of carbanion as the intermediate (Incorrect): In electrophilic addition reactions to alkenes, the intermediate formed is generally a carbocation (a positively charged carbon species), not a carbanion (a negatively charged carbon species). Carbanions are typically formed in reactions where a carbon atom is deprotonated or undergoes nucleophilic attack at an electron-deficient center.
E. Option A is correct

Q53. The basic structure of a crystalline substance is called;

A. Unit Cell✓
B. Molecule
C. Lattice
D. Matrix

Explanation: A crystal is formed by a large number of repetition of basic pattern of atoms, ions or molecules in space. The basic structural unit which when repeated in three dimensions generates the crystal structure is called a unit cell.

Why the other options are wrong

B. A crystal is formed by a large number of repetition of basic pattern of atoms, ions or molecules in space. The basic structural unit which when repeated in three dimensions generates the crystal structure is called a unit cell.
C. A crystal is formed by a large number of repetition of basic pattern of atoms, ions or molecules in space. The basic structural unit which when repeated in three dimensions generates the crystal structure is called a unit cell.
D. A crystal is formed by a large number of repetition of basic pattern of atoms, ions or molecules in space. The basic structural unit which when repeated in three dimensions generates the crystal structure is called a unit cell.

Q54. Which of the following atom has an ionic radius that is larger than its atomic radius?

A. Na
B. Al
C. Mg
D. Si
E. Cl✓

Explanation: (i) Ionic radii show that the cations have smaller radii than neutral atoms. On removal of an electron the effective charge on the nucleus increases and pulls the electrons more firmly e.g. radius of Na atom: 1.57Å radius of Na+ ion = 0.95 Å (ii) Anions have a larger radii than the neutral atoms as an excess of negative charge results in greater electron repulsion e.g. radius of Cl atom = 0.99 Å radius of Cl- ion = 1.81 Å(iii) In case of ions which have the same electronic configuration like Na+, Mg2+, Al3+, (1s2, 2s2, 2p6) the radii decrease with increasing nuclear charge. The ions which have the same electronic configuration, as given above, are called isoelectronic.

Why the other options are wrong

A. (i) Ionic radii show that the cations have smaller radii than neutral atoms. On removal of an electron the effective charge on the nucleus increases and pulls the electrons more firmly e.g. radius of Na atom: 1.57Åradius of Na+ ion = 0.95 Å(ii) Anions have a larger radii than the neutral atoms as an excess of negative charge results in greater electron repulsion e.g.radius of Cl atom = 0.99 Åradius of Cl- ion = 1.81 Å(iii) In case of ions which have the same electronic configuration like Na+, Mg2+, Al3+, (1s2, 2s2, 2p6) the radii decrease with increasing nuclear charge. The ions which have the same electronic configuration, as given above, are called isoelectronic.
B. (i) Ionic radii show that the cations have smaller radii than neutral atoms. On removal of an electron the effective charge on the nucleus increases and pulls the electrons more firmly e.g. radius of Na atom: 1.57Åradius of Na+ ion = 0.95 Å(ii) Anions have a larger radii than the neutral atoms as an excess of negative charge results in greater electron repulsion e.g.radius of Cl atom = 0.99 Åradius of Cl- ion = 1.81 Å(iii) In case of ions which have the same electronic configuration like Na+, Mg2+, Al3+, (1s2, 2s2, 2p6) the radii decrease with increasing nuclear charge. The ions which have the same electronic configuration, as given above, are called isoelectronic.
C. (i) Ionic radii show that the cations have smaller radii than neutral atoms. On removal of an electron the effective charge on the nucleus increases and pulls the electrons more firmly e.g. radius of Na atom: 1.57Åradius of Na+ ion = 0.95 Å(ii) Anions have a larger radii than the neutral atoms as an excess of negative charge results in greater electron repulsion e.g.radius of Cl atom = 0.99 Åradius of Cl- ion = 1.81 Å(iii) In case of ions which have the same electronic configuration like Na+, Mg2+, Al3+, (1s2, 2s2, 2p6) the radii decrease with increasing nuclear charge. The ions which have the same electronic configuration, as given above, are called isoelectronic.
D. (i) Ionic radii show that the cations have smaller radii than neutral atoms. On removal of an electron the effective charge on the nucleus increases and pulls the electrons more firmly e.g. radius of Na atom: 1.57Åradius of Na+ ion = 0.95 Å(ii) Anions have a larger radii than the neutral atoms as an excess of negative charge results in greater electron repulsion e.g.radius of Cl atom = 0.99 Åradius of Cl- ion = 1.81 Å(iii) In case of ions which have the same electronic configuration like Na+, Mg2+, Al3+, (1s2, 2s2, 2p6) the radii decrease with increasing nuclear charge. The ions which have the same electronic configuration, as given above, are called isoelectronic.

Q55. The rate law expression for the reaction N2 + 3H2 → 2NH3

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

Explanation: The rate equation for this reaction should be given by: Rate= k[N2][H2]3

Why the other options are wrong

A. The rate equation for this reaction should be given by:Rate= k[N2][H2]3
B. The rate equation for this reaction should be given by:Rate= k[N2][H2]3
C. The rate equation for this reaction should be given by:Rate= k[N2][H2]3
D. The rate equation for this reaction should be given by:Rate= k[N2][H2]3

Q56. The IUPAC name of the given compound is:

A. 6,7-Dimethyl-7-n-propyl nonane
B. 4,5-Dimethyl-4-ethyl decane✓
C. 3,4-Dimethyl-3-n-propyl nonane
D. 6,7-Dimethyl-7-ethyl decane
E. 6,7-Dimethyl-7-ethyl octane

Explanation: The structure can be simplified in the following way:

Why the other options are wrong

A. 4,5-Dimethyl-4-ethyl decane is correct
C. 4,5-Dimethyl-4-ethyl decane is correct
D. 4,5-Dimethyl-4-ethyl decane is correct
E. 4,5-Dimethyl-4-ethyl decane is correct

Q57. An inorganic compound forming an organic substance on heating is:

A. Ammonium Cyanate✓
B. Sodamide
C. Soda lime
D. Potassium Cyanate
E. Both ‘Ammonium Cyanate’ & 'Potassium Cyanate’

Explanation: Factual recall: “..in Germany, a young chemist Wohler in 1828, prepared urea in the laboratory from inorganic material, ammonium cyanate by boiling it with water. Urea is a waste product of metabolism and is excreted through urine and hence it is a typical organic compound.” (Sindh Textbook For Chemistry Grade 12, Chapter 6: Introduction to organic chemistry).

Why the other options are wrong

B. Factual recall: “..in Germany, a young chemist Wohler in 1828, prepared urea in the laboratory from inorganic material, ammonium cyanate by boiling it with water. Urea is a waste product of metabolism and is excreted through urine and hence it is a typical organic compound.” (Sindh Textbook For Chemistry Grade 12, Chapter 6: Introduction to organic chemistry).
C. Factual recall: “..in Germany, a young chemist Wohler in 1828, prepared urea in the laboratory from inorganic material, ammonium cyanate by boiling it with water. Urea is a waste product of metabolism and is excreted through urine and hence it is a typical organic compound.” (Sindh Textbook For Chemistry Grade 12, Chapter 6: Introduction to organic chemistry).
D. Factual recall: “..in Germany, a young chemist Wohler in 1828, prepared urea in the laboratory from inorganic material, ammonium cyanate by boiling it with water. Urea is a waste product of metabolism and is excreted through urine and hence it is a typical organic compound.” (Sindh Textbook For Chemistry Grade 12, Chapter 6: Introduction to organic chemistry).
E. Factual recall: “..in Germany, a young chemist Wohler in 1828, prepared urea in the laboratory from inorganic material, ammonium cyanate by boiling it with water. Urea is a waste product of metabolism and is excreted through urine and hence it is a typical organic compound.” (Sindh Textbook For Chemistry Grade 12, Chapter 6: Introduction to organic chemistry).

Q58. Halogenations of benzene can be done by:

A. FeCI3
B. FeBr3
C. AlCI3
D. AlBr3
E. All of the given options✓

Explanation: The halogenation of benzene requires a Lewis acid catalyst. All of these compounds are Lewis acids. For more context, please refer to the following extract from the Sindh Textbook for Chemistry, Grade 12, Chapter 7: Hydrocarbons.

Why the other options are wrong

A. The halogenation of benzene requires a Lewis acid catalyst. All of these compounds are Lewis acids. For more context, please refer to the following extract from the Sindh Textbook for Chemistry, Grade 12, Chapter 7: Hydrocarbons.
B. The halogenation of benzene requires a Lewis acid catalyst. All of these compounds are Lewis acids. For more context, please refer to the following extract from the Sindh Textbook for Chemistry, Grade 12, Chapter 7: Hydrocarbons.
C. The halogenation of benzene requires a Lewis acid catalyst. All of these compounds are Lewis acids. For more context, please refer to the following extract from the Sindh Textbook for Chemistry, Grade 12, Chapter 7: Hydrocarbons.
D. The halogenation of benzene requires a Lewis acid catalyst. All of these compounds are Lewis acids. For more context, please refer to the following extract from the Sindh Textbook for Chemistry, Grade 12, Chapter 7: Hydrocarbons.

Q59. The following reaction is called:C2H5ONa + C2H5Br —> (C2H5)2O + NaBr

A. Wurtz synthesis
B. Kolbe synthesis
C. Williamson synthesis✓
D. Hoffman reaction
E. Castner-Kellner reaction

Explanation: Williamson ether synthesis takes place as follows:

Why the other options are wrong

A. Wurtz Synthesis: This reaction is used to couple two alkyl halides in the presence of sodium metal and dry ether to form a higher alkane (e.g., 2 R-X + 2 Na → R-R + 2 NaX). It involves the formation of a new carbon-carbon bond, typically leading to a longer alkane chain. This is different from the given reaction, which produces an ether, not an alkane, and does not use sodium metal as a coupling agent in the same way.
B. Kolbe Synthesis: The term "Kolbe synthesis" can refer to two different reactions: * Kolbe Electrolysis: This is an electrolytic decarboxylation of carboxylic acid salts to produce symmetrical alkanes. * Kolbe-Schmitt Reaction: This is the carboxylation of phenols with carbon dioxide under pressure to form salicylic acid. Neither of these reactions matches the reactants or products of the given reaction.
D. Hoffman Reaction (Hoffmann Bromamide Degradation): This reaction is a method for converting a primary amide into a primary amine with one less carbon atom. It typically involves treating the amide with bromine and a strong base. The given reaction does not involve an amide as a reactant or produce an amine with carbon loss.
E. Castner-Kellner Reaction: This is an industrial electrochemical process used for the production of sodium hydroxide (caustic soda) and chlorine gas from the electrolysis of an aqueous solution of sodium chloride (brine) using a mercury cathode. This is an inorganic industrial process, entirely unrelated to the organic synthesis of ethers.

Q60. The radius of the second Bohr orbit is:

A. 0.529 Å
B. 0.529/4 Å
C. 0.529 x 22 Å✓
D. 0.529 x 2 Å

Explanation: The radius of nth Bohr orbit is given by (0.529 x n2) Å.

Why the other options are wrong

A. The radius of nth Bohr orbit is given by (0.529 x n2) Å.
B. The radius of nth Bohr orbit is given by (0.529 x n2) Å.
D. The radius of nth Bohr orbit is given by (0.529 x n2) Å.

Q61. Which of the following is a false statement concerning an ionic compound?

A. Greater the lattice energy, the greater the solubility✓
B. Higher the dielectric constant of the solvent, the greater is the solubility
C. Higher the dipole moment of the solvent, the greater is the solubility
D. Increase in temperature generally increases solubility
E. Have greater boiling and melting point

Explanation: Higher the lattice enthalpy lower will be solubility i.e., lattice enthalpy ∝ 1 Solubility

Why the other options are wrong

B. Higher the lattice enthalpy lower will be solubility i.e.,lattice enthalpy ∝ 1 /Solubility
C. Higher the lattice enthalpy lower will be solubility i.e.,lattice enthalpy ∝ 1 / Solubility
D. Higher the lattice enthalpy lower will be solubility i.e.,lattice enthalpy ∝ 1 /Solubility
E. Higher the lattice enthalpy lower will be solubility i.e.,lattice enthalpy ∝ 1 / Solubility

Q62. The term -d2x / dt in the rate expression refers to the:

A. Decrease in concentration of the reactant X✓
B. Instantaneous rate of reaction
C. Increase in concentration of the reactants
D. Increase in solubility of the reactants

Explanation: The negative sign is a clear indication that the quantity “x” is decreasing. It is the reactant concentration that decreases during a chemical reaction.

Why the other options are wrong

B. The negative sign is a clear indication that the quantity “x” is decreasing. It is the reactant concentration that decreases during a chemical reaction.
C. The negative sign is a clear indication that the quantity “x” is decreasing. It is the reactant concentration that decreases during a chemical reaction.
D. The negative sign is a clear indication that the quantity “x” is decreasing. It is the reactant concentration that decreases during a chemical reaction.

Q63. For the reaction 2HCl ⇌ H2 + Cl2 the Kc is 1.0 x 10-5 . If the equilibrium concentrations of H2 and Cl2 are 1.2 x 10-3 M and 1.2 x 10-4 M respectively, the concentration of HCl is:

A. 12 x 10-2 M✓
B. 12 x 10-4 M
C. 12 x 10-3 M
D. 12 x 10-5 M
E. 10 x 10-2 M

Explanation: The correct answer is A, according to the official NTS key.

Why the other options are wrong

Q64. Which of the following statements is wrong?

A. An endothermic reaction must absorb energy before it occurs
B. During exothermic reaction heat is evolved
C. After an endothermic reaction, there is no change in the temperature of the reaction mixture✓
D. None of the given options

Explanation: In an endothermic reaction, the energy absorbed by reactants during bond breaking exceeds that released by the products during bond formation. This results in a net decrease in the temperature of the surrounding.

Why the other options are wrong

A. In an endothermic reaction, the energy absorbed by reactants during bond breaking exceeds that released by the products during bond formation. This results in a net decrease in the temperature of the surrounding.
B. In an endothermic reaction, the energy absorbed by reactants during bond breaking exceeds that released by the products during bond formation. This results in a net decrease in the temperature of the surrounding.
D. Option A is correct

Q65. Which of the following series in H atom give spectral line of 4860 Å?

A. Lyman
B. Paschen
C. Balmer✓
D. Pfund
E. None of the given options

Explanation: This wavelength falls within the range of visible light (4000-7000 angstrom), hence this spectral line belongs to Balmer series.

Why the other options are wrong

A. The wavelength of first line of lyman series in hydrogen atom is 1216 angstrom.
B. The wavelength range of first line of paschen series in hydrogen atom is 8200A˚to 18750A˚ .
D. The wavelength range of first line of Pfund series in hydrogen atom is 18750A˚to 60000Ao.
E. This wavelength falls within the range of visible light (4000-7000 angstrom), hence this spectral line belongs to Balmer series.

Q66. In a compound X, all the bond angles are exactly 109.28˚, X is:

A. Chloroform
B. Iodoform
C. Chloromethane
D. Carbon tetrachloride✓

Explanation: The explanation is as follows:

Why the other options are wrong

A. Chloroform (CHCl3): Chloroform has a central carbon atom bonded to one hydrogen atom and three chlorine atoms. While its molecular geometry is tetrahedral, the presence of different atoms (H and Cl) around the central carbon causes slight distortions in the bond angles. The Cl-C-Cl bond angles are approximately 110.4°, and the H-C-Cl bond angles deviate from the ideal 109.28°.
B. Iodoform (CHI3): Similar to chloroform, iodoform has a central carbon atom bonded to one hydrogen atom and three iodine atoms. This difference in bonded atoms leads to a distorted tetrahedral geometry. The I-C-I bond angle is approximately 113.0°.
C. Chloromethane (CH3Cl): Chloromethane has a central carbon atom bonded to three hydrogen atoms and one chlorine atom. The differing electronegativity and size of hydrogen and chlorine atoms cause the bond angles to deviate from the ideal tetrahedral angle. For instance, the H-C-H bond angles are about 110.5°, and the H-C-Cl angles are around 109.1°.

Q67. Fat soluble vitamins are mainly stored in:

A. Liver✓
B. Skin
C. Stomach
D. Bones
E. Fat soluble vitamins cannot be stored

Explanation: Fat-soluble vitamins A, D, and E are stored primarily in the liver, kidney, and adipose tissues.

Why the other options are wrong

B. Fat-soluble vitamins A, D, and E are stored primarily in the liver, kidney, and adipose tissues.
C. Fat-soluble vitamins A, D, and E are stored primarily in the liver, kidney, and adipose tissues.
D. Fat-soluble vitamins A, D, and E are stored primarily in the liver, kidney, and adipose tissues.
E. Fat-soluble vitamins A, D, and E are stored primarily in the liver, kidney, and adipose tissues.

Q68. The rate of diffusion of a gas having molecular weight just double of N2 gas is 56 mL s-1 . The rate of diffusion of N2 will be:

A. 79.19 56 mL s-1✓
B. 56 mL s-1
C. 112.0 mL s-1
D. 90.0 mL s-1
E. 210 mL s-1

Explanation: R_N2 = Rate of diffusion of Nitrogen gas (N₂) * R_X = Rate of diffusion of the unknown gas (X) * M_N2 = Molar mass of Nitrogen gas (N₂) * M_X = Molar mass of the unknown gas (X)Given information: * R_X = 56 mL s⁻¹ * M_X = 2 * M_N2 (Molecular weight of the unknown gas is double that of N₂)Now, substitute these values into Graham's Law:R_N2 / R_X = √(M_X / M_N2)R_N2 / 56 = √((2 * M_N2) / M_N2)R_N2 / 56 = √2R_N2 = 56 * √2R_N2 = 56 * 1.4142R_N2 ≈ 79.1952 mL s⁻¹Rounding to two decimal places, R_N2 = 79.19 mL s⁻¹.

Why the other options are wrong

B. R_N2 = Rate of diffusion of Nitrogen gas (N₂) * R_X = Rate of diffusion of the unknown gas (X) * M_N2 = Molar mass of Nitrogen gas (N₂) * M_X = Molar mass of the unknown gas (X)Given information: * R_X = 56 mL s⁻¹ * M_X = 2 * M_N2 (Molecular weight of the unknown gas is double that of N₂)Now, substitute these values into Graham's Law:R_N2 / R_X = √(M_X / M_N2)R_N2 / 56 = √((2 * M_N2) / M_N2)R_N2 / 56 = √2R_N2 = 56 * √2R_N2 = 56 * 1.4142R_N2 ≈ 79.1952 mL s⁻¹Rounding to two decimal places, R_N2 = 79.19 mL s⁻¹.
C. R_N2 = Rate of diffusion of Nitrogen gas (N₂) * R_X = Rate of diffusion of the unknown gas (X) * M_N2 = Molar mass of Nitrogen gas (N₂) * M_X = Molar mass of the unknown gas (X)Given information: * R_X = 56 mL s⁻¹ * M_X = 2 * M_N2 (Molecular weight of the unknown gas is double that of N₂)Now, substitute these values into Graham's Law:R_N2 / R_X = √(M_X / M_N2)R_N2 / 56 = √((2 * M_N2) / M_N2)R_N2 / 56 = √2R_N2 = 56 * √2R_N2 = 56 * 1.4142R_N2 ≈ 79.1952 mL s⁻¹Rounding to two decimal places, R_N2 = 79.19 mL s⁻¹.
D. R_N2 = Rate of diffusion of Nitrogen gas (N₂) * R_X = Rate of diffusion of the unknown gas (X) * M_N2 = Molar mass of Nitrogen gas (N₂) * M_X = Molar mass of the unknown gas (X)Given information: * R_X = 56 mL s⁻¹ * M_X = 2 * M_N2 (Molecular weight of the unknown gas is double that of N₂)Now, substitute these values into Graham's Law:R_N2 / R_X = √(M_X / M_N2)R_N2 / 56 = √((2 * M_N2) / M_N2)R_N2 / 56 = √2R_N2 = 56 * √2R_N2 = 56 * 1.4142R_N2 ≈ 79.1952 mL s⁻¹Rounding to two decimal places, R_N2 = 79.19 mL s⁻¹.
E. R_N2 = Rate of diffusion of Nitrogen gas (N₂) * R_X = Rate of diffusion of the unknown gas (X) * M_N2 = Molar mass of Nitrogen gas (N₂) * M_X = Molar mass of the unknown gas (X)Given information: * R_X = 56 mL s⁻¹ * M_X = 2 * M_N2 (Molecular weight of the unknown gas is double that of N₂)Now, substitute these values into Graham's Law:R_N2 / R_X = √(M_X / M_N2)R_N2 / 56 = √((2 * M_N2) / M_N2)R_N2 / 56 = √2R_N2 = 56 * √2R_N2 = 56 * 1.4142R_N2 ≈ 79.1952 mL s⁻¹Rounding to two decimal places, R_N2 = 79.19 mL s⁻¹.

Q69. The most reactive state of H2 is:

A. Atomic Hydrogen✓
B. Molecular Hydrogen
C. Adsorbed Hydrogen
D. Hydrogen in compound
E. H2O

Explanation: The explanation is as follows:

Why the other options are wrong

B. B) Molecular Hydrogen: Molecular hydrogen (H2) is a diatomic molecule where two hydrogen atoms are bonded together. This bond is strong, making H2 relatively unreactive under standard conditions. It requires significant energy (like high temperatures) to break the bond and allow the hydrogen atoms to react.
C. C) Adsorbed Hydrogen: Adsorbed hydrogen refers to hydrogen molecules or atoms that are physically or chemically bound to a surface, typically a metal. While adsorbed hydrogen can be more reactive than free molecular hydrogen, especially in catalytic processes, it is still generally less reactive than isolated atomic hydrogen, which is in a much more unstable and energetic state.
D. D) Hydrogen in compound: When hydrogen is in a compound (e.g., water, methane, acids), it is already part of a stable chemical structure. Its reactivity is determined by the overall properties of the compound, not by the intrinsic reactivity of a free hydrogen atom. The hydrogen atoms are already bonded and do not behave as highly reactive individual species.
E. E) H2O: H2O is water, a very stable chemical compound. The hydrogen atoms in water are covalently bonded to oxygen atoms, making them unreactive in the context of being a "state of H2." While water itself can participate in various reactions, the hydrogen within the water molecule is not in a highly reactive, atomic state.

Q70. A parallel-plate capacitor with air between its plates has a capacitance of 2 x 10-6 F. What will be the capacitance if the capacitor is fitted with a dielectric whose dielectric constant is 3?

A. 3 x 10-6 F
B. 6 x 10-6 F✓
C. 9 x 10-6 F
D. 12 x 10-6 F
E. 15 x 10-6 F

Explanation: a capacitor, its capacitance increases by a factor equal to the dielectric constant εr.

Why the other options are wrong

A. The capacitance of a parallel-plate capacitor with air between its plates is given as C_air = 2 x 10^-6 F.When a dielectric material is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (C_dielectric) is related to the original capacitance (C_air) by the dielectric constant (k) using the formula:C_dielectric = k * C_airGiven: * Original capacitance (C_air) = 2 x 10^-6 F * Dielectric constant (k) = 3Substitute the values into the formula:C_dielectric = 3 * (2 x 10^-6 F)C_dielectric = 6 x 10^-6 F
C. The capacitance of a parallel-plate capacitor with air between its plates is given as C_air = 2 x 10^-6 F.When a dielectric material is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (C_dielectric) is related to the original capacitance (C_air) by the dielectric constant (k) using the formula:C_dielectric = k * C_airGiven: * Original capacitance (C_air) = 2 x 10^-6 F * Dielectric constant (k) = 3Substitute the values into the formula:C_dielectric = 3 * (2 x 10^-6 F)C_dielectric = 6 x 10^-6 F
D. The capacitance of a parallel-plate capacitor with air between its plates is given as C_air = 2 x 10^-6 F.When a dielectric material is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (C_dielectric) is related to the original capacitance (C_air) by the dielectric constant (k) using the formula:C_dielectric = k * C_airGiven: * Original capacitance (C_air) = 2 x 10^-6 F * Dielectric constant (k) = 3Substitute the values into the formula:C_dielectric = 3 * (2 x 10^-6 F)C_dielectric = 6 x 10^-6 F
E. The capacitance of a parallel-plate capacitor with air between its plates is given as C_air = 2 x 10^-6 F.When a dielectric material is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (C_dielectric) is related to the original capacitance (C_air) by the dielectric constant (k) using the formula:C_dielectric = k * C_airGiven: * Original capacitance (C_air) = 2 x 10^-6 F * Dielectric constant (k) = 3Substitute the values into the formula:C_dielectric = 3 * (2 x 10^-6 F)C_dielectric = 6 x 10^-6 F

Q71. One thousand watts of electric power are transmitted to a device by means of two wires, each of which has a resistance of 2 ohms. If the resulting potential difference across the device is 100 volts, the potential difference across the source supplying the power is:

A. 20 V
B. 40 V
C. 100 V
D. 140 V✓
E. 500 V

Explanation: P=VI I = (P/V) = (1000/100) = 10 A IR drop = IR = 20V 100 + 20 + 20 = 140 V

Why the other options are wrong

A. P=VII = (P/V) = (1000/100) = 10 A IR drop = IR = 20V100 + 20 + 20 = 140 V
B. P=VII = (P/V) = (1000/100) = 10 A IR drop = IR = 20V100 + 20 + 20 = 140 V
C. P=VII = (P/V) = (1000/100) = 10 A IR drop = IR = 20V100 + 20 + 20 = 140 V
E. P=VII = (P/V) = (1000/100) = 10 A IR drop = IR = 20V100 + 20 + 20 = 140 V

Q72. A 3-cm length of wire is moved at right angles across a uniform magnetic field with a speed of 2.0 m/s . If the flux density is 5.0 teslas, what is the magnitude of the induced e.m.f?

A. 0.03 V
B. 0.3 V✓
C. 0.6 V
D. 10 V
E. 20 V

Explanation: E=BVLE=5*2*0.03E=0.3V

Why the other options are wrong

A. This is incorrect. Make sure to multiply the magnetic flux density, speed, and the length of the wire correctly.
C. This is incorrect. Recheck your calculations; the length of the wire should be converted to meters as 0.03 m.
D. This is incorrect. The result is too large. Ensure you use the formula correctly and with the right conversion for length.
E. This is incorrect. The calculation does not account for the length of the wire correctly. Verify your use of the formula E = BVL.

Q73. The angular arrangement of the space positions of the atoms in a crystal is called;

A. Space Lattice
B. Lattice array
C. Lattice energy
D. Doping
E. Both A and B✓

Explanation: “The angular arrangement of the space positions of the atoms in a crystal is called a space lattice or lattice array.” (Sindh Textbook for Physics, Chapter 16: Electronics, Section 16.7)

Why the other options are wrong

A. “The angular arrangement of the space positions of the atoms in a crystal is called a space lattice or lattice array.” (Sindh Textbook for Physics, Chapter 16: Electronics, Section 16.7)
B. “The angular arrangement of the space positions of the atoms in a crystal is called a space lattice or lattice array.” (Sindh Textbook for Physics, Chapter 16: Electronics, Section 16.7)
C. “The angular arrangement of the space positions of the atoms in a crystal is called a space lattice or lattice array.” (Sindh Textbook for Physics, Chapter 16: Electronics, Section 16.7)
D. “The angular arrangement of the space positions of the atoms in a crystal is called a space lattice or lattice array.” (Sindh Textbook for Physics, Chapter 16: Electronics, Section 16.7)

Q74. For call bells in houses, a tranformer may be used to step the voltage down from _ volt.

A. 440 to 220
B. 220 to 110
C. 110 to 70
D. 220 to 4✓
E. 440 to 110

Explanation: The voltage usually supplied to domestic consumers is 220 V. This voltage must be stepped down to much smaller values if it is to be used in an appliance such as a doorbell.

Why the other options are wrong

A. The voltage usually supplied to domestic consumers is 220 V. This voltage must be stepped down to much smaller values if it is to be used in an appliance such as a doorbell.
B. The voltage usually supplied to domestic consumers is 220 V. This voltage must be stepped down to much smaller values if it is to be used in an appliance such as a doorbell.
C. The voltage usually supplied to domestic consumers is 220 V. This voltage must be stepped down to much smaller values if it is to be used in an appliance such as a doorbell.
E. The voltage usually supplied to domestic consumers is 220 V. This voltage must be stepped down to much smaller values if it is to be used in an appliance such as a doorbell.

Q75. _ is a useful device for quick measurement of resistance. It includes a sensitive galvanometer, adjustable resistor and a torch cell connected in series between two terminals.

A. Ohm meter✓
B. Post Office Box
C. Meter Bridge
D. Wheatstone Bridge
E. Ammeter

Explanation: A) Ohm meter: An ohmmeter is an instrument specifically designed for the quick and direct measurement of electrical resistance. Early and simplified ohmmeters often consist of an internal battery (like a "torch cell") connected in series with a sensitive galvanometer and an adjustable resistor. The adjustable resistor is used for calibration (e.g., to zero the meter or select ranges), and the galvanometer measures the current flowing through the unknown resistance, which is inversely proportional to the resistance. This setup allows for a quick reading of the resistance value.

Why the other options are wrong

B. B) Post Office Box: A Post Office Box is a compact form of the Wheatstone bridge. While it uses a galvanometer and adjustable resistors (resistance plugs) and a cell for operation to measure unknown resistance, it is typically employed for precise measurements using a null-deflection method (balancing the bridge), which is not usually described as "quick measurement" in the same way a direct-reading ohmmeter is.
C. * C) Meter Bridge: Similar to the Post Office Box, a Meter Bridge (also known as a slide wire bridge) operates on the principle of the Wheatstone bridge. It involves finding a null point by sliding a jockey along a wire, using a galvanometer, resistance box, and a cell. Its purpose is also for accurate determination of unknown resistance, rather than a "quick" direct reading.
D. D) Wheatstone Bridge: The Wheatstone bridge is a fundamental electrical circuit used for precise measurement of unknown electrical resistance by balancing two legs of the bridge. It comprises four resistors, a galvanometer, and a voltage source. While the other options (Post Office Box and Meter Bridge) are applications of the Wheatstone bridge, the Wheatstone bridge itself is a circuit configuration for precision measurement, not a standalone "device" for quick direct measurement in the context of the question.
E. E) Ammeter: An ammeter is an instrument used exclusively to measure electric current in a circuit. It does not directly measure resistance.Therefore, the device that fits the description of being useful for "quick measurement of resistance" and includes a "sensitive galvanometer, adjustable resistor and a torch cell connected in series between two terminals" is an Ohmmeter.

Q76. The length of a measuring rod is 2 when it is at rest. What will its length be if it is moving with a velocity one third of the speed of light?

A. 5.886 m
B. 4.886 m
C. 3.886 m
D. 2.886 m
E. 1.886 m✓

Explanation: According to the special theory of relativity, length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame. This contraction is usually only noticeable at a substantial fraction of the speed of light. Only option E satisfies this principle.

Why the other options are wrong

A. According to the special theory of relativity, length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame. This contraction is usually only noticeable at a substantial fraction of the speed of light. Only option E satisfies this principle.
B. According to the special theory of relativity, length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame. This contraction is usually only noticeable at a substantial fraction of the speed of light. Only option E satisfies this principle.
C. According to the special theory of relativity, length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame. This contraction is usually only noticeable at a substantial fraction of the speed of light. Only option E satisfies this principle.
D. According to the special theory of relativity, length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame. This contraction is usually only noticeable at a substantial fraction of the speed of light. Only option E satisfies this principle.

Q77. A material consisting of fissionable (or fissile) isotope is called as:

A. Moderators
B. Reactor fuel✓
C. Coolant
D. Control Material
E. Shielding

Explanation: Reactor fuel is material used in nuclear power stations to produce heat to power turbines. Heat is generated when reactor fuel undergoes nuclear fission.

Why the other options are wrong

A. Reactor fuel is material used in nuclear power stations to produce heat to power turbines. Heat is generated when reactor fuel undergoes nuclear fission.
C. Reactor fuel is material used in nuclear power stations to produce heat to power turbines. Heat is generated when reactor fuel undergoes nuclear fission.
D. Reactor fuel is material used in nuclear power stations to produce heat to power turbines. Heat is generated when reactor fuel undergoes nuclear fission.
E. Reactor fuel is material used in nuclear power stations to produce heat to power turbines. Heat is generated when reactor fuel undergoes nuclear fission.

Q78. Identify the application/s of laser;I. as surgical tools for “welding” detached retinaII. for telephone communications along optical fiversIII. for precision cutting of metals and other materials

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

Explanation: Following are some applications of a laser: Medical applications Welding and Cutting Surveying Garment industry Laser nuclear fusion Communication Laser printing CDs and optical discs Spectroscopy Heat treatment Barcode scanners Laser cooling

Why the other options are wrong

A. Following are some applications of a laser:Medical applications Welding and Cutting SurveyingGarment industry Laser nuclear fusion CommunicationLaser printing CDs and optical discs SpectroscopyHeat treatment Barcode scanners Laser cooling
B. Following are some applications of a laser:Medical applications Welding and Cutting SurveyingGarment industry Laser nuclear fusion CommunicationLaser printing CDs and optical discs SpectroscopyHeat treatment Barcode scanners Laser cooling
C. Following are some applications of a laser:Medical applications Welding and Cutting SurveyingGarment industry Laser nuclearfusion CommunicationLaser printing CDs and optical discs SpectroscopyHeat treatment Barcode scanners Laser cooling
D. Following are some applications of a laser:Medical applications Welding and Cutting SurveyingGarment industry Laser nuclear fusion CommunicationLaser printing CDs and optical discs SpectroscopyHeat treatment Barcode scanners Laser cooling

Q79. Gamma rays may lose only a part of its energy in an encounter with the electron of an atom, the phenomena is known as:

A. Photoelectric Effect
B. Compton’s Effect✓
C. Coulomb's Law
D. Ampere’s Law
E. Faraday’s Law

Explanation: Compton scattering, discovered by Arthur Holly Compton, is the scattering of a high frequency photon after an interaction with a charged particle, usually an electron.

Why the other options are wrong

A. Compton scattering, discovered by Arthur Holly Compton, is the scattering of a high frequency photon after an interaction with a charged particle, usually an electron.
C. Compton scattering, discovered by Arthur Holly Compton, is the scattering of a high frequency photon after an interaction with a charged particle, usually an electron.
D. Compton scattering, discovered by Arthur Holly Compton, is the scattering of a high frequency photon after an interaction with a charged particle, usually an electron.
E. Compton scattering, discovered by Arthur Holly Compton, is the scattering of a high frequency photon after an interaction with a charged particle, usually an electron.

Q80. The value 24.90 represents _ significant figure/s.

A. 1
B. 2
C. 3
D. 4✓
E. 5

Explanation: The digits 2, 4, 9 and 0 will be considered significant.

Why the other options are wrong

A. The digits 2, 4, 9 and 0 will be considered significant.
B. The digits 2, 4, 9 and 0 will be considered significant.
C. The digits 2, 4, 9 and 0 will be considered significant.
E. The digits 2, 4, 9 and 0 will be considered significant.

Q81. The speed of a train is reduced from 60 km/h to15 km/h, at the same time as it travels a distance of 450 m. If the retardation is uniform, find how much further it will travel (approx) before coming to rest?

A. 10 m
B. 30 m✓
C. 50 m
D. 70 m
E. 90 m

Explanation: Changing given speeds to SI unitsu = 60 *1000/3600 = 16.67 m/sv= 15*1000/3600 = 4.17 m/sTo find retardation we use the kinematic expressionv2-u2 = 2asnserting given values we gel(4.17)2 - (16.67)2 = 2*a*450a =-0.289 m/s2For the train to come to rest new equation becomes ( 02 - (4.17)2 = 2*(-0.289) s2Solving it we gets2 = (4.17)2/0.578 = 30 m

Why the other options are wrong

A. The correct answer is 30m.
C. The correct answer is 30m.
D. The correct answer is 30m.
E. The correct answer is 30m.

Q82. The velocity of the body moving in a straight line with a constant acceleration, is 10 m/s at a certain instant ‘t’. After 5 s the velocity become 20 m/s, the velocity 3 s before ‘t’ was:

A. 4 m/s✓
B. 6 m/s
C. 7 m/s
D. 8 m/s
E. 10 m/s

Explanation: The question can be solved using the first equation of motion, as shown below:

Why the other options are wrong

B. using v = u + atWhere: * v (final velocity at instant 't') = 10 m/s * u (initial velocity 3 seconds before 't') = u_initial (what we need to find) * a (acceleration) = 2 m/s² * t (time interval) = 3 sPlugging in the values:10 = u_initial + (2 * 3)10 = u_initial + 6u_initial = 10 - 6u_initial = 4 m/s
C. using v = u + atWhere: * v (final velocity at instant 't') = 10 m/s * u (initial velocity 3 seconds before 't') = u_initial (what we need to find) * a (acceleration) = 2 m/s² * t (time interval) = 3 sPlugging in the values:10 = u_initial + (2 * 3)10 = u_initial + 6u_initial = 10 - 6u_initial = 4 m/s
D. using v = u + atWhere: * v (final velocity at instant 't') = 10 m/s * u (initial velocity 3 seconds before 't') = u_initial (what we need to find) * a (acceleration) = 2 m/s² * t (time interval) = 3 sPlugging in the values:10 = u_initial + (2 * 3)10 = u_initial + 6u_initial = 10 - 6u_initial = 4 m/s
E. using v = u + atWhere: * v (final velocity at instant 't') = 10 m/s * u (initial velocity 3 seconds before 't') = u_initial (what we need to find) * a (acceleration) = 2 m/s² * t (time interval) = 3 sPlugging in the values:10 = u_initial + (2 * 3)10 = u_initial + 6u_initial = 10 - 6u_initial = 4 m/s

Q83. One of the two forces is double the other and their resultant is equal to the greater force. The angle between them is:

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

Explanation: The explanation is as follows.

Why the other options are wrong

A. Let the two forces be F₁ and F₂.According to the problem statement, one force is double the other. Let's assume F₂ = 2F₁.The resultant force (R) is equal to the greater force, so R = F₂ = 2F₁.The formula for the resultant of two forces is given by the law of cosines:R² = F₁² + F₂² + 2F₁F₂cosθwhere θ is the angle between the two forces.Substitute the given conditions into the formula:(2F₁)² = F₁² + (2F₁)² + 2(F₁)(2F₁)cosθ4F₁² = F₁² + 4F₁² + 4F₁²cosθ4F₁² = 5F₁² + 4F₁²cosθNow, rearrange the equation to solve for cosθ:4F₁² - 5F₁² = 4F₁²cosθ-F₁² = 4F₁²cosθDivide both sides by 4F₁²:cosθ = -F₁²/ (4F₁²)cosθ = -1/4Therefore, the angle between the two forces is θ = Cos⁻¹(-1/4).Why other options are wrong:The resultant force R will only be equal to 2F₁ when cosθ = -1/4. If the angle were different, the value of cosθ would change, leading to a different resultant force.
B. Let the two forces be F₁ and F₂.According to the problem statement, one force is double the other. Let's assume F₂ = 2F₁.The resultant force (R) is equal to the greater force, so R = F₂ = 2F₁.The formula for the resultant of two forces is given by the law of cosines:R² = F₁² + F₂² + 2F₁F₂cosθwhere θ is the angle between the two forces.Substitute the given conditions into the formula:(2F₁)² = F₁² + (2F₁)² + 2(F₁)(2F₁)cosθ4F₁² = F₁² + 4F₁² + 4F₁²cosθ4F₁² = 5F₁² + 4F₁²cosθNow, rearrange the equation to solve for cosθ:4F₁² - 5F₁² = 4F₁²cosθ-F₁² = 4F₁²cosθDivide both sides by 4F₁²:cosθ = -F₁²/ (4F₁²)cosθ = -1/4Therefore, the angle between the two forces is θ = Cos⁻¹(-1/4).Why other options are wrong:The resultant force R will only be equal to 2F₁ when cosθ = -1/4. If the angle were different, the value of cosθ would change, leading to a different resultant force.
C. Let the two forces be F₁ and F₂.According to the problem statement, one force is double the other. Let's assume F₂ = 2F₁.The resultant force (R) is equal to the greater force, so R = F₂ = 2F₁.The formula for the resultant of two forces is given by the law of cosines:R² = F₁² + F₂² + 2F₁F₂cosθwhere θ is the angle between the two forces.Substitute the given conditions into the formula:(2F₁)² = F₁² + (2F₁)² + 2(F₁)(2F₁)cosθ4F₁² = F₁² + 4F₁² + 4F₁²cosθ4F₁² = 5F₁² + 4F₁²cosθNow, rearrange the equation to solve for cosθ:4F₁² - 5F₁² = 4F₁²cosθ-F₁² = 4F₁²cosθDivide both sides by 4F₁²:cosθ = -F₁²/ (4F₁²)cosθ = -1/4Therefore, the angle between the two forces is θ = Cos⁻¹(-1/4).Why other options are wrong:The resultant force R will only be equal to 2F₁ when cosθ = -1/4. If the angle were different, the value of cosθ would change, leading to a different resultant force.
E. Let the two forces be F₁ and F₂.According to the problem statement, one force is double the other. Let's assume F₂ = 2F₁.The resultant force (R) is equal to the greater force, so R = F₂ = 2F₁.The formula for the resultant of two forces is given by the law of cosines:R² = F₁² + F₂² + 2F₁F₂cosθwhere θ is the angle between the two forces.Substitute the given conditions into the formula:(2F₁)² = F₁² + (2F₁)² + 2(F₁)(2F₁)cosθ4F₁² = F₁² + 4F₁² + 4F₁²cosθ4F₁² = 5F₁² + 4F₁²cosθNow, rearrange the equation to solve for cosθ:4F₁² - 5F₁² = 4F₁²cosθ-F₁² = 4F₁²cosθDivide both sides by 4F₁²:cosθ = -F₁²/ (4F₁²)cosθ = -1/4Therefore, the angle between the two forces is θ = Cos⁻¹(-1/4).Why other options are wrong:The resultant force R will only be equal to 2F₁ when cosθ = -1/4. If the angle were different, the value of cosθ would change, leading to a different resultant force.

Q84. Identify the example/s of Projectile motion;I. a missile shot from a gunII. a player making a long jumpIII. a ball rolling on smooth surfaceIV. a football kicked off by a player

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

Explanation: A projectile is an object upon which the only force acting is gravity. Only I, II and IV are examples of projectiles.

Why the other options are wrong

A. A projectile is an object upon which the only force acting is gravity. Only I, II and IV are examples of projectiles.
B. A projectile is an object upon which the only force acting is gravity. Only I, II and IV are examples of projectiles.
C. A projectile is an object upon which the only force acting is gravity. Only I, II and IV are examples of projectiles.
E. A projectile is an object upon which the only force acting is gravity. Only I, II and IV are examples of projectiles.

Q85. Equations like x= v0xt and y= v0yt - ½ gt2 are independent descriptions of motion: one involving coordinate x and the other the coordinate y, where x and y are both dependant upon a common variable, the time 't’. Such equations are called:

A. Quadratic equations
B. Simple equations
C. Parametric equations✓
D. Linear equations

Explanation: A parametric equation is a type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.

Why the other options are wrong

A. A parametric equation is a type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.
B. A parametric equation is a type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.
D. A parametric equation is a type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.

Q86. When a wave goes from one medium to another medium, which one of the following characteristics of the wave remains constant?

A. Velocity
B. Frequency✓
C. Wavelength
D. Phase

Explanation: The frequency of a wave remains constant as it moves from one medium to another. Frequency is the number of complete oscillations (cycles) of the wave that occur per unit time. It is a fundamental property of the wave and remains unchanged regardless of the medium it travels through.

Why the other options are wrong

A. The velocity of a wave changes when it goes from one medium to another. The velocity of a wave depends on the properties of the medium through which it travels. When a wave enters a different medium, its velocity may increase or decrease, depending on the properties of the new medium.
C. The wavelength of a wave does not remain constant when it goes from one medium to another. Wavelength is the distance between two successive points in a wave that are in phase (e.g., crest to crest or trough to trough). It is determined by the source of the wave and remains the same as it moves through different media.
D. The phase of a wave changes when it enters a different medium. Phase refers to the position of a point on a wave cycle relative to a reference point (e.g., crest to trough or any other point on the wave). The change in the medium can cause a phase shift in the wave.

Q87. A torque which produces counterclockwise rotation is considered to be _ and that producing a clockwise rotation is taken as _.

A. negative …. negative
B. positive …. positive
C. positive …. negative✓
D. negative …. Positive
E. None of the above

Explanation: In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.

Why the other options are wrong

A. In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.
B. In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.
D. In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.
E. In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.

Q88. Every point of the spinning object moves along an arc of the circle in a small interval of time and the centers of all these circles lie along a straight line. This straight line is known as:

A. Spinning point
B. Double spinning point
C. Axis of rotation✓
D. Centre of rotation
E. Spinning centre

Explanation: The axis of rotation is defined as the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.

Why the other options are wrong

A. The axis of rotation is defined as the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.
B. The axis of rotation is defined as the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.
D. The axis of rotation is defined as the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.
E. The axis of rotation is defined as the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.

Q89. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the gravitational force do on the book?

A. -30 J✓
B. -15 J
C. 0 J
D. 25 J
E. 40 J

Explanation: Work= Fdcosθ= 2 x 10 x 1.5(cos 180)= -30 J θ= 180° because the displacement is directed opposite to the force of gravity.

Why the other options are wrong

B. Work= Fdcosθ= 2 x 10 x 1.5(cos 180)= -30 Jθ= 180° because the displacement is directed opposite to the force of gravity.
C. Work= Fdcosθ= 2 x 10 x 1.5(cos 180)= -30 Jθ= 180° because the displacement is directed opposite to the force of gravity.
D. Work= Fdcosθ= 2 x 10 x 1.5(cos 180)= -30 Jθ= 180° because the displacement is directed opposite to the force of gravity.
E. Work= Fdcosθ= 2 x 10 x 1.5(cos 180)= -30 Jθ= 180° because the displacement is directed opposite to the force of gravity.

Q90. A force of 200 N is required to keep an object sliding at a constant speed of 2 m/s across a rough floor. How much power is being expended to maintain this motion?

A. 50 W
B. 100 W
C. 200 W
D. 400 W✓

Explanation: Power=Force x velocity Power= 200 x 2 = 400 watts

Why the other options are wrong

A. Incorect as per formulaPower=Force x velocityPower= 200 x 2 = 400 watts
B. Incorect as per formulae Power=Force x velocityPower= 200 x 2 = 400 watts
C. Incorect as per formulae Power=Force x velocityPower= 200 x 2 = 400 watts

Q91. Identify the examples of kinetic energy A moving ball can break a glass window A striking hammer can drive a nail A stone thrown upward can lift itself against the force of gravity

A. I ONLY
B. II ONLY
C. III ONLY
D. I, II AND III✓

Explanation: The correct answer is I, II, and III. Each scenario demonstrates kinetic energy due to the motion of the objects involved:Option I: The moving ball possesses kinetic energy, which can break a window upon collision.Option II: The hammer's motion imparts kinetic energy, allowing it to drive a nail.Option III: The stone, while initially moving upward, possesses kinetic energy. Even though this energy is converted to potential energy at the peak of its trajectory, the initial and final stages involve kinetic energy.Therefore, all options are correct as they all represent instances of kinetic energy due to motion.

Why the other options are wrong

A. A moving ball can break a glass window: The ball is in motion, which means it has kinetic energy. This energy can be transferred to the window upon impact, potentially breaking it.
B. A striking hammer can drive a nail: The hammer, in motion, possesses kinetic energy. This energy is transferred to the nail during the strike, driving it into a surface.
C. A stone thrown upward initially has kinetic energy due to its motion. Although this energy converts to potential energy as the stone reaches its peak, it initially demonstrates kinetic energy.

Q92. A particle travels in a circular path of radius 0.2 m with a constant kinetic energy of 4J. What is the net force on this particle?

A. 4 N
B. 16 N
C. 20 N
D. 40 N✓
E. Cannot be determined from the information given

Explanation: K.E= ½ mv2 4 = ½ mv2 mv2=8 Fcentripetal= mv²/r= 80.2 = 40 N

Why the other options are wrong

A. K.E= ½ mv24 = ½ mv2mv2=8Fcentripetal= mv²/r= 80.2 = 40 N
B. K.E= ½ mv24 = ½ mv2mv2=8Fcentripetal= mv²/r= 80.2 = 40 N
C. K.E= ½ mv24 = ½ mv2mv2=8Fcentripetal= mv²/r= 80.2 = 40 N
E. K.E= ½ mv24 = ½ mv2mv2=8Fcentripetal= mv²/r= 80.2 = 40 N

Q93. A strong, fixed at both ends, supports a standing wave with a total of 4 nodes. If the length of the string is 6 m, what is the wavelength of the wave?

A. 0.67 m
B. 1.2 m
C. 1.5 m
D. 3 m
E. 4 m✓

Explanation: A standing wave in the third harmonic will have four nodes, as shown below. To find the wavelength: 1.5 waves—> 6 m 1 wave—> 6/1.5 = 4 m

Why the other options are wrong

A. A standing wave in the third harmonic will have four nodes, as shown below. To find the wavelength:1.5 waves—> 6 m1 wave—> 6/1.5 = 4 m
B. A standing wave in the third harmonic will have four nodes, as shown below. To find the wavelength:1.5 waves—> 6 m1 wave—> 6/1.5 = 4 m
C. A standing wave in the third harmonic will have four nodes, as shown below. To find the wavelength:1.5 waves—> 6 m1 wave—> 6/1.5 = 4 m
D. A standing wave in the third harmonic will have four nodes, as shown below. To find the wavelength:1.5 waves—> 6 m1 wave—> 6/1.5 = 4 m

Q94. The length of a simple pendulum is 88.2 cm. The value of acceleration due to gravity is 9.8 m/s2. Its time period is:

A. 0.5 s
B. 5 s
C. 4 s
D. 3 s
E. 1.885 s✓

Explanation: T = 2π√(L/g)Where: * L = length of the pendulum * g = acceleration due to gravity * π ≈ 3.14159Given values: * Length (L) = 88.2 cm = 0.882 m (converting centimeters to meters) * Acceleration due to gravity (g) = 9.8 m/s²Now, let's plug these values into the formula:T = 2 * 3.14159 * √(0.882 / 9.8)T = 2 * 3.14159 * √(0.09)T = 2 * 3.14159 * 0.3T ≈ 1.885 s

Why the other options are wrong

A. T = 2π√(L/g)Where: * L = length of the pendulum * g = acceleration due to gravity * π ≈ 3.14159Given values: * Length (L) = 88.2 cm = 0.882 m (converting centimeters to meters) * Acceleration due to gravity (g) = 9.8 m/s²Now, let's plug these values into the formula:T = 2 * 3.14159 * √(0.882 / 9.8)T = 2 * 3.14159 * √(0.09)T = 2 * 3.14159 * 0.3T ≈ 1.885 s
B. T = 2π√(L/g)Where: * L = length of the pendulum * g = acceleration due to gravity * π ≈ 3.14159Given values: * Length (L) = 88.2 cm = 0.882 m (converting centimeters to meters) * Acceleration due to gravity (g) = 9.8 m/s²Now, let's plug these values into the formula:T = 2 * 3.14159 * √(0.882 / 9.8)T = 2 * 3.14159 * √(0.09)T = 2 * 3.14159 * 0.3T ≈ 1.885 s
C. T = 2π√(L/g)Where: * L = length of the pendulum * g = acceleration due to gravity * π ≈ 3.14159Given values: * Length (L) = 88.2 cm = 0.882 m (converting centimeters to meters) * Acceleration due to gravity (g) = 9.8 m/s²Now, let's plug these values into the formula:T = 2 * 3.14159 * √(0.882 / 9.8)T = 2 * 3.14159 * √(0.09)T = 2 * 3.14159 * 0.3T ≈ 1.885 s
D. T = 2π√(L/g)Where: * L = length of the pendulum * g = acceleration due to gravity * π ≈ 3.14159Given values: * Length (L) = 88.2 cm = 0.882 m (converting centimeters to meters) * Acceleration due to gravity (g) = 9.8 m/s²Now, let's plug these values into the formula:T = 2 * 3.14159 * √(0.882 / 9.8)T = 2 * 3.14159 * √(0.09)T = 2 * 3.14159 * 0.3T ≈ 1.885 s

Q95. The radius of 14th Newton’s ring is 1 mm, when light of wavelength 5.89 x 10-7 m is used. The radius of curvature of the lower surface of the lens is used is:

A. 75.7 mm
B. 85.7 mm
C. 95.7 mm
D. 105.7 mm
E. 125.7 mm✓

Explanation: Explanation attached below

Why the other options are wrong

A. The problem describes a Newton's rings experiment and asks for the radius of curvature of the lens. The given values are: * Radius of the 14th Newton's ring (r_n) = 1 mm = 1 × 10⁻³ m * Wavelength of light (λ) = 5.89 × 10⁻⁷ m * Ring number (n) = 14For Newton's rings, the formula for the radius of the nth bright ring is given by:r_n^2 = (n - \frac{1}{2})\lambda RWhere: * r_n is the radius of the nth bright ring * n is the ring number * \lambda is the wavelength of light * R is the radius of curvature of the lower surface of the lensWe need to solve for R:R = \frac{r_n^2}{(n - \frac{1}{2})\lambda}Now, substitute the given values into the formula:R = \frac{(1 \times 10^{-3} \text{ m})^2}{(14 - \frac{1}{2}) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6} \text{ m}^2}{(13.5) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6}}{7.9515 \times 10^{-6}} \text{ m}R \approx 0.12576 \text{ m}To convert meters to millimeters, multiply by 1000:R \approx 0.12576 \times 1000 \text{ mm}R \approx 125.76 \text{ mm}Rounding to one decimal place, we get 125.7 mm.
B. The problem describes a Newton's rings experiment and asks for the radius of curvature of the lens. The given values are: * Radius of the 14th Newton's ring (r_n) = 1 mm = 1 × 10⁻³ m * Wavelength of light (λ) = 5.89 × 10⁻⁷ m * Ring number (n) = 14For Newton's rings, the formula for the radius of the nth bright ring is given by:r_n^2 = (n - \frac{1}{2})\lambda RWhere: * r_n is the radius of the nth bright ring * n is the ring number * \lambda is the wavelength of light * R is the radius of curvature of the lower surface of the lensWe need to solve for R:R = \frac{r_n^2}{(n - \frac{1}{2})\lambda}Now, substitute the given values into the formula:R = \frac{(1 \times 10^{-3} \text{ m})^2}{(14 - \frac{1}{2}) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6} \text{ m}^2}{(13.5) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6}}{7.9515 \times 10^{-6}} \text{ m}R \approx 0.12576 \text{ m}To convert meters to millimeters, multiply by 1000:R \approx 0.12576 \times 1000 \text{ mm}R \approx 125.76 \text{ mm}Rounding to one decimal place, we get 125.7 mm.
C. The problem describes a Newton's rings experiment and asks for the radius of curvature of the lens. The given values are: * Radius of the 14th Newton's ring (r_n) = 1 mm = 1 × 10⁻³ m * Wavelength of light (λ) = 5.89 × 10⁻⁷ m * Ring number (n) = 14For Newton's rings, the formula for the radius of the nth bright ring is given by:r_n^2 = (n - \frac{1}{2})\lambda RWhere: * r_n is the radius of the nth bright ring * n is the ring number * \lambda is the wavelength of light * R is the radius of curvature of the lower surface of the lensWe need to solve for R:R = \frac{r_n^2}{(n - \frac{1}{2})\lambda}Now, substitute the given values into the formula:R = \frac{(1 \times 10^{-3} \text{ m})^2}{(14 - \frac{1}{2}) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6} \text{ m}^2}{(13.5) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6}}{7.9515 \times 10^{-6}} \text{ m}R \approx 0.12576 \text{ m}To convert meters to millimeters, multiply by 1000:R \approx 0.12576 \times 1000 \text{ mm}R \approx 125.76 \text{ mm}Rounding to one decimal place, we get 125.7 mm.
D. The problem describes a Newton's rings experiment and asks for the radius of curvature of the lens. The given values are: * Radius of the 14th Newton's ring (r_n) = 1 mm = 1 × 10⁻³ m * Wavelength of light (λ) = 5.89 × 10⁻⁷ m * Ring number (n) = 14For Newton's rings, the formula for the radius of the nth bright ring is given by:r_n^2 = (n - \frac{1}{2})\lambda RWhere: * r_n is the radius of the nth bright ring * n is the ring number * \lambda is the wavelength of light * R is the radius of curvature of the lower surface of the lensWe need to solve for R:R = \frac{r_n^2}{(n - \frac{1}{2})\lambda}Now, substitute the given values into the formula:R = \frac{(1 \times 10^{-3} \text{ m})^2}{(14 - \frac{1}{2}) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6} \text{ m}^2}{(13.5) \times (5.89 \times 10^{-7} \text{ m})}R = \frac{1 \times 10^{-6}}{7.9515 \times 10^{-6}} \text{ m}R \approx 0.12576 \text{ m}To convert meters to millimeters, multiply by 1000:R \approx 0.12576 \times 1000 \text{ mm}R \approx 125.76 \text{ mm}Rounding to one decimal place, we get 125.7 mm.

Q96. During each cycle, a heat engine absorbs 400 J of heat from its high-temperature source and discards 300 J of heat into its low-temperature sink. What is the efficiency of this engine?

A. 1/7
B. ¼✓
C. 3/7
D. 4/7
E. ¾

Explanation: Since the engine takes in 400 J of energy and produces only 400 J - 300 J =100 J of useful work, its efficiency is (400-300)/400 = 1/4.

Why the other options are wrong

A. Since the engine takes in 400 J of energy and produces only 400 J - 300 J =100 J of useful work, its efficiency is (400-300)/400 = 1/4.
C. Since the engine takes in 400 J of energy and produces only 400 J - 300 J =100 J of useful work, its efficiency is (400-300)/400 = 1/4.
D. Since the engine takes in 400 J of energy and produces only 400 J - 300 J =100 J of useful work, its efficiency is (400-300)/400 = 1/4.
E. Since the engine takes in 400 J of energy and produces only 400 J - 300 J =100 J of useful work, its efficiency is (400-300)/400 = 1/4.

Q97. An object 8 cm high is placed 20 cm from the converging lens whose focal length is 40 cm, the magnification in this case is:

A. 2✓
B. 3
C. 5
D. 7
E. 9

Explanation: The question can be solved using the lens equation

Why the other options are wrong

B. * Focal length (f) = +40 cm * Object distance (d₀) = 20 cmFirst, calculate the image distance (dᵢ) using the lens formula:1/f = 1/d₀ + 1/dᵢ1/40 = 1/20 + 1/dᵢ1/dᵢ = 1/40 - 1/201/dᵢ = 1/40 - 2/401/dᵢ = -1/40dᵢ = -40 cmThe negative sign for dᵢ indicates that the image is virtual and on the same side of the lens as the object.Next, calculate the magnification (M) using the magnification formula:M = -dᵢ / d₀M = -(-40 cm) / 20 cmM = 40 / 20M = 2
C. * Focal length (f) = +40 cm * Object distance (d₀) = 20 cmFirst, calculate the image distance (dᵢ) using the lens formula:1/f = 1/d₀ + 1/dᵢ1/40 = 1/20 + 1/dᵢ1/dᵢ = 1/40 - 1/201/dᵢ = 1/40 - 2/401/dᵢ = -1/40dᵢ = -40 cmThe negative sign for dᵢ indicates that the image is virtual and on the same side of the lens as the object.Next, calculate the magnification (M) using the magnification formula:M = -dᵢ / d₀M = -(-40 cm) / 20 cmM = 40 / 20M = 2
D. * Focal length (f) = +40 cm * Object distance (d₀) = 20 cmFirst, calculate the image distance (dᵢ) using the lens formula:1/f = 1/d₀ + 1/dᵢ1/40 = 1/20 + 1/dᵢ1/dᵢ = 1/40 - 1/201/dᵢ = 1/40 - 2/401/dᵢ = -1/40dᵢ = -40 cmThe negative sign for dᵢ indicates that the image is virtual and on the same side of the lens as the object.Next, calculate the magnification (M) using the magnification formula:M = -dᵢ / d₀M = -(-40 cm) / 20 cmM = 40 / 20M = 2
E. * Focal length (f) = +40 cm * Object distance (d₀) = 20 cmFirst, calculate the image distance (dᵢ) using the lens formula:1/f = 1/d₀ + 1/dᵢ1/40 = 1/20 + 1/dᵢ1/dᵢ = 1/40 - 1/201/dᵢ = 1/40 - 2/401/dᵢ = -1/40dᵢ = -40 cmThe negative sign for dᵢ indicates that the image is virtual and on the same side of the lens as the object.Next, calculate the magnification (M) using the magnification formula:M = -dᵢ / d₀M = -(-40 cm) / 20 cmM = 40 / 20M = 2

Q98. An electron is situated midway between two parallel plates 0.5 cm apart. One of the plates is maintained at a potential of 60 volts above the other. The force on the electron is:

A. 5.92 x 10-15 N
B. 4.92 x 10-15 N
C. 3.92 x 10-15 N
D. 2.92 x 10-15 N
E. 1.92 x 10-15 N✓

Explanation: Explanation attached

Why the other options are wrong

A. * Potential difference (V) = 60 V * Distance between plates (d) = 0.5 cm = 0.005 m (converted to meters) * Charge of an electron (q) = 1.602 × 10^-19 CCalculations: * Calculate the electric field (E): E = 60 V / 0.005 m = 12000 N/C (or V/m) * Calculate the force (F) on the electron: F = (1.602 × 10^-19 C) × (12000 N/C) F = 1.9224 × 10^-15 NTherefore, the force on the electron is approximately 1.92 × 10^-15 N.
B. * Potential difference (V) = 60 V * Distance between plates (d) = 0.5 cm = 0.005 m (converted to meters) * Charge of an electron (q) = 1.602 × 10^-19 CCalculations: * Calculate the electric field (E): E = 60 V / 0.005 m = 12000 N/C (or V/m) * Calculate the force (F) on the electron: F = (1.602 × 10^-19 C) × (12000 N/C) F = 1.9224 × 10^-15 NTherefore, the force on the electron is approximately 1.92 × 10^-15 N.
C. * Potential difference (V) = 60 V * Distance between plates (d) = 0.5 cm = 0.005 m (converted to meters) * Charge of an electron (q) = 1.602 × 10^-19 CCalculations: * Calculate the electric field (E): E = 60 V / 0.005 m = 12000 N/C (or V/m) * Calculate the force (F) on the electron: F = (1.602 × 10^-19 C) × (12000 N/C) F = 1.9224 × 10^-15 NTherefore, the force on the electron is approximately 1.92 × 10^-15 N.
D. * Potential difference (V) = 60 V * Distance between plates (d) = 0.5 cm = 0.005 m (converted to meters) * Charge of an electron (q) = 1.602 × 10^-19 CCalculations: * Calculate the electric field (E): E = 60 V / 0.005 m = 12000 N/C (or V/m) * Calculate the force (F) on the electron: F = (1.602 × 10^-19 C) × (12000 N/C) F = 1.9224 × 10^-15 NTherefore, the force on the electron is approximately 1.92 × 10^-15 N.

Q99. Consider the series circuit shown in the figure below. R1 and R2 are 20 and 30 ohms respectively are connected in series. The potential difference is 100 V. What is the current through each resistor?

A. 1 A
B. 2 A✓
C. 3 A
D. 4 A
E. 5 A

Explanation: Since this is a series, arrangement: Rnet= 20+30=50 ohm I= V/Rnet= 100/50= 2A The same current flows through each component connected in series.

Why the other options are wrong

A. Since this is a series, arrangement:Rnet= 20+30=50 ohmI= V/Rnet= 100/50= 2AThe same current flows through each component connected in series.
C. Since this is a series, arrangement:Rnet= 20+30=50 ohmI= V/Rnet= 100/50= 2AThe same current flows through each component connected in series.
D. Since this is a series, arrangement:Rnet= 20+30=50 ohmI= V/Rnet= 100/50= 2AThe same current flows through each component connected in series.
E. Since this is a series, arrangement:Rnet= 20+30=50 ohmI= V/Rnet= 100/50= 2AThe same current flows through each component connected in series.

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