Fmdc 2013 — Solved Past Paper with Answers

Q1. Which is not a past from of a verb.

• A. Was
• B. Had
• C. Looked
• D. Spoke
• E. Hear✓

Explanation: A past tense is a tense that is used to talk about something that has happened in the past, or the way something was in the past. It's one of the three main tenses alongside present and future.

Q2. Which contains an adjective?

• A. Old man✓
• B. On Tuesday
• C. She said
• D. And you
• E. Afternoon

Explanation: A. Old is an adjective here as it modifies the noun ‘man’.B. On Tuesday is an example of adverb of time.C. She said is an example of an intransitive verb.D. And you is just a sentence.E. Afternoon is a noun.

Q3. Which contains an adverb?

• A. Full house
• B. Three women
• C. Was dirty
• D. Very funny✓
• E. Early morning

Explanation: The correct answer is Very funny because 'very' is an adverb that modifies the adjective 'funny,' enhancing its meaning. In contrast, 'full,' 'three,' 'dirty,' and 'early' are used as adjectives in the other options, each describing a noun or noun phrase.

Q4. Which is not correct?

• A. In February
• B. On 5:00 o’clock✓
• C. At my house
• D. Near me
• E. On May 2

Explanation: The phrase 'On 5:00 o’clock' is incorrect because the preposition 'On' is used for dates and days, not for specific times. The correct preposition for expressing time is 'At,' hence 'At 5:00 o’clock' is appropriate. Other options are correct: 'In February' for months, 'At my house' for specific locations, 'Near me' for proximity, and 'On May 2' for specific dates.

Q5. He complains _ headache.

• A. At
• B. From
• C. Of✓
• D. To

Explanation: The correct answer is 'of' because it is the standard preposition used when someone is expressing a complaint about a symptom or condition, such as a headache. For example, 'He complains of back pain.' The other options, 'at', 'from', and 'to', do not convey the correct relationship between the subject and the complaint. 'At' typically refers to location or time, 'from' can denote origin but not in this context, and 'to' usually indicates direction or recipient.

Q6. Sharks differ from other fish in that Their skeleton are made of cartilage instead of bone.

• A. Differ from
• B. Other fish
• C. Are made✓
• D. Instead

Explanation: OPTION A: Differ from: This phrase is used to highlight the contrast or distinction between two or more things. In the sentence, it shows that sharks have a difference compared to other fish. OPTION B: Other fish: Refers to the group of fish that is being compared to sharks. It is used to specify the category of fish from which sharks are distinct. OPTION C: The correct answer is:"Sharks differ from other fish in that their skeleton is made of cartilage instead of bone."Are made: This is the correct verb form to use when referring to the subject "their skeleton," which is plural. OPTION D: Instead of: This phrase is used to indicate an alternative or substitution. It shows that sharks have cartilage as their skeleton material instead of bone.

Q7. Which one of the following four options is nearest in meaning to the word in capitals?STAGNATE

• A. Active
• B. Vegetate✓
• C. Vigorous
• D. Bewilder

Explanation: The correct answer is Option B: Vegetate. The term 'stagnate' means to cease to flow or move, becoming inactive or dull. 'Vegetate' is similar because it implies living in a passive, inactive manner. On the other hand, Option A: Active and Option C: Vigorous imply energy and movement, which are the opposite of stagnation. Option D: Bewilder refers to causing confusion and is unrelated to the concept of inactivity or lack of progress.

Q8. Which one of the following four options is nearest in meaning to the word in capitals? ASTOUND

• A. Shock✓
• B. Confer
• C. Condescend
• D. Strengthen

Explanation: The word ASTOUND means to cause surprise or shock due to something unexpected or extraordinary. The correct option is Shock as it closely aligns with this meaning, signifying a sudden and intense surprise.Option B: Confer means to discuss or exchange views, which is unrelated to the sense of surprise or shock.Option C: Condescend refers to an attitude of superiority, which does not match the meaning of astound.Option D: Strengthen is about making something stronger, which again does not relate to the concept of causing surprise or shock.

Q9. Which one of the following four options is nearest in meaning to the word in capitals? SCORN

• A. Cleans
• B. Mock✓
• C. Scrutinise
• D. Stifle

Explanation: The word 'scorn' denotes a feeling of contempt or disdain towards someone or something. The correct answer is B : ‘Mock’, as it involves making fun of someone in a scornful manner, aligning closely with the meaning of 'scorn'.

Q10. Which one of the following four options is nearest in meaning to the word in capitals?WISP

• A. Tuft✓
• B. Pack
• C. Small
• D. Spry

Explanation: The word Wisp refers to a small, thin, or twisted bunch, piece, or amount of something, often used to describe things like a wisp of smoke or a wisp of hair. Option A: Tuft is the correct choice as it also denotes a small bunch or collection, particularly of threads, grass, or hair, aligning closely with the meaning of 'wisp'.

Q11. A student is trying to determine the type of membrane transport occurring in a cell. She finds that the molecules to be transported are very large and when transported across the membrane, no ATP is used. Which of the following is the most likely mechanism of transport?

• A. Active transport
• B. Simple diffusion
• C. Facilitated diffusion✓
• D. Exocytosis

Explanation: The correct answer is facilitated diffusion. This process involves the passive movement of molecules across the cell membrane via specific transport proteins. It does not require energy, which is consistent with the observation that no ATP was used. Large molecules typically cannot pass through the membrane without assistance, which is why facilitated diffusion is necessary. Active transport is incorrect because it requires ATP to move molecules against their concentration gradient. Simple diffusion is not suitable for large molecules as it involves the movement of small or non-polar molecules directly through the lipid bilayer. Lastly, exocytosis is a process for expelling substances from the cell rather than bringing them in.

Q12. In the course of glycolysis:

• A. NADH is reduced to 𝑁𝐴𝐷+
• B. 𝑁𝐴𝐷+ is oxidized to NADH
• C. Glucose is degraded into two molecules of pyruvate✓
• D. Both NADH is reduced to 𝑁𝐴𝐷+ & 𝑁𝐴𝐷+ is oxidized to NADH

Explanation: In glycolysis, glucose is broken down into two molecules of pyruvate. This process involves the reduction of NAD+ to NADH, capturing energy released from glucose. The production of NADH is crucial for maintaining the flow of electrons needed for further energy-releasing reactions. Option A is incorrect because NADH is not reduced to NAD+ during glycolysis; instead, NAD+ is reduced. Option B is incorrect because NAD+ is reduced to NADH, not oxidized. Option D is incorrect because it inaccurately describes the processes occurring within glycolysis. The reduction of NAD+ to NADH happens in glycolysis, but oxidation back to NAD+ occurs in different stages of cellular respiration.

Q13. The epiglottis is to trachea as the lower esophageal (cardiac) sphincter is to the:

• A. Stomach✓
• B. Heart
• C. Small intestine
• D. Liver

Explanation: The cardiac sphincter is a sphincter between the esophagus and the stomach, opening at the approach of food that can then be swept into the stomach by rhythmic peristaltic waves.

Q14. Starch is hydrolyzed into maltose by:

• A. Salivary amylase
• B. Maltose
• C. Pancreatic amylase
• D. Both A & C✓

Explanation: The correct answer is Option D: Both A & C. During digestion, starch is initially broken down into maltose by salivary amylase in the mouth. This process is further continued by pancreatic amylase in the small intestine. Option A, salivary amylase, alone is not sufficient as it only begins the process. Option C, pancreatic amylase, alone does not account for the initial breakdown that occurs in the mouth. Option B, maltose, is incorrect because it is the product of the reaction, not an enzyme that facilitates it.

Q15. Which of the following best describes the residual volume of the lungs?

• A. The amount of air normally inhaled and exhaled with each breath.
• B. The maximum amount of air that can be forcibly inhaled and exhaled from the lungs.
• C. The volume of air that can still be forcibly exhaled following a normal exhalation.
• D. The volume of air that always remains in the lungs.✓

Explanation: This is also a book line.The volume of air that always remains in the lungs. Residual volume is the volume of air remaining in the lungs after maximum forceful expiration. In other words, it is the volume of air that cannot be expelled, thus causing the alveoli to remain open at all times. The residual volume remains unchanged regardless of the lung volume at which expiration was started.

Q16. The diagram shows the sequence of events occurring as an action potential arrives at a synapse.The numbered arrows represent movement of substances across the membranes.What are the substances moving across the membranes?

• A. A
• B. B
• C. C✓
• D. D

Explanation: During an action potential at the synapse, the presynaptic membrane is initially depolarized by the influx of sodium ions (1). It is then repolarized by the efflux of potassium ions (2). The arrival of the action potential opens calcium channels, allowing calcium ions to enter the synaptic knob (3), which prompts synaptic vesicles to release neurotransmitters like acetylcholine into the synaptic cleft (4). Acetylcholine then binds to receptors on the postsynaptic membrane, opening ion channels and causing sodium ions to enter (5), leading to depolarization. Options A, B, and D do not correctly describe this sequence or the involvement of these ions.

Q17. Arthropods can be characterized by all of the following EXCEPT:

• A. A hard exoskeleton
• B. A water vascular system✓
• C. Jointed appendages
• D. Molting
• E. Segmented body

Explanation: All are the characteristics of the phylum Arthropoda except option B. A water vascular system is a characteristic of the phylum Echinodermata, the "spinny-skinned" animals such as sea stars and sea urchins. Their water vascular system ends in tube feet that play a role in locomotion.

Q18. The role of decomposers in the nitrogen cycle is to:

• A. Fix Atmospheric Nitrogen Into Ammonia
• B. Incorporate nitrogen into amino acids and organic compounds
• C. Convert ammonia to nitrate, which can then be absorbed by plants
• D. Denitrify ammonia, thus returning nitrogen to the atmosphere
• E. Release ammonia from organic compounds, thus returning it into the soil✓

Explanation: The primary role of decomposers in the nitrogen cycle is to break down dead organic matter and release ammonia into the soil. This ammonia can then be used directly by plants or further converted into nitrates by other bacteria for plant absorption. This process is crucial for recycling nitrogen within ecosystems, ensuring a continuous supply of nitrogen for plant growth.Option A is incorrect because nitrogen fixation is not performed by decomposers. Option B is misleading because while decomposers break down nitrogen-containing compounds, their role is not to synthesize new ones. Option C refers to nitrification, a process carried out by other bacteria, not decomposers. Option D involves denitrification, which is not a function of decomposers but of denitrifying bacteria.

Q19. Black coat color in horses is caused by a dominant allele, while white coat color is due to the recessive allele. Two black horses produce a foal with a coat, if they were to produce a second foal, what would be the probability of the second foal having a black coat?

• A. 0
• B. 1/4
• C. ½
• D. 3/4✓
• E. 1

Explanation: If the black coat color in horses is the dominant phenotype, and both parental horses are black, then they must have at least one allele for the black coat color. Because the white coat color is the recessive phenotype, it can be caused only by a homozygous recessive genotype. The probability of these horses producing a foal with a black coat is 75%, or 3 out of 4.

Q20. Organisms that live in the intertidal zone might have which of the following characteristics?i.Ability to conduct photosynthesisii.Tolerance of periodic droughtiii. Tolerance of wide range of temperatures

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

Explanation: The organisms in the intertidal zone can conduct photosynthesis, have a tolerance to drought as well have a tolerance for a wide range of temperatures.

Q21. In floral formula “K” stands for:

• A. Corolla
• B. Calyx✓
• C. Perianth
• D. Androecium
• E. Gynoecium

Explanation: The floral formula uses standard symbols to denote different parts of a flower. 'K' stands for calyx, which is the collective term for the sepals that encase and protect the flower, especially during the bud stage. This is why K5 would indicate a calyx composed of five sepals. The other options refer to different parts of the flower: 'C' for corolla (petals), 'P' for perianth (when sepals and petals are indistinguishable), 'A' for androecium (male reproductive parts), and 'G' for gynoecium (female reproductive parts).

Q22. “Hordeum vulgare” is the botanical name of:

• A. Wheat
• B. Oats
• C. Rice
• D. Barley✓
• E. Bajra

Explanation: Hordeum vulgare is the botanical name of Barley.

Q23. The usual duration of luteal phase in the menstrual cycle of human female is:

• A. 4-6 days
• B. 8-10 days
• C. 12-14 days✓
• D. 10-12 days

Explanation: The usual duration of luteal phase in the menstrual cycle for the human female is 12-14 days.The luteal phase lasts from the day after ovulation until the day before your period starts. Experts say the average length of the luteal phase is 14 days, but there is a little difference that varies from one person to another.

Q24. Response to plants to touch is called:

• A. Geotropism
• B. Thigmotropism✓
• C. Nasticism
• D. Mechanoreception

Explanation: The correct answer is Thigmotropism. This term describes the movement or growth of a plant in response to touch or physical contact. 'Thigm' means touch, and plants like vines exhibit thigmotropism by wrapping around objects they come into contact with. Geotropism, on the other hand, deals with gravity, while nastic movements are non-directional and involve stimuli like light or temperature. Mechanoreception is more about detecting mechanical changes and is typically associated with animal responses.

Q25. Select the false statement:

• A. All fungi are saprophytic
• B. Mycology is the study of fungi
• C. Fungi are non coenocytic✓
• D. Puccinia is an obligate parasite

Explanation: The correct answer is Option C: Fungi are non coenocytic. This is false because many fungi, such as those in the group Zygomycetes, have coenocytic hyphae, which are characterized by a lack of septa, allowing cytoplasmic streaming and multinucleate cells.Option A is incorrect because not all fungi are saprophytic; some are parasitic (requiring a living host) or symbiotic, forming mutualistic relationships.Option B is a true statement; mycology is indeed the scientific study of fungi.Option D is also true; Puccinia species are obligate parasites, meaning they cannot complete their life cycle without a host organism.

Q26. Photosynthetic product from leaves to all parts of plant are disturbed through:

• A. Vascular bundles
• B. Phloem✓
• C. Xylem
• D. Stomata
• E. None of the above

Explanation: Sugars produced in sources, such as leaves, need to be delivered to growing parts of the plant via the phloem in a process called translocation, or movement of sugar. Hence, phloem is the correct answer.

Q27. In the 𝐹 generation of a dihybrid cross between yellow, round seeded and green, wrinkled seeded pea plants, 17 out of 254 seeds were green and wrinkled other seeds were:Yellow and roundGreen and roundYellow and wrinkledWhat do these results indicate?

• A. Crossing-over has occurred
• B. Green and wrinkled are both recessive characters✓
• C. The alleles for green and wrinkled are linked
• D. The allele for green is recessive but not the allele for wrinkled
• E. The allele for wrinkled is recessive but not the allele for green

Explanation: The correct answer is that both green and wrinkled are recessive characters. This conclusion is based on the fact that recessive traits appear only when both alleles are recessive (homozygous recessive). The observation of 17 green and wrinkled seeds out of 254 suggests these traits follow Mendel's laws of inheritance for recessive traits. Other options are incorrect because they either suggest genetic phenomena not supported by the data (crossing-over, linkage) or propose incorrect inheritance patterns for the traits involved.

Q28. Duckbill platypus and spiny anteater have internal fertilization and are:

• A. Ovoviviparous
• B. Viviparous
• C. Oviparous✓
• D. None of the above

Explanation: The duck-billed platypus and the spiny anteater, also known as the echidna, are unique among mammals because they are oviparous. Unlike most mammals that are viviparous, meaning they give birth to live young, these two species lay eggs. After internal fertilization, the eggs are laid by the mother and hatch outside her body. This differentiates them from ovoviviparous animals, which retain the eggs inside their body until they hatch, and viviparous animals, which do not lay eggs at all. Therefore, the correct option is 'Oviparous'.

Q29. Nematocysts are characteristics of:

• A. Porifera
• B. Protozoa
• C. Cnidarians✓
• D. Annelida
• E. Echinodermata

Explanation: Nematocysts are specialized stinging cells unique to the phylum Cnidaria, which includes jellyfish, corals, and sea anemones. These cells contain venomous coiled threads that can be deployed for protection or capturing prey. Porifera, Protozoa, Annelida, and Echinodermata do not possess these specialized structures, distinguishing Cnidarians as the correct group.

Q30. Which of the following is an acceptable nitrogen base composition for double stranded:

• A. 31% A, 19% T, 31% C, 19% G
• B. 36% A, 36% U, 24% C, 24% G
• C. 48% A, 48% T, 52% C, 52% G
• D. 31% A, 31% T, 19% C, 19% G✓
• E. 24% A, 24% U, 36% C, 19% G

Explanation: In double-stranded DNA, the nitrogen bases pair specifically: adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). Therefore, their percentages in the composition must match accordingly. In this case, the correct option shows equal percentages of A and T, and equal but lower percentages of C and G, balancing out the total to 100%. The incorrect options either misrepresent these pairings, include bases not found in DNA, or present impossible compositions.

Q31. The correct order of the structures through which air passes isI. Nasal cavityII. BronchiIII. LarynxIV. Air sacsV. Trachea

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

Explanation: Air enters the nose through nasal cavity,after this goes to the larynx and then to the trachea also called the windpipe and then it does to the bronchi that are cartilaginous,then to the bronchioles and at the end to the air sacs or alveoli.

Q32. Which of the following pathways outlines the order of events during aerobic cellular respiration?First ———-> Last

• A. Glucose -> Triose Phosphate -> Pyruvate -> Krebs Cycle -> CO2 + H2O + ATP✓
• B. Glucose -> Triose Phosphate -> Pyruvate -> Krebs Cycle -> CO2 + H2O + ADP + Pi
• C. Glucose -> Hexose Phosphate -> Pyruvate -> Krebs Cycle -> CO2 + H2O + ADP + Pi
• D. Glucose -> Hexose Phosphate -> Pyruvate -> Krebs Cycle -> Ethanol + CO2 + ATP

Explanation: The correct sequence for aerobic cellular respiration begins with glucose, which is phosphorylated and split into triose phosphate. This compound is converted into pyruvate, which enters the Krebs cycle. The cycle results in the production of CO2, H2O, and ATP through the electron transport chain. Other options either incorrectly represent products or suggest anaerobic processes like fermentation.

Q33. The diameter of a tree is reduced slightly during the day and increased at night. Which of the following changes in environmental conditions cause the greatest reduction in diameter?

• A. Increase in wind velocity, temperature, humidity, and light intensity
• B. Increases in temperature, humidity, and light intensity
• C. Increases in wind velocity, humidity, and light intensity
• D. Increases in wind velocity, temperature, and light intensity✓
• E. Increases in wind velocity, temperature, and humidity

Explanation: The correct answer is Option D, which includes increases in wind velocity, temperature, and light intensity. Higher wind velocity increases the rate of transpiration, leading to more water loss from the plant and thus a reduction in diameter during the day. Additionally, increased temperature and light intensity can elevate metabolic rates, further intensifying water loss. In contrast, the other options either include humidity, which counteracts water loss, or do not combine the factors effectively enough to maximize diameter reduction.

Q34. Why is there no glucose present in the distal of a nephron?

• A. Glucose molecules are too large to pass across the basement membrane.
• B. Glucose is removed by osmosis from the tubule.
• C. Glucose is passively absorbed by the cells lining the descending loop of Henle.
• D. Glucose is reabsorbed by the proximal tubule cells.✓

Explanation: Glucose is initially filtered through the glomerulus into the nephron, but it is actively reabsorbed by transport proteins in the proximal tubule cells. This mechanism ensures that glucose does not reach the distal parts of the nephron, making it absent there. Option A is incorrect because glucose is small enough to be filtered. Option B is incorrect because osmosis does not apply to glucose reabsorption. Option C is incorrect because glucose reabsorption does not occur in the descending loop of Henle.

Q35. Which of the following is the stage of meiosis during which pairs of homologous chromosomes align at the center of the cell?

• A. Anaphase II
• B. Metaphase II
• C. Prophase II
• D. Metaphase I✓
• E. Prophase I

Explanation: At the end of prometaphase I, meiotic cells enter metaphase I. Here, in sharp contrast to mitosis, pairs of homologous chromosomes line up opposite each other on the metaphase plate, with the kinetochores on sister chromatids facing the same pole. Pairs of sex chromosomes also align on the metaphase plate.Just remember it as the fact that alignment in Meiosis of homologous means formation of Metaphase plates and this occurs during Metaphase 1.

Q36. The tricuspid valve prevents backflow of blood from the

• A. Left ventricle into the left atrium
• B. Aorta into the left ventricle
• C. Pulmonary artery into the right ventricle
• D. Right ventricle into the right atrium✓

Explanation: The tricuspid valve is one of the heart's atrioventricular valves, situated between the right atrium and the right ventricle. Its primary function is to prevent the backflow of blood from the right ventricle into the right atrium when the ventricle contracts. This distinguishes it from other valves like the mitral valve on the left side, which prevents backflow from the left ventricle into the left atrium, the aortic valve, which prevents backflow into the left ventricle from the aorta, and the pulmonary valve, which prevents backflow into the right ventricle from the pulmonary artery. Therefore, option D is the correct answer, as it accurately describes the function of the tricuspid valve.

Q37. The liver:

• A. Decreases blood glucose levels
• B. Increases blood glucose levels
• C. Synthesizes glucose✓
• D. All of the above are functions of the liver

Explanation: The correct answer is that the liver synthesizes glucose. The liver plays a key role in glucose metabolism through processes like glycogenolysis, where it breaks down glycogen into glucose, and gluconeogenesis, where it creates glucose from non-carbohydrate sources. While the liver can release glucose into the bloodstream, it does not directly decrease or increase blood glucose levels like the pancreas does through insulin and glucagon. Thus, options A and B are incorrect as they describe functions related to pancreatic hormones more than liver functions. Option D is incorrect because not all the listed functions are roles of the liver.

Q38. At which two points of the menstrual cycle are the levels of oestrogen heights?

• A. Immediately before and after ovulation
• B. At ovulation and in the mid-luteal phase✓
• C. During the menstrual flow and pregnancy
• D. Pregnancy and after menopause

Explanation: Oestrogen levels peak twice during the menstrual cycle: first at ovulation (around day 12-14) and again in the mid-luteal phase. At ovulation, the surge in oestrogen triggers the release of luteinizing hormone (LH), leading to ovulation. The second peak during the mid-luteal phase occurs due to the corpus luteum's activity, which supports a potential pregnancy. Options mentioning pregnancy and menopause are not relevant to the menstrual cycle's regular hormonal patterns.

Q39. Herpes is a virus that enters the human body and remains dormant in the nervous system until it produces an outbreak, without any particular reason. Which of the following statements correctly describes herpes?

• A. The herpes virus can integrate its DNA into the host cell during its lysogenic cycle.
• B. During an outbreak, the herpes virus is in the lytic cycle, leading to symptoms.
• C. Herpes can remain dormant in the nervous system without showing symptoms for years.
• D. All of the above statements correctly describe the herpes virus.✓

Explanation: The correct answer is all of the above because:Option A: The herpes virus enters the lysogenic cycle by integrating its DNA into the host cell's DNA, allowing it to remain dormant.Option B: During an outbreak, the virus enters the lytic cycle, leading to rapid replication and symptoms.Option C: Herpes can indeed integrate into the host's DNA and remain dormant in the nervous system, evading the immune system's detection.Therefore, each statement provides a correct aspect of how the herpes virus operates within the body.

Q40. Which of the following statements could not be used to describe a species?

• A. A group of organisms showing distinctly similar autosomes
• B. A group of organisms showing analogous body structures✓
• C. A group of organisms capable of mating to produce viable offspring
• D. A group of organisms sharing the same eco- logical niche

Explanation: Analogous body structures refer to structures that serve similar functions but have different evolutionary origins. While this similarity in body structures can be observed in different species, it is not used to define a species. Species are typically defined based on factors such as genetic similarity, reproductive compatibility, and shared ancestry.

Q41. Which gaseous hydride most readily decomposes into its elements on contact with a hot glass rod?

• A. Ammonia
• B. Hydrogen chloride
• C. Hydrogen iodide✓
• D. Steam

Explanation: It is the answer because again the strength or thermal stability of hydrogen halides decrease down the group iodine is below both chlorine and bromine therefore its hydrogen halide thermal stability is the weakest because hydrogen iodide bond length is long and therefore less polar than HCl therefore decompose more readily , more easily.

Q42. A hydrocarbon, which is liquid at room temperature, decolourizes aqueous bromine. Which could be the molecular formula of the compound?

• A. C2H2
• B. C2H4
• C. C7H16
• D. C10H20✓

Explanation: Alkenes and Alkynes can decolorize bromine. so, the molecules must be either an alkyne or alkene. Since C5-C15 alkenes or C5-C12 alkynes are only liquids at room temperature, the molecules must fall within this range. Among the options Decene is an alkene with 10 carbons so can decolorize bromine as well as is a liquid at room temperature.

Q43. Bleaching powder is a good:

• A. Hydrating agent
• B. Oxidizing agent✓
• C. Dehydrating agent
• D. Reducing agent

Explanation: Bleaching powder is a good oxidizing agent because it can remove electrons from other substances. This property allows it to break down the chemical bonds of colored pigments, making it effective for bleaching fabrics. Additionally, it oxidizes and kills microorganisms, making it useful as a disinfectant. The other options are incorrect because:Hydrating agents add water to substances, which is not a function of bleaching powder.Dehydrating agents remove water, which is not relevant to the function of bleaching powder.Reducing agents donate electrons, which is opposite to the electron-removing function of an oxidizing agent like bleaching powder.

Q44. The value of the enthalpy change for the process represented by the equation:Na(s) —-> Na+(g) + e- is equal to:

• A. The first ionisation energy of sodium.
• B. The enthalpy change of vaporisation of sodium.
• C. The sum of the enthalpy change of the atomization and the first ionisation energy of sodium.✓
• D. The sum of the enthalpy change of atomization and the electron affinity of sodium.

Explanation: The correct answer is the sum of the enthalpy change of atomization and the first ionisation energy of sodium. This is because the process Na(s) —-> Na+(g) + e- involves two key steps: converting solid sodium to gaseous atoms (atomization) and then ionizing these gaseous atoms by removing an electron (ionisation energy). Option A is incorrect because it only accounts for the ionisation step. Option B is incorrect because it only considers the phase change to gas, not ionisation. Option D is incorrect because electron affinity involves electron gain, not loss, as described in the equation.

Q45. Which statement about one mole of metal is always correct?

• A. It contains the same number of atoms as 1 mole of hydrogen atoms.✓
• B. It contains the same number of atoms as 1/12 mole of 12 C.
• C. It has the same mass as 1 mole of carbon atoms.
• D. It is liberated by 1 mole electrons.

Explanation: One mole of any elememt contain same(Avagadro number)of particles1mol of Cu=6.02×10²³particles1mol of H=6.02×10²³particles

Q46. As the atomic number increases in a group, the chemical properties:

• A. Change
• B. Stay roughly the same✓
• C. Decrease
• D. Increase

Explanation: Chemical properties depend upon no. Of electrons in valance shell.All the elements of a group have similar chemical properties because they have same number of valence electrons in their outermost shell.

Q47. The crystals formed as a result of vander der Waals interactions are:

• A. Molecular crystals✓
• B. Covalent crystals
• C. Metallic crystals
• D. Ionic crystals
• E. None of these

Explanation: Molecular crystals have Vander Waal forces between them. Covalent crystals have strong covalent bonds between its structure. Ionic crystals are bound by ionic bonding. Metallic bonds are bound by strong electrostatic attraction between positively charged ions and electrons dispersed in the matrix.

Q48. All the following are the true statement concerning catalyst EXCEPT:

• A. A catalyst will speed up the rate determining step
• B. A catalyst will be used up in a reaction✓
• C. A catalyst may induce steric strain in a molecule to make it react more readily
• D. A catalyst will lower the activation energy of reaction

Explanation: The correct answer is Option B: A catalyst will be used up in a reaction. This statement is false because a catalyst is not consumed in the reaction; it remains unchanged and can be used repeatedly. Catalysts work by lowering the activation energy, thereby increasing the reaction rate, as correctly described in Options A and D. Option C is also true because catalysts can alter the spatial arrangement of molecules to make them more reactive, such as by inducing steric strain.

Q49. Which of the following processes is endothermic?

• A. The condensation of steam
• B. The electrolysis of water✓
• C. The freezing of water
• D. 𝐶𝑎(𝑠)+ 2𝐻2O (I)---> 𝐶𝑎2 (aq)+ 𝐻2 (g)
• E. 𝐻+(aq) + OH(aq) ---->𝐻2O (I)

Explanation: During electrolysis, heat is given to the water which gets absorbed by water to give oxygen and hydrogen and so the reaction is an endothermic reaction.

Q50. Which reagent gives a colorless homogeneous solution when added tp phenol?

• A. Aqueous bromine
• B. Aqueous sodium carbonate
• C. Aqueous sodium hydroxide✓
• D. Aqueous sodium hydroxide and benzoyl chloride

Explanation: Phenol reacts with sodium hydroxide solution to give a colourless solution containing sodium phenoxide. In this reaction, the hydrogen ion has been removed by the strongly basic hydroxide ion in the sodium hydroxide solution

Q51. Which substance has tetrahedral geometry?

• A. Benzene
• B. Methane✓
• C. Cyclohexane
• D. Both B and C

Explanation: The correct answer is Methane (Option B). Methane (CH4) has a central carbon atom bonded to four hydrogen atoms, and the molecule adopts a tetrahedral geometry with bond angles of about 109.5°. Benzene (Option A) is a planar molecule with a hexagonal shape, and cyclohexane (Option C) is a cyclic molecule that can adopt various conformations but is not a simple tetrahedral molecule. Therefore, Option D is incorrect as well.

Q52. The _ free radical takes part in the destruction of the ozone layer.

• A. Chlorine✓
• B. Helium
• C. Neon
• D. Xenon

Explanation: The chlorine free radical is pivotal in the degradation of the ozone layer. When chlorofluorocarbons, which contain chlorine, carbon, and fluorine, are exposed to ultraviolet radiation in the stratosphere, they release chlorine atoms. These chlorine atoms are highly reactive and can destroy ozone (O3) molecules through a catalytic cycle, ultimately forming chlorine monoxide (ClO) and molecular oxygen (O2). The chlorine atom is then free to repeat the process, leading to significant ozone depletion.In contrast, Helium, Neon, and Xenon are noble gases that are chemically inert. They do not form free radicals and are not involved in ozone layer depletion. Therefore, they are incorrect options in the context of this question.

Q53. How many atoms of carbon are present in 18 g of glucose, 𝐶6 𝐻12 𝑂6?

• A. 6.0 x 1022
• B. 3.6 x 1023✓
• C. 6.0 x 1023
• D. 3.6 x 1024
• E. 6.0 x 1024

Explanation: •18g of glucose =0 1 mol•No of moles of C in = 6mol1mol of glucose•moles of C in 0.1 =0.6molMol of glucose •no.of atoms of C = 0.6×6.02×10²³=3.6×10²³In 0.1 mol of Glucose ACCORDING TO ABOVE CALCULATION OPTION B is only correct

Q54. Which property of a gas affects the rate at which it spreads throughout a laboratory?

• A. Boiling point
• B. Molecular Mass✓
• C. Reactivity
• D. Solubility in water

Explanation: The correct answer is molecular mass. According to Graham's law of diffusion, the rate at which a gas spreads is inversely proportional to the square root of its molecular mass. This means lighter gases diffuse faster than heavier gases. For example, hydrogen (H2) with a molecular mass of 2, diffuses faster than oxygen (O2) with a molecular mass of 32.Other options, such as boiling point, reactivity, and solubility in water, pertain to different physical properties and chemical behaviors of substances and do not directly influence the rate of gas diffusion in air.

Q55. The bonding in sulphuric acid can be represented by the structure shown: What is the total number of electrons in the covalent bonds surrounding the sulphur atom?

• A. 4
• B. 6
• C. 8
• D. 12✓

Explanation: In sulphuric acid (H2SO4), sulphur is the central atom surrounded by four oxygen atoms. Sulphur forms two double bonds with two of the oxygen atoms and two single bonds with the other two oxygen atoms. Each covalent bond involves two electrons. Therefore, the total number of electrons in the covalent bonds around the sulphur atom is 12 (6 covalent bonds x 2 electrons each). Option D is correct because it accurately reflects this total. The other options (4, 6, 8) underestimate the number of electrons based on a misunderstanding of the bonding structure in sulphuric acid.

Q56. One mole of an organic compound is completely burnt in oxygen. Which compound produces exactly three moles of water?

• A. Butane C4H10
• B. Butanol C4H9OH
• C. Ethanol C2H5OH✓
• D. Propane C3H8

Explanation: C2H5OH + 3O2 = 2CO2 + 3H2O

Q57. Which pair of structures are isomers of each other?

• A. Ethane, Ethanol
• B. Ethanoic Acid, Ethanol
• C. Propanol, 2-Propanol✓
• D. Ethene, Butene

Explanation: The pair of structures that are isomers of each other is Propanol, 2-Propanol. Both have the molecular formula C3H8O, but differ in the position of the hydroxyl group. Propanol is a linear alcohol with the -OH on the first carbon, while 2-propanol is a secondary alcohol with the -OH on the second carbon. This structural difference classifies them as isomers. Other options are not correct because they involve compounds with different molecular formulas, which cannot be isomers by definition.

Q58. When must a substance be an alkane?

• A. When it burns in air or in oxygen
• B. When it contains carbon and hydrogen only
• C. When it has the general formula CnH2n+2✓
• D. When it is generally unreactive

Explanation: The correct answer is that a substance must have the general formula CnH2n+2 to be classified as an alkane. Alkanes are saturated hydrocarbons that consist entirely of single-bonded carbon and hydrogen atoms. This specific formula distinguishes alkanes from other types of hydrocarbons, such as alkenes (CnH2n) and alkynes (CnH2n-2). Option A is incorrect because combustion is not unique to alkanes. Option B is too broad, as other hydrocarbons also contain only carbon and hydrogen. Option D is misleading, as the concept of being generally unreactive is more closely associated with noble gases than with alkanes.

Q59. Which statement shows that diamond and graphite are allotropes of carbon?

• A. Both have giant molecular structures
• B. Complete combustion of equal masses of carbon dioxide as the only product
• C. Graphite conducts electricity, whereas diamond does not
• D. Under suitable conditions, graphite can be converted into diamond✓

Explanation: This statement directly proves they are allotropes because it shows they are two different physical forms of the same element that can be interconverted. Allotropes differ in the arrangement of atoms but are made of the same element. The ability to change from graphite to diamond under high temperature and pressure is clear evidence that both are pure carbon in different structural forms.

Q60. If Aufbau rule is not followed in filling of the sub shell then the block of which elements will change in the periodic table.

• A. K (19)✓
• B. Se (34)
• C. V (23)
• D. Ni (28)

Explanation: When the Aufbau rule is not followed, the electron configuration of potassium (K) changes as the last electron occupies the 3d sublevel instead of the 4s. This would classify potassium as a d block element, instead of its usual classification in the s block. In the case of other elements like Se, V, and Ni, their block classification remains unchanged because their electron configurations naturally place them in the p or d blocks, respectively. Only in potassium does the change in electron order affect its block classification.

Q61. Which gas shows real behaviour?

• A. 8 g O2 at STP occupies a volume of 5.6 liters.
• B. 1 g H2 in a 0.5-liter flask exerts a pressure of 24.63 atm at 300 K.
• C. 1 mole NH3 at 300 K and 1 atm occupies a volume of 22.4 liters.✓
• D. 5.6 liters of CO2 at STP has a mass of 11 g.

Explanation: The correct answer is 1 mole NH3 at 300 K and 1 atm occupies a volume of 22.4 liters. Although NH3 occupies the expected volume for an ideal gas, it is a polar molecule that experiences significant intermolecular hydrogen bonding. These interactions cause deviations from ideal gas behavior, making NH3 exhibit real gas characteristics.In contrast, O2 and CO2 are non-polar and show minimal deviation from ideal behavior under standard conditions. H2, despite the high pressure in the flask, is a small non-polar molecule that generally behaves ideally under normal conditions.

Q62. Heat of neutralisation of strong acid by strong base is constant value because:

• A. Salts formed do not hydrolyze
• B. Only H+ and OH- ions react in every case
• C. Strong base and strong acid react completely✓
• D. Strong base and strong acid react in aqueous solution

Explanation: The heat of neutralization of a strong acid by a strong base is constant because these substances react completely. Every molecule of the acid and base fully participates in the reaction, leading to a consistent energy change as no molecule is left unreacted or involved in side reactions. The other options, while mentioning true aspects of the reaction, such as lack of hydrolysis and the aqueous medium, do not address the fundamental reason for the constancy, which is the complete reaction of the reactants.

Q63. Compared with alkaline earth metals the alkali metals exhibit:

• A. Smaller Ionic Radii
• B. Greater Hardness
• C. High Boiling Point
• D. Lower Ionisation Energy✓

Explanation: Alkali metals exhibit lower ionisation energy compared to alkaline earth metals. This is because alkali metals have only one valence electron, which is further from the nucleus and experiences less nuclear charge, making it easier to remove. In contrast, alkaline earth metals have two valence electrons, which are more tightly bound due to the higher nuclear charge, resulting in greater ionisation energy. The other options are incorrect because alkali metals have larger ionic radii, are softer, and have lower boiling points compared to alkaline earth metals.

Q64. The number and types of bonds between two carbons atoms in 𝐶𝑎𝐶2 are:

• A. One sigma and one pi bond
• B. One sigma and two pi bonds✓
• C. Two sigma bonds
• D. One sigma bond

Explanation: The correct answer is that the bond between the two carbon atoms in CaC2 is one sigma (σ) and two pi (π) bonds, forming a triple bond. This is characteristic of acetylide ions (C22-), where two carbon atoms are bound by a triple bond. The sigma bond is formed by the head-to-head overlap of orbitals, whereas the pi bonds are formed by the side-to-side overlap of p orbitals. The calcium ions provide an ionic bond with the carbide ion, balancing the overall charge.Options A, C, and D are incorrect because they do not accurately describe the nature of the triple bond present in CaC2. A triple bond is always composed of one sigma bond and two pi bonds.

Q65. The alloy of copper and tin is called:

• A. Brass
• B. Bronze✓
• C. German silver
• D. Metal

Explanation: Bronze is the correct answer because it is an alloy of copper and tin, known for its durability and historical significance in making statues and tools. Brass is incorrect as it is made from copper and zinc. German silver is incorrect because it is an alloy of copper, zinc, and nickel, with no tin content. Metal is incorrect as it is a general term for metallic elements and not a specific alloy.

Q66. Which of the following is not an electrophile?

• A. 𝑁𝐻3✓
• B. 𝐵ℓ3
• C. AlCl3
• D. 𝐻𝑔2+

Explanation: Ammonia (𝑁𝐻3) is a nucleophile. The nitrogen atom in 𝑁𝐻3 has a lone pair of electrons making it an electron donor, hence it is not an electrophile. On the other hand, 𝐵ℓ3, AlCl3, and 𝐻𝑔2+ are all electrophiles. 𝐵ℓ3 and AlCl3 have incomplete octets that allow them to accept electrons, while 𝐻𝑔2+ has a positive charge that attracts electrons.

Q67. How many isomers are possible for the compound having molecular having molecular formula C3H5Br.

• A. 2
• B. 4✓
• C. 6
• D. 8

Explanation: The correct answer is 4. The compound with the molecular formula C3H5Br can form four isomers, which are structural isomers due to variations in the placement of the bromine atom and the types of carbon-carbon bonds:1-bromopropene: CH2=CH-CH2Br2-bromopropene: CH3-CH=CHBr3-bromopropene: CH3-CBr=CH21,2-dibromopropane: BrCH2-CH=CH2Each of these isomers represents a unique structural arrangement of the atoms in C3H5Br. The other options (2, 6, and 8) are incorrect because they either underestimate or overestimate the possible isomers based on the molecular structure constraints.

Q68. Which xylene is most easily sulphonated:

• A. Ortho
• B. Para
• C. Meta✓
• D. All at the same rate

Explanation: Meta-xylene is the most easily sulphonated due to the arrangement of the -CH3 groups, which enhances electron donation, making the ring more reactive to electrophiles. Ortho-xylene experiences steric hindrance due to adjacent methyl groups, reducing its reactivity. Para-xylene, while less hindered, is still less reactive than meta due to its electronic configuration. The idea that all xylenes sulphonate at the same rate is incorrect because steric and electronic factors vary with their molecular structure.

Q69. Which one of the following is likely to give a precipitate with 𝐴𝑔𝑁𝑂3 solution?

• A. 𝐶𝐶𝑙4
• B. 𝐶𝐻𝐶𝑙3
• C. 𝐶𝐻2=CHCl
• D. (𝐶𝐻3)3𝐶𝐶𝑙✓

Explanation: The correct answer is (𝐶𝐻3)3𝐶𝐶𝑙, or tert-butyl chloride, because its chloride ion is more likely to dissociate and react with AgNO3, forming a white precipitate of silver chloride (AgCl). The other options involve molecules where the chloride is covalently bonded in a manner that does not easily allow for dissociation: in CCl4 and CHCl3, the chloride is tightly bonded to carbon atoms, and in CH2=CHCl, the chloride is part of a stable double bond structure.

Q70. In 𝑆𝑝3𝑑 hybridization, the o orbital that participates in hybridization is:

• A. 𝑑𝑥2−𝑦2✓
• B. 𝑑𝑧2
• C. dxy
• D. Dxz

Explanation: In 𝑆𝑝3𝑑 hybridization, the 𝑑𝑥2−𝑦2 orbital is involved because it has the appropriate energy and symmetry to combine with the s and p orbitals. The 𝑑𝑧2 orbital is generally not involved in this type of hybridization as it doesn't align well with the directional properties required. Similarly, the dxy and Dxz orbitals do not participate in 𝑆𝑝3𝑑 hybridization because they form part of a different symmetry set that is not compatible for this hybridization.

Q71. A frictionless heat engine can be 100% efficient only its exhaust temperatures is:

• A. Double of its input temperature
• B. Half of its input temperature
• C. Equal to its input temperature
• D. 100%
• E. 0 K✓

Explanation: The correct answer is 0 Kelvin. The efficiency of a heat engine is given by the equation:η = 1 - T2/T1 where η is the efficiency, T2 is the exhaust temperature, and T1 is the input temperature. For a heat engine to be 100% efficient, the exhaust temperature (T2) must be 0 Kelvin. This is because the efficiency equation approaches 1 (or 100%) as T2 approaches 0. None of the other options reach 100% efficiency because their exhaust temperatures are not low enough to maximize the efficiency ratio.

Q72. The vector which only specifies the direction of a given vector is called:

• A. Free vector
• B. Position vector
• C. Null vector
• D. Unit vector✓

Explanation: Option A:Free vectors refers to a vector which is neither a point nor a line, and something that can move freely around the space though it has a fixed magnitude and fixed direction.Option B: Position vector specifies the given direction of a vector.Option C: A null vector is a vector having magnitude equal to zero. A null vector has no direction or it may have any direction. Generally a null vector is either equal to the resultant of two equal vectors acting in opposite directions or multiple vectors in different directions.Option D: A unit vector is one whose magnitude is equal to one. The "cap" symbol(^) is used to indicate unit vectors.

Q73. A ball is thrown vertically upward with velocity of 196 m/s. How high does the ball rise?

• A. 1960 metres✓
• B. 2960 metres
• C. 1000 metres
• D. 1100 metres

Explanation: As the ball is thrown upward the final velocity (vf) will be zero.2gh=vf2-vi2 =>h=0-(196)2/2×10 => h= 1920 m which is closer to 1960 m.

Q74. ‘’If there is no external force applied to a system, then the total momentum of that system remains constant”. This known as:

• A. Law of conservation of mass
• B. Elastic collision
• C. Law of conservation of momentum✓
• D. Momentum of a body

Explanation: This is the statement of the law of conservation of momentum.Option A: The mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another.Option B: Elastic collision is an elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.Option D:The momentum of the body is the product of the mass and velocity of the body. The factors on which the momentum of the body depends are mass and velocity.

Q75. A car travelling at a constant speed of 90 km/h rounds a curve of a radius 100m. What is its acceleration?

• A. 4.0 m/𝑠𝑒𝑐2
• B. 6.25 m/𝑠𝑒𝑐2✓
• C. 6.5 m/𝑠𝑒𝑐2
• D. 4.5 m/𝑠𝑒𝑐2
• E. 7.5 m/𝑠𝑒𝑐2

Explanation: To find the centripetal acceleration of a car traveling in a circular path, use the formula: a = v2/R. First, convert the speed from km/h to m/s: 90 km/h × (5/18) = 25 m/s. Then, apply the values to the formula: a = (25)2 / 100 = 6.25 m/𝑠𝑒𝑐2. Therefore, the correct answer is Option B. Options A, C, D, and E are incorrect because they do not use the correct calculations for the given speed and radius.

Q76. A boy on a 20 m high cliff drops a stone. One second later, he throws down another stone. Both the stones hit the ground simultaneously. Find the initial velocity of the second stone. (g=10/𝑠 2)

• A. 5 m/s2
• B. 10 m/s2
• C. 15 m/s2✓
• D. 20 m/s2
• E. 30 m/s2

Explanation: In case of drop initial velocity= 0Time will be given by h=1/2gt2t=√2h/g =√2×20/10 = 2 sec.Since both stones hit the ground simultaneously and the 2nd one was dropped 1 sec later, the time taken by the 2nd stone will be t' =2sec -1sec = 1sec .H'=vi×t+1/2gt'2 =>20= vi×1+½ ×10×1vi= 15 ms-1.

Q77. An elevator, in which a man is standing is moving upward with a constant speed of 10 m/𝑠 2. If the an of the drops a coin from a height of 2.5 m, find the time taken by it to reach the floor of the elevation (g=9.8 m/𝑠 2)

• A. 0.707 s✓
• B. 1.9 s
• C. 3.1 s
• D. 6.17 s
• E. 7.15 s

Explanation: Initial speed of the coin u=0 m/sAcceleration a=9.8 ms-2Initial height of coin from the floor of elevator h=2.5 m , vi=0H=vit+1/2gt2 => t=√2H/gt=√2×2.5/10 = 0.707 sec.

Q78. A 100 kg man runs up a hill through a height of 4m in seconds. How much work does he do against gravitational force?

• A. 2060 J
• B. 3920 J✓
• C. 5000 J
• D. 5290 J

Explanation: Work done = mghW= 100×9.8×4 =3920 J.

Q79. Which statement describes the electrical potential difference between two points in a wire carrying a current?

• A. The force required to move a unit positive charge between the points
• B. The ratio of the energy dissipated between the points to the current
• C. The ratio of the power dissipated between the points to the current✓
• D. The ratio of the power dissipated between the points to the charge moved
• E. None of the above

Explanation: As, P=VIV= P/I. Hence option C is correct.

Q80. Find the time period of a simple pendulum whose length is 88.2 cm. The value acceleration due to gravity is 9.8 m/𝑠 2 at the place where experiments is performed?

• A. 1.885 sec✓
• B. 1.233 sec
• C. 2.05 sec
• D. 4 sec

Explanation: The formula to calculate the time period of simple pendulum is T=2π √l/g where l=length of pendulum, g=acceleration due to gravity.By putting valuesFirst convert length from cm to m by dividing by 100 and we get 0.882 m. By putting values T=2(3.14) √0.882/9.8 = 1.88 sec.

Q81. A light bulb has resistance of 150 ohm, find the voltage while the current is 1.5A?

• A. 250 V
• B. 300 V
• C. 224 V
• D. 225 V✓

Explanation: Ohm's Law is given byV=IRBy putting valuesV= (1.5)(150) = 225 V.

Q82. An object that’s moving with constant speed travels around a circular path. Which of the following is/are true concerning this motion?I The displacement is zeroII. The average speed is zeroIII. The circulation is zero

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

Explanation: I-Travelling once around a circular path means that the final position coincides with the initial position. Therefore, the displacement is zero. II-The average speed, which is total distance travelled divided by elapsed time, cannot be zero.III-Circulation is the amount of force that pushes along a closed boundary or path. It's the total "push" you get when going along a path, such as a circle.Circulation will be zero when the body passes through the origin which in the given case is not true.Therefore, only I is correct.

Q83. A system absorbs 80 J through heating while doing 100 J of external work, what is the change in the internal energy of the system?

• A. -100 J
• B. -20 J✓
• C. +80 J
• D. +180 J

Explanation: Since the work is done by the system it is positive.And heat is absorbed by the system so it is also positive.As, Q=∆U+W where ∆U is change in internal energy.∆U=Q-W =80-100 = -20 J

Q84. A region around a charge body in which another charge experiences an electric force is called:

• A. Electric flux
• B. Electric field✓
• C. Electric potential
• D. Capacitance

Explanation: Option A: [17/12, 10:07 am] mariam: Electric flux is 'the total number of electric field lines passing a given area in a unit of time'.Option B: This is the definition of electric field.Option C: Electric potential is defined as work done in moving a charged particle from infinity( where potential is zero) to that point. It is basically a potential difference between two points.Option D:Capacitance is the ability of a capacitor to store charge.

Q85. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. the power of the combination in dioptres is:

• A. -1.5✓
• B. -6.5
• C. +6.5
• D. +6.67
• E. -7.7

Explanation: Given,f1=40cm , f2=-25cmPower, P=1/fThe power of combination is, P=1/f1+1/f2P=1/40+ 1/-25 = -1.5 D

Q86. A simple pendulum suspended from the ceiling of a train has a period ‘T’ when the train is at rest. When the train is accelerating with a uniform acceleration, the time period of simple pendulum will:

• A. Decrease✓
• B. Increase
• C. Remain unchanged
• D. Become infinite
• E. Insufficient information

Explanation: Understanding the concept:The time period of a simple pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. When the train accelerates, an additional pseudo force acts on the pendulum bob in the opposite direction of the train's acceleration. This pseudo force effectively increases the "effective gravity" experienced by the pendulum. 2. Calculating effective gravity:When the train accelerates horizontally with acceleration 'a', the pendulum bob experiences two accelerations: 'g' vertically downwards and 'a' horizontally. The effective acceleration (g_eff) is the vector sum of these two accelerations.g_eff = √(g² + a²) 3. Calculating the new time period:The new time period (T') will be:T' = 2π√(L/g_eff) = 2π√(L/√(g² + a²))4. Comparing T' and T:Since g_eff > g, the denominator in the equation for T' is larger than the denominator in the equation for T. Therefore, T' < T.Conclusion:When the train accelerates, the time period of the simple pendulum decreases.

Q87. Lenz’s law states that:

• A. The flow of a fluid in a medium under the same applied force experiences some sort of friction or resistance in its path
• B. A body remains at rest or continues to move with uniform velocity unless acted upon by an unbalanced force
• C. The induced current always flows in such a direction as to oppose the change which is giving rise to it✓
• D. When a particle bearing a charge q and moving with a velocity V enters the region of a uniform magnetic field of induction B, it is acted upon by a force

Explanation: This is the core of Lenz's Law. It's a consequence of the conservation of energy. If the induced current aided the change, you'd get a runaway effect, creating energy from nothing.

Q88. Two rail cargo cars are being hitched together. The first car has a mass of 15,750 kg and is moving at a speed of 4 m/s. What is the final velocity of the two cars? m2 is 19250

• A. 1.8 m/s✓
• B. 1.9 m/s
• C. 5.8 m/s
• D. 7.8 m/s
• E. 9.8 m/s

Explanation: The problem requires the application of the conservation of momentum principle. The equation for the conservation of momentum is given by:m1v1i + m2v2i = (m1 + m2)vfWhere m1 is the mass of the first car (15,750 kg) and v1i is its initial velocity (4 m/s). The second car is initially at rest, so v2i is 0 m/s, and m2 is 19,250 kg.Substituting the values, we get:vf = (15,750 kg * 4 m/s + 19,250 kg * 0 m/s) / (15,750 kg + 19,250 kg) = 1.8 m/sThis is the final velocity of the combined mass system. Option A is the correct answer. The other options are incorrect due to calculation errors or misapplication of the formula.

Q89. Two capacitors 𝐶1 = 2𝑢 F and 𝐶1 = 4𝑢 F are connected series across a 100 V supply. Find the effective capacitance.

• A. 1/2𝑢 F
• B. 3/2𝑢 F
• C. 4/3𝑢 F✓
• D. 7/2𝑢 F
• E. 9/2𝑢 F

Explanation: The total capacitance of capacitors in series is found using the formula: 1/Ctotal = 1/C1 + 1/C2. For C1 = 2μF and C2 = 4μF, we get:1/Ctotal = 1/2 + 1/4 = 2/4 + 1/4 = 3/4So, Ctotal = 4/3 μF.This shows that the correct answer is 4/3 μF, making Option C correct. The other options result from incorrect calculations or misunderstandings of the series capacitor formula.

Q90. Consider two equal resistors wired in parallel. What is the equivalent resistance of the two?

• A. 3R/2
• B. R/2✓
• C. R/3
• D. R/5
• E. R/7

Explanation: In order to find the equivalent resistance of resistors with identical resistances connected in parallel, we useReq= R/n , where n is number of resistors Req= R/2

Q91. A detector with a surface area of 1 square meter is placed 1 meter from an operating jackhammer. It measures the power of the jackhammer’s sounds as being 10−3 W and finds the intensity of the jackhammer.

• A. 10 W/m2
• B. 10-3 W/m2✓
• C. 10-9 W/m2
• D. 10-7 W/m2
• E. 10-9 W/m2

Explanation: Sound Intensity (SI) is defined as the power carried by sound waves per unit area in a direction perpendicular to that area. The units associated with SI is the watt per square metre (W/m2).I = P/A =10-3/1 I = 10-3 Wm-2.

Q92. Ultraviolet light is more likely to cause a photoelectric effect than visible light. This is because photons of ultraviolet light:

• A. Have a longer wavelength
• B. Have a higher velocity
• C. Are not visible
• D. Have a higher energy✓

Explanation: Photoelectric effect: When the light of a particular frequency is focused on metal, it ejects an electron from the metal's surfaceFor ejection of metal, the light should be of greater frequency than the threshold (minimum) frequency and it is different for different metals.From the electromagnetic spectrum, we know that the wavelength of UV light is low.Wavelength and frequency are inversely proportional.Therefore, UV light has more frequency and energy than visible light which makes it suitable for use in photoelectric effect.

Q93. If an object is released 19.6 metres above the ground, how long does it take the object to reach the ground? (g=9.8 𝑚/𝑠^2)

• A. 1 second
• B. 2 seconds✓
• C. 6 seconds
• D. 8 seconds
• E. 10 seconds

Explanation: To find the time it takes for an object to fall to the ground from a height of 19.6 meters with an acceleration due to gravity of 9.8 m/s2, we use the formula for distance in free fall: h = (1/2)gt2. Solving for t, we get:t = sqrt(2h/g) = sqrt(2 * 19.6 / 9.8) = sqrt(4) = 2 seconds.Therefore, the correct answer is 2 seconds. The other options are incorrect as they do not satisfy the equation derived from the physics of free fall motion.

Q94. Two tuning forks are sounded, one has a frequency of 250 Hz while the other has a frequency of 245 Hz. What is the frequency of the beats?

• A. 250 Hz
• B. 245 Hz
• C. 5 Hz✓
• D. 10 Hz
• E. 235 Hz

Explanation: Beat frequency i.e,number of beats per second is equal to the difference between the two combining frequencies. Hence, there will be 250-245= 5 Hz.

Q95. A rock is dropped from a high bridge at the end of 3 seconds of free fall the speed of the rock in cm/s:

• A. 30
• B. 100
• C. 500
• D. 1000
• E. 2940✓

Explanation: As, vf= vi+atIn case of drop initial velocity=0 vf=0+9.8×3vf=29.4m/s => vf=29.4×100 cm/svf= 2940 cm/s.

Q96. A body rolling freely on the surface of the earth eventually comes to rest because:

• A. It has mass
• B. It suffers friction✓
• C. It has inertia of rest
• D. It has momentum
• E. It is gravitation less because it is already on the surface

Explanation: The correct answer is: It suffers friction.When a body rolls on a surface, friction acts between the contact points of the body and the surface. This force opposes the body's motion, gradually converting its kinetic energy into heat, which slows it down until it comes to rest.Other options are incorrect because:Mass is not responsible for stopping the body; it influences inertia but not the cessation of motion.Inertia of rest only applies to objects initially at rest, not moving ones.Momentum describes motion but does not explain why an object stops; external forces like friction are required to change it.The notion of being "gravitation-less" is irrelevant; gravity acts on all objects on Earth's surface.

Q97. A 1000 kg car can accelerate from rest to a speed of 25 m/s in 10s. what average power (in kilowatts) must the engine of the car produce in order to cause this acceleration? Neglect the friction loss.

• A. 33.25
• B. 3625
• C. 48.44
• D. 3125
• E. 31.25✓

Explanation: Kinetic energy gained by the car is equal to (1/2mv2)K.E=½ x 1000 x 252 = 312500 JMinimum power required by the car = K.E/t P= 312500/10=31250 watt.And in kilowatt, it is 31.25 kwatt. So E is correct.

Q98. The kinetic energy of a projectile at the highest point is half of its kinetic energy. The angle of projection is:

• A. 0 degrees
• B. 30 degrees
• C. 60 degrees
• D. 45 degrees✓
• E. 90 degrees

Explanation: The correct answer is 45 degrees. When a projectile is launched at 45 degrees, its initial kinetic energy is distributed equally between the horizontal and vertical components. At the highest point, the vertical component is zero, and only the horizontal component contributes to the kinetic energy, which is half of the initial total kinetic energy. The other options either lack the correct distribution of kinetic energy or do not satisfy the condition described in the problem.

Q99. A small and a large raindrops are falling through air

• A. The small drop will evaporate
• B. The large drop moves faster✓
• C. The small drop moves faster
• D. Both move with the same speed
• E. No conclusion can be drawn unless the exact sizes of the drops are known

Explanation: vt=2r2ρg/9ηWhere ρ is the density of water and η the viscosity of air. As the terminal velocity of the sphere is directly proportional to the square of radius. So a larger drop will have larger terminal velocity and hence will move faster.

Q100. A container is divided into two equal portions. One portion contains an ideal gas at pressure P and temperature T while the other portion is a perfect vacuum. If a hole is opened between the two portions:

• A. There will be a change in internal energy
• B. There will be a change in temperature
• C. There will be no change in internal energy✓
• D. The external pressure will increase
• E. The external pressure will decrease

Explanation: When the hole is opened between the two portions of the container, the ideal gas undergoes free expansion into the vacuum. In this process, no work is done by or on the system, and there is no heat exchange with the surroundings. Therefore, the internal energy, which for an ideal gas depends solely on temperature, remains constant. Option C is correct because the internal energy of the gas does not change.Other options are incorrect because they misunderstand the concept of an ideal gas's internal energy and external pressure interactions. The temperature remains constant because the internal energy does not change, and external pressure is unaffected by the internal changes in the container.