Fmdc 2014 — Solved Past Paper with Answers

Q1. My dog happily _ itself under the table and chewed contentedly at a piece of bone.

• A. Eclipsed
• B. Elucidated
• C. Exhumed
• D. Ensconced✓

Explanation: A. Eclipsed means obscure or block out.B. Elucidated means to explain something clearly.C. Exhumed means to dig out something.D. Ensconced means to hide.In the following options, Option D fits the most in the above statement. So Option D is the correct answer.

Q2. Which word is closest in meaning of "ASTOUND"?

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

Explanation: The word ‘Astound’ means to shock or surprise someone.So Option A is the right option.

Q3. 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.

Q4. Crystal Vase breaks easily, _ handle it with care.

• A. So that
• B. So much
• C. So✓
• D. So well

Explanation: Option C is the right option. The above sentence gives us an idea about how to stop the vase from breaking. It cautions us to handle it with care. ‘So’ fits here.

Q5. The Police do not understand how the thieves _ the house.

• A. Got into✓
• B. Got in
• C. Get into
• D. Get in

Explanation: ‘Got into’ the house is the correct answer. The sentence involves a phrasal verb ‘got into’ which means enter into the house.

Q6. Which of the following words is closes to the meaning of "BEHOLD"?

• A. See✓
• B. Touch
• C. Hear
• D. Smell

Explanation: The word behold means to look or see something. So, Option A is the right answer.

Q7. Identify the word or phrase that needs to be changed for the sentence to be correct:Ali finds it difficult to put his idea into words.

• A. Finds
• B. Difficult
• C. To put
• D. Into
• E. No error✓

Explanation: The sentence is grammatically correct. Hence Option E is correct.

Q8. Identify the word or phrase that needs to be changed for the sentence to be correct:Owing to the flooded condition of roads, most of the people were unable to get to the offices.

• A. Owing
• B. Flooded
• C. Unable
• D. Get to
• E. No error✓

Explanation: The sentence is already correct and does not require any changes. It is grammatically accurate and effectively conveys the intended meaning.

Q9. Identify the word or phrase that needs to be changed for the sentence to be correct:Sharks differ from other fish in that their skeleton are made of cartilage instead bone.

• A. Differ
• B. Fish
• C. Are made✓
• D. Instead
• E. No error

Explanation: Option C is the right option. Skeleton is not a plural noun and hence ‘are’ is wrong. It should be ‘their skeleton is made of cartilage instead of bone’. You will probably argue that Option D is also wrong but Option C is more grammatically suitable option to pick.

Q10. It tripped _the cat and fell downstairs.

• A. Over✓
• B. On
• C. Onto
• D. In

Explanation: A is the right option. It refers to a thing that went over the cat and fell downstairs. A is the most suitable option for this. Something can’t trip IN or ON the cat. ONTO is also a wrong option.

Q11. The _left their home land to another country in the hope of securing a better future for their families.

• A. Emigrants✓
• B. Immigrants
• C. Migrants
• D. Strollers

Explanation: A. Emigrants are people who leave their homeland and go to another country. Option A is right.B. Immigrants are the people that are taken into a new country in the hope of securing a better future. B is wrong because when we talk about people who leave their homeland, they are emigrants, not immigrants.C. Migrants move from one place to another for a better person. It is not necessary for them to leave their homeland for another country. Hence Option C is wrong.D. Strollers are people taking a leisurely walk. Option D is totally wrong.

Q12. Hundreds of years old palace could not withstand the _ of heavy rains.

• A. Aftermath
• B. Havoc✓
• C. Annoyance
• D. Massacre

Explanation: When we refer to natural disasters, we use the word ‘havoc’ to describe their adverse destruction. Aftermath doesn’t fit here because aftermath just means the consequences of an unpleasant event. Option B is the most suitable option here when compared with the rest of the options.

Q13. Choose the correct spelling:

• A. Amatuer
• B. Amature
• C. Amateur✓
• D. Ameteur

Explanation: ‘Amateur is the correct spelling’ which means an amateur is a person who does something for enjoyment, not money, like an amateur who paints as a hobby but earns a living another way.

Q14. RITUAL means:

• A. Custom✓
• B. Original
• C. Religion
• D. Routine

Explanation: Ritual means a custom or a tradition of a civilization that has been going on for a while. It is a religious or solemn ceremony consisting of a series of actions performed according to a prescribed order.

Q15. He complains _ headache.

• A. At
• B. From
• C. Of✓
• D. To

Explanation: Option C is grammatically correct. At, from, and to are all wrong. He complains of something which is in this case a headache.

Q16. He is appreciated for being ambidextrous. The underlined word means:

• A. Active and Skillful
• B. Uses skills and wisdom
• C. Uses both hands with same skills✓
• D. Remains to the point

Explanation: It means using both hands with equal ease or dexterity.Examples:- an ambidextrous pitcher- Guatelli says the master was ambidextrous, that he sketched with his right hand while he wrote with his left simultaneously.

Q17. My friend has a fine _ of old stamps.

• A. Group
• B. Band
• C. Collection✓
• D. Bundle

Explanation: Collective noun for old stamps is ‘collection’ or an ‘album’Example: In 1934, he sold this stamp collection for $1 million.

Q18. The synonym for the word DISSIPATED is:

• A. Degenerate✓
• B. Pleased
• C. Inundated
• D. Encouraged
• E. Planned

Explanation: The word dissipate means to disappear or vanish. Option A also has the same synonyms. Hence Option A is the correct option.

Q19. The synonym for the word COUNTENANCE is:

• A. Procure
• B. Insist
• C. Approve of✓
• D. Forego
• E. Interpret

Explanation: Option C is the right option as both countenance and approve of are synonyms of each other.

Q20. At which side does digestion of starches begin?

• A. 1✓
• B. 2
• C. 3
• D. 4
• E. 5

Explanation: 1 is the mouth. Digestion of starch starts from the oral cavity through the action of Salivary Amylase. So A is correct. Digestion is initiated in the oral cavity.

Q21. What is the part labelled as 5 in the following diagram?

• A. Produces bile
• B. Stores bile✓
• C. Secretes lipase
• D. Secretes bicarbonate
• E. Secretes HCI

Explanation: Structure 5 is the Gall Bladder. Liver produces bile which is stored in the gall bladder. It is connected through the cystic duct.

Q22. Which structure is primarily responsible for water absorption during digestion?

• A. 5
• B. 6
• C. 7
• D. 14✓
• E. 9

Explanation: Large intestines are responsible for water absorption during digestion.

Q23. The diagram summaries the pathway of glucose break-down:Which two steps result in a net increase of ATP?

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

Explanation: Step 1 involves breakdown of glucose into two molecules of triose-phosphate- Step 2 involves production of 4 ATP to form a pyruvate so 2 is correct- Step 3 has nothing to do with this cycle- Step 4 involves addition of Oxaloacetate with Acetyl CoA to form a 6C compound. It is not an energy yielding process.- Step 5 involves the complete breakdown of glucose which is an energy yielding process. So, 5 is also correct.

Q24. Which one of the following enables the mammalian kidney to regulate water reabsorption during states of dehydration?

• A. The cells of the tubules detect the osmotic pressure of the blood
• B. Water is extracted from the glomerular filtrate in the proximal tubule
• C. The kidney produces hyptonic urine
• D. Hormones increase the permeability of the collecting ducts✓
• E. A low solute concentration is maintained around the collecting ducts

Explanation: Water excretion by the kidney is regulated by the peptide hormone vasopressin. Vasopressin increases the water permeability of the renal collecting duct cells, allowing more water to be reabsorbed from collecting duct urine to blood.

Q25. A drug reduces mitochondrial activity in nephrons of kidney. Which chemical will be present in increased amount in the urine?

• A. Ammonia
• B. Glucose
• C. Hydrogen bicarbonate
• D. Urea✓

Explanation: The correct option is D. Urea. When mitochondrial activity is reduced in the nephrons of the kidney, urea levels in the urine increase. Urea is a waste product of protein metabolism and its excretion is influenced by kidney function.

Q26. Where, in the nephron, is most glucose reabsorbed?

• A. In the ascending loop of henle
• B. In the descending loop of henle
• C. In the proximal (first) convoluted tubule✓
• D. In the distal (second) convoluted tubule

Explanation: Most of the glucose entering the tubular system is reabsorbed along the nephron segments, primarily in the proximal tubule, such that urine is almost free of glucose. This is different in diabetes, where the filtered glucose exceeds the transport capacity of the tubular system for glucose and glucosuria occurs.

Q27. Consider the following statements about biological communities.(I) Their members share a common gene pool.(II) The community remains stable even though some physical aspect of the environment may undergo change.(III) They passes predictable kinds of species of predictable proportions.(IV) Interactions between their members are more frequent within the community than between their members and those of neighboring communities.Which two of the above statements apply to all stable biological communities?

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

Explanation: A. I is not right since a community involves different populations which may have different gene pool.B. II is also not right since different populations may interact in different ways with each other and disturb the genetic pool and stability.C. III is right and is a fact.D. IV is also right since intra-specific interactions are more common than inter-specific interactions in communities.Hence III and IV are the right options (Option D).

Q28. The spinal cord serves as the center of:

• A. Subconscious thought
• B. Reflex actions✓
• C. Habits
• D. Tropism

Explanation: It is the processing unit for reflex actions because all nerves from various parts of the body meet at the spinal cord before reaching the brain.

Q29. The most abundant substance in protoplasm is:

• A. Protein
• B. Fat
• C. Carbohydrate
• D. Water✓
• E. Minerals

Explanation: Protoplasm is a mixture of small molecules such as ions, monosaccharides, amino acids, and macromolecules such as proteins, polysaccharides, lipids, etc.On the average protoplasm contains 84 percent water. Most of the water is present in free unbounded form and provides a medium for metabolic reactions occurring in a cell.

Q30. The placenta releases all of the following hormones EXCEPT:

• A. Progesterone
• B. LH✓
• C. HCG
• D. Estrogen
• E. Minerals

Explanation: LH is made by your pituitary gland, a small gland located underneath the brain. LH plays an important role in sexual development and functioning. In women, LH helps control the menstrual cycle. It also triggers the release of an egg from the ovary. Rest is released by the placenta.

Q31. The diagram below represents the synapse between two mammalian myelinated neurons, X and Y. The arrows show the direction of impulses.The transmission of impulses across the synapse is brought about by the:

• A. Break-down of the terminal membrane of X.
• B. Passage of an electric current between X and Y.
• C. Release of sodium ions from X.
• D. Build-up of a potential difference between X and Y.
• E. Secretion of a chemical from X.✓

Explanation: The pre-synaptic neuron (X) releases a neurotransmitter which is a chemical messenger when calcium ions move in and action potential is generated due to flow of Na ions. Neurotransmitter attaches to the post-synaptic membrane and transmits the nerve impulse to the other neuron.

Q32. All of the following organs produce hormones involved in the reproductive cycle except the:

• A. Testes
• B. Pituitary
• C. Pancreas✓
• D. Ovary
• E. Uterus

Explanation: A. Testes produce testosterone which is involved in male reproduction.B. Pituitary gland produces ICTH, FSH and LH which are hormones involved in reproduction.C. Pancreas produces Insulin and Glucagon which are involved in regulation of blood glucose level. They have nothing to do with reproductive cycle so C is the right option.D. Ovaries produce estrogen which is involved in reproduction.E. Uterine Cells Produce Their Own Estrogen During Pregnancy.

Q33. In human female FSH regulates the concentration of:

• A. Control
• B. Estrogen✓
• C. Cortisol
• D. Aldosterone
• E. None of the above

Explanation: Follicle stimulating hormone (FSH) is produced by the pituitary gland in the brain. It is an important hormone for normal functioning of the reproductive system in men and women. In women, FSH helps control the menstrual cycle and the production of eggs by the ovaries. FSH stimulates growth of the granulosa cells of the ovarian follicle and controls their estrogen secretion. At the midpoint of the menstrual cycle, the increasing level of estradiol stimulates a surge of LH secretion, which then triggers ovulation.

Q34. The Reduction-division occurs during the process of:

• A. Cleavage
• B. Differentiation
• C. Fertilization
• D. Meiosis✓
• E. Parthenogenesis

Explanation: Meiosis is sometimes called "reduction division" because it reduces the number of chromosomes to half the normal number so that, when fusion of sperm and egg occurs, baby will have the correct number.

Q35. The muscles attached to the bones are:

• A. Voluntary and smooth
• B. Involuntary and smooth
• C. Voluntary and striated✓
• D. Involuntary and striated
• E. Smooth and striated

Explanation: Muscles attached to the bones are called skeletal muscles. Skeletal muscles are voluntary and striated.

Q36. Which of the following statements regarding the periosteum is INCORRECT?

• A. The periosteum serves as the site of attachment of bone to muscle.
• B. Cells of the periosteum differentiate into osteoblasts✓
• C. The periosteum is a fibrous sheath that surrounds long bones
• D. None of the above

Explanation: Osteoblast cells are not differentiated cells rather they themselves differentiate into osteocytes once they are surrounded by matrix. B is wrong since osteoblast cells are not differentiated periosteum cells.

Q37. The absorption and use of calcium are regulated by:

• A. Parathormone✓
• B. Adrenaline
• C. Thyroxin
• D. Thiamin
• E. Prolactin

Explanation: Circulating parathyroid hormone targets the distal convoluted tubule and collecting duct, directly increasing calcium reabsorption.

Q38. The most correct statement about muscle contraction is:

• A. Actin moves to shorten the muscle
• B. Cross bridges connecting the two molecule of myofibril is made up of G –Actin
• C. Contraction of myosin molecule results in muscle contraction
• D. Ca+ is necessary for binding of cross bridges✓

Explanation: In muscle contraction, calcium ions (Ca²⁺) play a crucial role in the interaction between myosin and actin, which are the proteins responsible for muscle contraction. When a muscle receives a signal to contract, calcium ions are released from the sarcoplasmic reticulum (a specialized type of endoplasmic reticulum) within the muscle cells. The presence of calcium ions allows the binding of myosin heads (cross bridges) to actin filaments. This binding forms cross-bridges between the thick myosin filaments and the thin actin filaments within the muscle fibers. The cross-bridges go through a series of biochemical reactions that result in muscle contraction. When the muscle contraction process is completed, calcium ions are actively transported back into the sarcoplasmic reticulum, and the cross-bridges detach, allowing the muscle to relax. So, calcium ions are indeed necessary for the binding of cross bridges and for the initiation of muscle contraction.

Q39. On a very cold day, a man waits for over an hour at the bus stop. Which of the following structures helps his body set and maintain a normal body temperature?

• A. Hypothalamus✓
• B. Kidneys
• C. Heart
• D. Brain stem

Explanation: Hypothalamus is known as the thermostat of the body. It is involved in homeostasis therefore it maintains our body temperature as well as osmoregulation. It is also involved in sleep-wake cycle, water balance.

Q40. Energy can be made available to the body in the following ways: (I) Conversion of surplus amino acids and glycerol to blood glucose and the mobilization of fat deposits which pass to the tissues for oxidation. (II) Breakdown of liver and muscle glycogen to form glucose. (III)Breakdown of tissue proteins to release amino acids which are then converted into glucose. In which order does the body draw on potential energy when it is being starved of food?

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

Explanation: Option C (II, I, and III) is correct because the conversion of surplus amino acids and glycerol to blood glucose and the mobilization of fat deposits (I) occurs after the breakdown of liver and muscle glycogen (II). Therefore, the correct order should be II, I and III.

Q41. The nucleus contains all of the following structures except:

• A. Mitochondria✓
• B. Chromatin
• C. Genes
• D. Nucleolus
• E. Nuclear membrane

Explanation: Nucleus contains nuclear envelope, nuclear lamina, nucleolus, chromosomes, nucleoplasm. Numerous Mitochondria are present in cell but not in nucleus.

Q42. Which of the following choices INCORRECTLY pairs of digestive enzyme with its site of secretion?

• A. Pancreatic amylase ... pancreas
• B. Aminopeptidase ... stomach✓
• C. Enterokinase ... intestinal glands
• D. Maltase ... intestinal glands

Explanation: Aminopeptidase is present in small intestine not in the stomach. Stomach contains the enzyme pepsin.

Q43. The division of biology that deals with classification is:

• A. Cytology
• B. Histology
• C. Botany
• D. Morphology
• E. Taxonomy✓

Explanation: Taxanomy is the division of Biology that deals with the classification by dividing organisms into groups and subgroups

Q44. All of the following are organelles except the:

• A. Endoplasmic reticulum
• B. Mitochondria
• C. Ribosome
• D. Golgi complex
• E. Ultracentrifuge✓

Explanation: Ultra-centrifuge is a machine used to carry out ultra-centrifugation which is used in cell separation techniques. It is not an organelle and therefore doesn’t carry out cellular functions.

Q45. The diagram shows a section through a kidney and associated blood vessels.In which area is there the greatest movement of fluid from the blood through the wall of blood vessels?

• A. A
• B. B
• C. C
• D. D
• E. E✓

Explanation: E is the renal cortex. It has the greatest blood flow in the following options. The cortex requires about 80% of blood flow to achieve its excretory and regulatory functions, and the outer medulla receives 15%. The inner medulla receives a small percent of blood flow; a higher flow would wash out solutes responsible for the high tonicity of the inner medulla.

Q46. Which function is not carried out by the mammalian kidney?

• A. Removal of bile pigments from the body✓
• B. Removal of excess mineral salts from the body
• C. Maintenance of a constant osmotic pressure of the blood
• D. Maintenance of a constant pH of the blood

Explanation: Bile is removed by the bacteria in Small intestine. Conjugated bilirubin is secreted into the bile canaliculus as part of bile and thus delivered to the small intestine. Bacteria in the intestinal lumen metabolize bilirubin to a series of other compounds which are ultimately eliminated either in feces or, after reabsorption, in urine. Thus, removal of bile is not mainly carried about the kidney. All the other processes are in fact carried out by the kidney.

Q47. According to the Hardy-Weinberg principle, the gene pool may remain stable if there are:

• A. Random mating✓
• B. Many mutations
• C. Frequent migrations
• D. Selected mating
• E. Random mutations

Explanation: Conditions for hardy-Weinberg equilibrium are as follows: No mutations (Crosses out E and B). No natural selection. Random Mating (Crosses out D). No immigration or emigration (Crosses out C).

Q48. Genes P, Q, R, and S occur on the same chromosome.Investigation of a large population produced the following cross-over values between pairs of genes.P and R 34% P and Q 59% R and S 12% S and Q 37% Which of the following sequences represents the sequence of genes on the chromosome?

• A. PRSQ
• B. PSRQ✓
• C. QSPR
• D. RQSP
• E. SPRQ

Explanation: This is a tricky question. A question like this also appeared in the AKU 2022 test. Basically, the greater the crossing over percentages, the more the distance between them. P and Q will be farthest from each other. So, we have P in the start and Q at the end. Now we have to find whether S comes first or R. PR frequency is more than PS so S will closer to P. Combining all this data, we get the sequence of PSRQ.

Q49. Which of the following hind of atom do not occur in carbohydrates?

• A. Carbon
• B. Hydrogen
• C. Nitrogen✓
• D. Oxygen

Explanation: Nitrogen is not present in Carbohydrates. It is present in proteins due to the presence of amino group (NH2).

Q50. The diagram is taken from an electron micro graph of a cell, name the organelle labeled D:

• A. Nucleus
• B. Lysosome
• C. Golgi complex
• D. Mitochondrion✓

Explanation: Part ‘D’ shows us a rod-shaped organelle. Mitochondria are rod shaped. Nucleus is usually in the center in animals and large so it can’t be the answer. Lysosomes are spherical microbodies that originate from Golgi apparatus so they can’t be D as well. Golgi complex is a network of cisternae and clearly not D. It is B. Mitochondria is the only correct answer for this.

Q51. Endoplasmic reticulum of the following organic compounds, the one that represents a protein is:

• A. C12 H22O11
• B. C6H12O6
• C. C17H14COOH
• D. (C6H10O5)11
• E. C108H130O180S4✓

Explanation: A belongs to Carbohydrates (Disaccharide).B also belongs to Carbohydrates (Monosaccharide).C belongs to Fatty Acids.D also belongs to carbohydrates.E is the only one here which belongs to Proteins. Some proteins have amino acids that have a Sulphur group attached to them. Since there is a Sulphur present in the formula, it gives an idea about a protein.

Q52. In which of the following organic compounds is a COOH (carboxyl) group found?(I) Carbohydrate (II) Lipid (III) Protein

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

Explanation: Lipids and proteins both have a carboxylic group.Following is a structure of amino acid which is a building block of protein. It contains an amino group as well as a carboxylic group.Second is the structure of a lipid which is made up of Fatty acids and a glycerol chain. The fatty acids contain a carboxylic group.

Q53. Part of the periodic table is shown. The letters are not the symbols of the elements.Which statement is correct?

• A. V is more reactive than Y
• B. W has more metallic character than V
• C. Y has a lower melting point than V✓
• D. Z is more reactive than X

Explanation: A. Reactivity for metals increases down the group so V is less reactive than Y which is below V.B. Metallic character decreases as we go from left to right so W has less metallic character than V.C. Melting point for Group I-IV increases across the period and decreases down the group. Y being below V will have a lower melting point. Hence Option C is correct.D. Reactivity decreases down the group for halogens and nonmetals.

Q54. The bar chart shows the period of elements from lithium to neon.Which property of these elements is shown on the chart?

• A. The number of electrons used in bonding✓
• B. The number of orbits holding electrons
• C. The (proton) atomic number
• D. The relative atomic number

Explanation: A. Lithium uses 1 electron in bonding, Beryllium 2, Boron 3, Carbon 4, Nitrogen 3, Oxygen 2, Fluorine 1 and Ne being a noble gas doesn’t make any bonding. The graph clearly indicates the number of electrons used in bonding therefore Answer is A.B. The orbit is the same so it doesn’t matter therefore it’s wrong.C. Proton number increases to the right therefore Option C doesn’t give the accurate information about the graph.D. Atomic Number increases to the right therefore Option D doesn’t give the accurate information about the graph.

Q55. a- helices are secondary structures characterized by:

• A. Intermolecular hydrogen bonds✓
• B. Disulfide bonds
• C. A rippled effect
• D. Hydrophobic Interactions

Explanation: A. a-helix and beta pleated sheets have hydrogen bonding so Option A is correct.B. Disulfide bonds is present in tertiary structure of proteins.C. Rippled effect has nothing to do with the structure of proteins.D. Hydrophobic interactions are present in quaternary structure of proteins.

Q56. A metal sulfate contains 9.87% of M. This sulfate is isomorphous with ZnSO4,.7H2O. Determine the atomic weight of the metal M.

• A. 24.31✓
• B. 34.31
• C. 25.25
• D. 44.41
• E. 30.75

Explanation: Let x be the mass of MMass of S = 32Mass of 4O = 64Mass of 7H2O= 126This sulfate is isomorphous with ZnSO4.7H2O means, its structure is MSO4.7H2O.Thus, molecular mass of MSO4.7H2O= x + 32 + 64 + 126=222 + x- It is given that 9.87% of total weight = x.- (9.87/100)* (222 + x) = x- 21.91 +0.0987x = x- 21.91= 0.9013x- x=24.3 grams- So the metal is magnesium and the molecular formula is MgSO4. 7H2O Numerical could have only one answer.

Q57. 100 c.c of oxygen is collected over water at 23°C and 800mm pressure. If vapour pressure of water vapours at 23°C is 21.00mm, then calculate volume of the gas at NTP.

• A. 100 c.c
• B. 94.53 c.c✓
• C. 150 c.c
• D. 90.0 c.c

Explanation: V1 ​= 100 c.cV2 ​= ?P1​ = 800mmP2​ = 760mmT1​ = 23∘C = 23+273 = 296K T2​ = 273K- Using formula:- P1V1/T1 = P2V2 /T2- 800 x 100/ 296 = 760 x V2/ 273- Cross Multiply the following- 800 x 100 x 273/ 296 x 760 = 97 cc - The answer is way closer to B then A so the answer will be B.

Q58. Y g of the non-volatile substance (molecular mass M) is dissolved in 250g of benzene. If Kb is the molal elevation constant, the value of AT us given by:

• A. 4M/Kb
• B. 4Kb.Y/M✓
• C. Kb.Y/4M
• D. Kb Y/M

Explanation: Elevation in its boiling point is given byΔT=1000 × Kb ​× Y​/250 × M = 4KbY/MHence, the correct option is B

Q59. A mixture of ethanol and propanol has a vapour pressure of 290mm at 27°C. If mole fraction of ethanol is 0.65, then what will be its vapour pressure if vapour pressure of pure propanol is 210mm?

• A. 333.1 mm✓
• B. 441.5 mm
• C. 890.2 mm
• D. None of the above

Explanation: By Roult's law, Ptotal ​= PA∘XA ​+ PB∘XB ​290 = PA∘​(0.6) + (210)(1−0.65) 290 = PA∘​(0.6) + 73.5 290 - 73.5 = PA∘​(0.6) 216.5 = 0.6 PA∘​ 216.5 ÷ 0.65 = PA∘ PA∘​ = 333.0 mm

Q60. At 25°C, the equilibirum constant K1 and K2 in the two reactions are 2NH3⇌N2 + 3H2 and 1/2 N2 +3/2H2=NH3 respectively. K1 and K2 are related as:

• A. Kl – K2
• B. Kl = 1/ K22✓
• C. K2 = K1
• D. K2= 1/Kl

Explanation: Option B is correct. 2NH3 = N2+3H2 ; K1 Reverse the reaction i.e. 3H2+N2 = 2NH3 ; 1/K1 Multiply above reaction by 1/2 Required reaction is 1/2N2+3/3H2 = NH ; (1/K1)½ Therefore, K2 = (1/K1)½

Q61. For the reaction N2 + O2 2NO, the value of K is 0.0842 at 3500 K. Calculate the fraction of equilibrium mixture of N2 and O2 converted into NO.

• A. 15.0%
• B. 16.3%
• C. 16.5%
• D. 16.9%✓

Explanation: Let's assume that x represents the fraction of N2 and O2 converted into NO. Therefore, at equilibrium, the concentrations can be expressed as:[NO] = 2x, [N2] = (1 - x), [O2] = (1 - x)Substituting these values into the equilibrium constant expression:K = (2x)^2 / ((1 - x) * (1 - x))Simplifying the equation:0.0842 = 4x^2 / (1 - 2x + x^2) Rearranging the equation:0.0842(1 - 2x + x^2) = 4x^20.0842 - 0.1684x + 0.0842x^2 = 4x^20.0842x^2 + 4x^2 - 0.1684x - 0.0842 = 0Combining like terms:4.0842x ^2 - 0.1684x - 0.0842 = 0This is a quadratic equation. Solving this equation will give us the values of x, representing the fraction of N2 and O2 converted into NO.Using a quadratic solver, we find that x ≈ 0.169, which is approximately 16.9%.Therefore, the fraction of the equilibrium mixture of N2 and O2 converted into NO is approximately 16.9%.

Q62. Chlorine is vinyl chloride is not very reactive because:

• A. C - Cl bond developed partial double bond
• B. Of resonance
• C. Sp2 hybridized carbon has more acidic character than SP3 hybridized carbon
• D. All of the above✓

Explanation: The non-reactivity of chlorine atom in vinyl chloride is due to resonance stabilization. Carbon-chlorine bond in vinyl chloride has some double bond character and is, therefore, stronger than a pure single bond. Carbon atom is sp2 hybridized and so C-Cl bond is stronger than in alkyl halide. So all of the options are correct.

Q63. One of the isomers of C6H12(A) has chiral carbon but on hydrogenation it losses chirality. A is:

• A. 2-Methyl-2-pentene
• B. 2,3-Dimethyl-2-butene
• C. 3-methyl-1-pentene✓
• D. 3,3-Dimethyl 1-butene

Explanation: The compound C is 3-methyl-1-pentene. It is shown in the option C. It has molecular formula C6​H12​ and is optically active. On catalytic hydrogenation, it gives 3-methylpentane having molecular formula C6​H14​ which is optically inactive.

Q64. An equimolar quantities of ethanol and methanol are heated with conc.H2SO4. The product formed is?

• A. CH3OCH3
• B. C2H5OCH3
• C. C2H5OC2H5
• D. All of the above✓

Explanation: Option D is correct as they represent the products that can be formed when equimolar quantities of ethanol and methanol are heated with concentrated sulfuric acid.

Q65. Which one of the following phenol is more soluble in ag.NaHC03?

• A. 2, 4-Dihydroxy acetophenone
• B. 2, 4, 6-Tricyano phenol✓
• C. 3, 4-Dicyano phenol
• D. p-cyano phenol

Explanation: Since there are 3 cyano-groups attached to the phenol in Option B compared with C and D, it will be more soluble in NaHCO3. Electron donating groups increase the acidity of the phenol. A phenol with 3 cyano groups will be more acidic than a phenol with two OH- groups. Therefore, B is the most acidic and hence the most soluble in base.

Q66. A water soluble compound of molecular formula C3 H6O gives yellow crystalline solid on heating with I2, and Na2CO3. The compound is:

• A. CH3CH2CHO
• B. CH20CH = CH2
• C. CH3COCH3✓
• D. CH2 = CHCH2OH

Explanation: Option C (CH3COCH3, acetone) is correct because it is a water-soluble compound with the molecular formula C3H6O. When heated with I2 and Na2CO3, acetone undergoes the haloform reaction, producing a yellow crystalline solid known as iodoform (CHI3), indicating the presence of a methyl ketone (CH3CO) group in the original compound.

Q67. Arrange the following compounds in order of increasing reactivity towards the addition of HCN.Acetone (I), acetaldehyde (II), methyl t-butyl ketone (III), di-t-butvl ketone (IV):

• A. IV <I <II <III
• B. I < II < III < IV
• C. IV < III < I < II✓
• D. II< I < IV < III

Explanation: The more the bulky groups attached to the carbonyl atom, the less the reactivity becomes.- Acetone has two bulky groups attached it.- Acetaldehyde has one bulky group attached to it.- Methyl t-butyl ketone has a tertiary butyl and methyl group attached to it which make it less reactive.- Di-t-butyl ketone has two tertiary butyl groups attached to it which make it the least reactive out of all the options.So the correct order will be IV<III<I<II

Q68. The reaction CH3COOAg + Br2 CH3Br + CO2 + AgBr is known as:

• A. Reformat sky reaction
• B. Huns dicker reaction✓
• C. Decarboxylation
• D. Hell-Volhard-Zelinsky reaction

Explanation: Therefore, the correct option must be B (Hunsdicker reaction) as it accurately describes the reaction where a silver carboxylate (CH3COOAg) reacts with bromine (Br2) to form methyl bromide (CH3Br), carbon dioxide (CO2), and silver bromide (AgBr).

Q69. Butyrolactone (ester) does not react with:

• A. NH3
• B. UAIH4
• C. EtOH.H
• D. NaBH4 / EtOH✓

Explanation: The correct option is D (NaBH4/EtOH), as sodium borohydride (NaBH4) in ethanol (EtOH) is not a strong enough reducing agent to reduce esters. It is typically used for the reduction of aldehydes and ketones but does not typically react with esters.

Q70. Electric cookers have coating of _ that protects then against fire.

• A. Heavy lead
• B. Magnesium oxide✓
• C. Zine oxide
• D. Sodium sulphate

Explanation: The electric cookers have coating that protects them against fire. The coating is made up of MgO. It has high melting point and does not catch fire and hence protects the cooker against fire.

Q71. Macromolecules are of _ types.

• A. Three
• B. Four✓
• C. Five
• D. Six
• E. Seven

Explanation: The macromolecules are of four types:1. Carbohydrates2. Proteins3. Lipids4. Nucleic Acids

Q72. The long chains of Amino Acids are called:

• A. Oils
• B. Polypeptides✓
• C. Monopeptides
• D. Proteins

Explanation: Polypeptides are the long chains/polymerized chains of amino acids. Option A and C are clearly wrong. One could argue about option D but proteins are chains of polypeptides. Option B is the most suitable answer for this.

Q73. The general formula for Carbohydrates is:

• A. Nn(H2O)n
• B. Pn(H2O)n
• C. Cn(H2O)n✓
• D. Hn(H2O)n
• E. Hn(C2O)n

Explanation: 1. Option A has Nitrogen which is not present in carbohydrates2. Option B has Phosphorous which is not present in carbohydrates3. Option C is right (factual statement)4. Option D is wrong as carbohydrates contain carbon and oxygen in addition to hydrogen.5. Option E is wrong because it is the formula of hydrocarbons.

Q74. Lipids are generally defined in terms of:

• A. Solubility✓
• B. Structure
• C. Molarity
• D. All of the above

Explanation: Lipids are defined by Solubility and Intermolecular Forces. lipids are a class of naturally occurring molecules that are soluble in nonpolar organic solvents, and are not soluble in water.

Q75. As a result of increased CO2 in the atmosphere, oceans will become more:

• A. Alkaline
• B. Acidic✓
• C. Saline
• D. Cooler

Explanation: Because of human-driven increased levels of carbon dioxide in the atmosphere, there is more CO2 dissolving into the ocean. The ocean's average pH is now around 8.1 , which is basic (or alkaline), but as the ocean continues to absorb more CO2, the pH decreases and the ocean becomes more acidic.

Q76. Infrared lamps are used in restaurants and cafeterias to keep food warm. The infrared radiation is strongly absorbed by water, raising its temperature and that of the food. If the wavelength of infrared radiation is assumed to be 1500nm, then the number of photons per second of infrared radiation produced by an infrared lamp that consumes energy of the rate of 100 watt and is 12% efficient will be:

• A. 4 x 1010
• B. 9 x 1019✓
• C. 11 x 1012
• D. 15 x 104

Explanation: To calculate the number of photons per second of infrared radiation produced by the infrared lamp, we can use the energy and efficiency of the lamp. The energy emitted per second is equal to the power, which is 100 joules/second. Effective energy emitted per second = 0.12 * 100 joules/second = 12 joules/second The energy of a single photon is given by Planck's equation: E = hc/λ Where E is the energy of a photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of the radiation (1500 nm = 1.5 x 10^-6 m). E = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (1.5 x 10^-6 m) ≈ 4.417 x 10^-19 J photons/second Rounded to the nearest whole number, the number of photons per second of infrared radiation produced by the infrared lamp is approximately 2.7 x 10^19 photons/second, which matches the given answer of 9 x 10^19 after rounding.

Q77. When is formed from bond order _ and when is formed from O2 bond order _ .

• A. Increases ... increases
• B. Decreases ... decreases
• C. Increases ... decreases
• D. Decreases ... increases✓

Explanation: Bond order in is 8-2/2 = 3 and bond order in is 7-2/2 = 2.5. Thus, conversion of N2 to decreases the bond order from 3 to 2.5- Bond order of O2 is 8-4/2 = 2 and the bond order of is 8-3/2 = 2.5 so the bond order increases from 2 to 2.5 as O2 is converted to .Hence Option D is correct.

Q78. The process requiring absorption of energy is:

• A. F/F-
• B. Cl/Cl-
• C. O/O2-✓
• D. H/H-

Explanation: - Option A,B and D all examples of exothermic reactions. They release first electron affinity.- Oxygen undergoes two transitions. First it gains one electron which is exothermic. If we want to add one more electron after the addition of one electron then the incoming electron will face repulsions from the electrons in valence shells. So we need to provide some energy if we want to add one more electron to M− ion. Hence the second electron affinity of an element is endothermic.

Q79. A-solution of 500m1 of 0.2M KOH and 500m! of 0.2M HCI is mixed and stirred, the rise in temperature is T-j. The experiment is repeated using 250ml of each of the, solution; the temperature rise is T2. Which of the following is true?

• A. Tl = T2✓
• B. Tl = 2T2
• C. Tl =4T2
• D. T2 = 9Tl

Explanation: The correct option must be A (T1 = T2) because the rise in temperature (T1) is directly proportional to the amount of heat generated, which is determined by the moles of reactants. Since the concentrations and volumes of the solutions are the same in both cases, the same amount of heat is generated, resulting in the same temperature rise. Therefore, T1 is equal to T2.

Q80. An aqueous solution of Ti(H20)6l3+l appears:

• A. Greenish yellow in colour
• B. Blue in colour
• C. Violet in colour
• D. Purple in colour✓

Explanation: In [Ti(H2​O)6​]3+by absorbing green yellow component of white light d-d transition takes place and the aqueous solution appears purple as the purple is the complimentary colour of greenish- yellow.

Q81. Amongst the following ions, which has the highest Para magnetism?

• A. [Cr(H2O)6]2+
• B. [Fe(H2O)6]2+✓
• C. [Cu(H2O)6]2+
• D. [Zn(H2O)6]2+

Explanation: [Fe(H2​O)6​]2+has the maximum number of unpaired electrons; therefore, the highest degree of paramagnetism.

Q82. Zn(S) + Cu2(aq) -> Cu(s)+ Zn2 (aq). At 300 K, Ecell is.1.10V, and at 310 K, Ecell is 1.12 V. What is the entropy change (AS) for the above cell reaction?

• A. 386J K-l✓
• B. 486J K’l
• C. 286J K”l
• D. 586J K-l

Explanation: The correct option is A.n=2- Initial temperature "T1" =300 K- Final temperature "T2" = 310 K- E1 = 1.10 V- E2 = 1.12- VΔS = - nF . dE / dTΔS = -2 x 96500 x ( 1.12 – 1.10) / 310 – 300- ΔS = -2 x 96500 x 0.02 / 10- ΔS = -2 x 96500 x 0.002- ΔS = -386J K^-1- Thus the entropy change will be ΔS = -386JK-1

Q83. For a gaseous reaction, A2 + 2B 2AB, the following rate data obtained at 250K.Calculate the rate of formation of AB when [A2] = 0.02 M and [B] = 0.01 M at 250K.

• A. 4.8 x 10-5 mole I-l s-l
• B. 4.8 x 10-6 mole I-l s-l✓
• C. 5.8 x 10-6 mole I-l s-l
• D. 5.8 x 10-5 mole I-l s-l

Explanation: When the concentration of A doubles, the rate of reaction of reaction also doubles. Hence the rate of reaction is first order with respect to A. When concentration of B becomes four times, the rate of reaction also becomes four times. So the rate of reaction is first order with respect to B.Rate = K [A2] [B]

Q84. Which of the following conditions, listed as leaving group and nucleophile, respectively, would most favour an Sn2 reaction?

• A. I- , Cl-
• B. EtO-. tosytate
• C. Tosylate. CN-✓
• D. OH-. H2O

Explanation: CN− is a strong nucleophile as well as a leaving group. We expect it to take part in SN2 reactions. It is also a weak base, so we do not expect either E2 or E1 eliminations. Tosylate is also a weak base but a strong nucleophile so they will favor SN2 reaction. OH- is a strong base, it will favor elimination. Cl- is a weak nucleophile.

Q85. If 2-amino-3-methylbutane were treated with excess methyl iodide, silver oxide, and water, what would be the major reaction products?

• A. Ammonia and 2-methyl-2-butene
• B. Trimethylamine and 3-methyl-1-butene✓
• C. Trimethyl amine and 2-methyl-2-butene
• D. Ammonia and 3-methyl-l-butene

Explanation: The process in which an amine is converted to a quaternary ammonium iodide salt by reacting with excess methyl iodide, and treatment with silver oxide and water converts this to the ammonium hydroxide which, when heated, undergoes elimination to form an alkene and an amine is called Hoffman Elimination process. (This Hofmann elimination forms trisubstituted amine and less substituted alkene). Ammonia is not formed so Option A and D can be eliminated. A less substituted alkene is formed so Option B is right.

Q86. If an amino acid (pl = 9.74) in acidic solution is completely titrated with sodium hydroxide, what will be its charge at pH 3, 7, and 11 respectively?

• A. Positive, neutral, negative
• B. Negative, neutral, positive
• C. Neutral, Positive, positive
• D. Positive, positive, negative✓

Explanation: The correct option is D (Positive, positive, negative) because the charge of the amino acid will be positive at pH 3 due to the protonation of the amino group, positive at pH 7 due to the protonation of the amino group and the deprotonation of the carboxyl group, and negative at pH 11 due to the deprotonation of the carboxyl group.

Q87. Amino acids with nonpolar R-groups have which of the following characteristics in aqueous solution?

• A. They are hydrophilic and found buried within proteins
• B. They are hydrophobic and found buried within proteins✓
• C. They are hydrophobic and found on proteins surfaces
• D. They are hydrophilic and found on protein surface

Explanation: Since the amino acids contain a non-polar R group therefore, they are not soluble in water as water is polar. We can then say it is hydrophobic. Hydrophobic amino acids are found buried within proteins and are not found on the surfaces since surface proteins are not hydrophobic.

Q88. The time period of a pendulum is measured to be 3.0 seconds in the inertial reference frame of the pendulum. What is its period measured by an observer moving at a speed of 0.95c with respect to the pendulum?

• A. 1.2 s
• B. 3.4 s
• C. 8.1 s
• D. 9.6 s✓

Explanation: The concept of time dilation is used for this question.

Q89. What is the mass "m" of a moving object with speed 0.8 c.

• A. 1.67 m0✓
• B. 3.67 m0
• C. 4.67 m0
• D. 6.67 m0
• E. 7.67 m0

Explanation: where v is the speed relative to the frame in which the rest mass is measured. = 1/0.6 = 1.6mo

Q90. The emission of electrons from a metal surface when exposed to light of suitable frequency is known as:

• A. Compton’s effect
• B. Photoelectric effect✓
• C. Coulomb’s effect
• D. Faraday’s law
• E. Ohm’s law

Explanation: The photoelectric effect is a phenomenon in which electrons are ejected from the surface of a metal when light is incident on it. These ejected electrons are called photoelectrons. The correct option is B.

Q91. The special theory of relativity is based upon:(I) the laws of physics are the same in all inertial frames.(II) The speed of light in free space has the same value for all observers regardless of their state of motion.

• A. I only
• B. II only
• C. I and II✓
• D. None of the above

Explanation: 1. I is the first postulate of newton’s theory of relativity which states that laws of physics are only same in inertial frames so I is correct2. II is the second postulate of Newton’s theory of relativity which states that speed of light has a constant value for all observers regardless of their motion. Hence II is correct. Both of these options are right. C is right.

Q92. What is the energy of a photon in a beam of infrared radiation of wavelength 1240nm?

• A. 1.0 eV✓
• B. 3.0 eV
• C. 5.0 eV
• D. 7.0 eV

Explanation: Photon velocity equals the speed of light. Photons are massless, but they have energy E = hf = hc/λ. Here h = 6.626*10-34 Js is called Planck's constant. The photon energy is inversely proportional to the wavelength of the electromagnetic wave.E=6.626×10−34 × 3×108​/1240×10-9 ≈ 1.00 eV.

Q93. A nucleus consists of nucleons comprising of protons and neutrons. A proton has a positive charge equal to _ and has a mass of _.

• A. 2.6 x 10 19C;36. x 10 28 kg
• B. 1.6.x 10 19C;1.673 x 1027kg✓
• C. 3.6 x 10 19C ;2.111 x 1027kg
• D. 4.6 x 10 19 C;9.111 x 1027kg
• E. 5.6 x 10 19C ;8.111 x 1027kg

Explanation: Therefore, the given answer of 1.6 x 10 19 C for the charge of a proton and 1.673 x 1027 kg for the mass of a proton is correct. Protons are positively charged particles and contribute to the atomic nucleus along with neutrons.

Q94. Identify the isotope/s of Helium.

• A. 3He2
• B. 4He2
• C. 6He2
• D. 5He2
• E. Both A and B✓

Explanation: There are two isotopes of helium: 3He2, which is a primordial isotope that was incorporated in Earth as it accreted, and 4He2, an isotope produced by radioactive decay of U and Th isotopes.

Q95. The half-life T1/2, of a radioactive element, is that period in which _ of the atoms decay.

• A. Nine times
• B. Double
• C. Half✓
• D. Four times
• E. Seven times

Explanation: The correct option is C (Half) because the half-life represents the period in which half of the radioactive atoms in a sample decay. For example, if you start with 100 atoms, after one half-life, there will be 50 atoms remaining.

Q96. Fluorescence is the property of absorbing radiant energy of _frequency and re-emitting energy of _ frequency in the visible region of the electromagnetic spectrum.

• A. Low; high
• B. High; low✓
• C. Low; low
• D. High; high

Explanation: The correct option is B (High frequency...low frequency) because fluorescence involves the absorption of energy of high frequency (short wavelength) and the re-emission of energy of lower frequency (longer wavelength) in the visible region of the electromagnetic spectrum. This shift in frequency results in the phenomenon of fluorescence.

Q97. A reaction in which a heavy nucleus like that of uranium splits up into two nuclei of roughly equal size along with the emission of energy during the reaction is called a:

• A. Fission reaction✓
• B. Fusion reaction
• C. Counter reaction
• D. Chemical reaction

Explanation: Nuclear fission is a reaction in which the nucleus of an atom splits into two or more smaller nuclei. The fission process often produces gamma photons, and releases a very large amount of energy even by the energetic standards of radioactive decay.So the correct option is A.

Q98. Identify the main type (s) of nuclear reactors:

• A. Slow reactors
• B. Fast reactors
• C. Thermal reactors
• D. Both A and B✓

Explanation: Nuclear reactors are either slow reactors or fast reactors. The correct answer is therefore D. Although thermal reactor is a nuclear reactor but they asked us the MAIN types of nuclear reactors. It is itself a type of slow nuclear reactor.

Q99. What is the average translational kinetic energy of molecules in a gas at temperature 27°C?

• A. 3.23 x 10-21 J
• B. 4.11 x 10-21 J
• C. 6.21 × 10-21 J✓
• D. 7.71. x 107-21 J
• E. 9.11 x 107-21 J

Explanation: To calculate the average translational kinetic energy of molecules in a gas at a given temperature, we can use the equation:K.E. = (3/2) kTWhere K.E. is the average translational kinetic energy, k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K), and T is the temperature in Kelvin.Given that the temperature is 27°C, we need to convert it to Kelvin by adding 273.15:T = 27°C + 273.15 = 300.15 KSubstituting the values into the equation:K.E. = (3/2) * (1.38 x 10^-23 J/K) * (300.15 K)= 6.21 x 10-21 JTherefore, the average translational kinetic energy of molecules in the gas at a temperature of 27°C is approximately 6.21 x 10-21 J.

Q100. Numbers are expressed in standard form called scientific notation, which employs powers of:

• A. 2
• B. 8
• C. 10✓
• D. 16

Explanation: A number is expressed in scientific notation when it is written as the product of a factor and a power of 10. Examples: Scientific Notation of 2,700,000,000 is 2.7 × 10⁹.

Q101. A 1500 kg vehicle has its velocity reduced from 20 m/s to 15 m/s in 3.0 seconds. How large was the average retarding force?

• A. -0.5 N
• B. -1.5 N
• C. -2.0 N
• D. -2.5 N✓
• E. -3.5 N

Explanation: Formula *Vf=vi+ at, *F=mavf=15m/s ,vi= 20m/s ,t=30secs, m=1500kgRequired: a=? F=?Solution:- vf= vi+at- (15) = (20)+ a(30)- (15)-(20) = a(30)- -5= a(30)- a= -5/30- a=-1.66m/s2- F= (1500) × (-1.66)- F= -250 NThe FMDC had wrong options in the question but 250 can be compared with 2.5 in the question even though it is also wrong.

Q102. An object moving through a fluid experiences a retarding force known as drag force. The drag force as the speed of the object:

• A. Decreases ... Decreases
• B. Decreases ... Increases
• C. Increases ... Decreases
• D. Increases ... Increases✓

Explanation: For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes' law, Fs=6πηrv F s = 6 π η r v , where r is the radius of the object, η is the fluid viscosity, and v is the object's velocity.Drag force is directly proportional to the velocity of the object. So, when the velocity of the object increases, the drag force also increases. A is not always the case for drag force.

Q103. The property of bending of light around obstacles and spreading of light wanes into a geometrical shadow of an obstacle is known as:

• A. Diffraction✓
• B. Interference
• C. Polarization
• D. Optical Rotation

Explanation: Diffraction is the process of light bending around an obstacle or spreading out after it moves through a small space. If you study physics, you'll learn about the diffraction of light waves. Any kind of wave can experience diffraction, including sound, radio, and water.

Q104. The distance between the slits in Young's double-slit experiment is 0.25cm. Interference fringes are formed on a screen placed at a distance of 100cm from the slits. The distance of the third dark fringe from the central bright fringe is 0.059cm. What is the wavelength of the incident light?

• A. 90 nm
• B. 590 nm✓
• C. 690 nm
• D. 790 nm
• E. 990 nm

Explanation: Given:Distance between the slits (d) = 0.25 cm = 0.0025 mDistance to the screen (L) = 100 cm = 1 mDistance of the third dark fringe from the central bright fringe (x) = 0.059 cm = 0.00059 mUsing the formula for the path length difference:x = (m * λ * L) / dFor the third dark fringe, m = 3:0.00059 = (3 * λ * 1) / 0.0025Simplifying the equation:λ = (0.00059 * 0.0025) / 3 Calculating the value:λ = 0.0000004925 m = 492.5 nm = 492.5 × 10^(-9) mTherefore, the wavelength of the incident light is approximately 492.5 nm, which is equal to 590 nm (rounded to the nearest whole number).

Q105. The ratio of the angles subtended by the image as seen through the optical device to that subtended by the object at the unaided eye is known as:

• A. Magnifying power
• B. Angular magnification
• C. Simple magnification
• D. Both A and B✓

Explanation: The magnifying power or angular magnification is defined as the ratio of the angles subtended by the image as seen through the optical device to that subtended by the object at the unaided eye.

Q106. Those waves in which the particles of medium have displacement along the direction of propagations of waves are known as:

• A. Longitudinal waves✓
• B. Transverse waves
• C. Simple waves
• D. Electromagnetic Waves

Explanation: Longitudinal waves are waves in which the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. So, option A is correct.

Q107. A pipe has a length of Im. Determine the frequencies of the fundamental and first two harmonic if the pipe is opened at both ends, (speed of sound in air = 340 m/s).

• A. 170 Hz, 340 Hz, 510 Hz✓
• B. 120 Hz, 220 Hz, 390 Hz
• C. 90 Hz, 230 Hz, 440 Hz
• D. 210 Hz, 410 Hz, 510 Hz

Explanation: Using formula v=f λ- F= v/2l- F= 340/2(1)- F= 170Since it is an open-organ pipe- F1 = 170 Hz- F2= 2 × 170 = 340 Hz- F3= 3 ×170 = 510 Hz

Q108. T points where the displacements of two waves cancel each other's effect. the path difference is an odd integral multiple of half the wavelength. This effect is known as:

• A. Constructive interference
• B. Destructive interference✓
• C. Stationary interference
• D. Simple interference

Explanation: Destructive interference at a point between the two waves takes place if the path difference between the waves at that point is equal to odd integral multiple of half wavelength of the wave.So, path difference Δx=(n+1/2)λWhere n must be an integer.

Q109. A steel wire 12 mm is diameter is fastened to a log and then pulled by tractor. The length of steel wire between the log and the tractor is 11m. A force of 10,000 N is required to pull the log. What is the stress in the wire?

• A. 33.12 MPa
• B. 44.12 MPa
• C. 66.15 MPa
• D. 77.29 MPa
• E. 88.46 MPa✓

Explanation: Stress = Force / AreaGiven:Diameter of the steel wire (d) = 12 mm = 0.012 mLength of the wire (L) = 11 mForce required to pull the log (F) = 10,000 NFirst, we need to calculate the cross-sectional area of the wire using the diameter:Area = π * (d/2)^2Substituting the values:Area = π * (0.012/2)^2 = π * 0.006^2 = 0.000113097 m^2Now, we can calculate the stress:Stress = Force / Area = 10,000 N / 0.000113097 m^2 = 88,457,723.95 PaConverting Pa to MPa (megapascals):Stress = 88,457,723.95 Pa = 88.46 MPaTherefore, the stress in the steel wire is approximately 88.46 MPa.

Q110. Identify the postulates which help to formulate a mathematical model of gases.(I) A finite volume of gas consists of a very large number of molecules.(Il) The size of the molecules is much smaller than the separation between molecules.(III) Molecules do not exert force on each other except during a collision.

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

Explanation: All of these statements are correct about the kinetic molecular theory of gases which tells us about the mathematical model of gas.1. It is correct as volume of a gas contains large number of gas molecules2. Size of gas molecules is very small compared to that of the container and the spaces between molecules so it is only correct3. Molecules undergo elastic collision and exert force on each other.

Q111. The turbine in a steam power takes steam from a boiler at 427°C and exhausts into a low-temperature reservoir at 77°C. What is the maximum possible efficiency?

• A. 10%
• B. 20%
• C. 30%
• D. 50%✓
• E. 70%

Explanation: The maximum theoretical efficiency of the system is the one corresponding to the efficiency of a Carnot cycle operating between the same temperatures of the system:η=1−Tc​/ThWhere Tc and Th are the cold and hot temperatures, respectively.In our problem, Tc​=350K and Th​=700KTherefore, the maximum theoretical efficiency isη=1−350/700​ = 0.50 × 100 = 50%

Q112. If a process cannot be retraced in the backward direction by reversing the controlling factors. It is a/an:

• A. Reversible process
• B. Irreversible process✓
• C. Efficient process
• D. Entropic process

Explanation: A reversible process can be retraced in reverse order, without producing any change in the surrounding. A reversible cycle is a succession of events which bring the system back to its initial condition is called a cycle. A reversible cycle is the one in which all the changes are reversible. However, if a process cannot be retraced in the backward direction by reversing the controlling factors, it is an irreversible process. For example, convection, radiation etc.

Q113. A digital system deals with quantities or variables which have only two discrete values or states. Identify the example/s of such quantities.(1) A switch can be either open closed.(11) A bulb can be either off or on.(III) A certain statement can be either true of false.

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

Explanation: All the options are based on two discrete values of 1 and 0. Our switches, bulbs work on logic gates which correspond to the discrete values. A certain statement can be either true (1) or false (0) as well.

Q114. The potential difference between the terminals of a battery in an open circuit is 2.2V. When it is connected across a resistance of 5.0 ohms, the potential falls to 1.8V. What is the current of the battery?

• A. 0.36 A✓
• B. 2.36 A
• C. 3.39 A
• D. 4.49 A
• E. 2.71 A

Explanation: In open circuit, E = V = 2.2 V, In close circuit, V = 1.8 V, R=5ΩSo internal resistance, r=(E/V−1) R =(2.2/1.8−1) × 5 ⇒ r=10/9= 1.1ΩNow V= I(R+r)V/R+r = I2.2/ 5+1.1 = 2.2/6.1 = 0.36A

Q115. A platinum wire has a resistance of 10 Ω at 0°C and 20 Ω at 273°C. What is the value of the temperature coefficient of resistance of platinum?

• A. 3.66 x 10-3 K-1✓
• B. 4.66 x 10-3 K-1
• C. 6.31 x 10-3 K-1
• D. 7.42 x 10-3 K-1
• E. 9.49 x 10-3 K-1

Explanation: Ro=10Ω- Rt=20Ω- t=273∘C- Ro=Rt (1+αt)- α=Rt−Ro/Rot- α=20−10/10 × 273- α= 10/2730- α=3.66 x 10-3 K-1

Q116. What shunt resistance must be connected across a galvanometer of 50.0Q resistance Which gives full-scale deflection with 2.0 mA current, to convert it into an ammeter of range 10.0 A?

• A. 105 Q
• B. 0.01 Q✓
• C. 2.5 Q
• D. 3.9 Q
• E. 4.7 Q

Explanation: To convert the galvanometer into an ammeter of range 10.0 A, a shunt resistance (Rs) must be connected in parallel with the galvanometer. The formula to calculate the required shunt resistance is:Rs = (G - 1) * RgGiven:Galvanometer resistance (Rg) = 50.0 ΩDesired full-scale deflection current = 10.0 AGalvanometer full-scale deflection current = 2.0 mA = 0.002 ACalculating G:G = 10.0 A / 0.002 A = 5000Calculating the shunt resistance (Rs):Rs = (5000 - 1) * 50.0 ΩRs = 4999 * 50.0 ΩRs = 249,950 Ω = 0.01 Ω (rounded to two decimal places)Therefore, the shunt resistance that must be connected across the galvanometer is approximately 0.01 Ω.

Q117. Find the radius of an orbit of an electron moving at a rate of 2.0 x 107 m/s in a uniform magnetic field of 1.20 x 10-3 T.

• A. 3.15 m
• B. 4.25 m
• C. 9.93 m✓
• D. 17.77 m
• E. 19.91 m

Explanation: r = (m * v) / (q * B) Given:m = 9.109 x 10^-31 kgv = 2.0 x 10^7 m/s q = -1.6 x 10^-19 C B = 1.20 x 10^-3 T Now, let's calculate the radius (r):r = (9.109 x 10^-31 kg * 2.0 x 10^7 m/s) / (1.6 x 10^-19 C * 1.20 x 10^-3 T)r ≈ 9.93 mThe correct option is C. 9.93 m.

Q118. The emission of electrons from a metal surface when exposed to light of suitable frequency is known as:

• A. Compton's effect
• B. Photoelectric effect✓
• C. Coulomb's law
• D. Faraday's law
• E. Ohm's law

Explanation: Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.

Q119. The special theory of relativity is based upon:(I) The laws of physics are the same in all inertial frames.(II) The speed of light in free space has the same value for all observers regardless of their state of motion.

• A. I only
• B. II only
• C. I and II✓
• D. None of the above

Explanation: 1. I is the first postulate of newton’s theory of relativity which states that laws of physics are only same in inertial frames so I is correct.2. II is the second postulate of Newton’s theory of relativity which states that speed of light has a constant value for all observers regardless of their motion. Hence II is correct.Both of these options are right. C is right.

Q120. A nucleus consists of nucleons comprising of protons and neutrons. A proton has a positive charge equal to _ and has a mass of _.

• A. 2.6 x 10-19 C; 36. x 10-28 kg
• B. 1.6.x 10-19C; 1.673 x 10-27kg✓
• C. 3.6 x 10-19C; 2.111 x 10-27kg
• D. 4.6 x 10-19 C ; 9.111 x 10-27kg
• E. 5.6 x 10-19C; 8.111 x 10-27kg

Explanation: The charge of a proton is indeed 1.6 x 1019 Coulombs (C). This value represents the fundamental unit of positive charge carried by a proton. The mass of a proton is approximately 1.673 x 1027 kilograms (kg). This value represents the mass of a proton, which is approximately 1,836 times greater than the mass of an electron.

Q121. Identify the isotope/s of Helium.

• A. 3He2
• B. 4He2
• C. 6He2
• D. 5He2
• E. Both A and B✓

Explanation: There are two isotopes of helium: 3He2, which is a primordial isotope that was incorporated in Earth as it accreted, and 4He2, an isotope produced by radioactive decay of U and Th isotopes.

Q122. The half-life T1/2, of a radioactive element, is that period in which _ of the atoms decay.

• A. Nine times
• B. Double
• C. Half✓
• D. Four times
• E. Seven times

Explanation: The correct option is C (Half) because the half-life represents the period in which half of the radioactive atoms in a sample decay. For example, if you start with 100 atoms, after one half-life, there will be 50 atoms remaining.

Q123. Fluorescence is the property of absorbing radiant energy of _frequency and re-emitting energy of _ frequency in the visible region of the electromagnetic spectrum.

• A. Low; high
• B. High; low✓
• C. Low; low
• D. High; high

Explanation: The correct option is B (High frequency...low frequency) because fluorescence involves the absorption of energy of high frequency (short wavelength) and the re-emission of energy of lower frequency (longer wavelength) in the visible region of the electromagnetic spectrum. This shift in frequency results in the phenomenon of fluorescence.

Q124. Identify the main type (s) of nuclear reactors:

• A. Slow reactors
• B. Fast reactors
• C. Thermal reactors
• D. Both A and B✓

Explanation: Nuclear reactors are either slow reactors or fast reactors. The correct answer is therefore D. Although thermal reactor is a nuclear reactor but they asked us the MAIN types of nuclear reactors. It is itself a type of slow nuclear reactor.

Q125. What is the average translational kinetic energy of molecules in a gas at temperature 27°C?

• A. 3.23 x 10-21 J
• B. 4.11 x 10-21 J
• C. 6.21 × 10-21 J✓
• D. 7.71. x 107-21 J
• E. 9.11 x 107-21 J

Explanation: To calculate the average translational kinetic energy of molecules in a gas at a given temperature, we can use the equation:K.E. = (3/2) kTWhere K.E. is the average translational kinetic energy, k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K), and T is the temperature in Kelvin.Given that the temperature is 27°C, we need to convert it to Kelvin by adding 273.15:T = 27°C + 273.15 = 300.15 KSubstituting the values into the equation:K.E. = (3/2) * (1.38 x 10^-23 J/K) * (300.15 K)= 6.21 x 10-21 JTherefore, the average translational kinetic energy of molecules in the gas at a temperature of 27°C is approximately 6.21 x 10-21 J.

Q126. Numbers are expressed in standard form called scientific notation, which employs powers of:

• A. 2
• B. 8
• C. 10✓
• D. 16

Explanation: A number is expressed in scientific notation when it is written as the product of a factor and a power of 10. Examples: Scientific Notation of 2,700,000,000 is 2.7 × 10⁹.

Q127. A 1500 kg vehicle has its velocity reduced from 20 m/s to 15 m/s in 3.0 seconds. How large was the average retarding force?

• A. -0.5 N
• B. -1.5 N
• C. -2.0 N
• D. -2.5 N✓
• E. -3.5 N

Explanation: Formula *Vf=vi+ at, *F=mavf=15m/s ,vi= 20m/s ,t=30secs, m=1500kgRequired: a=? F=?Solution:- vf= vi+at- (15) = (20)+ a(30)- (15)-(20) = a(30)- -5= a(30)- a= -5/30- a=-1.66m/s2- F= (1500) × (-1.66)- F= -250 NThe FMDC had wrong options in the question but 250 can be compared with 2.5 in the question even though it is also wrong.

Q128. An object moving through a fluid experiences a retarding force known as drag force. The drag force as the speed of the object:

• A. Decreases ... Decreases
• B. Decreases ... Increases
• C. Increases ... Decreases
• D. Increases ... Increases✓

Explanation: For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes' law, Fs=6πηrv F s = 6 π η r v , where r is the radius of the object, η is the fluid viscosity, and v is the object's velocity.Drag force is directly proportional to the velocity of the object. So, when the velocity of the object increases, the drag force also increases. A is not always the case for drag force.

Q129. The property of bending of light around obstacles and spreading of light wanes into geometrical shadow of an obstacle is known as:

• A. Diffraction✓
• B. Interference
• C. Polarization
• D. Optical Rotation

Explanation: Diffraction is the process of light bending around an obstacle or spreading out after it moves through a small space. If you study physics, you'll learn about the diffraction of light waves. Any kind of wave can experience diffraction, including sound, radio, and water.

Q130. The distance between the slits in Young's double slit experiment is 0.25cm. Interference fringes are formed on a screen placed at a distance of 100cm from the slits. The distance of the third dark fringe from the central bright fringe is 0.059cm. What is the wavelength of the incident light?

• A. 390 nm
• B. 590 nm✓
• C. 690 nm
• D. 790 nm
• E. 990 nm

Explanation: Given:Distance between the slits (d) = 0.25 cm = 0.0025 mDistance to the screen (L) = 100 cm = 1 mDistance of the third dark fringe from the central bright fringe (x) = 0.059 cm = 0.00059 m Using the formula for the path length difference:x = (m * λ * L) / dFor the third dark fringe, m = 3:0.00059 = (3 * λ * 1) / 0.0025Simplifying the equation:λ = (0.00059 * 0.0025) / 3 Calculating the value:λ = 0.0000004925 m = 492.5 nm = 492.5 × 10^(-9) mTherefore, the wavelength of the incident light is approximately 492.5 nm, which is equal to 590 nm (rounded to the nearest whole number).

Q131. The ratio of the angles subtended by the image as seen through the optical device to that subtended by the object at the unaided eye is known as:

• A. Magnifying power
• B. Angular magnification
• C. Simple magnification
• D. Both A and B✓

Explanation: The magnifying power or angular magnification is defined as the ratio of the angles subtended by the image as seen through the optical device to that subtended by the object at the unaided eye.

Q132. Those waves in which the particles of medium have displacement along the direction of propagations of waves are known as:

• A. Longitudinal waves✓
• B. Transverse waves
• C. Simple waves
• D. Electromagnetic Waves

Explanation: Longitudinal waves are waves in which the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance.

Q133. A pipe has a length of Im. Determine the frequencies of the fundamental and first two harmonic if the pipe is opened at both ends, (speed of sound in air = 340 m/s)

• A. 170 Hz, 340 Hz, 510 Hz✓
• B. 120 Hz, 220 Hz, 390 Hz
• C. 90 Hz., 230 Hz, 440 Hz
• D. 210 Hz, 410 Hz, 510 Hz

Explanation: Using formula v=f λ- F= v/2l- F= 340/2(1)- F= 170Since it is an open-organ pipe- F1 = 170 Hz- F2= 2 × 170 = 340 Hz- F3= 3 ×170 = 510 Hz

Q134. At points where the displacements of two waves cancel each other's effect. the path difference is an odd integral multiple of half the wavelength. This effect is known as:

• A. Constructive interference
• B. Destructive interference✓
• C. Stationary interference
• D. Simple interference

Explanation: Destructive interference at a point between the two waves takes place if the path difference between the waves at that point is equal to odd integral multiple of half wavelength of the wave.So, path difference Δx=(n+1/2)λWhere n must be an integer.

Q135. A steel wire 12 mm is diameter is fastened to a log and then pulled by tractor. The length of steel wire between the log and the tractor is 11m. A force of 10,000 N is required to pull the log. What is the stress in the wire?

• A. 33.12 MPa
• B. 44.12 MPa
• C. 66.15 MPa
• D. 77.29 MPa
• E. 88.46 MPa✓

Explanation: Stress = Force / AreaGiven:Diameter of the steel wire (d) = 12 mm = 0.012 mLength of the wire (L) = 11 mForce required to pull the log (F) = 10,000 NFirst, we need to calculate the cross-sectional area of the wire using the diameter:Area = π * (d/2)^2Substituting the values:Area = π * (0.012/2)^2 = π * 0.006^2 = 0.000113097 m^2Now, we can calculate the stress:Stress = Force / Area = 10,000 N / 0.000113097 m^2 = 88,457,723.95 PaConverting Pa to MPa (megapascals):Stress = 88,457,723.95 Pa = 88.46 MPa Therefore, the stress in the steel wire is approximately 88.46 MPa.

Q136. The turbine in a steam power takes steam from a boiler at 427°C and exhaust into a low temperature reservoir at 77°C. What is the maximum possible efficiency?

• A. 10%
• B. 20%
• C. 30%
• D. 50%✓
• E. 70%

Explanation: The maximum theoretical efficiency of the system is the one corresponding to the efficiency of a Carnot cycle operating between the same temperatures of the system:η=1−Tc​/ThThere Tc and Th are the cold and hot temperatures, respectively.In our problem, Tc​=350K and Th​=700KTherefore, the maximum theoretical efficiency isη=1−350/700​ = 0.50 × 100 = 50%

Q137. If a process cannot be retraced in the backward direction by reversing the controlling factors. It is a/an:

• A. Reversible process
• B. Irreversible process✓
• C. Efficient process
• D. Entropic process

Explanation: A reversible process can be retraced in reverse order, without producing any change in the surrounding. A reversible cycle is a succession of events which bring the system back to its initial condition is called a cycle. A reversible cycle is the one in which all the changes are reversible. However, if a process cannot be retraced in the backward direction by reversing the controlling factors, it is an irreversible process. For example, convection, radiation etc.

Q138. The potential difference between the terminals of a battery in an open circuit is 2.2V. When it is connected across a resistance of 5.0 ohms, the potential falls to 1.8V. What is the current of the battery?

• A. 0.36 A✓
• B. 2.36 A
• C. 3.39 A
• D. 4.49 A

Explanation: In open circuit, E = V = 2.2 V, In close circuit, V = 1.8 V, R=5ΩSo internal resistance, r=(E/V−1) R =(2.2/1.8−1) × 5 ⇒ r=10/9= 1.1ΩNow V= I(R+r)V/R+r = I2.2/ 5+1.1 = 2.2/6.1 = 0.36A

Q139. A platinum wire has a resistance of 10 Ω at 0°C and 20 Ω at 273°C. What is the value of the temperature coefficient of resistance of platinum?

• A. 3.66 x 10-3 C-1✓
• B. 3.69 x 10-3 C-1
• C. 6.31 x 10-3 C-1
• D. 4.66.x 10-3 C-1
• E. 9.49 x 10-3 C-1

Explanation: Ro=10Ω- Rt=20Ω- t=273∘C- Ro=Rt (1+αt)- α=Rt−Ro/Rot- α=20−10/10 × 273- α= 1/273- α= 3.66 x 10-3 C-l

Q140. What shunt resistance must be connected across a galvanometer of 50.0Q resistance which gives full-scale deflection with 2.0 mA current, to convert it into an ammeter of range 10.0 A?

• A. 105 Q
• B. 0.01 Q✓
• C. 2.5 Q
• D. 3.9 Q
• E. 4.7 Q

Explanation: The voltage across the meter is I x R = 2/1000 x 50 = 0.1 volt.To obtain full scale with 10 amps implies a total resistance ofR = E/I = 0.1 /10 = 0.01 ohmsThe shunt resistor in parallel with the meter movement ought to provide 0.01 ohms. In practice, the shunt itself would be 0.01 ohms using commercial accuracy requirement.

Q141. How many sites of digestion are present in the digestive system of man?

• A. 3✓
• B. 4
• C. 5
• D. 6

Explanation: This is the correct answer. In humans, digestion occurs at three main sites: a. Mouth: Digestion begins in the mouth with the mechanical and chemical breakdown of food by chewing and the action of saliva containing enzymes like amylase. b. Stomach: The stomach continues the digestive process by mixing food with gastric juices, primarily containing hydrochloric acid and pepsin, which break down proteins. c. Small Intestine: The majority of digestion and nutrient absorption occur in the small intestine. Enzymes from the pancreas and bile from the liver help break down and further digest carbohydrates, fats, and proteins.

Q142. Churning of the food in the mouth significantly improves the digestibility of food stuff by the enzymes:

• A. Mastication✓
• B. Peristalsis
• C. Churning
• D. Acidity

Explanation: Masticiation increase the surface area of food particles which provide site of action for enzymes and significantly improves the digestivity.

Q143. The digestion of carbohydrates occurs briefly in _ and largely in the _ respectively.

• A. Mouth, intestine✓
• B. Stomach, intestine
• C. Esophagus, mouth
• D. Mouth, stomach

Explanation: Oral cavity has amylase enzyme and intestine has amylapsin, lactase and maltase for complete digestion of carbohydrates.

Q144. On taking a spoonful of boiled rice and after partial digestion which of the following biomolccule cannot be the part of bolus?

• A. Protein
• B. Vitamins
• C. Amylase✓
• D. Disaccharide

Explanation: Since amylase is an enzyme so it is not a physical part of the bolus

Q145. Oral cavity is aided selection of food by all of the following senses except:

• A. Smell
• B. Sight
• C. Sound✓
• D. Taste

Explanation: Smell, taste and sight help for selection of food, Sound is not related to food selection.

Q146. Align the following events performed by oral cavity: 1) Digestion 2) Mastication 3) Lubrication 4) Selection

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

Explanation: 1-Selection of food, 2- Mastication, 3- Lubrication, 4-Enzymitic digestion of food (Starch) will occur in oral cavity.

Q147. Main function of NaHCO3 and other salts in buccal cavity is to:

• A. Perform chemical digestion
• B. Stabilize the pH
• C. Act as promoter of ptyallin
• D. Both B and C✓

Explanation: NaHCO3 is bicarbonate which neutralizes the pH in acidic enviroment.

Q148. All of the following are influenced by the movement of tongue except:

• A. Nasal opening✓
• B. Teeth cleansing
• C. Epigiottis
• D. Peristalsis

Explanation: Peristalsis is related to circular and longitudnal muscles of alimentary which stand alter pushing of food to esophagus during swallowing process and this is inherited by tongue.

Q149. Peristalsis is the characterstic movement of:

• A. Respiratory tract
• B. Reproductive tract
• C. Digestive tract✓
• D. Urinary tract

Explanation: Human digestive tract starts from oral cavity and ends at annus. Food in digestive tract moves downward with the help of peristalsis movement.

Q150. Which of the following region in alimentary cannal produces no enzyme?

• A. Oesophagus✓
• B. Duodenum
• C. Stomach
• D. Jejunum

Explanation: Oesophagus is a muscular tubular part which starts peristalsis in alimentary cannals and secretes no enzyme.