Sindh Mdcat Mock Paper 5 — Solved Past Paper with Answers

Q1. Identify the word or phrase that needs to be changed for the sentence to be correct:

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

Explanation: The send version of the verb is only used in the present tense in request form e.g Send me the email, send that version of the letter please etc. In the sentence past is being used so ‘sent’ will be used instead, so option B is the answer.

Q2. Choose the lettered word or phrase that is most nearly opposite in meaning to the word in capital letters. GLOOMY:

• A. Dark
• B. Bright✓
• C. Unexciting
• D. Faint

Explanation: "Gloomy" refers to a state or atmosphere that isdim, dark, or lacking in light. It often implies a feeling of sadness,melancholy, or a general lack of hope or optimism. Bright means giving out or reflecting much light; shining, so the total opposite. So option B is the answer as bright has the opposite meaning of gloomy.

Q3. Read the passage and answer the following question:“What a silly boy you are!” cried the miller. “I really don’t know what is the use of sending you to school. You seem not to learn anything. If little Hans came up here and saw our warm fire, our good supper, and our great cask of red wine, he might get envious. Envy is the most terrible thing and would spoil anybody’s nature. I certainly will not allow Hans’s nature to be spoiled. I am his best friend, and I will always watch over him, and see that he is not led into any temptations. Besides, if Hans came here, he might ask me to let him have some flour on credit, and that I could not do. Flour is one thing and friendship is another and they should not be confused. Why? The words are spelt differently and mean quite different things. Everybody can see that.” What does the statement “ Flour is one thing, and friendship is another” tell about the nature of miller?

• A. Very kind
• B. Selfish✓
• C. Friendly
• D. Caring

Explanation: If the miller was kind, he would have lent the flour.If the miller was friendly and caring, as he claims in the previous lines by saying “I will not allow Hans’s nature to be spoiled”, he would certainly let Hans have some flour on credit if he asked him for it.Since he is saying that he will not give flour on credit even to someone he cares for, this makes him selfish

Q4. Complete the sentences by choosing the most appropriate option, from the given lettered choices. Two airplanes crashed _ the fog yesterday?

• A. On
• B. Of
• C. In✓
• D. To

Explanation: The correct preposition to complete the sentence is "in." Using "in" suggests that the crash happened within the foggy conditions, effectively describing the environment of the crash. "On" and "of" do not correctly convey this spatial relationship—"on" implies a surface, and "of" suggests possession. "To" indicates movement towards something, which is irrelevant to this context. Therefore, "in" is the most suitable choice.

Q5. Identify the word or phrase that needs to be changed for the sentence to be correct.It’s too late to go for a walk now; besides, it’s beginning to rain.

• A. It's
• B. Go
• C. Besides
• D. Beginning
• E. No Error✓

Explanation: The given sentence is an example of a compound sentence. A compound sentence contains at least two independent clauses. These two independent clauses can be connected with a comma and a coordinating conjunction or with a semicolon.

Q6. Choose the word nearly opposite in meaning to the capitalized one. MUFFLED

• A. Loud✓
• B. Quiet
• C. Soft
• D. Noiseless

Explanation: The word 'muffled' describes a sound that is muted or not clear due to some form of obstruction. The correct opposite of 'muffled' is 'loud', which denotes a sound that is strong and easily heard.

Q7. Identify errors in the sentences:Because of their identical appearance and dress, the twins were often mistaken for each other.

• A. Because of
• B. Identical
• C. And dress
• D. Were
• E. No error✓

Explanation: There is no error in the sentence above according to grammar.

Q8. Complete the sentences by choosing the most appropriate option, from the given lettered choices below. We must _ back by six o’clock.

• A. Be✓
• B. Can
• C. Has
• D. Have

Explanation: The sentence requires a word that fits grammatically and logically in the expression 'must _ back by six o’clock.' The correct choice is 'be' because 'must be back' is a standard expression indicating an obligation or necessity to return by a specified time. The other options ('can,' 'has,' 'have') do not form correct or meaningful expressions in this context.

Q9. Complete the sentences by choosing the most appropriate option, from the given lettered choices (A to D) below each. The recent discoveries of medical science have_ life and health to millions of people.

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

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

Q10. Root pressure develops due to

• A. Passive absorption
• B. Active absorption✓
• C. Increase in transpiration
• D. Low osmotic potential in soil

Explanation: Root pressure develops due to active absorption of mineral ions by root cells. This lowers the water potential inside the roots, causing water to enter by osmosis. The inward movement of water generates a positive pressure called root pressure.

Q11. Keratin is an example of:

• A. Transport protein
• B. Fibrous protein✓
• C. Catalytic protein
• D. Globular protein

Explanation: (b) Fibrous protein (Correct Answer) – Keratin is a fibrous structural protein that provides strength and protection in hair, nails, and skin.

Q12. It is an example of keto-pentose:

• A. Ribulose✓
• B. Fructose
• C. Ribose
• D. Sedoheptulose

Explanation: (a) Ribulose (Correct Answer) – Ribulose is a five-carbon sugar (pentose) with a ketone functional group, making it a keto-pentose.

Q13. Which part of the tobacco plant is infected by Meloidogyne incognita?

• A. Stem
• B. Root✓
• C. Flower
• D. Leaf

Explanation: Meloidogyne incognita, commonly referred to as the root-knot nematode, infects the roots of the tobacco plant. The nematode is so named because it causes the formation of galls or knots on the roots, disrupting the plant's ability to absorb water and nutrients efficiently. This results in stunted growth and can significantly diminish crop yield. While the stem, leaves, and flowers may show signs of stress due to root damage, they are not directly infected by the nematode. Thus, the root is the primary site of infection.

Q14. The deficiency of which of the following causes severe combined immunodeficiency?

• A. Hypoxanthine-guanine transferase
• B. Xanthine oxidase
• C. PRPP synthetase
• D. Adenosine deaminase✓

Explanation: ADA deficiency leads to SCID, a group of rare, inherited disorders characterized by a severe impairment of both the T and B lymphocytes of the immune system. Adenosine deaminase is involved in the breakdown of purines, and its deficiency results in the accumulation of toxic metabolites. that are particularly harmful to developing immune cells.

Q15. Which one of the following technique rapidly replicates specific target fragment of DNA without cloning?

• A. DNA sequencing
• B. Genetic probe
• C. Gel electrophoresis
• D. Polymerase chain reaction✓

Explanation: PCR involves using short synthetic DNA fragments called primers to select a segment of the genome to be amplified, and then multiple rounds of DNA synthesis to amplify that segment.

Q16. Nucleotides sequence that is identical to its complementary strand when each is read in the same chemical direction for example i-e; 5'GATC3' 3'CATG5' these are called as:

• A. Flanking sequences
• B. Nucleotide order
• C. Palindromic sequences✓
• D. Antagonistic sequences

Explanation: Palindromic sequences occur when a sequence of base pairs reads the same in both directions on complementary strands, like 5' GATC 3' and 3' CTAG 5'. Such sequences play a crucial role in genetic engineering, particularly in restriction enzyme recognition and cutting of DNA. The other options do not fit this definition: Flanking sequences are adjacent to DNA regions of interest, Nucleotide order refers to the sequence without implying symmetry, and Antagonistic sequences is not a concept in DNA sequence terminology.

Q17. In case of cystic fibrosis, the solution containing gene coated liposomes is sprayed into patients:

• A. Oral cavity
• B. Nostrils✓
• C. Thoracic cavity
• D. Mouth

Explanation: For treatment of cystic fibrosis, a solution containing gene coated liposomes is sprayed into patient's nostrils.

Q18. In vivo method of treatment is being tried in:

• A. SCIDS
• B. Cystic fibrosis✓
• C. Hypercholesterolemia
• D. Familial hypercholesterolemia

Explanation: The gene is introduced into the nasal sinus by coating it on liposomes which are sprayed just like inhalers in the nasal sinuses.

Q19. Liver cells are infected with retrovirus containing normal gene for the receptor for the treatment of:

• A. SCIDS
• B. Cystic fibrosis
• C. Hypercholesterolemia✓
• D. Cardiovascular disorder

Explanation: In this gene therapy, liver cells are infected with a retrovirus carrying the normal LDL receptor gene. This helps remove excess LDL cholesterol from the blood

Q20. For the treatment of SCIDS, bone marrow stem cells are removed from the blood and are infected with a:

• A. Bacteriophage
• B. Retrovirus✓
• C. Lambda virus
• D. Bacterium

Explanation: The retrovirus acts as a vector for gene therapy.

Q21. There are two main methods used for gene therapy i.e:

• A. Ex – vivo and in vivo✓
• B. Ex vivo – Ex vitro
• C. Ex – vivo and in vitro
• D. In vitro – Ex vitro

Explanation: In ex-vivo method cells or tissurs are taken out of the body and after insersion of normal genes they are implanted in the body. However, in in-vivo method gene/genes is/are inserted into the cells or at their original site in the body.

Q22. A decrease in blood pressure/volume will not cause the release of:

• A. Atrial natriuretic factor✓
• B. Aldosterone
• C. ADH
• D. Renin

Explanation: The correct answer is Atrial natriuretic factor (ANF). ANF is secreted by the atria of the heart in response to increased blood volume and pressure, not a decrease. It works to reduce blood volume and pressure by promoting sodium excretion and decreasing the release of aldosterone. On the other hand, Aldosterone, ADH, and Renin are all released in response to a decrease in blood pressure or volume. Aldosterone increases sodium reabsorption, ADH reduces water excretion, and renin triggers the renin-angiotensin-aldosterone system to increase blood pressure.

Q23. The part of the nephron involved in active reabsorption of sodium is:

• A. Distal convoluted tubule
• B. Ascending loop of Henle✓
• C. Bowman’s capsule
• D. Descending limb of Henle’s loop

Explanation: The correct answer is the ascending loop of Henle, which is crucial for the active reabsorption of sodium ions from the filtrate back into the bloodstream. This segment utilizes active transport mechanisms to move sodium ions against their concentration gradient, which is essential for maintaining osmotic balance and blood pressure.The distal convoluted tubule causes gradual osmotic outlow of water from the iltrate back to kidney as it passes downward in the descending loop of Henle. Bowman's capsule serves a different function by filtering blood but does not participate in sodium reabsorption. Finally, the descending limb of Henle's loop is primarily involved in the passive reabsorption of water, making it ineffective for sodium reabsorption.

Q24. Which of the following does NOT favour the formation of large quantities of dilute urine?

• A. Renin✓
• B. Atrial Natriuretic Factor (ANF)
• C. Alcohol
• D. Caffeine

Explanation: Renin promotes sodium/water retention, reducing dilute urine formation.

Q25. The maximum amount of electrolytes and water (70 – 80 %) from the glomerular filtrate is reabsorbed in which part of the nephron?

• A. Ascending limb of loop of Henle
• B. Distal convoluted tubule
• C. Proximal convoluted tubule✓
• D. Descending limb of loop of Henle

Explanation: During glomerular filtration in kidneys, ultrafiltration of blood occurs i.e: almost all the constituents of plasma except proteins pass onto the Bowman’s capsule.Then nearly 99% of the filtrate is reabsorbed by renal tubules.The proximal convoluted tubules show maximum reabsorption and nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment.

Q26. A fall in glomerular filtration rate (GFR) activates:

• A. Juxtaglomerular cells to release renin✓
• B. Adrenal cortex to release aldosterone
• C. Adrenal medulla to release adrenaline
• D. Posterior pituitary to release vasopressin

Explanation: Low GFR triggers renin release to maintain blood pressure.

Q27. Which one of the following is not a part of a renal pyramid?

• A. Peritubular capillaries
• B. Convoluted tubules✓
• C. Collecting ducts
• D. Loop of Henle

Explanation: Convoluted tubules are in renal cortex, not in pyramids.

Q28. Which one of the following statements is correct with respect to kidney function regulation?

• A. When someone drinks a large amount of water, ADH release is suppressed.✓
• B. Exposure to cold temperatures stimulates ADH release.
• C. An increase in glomerular blood flow stimulates formation of angiotensin II.
• D. During summer when the body loses a lot of water due to evaporation, the release of ADH is suppressed.

Explanation: Antidiuretic hormone (ADH) or vasopressin increases the reabsorption of water in the distal convoluted tubule, collecting tubules and collecting ducts of the nephrons of the kidneys. As a result, the reabsorption of water from the glomerular filtrate is increased. When someone drinks lot of water, requirement of absorption of water decreases, so ADH release is suppressed.

Q29. Uricotelic mode of excreting nitrogenous wastes is found in:

• A. Reptiles and birds✓
• B. Birds and annelids
• C. Amphibians and reptiles
• D. Insects and amphibians

Explanation: The correct answer is Reptiles and birds. These animals are uricotelic, meaning they excrete nitrogenous wastes as uric acid, which is advantageous in conserving water due to its low solubility. This adaptation is particularly beneficial for survival in arid and water-scarce environments. On the other hand, while insects also excrete uric acid, amphibians usually excrete urea or ammonia, reflecting their more aquatic lifestyles. Annelids excrete ammonia or urea, not uric acid, which is why options involving these animals are incorrect.

Q30. A person who is on a long hunger strike and is surviving only on water, will have:

• A. Less amino acids in his urine
• B. More glucose in his blood
• C. Less urea in his urine✓
• D. More sodium in his urine.

Explanation: During starvation, the body must rely on body proteins for the amino acids. On high protein diets the carbon skeletons of the amino acids are oxidized for energy or stored as fat and glycogen, but the amino nitrogen must be excreted. To facilitate this process, enzymes of the urea cycle are controlled at the gene level. When dietary proteins increase significantly, enzyme concentrations rise. On return to a balanced diet, enzyme levels decline. Under conditions of starvation, enzyme levels rise as proteins are degraded and amino acid carbon skeletons are used to provide energy, thus increasing the quantity of nitrogen that must be excreted in the form of urea.

Q31. If the loop of Henle were absent from the mammalian nephron, which one of the following is to be expected?

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

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

Q32. A condition of failure of kidney to form urine is called:

• A. Anuria✓
• B. Deamination
• C. Uremia
• D. Oliguria

Explanation: Anuria is the medical term for the complete cessation of urine production by the kidneys, often resulting from severe kidney damage or failure. In contrast, 'Deamination' refers to a metabolic process unrelated to kidney function. 'Uremia' describes a condition where waste products accumulate in the blood due to impaired kidney function, but does not specifically indicate the absence of urine production. Therefore, Anuria is the correct choice as it directly addresses the failure to form urine.

Q33. In sexual reproduction, sex cells contain _ chromosomes when compared to other cells of the body:

• A. Same
• B. Half✓
• C. Double
• D. All of the above

Explanation: In sexual reproduction, sex cells (gametes) are formed through a process called meiosis, which reduces the chromosome number by half. This is crucial because it ensures that when two gametes fuse during fertilization, the resulting offspring has the correct diploid number of chromosomes. Option B is correct because sex cells contain half the number of chromosomes compared to somatic cells. Option A is incorrect because gametes do not have the same number of chromosomes as body cells. Option C is incorrect because gametes do not contain double the chromosomes. Option D is incorrect as only one option accurately describes the chromosome number in sex cells.

Q34. Pituitary gland releases _ hormone and _ hormone while ovaries produce _ and progesterone.

• A. Follicle-stimulating and luteinizing, estrogen✓
• B. Estrogen and follicle-stimulating, luteinizing
• C. Luteinizing and estrogen, follicle-stimulating
• D. Follicle-stimulating and estrogen, luteinizing

Explanation: The pituitary gland is responsible for releasing the follicle-stimulating hormone (FSH) and luteinizing hormone (LH), both of which play key roles in regulating the reproductive processes. The ovaries, on the other hand, produce estrogen and progesterone, which are crucial for the development and regulation of the female reproductive system. Options B, C, and D incorrectly assign estrogen to the pituitary gland or misplace the order of the hormones.

Q35. Which one is NOT an involuntary function:

• A. Breathing
• B. Pumping of blood
• C. Skeletal muscle movement✓
• D. Blinking of eyes

Explanation: Involuntary functions are those that occur without our conscious control.Skeletal muscle movement is not an involuntary function;it is voluntary/intentional.Blinking of eyes, heart-pumping of blood, and breathing are all involuntary.

Q36. The passageways of the respiratory system have mucous secreting cells called:

• A. Tracheal cells
• B. Goblet cells✓
• C. Surfactant cells
• D. Pleural cells

Explanation: Goblet cells are the specialized cells within the respiratory tract responsible for secreting mucus, which helps trap dust and microbes.

Q37. A man who has type AB blood cannot father a child with type _ blood, because he would pass on either the _ or the B allele to all of his offspring.

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

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

Q38. Identify purine and pyrimidines from the following figures

• A. a and b purines, c, d, and e pyrimidines
• B. d and b purines, c, a and e pyrimidines
• C. a and e purines, c, d, and b pyrimidines
• D. a and c purines b, d, and e pyrimidines.✓

Explanation: Purines are those bases that are double-ringed. Pyrimidines have a single ring. As we can see, a and c are double-ringed so they are purines. b,c and d are single rings so they are pyrimidines.

Q39. A cross between a black cat and a tan cat produces a tabby pattern (black and tan fur together). What percent of kittens would have tan fur if a tabby cat is crossed with a black cat?

• A. 100%
• B. 50%
• C. 25%
• D. 0%✓

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

Q40. Haemophilia is a sex-linked _ trait.

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

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

Q41. The "d" and "D" alleles are used for lighter and darker skin colour in humans, respectively. By keeping in view the inheritance pattern of skin colour in human beings(controlled by 3 genes), choose which combination shows medium skin colour from the following picture:

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

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

Q42. A/An _ is mostly a non-protein chemical compound that is required for the protein’s biological activity.

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

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

Q43. A group of biologically active molecules formed from amino acids which interact with the surface of the lipid bilayer of cell membranes are called:

• A. Integral Proteins
• B. Peripheral proteins✓
• C. Cell wall
• D. Plasmodesmata

Explanation: Peripheral proteins, or peripheral membrane proteins, are a group of biologically active molecules formed from amino acids which interact with the surface of the lipid bilayer of cell membranes. Unlike integral membrane proteins, peripheral proteins do not enter into the hydrophobic space within the cell membrane. Instead, peripheral proteins have specific sequences of amino acids which allow them to attract to the phosphate heads of the lipid molecules or to integral proteins.

Q44. Read the passage below to answer the question:Hemophilia is a disorder in which blood fails to clot. Saad, a male hemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Hemophilia. Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmeds or Ali’s children, male or female, display symptoms of hemophilia, but the sons of Alia and Ayesha display symptoms of hemophilia while the daughters of Alia and Ayesha do not. Which of the following individuals are heterozygous for hemophilia:

• A. Saad, Ahmed, and Ali
• B. Ahmed, Ali, Alia, and Ayesha
• C. Saad and Sara
• D. Alia and Ayesha✓
• E. Ahmed and Ali

Explanation: Hemophilia is an X-linked recessive disorder, meaning that males (XY) need only one affected X chromosome to express the disorder, while females (XX) need two affected X chromosomes to have the disease.Saad (hemophiliac male) has only one X chromosome, which carries the hemophilia mutation, and a Y chromosome.Sara (normal female) has two normal X chromosomes.Ahmed and Ali (sons of Saad and Sara) inherit their Y chromosome from Saad and a normal X chromosome from Sara, so they are unaffected.Alia and Ayesha (daughters of Saad and Sara) inherit one affected X chromosome from Saad and one normal X chromosome from Sara, making them carriers (heterozygous) for hemophilia.Since Alia and Ayesha are carriers, they do not show symptoms themselves but can pass the affected X chromosome to their sons. Their sons receive their Y chromosome from their fathers and have a 50% chance of inheriting the affected X chromosome from their mothers, leading to hemophilia.Thus, Alia and Ayesha are heterozygous for hemophilia.

Q45. Given below is the ECG of a normal human. Which one of its components is correctly interpreted below?

• A. Complex QRS - one complete pulse
• B. Peak T initiation of total cardiac contraction
• C. Peak P and peak R together ­systolic and diastolic blood pressures
• D. Peak P­ initiation of left atrial contraction only✓

Explanation: P wave represents atrial depolarization, which initiates contraction of the atria.QRS complex represents ventricular depolarization, not “one complete pulse.”T wave represents ventricular repolarization, not initiation of total cardiac contraction.Blood pressures are not directly represented by peaks on the ECG.

Q46. If due to some injury, the chordae tendinae of the tricuspid valve of the human heart is partially non-­functional. What will be the immediate effect?

• A. The flow of blood into the aorta will be slowed down
• B. The 'pacemaker' will stop working
• C. The blood will tend to flow back into the left atrium
• D. The flow of blood into the pulmonary artery will be reduced✓

Explanation: The tricuspid valve is located on the right side of the heart and prevents backflow of blood from the right ventricle into the right atrium. If the chordae tendinae are partially non-functional, the tricuspid valve may not close properly, leading to regurgitation of blood from the right ventricle into the right atrium. This can result in a reduction of blood flow into the pulmonary artery, which carries blood from the right ventricle to the lungs.

Q47. Which two of the following changes (i-iv) usually tend to occur in the plain dwellers when they move to high altitudes (3,500 m or more)? (i) Increase in red blood cell size (ii) Increase in red blood cell production (iii) Increased breathing rate (iv) Increase in thrombocyte count Changes occurring are?

• A. (ii) and (iii)✓
• B. (iii) and (iv)
• C. (i) and (iv)
• D. (i) and (ii)

Explanation: (ii) Increase in red blood cell production: When individuals move to high altitudes, there is a decrease in the availability of oxygen. As a response, the body increases the production of red blood cells to enhance oxygen-carrying capacity and improve oxygen delivery to tissues. (iii) Increased breathing rate: At higher altitudes, the partial pressure of oxygen in the air is lower, leading to reduced oxygen uptake in the lungs. To compensate for this, the body increases the breathing rate, allowing more oxygen to be taken in and delivered to the tissues.

Q48. The fastest distribution of some injectable material/medicine and with no risk of any kind, can be achieved by injecting it into the:

• A. Capillaries
• B. Arteries
• C. Veins✓
• D. Lymph vessels

Explanation: Yes, the fastest distribution of an injectable material or medicine is achieved by injecting it into the veins. This is because veins are blood vessels that carry blood back to the heart. The blood in the veins is under pressure, which helps to push the medicine throughout the body quickly. Veins are blood vessels that carry blood back to the heart. They are typically thin-walled and have valves that prevent blood from flowing backwards.

Q49. Given below are four statements (i-­iv) regarding human blood circulatory system. (i) Arteries are thick-­walled and have narrow lumen as compared to veins. (ii) Angina is acute chest pain when the blood circulation to the brain is reduced. (iii) Persons with blood group AB can donate blood to any person with any blood group under ABO system. (iv) Calcium ions play a very important role in blood clotting. Which two of the above statements is/are correct?

• A. (i) and (iv)✓
• B. (i) and (ii)
• C. (ii), (iii) and (iv)
• D. (iii), (iv) and (i)

Explanation: The correct statements (i)and (iv) are correct.(i) Arteries are thick-walled and have narrow lumen as compared to veins. Arteries are blood vessels that carry oxygenated blood away from the heart to various parts of the body. They have thick walls containing smooth muscle and elastic fibers, which allow them to withstand the high pressure generated by the heart's pumping action. Arteries also have a smaller lumen (the central opening through which blood flows) to maintain the blood pressure and ensure proper blood flow to the tissues. (iv) Calcium ions play a very important role in blood clotting.The statement about calcium ions is correct. Calcium ions play a crucial role in the process of blood clotting or coagulation. When there is an injury or damage to blood vessels, a series of complex reactions involving various clotting factors take place, ultimately leading to the formation of a blood clot to prevent excessive bleeding. Calcium ions are essential for the activation of clotting factors and the formation of a stable blood clot.

Q50. The propulsive movement of the gastrointestinal tract (GI tract) is:

• A. Peristalsis✓
• B. Epiglottis
• C. Antiperistalsis
• D. Anus
• E. None of these options

Explanation: Peristalsis is the contraction of muscle tissue that helps move and break down foodstuffs. The walls of the alimentary canal include layers of smooth muscle controlled by the autonomic nervous system. The alternating contraction and relaxation of these muscles is called peristalsis.

Q51. The following sequence of events occurs at the neuromuscular junction. Nerve impulse -> release of V -> end plate potential -> W produced in muscle fibre -> X release from sarcoplasmic reticulum -> Formation of Y -> muscle contraction. Which one of the following shows the correct sequence from V -> Y?

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

Explanation: Neurotransmitters called acetylcholine (V) are released, and an action potential (W) is generated. After an action potential is generated in the muscle fibre, calcium ions (X) are released from the sarcoplasmic reticulum. Ca ions act on troponin and displace it so that myosin can attach to the sites on the actin molecule, forming an actomyosin (Y) complex for muscle contraction.

Q52. The diagram shows some organs of the digestive system. Where is amylase present?

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

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

Q53. Which of the following is correctly matched?

• A. AF, BG, CJ, DI, EH
• B. AH, BI, DF, CG, EJ✓
• C. AJ, BI, CH, DG, EF
• D. AH, BF, DG, CI, EJ

Explanation: To solve this quickly just match one or two organelles to their correct function,crossing out the wrong options while doing so. Peroxisomes carry out detoxification of alcohol which is only in option B so we can easily choose it,no other options have DF. To check others, we know that smooth ER deals with cholesterol and fats that is why E goes with J.Ribosomes carry out protein synthesis.Centrioles form astral rays.Lysosomes destroy foreign particles.Hence,AH, BI, DF, CG, EJ is correct.

Q54. Which of the following statement(s) is/are correct regarding a nucleus? I.Stores wastes and other SubstancesII.Contains genetic materialIII.Helps in cellular transport systemIV.Control centre of the cell

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

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

Q55. In the pedigree of a family shown above, brown eyes are indicated as white circles and blue eyes as dark circles. Javeria and Juhi are twins. From this chart, it can be determined that:

• A. Tahir and Mary are homozygous for brown eyes
• B. Jawaria and Juhi are identical twins
• C. Juhi is heterozygous for blue eyes
• D. Juhi is homozygous for blue eyes✓
• E. Jawaria and Saad are homozygous for brown eyes

Explanation: It is imperative for a recessive trait present in a homozygous state to show up in phenotype so answer D is correct.

Q56. Choose the correct combinations of labeling the carbohydrate molecule involved in the Calvin cycle.

• A. A - RuBP, B - Triose phosphate, C - PGA
• B. A - PGA, B - RuBP, C - Triose Phsophate
• C. A - PGA, B - Triose Phosphate, C - RuBP
• D. A - RuBP, B - PGA, C - Triose Phosphate✓
• E. A - Triose Phosphate, B - PGA, C - RuBP

Explanation: The Calvin cycle is a metabolic pathway found in the stroma of the chloroplast in which carbon enters in the form of CO2 and leaves in the form of sugar. The cycle spends ATP as an energy source and consumes NADPH2 as reducing power for adding high energy electrons to make the sugar. In the first phase, carboxylation of RUBP takes place, which leads to the formation of 1,3-bis phosphoglycerate (PGA). PGA then gets reduced to triose phosphate (glyceraldehyde phosphate). Thus, the labelled molecules in the diagram are A- RUBP, B- PGA, C- triose phosphate. Thus, the correct answer is option D.

Q57. The prolactin hormone, responsible for the activation of mammary glands to start producing milk, is a hormone of the:

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

Explanation: In humans, prolactin is produced both in the front portion of the pituitary gland (anterior pituitary gland) and in a range of sites elsewhere in the body. Lactotroph cells in the pituitary gland produce prolactin, where it is stored and then released into the bloodstream.

Q58. The Law of Dominance Is illustrated in the garden pea by:

• A. Homozygous tall x heterozygous tall
• B. Heterozygous tall x heterozygous tall
• C. Homozygous tall x homozygous tall
• D. Pure short x pure short
• E. Homozygous tall x pure short✓

Explanation: The Law of Dominance states that in a heterozygous individual, the dominant allele will determine the phenotype while the recessive allele is masked. In this case, the cross between a homozygous tall plant (TT) and a pure short plant (tt) clearly illustrates this law, as all offspring (Tt) will be tall, demonstrating that the dominant allele 'T' expresses itself over the recessive 't'.The other options do not adequately illustrate the Law of Dominance: the first option shows both parents expressing the tall phenotype, while the second option leads to a mix of phenotypes. The third option results in only tall offspring, but both parents are homozygous dominant, failing to demonstrate dominance. The fourth option produces only short offspring, which lacks any dominant allele.

Q59. Identify the incorrect statement about Charles Darwin’s theory:

• A. The individual of species have variations among them.
• B. There is always a tendency of over reproduction in a species.
• C. Vast gradual changes result in the origin of a new species.
• D. Favorable variations survive and unfavorable will be exterminated.
• E. Intra-specific competition occurs between different species and interspecific competition occurs among the individuals in a species.✓

Explanation: Option E has a factually incorrect statement. “Intraspecific” competition means competition between individuals of the same species.” Interspecific” competition means competition between two different species.

Q60. In which model do proteins float in membrane like icebergs in sea:

• A. Lock and key model
• B. Induced fit model
• C. Fluid mosaic model✓
• D. Lotka volterra model
• E. All of these

Explanation: As per the fluid mosaic model cell membranes have proteins as icebergs in a sea of lipids (lipid). Hence option C is correct.

Q61. In the Hardy - Weinberg principle, p2 + 2pq + q2 = 1. q2 represents the frequency of the:

• A. Homozygous dominant
• B. Heterozygous dominant
• C. Heterozygous recessive
• D. Homozygous recessive✓
• E. Blended genes

Explanation: q2 represents the homozygous recessive trait in the Hardy-Weinberg principle.It is calculated by squaring the frequency of the recessive allele q, where q is the frequency of the recessive allele and p is the frequency of the dominant allele. The entire equation is p2+2pq+q2 where p2 is the homozygous dominant frequency and 2pq is the heterozygous frequency.

Q62. Ileum is about _ long:

• A. 3.6 centimeters
• B. 3.6 millimeters
• C. 3.6 inches
• D. 3.6 meters✓
• E. 3.6 kilometers

Explanation: The ileum is about 3.6 meters (11.5 feet) long (or about three-fifths the length of the small intestine) and extends from the jejunum (the middle section of the small intestine) to the ileocecal valve, which empties into the colon (large intestine).

Q63. Haemoglobin carries more oxygen than plasma by:

• A. 50 times
• B. 20 times
• C. 70 times✓
• D. 100 times

Explanation: Hemoglobin significantly enhances the oxygen-carrying capacity of blood by approximately 70 times compared to plasma. Hemoglobin, found in red blood cells, binds and transports oxygen efficiently, with each gram carrying about 1.34 mL of oxygen. Given the typical hemoglobin concentration of 15 g/dL in blood, this results in roughly 20 mL of oxygen per 100 mL of blood, far surpassing the minimal oxygen dissolved in plasma.

Q64. Deficient production of hormones by adrenal glands results in:

• A. Cushing's syndrome
• B. Addison's disease✓
• C. Diabetes Mellitus
• D. Goiter
• E. Epilepsy

Explanation: Addison's disease is the result of insufficient production of hormones from the adrenal glands, particularly cortisol and aldosterone. This condition leads to symptoms such as fatigue, muscle weakness, and low blood pressure. Cushing's syndrome, in contrast, is caused by excessive cortisol. Diabetes Mellitus involves insulin issues and not adrenal hormones. Goiter relates to thyroid gland issues, and epilepsy is a neurological disorder, both unrelated to adrenal insufficiency.

Q65. In the pea plants, the factors round and yellow seeds (RRYY) is dominant over wrinkled and green seeds (rryy). Keeping this in mind, identify the seed that has the genotype round and green.

• A. RRyy✓
• B. rrYy
• C. RrYy
• D. RRYy

Explanation: The genotype RRyy produces round and green seeds. The 'RR' indicates the presence of the dominant trait for round seeds, while 'yy' indicates the recessive trait for green seeds. Other options either contain the recessive 'rr' indicating wrinkled seeds or 'YY' or 'Yy' indicating yellow seeds, which does not match the requirement for round and green seeds.

Q66. Lysosomes function in:

• A. Protein synthesis
• B. Processing of cellular materials
• C. Intracellular digestion✓
• D. Lipid synthesis
• E. Carbohydrate synthesis

Explanation: Lysosomes are specialized organelles within the cell that contain hydrolytic enzymes capable of breaking down macromolecules like proteins, nucleic acids, lipids, and carbohydrates. Their primary function is intracellular digestion, where they degrade and recycle cellular waste, damaged organelles, and material ingested by the cell. Unlike ribosomes, lysosomes do not synthesize proteins, and they are not involved in lipid or carbohydrate synthesis, which occur in other organelles. Options such as 'protein synthesis', 'lipid synthesis', and 'carbohydrate synthesis' refer to processes that take place outside of lysosomes, highlighting their specialized role in cellular maintenance through digestion.

Q67. A _ is a junction between two neurons or between a motor neuron and a muscle cell.

• A. Impulse
• B. Synapse✓
• C. Axon
• D. Cleft

Explanation: In the nervous system, a synapse is a structure that permits a neuron (or nerve cell) to pass an electrical or chemical signal to another neuron or to the target effector cell. It can be visualized as a gap between two neurons; however, that gap isn't referred to as a synapse, but rather the gap, also known as the synaptic cleft, alongside the end of the pre-synaptic neuron and the start of the post-synaptic neuron is collectively known as the synapse. Option B is hence correct.

Q68. The lymph vessels empty into:

• A. Arteries
• B. Veins✓
• C. Capillaries
• D. None of the given options

Explanation: Lymph drains into subclavian veins, returning lymph to the bloodstream, maintaining fluid balance and immunity.

Q69. Which of the following is not the function of the cerebrum?

• A. Voluntary digestion✓
• B. Thinking
• C. Intelligence
• D. Skeletal muscles

Explanation: Functions of the cerebrum include initiation of movement, coordination of movement, temperature, touch, vision, hearing, judgment, reasoning, problem-solving, emotions, and learning, which do not involve digestion. Digestion is controlled by the autonomic nervous system- it is not a voluntary function!

Q70. The spinal nerves are functionally:

• A. Sensory nerves
• B. Motor nerves
• C. Mixed nerves✓
• D. Unknown

Explanation: The cell bodies of the sensory neurons are in the dorsal root ganglion, but the motor neuron cell bodies are in the grey matter. The two roots join to form the spinal nerve just before the nerve leaves the vertebral column. Because all spinal nerves have both sensory and motor components, they are mixed nerves.

Q71. What are the functions of the inter, motor and sensory neurons in a reflex response?

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

Explanation: To relay nerve impulses from within the CNS, interneurons are used. To conduct the nerve impulses from the CNS to the effector, motor neurons are used. To conduct the nerve impulses from the receptors to the CNS, sensory neurons are used.

Q72. How many meninges cover the human brain?

• A. 5
• B. 4
• C. 3✓
• D. 2
• E. 1

Explanation: There are three layers of meninges around the brain and spinal cord. The outer layer, the dura mater, is tough, white fibrous connective tissue. The middle layer of meninges is the arachnoid, a thin layer resembling a cobweb with numerous threadlike strands attaching it to the innermost layer.

Q73. The autonomic nervous system controls all of the following activities except:

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

Explanation: The autonomic nervous system does not directly control thought processes. Thought is primarily regulated by the central nervous system, which includes the brain and spinal cord. While the autonomic nervous system can be influenced by certain thought patterns, such as stress or relaxation, it does not directly control the generation or content of thoughts.

Q74. Which of the following neurotransmitters function, both as neurotransmitter and hormones, decreasing our perception of pain?

• A. Epinephrine
• B. Serotonin
• C. Dopamine
• D. Endorphins✓

Explanation: Neurotransmitters are chemical messengers that transmit a message from a nerve cell across the synapse to a target cell. Hormones are chemical messengers that are secreted directly into the blood, which carries them to organs and tissues of the body to exert their functions. All the options are both, a hormone as well neurotransmitters. Epinephrine, commonly referred to as Adrenaline, relaxes the airway muscles, making it easier for the person to breathe. At the same time, it constricts blood vessels, helping to improve blood pressure. It also increases the heart rate. However, it does not decrease the response to pain. Hence Option A is rejected. Serotonin, on the other hand, is used by the body to send messages between nerve cells. It appears to play a role in mood, emotions, appetite, and digestion. As the precursor for melatonin, it helps regulate sleep-wake cycles and the body clock. Since it is not used to decrease pain response, Option B is rejected. Dopamine is a neurotransmitter that plays a role in how we feel pleasure. It's a big part of our unique human ability to think and plan. However, it plays no role in the pain response. Hence Option C is rejected. Endorphins are substances produced by the nervous system to cope with pain or stress. They are often called “feel-good” chemicals because they can act as pain relievers and happiness boosters. Hence, Option D is correct.

Q75. The following sequence of events occurs at the neuromuscular junction. nerve impulse -> release of V -> end plate potential -> W produced in muscle fibre -> X release from sarcoplasmic reticulum -> Formation of Y -> muscle contraction.Which one of the following shows the correct sequence from V -> Y?

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

Explanation: The correct sequence of events at the neuromuscular junction is as follows:Nerve impulse triggers the release of Acetylcholine (V) from the presynaptic neuron.Acetylcholine binds to receptors, generating an End Plate Potential, leading to an Action Potential (W) in the muscle fiber.The Action Potential triggers the release of Calcium ions (X) from the sarcoplasmic reticulum.Calcium ions bind to regulatory proteins, allowing the formation of Actomyosin (Y) cross-bridges, leading to muscle contraction.V = Acetylcholine, W = Action Potential, X = Calcium ions, Y = Actomyosin ✅ (Correct)

Q76. The phytohormones controlling the plant movements are?

• A. Auxins
• B. Gibberellins
• C. Abscisic acid
• D. Cytokinins
• E. All of the above✓

Explanation: Phytohormones are critical for plant growth, development, and responses to stimuli. Auxins are directly involved in tropisms, which are movements towards or away from stimuli. Gibberellins, while primarily growth hormones, also influence tropic movements by modulating growth patterns. Abscisic acid is more related to stress responses but contributes to movement by regulating stomatal closure. Cytokinins affect growth and cell division, indirectly influencing movement. Thus, all these hormones together play a role in plant movements and responses.

Q77. Semilunar valves are not present:

• A. At the base of the pulmonary trunk
• B. At the base of the aorta
• C. In veins
• D. In the coronary arteries✓

Explanation: Semilunar valves exist only at the pulmonary trunk and aorta to prevent backflow. Coronary arteries and veins do not have valves.

Q78. Which of the following occurs in the body in response to the secretion of Glucagon

• A. Conversion of glucose to glycogen in liver cells
• B. Decrease in blood glucose concentration
• C. Increased uptake of glucose by muscle cells
• D. Production of cyclic AMP in target cells✓

Explanation: Option D is correct. 'Glucagon' has the opposite role to 'insulin', i.e., it mobilizes blood glucose in the plasma when you’re hypoglycemic. When glucagon is released, it can perform the following tasks: Stimulating the liver to break down glycogen to be released into the blood as glucose. Activating gluconeogenesis is the conversion of amino acids into glucose. Break down stored fat (triglycerides) into fatty acids for use as fuel by cells. All of the other options, except D, explain the role of insulin and not glucagon. By simply using an elimination method, we can deduce that D is the right answer. Just a piece of additional information: The chief role of cyclic AMP in several tissues seems to be to facilitate or promote the mobilization of glucose and fatty acid reserves. In the liver, glucagon and catecholamines cause an increase in the intracellular level of cyclic AMP by stimulating adenylyl cyclase.

Q79. _ is sometimes given by injection as an emergency treatment in cardiac arrest.

• A. Acetylcholine
• B. Dopamine
• C. Serotonin
• D. Epinephrine✓

Explanation: When CPR does not generate coronary perfusion of more than 15-20 mmHg, return of spontaneous circulation (ROSC) rarely or never occurs. After more than a few minutes of cardiac arrest, arterial tone collapses, and epinephrine or another vasoconstrictor is essential for restoration of cardiac activity.

Q80. The massive accumulation of blood within a tissue is called:

• A. Hemorrhage
• B. Haematoma✓
• C. Hepatoma
• D. Haemocoel
• E. Haematemesis

Explanation: A hematoma is a localized collection of blood outside vessels, usually due to vessel injury.

Q81. Growth movement caused in response to gravitational stimulus is called:

• A. Nutation
• B. Geotropism✓
• C. Nastic movement
• D. Tropic movement
• E. Turgor movement

Explanation: Growth movements induced by the stimulus of gravity are said to be geotropism.Nutation is a rocking, swaying, or nodding motion in the axis of rotation of a largely axially symmetric object. Nastic movements are non-directional responses to stimuli (e.g. temperature, humidity, light irradiance), and are usually associated with plants. The movement can be due to changes in turgor or changes in growth. One can define tropic movement or tropism as the directional movement of particular parts of a plant in response to environmental stimuli.Turgor movement is due to the difference of turgidity of the cells in the lower half and upper half of pulvinus (petiole of leaf).

Q82. The movement of plants in response to touch stimulus is called:

• A. Hydrotropism
• B. Chemotropism
• C. Geotropism
• D. Thigmotropism✓
• E. Phototropism

Explanation: Thigmotropism is a directional growth movement that occurs as a mechanosensory response to a touch stimulus. It is typically found in twining plants and tendrils; however, plant biologists have also found thigmotropic responses in flowering plants and fungi. Phototropism is movement in response to light, geotropism is movement due to gravitational forces, and chemotropism is movement in response to chemicals.

Q83. Steroids consist of _ 6 membered carbon rings and one 5 membered carbon ring.

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

Explanation: The steroid core structure is typically composed of seventeen carbon atoms, bonded in four "fused" rings: three six-member cyclohexane rings and one five-member cyclopentane ring.

Q84. Closely related species are grouped together into _ .

• A. Families
• B. Orders
• C. Phyla
• D. Kingdom
• E. Genus✓

Explanation: Closely related species are grouped together into a “Genus”.

Q85. The cell wall of most of the bacteria have a unique macromolecule called_.

• A. Cellulose
• B. Chitin
• C. Fibres
• D. Fats
• E. Peptidoglycan✓

Explanation: A cell wall is a layer located outside the cell membrane found in plants, fungi, bacteria. Plant cell is made up of cellulose, fungi cell of chitin and bacteria cell is made up of peptidoglycan.

Q86. The process of replacing or supplementing the defective allele with a functional, normal allele is known as _.

• A. Allele transplant
• B. Physiotherapy
• C. Gene therapy✓
• D. Mutation
• E. Cloning

Explanation: Gene therapy is an experimental technique that uses an allele to treat or prevent disease,by replacing the defective allele. In the future, this technique may allow doctors to treat a disorder by inserting a gene into a patient's cells instead of using drugs or surgery.

Q87. In cats, the genes controlling coat color are co-dominant and are carried on the X chromosomes. When a black female was mated with a ginger male the resulting litter consisted of black male and tortoise-shell female kittens. What phenotypic ratio would be expected in the F2 generation?

• A. 1 black male : 1 ginger male : 2 tortoise - shell females
• B. 1 black male : 1 ginger male : 1 tortoise - shell female : 1 black female✓
• C. 2 black males : 1 tortoise - shell female : 1 ginger female
• D. 1 black male : 1 tortoise - shell female : 1 ginger female : 1 black female
• E. 2 black males : 1 tortoise - shell female : 1 black female

Explanation: A black male has a genotype of XbY Tortoise-shell female kittens will have: XbXg XbY and XbXg XbXb; XbXg; XbY; XgY Hence: 1 Black-female; 1 tortoiseshell female; 1 black male; 1 ginger male.

Q88. A bean seed contains all of the following EXCEPT:

• A. A seed coat
• B. An epicotyl
• C. A hypocotyl
• D. A hypha✓
• E. Cotyledon

Explanation: A seed coat is the outer protective layer of a seed. An epicotyl is the part of the embryo that grows above the ground. A hypocotyl is the part of the embryo that grows below the ground. A cotyledon is a leaf-like structure that stores food for the embryo. A hypha is a thread-like filament that is found in fungi.

Q89. Highly condensed portions of the chromatin are called:

• A. Heterochromatin✓
• B. Nucleosome
• C. Super coils
• D. None of these

Explanation: Heterochromatin represents the highly condensed portions of chromatin. This dense packing makes the DNA inaccessible to the transcription machinery, leading to a transcriptionally inactive state. In contrast, euchromatin is less condensed and allows for active gene transcription. Nucleosomes are the structural units of chromatin, and supercoiling refers to the additional twisting of the DNA helix to aid in its compaction, not its condensation state. Therefore, heterochromatin is the best answer to describe highly condensed chromatin.

Q90. All of the following are sexually transmitted diseases except:

• A. Syphilis
• B. Gonorrhea
• C. Alzheimer's disease✓
• D. Genital herpes
• E. AIDS

Explanation: Alzheimer's disease is not a sexually transmitted disease. It is a progressive neurodegenerative disorder unrelated to sexual activity, primarily influenced by genetic and age-related factors. In contrast, syphilis, gonorrhea, genital herpes, and HIV/AIDS are all conditions that are transmitted through sexual contact. Identifying Alzheimer's disease as the exception is key to answering this question correctly.

Q91. Electronic configuration of M2+ ion is 2, 8, 14 and its atomic mass is 56 amu. The number of neutrons in its nucleus are:

• A. 30✓
• B. 32
• C. 42
• D. 52

Explanation: Total number of electrons in M2+ ions are 2+8+14=24, since they lost two electrons to become this ion that means neutral M will have 26 electrons. Since neutral atoms have the same numbers of electrons and protons so the number of protons in M is also 26. As the question states that the mass number of M2+ is 56. Mass number, being the sum of protons and neutrons, can be used to calculate neutrons, by subtracting protons from it(56-26=30)

Q92. The quantum numbers +½ and -½ for the electron spin represent:

• A. Rotation of the electron in clockwise and anticlockwise direction respectively
• B. Rotation of the electron in anticlockwise and clockwise direction respectively
• C. Magnetic movement of the electron pointing up and down respectively
• D. Two quantum mechanical spin states which have no classical analogue✓

Explanation: This is the most suitable option as it covers all aspects of describing the spin quantum number and its states.The two states +1/2 and -1/2, describes the spin of an electron. It is the term used for denoting the property of an electron only in Quantum physics and it does not have any proof in classical physics. That's why the term "no classical analogue" is used.

Q93. The diagram below represents the process of:

• A. Hydrolysis
• B. Condensation✓
• C. Neutralization
• D. Metabolism

Explanation: When two monomers join together to form a single product, with the removal of H2O, the process is called the condensation process. Hence, option B is correct.

Q94. The basic functional unit of human kidney is

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

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

Q95. The electronic configuration of iron is:

• A. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d5
• B. 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
• C. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6✓
• D. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d3
• E. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d2

Explanation: The electronic configuration of iron (Fe) with 26 electrons is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6. This follows the Aufbau principle, where electrons fill orbitals starting from the lowest energy level. The 4s subshell is filled before the 3d subshell. Options A, B, D, and E account for fewer electrons than iron's 26, each missing one or more electrons in their configuration.

Q96. IUPAC Nomenclature of: CH3 - CH = CH - CH2 - CH = CH2 is:

• A. 2 - pentene
• B. 1, 4 - hexadiene✓
• C. 3 - methyl butane
• D. 1, 3, 5 - heptatriene
• E. 1, 3, 6 - hexatriene

Explanation: The correct IUPAC name for the compound CH3 - CH = CH - CH2 - CH = CH2 is 1,4-hexadiene. This is because the compound has a longest chain of 6 carbon atoms, indicated by the prefix 'hex'. The double bonds are located at the first and fourth carbon atoms, which is specified in the name as 1,4-diene. The other options are incorrect because they either miscount the number of carbon atoms, suggest incorrect branching, or provide inappropriate locations for the double bonds.

Q97. For the equilibrium reaction 2NO2 ⇌ N2O4 (g) (-61 kJ), increase of temperature would:

• A. Favour the Formation of N2O4
• B. Favour the Decomposition of N2O4✓
• C. No Effect on Equilibrium
• D. Stop the Reaction

Explanation: The reaction is exothermic (ΔH = -61 kJ) in the forward direction (2NO₂ → N₂O₄). Increasing temperature adds heat, which the system counteracts by shifting toward the endothermic reverse reaction. Thus, the equilibrium shifts to decompose N₂O₄ into NO₂.

Q98. The chief ore of aluminium is:

• A. Na3AlF6
• B. Al2O3.nH2O
• C. Al2O3
• D. Al2O3.2H2O✓

Explanation: The chief ore of aluminium is bauxite, with the chemical formula Al2O3.2H2O. Bauxite is primarily composed of aluminium hydroxide minerals, which are processed to produce pure aluminium. The other options are related to aluminium but are not the primary ore. Cryolite (Na3AlF6) is used in the extraction process, not as an ore. Alumina (Al2O3) is a refined product derived from bauxite, while Al2O3.nH2O refers to various alumina hydrates but not specifically to bauxite.

Q99. Sp hybrid orbitals are formed by the mixing of:

• A. One s and two p
• B. One s and three p
• C. One s and one p✓
• D. Two s and two p

Explanation: In sp hybridization, one s orbital and one p orbital hybridize to form two sp orbitals, each consisting of 50% s character and 50% p character. This type of hybridization is required whenever an atom is surrounded by two groups of electrons.

Q100. Ethers show the phenomenon of:

• A. Position isomerism
• B. Functional group isomerism
• C. Metamerism✓
• D. Cis-trans isomerism

Explanation: Metamerism is a type of isomerism in which molecules have the same molecular formula but different alkyl groups around the functional group.For example,CH3-CH2-O-CH2-CH3 and CH3-O-CH2-CH2-CH3 are two metamers of each other.

Q101. The net heat change in a chemical reaction is the same whether it is brought about in two or more different ways in one or several steps. This is known as:

• A. Henry's law
• B. Hess's law✓
• C. Joule's principle
• D. Dalton's law

Explanation: Hess's Law is a fundamental principle in thermochemistry, asserting that the total enthalpy change during a chemical reaction is the same regardless of the number of steps in which the reaction is carried out, because enthalpy is a state function. This means that it depends only on the initial and final states of the system, not on the path taken.Henry's law, Dalton's law, and Joule's principle all deal with different principles: Henry's law is about gas solubility in liquids, Dalton's law is about partial pressures in gas mixtures, and Joule's principle is about the conversion of electrical energy to heat in conductors. None of these involve the calculation of enthalpy changes in chemical reactions.

Q102. Metallic character of the elements:

• A. Decreases down the groups
• B. Increases down the groups✓
• C. Remains the same across the periods
• D. Increases across the periods

Explanation: The correct answer is that metallic character increases down the groups. This is because as we move down a group in the periodic table, the atomic size increases due to the addition of electron shells. As a result, the valence electrons are further from the nucleus and experience less effective nuclear charge, making them easier to lose and thus increasing metallic character.Option A is incorrect because metallic character increases, not decreases, down a group. Option C is incorrect because metallic character decreases as we move across a period from left to right, due to increased nuclear charge attracting electrons more strongly. Option D is incorrect because metallic character does not increase across a period for the same reason.

Q103. Tritium, an isotope of hydrogen contains:

• A. Equal number of electrons and neutrons
• B. Equal number of electrons, protons and neutrons
• C. Number of neutrons is double than the number of protons✓
• D. Number of neutrons is half than the number of protons

Explanation: Tritium is the isotope of the hydrogen having mass number 3 and proton number 1.Number of neutrons are mass number-number protons which turn out to be 2. So we can say that the number of neutrons is twice the number of protons.

Q104. The chemical name of K2[Zn(OH)4] is:

• A. Potassium tetrahydroxyzinc (II)
• B. Potassium tetrahydroxozincate (II)✓
• C. Potassium tetrahydroxyzincate (VI)
• D. Potassium hydroxyzinc (II)

Explanation: The chemical name of K2[Zn(OH)4] is derived by naming the cation first, which is Potassium. The complex ion is named by first identifying the central metal, Zinc, which has an oxidation state of +2, indicated in Roman numerals as (II). The presence of four hydroxide ions is denoted by 'tetrahydroxo'. Therefore, the correct name for the compound is Potassium tetrahydroxozincate (II), reflecting both the coordination number and the oxidation state of zinc. Other options are incorrect because they either misrepresent the oxidation state or the number of hydroxide ions in the compound.

Q105. Paramagnetic elements contain:

• A. All paired electrons
• B. All unpaired electrons
• C. Few unpaired electrons.✓
• D. Unequal electrons and protons

Explanation: Paramagnetic elements are those with one or more unpaired electrons. These unpaired electrons create a net magnetic moment, making the element attracted to an external magnetic field. Option C is correct because it acknowledges the presence of unpaired electrons, which is the primary cause of paramagnetism. Option A is incorrect because paired electrons result in diamagnetism, not paramagnetism. Option B is misleading as it implies that all electrons must be unpaired for paramagnetism, which is not true. Even one unpaired electron can result in paramagnetism. Option D is unrelated to paramagnetism; it describes ions, which have an imbalance of electrons and protons, resulting in a charge.

Q106. The mass of an electron is:

• A. 1.008 amu
• B. 1.009 amu
• C. 0.000550 amu✓
• D. 0.5500 amu

Explanation: The mass of an electron is approximately 0.000550 amu, which is much smaller compared to the masses of protons and neutrons (approximately 1 amu each). This small mass is due to the electron's relatively lower contribution to the overall mass of an atom. The incorrect options suggest masses closer to those of protons and neutrons, highlighting the importance of distinguishing between the masses of different subatomic particles.

Q107. SN2 reactions can be best carried out with:

• A. Primary alkyl halides✓
• B. Secondary alkyl halides
• C. Tertiary alkyl halides
• D. Both primary and tertiary alkyl halides

Explanation: SN2 reactions involve a nucleophilic attack on the electrophilic carbon, leading to the inversion of configuration. The reaction is sensitive to steric hindrance around the electrophilic center. Therefore, primary alkyl halides are most suitable for SN2 reactions as they present the least steric congestion, allowing for efficient backside attack by the nucleophile. In contrast, secondary alkyl halides present moderate steric hindrance, making SN2 reactions possible but less favorable compared to primary halides. Tertiary alkyl halides are too sterically hindered for SN2 reactions and instead undergo SN1 reactions. Thus, primary alkyl halides are the best choice for SN2 reactions.

Q108. Which one of the following is NOT a nucleophile?

• A. H2S
• B. BF3✓
• C. NH3
• D. CN-

Explanation: A nucleophile is an electron-rich species capable of donating electron pairs to electron-deficient species. In contrast, electrophiles are electron-deficient and seek electrons. BF₃ is not a nucleophile because it is electron-deficient and acts as an electrophile, accepting electron pairs into its vacant orbital. H₂S, NH₃, and CN⁻ all have lone pairs of electrons that enable them to act as nucleophiles by donating electrons in reactions.

Q109. The oxidation number of Nitrogen in HNO3 is:

• A. +4
• B. +2
• C. +6
• D. +5✓
• E. +7

Explanation: In HNO₃, hydrogen is +1 and each oxygen is –2. Total must equal zero: +1+x+3(−2)=0. Solving gives x=+5, so nitrogen has oxidation number +5.

Q110. The unit of rate of reaction is:

• A. Mole (dm3) sec
• B. Mole (dm3) sec-1
• C. Mole(dm-3) sec -1✓
• D. Mole(dm3) sec-2
• E. Mole (dm3)-1 sec -2

Explanation: Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per decimeter cube per unit time [Mole (dm-3) sec-1 ].

Q111. The chemical properties of Sulphuric Acid is/are:

• A. Acidic properties
• B. Oxidizing properties
• C. Dehydrating properties
• D. Sulfonating properties
• E. All of these✓

Explanation: Sulfuric acid is a multi-functional chemical with a range of properties:Acidic properties: As a strong acid, sulfuric acid dissociates completely in water, releasing H+ ions, which are responsible for its acidic nature.Oxidizing properties: In redox reactions, sulfuric acid can act as an oxidizing agent, accepting electrons from other substances.Dehydrating properties: Sulfuric acid can remove water from other compounds, a process evident in its interactions with organic materials like carbohydrates.Sulfonating properties: Its ability to add sulfonate groups to organic compounds is crucial for producing sulfonated derivatives.Therefore, the correct answer is All of these, as sulfuric acid exhibits all these chemical properties. Each of the individual options represents just one aspect of its chemical behavior.

Q112. Benzene can be prepared from:

• A. From Petroleum
• B. From Coal
• C. From Acetylene
• D. From Phenol
• E. All of these✓

Explanation: The correct answer is that benzene can be prepared from all the mentioned sources. Petroleum: Benzene is obtained through catalytic reforming and steam cracking of naphtha, which are standard industrial processes for hydrocarbon refinement. Coal: During the carbonization of coal, benzene is one of the volatile compounds distilled from coal tar. Acetylene: Acetylene undergoes cyclic polymerization at high temperatures, forming benzene as a result. Phenol: Reduction of phenol using zinc dust is another chemical method to synthesize benzene, with zinc oxide as a byproduct. Thus, each method demonstrates a valid approach to benzene production.

Q113. - ROR is the nucleophile of

• A. Alcohols
• B. Esters
• C. Cyanides
• D. Ethers✓
• E. Aldehydes

Explanation: We have to look at the general form of given functional groups.

Q114. Primary Alcohol is produced by reactions of Grignard's reagent with _ followed by hydrolysis in an acidic medium.

• A. Carbon dioxide
• B. Formaldehyde✓
• C. Acetaldehyde
• D. Ketone
• E. Methyl chloride

Explanation: The Grignard Reaction involves the addition of an organomagnesium halide (Grignard reagent) to carbonyl compounds. When Grignard reagents react with formaldehyde, a primary alcohol is produced upon hydrolysis in an acidic medium. Formaldehyde is unique as it is the simplest aldehyde, thus allowing the formation of a primary alcohol. Other aldehydes and ketones lead to secondary and tertiary alcohols, respectively. Carbon dioxide reacts with Grignard reagents to form carboxylic acids, not alcohols, and methyl chloride does not react to form an alcohol.

Q115. If the reaction, P + Q → R + S, is described as being of zero-order with respect to P, it means that:

• A. P is a catalyst in this reaction
• B. No P molecules possess sufficient energy to react
• C. The concentration of P does not change during the reaction
• D. The rate of reaction is independent of the concentration of P✓
• E. The rate of reaction is proportional to the concentration of Q

Explanation: If the order of reaction with respect to P is 0 (zero), this means that the concentration of P doesn't affect the rate of reaction. Mathematically, any number raised to the power of zero (x0) is equal to 1. That means that that particular term disappears from the rate equation.

Q116. The oxidation number of all the elements in the free state is:

• A. 1
• B. 3
• C. 0✓
• D. -1
• E. -3

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

Q117. 2 grams of H2 molecules contain _ molecules.

• A. 12.04 x 1023
• B. 6.02 x 1023✓
• C. 3.01 x 1023
• D. 1.008
• E. 2

Explanation: 1 mole of hydrogen contains = 2×1 = 2g2g of H2 contain = 2/2 × avogadro number = 1 × 6.022 × 10²³ = 6.022 × 10²³ molecules.

Q118. What will be the product of the following reaction?

• A. A
• B. B
• C. C
• D. D✓

Explanation: Step 1: Starting compoundWe have p-cresol (4-methylphenol).Step 2: Reaction with CHCl₃/NaOHThis is the Reimer–Tiemann reaction.In this reaction, an aldehyde group –CHO gets introduced at the ortho position to the –OH group.So product A is o-formyl-p-cresol (4-methyl-2-hydroxybenzaldehyde).Step 3: Reaction with HCNAn aldehyde reacts with HCN to give a cyanohydrin.The –CHO group becomes –CH(OH)CN.So compound B is o-(CH(OH)CN)-p-cresol.Step 4: Hydrolysis with H₃O⁺On hydrolysis, –CN group converts to –COOH.So the final product C is o-(CH(OH)COOH)-p-cresol.Step 5: Match with given optionsLooking at the options:(a) has –CH₂COOH (wrong, no hydroxyl group)(b) also has –CH₂COOH (wrong)(c) has –CH(OH)COOH but with two methyls (wrong position)(d) has –CH(OH)COOH correctly at the ortho position to OH, with a para –CH₃ substituent (correct).

Q119. The product for the given reaction is:

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

Explanation: Structure: Cyclohexene with a =CH₂ group (i.e., methylene cyclohexane).Explanation:This is the correct product.The –CH₂OH group loses water and forms a double bond at the exocyclic position.This is called an exocyclic alkene (double bond between ring carbon and exo carbon).Matches expected product of alcohol dehydration

Q120. Which of the following values of heat of formation indicates that the product is least stable?

• A. - 94 kcal
• B. + 21.4 kcal
• C. - 231 kcal
• D. + 64.8 kcal✓

Explanation: In the context of heat of formation (ΔH), a negative value indicates that the formation of the compound is exothermic, meaning the compound is more stable compared to its constituent elements in their standard states.So, the least stable compound among the options provided would have the most positive (ΔH) value and option d has the most positive value ,hence that is the correct option.

Q121. Most abundant salt of sodium in nature is:

• A. NaNO3
• B. Na2SO4
• C. NaOH
• D. NaCl✓

Explanation: The most abundant salt of sodium in nature is sodium chloride (NaCl).The concentration of salt in seawater (its salinity) is about 35 parts per thousand; in other words, about 3.5% of the weight of seawater comes from the dissolved salts.

Q122. What is the volume in cm3 of 3.01 x 10^23 molecules of O2 gas at S.T.P.?

• A. 1000 cm3
• B. 11000 cm3
• C. 1120 cm3
• D. 11200 cm3✓

Explanation: 3.01*10^23 /6.02*10^23 = 0.5 mole at STP,1 mole of gas –––> 22,4 L 0,5 mole –––––––> x x= 11,2 L 11,2 L = 11200 cm3 As this is a numerical, it can only have one correct answer which is option D

Q123. In the reaction 2Fe+Cl2 -> 2FeCl3

• A. Fe is reduced
• B. Fe is oxidized✓
• C. Cl2 oxidized
• D. None of the above

Explanation: The Oxidation state of iron(Fe) is changed from "0" to "+3". Hence it is oxidised.

Q124. Given the reaction: C3H8 + 5O2 → 3CO2 + 4H2O At STP , how many litres of O2 are needed to completely burn 5.0 litres of C3H8?

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

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

Q125. For the reaction: N2 + 3H2 ⇌ 2NH3 .The production of NH3 will be favored at:

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

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

Q126. Catenation is a process in which carbon shows the properties of:

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

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

Q127. The maximum number of electrons that an orbital can accommodate is/are:

• A. 0
• B. 1
• C. 2✓
• D. 3
• E. 4

Explanation: The maximum number of electrons in a single orbital is two. The maximum number of electrons an orbital can accommodate is 2. As in an orbital two electrons can never have the same values of all 4 quantum no.s according to Pauli’s exclusion principle. So to follow this rule, an orbital can accommodate 2 electrons only.

Q128. The general formula for an aldehyde is:

• A. R-OH
• B. R-COOH
• C. R-CO-R
• D. R-X
• E. R-CHO✓

Explanation: Aldehydes and ketones have a carbonyl group (C=O) as a functional group. A ketone has two alkyl or aryl groups attached to the carbonyl carbon (RCOR’). The simplest ketone is acetone, which has two methyl groups attached to the carbonyl carbon (CH3COCH3).An aldehyde is similar to a ketone, except that instead of two side groups connected to the carbonyl carbon, they have at least one hydrogen (RCOH or R-CHO). The simplest aldehyde is formaldehyde (HCOH), as it has two hydrogens connected to the carbonyl group. All other aldehydes have one hydrogen bonded to the carbonyl group, like the simple molecule acetaldehyde, which has one hydrogen and one methyl group (HCOCH3).The carbonyl carbon in both aldehydes and ketones is electrophilic, meaning that it has a dipole due to the electronegativity of the attached oxygen atom. This makes the carbonyl carbon an ideal target for nucleophiles in a nucleophilic addition reaction. During this reaction, the nucleophile, or electron donor, attacks the carbonyl to form the tetrahedral intermediate. The negatively charged oxygen accepts a hydrogen ion to form a hydroxyl group.Typically, nucleophiles possess a negative charge or lone pair on a heteroatom, which can take several forms (OH-, RO-, CN-, R3C-, RNH2, ROH). For primary amines (RNH2), the reaction does not stop at the formation of the tetrahedral intermediate with a hydroxyl group. Rather, an elimination reaction occurs that produces a double-bonded carbon and nitrogen functional group known as an imine. Understanding the reactions that aldehydes and ketones can undergo provides a way to differentiate between these similar organic compounds.

Q129. When an element exists in more than one crystalline form, the phenomenon is termed as:

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

Explanation: Allotropy is the correct answer because it describes the ability of an element to exist in multiple forms or structures in the same physical state, such as carbon which can be found as diamond, graphite, and graphene.Isomorphism is incorrect as it refers to different substances with the same crystal structure. Isomerism is about different structural arrangements in organic compounds, not crystalline forms. Anisotropy deals with directional properties and not multiple forms of an element. Enthalpy involves thermodynamic measurements, not crystalline structures.

Q130. Which of the following statements about H2S is false:

• A. It is a covalent compound
• B. It is a gas with a bad smell
• C. It is a strong reducing agent
• D. It is a molecule which is non linear
• E. It is a weak base in water✓

Explanation: H2S is a weak acid, donating two hydrogen ions in neutralization reactions, forming HS- and S-2 ions, not a weak base. Hence option E is the correct answer.H2S is a covalent compound with a rotten egg smell. It is a strong reducing agent with a bent (non-linear) geometry.

Q131. When 18 ×10-3 moles/dm3 of acetic acid react with 22×10-3 moles/dm3 of ethyl alcohol to form 40×10-3 moles/dm3 of ethyl acetate and 40×10-3 moles/dm3 of waterFind the value of equilibrium constant (Kc)

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

Explanation: Kc= Concentration of products / Concentration of Reactants. Reactants = acetic acid and ethyl alcohol Products = ethyl acetate and water Multiplication product of products (40 × 40 =1600) Divide by multiplication product of reactants (18× 22= 396) 1600/396= 4.04.

Q132. Sigma bond is formed by:

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

Explanation: The sigma bond, the strongest type of covalent bond, is formed by the head-on overlapping of atomic orbitals. This overlap allows for maximum electron density between the two bonding nuclei, resulting in a stable bond. In contrast, ionic bonds are formed by the transfer of electrons (Option A), polar covalent bonds arise from mutual but unequal sharing of electrons (Option C), and pi bonds result from the parallel overlapping of atomic orbitals (Option D). Thus, Option B is the only correct description for sigma bond formation.

Q133. The unit of viscosity is:

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

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

Q134. The maximum possible number of electrons a shell 'n' can accommodate is given by:

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

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

Q135. Cl + e ----> Cl-ΔH = -348kJ/molThe value -348KJ/mol in this case will be:

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

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

Q136. Which of the following is an instrument for monitoring radiations:

• A. GM tube
• B. Radar
• C. Wilson cloud chamber
• D. Both A and C✓

Explanation: The Geiger-Mueller (GM) detector is a common portable instrument choice for a general laboratory radioactive material survey. GM detectors are capable of detecting alpha, beta, and gamma radiation. However, this instrument is limited by its ineffectiveness in detecting lower energy beta and gamma emitters.Radars send out electromagnetic waves as short pulses which may be reflected by objects in their path, in part reflecting to the radar. When these pulses intercept precipitation, part of the energy is scattered back to the radar.A cloud chamber, also known as a Wilson cloud chamber, is a particle detector used for visualizing the passage of ionizing radiation.

Q137. The half-life of a radioactive element depends upon:

• A. Amount of element present
• B. Pressure
• C. Temperature
• D. None of the above options are correct✓

Explanation: The correct answer is Option D. The half-life of a radioactive element is a fundamental property of that element, determined by the stability of its nucleus, and does not change with external factors like the amount of element present, temperature, or pressure. Options A, B, and C are incorrect because they suggest that half-life could be influenced by these external conditions, which it is not.

Q138. Radiation can cause:

• A. leukemia
• B. radiation sickness
• C. skin cancer
• D. gene mutations
• E. all of the above✓

Explanation: Radiation can lead to multiple adverse health effects, including leukemia, radiation sickness, skin cancer, and gene mutations. Each of these outcomes stems from the ability of radiation to damage biological tissues and DNA. Leukemia is linked to radiation's impact on bone marrow, radiation sickness results from acute high-level exposure, skin cancer is associated with UV radiation, and gene mutations can occur due to ionizing radiation disrupting DNA. Therefore, the most comprehensive answer is all of the above, as radiation can indeed cause all these effects.

Q139. The shortest wavelength photon emitted in the Balmer series is:

• A. 1.0974 nm
• B. 4 nm
• C. 364.6 nm✓
• D. 3.64 m
• E. 480 nm

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

Q140. One complete circle is equal to:

• A. 2π radian✓
• B. 3 radian
• C. 5 radian
• D. 9 radian

Explanation: There are 2π radians in a full circle. So 2π radians should equal 360°.

Q141. The frequency of a man's voice is 300 Hz. If velocity of sound is 336 m/s, the wavelength of the sound is:

• A. 1.12 m✓
• B. 300 m
• C. 330 m
• D. 112 m
• E. 11201 m

Explanation: We are given thatfrequency=300Hz, and velocity v=336ms−1From the relation v=f×λ,λ=v/f=336/300=1.12mOption "A" is correct.

Q142. The physical quantity which produces angular acceleration in the body is

• A. Force
• B. Moment of inertia
• C. Impulse
• D. Torque✓

Explanation: Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration

Q143. A jeep travelling at a constant speed of 108 km/hr rounds a curve of radius 100 m. Find its acceleration.

• A. 3 m/s2
• B. 6 m/s2
• C. 7 m/s2
• D. 8 m/s2
• E. 9 m/s2✓

Explanation: The correct answer is 9 m/s2. To find the acceleration of the jeep as it rounds the curve, we use the formula for centripetal acceleration: a = v2 / r, where 'v' is the velocity and 'r' is the radius of the curve. The velocity is given as 108 km/hr, which must be converted to meters per second by dividing by 3.6, resulting in 30 m/s. Substituting into the formula: a = (30 m/s)2 / 100 m = 9 m/s2. The other options are incorrect because they arise from either incorrect unit conversion or misapplication of the centripetal acceleration formula.

Q144. In what form is the energy stored in a charged capacitor?

• A. Gravitational
• B. Electrical potential✓
• C. Magnetic field
• D. Potential

Explanation: Energy is stored in a form of electrostatic field between plates of capacitor. This energy can be expressed in the form of work done by the battery.

Q145. A body moves on a circular path subtending an angle of 2 radians at the center of a circle of radius 3.5 meters. The linear displacement of the body is _.

• A. 6.5 m
• B. 7 m✓
• C. 7.5 m
• D. 1.75 m

Explanation: The linear displacement of a body moving along a circular path can be determined using the formula:Linear Displacement = 2 × Radius × sin(θ/2)Here, the radius (r) is 3.5 meters, and the angle (θ) is 2 radians. Plugging in the values, we apply:Linear Displacement = 2 × 3.5 × sin(2/2) = 2 × 3.5 × sin(1) ≈ 7 metersTherefore, the correct linear displacement of the body is 7 meters. All other options (6.5 m, 5.8 m, 7.5 m, and 1.75 m) do not match the computed value from the provided formula and parameters.

Q146. A car starts from rest and moves with a constant acceleration. During the 5th second of its motion, it covers a distance of 36 meters. What is the acceleration of the car?

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

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

Q147. Projectile must be launched at which angle with the horizontal to attain maximum range?

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

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

Q148. Law of conservation of momentum states that:I. there is no external force applied to a system, then the total momentum of that system remains constant II. if there is an external force applied to a system, then the total momentum of that system remains constant III. if there is no external force applied to a system, then the total momentum of that system keeps changing

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

Explanation: If there is no external force applied to the system, then the total momentum of that system remains constant. If an external force is applied then the total momentum will change.The law of conservation of momentum states that, for two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed and will stay constant.

Q149. The half-life of C14 is about:

• A. 1637 years
• B. 892 years
• C. 10 years
• D. 100 years
• E. 5730 years✓

Explanation: Carbon-14 has a half-life of 5,730 ± 40 years—i.e. half the amount of the radioisotope present at any given time will undergo spontaneous disintegration during the succeeding 5,730 years.

Q150. Sodium nucleus consists of 11 protons and 12 neutrons. The conventional symbol of this nucleus is?

• A. 11Na11
• B. 12Na12
• C. 11Na23✓
• D. 12Na23
• E. 12Na11

Explanation: The conventional symbol of the element contains the proton number(number of protons) and the mass number(proton number+number of neutrons), which is 11 and (11+12=)23, in this case. So the correct option is C. The conventional symbol of a sodium nucleus is 11Na23.

Q151. The process of natural decay of some heavy nuclides is because of the phenomenon:

• A. Emission Spectra
• B. Nuclear fusion
• C. Nuclear fission
• D. Radioactivity✓

Explanation: Radioactivity is the spontaneous emission of radiation in the form of particles or high-energy photons resulting from a nuclear reaction. It is also known as radioactive decay, nuclear decay, nuclear disintegration, or radioactive disintegration. Alpha, beta, and gamma decay are three common radioactivity processes

Q152. The deviation of I-V graph from straight line is due to:

• A. Decrease in temperature and decrease in resistance
• B. Increase in temperature and increase in resistance✓
• C. Decrease in temperature and increase in resistance
• D. Increase in temperature and decrease in resistance

Explanation: The most common scenario leading to a deviation from a straight-line I-V graph is an increase in temperature and an increase in resistance. This is often observed in conductors and is known as a positive temperature coefficient of resistance. As temperature rises, the resistance increases, causing a departure from the ideal linear behavior in the I-V graph.

Q153. A battery whose emf is 40 V has an internal resistance of 5 Ohms. If this battery is connected to a 15 Ohms resistor ‘R’, what will be the voltage drop across ‘R’:

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

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

Q154. What is the flux density at a point 3 cm from the current carrying conductor, when there is a current of 25 A in a wire: (u0= 4𝜋 x10-7)

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

Explanation: The correct answer is 1.67x10-4T. To find the magnetic flux density (B) at a distance (r) from a long straight wire carrying current (I), you can use the formula:B = (μ0 * I) / (2 * π * r).Substituting in the values: μ0 = 4π x 10-7 T·m/A, I = 25 A, and r = 0.03 m (3 cm), we get:B = (4π x 10-7 * 25) / (2 * π * 0.03) = 1.67 x 10-4T.Other options are incorrect because:0.23x10-1T is too high, not matching the expected results.2.99x10-6T is too low compared to the calculations.3.63x10-8T is orders of magnitude smaller than expected.999x10-7T converts to an unreasonably high value of 0.0999 T.

Q155. A battery has an e.m.f of 6.0 volts and an internal resistance of 0.4 ohms. It is connected to a 2.6 ohms resistor through a SPST (single pole, single throw switch). When the switch is open, the potential difference between the terminal of the battery is:

• A. 0 V
• B. 0.8 V
• C. 2.6 V
• D. 5.2 V
• E. 6.0 V✓

Explanation: The emf of a battery is generated by the chemical action in the battery. When no current is being drawn from the battery, the difference in potential between the terminals is equal to the emf.

Q156. A coil of 600 turns is threaded by a flux of 8x10-5 webers if this flux is reduced to 3x10-5 webers in 0.015 seconds. The average induced e.m.f. is:

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

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

Q157. A semiconductor photodiode is a:

• A. reverse biased junction diode✓
• B. forward biased junction diode
• C. half wave rectifier
• D. full wave rectifier
• E. Transistor

Explanation: The photodiode is reverse biased for operating in the photoconductive mode. As the photodiode is in reverse bias the width of the depletion layer increases. This reduces the junction capacitance and thereby the response time. In effect, the reverse bias causes faster response times for the photodiode.

Q158. A circuit in which there is a current of 5 amp is changed so that the current falls to zero in 0.1s. If an average e.m.f of 200 volts is induced, what is the self-inductance of the circuit?

• A. 4 henrys✓
• B. 8 henrys
• C. 12 henrys
• D. 16 henrys
• E. 20 henrys

Explanation: Initial currect= I1 = 5.0 A Final current= I2= 0 A change in current , dI= I1 -I2 =5A Time taken for the change= 0.1s Avg emf= 200V e=L di/dt L= e /di/dt 200/5/01 = 4H Hence the self inductance of the coil is 4H.

Q159. The atomic spectra deals with the measurement of:

• A. Wavelengths
• B. Intensities of electromagnetic radiations emitted by atoms
• C. Intensities of electromagnetic radiations absorbed by atoms
• D. All of these options✓
• E. Both B and C

Explanation: The correct answer is D: All of these options. Atomic spectra encompass the study of both the wavelengths and the intensities of electromagnetic radiation emitted and absorbed by atoms. When electrons transition between energy levels, they emit or absorb photons, whose characteristics—like wavelength and intensity—can be measured. Options A, B, and C all address partial aspects of these measurements, but the full scope of atomic spectra includes all these components.

Q160. What will be the energy in Joules, when an electron acquires a speed of 10 m/s?

• A. 3.61 x 10-19 J
• B. 4.55 x 10-29 J✓
• C. 5.13 x 10-19 J
• D. 6.13 x 10-19 J
• E. 8.71 x 10-19 J

Explanation: KE = 1/2mv2KE = ½(9.11 x 10-31)(10)2KE = 4.55 x 10-29 J

Q161. In a certain circuit, a transistor has a collector current of 10mA and a base current of 40μA. What is the current gain of the transistor?

• A. 150
• B. 200
• C. 250✓
• D. 300
• E. 350

Explanation: To determine the current gain (β) of a transistor, use the formula: β = IC/IB. Here, the collector current (IC) is 10mA (10 × 10-3 A) and the base current (IB) is 40μA (40 × 10-6 A). Calculating gives:Current gain = (10 × 10-3 A) / (40 × 10-6 A) = 250.Therefore, the correct answer is Option C: 250. The other options (150, 200, 300, 350) do not match the calculated current gain of 250, making them incorrect.

Q162. Electromagnetic wave theory was proposed by:

• A. Huygen
• B. Fraunhofer
• C. Fresnel
• D. Coulomb
• E. Maxwell✓

Explanation: James Clerk Maxwell is credited with formulating the electromagnetic wave theory in 1864. He proposed that light is a form of electromagnetic radiation, which was a groundbreaking unification of previously separate fields: electricity, magnetism, and optics. His set of equations, known as Maxwell's equations, describe how electric and magnetic fields interact and propagate as waves through space.Other options are incorrect as they contributed to different aspects of physics: Huygens with wave theory of light, Fraunhofer with spectral lines in light, Fresnel with wave optics, and Coulomb with electrostatics.

Q163. If the speed of waves is 10m/sec and its frequency is 5Hz, find its wavelength:

• A. 1
• B. 2✓
• C. 4
• D. 6

Explanation: V=F λ λ = v/f λ =10/5 λ = 2m

Q164. For the adiabatic process, the first law of thermodynamics is:

• A. w = ∆u = Q
• B. Q = -w
• C. Q = w
• D. w = -∆u✓

Explanation: An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. This condition can be used to derive the expression for the work done during an adiabatic process. When a gas expands, it does work on the surroundings; compression of a gas to a smaller volume similarly requires that the surroundings perform work on the gas. If the gas is thermally isolated from the surroundings, then the process is said to occur adiabatically. The first law is; ΔQ=ΔU+w In an adiabatic change, q = 0, so the First Law becomes 0=ΔU+w -ΔU=w or ΔU=-wSince the temperature of the gas changes with its internal energy, it follows that adiabatic compression of a gas will cause it to warm up, while adiabatic expansion will result in cooling.

Q165. If the body is rotating with uniform angular velocity, then its torque is:

• A. Zero✓
• B. 90
• C. 1
• D. -1

Explanation: Torque will be zero because the force on the object is zero, since it is not accelerating. And also if the body is at rest or rotating with uniform angular velocity the angular acceleration will be zero in this case the torque acting on a body will be zero.Torque is given by solution i.e. torque = r.F sin (theta)Here force = maAcceleration is given by change in velocity/timeIf the change in velocity is zero, acceleration is zero, force is zero, thus torque is also zero. Any other value is not remotely possible

Q166. According to the first law of thermodynamics, ΔU = Q + W, where ΔU Is the increase in internal energy of the system, Q is the heat transferred to the system and W is the external work done by the system.Which of the following is NOT a correct expression?

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

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

Q167. Two capacitors C1 = 2µ and C2 = 4µ F are connected in series across in a 100V supply. Find the effective capacitance.

• A. ½ µF
• B. 3/2 µF
• C. 5/2 µF
• D. 4/3 µF✓
• E. 2 µF

Explanation: Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)Therefore,1/Ceffective = 1/C1 + 1/C21/Ceffective = ½ +¼ 1/Ceffective = ¾C effective= 4/3 µF Hence option D is the correct answer.

Q168. Example(s) of spin motion is/are:

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

Explanation: Spin motion is the movement of an object about its own axis, which is depicted by;(i) the daily motion of the earth(ii)flywheel rotation about its own axis.

Q169. The following question belongs to the topic of momentum:

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

Explanation: The correct answer is Angular momentum. In three-dimensional space, the angular momentum of a point particle is defined as the cross product of its position vector (r) with respect to a given origin and its linear momentum vector (p). This forms a pseudovector, which is essential in describing rotational motion in physics. The other options are incorrect because they do not correctly describe the properties or definitions related to momentum or angular momentum. 'Current' relates to electric charge flow, 'Length' and 'Displacement' relate to measures of distance, and 'Torque' is a measure of rotational force, not a type of momentum.

Q170. The amount of heat provided to a system at constant pressure (qp) is equal to _.

• A. Change in internal energy (ΔU)
• B. Change in enthalpy (ΔH)✓
• C. Change in free energy (ΔG)
• D. Change in temperature only (ΔT)
• E. Change in pressure only (ΔP)

Explanation: At constant pressure, the heat added to a system is equal to the change in enthalpy (ΔH). This is because enthalpy is defined as the sum of the internal energy and the product of pressure and volume (H = U + PV). Therefore, when heat is added at constant pressure, it primarily changes the enthalpy. Option B is correct. Option A is incorrect because internal energy change (ΔU) does not account for the work done by the system, while Option C (ΔG) involves entropy and is related to the spontaneity of reactions, not just heat change. Options D and E are unrelated to the direct measure of heat at constant pressure.

Q171. A 10 nanofarad (10 x 10-9 F) parallel plate capacitor holds a charge of magnitude 50 µC on each plate.If the plates are separated by a distance of 0.885mm, what is the area of each plate?

• A. 1.0 m2✓
• B. 3.0 m2
• C. 5.5 m2
• D. 7.5 m2

Explanation: To find the area of each plate, we use the formula for the capacitance of a parallel plate capacitor: C = ε0 * (A/d), where C is the capacitance, ε0 is the permittivity of free space (approximately 8.85 x 10-12 F/m), A is the area of the plates, and d is the distance between the plates. Given C = 10 x 10-9 F and d = 0.885 x 10-3 m, we rearrange the formula to solve for A: A = (C * d) / ε0. Plugging in the values gives A = (10 x 10-9 F * 0.885 x 10-3 m) / 8.85 x 10-12 F/m = 1.0 m2. Thus, Option A is correct. The other options provide areas that are either too large or too small when calculated properly with the given values.

Q172. If '-' means'+', '+' means '-', '*' means '÷' and '÷' means '*' then which of the following will be the correct equation?

• A. 30 + 5 +14 -10 * 15 = 122
• B. 10 + 5 - 14 + 10 * 15 = 158
• C. 30 - 5 + 14 +10 * 15 = 162
• D. 30 * 5 - 4 + 10 - 30 = 30✓

Explanation: Option A: 30 + 5 +14 -10 * 15 = 122= 30 - 5 - 14 + 10 ÷ 15 (substituting the new symbols)= 30 - 5 - 14 + 0 (10 ÷ 15 = 0)= 11Option B: 10 + 5 - 14 + 10 * 15 = 158= 10 - 5 + 14 ÷ 10 - 15 (substituting the new symbols)= 10 - 5 + 1 - 15 (14 ÷ 10 = 1)= -9Option C: 30 - 5 + 14 +10 * 15 = 162= 30 + 5 - 14 - 10 ÷ 15 (substituting the new symbols)= 30 + 5 - 14 - 0 (10 ÷ 15 = 0)= 21 Option D: 30 * 5 - 4 + 10 - 30 = 30= 30 ÷ 5 + 4 - 10 + 30 (substituting the new symbols)= 6 + 4 - 10 + 30 (30 ÷ 5 = 6)= 30

Q173. Statement: I) No rabbit is lion. II) Some horses are lions. III) All rabbits are tables.Conclusions:I) Some tables are lions.II) Some horses are rabbits.III) No lion is table.

• A. Only I and II
• B. Only III follows
• C. Only II and III follow
• D. None follows✓

Explanation: To determine which conclusions logically follow from the premises, we analyze each statement:- Conclusion I: Some tables are lions. This cannot be determined as the statements do not establish any direct or indirect connection between tables and lions.- Conclusion II: Some horses are rabbits. This is contradicted by statement I, which says no rabbits are lions; since some horses are lions, they cannot be rabbits.- Conclusion III: No lion is table. While it may seem logical since all rabbits are tables and no rabbits are lions, the premises do not explicitly support this conclusion. The relationship between lions and tables is not defined in the provided statements.Therefore, none of the conclusions logically follow from the premises, making Option D the correct choice.

Q174. Statement: A large number of students are reported to be dropping out of school in villages as their parents want their children to help them in farms.Courses of Action:I. The government should immediately launch a programme to create awareness among the farmers about the value of education.II. The government should offer incentives to those farmers whose children remain in schools.III. Education should be made compulsory for all children up to the age of 14 and their employment banned.

• A. Only I and II follow
• B. Only II and III follow
• C. Only I and III follow
• D. All follow✓

Explanation: Literacy at basic level is the utmost need to prepare good future citizens. So, all children need to be educated. This can be achieved by creating awareness, providing incentives, enforcing education and banning employment of children. Thus, all the three courses follow.

Q175. Statement: Most of the steel producing companies in the country have made considerable profit during the last financial year. Many Asian countries have been importing huge quantities of steel from Pakistan

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

Explanation: The correct answer is that both statements I and II are effects of some common cause. Statement I talks about steel companies making profit, which is likely the effect of the cause mentioned in statement II - many Asian countries importing steel from Pakistan. The explanation shows how the import of steel from Pakistan leads to increased profits for the country's steel companies. The other options are incorrect as they do not capture the cause-effect relationship between the statements accurately.

Q176. Statement:The Management of School M has decided to give free breakfast from next academic year to all the students in its primary section through its canteen even though they will not get any government grant.Courses of Action:I. The school will have to admit many poor students who will seek admission for the next academic year.II. The canteen facilities and utensils have to be checked and new purchases to be made to equip it properly.III. Funds will have to be raised to support the scheme for years to come.

• A. Only II and III follows✓
• B. Only III and I follow
• C. Only I and II follow
• D. Only I follows

Explanation: Since the school has introduced the scheme without receiving any grant, it needs to pool up the funds and make adequate arrangements to keep the scheme going. So, only II and III follow.

Q177. What should come next: save, secure, protect, -----

• A. Guard✓
• B. Lock
• C. Conserve
• D. Humble

Explanation: Save is a synonym of secure so to follow the pattern the answer needs to be a synonym of protect, which is guard. So, the correct option is A.

Q178. Pick the odd word out:

• A. Just
• B. Fair
• C. Equitable
• D. Biased✓

Explanation: The words "just," "fair," and "equitable" are all synonyms related to impartiality and justice. "Biased" is an antonym, meaning to show prejudice for or against something, making it the odd one out.

Q179. Each problem consists of three statements. Based on the first two statements, the third statement may be true, false, or uncertain.Blueberries cost more than strawberries.Blueberries cost less than raspberries.Raspberries cost more than both strawberries and blueberries.If the first two statements are true, the third statement is:

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

Explanation: The first two statements provide a clear hierarchy of fruit prices: blueberries are more expensive than strawberries, and raspberries are more expensive than blueberries. Therefore, raspberries must be more expensive than both strawberries and blueberries, making the third statement true. Option A is correct as it accurately reflects this deduction. Options B and C are incorrect because the cost relationships provided lead to a definite conclusion about the third statement. Option D is also incorrect as it suggests there is no correct option, which contradicts the evidence given.

Q180. Passage: A, P, R, X, S and Z are sitting in a row. S and Z are in the centre, and A and P are at the ends. R is sitting on the left of A. Then who is sitting on the right of P?

• A. A
• B. S
• C. X✓
• D. Z

Explanation: The statement "R is sitting on the left of A" establishes the orientation. If A is at one end and R is to his left, A must be at the right end. This makes P the left end. With S and Z in the center, the only person left to sit between P and the center is X. The order is P, X, (S, Z), R, A.