Sindh Mcat Nts 2016 Duhs And Jsmu — Solved Past Paper with Answers

Q1. Complete the sentences by choosing the most appropriate option, from the given lettered choices below each.The doctor is going _ vaccinate me tomorrow.

• A. Of
• B. To✓
• C. On
• D. As

Explanation: In this specific sentence, "to" is the correct preposition to use with the verb "vaccinate" because it indicates thedirection or purpose of the action. The verb "vaccinate" typically takes the preposition "to" to show that someone is receiving the vaccination. Therefore, "to" is the most appropriatepreposition in this context to indicate that the doctor will be administeringthe vaccination to you.

Q2. 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.

Q3. 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.

Q4. Choose the word most similar in meaning to the capitalized one. PREVENT:

• A. Stop✓
• B. Permit
• C. Verify
• D. Confirm

Explanation: Prevent means to keep (something) from happening. Stop means to prevent or dissuade (someone) from continuing in an activity or achieving an aim. "Stop" and "prevent" both convey the idea of halting or hindering an action from happening or continuing. They both involve taking measures to avoid or put an end to something undesirable.

Q5. Choose the word most similar in meaning to the capitalized one.WICKED:

• A. Good
• B. Evil✓
• C. Excellent
• D. Luxury

Explanation: Wicked means evil or morally wrong. Evil means profoundly immoral and wicked, so very similar. So option B is correct.

Q6. 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.

Q7. Choose the word opposite in meaning to the word in capital letters.DEMONSTRATE:

• A. Show
• B. Display
• C. Conceal✓
• D. Expose

Explanation: Demonstrate means to clearly show the existence or truth of (something) by giving proof or evidence. Conceal means to not allow to be seen; hide, so totally opposite, hence option C is correct.

Q8. Read the passage to answer the question:The names of three men - an Englishman, a German, and an Italian stand out from the many who have opened up for us this new path of progress: James Clerk Maxwell, Heinrich Hertz, and Guglielmo Marconi. Maxwell prophesied wireless, discovering its principles. Hertz discovered and demonstrated those waves which are its secret, and Marconi invented the instruments which put these ideas to practical use. A crowd of other brilliant men have made their different contributions. Sir Oliver Lodge came very near to doing what Marconi did; indeed, a year before Marconi invented his instrument, Lodge demonstrated the possibility of sending a signal by these Hertzian waves but turned aside under pressure from other work. Professor Righi, Marconi's science master, experimented in the laboratory and showed young Marconi the potentialities Oliver Heaviside, the English telegraphic engineer, subsequently revealed to us the amazing fact that these wireless waves are thrown back to the earth from the two curious 'mirrors' which, moving in the depth of the sky hundreds of miles from the surface of the earth, act as a kind of double sounding board. These are but a few of the men whose work all over the earth has brought the miracle of wireless to its present stage. But the three names stand pre-eminent.Which of the following could be the best title of the paragraph?

• A. Contribution of Sir Oliver Lodge
• B. Role of Technology in the Modern World
• C. Discovery of Wireless✓
• D. The Double Sounding Board

Explanation: The paragraph primarily discusses the contributions of James Clerk Maxwell, Heinrich Hertz, and Guglielmo Marconi to the field of wireless communication. It highlights Maxwell's prophesying of wireless, Hertz's discovery and demonstration of the principles behind it, and Marconi's invention of instruments to put these ideas into practical use. While the paragraph briefly mentions the work of other individuals, it emphasizes the significance of these three men in the development of wireless technology.

Q9. Read the passage to answer the question:The names of three men - an Englishman, a German, and an Italian stand out from the many who have opened up for us this new path of progress: James Clerk Maxwell, Heinrich Hertz, and Guglielmo Marconi Maxwell prophesied wireless, discovering its principles. Hertz discovered and demonstrated those waves which are its secret, and Marconi invented the instruments which put these ideas to practical use. A crowd of other brilliant men have made their different contributions. Sir Oliver Lodge came very near to doing what Marconi did; indeed, a year before Marconi invented his instrument, Lodge demonstrated the possibility of sending a signal by these Hertzian waves but turned aside under pressure from other work. Professor Righi, Marconi's science master, experimented in the laboratory and showed young Marconi the potentialities Oliver Heaviside, the English telegraphic engineer, subsequently revealed to us the amazing fact that these wireless waves are thrown back to the earth from the two curious 'mirrors' which, moving in the depth of the sky hundreds of miles from the surface of the earth, act as a kind of double sounding board. These are but a few of the men whose work all over the earth has brought the miracle of wireless to its present stage. But the three names stand pre-eminent.Which of the following is mentioned in the paragraph as the contribution of Heinrich Hertz?

• A. Invented the instruments for the practical use of wireless waves
• B. Demonstrated the possibility of sending signal
• C. That the waves are thrown back to the earth
• D. Discovered and demonstrated the wireless waves✓

Explanation: The paragraph states that “Hertz discovered and demonstrated those waves which are its secret”, which shows option D is correct. Instruments to use wireless waves were invented by Gugliclmo Marconi Maxwell.

Q10. Leukaemia is associated with an increased number of _ in blood:

• A. Erythrocytes
• B. Basophils
• C. Eosinophils
• D. Leucocytes✓

Explanation: Leukemia is a type of cancer that affects the blood and bone marrow, leading to the overproduction of abnormal white blood cells. It is caused by genetic mutations in the DNA of blood cells, which disrupt the normal process of cell growth and division.

Q11. Scurvy is caused by deficiency of vitamin _.

• A. A
• B. B
• C. C✓
• D. D

Explanation: Scurvy is caused due to a deficiency of vitamin C (ascorbic acid). Vitamin C plays a crucial role in the synthesis of collagen, which is essential for the proper functioning of blood vessels, connective tissues, and wound healing. The deficiency of vitamin C leads to the breakdown of collagen, resulting in symptoms of scurvy such as fatigue, weakness, joint pain, and gum bleeding. However, deficiencies of vitamins A, B, and D do not result in scurvy. Vitamin A is essential for vision and immune function, vitamin B encompasses a group of vitamins involved in various aspects of metabolism and nervous system function, and vitamin D is crucial for calcium absorption and bone health. While deficiencies of these vitamins can cause other health issues, they do not directly contribute to the development of scurvy.

Q12. The optimum pH for the action of pepsin and salivary amylase is

• A. 3.5 and 7.7 respectively
• B. 1.8 and 6.8, respectively✓
• C. 6.8 and 7.4 respectively
• D. 1.8 and 7.4 respectively

Explanation: The stomach has a pH between 1.5 and 3.5 generally, and this is due to the cells in the stomach releasing hydrochloric acid, which is where pepsin is present, so it has a low optimum pH of around 2, so options A and C are incorrect. Saliva has a pH normal range of 6.2-7.6, with 6.7 being the average pH, which means salivary Amylase should have an optimum pH around the same range. Amylase has an optimal pH between 6.4 and 7.0.

Q13. The diagram shows an apparatus set up to investigate the effect of changing the concentration of glucose in the surrounding solution on the movement of molecules through a selectively permeable membrane (Visking tubing) in 15 minutes. As the concentration of glucose solution in the surrounding solution increases, which of the following statements are correct? Net diffusion of water increases. Glucose molecules reach an equilibrium quicker. There is less change in the volume of the surrounding solution. Net diffusion of glucose increases.

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

Explanation: Looking at the diagram, it is apparent that increasing glucose concentration in the surrounding tube will decrease the difference in the concentration of glucose between the surrounding liquid and inside the visking tubing. As there is less concentration gradient, the net diffusion of water would decrease so statement I is incorrect. The time to reach equilibrium would decrease as there is less difference in concentration of glucose, so statement II is correct. As the concentration gradient is less, less net movement of water would be needed to reach equilibrium, so volume would also change less, hence statement III is correct. There is no movement of glucose molecules, so statement IV is incorrect.

Q14. Cell X contains 24 chromosomes. It divides by mitosis to produce cells Y and Z. How many chromosomes does cell Z contain?

• A. 12
• B. 24✓
• C. 46
• D. 48

Explanation: Mitosis is a process of cell duplication, in which one cell divides into two genetically identical daughter cells. In the various stages of mitosis, the cell's chromosomes are copied and then distributed equally between the two new nuclei of the daughter cells. So cells Y and Z will have the same amount of chromosomes as the parent cell i.e. Cell X. So option B is correct and options A, C, and D are incorrect. Option A would be correct if cell X had undergone meiosis which divides the number of chromosomes into two. Option D would be after DNA replication but before mitosis, where the double amount of chromosomes is present temporarily.

Q15. One type of congenital heart defect is called atrial septal defect (ASD). Which of the following rows describes the effects of ASD on blood pressure and oxygenation?

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

Explanation: In ASD, blood moves from left to right atrium, increasing pulmonary artery pressure, decreasing systemic aorta pressure, and increasing oxygen saturation in pulmonary artery.

Q16. The condition due to loss of appetite for food is termed as:

• A. Anorexia nervosa✓
• B. Bulimia nervosa
• C. Colour blindness
• D. Retarded growth

Explanation: Anorexia nervosa is a serious eating disorder characterized by an abnormally low body weight, loss of appetite, an intense fear of gaining weight, and a distorted body image. People with anorexia nervosa place a high value on controlling their weight and shape, using extreme efforts that tend to significantly interfere with their lives.

Q17. The diagram shows some of the organs of the human body. In which organs does the digestion of carbohydrates take place?

• A. P and Q
• B. P and R✓
• C. Q and R
• D. Q and S

Explanation: Digestion of carbohydrates takes place in the mouth (P) with the help of salivary amylase which converts starch to maltose. It also occurs in the intestine(R) with the help of pancreatic amylase, which yields disaccharides from starch by digesting the glycosidic bonds. The disaccharides produced (maltose, maltotriose, and α-dextrins) are all converted to glucose by brush border enzymes. Q is the stomach where protein digestion starts. S is the liver where digestion does not take place, although it does aid in digestion by producing bile and enzymes. So option B is correct (P & R)(Mouth and small intestine).

Q18. Which of the following are animals with pointed ends and a tube-like digestive tract?

• A. Annelida
• B. Arthropoda
• C. Nematoda✓
• D. Mollusca

Explanation: Nematodes are known for having tapered ends (i.e. pointy ends). Arthropods, mollusca and annelids are not known or identified by their shape having pointed ends (and organisms under these generally do not have tapered ends). So Option C is correct.

Q19. The dark reactions of photosynthesis are characterized by:

• A. Synthesis of ATP, O2 and NADH
• B. Utilization of ATP, CO2 and NADPH✓
• C. Electron transfer from NADPH to RuBP
• D. Carbon dioxide transfer from RuBP to glucose

Explanation: Dark reaction is also called carbon-fixing reaction. It is a light-independent process in which sugar molecules are formed from carbon dioxide, water molecules, ATP, and NADPH(products from the light-dependent reaction). The dark reaction occurs in the stroma of the chloroplast, where they utilize the products of the light reaction.

Q20. Inheritance of acquired characteristics is based on:

• A. Genetic role in reproduction
• B. Use and disuse of organ✓
• C. Survival of the fittest
• D. Mutations

Explanation: Inheritance of acquired traits is when a parent uses a organ more/less than usual, hence changing certain aspects of it (for e.g. Giraffes stretched their necks to reach leaves on a tall tree, which was inherited by the offsprings leading to giraffes having longer necks), so inheritance of acquired characteristics is related to the use or disuse of organs, so Option B is correct. Physical traits rarely have a genetic role in inheritance (i.e. the genes that give rise to that trait are passed on and so play a part in reproduction but the trait itself does not have a genetic role). Survival of the fittest is related to natural selection and how frequently the trait is passed on to offsprings. Mutations are random changes in DNA sequence and so is not an acquired trait. So Option A, C and D are incorrect.

Q21. The family tree shows the inheritance of the ability to taste a certain substance. The allele for the ability to taste this substance is dominant to the allele for the inability to taste it. What percentage of children of Pasha and Hina would be ‘non-tasters’?

• A. 25%✓
• B. 50%
• C. 75%
• D. 100%

Explanation: The ability to taste the substance is dominant (T) and the inability to taste is recessive (t). If Pasha and Hina are both heterozygous (Tt), their children can have the following genotypes: TT → tasterTt → taster tT → taster (same as Tt)tt → non-tasterThe genotype ratio is 1 TT : 2 Tt : 1 tt, so the phenotype ratio is 3 tasters : 1 non-taster.Percentage of non-tasters: 25%.Conclusion: 25% of Pasha and Hina’s children are expected to be non-tasters.

Q22. The first step in nitrogen cycle is:

• A. Nitrification
• B. Ammonification✓
• C. Oxidation
• D. Denitrification

Explanation: While there is no technical ‘start’ of a ‘cycle’, looking at the steps in the choices, one can say that Ammonification needs to happen first, before all the others. First Ammonification would occur first step to make nitrogen usable by plants, to make ammonia, which then undergoes nitrification, forming nitrates. Then, denitrification occurs, in which nitrogen is formed back from nitrates, and so nitrogen is returned to the atmosphere. Oxidation is the same as nitrification. So the correct answer is B.

Q23. Which of the following enzymes are used to join bits of DNA?

• A. Ligase✓
• B. Primase
• C. DNA polymerase
• D. Endonuclease

Explanation: DNA ligase is a type of enzyme that catalyses the formation of a phosphodiester bond between two DNA molecules. This is the same type of bond that links the nucleotides in DNA. DNA ligase is used in a variety of cellular processes, including DNA replication, DNA repair, and recombinant DNA technology.Primase, DNA polymerase, and endonuclease are all involved in DNA replication, but they are not involved in joining bits of DNA.Primase is an enzyme that synthesises short RNA primers, which are needed to initiate DNA replication.DNA polymerase is an enzyme that synthesises DNA molecules from a DNA template.An endonuclease is an enzyme that cuts DNA molecules at specific sites.

Q24. Which of the following correctly describes the thermoregulation in hot temperature?

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

Explanation: In hot temperatures, the body needs to cool down, which is done by increased blood flow to the skin, so that more heat is released from the blood as well as sweat production is increased, so that when the sweat evaporates it cools down the body, hence option A and C is incorrect. The blood flow is increased by dilating the vessels which increases their diameter hence allowing more blood to flow, so options A and B are incorrect. The option with the correct combination is option D.

Q25. Which row correctly shows the areas of the gas exchange system that contain cartilage, ciliated epithelium and goblet cells?

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

Explanation: It is a fact that bronchioles don't have cartilage, so options A and C are incorrect. Goblet cells are mucus secretory cells in the gas exchange system, and since alveoli don't contain mucus they have no need for goblet cells, so option B is also incorrect. This leaves option D which is the correct answer.

Q26. Match the hormones listed under column I with their functions listed under the column. Choose the answer which gives the correct combination of the alphabets of the II columns.

• A. A = s, B = r, C = p, D = q✓
• B. A = t, B = p, C = s, D = r
• C. A = s, B = q, C = r, D = t
• D. A = t, B = r, C = p, D = s

Explanation: One of the functions of oxytocin is to stimulate uterine contractions in labor and childbirth, so the correct letter is S for A (A=S), so options B and D are incorrect. Prolactin causes the breasts to grow and make milk during pregnancy and after birth, so the correct letter is R for B (B=R), so option C is also incorrect. LH triggers the release of an egg from the ovary i.e. ovulation. Progesterone helps to prepare the body for pregnancy by stimulating glandular development and the development of new blood vessels. This provides a good environment for implantation by a fertilized egg, so it aids in implantation. So option A is correct.

Q27. During which stage of meiosis do centromeres divide?

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

Explanation: Centromeres divide in Anaphase II of meiosis, allowing sister chromatids to separate and move to opposite poles. This process is crucial for producing genetically unique gametes. In contrast, during meiosis I, homologous chromosomes are separated, and centromeres remain intact. Therefore, options A (Prophase I) and B (Metaphase I) are incorrect as centromere division does not occur in meiosis I. Option C (Prophase II) is also incorrect because it precedes the division of centromeres; it is the stage where chromosomes prepare for final separation.

Q28. Phenylketonuria is:

• A. Sex linked dominant trait
• B. Sex linked recessive trait
• C. Autosomal dominant trait
• D. Autosomal recessive trait✓

Explanation: Phenylketonuria (PKU) is a rare inherited disorder that causes an amino acid called phenylalanine to build up in the body. PKU is caused by a change in the phenylalanine hydroxylase (PAH) gene. This gene helps create the enzyme needed to break down phenylalanine. PKU is inherited from a person's parents. The disorder is passed down in a recessive pattern, which means that for a child to develop PKU, both parents have to contribute a mutated version of the PAH gene. So option A and C are incorrect. It is found in chromosome 12 i.e. it is autosomal, so Option B is incorrect, and option D is correct.

Q29. Pseudopodia are:

• A. False feet developed in some unicellular organisms✓
• B. Small hairlike structures present on the unicellular organisms
• C. Long, tube like structures coming out of the mouth
• D. Suckers which are attached to the walls of intestines

Explanation: A pseudopod or pseudopodia is a temporary arm-like projection of a eukaryotic cell membrane that emerges in the direction of movement. The hair-like projections on unicellular organisms are called Cilia. The long tube-like structure connecting the mouth and the stomach is called the oesophagus. The head of some parasites contains structures, suckers which enable the parasite to attach to the gut wall. So option A is correct

Q30. Two animals are mated. One is homozygous dominant for one character and homozygous recessive for another. The other animal Is heterozygous for both characters. How many phenotypes are expected in the offspring of this cross?

• A. 1
• B. 2✓
• C. 3
• D. 4

Explanation: To understand and solve this, you can divide your thinking into two. For character 1. When genes are passed down to offspring, when the genes of this character pass down, the Dominant gene will always be present, so only one phenotype is possible from this character. For character 2. Three genotypes are possible, RR, Rr, and rr. RR and Rr are one phenotype and rr another. Total 2 So 2*1=2 total phenotypes

Q31. Part of the amino acid sequences in normal and sickle cell hemoglobin are shown. Normal haemoglobin (Thr - pro - glu - glu)Sickle cell haemoglobin (thr - pro - val - glu)The mRNA codons for these amino acids are Glutamine (glu) GAA GAG, Threonine (thr) ACU ACC, Proline (pro) CCU CCC,Valine (val) GUA GUG. Which transfer RNA molecule is Involved in the formation of this part of the sickle cell haemoglobin?

• A. GUG
• B. CAU✓
• C. UGC
• D. GAG

Explanation: The transfer RNA has complementary base pairs to mRNA. The unique amino acid in sickle cell is valine, the mRNA codons of valine are GUG and GUA, the complementary bases are CAC and CAU (Guanine to Cytosine, Uracil to Adenine and vice versa). So option B is correct.

Q32. Which of the following statements correctly describes homologous chromosomes?

• A. They are formed during meiosis.
• B. They are held together by centromeres.
• C. They are chromatids of the same chromosome.
• D. They carry both morphologically similar members with same set of genes.✓

Explanation: The homologous chromosomes are made before meiosis, so option A is incorrect. Early in prophase I, homologous chromosomes come together to form a synapse. The chromosomes are bound tightly together and in perfect alignment by a protein lattice, so option B is incorrect. The question talks about chromosomes, not chromatids, so option C is incorrect. Homologous chromosomes carry the same genes and hence are morphologically similar. So option D is correct.

Q33. Movement of ions and large molecules with the help of protein molecules in and out of the cell is called _

• A. Diffusion
• B. Facilitated diffusion✓
• C. Passive transport
• D. Osmosis

Explanation: Diffusion is the passive movement of molecules down the concentration gradient whereas osmosis is the passive movement of water down the water potential gradient across a semi-permeable membrane. On the contrary, facilitated diffusion is the passive movement of molecules down the concentration gradient with the help of transport proteins (as these molecules/ions are incapable of moving through cell membranes directly).

Q34. Molluscs have an exoskeleton made up of:

• A. Protein
• B. Silica
• C. CaCO3✓
• D. Cuticles

Explanation: While mollusca exoskeleton does indeed contain proteins as well as calcium carbonate(CaCO3), they are composed mostly of calcium carbonate with only a small quantity of protein--no more than 2 percent. These shells, unlike typical animal structures, are not made up of cells (so no cuticle). So options A, B, and D are incorrect as it does not contain silica either. Option C is correct.

Q35. Where does the Calvin Cycle occur?

• A. Stroma✓
• B. Cytosol
• C. Thylakoid Membrane
• D. No specific location

Explanation: The Calvin Cycle, also known as the dark reactions or light-independent reactions of photosynthesis, occurs in the stroma of the chloroplasts. The chloroplast is an organelle found in plant cells and some other eukaryotic cells, and it is the site of photosynthesis.

Q36. Match the parts of the human brain listed under Column I with the functions given under Column II. Choose the answer which gives the correct combination of alphabets of the two columns.

• A. A=r, B=q, C=p, D=s
• B. A=r, B s, C=q, D=t,
• C. A=t, B=p, C=q, D=r✓
• D. A=t, B=q, C=p, D=s

Explanation: Your cerebral hemisphere plays a key role in memory, intelligence, thinking, learning, reasoning, problem-solving, emotions, consciousness, and functions related to your senses. (so A=T, hence options A and B are incorrect). The thalamus is your body's information relay station. All information from your body's senses (except smell) must be processed through your thalamus before being sent to your brain's cerebral cortex for interpretation (so B=P, hence options A, B, and D are incorrect). The cerebellum controls voluntary movements such as walking, posture, balance, coordination, eye movements, and speech (C=Q). Neurons in the medulla also control arousal and sleep (D=R).

Q37. Five words are shown below:Labium, Maxillae, Labrum, Mandibles and AntennaeThese words can be used in the spaces P, Q, R, S, and T to complete the sentence below.Cockroaches are omnivorous and can eat any kind of organic matter. They search for their food by their ..P.. Their digestive system is tubular, having a straight slightly coiled digestive tube opening at both the ends. Hence, the digestive system is complete. The mouth lies at the base of the pre-oral cavity which is bounded by the mouth parts, ..Q.. (upper lip), ..R.. (lower lip), mandibles, and maxillae. The ..S.. pick up and bring food to the ..T.. for mastication.

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

Explanation: Roaches use their antennae to locate food and estimate the distance and height of any obstacles in front of them. The labrum is also known as the upper lip. It consists of a flattened cuticle and helps to hold the food in the correct position when the cockroach feeds. The labium is often called an insect's lower lip. The labium is positioned towards the back of the head and helps hold food in place when the insect feeds. Maxillae are used to hold and manipulate food so that it can be chewed or sliced by the mandibles. The function of mandibles is to grasp, crush, or cut the food (mastication means to grind or crush (food) with or as if with the teeth). Using this information, the only correct option is option B.

Q38. RuBP + O2 ---Rubisco---> PGA + _ .

• A. Phosphoglycerate
• B. Phosphoglycolate✓
• C. Glycerate
• D. Glycolate

Explanation: When RuBP reacts with oxygen instead of CO₂ in a process called photorespiration, RuBisCO catalyzes the formation of one molecule of 3-phosphoglycerate (PGA) and one molecule of 2-phosphoglycolate (Phosphoglycolate). This process occurs when plants have less CO2 but high O2.

Q39. What volume in dm3 of KCl is obtained in the following equation if the volume of KCIO3 at the beginning of the reaction is 20 dm3? 2KClO3 → 2KCl + 3O2

• A. 2 dm3
• B. 74.5 dm3
• C. 50 dm3
• D. 40 dm3
• E. 20 dm3✓

Explanation: To determine the volume of KCl obtained in the given equation, we need to know the stoichiometric ratio between KCIO3 and KCl. From the balanced equation: 2KClO3 → 2KCl + 3O2 We can see that for every 2 moles of KCIO3, we obtain 2 moles of KCl. Therefore, the ratio is 1:1. Given that the volume of KCIO3 at the beginning of the reaction is 20 dm3, and the stoichiometric ratio is 1:1, the volume of KCl obtained will also be 20 dm3. Therefore, the volume of KCl obtained in the given equation is 20 dm3.

Q40. J.J Thomson determined the:

• A. Charge of an electron
• B. Mass of an electron
• C. e/m value of an electron✓
• D. Charge of a proton
• E. Mass of a proton

Explanation: Thomson was able to deflect the cathode ray towards a positively charged plate deduce that the particles in the beam were negatively charged. Then Thomson measured how much various strengths of magnetic fields bent the particles. Using this information Thomson determined the mass to charge ratio of an electron.All other options are incorrect because those were not discovered by J.J thomson.

Q41. The number of electrons present in n = 2, l = 1 and m = -1, 0, +1 are:

• A. 6✓
• B. 2
• C. 8
• D. 10
• E. 18

Explanation: n is the principal quantum number and indicates the number of shells. n = 2 indicates that the second shell is being discussed. l is the azimuthal quantum number and indicates the number of subshells. l = 0 means s subshell, l = 1 means p subshell, and so on. m is the magnetic quantum number and indicates the orientation of orbitals in the subshells. m ranges from -1 to +1 through 0. Since m = -1, 0, +1, this means that all three sub-orbitals; 2px, 2py, 2pz are taken into consideration. Since all of these have 2 electrons each, the total number of electrons in them is 6.

Q42. If the difference between electronegativity values between two atoms is less than 1.7 , the bond is necessarily:

• A. Ionic
• B. Covalent✓
• C. Electrovalent
• D. Polar
• E. Non-polar

Explanation: If the electronegative difference is >1.7 then its nature is ionic and <1.7 then its nature is covalent and if its equal to 1.7 then nature of bond will be 50% ionic and 50% covalent. As here, it is less than 1.7, nature is covalent.

Q43. Which one of the following has the lowest % ionic character?

• A. CCl4✓
• B. BCl3
• C. BeCl2
• D. LiCl
• E. HCl

Explanation: Electronegativity increases as we go from left to right in a period. So, the electronegativity difference between elements and chlorine decreases. Ionic character decreases with a decrease in the value of electronegativity difference between elements. thus the correct order is: CCl4<BCl3<BeCl2<LiCl.

Q44. Consider the following reactionN2 (g) + O2 (g) < - > 2NO (g) Kc - 0.1 at 2000 CIf the original concentration of N2 and O2 were 0.1 M each. Calculate the concentrations of NO at equilibrium.

• A. 0.028 M✓
• B. 0.0012 M
• C. 0.18 M
• D. 0.0018 M
• E. 0.002 M

Explanation: Explanation is given below.

Q45. The geometry of H2S and its dipole moment are:

• A. Angular and non-zero✓
• B. Angular and zero
• C. Linear and zero
• D. Linear and non-zero

Explanation: H2S has a bent geometry according to VSEPR. It has 2 lone pairs and 2 bond pairs and a bond angle of 92o degrees. The dipole moment is non-zero because sulfur is more electronegative than hydrogen and makes the molecule slightly polar. Thus, the geometry of H2S and its dipole moment are Angular and non-zero.

Q46. Which one of the following is most abundant in the earth crust?

• A. Al✓
• B. B
• C. In
• D. Ga

Explanation: Al is the most abundant element among the elements mentioned above.

Q47. Which of the following has the maximum number of unpaired d electrons?

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

Explanation: The correct answer is Fe2+. The electron configuration for Fe2+ is [Ar] 3d6, with the 3d subshell holding six electrons: four of these are unpaired due to Hund's rule, which states that electrons fill degenerate orbitals singly first. Mg2+ lacks d electrons entirely, Ti3+ has only one unpaired d electron ([Ar] 3d1), and V3+ has two unpaired d electrons ([Ar] 3d2).

Q48. CH3CH2CH2COCH3 is the functional isomer of:

• A. CH3CH2CH2CH2CHO✓
• B. CH3CH2CH2CH2OH
• C. CH3CH2COCH2CH3
• D. CH3 - (CH)2 COCH2CH3

Explanation: CH3CH2CH2COCH3 and CH3CH2CH2CH2CHO have the same number of atoms for the constituent elements, Carbon, Oxygen, and Hydrogen. However, the second requirement of a 'Functional Isomer' is that the functional groups, of the two isomers, must be different and since, in this case, we have a ketone functional group and an aldehyde functional, both compounds can be positively termed as 'Functional Isomers' of each other.Thus, option A is correct.

Q49. IUPAC name of this is (CH3)2 - CH - CH (C2H5) - C(CH3)3 :

• A. 3-ethyl-2,2,4-trimethylpentane✓
• B. 4-ethyl-2,2,4-trimethylpentane
• C. 5-ethyl-2,2,4-trimethylpentane
• D. 2-ethyl-2,2,4-trimethylpentane
• E. 1-ethyl-2,2,4-trimethylpentane

Explanation: The IUPAC name for the given compound (CH3)2-CH-CH(C2H5)-C(CH3)3 is "3-ethyl-2,2,4-trimethylpentane." Here's how the name is derived:1. Identify the longest continuous carbon chain, which in this case is a five-carbon chain.2. Number the carbon chain to give the substituents the lowest possible locants. In this case, we start numbering from the end nearest to the ethyl group, so the ethyl group is located on carbon 3.3. Assign substituent names and locants. The methyl groups attached to carbon 2 and carbon 4 are called "2,2-dimethyl" and "2,4-dimethyl," respectively. The ethyl group attached to carbon 3 is named "ethyl."4. Combine all the parts to form the complete IUPAC name: "3-ethyl-2,2,4-trimethylpentane."Therefore, the compound (CH3)2-CH-CH(C2H5)-C(CH3)3 is named 3-ethyl-2,2,4-trimethylpentane.

Q50. Metaboric acid when heated produces:

• A. B2O3
• B. HBO2
• C. H2B4O7✓
• D. H3BO3

Explanation: This is the following solution:

Q51. Amylopectin has _ linkage.

• A. α 1 - 4 glycosidic
• B. β 1 - 4 and 1 - 6 glycosidic
• C. α 1-4 and α 1 - 6 glycosidic✓
• D. β 1 - 4 glycosidic

Explanation: Amylopectin has both α 1-4 and 1 - 6 glycosidic bond linkages.

Q52. Which one is not a monosaccharide?

• A. Galactose
• B. Fructose
• C. Mannose
• D. Lactose✓

Explanation: Lactose is not a monosaccharide but a disaccharide,as shown below:

Q53. All are enzymes involved in fermentation process except one:

• A. Zymase
• B. Invertase
• C. Lipase✓
• D. Diastase

Explanation: Fermentation mainly involves enzymes that break down carbohydrates, such as zymasa, invertase, and diastase. Lipase acts on fats, not sugars. Therefore, lipase is not typically involved in the fermentation process.

Q54. For which mechanism the first step involved is the same?

• A. E1 and E2
• B. SN1 and E2
• C. E1 and SN2
• D. E1 and SN1✓

Explanation: In the first step, the bond between the carbon atom and the leaving group breaks to produce a carbocation and, most commonly, an anionic leaving group. This is the same in SN1 and E1 reactions.

Q55. All are dehydrating agents for alcohols to produce alkene except:

• A. Al2O3
• B. P2O5
• C. Dilute H2SO4✓
• D. H3PO4

Explanation: Dilute H2SO4, on the other hand, generally does not promote dehydration; instead, concentrated H2SO4 is the form that acts as a strong dehydrating agent.

Q56. At the end of the periodic table _ are found.

• A. S block elements
• B. P block elements
• C. D block elements
• D. Noble gases✓

Explanation: At the end of the periodic table noble gases are found as shown below:

Q57. Which of the following elements is needed to add in the given diagram to make it aromatic benzene?

• A. -H✓
• B. -OH
• C. -CH
• D. -H2

Explanation: There is -H missing on one of the carbon atoms to make it a benzene ring.

Q58. An example of a solution not in equilibrium is:

• A. Chemical pH indicator
• B. Acid/base buffer
• C. Anhydrous solution
• D. Hypotonic solution✓

Explanation: A hypotonic solution can cause water to move across a semipermeable membrane (like into a cell), creating an osmotic imbalance. This movement continues until equilibrium is reached, so initially, it is not in equilibrium

Q59. The radius of the third Bohr orbit is:

• A. 0.529 Å
• B. 0.529 / 4 Å
• C. 0.529 x 2 Å
• D. 0.529 x 32 Å✓

Explanation: Radius of the 3rd Bohr Orbit,r = 0.529n2 Å = 0.529×32 Å

Q60. The change in concentration of reactant or product per unit time is called:

• A. Rate constant
• B. Rate of reaction✓
• C. Rate equation
• D. Rate law
• E. Both A & D

Explanation: The rate of reaction measures how fast reactants are consumed or products are formed. It is defined as the change in concentration per unit time.Thus, option B is the correct definition.

Q61. The internal energy of an ideal gas depends upon its:

• A. Molecular size
• B. Pressure only
• C. Volume only
• D. Temperature only✓

Explanation: In an ideal gas, the inter-molecular collisions are assumed to be absent and the collisions are perfectly elastic. Thus, the gas possesses only translational kinetic energy and hence the internal energy of the ideal gas depends only on temperature.

Q62. Indicate the number of chain isomers that can be obtained from the C6H14?

• A. 7
• B. 6
• C. 5✓
• D. 4

Explanation: Hexane (C6H14) is an alkane with six carbon atoms. Alkanes are hydrocarbons with only single bonds between carbon atoms. The general formula for an alkane with n carbon atoms is CnH2n+2. The correct answer is C, as five different chain isomers can be obtained from C6H14 (hexane). These isomers arise from the varying arrangement of carbon atoms in the chain, including straight-chain and branched-chain isomers. Remember that the no. of isomers for carbon 1 and 2 of alkane is 1. The trick to finding the total possible number of chain isomers for Alkanes up to C10 is given below.

Q63. The ground state of an atom corresponds to a state of :

• A. Maximum energy
• B. Minimum energy✓
• C. Negative energy
• D. Positive energy

Explanation: The nucleus of an atom is surrounded by electrons that occupy shells or orbitals of varying energy levels. The ground state of an electron, the energy level it normally occupies, is the state of lowest energy for that electron.

Q64. A sample of gas weighs 1.25 g, at 28 C occupying a volume of 2.50 x 102 ml and its pressure is 715 torr. What is the molar mass in grams?

• A. 131.1 g/mol✓
• B. 1.311 g/mol
• C. 1.212 g/mol
• D. 122.1 g/mol

Explanation: Mass of gas = 1.25 gThe volume of gas = 2.50 x 102 ml= 0.25 LPressure of gas = 715 torrThe temperature of gas = Using the ideal gas law,PV=nRTPV=(m/M)RTM=mRT/PVwhere,M = Molar mass of gasW = Mass of gasR = gas constantT = temperature of the gasP = Pressure of gasV = Volume of gasNow put all the given values in the above formula, and we get the value of the molar mass of gas.M=(1.25)(62.364)(1.25)/(715)(0.25)M=131.1g/mol

Q65. Common name of Octadecanoic acid is:

• A. Propionic acid
• B. Stearic acid✓
• C. Palmitic acid
• D. Formic acid
• E. Acetic acid

Explanation: Stearic acid (IUPAC systematic name: octadecanoic acid is one of the useful types of saturated fatty acids that comes from many animal and vegetable fats and oils.

Q66. Shape of the orbital occupied by an electron is defined by:

• A. Azimuthal quantum number✓
• B. Principal quantum number
• C. Magnetic quantum number
• D. Spin quantum number

Explanation: This is the correct answer. The azimuthal quantum number, also known as the angular momentum quantum number, determines the shape of an atomic orbital. It is denoted by the letter l. The possible values of l depend on the principal quantum number (n). For a given n, l can have values from 0 to n-1. Each value of l corresponds to a different orbital shape: l = 0: s orbital (spherical) l = 1: p orbital (dumbbell-shaped) l = 2: d orbital (cloverleaf-shaped)l = 3: f orbital (more complex shapes)

Q67. Which structures show primary alcohol that cannot be dehydrated to form an alkene? CH3OH CH3CH2OH CH3CH(OH)CH3

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

Explanation: For alkene, we need to have a double bond between two carbon atoms. CH3OH has only one carbon atom, so even if it is dehydrated it can not form an alkene. Hence it is our answer.

Q68. A catalyst:

• A. Increases the rate of the forward reaction only
• B. Increases the rate of both forward and reverse reactions✓
• C. Changes the equilibrium position
• D. Increases the rate of the reverse reaction only

Explanation: A catalyst increases the rate of both forward and reverse reaction, making the system reach equilibrium faster.

Q69. In SI units, the unit of Candela is used for the measurement of:

• A. Heat intensity
• B. Light intensity✓
• C. Energy/time
• D. Energy/area

Explanation: The Candela is specifically a unit for measuring luminous intensity, which is the amount of light emitted in a particular direction. This measurement takes into account the sensitivity of the human eye to different wavelengths of light. Luminous intensity is important in applications where the visual perception of light is crucial, such as lighting design, optics, and photometry.

Q70. What is the projection of E = 2i-3j+6k onto the direction of vector F=i+2j+2k?

• A. 1/2
• B. 8/3✓
• C. 3/5
• D. 5/7
• E. 7/9

Explanation: The explanation is as follows:To find the projection of vector E onto the direction ofvector F, we can use the formula: Projection of E onto F = (E dot F) / |F| First, let's calculate the dot product of E and F: E dot F =(2)(1) + (-3)(2) + (6)(2) = 2 - 6 + 12 = 8 Next, we need to find the magnitude of vector F: |F| = √(1² + 2² + 2²) = √9 = 3 Now we can calculate the projection: Projection of E onto F= (E dot F) / |F| = 8 / 3 ≈ 2.67 Therefore, the projection of vector E onto the direction ofvector F is approximately 2.67.

Q71. Two bodies 'X' and 'Y' are attached to the ends of a string which passes over a pulley so that the two bodies hang vertically. If the mass of the body 'X' is 10 kg and that of 'Y' is 9.8 kg. Find the acceleration? (g = 9.8 m/s2)

• A. 0.2 m/s2
• B. 2.2 m/s2
• C. 3.2 m/s2
• D. 4.2 m/s2
• E. 0.10 m/s2✓

Explanation: Exp;a= (m1-m2)g/m1+m2a=(10-9.8)(9.8) / 10+9.8a=0.09a=0.10m/s^2

Q72. A 400 gram ball is tied to the end of a cord and whirled in à horizontal circle of radius 0.6 m. If the ball makes five complete revolutions in 2 s, what is the ball's linear speed?

• A. 4.42 m/s
• B. 5.42 m/s
• C. 7.42 m/s
• D. 8.42 m/s
• E. 9.42 m/s✓

Explanation: Linear speed = Distance / TimeDistance = 5 × 2 × pi × r = 5 × 2 × 22/7 × 0.6 = 132 / 7 = 18.85 mspeed = 18.85 / 2 = 9.42 m/s

Q73. Find the gravitational force of attraction between two balls each weighing 10 kg, when placed at a distance of 1 meter apart.

• A. 6.674 x 10-9 N✓
• B. 7.71 x 10-9 N
• C. 8.91 x 10-9 N
• D. 9.91 x 10-9 N
• E. 10.31 x 10-9 N

Explanation: F = (G * m1 * m2) / r2 where F is the gravitational force, G is the gravitationalconstant (approximately 6.674 × 10-11 N m2/kg2), m1 and m2 are the masses ofthe objects and r is the distance between their centers. Given that both balls weigh 10 kg and are placed 1 meter apart,we can substitute the values into the formula: F = (6.674 × 10-11 N m2/kg2 * 10 kg * 10 kg) / (1 m)2 F = (6.674 × 10-11 N m2/kg2 * 100 kg2) / 1 m2 F = 6.674 × 10-9 N Therefore, the gravitational force of attraction between thetwo balls weighing 10 kg each, when placed 1 meter apart, is approximately6.674 × 10-9 Newtons.

Q74. When a 7000 N elevator moves from street level to the top of a building 300 m above the street level, what is the change in gravitational potential energy?

• A. 1.1 x 106 J
• B. 2.1 x 106 J✓
• C. 3.1 x 106 J
• D. 4.1 x 106 J
• E. 5.1 x 106 J

Explanation: Change in potential energy=Work doneChange in potential energy= F.d=7000x300=2100000 J2.1x106 J

Q75. A body of mass 4 kg attached to a spring is displaced through 0.04 m from its equilibrium position and then released. If the spring constant is 400 N/m, find the time period of the vibration?

• A. 4.567
• B. 3.416
• C. 2.315
• D. 1.325
• E. 0.628✓

Explanation: This is the solution to this question: The time period of vibration can be calculated using the formula: T = 2π * √(m/k) where T is the time period, m is the mass of the body and kis the spring constant. In this case, the mass of the body is 4 kg, and the springconstant is 400 N/m. T = 2π * √(4 kg / 400 N/m) T = 2π * √(0.01 s^2/kg) T = 2π * 0.1 s T ≈ 0.628 s Therefore, the time period of vibration for the bodyattached to the spring is approximately 0.628 seconds.

Q76. What will be the position of the object, when a convex lens of focal length 20 cm, is used to form an erect image which is twice as large as the object.

• A. 10 cm✓
• B. 20 cm
• C. 30 cm
• D. 40 cm

Explanation: This is the correct answer. To form an erect image twice as large as the object using a convex lens, the object must be placed between the focal point and the optical center of the lens. Focal length (f): 20 cm Magnification (m): 2 (since the image is twice as large as the object)Using the lens formula:1/v - 1/u = 1/fWhere: v is the image distance u is the object distanceSince the magnification is positive for an erect image:m = v/u = 2Therefore: v = 2uSubstituting this into the lens formula: 1/(2u) - 1/u = 1/20 cmSolving for u:u = 10 cmTherefore, the object must be placed 10 cm from the lens to form an erect image twice as large as the object.

Q77. A conductor carrying a current 'I' has length 'L'. When it is placed in a magnetic field 'B' at 90°, it experiences a force:

• A. BIL✓
• B. B²LI
• C. BLI²
• D. Zero
• E. Infinity

Explanation: The formula for the force experienced by a current-carrying conductor is BILsin(θ), where theta represents the angle between the current, I, and the magnetic field, B, hence sin(90)=1 and the value of the force will be BIL (maximum).

Q78. Find the heat transferred to or from the system when a thermodynamics system undergoes a process in which its internal energy decreases by 300 J. If at the same time, 120 J of work is done by the system.

• A. -120 J
• B. -220 J
• C. -320 J
• D. -180 J✓
• E. -520 J

Explanation: U=W+q -300=q-120 q= -180 J

Q79. 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

Q80. An object whose mass is 100 g starts from rest and moves with constant acceleration of 20 cm/s2. At the end of 8 s its momentum is _ in g cm/s.

• A. 500
• B. 8000
• C. 16000✓
• D. 33000
• E. 64000

Explanation: First, we convert the mass from grams to kilograms: 100 g = 0.1 kg. Then, using the equation v=u+at, we find the final velocity to be 1.6 m/s. Calculating the momentum using momentum = mass * velocity, we get 0.1 * 1.6 = 0.16 kg m/s = 16000 gram cm/s. Therefore, the correct momentum at the end of 8 seconds is 16000 gram cm/s. Option C is the correct answer.

Q81. A 3-cm length of wire is moved at right angles across a uniform magnetic field with a speed of 2.0 m/s . If the flux density is 5.0 teslas, what is the magnitude of the induced e.m.f?

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

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

Q82. Two capacitors C1 ( 12 uf ) and C2 (24 uf) are in series connected across a 360 volts d.c. supply. Calculate the charges on C1 and C2 respectively.

• A. 2880 x 10-6 C , 2880 x 10-6✓
• B. 4770 x 10-6 C, 4770 x 10-6
• C. 5810 x 10-6 , 6610 x 10-6
• D. 7170 x 10-6 , 8140 x 10-6
• E. 9090 x 10-6 , 8880 x 10-6

Explanation: Charge throughout the circuit is the same so either the answer is A or B, as only these options have the same charge throughout.Q=CVCapacitance in series can be calculated as:1/C=1/C1 + 1/C21/C=1/12 + 1/241/C=2+1/241/C=⅛C=8 microfarads.->Q=CVQ=(8x10^-6)(360) Q=2880 x 10-6 C

Q83. When a transformer is connected to 120-volt ac, it supplies 3000 V to a device. The current through the secondary winding then is o.06 A and the current through the primary is 2 A. The number of turns in the primary winding is 400. The number of turns in the secondary winding is:

• A. 16
• B. 30
• C. 1000
• D. 2000
• E. 10000✓

Explanation: Turns(primary)/V(primary)=Turns(secondary)/V(secondary)400 turns / 120 V = X turns / 3000 VX = 3000 * 400 / 120 = 10,000 turns

Q84. A car going around a certain curve at a speed of 25 km/h has centripetal force of 100 N acting on it. If the speed of the car is doubled, the centripetal force:

• A. Is quadrupled✓
• B. Is doubled
• C. Is multiplied by √2
• D. Is reduced to 1/2 of the original value
• E. Is reduced to 1/6 of the original value

Explanation: The formula that will be used here will be: F=(m)(v)2/r. By doubling the speed of the car, the centripetal force should be quadrupled due to the power of 2.

Q85. A handball is tossed vertically upward with a velocity of 19.6 m/s. Approximately how high will it rise?

• A. 15.8 m
• B. 19.6 m✓
• C. 25.6 m
• D. 30 m
• E. 60 m

Explanation: Initial velocity = 19.6 m/s Equations of motion = v = u - gt Height h = ut - 1/2 * g * t^2 v = final velocity at max. height = 0 g = acceleration due to gravity = 9.8 m/s^2 t = time of motion Hence we have, 0 = 19.6 - 9.8t t = 2 s h = 19.6 * 2 - 1/2 * 9.8 * 4 = 19.6 m

Q86. Upon signal of starting the race, the athlete starts from rest with a constant acceleration of 8 m/s2. At the same time, a cyclist traveling at a constant speed of 72 km/h passes the athlete. How far beyond the starting point will the athlete overtake the cyclist and what will be the speed of the athlete at the time when it overtakes the cyclist respectively?

• A. 100 m, 40 m/s✓
• B. 110 m, 50 m/s
• C. 120 m, 60 m/s
• D. 130 m, 70 m/s
• E. 140 m, 80 m/s

Explanation: Working is attached below:

Q87. A stone is thrown horizontally from 2.4 m above the ground at 35 m/s. The wall is 14 m away and 1 m high. At what height will the stone reach when at the wall? Where will the stone land?

• A. 1.62 m, 24.5 m✓
• B. 2.22 m, 31 m
• C. 3.22 m, 41 m
• D. 4.22 m, 51 m
• E. 5.22 m, 61 m

Explanation: A stone is thrown horizontally from 2.4m above the ground with a speed of 35 m/s. time is taken to reach a point which is located 14m away from the stone, t = 14/35 = 0.4s vertical displacement covered by a stone during 0.4 sec, here, so, y = 0 - 1/2 × g × (0.4)² if g = 10 m/s² then, y = -1/2 × 10 × 0.16 = -0.8m so, position of stone from the ground = 2.4m - 0.8m = 1.6 m but it is given that the height of the wall is just 1m. so, the stone doesn't hit the wall and the stone is 1.6m above the ground, where the wall stands. now again applying the formula, where, y = -2.4 m , or, -2.4 m = 0 - 1/2 × 10 × t² or, t² = 0.48 ⇒t = 0.7sec so, x = ut = 35m/s × 0.7s = 24.5 m Hence, the stone hits the ground 24.5 m away from the initial position of it.

Q88. All the heat supplied to a system is converted into work in the _ process.

• A. Isochoric
• B. Isobaric
• C. Isothermal✓
• D. Isentropic

Explanation: In the Isothermal process, the temperature is constant. Internal energy is a state function dependent on temperature. Hence, the internal energy change is zero and thus all the heat energy supplied is converted to work done.

Q89. According to Bohr’s theory of the Hydrogen atom, only those atomic orbits with radii ‘r’ and atomic shell ‘n’ around the nucleus are allowed which have the angular moments:

• A. rh / 2π
• B. nh / 2π✓
• C. 2πn / h
• D. 2πr / h

Explanation: nh / 2π is the answer to this, just a factual recall. This is the correct expression representing Bohr's quantisation of angular momentum, where n is the principal quantum number, and h is Planck's constant.

Q90. Minimum frequency below which no electrons are emitted from the metal surface is called _ frequency.

• A. Minimum
• B. Angular
• C. Maximum
• D. Threshold✓

Explanation: The correct answer is the 'threshold frequency', which is the minimum frequency of electromagnetic radiation necessary to dislodge electrons from the surface of a metal. This concept is fundamental to the photoelectric effect, where no electrons are emitted if the incident light's frequency is below this threshold, regardless of its intensity. The other options, 'minimum', 'angular', and 'maximum', either lack specificity or pertain to different contexts within physics.

Q91. In an npn transistor, the current that flows in the emitter circuit is:

• A. IC + IB✓
• B. IC - IB
• C. IC x IB
• D. Zero

Explanation: The emitter current (IE) in an NPN transistor is the sum of the collector current (IC) and the base current (IB). This is because the emitter is the source of charge carriers that flow to the collector and base. Therefore, option A, IC + IB, correctly represents this relationship. Option B incorrectly suggests subtraction, option C incorrectly suggests multiplication, and option D incorrectly suggests the emitter current is zero, which is not possible when IC and IB are present.

Q92. In photodiodes, the bias voltage is applied in _ bias form.

• A. Forward
• B. Resistance
• C. Reverse✓
• D. Converse

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.

Q93. Gamma rays are emitted from nuclei of radioactive atoms. They are nothing but:

• A. X rays
• B. High energy electromagnetic radiations✓
• C. Electrons emitted from nuclei
• D. Ultraviolet rays
• E. Infrared rays

Explanation: Gamma rays are high-energy electromagnetic radiation emitted by radioactive elements. They possess energy but no mass. Unlike α or β radiation, the loss of a γ particle does not change the composition of the nucleus, although it does lose energy.

Q94. Five words are shown below:Farthest Universe Spontaneously Photon InfiniteThese words can be used in the spaces P, Q, R, S, and T to complete the sentences below.The _P_ is a stable particle and therefore it does not decay _Q_ into any other particle. Its lifetime is, therefore, _R_ so long it does not undergo interaction with other particles, and is why photons are supposed to be reaching our earth from _S_ distances of the universe. Thus most of our information regarding the _T_ is carried by photons.

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

Explanation: The answer to this question can be reached just by finding what P is. P should be a ‘‘particle’’, in options having only a photon is the particle. · P: Photon · Q: Spontaneously · R: Infinitely · S: Farthest · T: Universe This is the correct order.

Q95. During the process of nuclear disintegration, when beta particle emission occurs, atomic no. of the atom changes by _ and the mass number changes by _.

• A. One unit ; one unit
• B. One unit ; no units✓
• C. No units ; one unit
• D. No units ; no units

Explanation: Option A: This option is incorrect. When a beta particle is emitted, the atomic number of the atom changes by one unit, but the mass number of the atom does not change. Option B: The correct answer is one unit; no units. When a beta particle is emitted, a neutron in the nucleus decays into a proton and an electron. The electron is emitted as a beta particle, and the proton remains in the nucleus. This changes the atomic number of the atom by one unit, since the number of protons is one more than the number of neutrons. The mass number of the atom does not change, since the mass of the electron is very small compared to the mass of the neutron or proton. Option C: This option is incorrect. The atomic number of the atom changes when a beta particle is emitted, so this option cannot be correct. Option D: This option is incorrect. The mass number of the atom changes when a beta particle is emitted, so this option cannot be correct.

Q96. The resistivity in SI units is measured in:

• A. Ω x m✓
• B. Ω - m2
• C. Ω / m
• D. Ω / m2

Explanation: The SI unit of electrical resistivity is the ohm-meter (Ω⋅m). For example, if a 1 m × 1 m × 1 m solid cube of material has sheet contacts on two opposite faces, and the resistance between these contacts is 1 Ω, then the resistivity of the material is 1 Ω⋅m.

Q97. One kilowatt - hour is equivalent to a factor _ in Joules.

• A. 1.2 x 105
• B. 1.2 x 106
• C. 3.6 x 105
• D. 3.6 x 106✓

Explanation: 1 kW =1000J/s 1 h =3600 s ∴ 1 kWh =1000 J/s×3600 s=36×10^5J = 3.6 X 106

Q98. Identify the schematic representation of Fission Process from the following options?

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

Explanation: The correct answer is Option A because it correctly depicts the fission process, where a large atomic nucleus splits into two or more smaller nuclei, releasing energy and neutrons. This process does not necessarily result in identical fragments, allowing for variation in the sizes of the resulting nuclei. Options B and C are incorrect because they imply that the fission fragments must be identical, which is not a requirement. Option D is incorrect because it inaccurately states that energy is absorbed in the fission process, whereas energy is, in fact, released.

Q99. The police officers _ to the theatre and every nook and corner was completely checked.

• A. Sent to
• B. Were sent✓
• C. Sent
• D. Would have sent to

Explanation: The correct answer is B: 'were sent'. This choice accurately reflects the past tense passive voice. In this sentence, the police officers were the ones receiving the action of being sent, which is typical of a passive construction.Option A lacks the auxiliary verb needed for passive voice.Option C presents an active voice structure without the required auxiliary verb for passive construction.Option D uses the conditional perfect tense, which is inappropriate given the sentence's need for a simple past tense passive construction.