

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
1. |
The acidity of compounds I-IV in waterI. ethanolII. acetic acidIII. phenolIV. acetonitrilefollows the order(a) IV < I < III < II(b) I < II < III < IV(c) IV < I < II < III(d) IV < III < I < II |
Answer» Answer is : (a) IV < I < III < II The acidity of compounds in water depends upon the ease with which it can lose H+ ions. Acetic acid is the strongest acid as the negative charge on carboxylate ion (conjugate base) is delocalised over two oxygen atoms. Hence, H+ ion can be easily lost. The next strongest acidic compound phenol. This is because the phenoxide ion is resonance stabilised. This easily allows the H to leave as H+ ion. Among acetonitrile and ethanol, ethanol is more acidic, this is because in ethanol the H-atom is directly attached to more electronegative atom, O. Thus, the correct order of acidity of compound I-IV in water will be IV < I < III < II. |
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2. |
Forelimbs of whales, bats, humans and cheetah are examples of which of the following processes ?(A) Divergent evolution(B) Convergent evolution(C) Adaptation(D) Saltation |
Answer» Correct option (A) Divergent evolution Explanation: Fore limbs of whale, bat, human & cheetah are homologous organ which represents divergent evolution. |
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3. |
The two alleles that determine the blood group AB of an individual are located on(A) two different autosomes(B) the same autosome(C) two different sex chromosomes(D) one on sex chromosome and the other on an autosome |
Answer» Correct option (B) the same autosome Explanation: Allele of the same gene are present at same gene locus on homologous chromosome |
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4. |
In biotechnology applications, a selectable marker is incorporated in a plasmid (A) to increase its copy number(B) to increase the transformation efficiency(C) to eliminate the non-transformants(D) to increase the expression of the gene of interest |
Answer» Correct option (C) to eliminate the non-transformants Explanation: Selectable marker is used to eliminate the non transformant from the transformants |
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5. |
Which one of the following is detrimental to soil fertility?(a) Saprophytic bacteria(b) Nitrosomonas(c) Nitrobacter(d) Pseudomonas |
Answer» Answer is : (d) Pseudomonas Pseudomonas is denitrifying bacteria. Denitrifying bacteria are microorganisms whose action results in the conversion of nitrates in soil to free atmospheric nitrogen, thus depleting soil fertility and reducing agricultural productivity. |
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6. |
If you are seeing mangroves around you, which part of India are you visiting?(a) Western Ghats(b) Thar desert(c) Sunderbans(d) Himalayas |
Answer» Answer is : (c) Sunderbans The Sundarbans delta is the largest mangrove forest in the world and is intersected by a complex network of tidal waterways, mudflats and small islands of salt-tolerant mangrove forests. It lies at the mouth of the Ganges. A mangrove is a shrub or small tree that grows in coastal saline or brackish water. |
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7. |
The diagram shows some chromosomes at late prophase of mitosis.How many chromosomes would be present in one nucleus at telophase-II of meiosis? (a) 6 (b) 12 (c) 18 (d) 24 |
Answer» Correct option is (a) 6 There are 12 chromosomes present. In a mitotic division, the number of chromosomes remains the same, i.e. 2n → 2n. In a meiotic division, the number is halved, i.e., 2n → n so, there should only be 6 chromosomes in the nucleus. |
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8. |
Pyrimidine bases contain four carbon atoms and purine bases contain 5. How many carbon atoms are there in a nucleotide containing cytosine? (a) 8 (b) 9 (c) 10 (d) 11 |
Answer» Correct option is (b) 9 Phosphate group, ribose or deoxyribose group and cytosine group form the nucleotide. Phosphate (H3PO4) Ribose (C5H10O5) Deoxyribose (C5H10O4) Cytosine has 4 carbon atoms, as it is a pyrimidine. Therefore, the nucleotide should have 5C + 4C = 9C |
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9. |
The major nitrogenous excretory product in mammals is(a) amino acids(b) ammonia(c) urea(d) uric acid |
Answer» Answer is : (c) urea The major nitrogenous excretory product in mammals is urea. Nitrogenous wastes in the body of animals tend to form toxic NH3, which must be excreted. NH3 is converted to urea in hepatocytes of the body. |
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10. |
In which of the following types of glands is the secretion collected inside the cell and discharged by disintegration of the entire gland ?(A) Apocrine(B) Merocrine(C) Holocrine (D) Epicrine |
Answer» Correct option (C) Holocrine Explanation: Holocrine gland distingrate completely for discharge of secretion. |
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11. |
The pulmonary artery carries(a) deoxygenated blood to the lungs(b) oxygenated blood to the brain(c) oxygenated blood to the lungs(d) deoxygenated blood to the kidney |
Answer» Answer is : (a) deoxygenated blood to the lungs Pulmonary artery arises from left ventricle and carries deoxygenated blood to the lungs. The pulmonary artery begins in the heart at the base of the right ventricle. |
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12. |
The pulmonary artery carries (A) deoxygenated blood to the lungs (B) oxygenated blood to the brain (C) oxygenated blood to the lungs (D) deoxygenated blood to the kidney |
Answer» Correct option (A) deoxygenated blood to the lungs Explanation: Pulmonary artery arises from left ventricle & carry deoxygenated blood to the lungs. |
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13. |
The element X which forms a stable product of the type XCl4 is(a) Al(b) Na(c) Ca(d) Si |
Answer» Answer is : (d) Si As element X is forming a stable product of the type XCl4, so X must be tetravalent. Among the given elements Al is trivalent, Na is monovalent, Ca is divalent and Si is tetravalent. |
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14. |
Frosted glass is widely used for translucent windows. The region, where a transparent adhesive tape is stuck over the frosted glass becomes transparent. The most reasonable explanation for this is(a) diffusion of adhesive glue into glass(b) chemical reaction at adhesive tape-glass interface(c) refractive index of adhesive glue is close to that of glass(d) adhesive tape is more transparent than glass |
Answer» Answer is : (c) refractive index of adhesive glue is close to that of glass Frosted glass has a rough layer which causes irregular refraction and makes glass translucent. When a transparent tape which has refractive index close to that of glass is pasted over the rough surface of glass, the tape glue fills the roughness of glass. This makes glass surface more smooth and so refraction is more regular. This makes region of tape transparent. |
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15. |
Which would be worse, the burn you get from holding your hand in a jet of steam or that from boiling water both at 100°C ? Give a reason for your answer. |
Answer» Steam causes more serious burns than that of water at 100°C because 1g of steam at 100°C imparts 2260 J of extra heat than the heat imparted by 1g of water at 100°C. |
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16. |
Which of the following biological phenomena involves a bacteriophage?(a) Transformation(b) Conjugation(c) Translocation(d) Transduction |
Answer» Answer is : (d) Transduction Transduction is a method of gene transfer in bacteria from donor to recipient using bacteriophage. In transduction, at first bacteriophage infects donor bacteria and then carries some part of donor genome with it. When this bacteriophage infects new bacterial cell, it transfers that DNA into recipient cell. |
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17. |
Acetic acid reacts with sodium metal at room temperature to produce(A) CO2 (B) H2 (C)H2O (D)CO |
Answer» Correct option(b) Explanation: 2CH3COOH + 2Na→2CH3COONa +H2 ↑ |
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18. |
Which of the following metals will precipitate copper from copper sulphate solution?(a) Hg(b) Sn(c) Au(d) Pt |
Answer» Answer is : (b) Sn More reactive metal than Cu can precipitate copper from copper sulphate solution. The increasing order of reactivity of given element is Au < Hg < Cu < Sn As Sn is more reactive than Cu, so it would precipitate Cu from CuSO4. |
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19. |
For streamlined flow of water, consider the following statements. I. Two streamlines does not cross each other. II. Streamlines are straight. III. Streamlined flow is more likely for fluids with low density and high viscosity. IV. Streamlined flow is more likely for liquids with high density and low viscosity.Which of the above statements are correct? (a) Statements I and III are correct (b) Statements II, III and IV are correct (c) Statements III and IV are correct (d) Statements I, III and IV are correct |
Answer» Correct option is (a) Statements I and III are correct Streamlines may be straight or curved. In fluids, with low viscosity streamlined flow occurs only at low flow speed. |
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20. |
A prime number p is called special if there exist primes p1, p2, p3, p4 such that p = p1 + p2 = p3 - p4. The number of special primes is(a) 0(b) 1(c) more than one but finite(d) infinite |
Answer» Answer is : (b) 1 It is given that for prime numbers p1, p2, p3, p4 the special prime number p = p1 + p2 = p3 - p4 Case I If all p1, p2, p3, p4 are odd, then (p1 + p2 ) and (p3 - p4) are even, which is not possible. Case II If one of p1 and p2 is even, say p2 is 2 and p4 must be 2. So, p = p1 + 2 = p3 - 2 the above equation is satisfied only if p = 5, p1 = 3 and p3 = 7 So, the number of special prime p is 1. |
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21. |
The peptidoglycans of bacteria consist of (A) sugars, D-amino acids and L-amino acids(B) sugars and only D-amino acids(C) sugars and only L-amino acids(D) sugars and glycine |
Answer» Correct option (A) sugars, D-amino acids and L-amino acids |
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22. |
A closed bottle containing water at 30ºC is open on the surface of the moon. Then -(A) the water will boil(B) the water will come as a spherical ball(C) the water will freez(D) the water will decompose into hydrogen and oxygen |
Answer» Correct option (A) the water will boil Explanation: Because on earth there is no atmosphere. So water will boil. (At Boiling point vapour pressure = Atmospheric pressure, in open vessel) |
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23. |
A simple pendulum of length l is made to oscillate with an amplitude of 45 degrees. The acceleration due to gravity is g. Let T0 = 2 π √l/g. The lime period of oscillation of this pendulum will be (A) T0 irrespective of the amplitude. (B) slightly less than T0.(C) slightly more than T0.(D) dependent on whether it swings in a plane aligned with the north-south or cast-west directions. |
Answer» The correct option is (A) T0 irrespective of the amplitude. Explanation: T0 = 2 π √l/g If θ is greater, then a < gθ Hence time-period T' > T0. |
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24. |
Among the following atomic orbital overlap, the non-bonding overlap is |
Answer» Correct option (A) Explanation: It has +ve & –ve overlap both simultaneous. So it leads to non-bonding overlap. |
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25. |
The diagram represents a reaction with and without an enzyme. What is the activation energy of the enzyme catalysed reaction?(a) A (b) B (c) C (d) D |
Answer» Correct option is (b) B The reaction that is enzyme-catalysed has lower activation energy and is represented by the dotted line. The energy input required to raise the energy of the reactants to a certain level before the reaction is triggered is called the activation energy. This is represented by the increase in energy of the reactants to the top of the ‘hill’, B. |
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26. |
Treponema pallidum pathogen is a cause of (a) leprosy (b) plague (c) syphilis (d) pertussis |
Answer» Correct option is (a) leprosy Treponema pallidum pathogen is a cause of syphilis. It is a sexually transmitted infection that causes infected sores, blisters or ulcers on your genitals, anus (bottom) or mouth. |
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27. |
How many moles of magnesium phosphate Mg3 (PO4)2 will contain 0.25 mole of oxygen atom?(a) 0.02 (b) 3.125 x 10-2 (c) 1.25 x 10-2(d) 2.5 × 10-2 |
Answer» Correct option is (b) 3.125 x 10-2 |
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28. |
If 14CO2 is added to a suspension of photosynthesizing chloroplasts, which of the following will be the first compound to be radioactive ?(A) ATP(B) NADPH(C) NADH(D) 3-phosphoglycerate |
Answer» Correct option (D) 3-phosphoglycerate Explanation: During Calvin cycle, the 14CO2 is incorporated with a 5C compound Ribulose 1-5 Biphosphate and forms 2 molecules of 3C compound 3-phosphoglycerate. |
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29. |
In 1911, the physical Ernest Rutherford discovered that atoms have a tiny , dense nucleus by shooting positively charged particles at a very thin gold foil. A key physical property which led Rutherford to use gold was that it was (A) electrically conducting (B) highly malleable (C) shiny (D) non-reactive |
Answer» Correct option (B) highly malleable Explanation: Very thin foil can be made only of highly malleable material. |
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30. |
A large number of random snap shots using a camera are taken of a particle in simple harmonic motion between x = – x0 and x = + x0 with origin x = 0 as the mean position . A histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble |
Answer» Correct option (C) Explanation: Speed of particle doing SHM decrease as it go away from mean position. Time during which particle remain in extreme position will be longer. |
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31. |
A charged particle initially at rest at O, when released follows a trajectory as shown. Such a trajectory is possible in the presence of (A) electric field of constant magnitude and varying direction. (B) magnetic field of constant magnitude and varying direction (C) electric field of constant magnitude and constant direction. (D) electric and magnetic fields of constant magnitudes and constant directions which are parallel to each other |
Answer» (A) electric field of constant magnitude and varying direction. |
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32. |
Two charges +q and –q are placed at a distance b apart as shown in the figure below.The electric field at a point P on the perpendicular bisector as shown as : (A) along vector →A (B) along vector →B(C) along vector →C (D) Zero |
Answer» Correct option: (A) along vector →A |
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33. |
A sequence of amino acids may end in either an amino group (—NH2) or a carboxyl group (—COOH). What is the theoretical number of chemically different dipeptides that may be assembled from 20 different amino acids?(a) 40(b) 80(c) 160(d) 400 |
Answer» Answer is : (d) 400 A dipeptide is made up of 2 amino acids which may be same or different. The total possible number of different dipeptides that may be assembled from 20 different amino acids will thus be 20n = 202 = 20 x 20 = 400 |
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34. |
Horse-shoe crabs belong to the group (a) Onychophora (b) Chelicerata (c) Uniramia(d) Crustacea |
Answer» Correct option is (b) Chelicerata Horse-shoe crabs are marine and brackish water arthropods. They resemble crustaceans but belong to separate subphylum of the arthropods, i.e. Chelicerata. The entire body of the horse-shoe crab is protected by a hard carapace. |
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35. |
Among K, Mg, Au and Cu, the one which is extracted by heating its ore in air is(a) K(b) Mg(c) Au(d) Cu |
Answer» Answer is : (d) Cu Metal with low reactivity can be extracted by heating ore in air. This process of extraction is known as roasting. Among the given metals, Cu has least reactivity, so it can be directly extracted by heating its ore in air. 2CU2S + 3O2 → 2Cu2O + 2SO2↑ |
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36. |
Among P2O5, As2O3, Sb2O3 and Bi2O3, the most acidic oxide is(a) P2O5(b) As2O3(c) Sb2O3(d) Bi2O3 |
Answer» Answer is : (a) P2O5 P2O5, As2O3 are acidic oxides, Sb2O3 is an amphoteric while Bi2O3 is basic oxide but P2O5 is most acidic among them all. This is because down the group, metallic character increases and metal oxides are more basic, thus the basicity of metallic oxides also increases. |
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37. |
The oxide, which is neither acidic nor basic is(a) As2O3(b) Sb4O10(c) N2O(d) Na2O |
Answer» Answer is : (c) N2O N2O is a neutral oxide, which is neither acidic nor basic. |
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38. |
Among the following, the most basic oxide is(a) Al2O3(b) P2O5(c) SiO2(d) Na2O |
Answer» Answer is : (d) Na2O All the central atom of the given oxide belong to 2nd period, as we move from left to right in a period the basicity of oxide decreases. ∴ The order of increasing basicity of oxide would be P2O5 < SiO2 < Al2O3 < Na2O Thus, the most basic oxide would be Na2O. |
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39. |
By dissolving 0.35 mole of sodium chloride in water, 1.30 L of salt solution is obtained. The molarity of the resulting solution should be reported as(a) 0.3 M(b) 0.269 M(c) 0.27 M(d) 0.2692 M |
Answer» Answer is : (c) 0.27 M Given, Number of moles of solute NaCl = 0.35 mole Volume of the solution = 1.30 L Molarity = (Number of moles of solute/Volume of solution in litres) = 0.35/1.30 = 0.269 ≈ 0.27 M |
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40. |
Among the quantities, density (ρ), temperature (T), enthalpy (H), heat capacity (Cp), volume (V) and pressure (p), a set of intensive variables are(a) (ρ, T, H)(b) (H, T, V)(c) (V, T, Cp)(d) (ρ, T, p) |
Answer» Answer is : (d) (ρ, T, p) Intensive variables are those variables which don’t depend upon the quantity or size of matter. Among the given quantities density, temperature and pressure are intensive variables whereas heat capacity, enthalpy and volume are an extensive variables. |
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41. |
Among the quantities, boiling point (I), entropy (II), pH (III) and emf of a cell (IV), intensive properties are(a) Both I and II(b) I, II and III(c) I, III and IV(d) All of these |
Answer» Answer is : (c) I, III and IV Intensive properties are those properties which do not depend upon the quantity or size of matter. Among the given quantities, boiling point (I), pH (III) and emf (IV) are intensive properties whereas entropy (II) is an extensive property. |
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42. |
Filariasis is caused by(a) Entamoeba histolytica(b) Plasmodium falciparum(c) Trypanosoma brucei(d) Wuchereria bancrofti |
Answer» Answer is : (d) Wuchereria bancrofti Filariasis is caused by Wuchereria bancrofti. It lives in lymphatic vessels and causes swelling of lower limbs and scrotum. Entamoeba histolytica causes amoebiasis. Plasmodium falciparum causes malaria. Trypanosoma brucei causes African sleeping sickness. |
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43. |
Which of the following transformations can be carried out by using HI as a reducing agent, under acidic conditions?[Given : I2(s) → 2I−; E° = 0.54 V](i) Cu+ → Cu(s); E° = 0.52 V(ii) Cr3+ → Cr2+; E° = − 0.41 V(iii) Fe3+ → Fe2+; E° = 0.77 V(iv) Fe2+ → Fe(s); E° = − 0.44 V(a) (i) and (iii)(b) (ii) and (iv)(c) Only (iii)(d) Only (ii) |
Answer» Answer is : (c) Only (iii) The more positive E° value of metal, feasible transformation can be carried out by using HI as reducing agent under acidic conditions. As E° = 0.77 V(Fe3+ → Fe2+) is more positive than, E° = 0.54(I2(s) → 2I−) thus can be used for carrying out transformation as it is the strongest oxidising agent among the other given options. |
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44. |
The cell organelle responsible for extracting energy from carbohydrates to form ATP is (a) lysosome (b) ribosome (c) chloroplast (d) mitochondrion |
Answer» Correct option is (d) mitochondrion Mitochondria is referred as powerhouse of the cell. It contains the enzymes for cellular respiration. It oxidises carbohydrate to produce ATP molecules in the process of aerobic respiration. |
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45. |
An electric heater consists of a nichrome coil and runs under 220 V, consuming 1 kW power. Part of its coil burned out and it was reconnected after cutting off the burnt portion. The power it will cunsume now is(a) more than 1 kW(b) less than 1 kW, but not zero(c) 1 kW(d) 0 kW |
Answer» Answer is : (a) more than 1 kW Power P = V2/R as V remains constant in given condition, while current changes as resistance is reduced. Power consumed by heater coil P ∝ 1/R. On cutting part of coil, its resistance decreases. So, power consumed by coil increases and it will be more than 1 kW. |
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46. |
A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The correct representation of the variation of the frictional force f, exerted by the table on the block with time t is given by |
Answer» Answer is : (a) Force of friction is self adjusting. When applied force is increased friction, first increases till it reaches a maximum value called limiting friction. If applied force is further increased, then friction does not increases further and body begins to move. Force of friction then remains constant and its value is given by f = µN |
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47. |
The shortest distance from the origin to a variable point on the sphere (x – 2)2 + (y – 3)2 + (z – 6)2 = 1 is (A) 5(B) 6(C) 7(D) 8 |
Answer» Correct option (B) 6 Explanation: Sphere x2 + y2 + z2 – 4x – 6x – 12z + 48 = 0 Centre (2, 3, 6) radius = √(4 + 9 + 36 - 48) = 1 distance between centre and origin = √(4 + 9 + 36) = 7 shortest distance = 7 – 1 = 6 (Origin lies outside the sphere) |
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48. |
Match the type of cells given in Column I with organisms given in Column II. Choose the appropriate combination from the options below.Column I Column II(P) Flame cells (i) Sponges (Q) Collar cells (ii) Hydra(R) Stinging cells (iii) Planaria (A) P-iii, Q-i, R-ii(B) P-iii, Q-ii, R-i(C) P-i, Q-ii, R-iii(D) P-ii, Q-iii, R-i |
Answer» Correct option (A) P-iii, Q-i, R-ii Explanation: Flame cells are for secretion and osmoregulation in flatworms. Collar cells line spongocoel and cavars in sponges and creat water amount. Cnidocytes are stringing cells and help in defense, affiance and capture of prey these cells also help in attachment with substrate |
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49. |
The geometry and the number of unpaired electron(s) of [MnBr4] 2– , respectively, are (A) tetrahedral and 1 (B) square planar and 1 (C) tetrahedral and 5 (D) square planar and 5 |
Answer» Answer: tetrahedral and 5 Explanation: Mn2+ : [Ar]3d5 Br– is weak field ligand. |
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50. |
The numbers of lone pairs and bond pairs in hydrazine are, respectively(a) 2 and 4(b) 2 and 6(c) 2 and 5(d) 1 and 5 |
Answer» Answer is : (c) 2 and 5 The molecular formula of hydrazine is NH2NH2. From the structure, it is clear that it has 2 lone pairs and 5 bond pairs (4 N — H and 1 N–N). |
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