This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Given the following reaction and equilibrium constant `CO(g)+(1)/(2)O_(2)(g)hArrCO_(2)(g),K_(c)=1.1xx10^(11)` So at equilibrium select the correct response from below:-A. The net driving force for the reaction will always be left to right because the equilibrium constant is very largeB. The rate of forward reaction will always be greater than the rate of the reverse reaction because the equilibrium constant is so largeC. For a reaction mixture containing equilibrium concentrations of the three components, the forward and reverse rates will both go to zeroD. All responses above are incorrect |
| Answer» Correct Answer - D | |
| 2. |
In aqueous solution an amino acid exits as a). Cation b). Anion c). Zwitterion d). Neutral molecule |
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Answer» c). Zwitterion |
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| 3. |
What is fibrous protein |
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Answer» A fibrous protein is a thread like which may occur singly or in groups. They are tough, non-enzymatic and structural proteins. It is insoluble in water e.g., Keratin. Some fibrous protein is contractile protein found in myosin of muscles and connective tissue (tendons, bones, and cartilage) are collagen. |
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| 4. |
Approximate p.H,0.1M aqueous `H_(2)S` solution when `K_(1)` and `K_(2)` for `H_(2)S` at `25^(@)C` are `1xx10^(-7)` and `1.3xx10^(-13)` respectively:-A. 4B. 5C. 6D. 8 |
| Answer» Correct Answer - A | |
| 5. |
In which of the following C = O group is present ?(A) Ether (B) Alcohol (C) Ketone (D) Amine |
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Answer» Correct answer is (C) Ketone |
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| 6. |
Glucose on oxidation with Br2 water gives a). Gluconic acid b). Tartaric acid c). Saccharic acid d). Meso ‐oxalic acid |
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Answer» a). Gluconic acid |
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| 7. |
The acidic hydrolysis of ether (X) shown below is fastest when(A) one phenyl group is replaced by a methyl group. (B) one phenyl group is replaced by a para-methoxyphenyl group. (C) two phenyl groups are replaced by two para-methoxyphenyl groups. (D) no structural change is made to X |
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Answer» (C) two phenyl groups are replaced by two para-methoxyphenyl groups. When two phenyl groups are replaced by two para methoxy group, carbocation formed will be more stable |
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| 8. |
Matching of polymers:(Column I)(Column II)(A) Nylon(i) Thermosetting(B) Bakelite(ii) Thermoplastic(C) Polythene(iii) Elastomer(D) Natural rubber(iv) Fibers(a) A → (i); B → (iii); C → (ii); D → (iv)(b) A → (iv); B → (i); C → (ii); D → (iii)(c) A → (iii); B → (ii); C → (iv); D → (1)(d) A → (ii); B→ (i); C → (iv); D → (iii) |
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Answer» (b) A → (iv); B → (i); C → (ii); D → (iii) Nylon ⇒ Fibers Bakelite ⇒Thermosetting Polythene ⇒Thermoplastic Natural rubber ⇒ Elastomer |
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| 9. |
When 2 moles of HCl is added to 1 lit. of an acidic buffer solution, its pH changes from 3.4 to 3.9. Find its buffer capacity. |
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Answer» Buffer capacity = 2/0.5 = 4 |
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| 10. |
How to calculate pH of different types of solutions? |
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Answer» Different Types of solutions : (a) Strong acid solution : (i) If concentration is greater than 10–6 M. In this case H+ ions coming from water can be neglected, so [H+] = normality of strong acid solution (ii) If concentration is less than 10–6 M In this case H+ ions coming from water cannot be neglected. So [H+] = normality of strong acid + H+ ions coming from water in presence of this strong acid. (b) Strong base solution : Calculate the [OH– ] which will be equal to normality of the strong base solution and then use Kw = [H+ ] × [OH–] = 10–14 , to calculate [H+] |
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| 11. |
On reaction with stronger oxidizing agent like KIO4, hydrogen peroxide oxidizes with the evolution of O2. The oxidation number of I in KIO4 changes to ______. |
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Answer» Correct answer is 5 \(IO^-_4+H_ 2O_ 2→IO^-_ 3+O_2\) |
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| 12. |
Which set of quantum number represent degenerate orbital(a) n = 3, l = 2, m = 0, s = \(\frac{-1}2\) & n = 3, l = 2, m = -1 s = \(\frac{+1}2\)(b) n = 2, l = 1, m = 1, s = \(\frac{-1}2\) & n = 3, l = 1, m = 1 s = \(\frac{+1}2\)(c) n = 4, l = 2, m = -1, s = \(\frac{1}2\) & n = 3, l = 2, m = -1 s = \(\frac{1}2\)(1) a(2) b(3) c(4) a, b |
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Answer» Correct option is (1) a The orbitals with same n & l value but with different m value are degenerate. |
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| 13. |
The boiling point of a 0.2 m solution of a non-electrolyte in water is : (Kb for water = 0.52 K kg mol-1)(A) 100°C(B) 100.52°C(C) 100.104°C(D) 100.26°C |
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Answer» Option : (C) 100.104°C We know that, ΔTb = Kbm m = molality. ΔTb = 0.52 kg mol-1 x 0.2 mol/kg ΔTb = 0.104 K It means - Boiling point increases by 0.104 K or 0.104°C Hence, The boiling point of 0.2 m solution of a non-electrolyte in water will be 100.104°C |
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| 14. |
Nucleic acids are polymer of :(A) amino acids(B) nucleosides(C) nucleotides(D) glucose |
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Answer» Option : (C) nucleotides Nuclelic acids are polymer of Nucleotides. |
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| 15. |
Name the metal refined by Mond’s process. |
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Answer» The metal refined by Mond’s process is Nickel. |
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| 16. |
Buffering action of a mixture of CH3COOH and CH3COONa is maximum when the ratio of salt to acid is equal to : (A) 1.0 (B) 100.0 (C) 10.0 (D) 0.1 |
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Answer» Correct option: (A) 1.0 Explanation: The buffer action of a buffer mixture is effective in the pH range pKa ± 1. It is maximum when pH = pKa |
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| 17. |
Carded slivers are subjected to combing process, which is further delivered to ________ and here diameter of material is reduced and length of the material is increased. a. Roving frame b. Ring spinning frame c. Drawing frame d. Spinneret |
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Answer» a. Roving frame Carded slivers are subjected to combing process, which is further delivered to Roving frame and here diameter of material is reduced and length of the material is increased. |
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| 18. |
When a water is heated from 0 degree celcius to 4 degree celcius then cp=cv or cp |
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Answer» I think cp= cv ( cp- cv= nR , at 4 degree Celsius , water has a maximum density so n=0) |
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| 19. |
Why does hydrogen occur in a diatomic from rather than in a monoatomic form under normal conditions? |
| Answer» Hydrogen atom has only one elecron and thus of attain stable inert gas configuration of helium, it needs one more electron. Hence, to achieve stable inert has configuration of helium, it shares its single electron with electron of other hydrogen and thus form a stable diatomic molecule. | |
| 20. |
Which of the following lanthanoid ions, has zero magnetic moment ? [Ce (z = 58), Sm (z = 62), Eu (z = 63), Yb (z = 70)]`A. `Sm^(2+)`B. `Eu^(2+)`C. `Yb^(2+)`D. `Ce^(2+)` |
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Answer» Correct Answer - C `{:(Sm^(2+) [(Xe) 4F^(6)],,Eu^(2+) [(Xe) 4F^(7)],),(Yb^(2+)[(Xe)4F^(14)],,Ce^(2+)[(Xe)4F^(1)5d^(1)],):}` |
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| 21. |
Which molecule has longest carbon chain ? (a) Neopentane (b) Isopentane (c) Neohexane (d) n- pentane |
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Answer» n- pentane molecule has longest carbon chain. |
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| 22. |
Which molecule all the least C-C distance ? (a) C2H6 (b) C2H4 (c) C2H2 (d) C4H8 |
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Answer» C2H2 molecule all the least C-C distance. |
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| 23. |
What is the value of C – C bond length in ethyne ? (a) 154 pm (b) 139 pm (c) 134 pm (d) 120 pm |
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Answer» 120 pm value of C – C bond length in ethyne. |
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| 24. |
Difference between Hard Bases and Soft Bases. |
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Answer» Hard Bases : (i) The ligands which preferably combine with hard acids are called hard bases. Ex: NH3, R3N, H2O and F– etc. (ii) Hard bases (ligands) are also small, not very polarisable and having high electronegativity. Soft Bases : (i) The ligands which preferably combine with soft acids are called soft bases. Ex: R3P (phospines), R2S (thioethers), CO, CN–, H– etc. (ii) Soft bases (ligands) are large sized, more polarisable and having low electronegativity. |
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| 25. |
what is the chemical formula of calomel? |
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Answer» Hg2Cl2 is he chemical formula of calomel Calomel with chemical formula Hg2Cl2, also called mercurous chloride or mercury(I) chloride is a very heavy, soft, white, odourless, and tasteless halide mineral formed by the alteration of other mercury minerals, such as cinnabar or amalgams. |
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| 26. |
State the properties of water. |
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Answer» Amphoteric (amphiprotic) Acid/Base nature : Water - an acid as well as base according to Bronsted - Lowry theory but according to Lewis concept it can only be taken as base only. In pure water [H+] = [OH–] so it is neutral. |
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| 27. |
How will the pH of milk change as it turns into curd? |
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Answer» When milk is turned into curd then its pH value will decrease and will become less than 7 due to the production of lactic acid in curd which is acidic in nature. |
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| 28. |
In a bakery, baking powder was not added while preparing cake. The cake obtained was hard and small in size. What is the reason for this. |
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Answer» Baking powder is used for baking cakes. It contains sodium hydrogen carbonate, which breaks down when heated to form CO2 gas. The CO2 help to make the cakes light and fluffy. |
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| 29. |
Trachea divide at the level of :-A. `T_(3)`B. `L_(3)`C. `T_(5)`D. `L_(5)` |
| Answer» Correct Answer - A | |
| 30. |
An ore like zinc blends is concentrated byA. Froth floatationB. Magnetic separationC. LeachingD. Washing with water |
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Answer» Correct Answer - A Froth floatation because it is sulphide ore `(ZnS)` |
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| 31. |
Copper pyrites are concentrated byA. Electromagnetic methodB. Gravity methodC. Froth floatation processD. All of the above methods |
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Answer» Correct Answer - C Sulphides aore are always concentrated by froth floatation process |
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| 32. |
Select the incorrect statement :-A. External nostrils leads to nasal chamber through the Nasal passage.B. Larynx is a cartilaginous sound box.C. The pharynx open through the larynx region into the trachea.D. During swallowing glottis can be covered thin collagenous flap called epiglottis. |
| Answer» Correct Answer - A | |
| 33. |
Moist skin help in respiration in :-A. AnnelidsB. AmphibiansC. Both (1) and (2)D. Mammals |
| Answer» Correct Answer - A | |
| 34. |
Simple organisms like sponges and cnideria circulate_____from their surrounding through their body cavities.A. LymphB. BloodC. Colourless plasmaD. Water |
| Answer» Correct Answer - A | |
| 35. |
Vascularized bags present in Terrestrial animals helps in :-A. Pulmonary respirationB. Branchial respirationC. Cutaneous respirationD. Pharyngeal respiration |
| Answer» Correct Answer - A | |
| 36. |
Special fluid helps in transportation of substances in complex animals is/are :-A. Blood and lymphB. Only bloodC. Only lymphD. RBC, WBC and platelets |
| Answer» Correct Answer - A | |
| 37. |
Any material entering the trachea causes :-A. vimitingB. CoughingC. SneezingD. Regurgitation |
| Answer» Correct Answer - A | |
| 38. |
To balance the bicarbonate ion loss in RBC the chloride ions move into RBC, it is known as :-A. Chloride shiftB. Hamburger phenomenonC. Bohr effectD. Both (1) and (2) |
| Answer» Correct Answer - A | |
| 39. |
Which compounds does not contain OH group |
Answer»
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| 40. |
Bohr effect is :-A. The effect of `Co_(2)` on haemoglobinB. The effect of `CO_(2)` on oxyhaemoglobinC. The effect of `CO_(2)` on RBCD. The effect of `O_(2)` on haemoglobin |
| Answer» Correct Answer - A | |
| 41. |
Which of the following pairs of reaction occurs most frquently in the blood surrounding alveoli ?A. `Hb + O_(2) rarr HbO_(2) and HHb rarr Hb + H^(+)`B. `HbO_(2) rarr Hb + O_(2)and Hb + H^(+) rarr HHb`C. `H^(+) + Hb rarr HHb and H_(2)CO_(3) rarrHCO_(3) + H^(+)`D. `CO_(2) + H_(2)O rarr H_(2)CO_(3) and H_(2)CO_(3) rarr HCO_(3) + H^(+)` |
| Answer» Correct Answer - A | |
| 42. |
Vocal cords occur in :-A. TracheaB. LarynxC. GlottisD. Bronchiai tube |
| Answer» Correct Answer - A | |
| 43. |
Trachea is supported by :-A. Hyoid apparatusB. PalateC. CartilageD. All |
| Answer» Correct Answer - A | |
| 44. |
Pulmonary surfactant is :-A. Amino acidB. SteriodC. PhospholipidD. Glycolipid |
| Answer» Correct Answer - A | |
| 45. |
Which of the following has value 1:A. `tan45^circ`B. `sin90^circ`C. `cos90^circ`D. `cos0^circ` |
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Answer» Correct Answer - A::B::D `tan45^@=1` `sin90^@=1` `cos0^@=1` |
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| 46. |
Which of the following has value zero:A. `sin0^circ`B. `tan0^circ`C. `cos0^circ`D. `cot0^circ` |
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Answer» Correct Answer - A::B `sin^@=0` `tan0^@=0` |
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| 47. |
Selecthe correct statement(s):A. Physical chemistryB. Physical ChemistryC. Cation and anion are called basic and acidic radical respectivelyD. `[NiCl_4]^(2-)` is alow spin complex. |
| Answer» Correct Answer - C | |
| 48. |
The end product of the following reactions are A. yellow ppt. of `CHl_(3)`, B. yellow ppt of `CHl_(3)`, C. yellow ppt of `CHl_(3)`, D. yellow ppt of `CHl_(3)`, |
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Answer» Correct Answer - A::B Ionization energy : `Ga lt Tl` Electron gain enthalpy : `S gt Se gt O` |
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| 49. |
A buffer solution can be prepared from a mixture ofA. `NH_(4)Cl` and NaOH in 2 : 1 mole ratioB. `CH_(3)COONa` and HCl in 1 : 1 mole ratioC. `CH_(3)COONa` and NaOH in 2 : 1 mole ratioD. `CH_(3)COONa` and HCl in 1 : 2 mole ratio |
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Answer» Correct Answer - C `[H^(+)=alphaC=10^(-2)` `alpha=0.1` `P = i " "c " "R" " T = 1.1 xx0.1xx24.63 = 2.7 " atm"` |
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| 50. |
When 1-butyne is treated with aqueous `H_(2)SO_(4)` in presence of `HgSO_(4)`, the major product isA. ButanalB. ButanoneC. ButenolD. Buten-2-ol |
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Answer» Correct Answer - D `{:(La^(3+):[Xe]5d^(0)6s^(0)),(Lu^(3+):[Xe]4f^(14)5d^(0)6s^(0)),(Ce^(4+):[Xe]4f^(0)5d^(0)6s^(0)):}}mu_("eff")=0` `Nd^(3+):[Xe]4f^(3)}mu_("eff")ne 0` |
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