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. |
The degree of dissociation of acetic acid in a 0.1 M solution is `1.32xx10^(-2)`, find out the pKa :-A. 5.75B. 3.75C. 4. 00D. 4. 75 |
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Answer» Correct Answer - D `Ka=Calpha^(2)0.1xx(1.32xx10^(-2))^(2)=1.74xx10^(-5)` `pKa=5-log1.74=5-0.26` =4.74 |
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| 2. |
Calculate the degree of hydrolysis of the 0.01 M solution of salt` (KF)(Ka(HF)=6.6xx10^(-4))` :-A. `3.87xx10^(-6)`B. `3.87xx10^(-5)`C. `3.87xx10^(-2)`D. None of these |
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Answer» Correct Answer - B `{:(F^(-)+H_(2)O,hArr,OH^(-)+HF,),(0.01(1-h),, 0.01h0.01h,):}` |
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| 3. |
`B^(3+)` cannot exist in aqueous solution because of its :A. Strong reducing abilityB. Strong oxidizing abilityC. Small size and large chargeD. Large size and small charge |
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Answer» Correct Answer - C Being covalent in nature because of small size and large charge it gets hydrolysed in aqueous solution according to the following reaction `BCl_3 +3H_2O overset("Hydrolysis")toH_3BO_3 + 3HCl` |
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| 4. |
(i) Why are artiflcial sweetening agents harmless when taken ?(ii) Name one such artificial sweeting agent.(iii) Why is the use of aspartame as an artificial sweetener limited to coldfoods ? |
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Answer» [Hint : (i) Because they are not metabolized by body and excreted from the body in urine unchanged. |
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| 5. |
Pick out the odd one amongst the following on the basis of their medicinal properties. Give suitable reason :(i) Luminal, seconal, terfenadine, equanil.(ii)Chloroxylenol, phenol, chloamphenicol, bithional.(iii) Sucralose, aspartame, alitame, sodium benzoate. |
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Answer» [Hint : (i) Terfenadine is antihistamine other three are used as tranquilizers. |
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| 6. |
Evaluate the following.1. 10C42. 21C33. 19C154. 31C29 |
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Answer» 1. 10C4 = \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4} \) = 10 × 3 × 7 = 210 2. 21C3 = \(\frac{21 \times 20 \times 19}{1 \times 2 \times 3}\) = 7 × 10 × 19 = 1330 3. 19C15 = \(\frac{19 \times 18 \times 17 \times 16}{1 \times 2 \times 3 \times 4}\) = 19 × 3 × 17 × 4 = 3876 4. 31C29 = \(\frac{31 \times 30}{1 \times 2}\) = 465. |
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| 7. |
Which is heavier: water or ice? |
| Answer» Correct Answer - Water. | |
| 8. |
What type of bonds are broken when water evaporates? |
| Answer» Correct Answer - Hydrogen bonds. | |
| 9. |
The degree of hardness of a given sample of hard water is `40 p p m`. If the entire hardness is due to `MgSO_(4)`, how much of `MgSO_(4)` is present per `kg` of water? |
| Answer» Correct Answer - `48 mg`. | |
| 10. |
Which of the following is generally used for induced mutagenesis in crop plantsA. x raysB. uv (260 nm)C. gamma rays (from cobalt 60)D. alpha paricles |
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Answer» Correct Answer - C Cobalt 60 is the synthetic radioactive isotope of cobalt .Ggamma rays are produced when an undstable atomic nulceus like coblat 60 releases energy to gain stability sharbati sonora and pusa lerma are the two imoprtnt varieties of wheat that are produced by gamma rays treatment of sonara 64 and lerma rojo 64 which are mexican dwarf wheat varities |
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| 11. |
why is vivipary an undesirable character for annual crop plants?A. it reduces the vigour of plantB. the seeds cannot be stroed under normal conditions for the next seasonC. the seeds exhibit long dormancyD. it adversely affects the fertility of the plant |
| Answer» vivipary is the condition when seeds are germinated on the plant it is an undersirable charcter for annual crop plants because germinated seeds cannot be stored under normal conditions for the next season ,. | |
| 12. |
Give 4 types for Collenchyma. |
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Answer» Collenchyma Has Four Major Types
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| 13. |
Conjugation between bacteria occur by |
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Answer» Bacterial conjugation is the transfer of genetic material between bacterial cells by direct cell-to-cell contact or by a bridge-like connection between two cells. This takes place through a pilus. It is a parasexual mode of reproduction in bacteria. |
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| 14. |
Consider the following carbocations, `undersetI((CH_3)_3C overset+CH_2), underset(II)((CH_3)_3C^(+)), CH_(3)underset(III)CH_(2)overset(+)CH_(2), CH_(3)underset(IV)overset(+)CH-CH_(3)` The correct order for the stability of the above carbocations isA. `IltIIIltIVltII`B. `IIIltIVltIltII`C. `IVltIIIltIIltI`D. `IIltIVltIIIltI` |
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Answer» Correct Answer - A Larger the number of `alpha-H` atoms linked toa carbocation, more will be stable carbocation. On the basis of hyperconjugation, the correct order of stability is `I lt III lt IV lt II` |
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| 15. |
What measures should be opted to prevent from contracting sexually transmitted diseases. |
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Answer» For prevention of sexually transmitted diseases following actions should be opted:– (i) Avoid sex with unknown partner/multiple partners. (ii) Always use condoms during coitus. (iii) In case of doubt, should go to a qualified doctor for early detection and get complete treatment if diagnosed with disease. |
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| 16. |
Discuss concept of productivity. What is primary productivity? Write about the factors affecting it. |
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Answer» The rate of synthesis of organic matter, or biomass produced at any trophic level during a given period of time is called productivity. For productivity three things are necessary – (i) Rate of production, (ii) Per unit area and (iii) Unit time. Productivity is of two types– (1) Primary productivity and (2) Secondary Productivity. It is measured as weight (Ex- g/m/yr) or energy (Ex-Kcal/m2/yr). Primary Productivity – Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis. It is expressed in terms of weight (g–2) or energy (K cal m–2). It can be divided into two types – (a) gross primary productivity and (b) net primary productivity. (a) Gross Primary Productivity – Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration. It is equal to addition weight of stored organic material present in body of producer and loss during different activity happens in its body. (b) Net Primary Productivity – Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores and decomposers). Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP). G.P.P – R = N.P.P. Primary productivity depends on the plant species inhabitating a particular area. It also depends on a variety of environmental factors, availability of nutrients and photosynthetic capacity of plants. Therefore, it varies in different ecosystems. The annual net primary productivity of the whole biosphere is approximately 170 billion tons (dry weight) of organic matter. Of this, despite occupying about 70 percent of the surface, the productivity of the oceans are only 55 billion tons. |
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| 17. |
The geometry of `overset(dot)CH_(3)` isA. pyramidalB. linearC. tetrahedralD. planar |
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Answer» Correct Answer - D The free radical `overset(*)CH_(3)` is planar in geometry. |
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| 18. |
Which one of the following species is not an electrophile?A. `overset(+)NO_(2)`B. `H_(3)O^(+)`C. `Cl^(+)`D. `BH_(3)` |
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Answer» Correct Answer - B The species which posess an incomplete octet can behave as an electrophile as it posses a tendency to accept the lone pair of electron. But `H_(3)O^(+)` contains a complete octet of O, therefore, it does not as an electrophile. |
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| 19. |
`CH_(3)underset(CH_(3))underset(|)(CHCH)=CH_(2)+HBrtoA,A` (predominant isA. `CH_(3)underset(CH_(3))underset(|)(CHCH)underset(Br)underset(|)CH_(2)`B. `CH_(3)underset(CH_(3))underset(|)(CHCH_(2)CH_(2)Br`C. `CH_(3)underset(CH_(3))underset(|)overset(Br)overset(|)C CH_(2)CH_(3)`D. None of these |
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Answer» Correct Answer - C `CH_(3) underset(CH_(3))underset(|)(CH)CH=CH_(2) overset(H^(+)) to CH_(3) underset(2^(@)"Carbocation") underset(CH_(3)) underset(|)(CH)overset(+)C HCH_(3) overset(1,2H^(-)"shift") to CH_(3) underset(B) underset(3^(@)"carbocation")underset(CH_(3)) underset(|) overset(+)CH_(2)CH_(3) overset(Br^(-)) toCH_(3) underset(CH_(3)) underset(|) overset(Br) overset(|)C CH_(2)CH_(3)` |
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| 20. |
The correct stability order of the following free radicals isA. `(CH_(3))_(2)overset(*)CHlt(CH_(3))_(3)overset(*)Clt(C_(6)H_(5))_(2)overset(*)CH`B. `(C_(6)H_(5))_(2)overset(*)CHlt(C_(6)H_(5))_(3)overset(*)Clt(CH_(3))_(3)overset(*)Clt(CH_(3))_(2)overset(*)CH`C. `(C_(6)H_(5))_(3)overset(*)CHlt(C_(6)H_(5))_(2)overset(*)CHlt(CH_(3))_(3)overset(*)(C)lt(CH_(3))_(2)overset(*)CH`D. `(CH_(3))_(2)overset(*)CHlt(CH_(3))_(3)overset(*)CHlt(C_(6)H_(5))_(3)overset(*)C` |
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Answer» Correct Answer - D Larger the number of phenyl groups linked to a free radical. More the number of resonating structures and greater is the stability of free radical is `(CH_(3))_(2)overset(*)CH lt (CH_(3))_(3) overset(*)C lt (C_(6)H_(5))_(2)overset(*)C H lt (C_(6)H_(5))_(3) overset(*)C` |
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| 21. |
Identify each of the following as carbon-intermediates: `{:((a)CH_(3)-CH=overset(+)CH,(b)(CH_(3))_(3)overset(.)C),((c)(CH_(3))_(3)bar(C),(d)C_(2)H_(5)overset(.)"CC"H_(3)),((e)CH_(3)-bar(C)H-CH_(3),(d)C_(2)H_(5)overset(.)"C"CH_(3)),((e)CH_(3)-bar(C)H-CH_(3),(f)C_(2)H_(5)overset(..)C-H),((g)C_(6)H_(5)-overset(..)underset(..)N,(h):"CCl"_(2)):}` |
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Answer» (a) Carbocation, (b) Free radical, (c) Carbanion, (d) Free Radical, (e) Carbanion, (f) Singlet carbene, (g) Nitrene, (h) Carbene. |
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| 22. |
Answer the following : (a) How many types of fission are possible of a covalent bond? (b) how many types of ions are formed by heterolytic fission and what are their names? (c) What are the names and nature of the parts formed by homolytic fission? (d) How many electrons are present in the valence shell of the carbon atom of the carbonium ion? (e) How many electrons are present in the valence shell of carbon atom of the carbanion ion? (f) Name the attacking reagents and give their nature. (g) Write the names of the parts obtained when `C-Cl` bond of ethyl chloride undergoes heterolytic fission. (h) What is the major factor that influences the relatives stabilities of carbocations, carbanions and free radicals. |
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Answer» (a) Two, homolytic and heterolytic. (b) Two, carbocation and carbanion. (c) Parts are free radicals. These are electrically neytral. (d) Six electrons. (e) Eight electrons. (f) Electrophilic reagents-Positively charged ions or neutral molecules which have affinity for electrons. Nucleophilic reagents- Negatively charged ions or neutral molecules which have tendency to lose an electron pair. (g) `underset("Ethyl chloride")(CH_(3)CH_(2)-Cl)rarrunderset("Ethyl carbocation")(CH_(3)overset(+)CH_(2))+underset("Chloride ion")(overset(-)Cl)` (h) Any diminution of `+` or `-` charge or of electron deficiency on the carbon stabilizes the intermediate. |
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| 23. |
1) Chemical nomenclature of procaine is? |
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Answer» It is a benzoate ester, a substituted aniline and a tertiary amino compound. It derives from a 2-diethylaminoethanol and a 4-aminobenzoic acid. It is a conjugate base of a procaine(1+). |
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| 24. |
Which of the following is not an action of insulin |
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Answer» Answer: Initiates the conversion of glycogen to glucose Glucagon is a peptide hormone, produced by alpha cells of the pancreas. It works to raise the concentration of glucose in the bloodstream. Its effect is opposite that of insulin, which lowers the glucose. The pancreas releases glucagon when the concentration of glucose in the bloodstream falls too low. Glucagon causes the liver to convert stored glycogen into glucose, which is released into the bloodstream. High blood-glucose level stimulates the release of insulin. Insulin allows glucose to be taken up and used by insulin-dependent tissues. Thus, glucagon and insulin are part of a feedback system that keeps blood glucose levels at a stable level. |
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| 25. |
Class boundaries are also called: (a) Mathematical limits (b) Arithmetic limits (c) Geometric limits(d) Qualitative limits |
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Answer» (a) Mathematical limits |
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| 26. |
The series is categorized as: Weights(pounds)15-2020-2525-3030-3535-0No. of items101530105(a) Continuous series (b) Discrete series (c) Time series (d) Geometric series |
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Answer» (a) Continuous series |
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| 27. |
A series of data with exclusive classes along with the corresponding frequencies is called: (a) Discrete frequency distribution (b) Continuous frequency distribution(c) Percentage frequency distribution (d) Cumulative frequency distribution |
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Answer» (b) Continuous frequency distribution |
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| 28. |
Calculate mass % of He in a mixture of `O_(2)` and He gas at 3 atm and `27^(@)C` havin density `(5)/(3)` gm/litre [Take : R=0.08 atm litre `mol^(-1)k^(-1)`] |
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Answer» PM=dRT for mixture `3xxM_("avg")=((5)/(3))xx0.08xx300` `M_("avg")=(5xx8xx3)/(3xx3)=(40)/(3)` `M_("avg")=(100)/((x)/(4)+(100-x)/(32))=(40)/(3)rArr(100xx32xx4)/(32x+400-4x)=(40)/(3)` `32xx10xx3=28x+400` `=28x=960-400=560` x=20% `(20)/(10)=2` |
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| 29. |
The average of lower and upper class limits is called: (a) Class boundary (b) Class frequency (c) Class mark (d) Class limit |
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Answer» (c) Class mark |
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| 30. |
If the number of workers in a factory is 128 and maximum and minimum hourly wages are 100 and 20 respectively. For the frequency distribution of hourly wages, the class interval is: (a) 8 (b) 9 (c) 10 (d) 80 |
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Answer» Correct option is (c) 10 |
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| 31. |
Complete the tablePareulalFirst generationSecond generation (Self Pollination)Approsimate RatioPosition of flowersAxial x Teminal.....(a).....651 (Axial) 207 (Terminal)..........(b).......Shape of seedsRound x Wrinkled......(c)......882 (Round) 299 (Wrinkled).........(d)...... |
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Answer» a) Axial b)3:1 c) Round d) 3:1 Inferences of Mendel:
Gene – Allele: The gene present in the chromosome of the nucleus determines the character. A gene that controls a trait has different forms. The different forms of a gene are called alleles. Generally, a gene has two alleles. When. we illustrate hybridization experiment, the allele that controls the dominant character that is expressed in the first generation is indicated by a capital letter and the allele that controls recessive character is indicated by a small letter. |
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| 32. |
Complete the illustration of the second generation obtained from the hybridization in which two traits of a plant are considered.Indicators:Dominant character – Tallness, red color of flower Recessive character – Dwarfness, white color of flower |
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Answer» A. TTRr B. TtRr C.TTRr D. TtRr E. TtRr F. ttRR G. ttRr H. TtRr |
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| 33. |
A portion of DNA molecule is shown below. Find out the missing nitrogen base pair from those given below. |
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Answer» Correct option: (b) C – G |
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| 34. |
In maize, coloured endosprem (C) is dominant over colourless (c) , and full endosperm (R) is dominant over shrunken (r). When a dihybrid of `F_(1)` generation was test crossed, it produced four phenotypes in the following percentage : Coloured full - 48 % Coloured shrunken - 5 % Coloured full - 7 % Colourless shrunken - 40 % From this data, what will be the distance between two non-allelic genes ?A. 48 unitsB. 5 unitsC. 7 unitsD. 12 units |
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Answer» Correct Answer - D Given that recombinat percentage is 7 % and 5 % therefore, total recombinants would be `7 + 5 = 12 %`. It is known that one map unit is the distance that yields `1 %` recombinant chromosomes. Hence distance between two non-allelic genes is 12 map units. |
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| 35. |
Rearrange B and C according to the data given in A.A Nucleic AcidsB SugarC Nitrogen Basei) DNARibosomeUracilii) RNADeoxyriboseAmeninRiboseThymine |
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Answer»
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| 36. |
Observe the nucleotide strands given below and answer the questions.a) Identify the strand that is found in DNA only. b) Identify the strand that can be found in both DNA and RNA. c) What is a nucleotide? |
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Answer» a) B b) A c) A unit of sugar, phosphate and nitrogen base / Component of nucleic acid |
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| 37. |
A part of the article. Variations in ourself is given below: The features seen in offspring that are different form their parents are called variations. Certain processes taking place in the initial phase of meiosis are responsible for such variations.a) Which process, as mentioned in the article, is responsible for variations?b) How does this process bring about variations? |
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Answer» a) Crossing over of chromosomes b) 1. Part of a DNA crosses over to become the part of another DNA. 2. This causes difference in the distributions of genes 3. When these chromosomes are transferred to the next generation, new characters are expressed. |
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| 38. |
What is the possible ratio for the birth of a male or a female child? What is the reason for this? |
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Answer» The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome in female the offspring will be male. |
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| 39. |
What is the possibility for the birth of a male or a female child? Discuss. |
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Answer» The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome In female the offspring will be male. |
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| 40. |
Observe the flow chart given below.i) RNA is synthesized ↓ii) DNA unwinds ↓iii) RNA combines with ribosome ↓iv) RNA comes out through nuclear membrane. ↓v) Protein molecule is formed. ↓vi) Different amino acids are formed. ↓ a) Arrange the flow orderly. b) Identify the process. |
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Answer» a) DNA unwinds ↓ RNA Is synthesized ↓ RNA comes out through nuclear membrane. ↓ RNA combines with ribosome ↓ Different amino acids are formed. ↓ Protein molecule is formed b) Protein synthesis or gene action |
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| 41. |
The practice of blaming those mothers who give birth to girl children exists even today.a) As a science student, how will you respond to this situation? Substantiate. |
Answer»
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| 42. |
The process of crossing over of chromosomes that takes place in the initial phase of meiosis is illustrated below. Analyse-it and answer the questions.a) Arrange the stages appropriately.b) This process brings about variations in offspring. How? |
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Answer» a) C, A, B b) 1. Part of a DNA crosses Over to become the part of another DNA. 2. This causes difference in the distributions of genes. 3. When these chromosomes are transferred to the next generation, new characters are expressed. |
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| 43. |
Is there any difference in the number of chromosomes in male and female. |
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Answer» No. In male and female, there is not any difference in the number of chromosomes. In men and women 46 chromosomes are present. |
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| 44. |
The chromosomes from the father determine whether the child is male or female. Evaluate this statement on a scientific basis. |
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Answer» Males have 2 types of sex chromosomes. (X, Y) Females have only one type of sex chromosome (X, X) Sex determination is based on the type of male gamete that fuses with the egg. If the male gamete with Y chromosome fuses with the egg, then male child is born, if the male gamete with X chromosome fuses with the egg, then female child is born. |
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| 45. |
Complete the flowchart illustrating the location of gene by using the information given in the box:nucleus, gene, DNA, cell, chromosome |
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Answer» A – Cell B – Nucleus C – Chromosome D – DNA |
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| 46. |
Fill in the blanks in the illustration. |
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Answer» A. ggww B. GW C. Green colored round seed |
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| 47. |
Observe the illustration given below and answer the questions.a) Identify the dominant character b) How does the parental plant with green colored seed and the plant in the first generation differ in their alleles. c) Describe alleles. |
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Answer» a) Green b) Alleles in parental plant – G, G Allele in the first generation – G, g c) Different forms of a gene |
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| 48. |
Identify the word pair relationship and fill in the blanks:Female : 44 + XXMale: …… |
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Answer» Male: 44 + XY. |
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| 49. |
Fill in the blanks in the illustration given below. |
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Answer» A) tt B) t C) Tt D) dwarf |
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| 50. |
Which one of the following is major source of air pollution? (a) Automobiles (b) Industries (c) Burning coal (d) All of these |
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Answer» (d) All of these |
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