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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.
| 51. |
An anther with malfunctioning tapetum often fails of produce viable male gametophytes. Give one reason. |
| Answer» Malfunctioning tapetum does not provide enough nourishment to the developing male gametophytes. | |
| 52. |
What is it that prevents a child to suffer from a disease he/she is vaccinated against? Give one reason. |
| Answer» In vaccination, a prepartion of antigenic proteins of pathogen or inactivated/weakened pathogen are introduced in the body. The antibodies produced against these antigens neutralise the microbes in future infections and produce memory-B and T cells as well. | |
| 53. |
(a) " Oraganisms may be conformers or regulators . " Explain this statement and give one example of each. (b) Why are there more conformers than regulators in the animals world ? |
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Answer» Regulators : Some organisms are able to maintain homeostasis by physiological and behavioral means which ensures constant body temperature, constant osmotic concentration etc. All birds and mammals and a very few lower vertebrate and invertebrate species are capable of such regulation. The success of mammals is lagely due to their ability to maintain a constant body temperature and thrive whether they live Antarctica or in the Sahara desert. Conformers : The body temperature of animals and plants changes with the surrounding temperature. In aquatic animals, the osmotic concentration of the body fluids changes with that of surrounding water osmotic concentration. These animals and plants are simply conformers. This is because, these organisms can regulate only over a limited range of environmental conditions beyond which they only conform. |
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| 54. |
During the studies on genes in Drosopila that were sex-linked T.H. Morgan found F2-populabon phenotypic ratios deviated from expected 9 : 3 : 3 : 1. Explain the conclusion he arrived at. |
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Answer» 1. T. H . Morgan worked on fruit flies (Drosphila Melanogaster). 2. They are found suitable for studies in genetics for the following reasons : - They could be grown on simple synthetic medium in the laboratory. - The flies complete their life cycle in about two weeks. 3. Morgan carried our many dihybrid crosses in Drosophila, with the genes that were sex-lined, i.e., the genes are present on the X-chromosome. 4. He observed that the two genes under consideration in his experiments, did not segregate independently as in the case of characters studied by Mendel. |
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| 55. |
Are humming birds and fish regulators or conformers ? Give reasons in support of your answer. |
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Answer» Humming birds are conformers because small animals have a larger surface area ralative to their volume, they tend to body heat very fast when it is cold outside, then they have to expend much energy to generate body heat through metabolism. Whereas, fish (mammals) is regulator because they can maintain a constant body temperature. |
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| 56. |
Enumerate any six essentials of good ,effective Dairy Farm Management Practices. |
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Answer» Choosing the breeds that have high yielding potential is essential. ltbr. 2. Proper accommadation and adequate water are essential for the care of cattle. 3. Periodic visit by a verterinary doctor should be adequate in quantity and of good health of cattle 4. The food given to the cattle should be adequate in quantity and of good quality 5. The procedure of milking should be hygienic ,emphasis should be laid on storage and transportation of milk , so that the quality of milk is not affected in any way 6. Regular inspection of dairy farm should be done by appointed officials to ensure that all the instructions are being strictly followed . |
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| 57. |
(a) What is green revolution ? Mention the steps that led to it (b) Name the scientist whose contribution led to the development of semi-dwarf wheat varieties in India. |
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Answer» (a) Green revolution refers to a set of research and development of technology that increased agricultural production worldwide, particularly in the developing world. The initiatives resulted in : (i) High-yielding and disease rasistant varieties in wheat, rise, maize etc. (ii) Expansion of irrigation infrastructure, modernization of management techniques, distribution of hybridized seeds, synthetic fertilizers etc. (b) Norman Borlaug, the Nobel Laureate contributed for wheat and maize improvement in Mexio and developed semi-dwarf wheat varieties. |
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| 58. |
Name the specific enzyme responsible for nucleotide polymerisation in DNA replication. Write characteristic features of this enzyme. Name the region on E. coli DNA where this enzyme can initiate replication. |
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Answer» DNA dependent DNA polymerase enzyme is responsible for polymerisation in DNA. Characteristic features of this enzyme: These enzymes are highly efficient as they catalyze polymerisation of a large no. of nucleotides in very short time in E. coli. High degree of accuracy : There is a definite region in E. coli DNA where the replication originates. Such regions are termed as origin of replication cori. |
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| 59. |
Morgan carried out several dihybrid crosses in Drosophila and found `F_(2)`-rations deviated very significantly from the expected Mendelian ratio. Explain his findings with the help of and example. |
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Answer» (i) Morgan hybridised yellow bodied, white eyed female to brown to brown bodied red eyed males. He then intercrossed the `F_(1)` progeny and noticed that the `F_(2)` ratio deviated from the mendel dihybrid ratio of `9 : 3 : 3 : 1`. (ii) Morgan observed that during his study the proportion of parental gene combinations was much higher than non-parental type. since Morgan knew that his genes of study were located on the X-chromosome, he concluded that physical association or linkage of genes was responsible for this observation. (iii) Morgan further observed that the formation of non parental types was related to the strength of linkage. The Recombination frequency was much more for genes less tightly linked than got the genes more tightly linked. |
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| 60. |
State the objective with which a dairy farm is set up .Describe the essential steps to be followed for dairy farm management. |
| Answer» In dairy farm management , we deal with processes and systems that increase yield and improve quality of milk . Milk yield is primarily dependent on the quality of milk. For the yield potential to be realised, the cattle have to be well looked after , they have to be housed well , should have adequate water and be maintained disease free. The feeding of cattle should be carried out in a scientific manner. Stringent cleanliness and hygiene while milking , stroage and transport of the milk and its products. | |
| 61. |
Why is molecular diagnosis preferred over conventional methods ? Name any two techniques giving one use of each. |
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Answer» Molecular diagnosis in preferred over conventional methods because it allows early detection of disease the two techniques of molecular diagnosis are : (i) Polymerase Chain Reaction (PCR) : To detect HIV in suspected AIDS patients. (ii) Enzyme Linked Immunosorbent Assay (ELISA) : To detect antigens (proteins, glycoproteins etc., ) of pathogens or antibodies synthesized against the pathogen. |
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| 62. |
Write the IUPAC name of `CH_3CH=CH-underset(Br)underset(|)overset(CH_3)overset(|)C-CH_3` |
| Answer» 4 - Bromo - 4 - methylpent - 2 - ene. | |
| 63. |
Write the IUPAC name of the following compound : `CH_(3)-O-CH_(2)-CH-underset(CH_(3))underset(|)(CH_(2))-CH_(3)` |
| Answer» `{:(" "1" "2" " 3" "4),(CH_(3)-O-CH_(2)-CH-CH_(2)-CH_(3)),(" "|),(" "CH_(3)),(" "1-"Methoxy-3-Methylbutne"):}` | |
| 64. |
Which the IUPAC name of the following compound : `CH_(3)underset(Cl)underset(|)(CH) underset(Br)underset(|)(CH) CH_(3)` |
| Answer» 2 - Bromo - 3 - Chlorobutane. | |
| 65. |
Write the IUPAC name of the given compound : `CH_(3)-underset(CH_(3))underset(|)CH-CH_(2)-O-CH_(2)-CH_(3)` |
| Answer» 1-Ethoxy-2-Methyl propane | |
| 66. |
Give the IUPAC name of `H_(2)N-CH_(2)-CH_(2)-CH=CH_(2)`. |
| Answer» Correct Answer - But-3-en-1- amine. | |
| 67. |
Explain the following giving an exmaple in each case: (i) Linkage isomersion. (ii) An outer orbital complex .. (iii) A bidentate ligand. |
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Answer» (i) Linkage isomerism: This type of the isomerism arises due to the presence of an ambidentae Ligand in a coordination compound e.g., `[Co(NH_(3))_(5) NO_(2)]Cl_(2)`. (ii) An outer orbital complex. When ns,np and nd orbitals are involved in hybridistaion, outer orbital complex is formed e.g., `[CoF_(6)]^(2-)` (iii) Bidentate ligand : When a Ligand is bound to a metal ion thorugh two donor atoms it is said to be didentate ligand,. e.g., `H_(2)N-CH_(2)-CH_(2)-NH_(2)`. |
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| 68. |
Linkage and crossing-over of genes are alternatives of each other. Justify with the help of an example. |
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Answer» Crossing over : The breaking during meiosis of one maternal and one paternal chromosome, the exchange of corresponding sections of DNA, and the rejoining of the chromosomes. This process can result in an exchange of alleles between chromosome. Linkage in Fruit Files : It is the Proximity of two or more genes on a chromosome , the closer the ganes, the lower the probability that they will be separated during meiosis and hence the greater the probability that they will be inherited together. An example of using linkage to explore gene position is provided by inheritance of eye color and baby color in fruit flies, both of which are located on the X chromosome. This example begins with purebred (homozygous) parents, one yellow-bodied and red-eyed, the other grey-bodied and white-eyed. They mate to product all hetrozygous daughters, who carry the yellow-red combination on one homologous chromosome and the grey-white combination on the other. When the heterozygotes create gametes, the eye-color allles cannot assort independently from the body- color alleles because they are linked. |
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| 69. |
For a certain chemical reaction variation in concentration [A] vs. time (s) plot is given below : (i) Predict the order of the given reaction ? (ii) What does the slope of the time and intercept indicate ? (iii) What is the unit of rate constant k? |
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Answer» (i) Zero order reaction. (ii) Slope represents -k, Intercept represents `[R]_(0)` (iii) `"mol "L^(-1)s^(-1)`. |
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| 70. |
A solution of aqueous of KO hydrolysis `CH_(3)CHClCH_(2)CH_(3) and CH_(3)CH_(2)CH_(2)CH_(2)Cl`. Which one of these is more easily hydrolysed.? |
| Answer» `CH_(3)CHClCH_(2)CH_(3)` | |
| 71. |
(a) Account for the following observations : (i) `SF_(1)` is easily hydrolysed whereas `SF_(6)` is not easily hydrolysed. (ii) Chlorine water is a powerful bleaching agent. (iii) Bi (V) is a stronger oxidising agent than Sb(V) (b) What happens when : (i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of `CO_(2).` (ii) `XeF_(6)` undergoes partial hydrolysis. (Give the chemical equations involved). |
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Answer» (a) (i) S atom in `SF_(4)` is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in `SF_(6)` is protected by six F atoms. Thus attack by water molecules cannot take place easily. (ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action. `CH_(2) + H_(2)O to 2HCI + [O]` (iii) Due to inert pair affect Bi(V) can accept a pair of electrons to form more stable Bi (III). (+ 3 oxidation state of Bi is more stable than its + 5 oxidation state). (b) (i) Phosphorus undergoes disproportionation reaction to form phosphine gas. `P_(4) + 3NaOH + 3H_(2)O to PH_(3) + 3NaH_(2) PO_(2)` (ii) On partial hydrolysis `XeF_(6)` gives oxyfluoride `XeOF_(4)` and HF. `XeF_(6) + H_(2)O to XeOF_(4) + 2HF` |
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| 72. |
(b) Describe a chemical test to distinguish between (i) Ethanal and propanal (ii) Benzldehyde and Acetophenone (iii) Pronepan 2 one and propan 3 one |
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Answer» (i) See Q 30 (b)(i) Set I (D.B)-2013 (ii) See Q 30 or (ii) Set I (O.D)-2011 (iii) See Q 30 (ii) Set -I (O.D)-2013 |
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| 73. |
(a) What inspired N.Bartlett for carrying out reaction between Xe and `PtF_(6)` ? (b) Arrange the following in the order of property indicated against each set : (i) `F_(2), I_(2), Br_(2), CI_(2)` (increasing bond dissociation enthalpy) (ii) `NH_(3), AsH_(3), SbH_(3), BiH_(3), PH_(3)` (decreasing base strength) (c) Complete the following equations : (i) `CI_(2) + NaOH ("cold and dilute") to` (ii) `Fe^(3+) SO_(2) + H_(2)O to` |
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Answer» (a) N.Barlett first prepared a red compound `O_(2) ""^(+)PtF_(6)^(-)`. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So, he carried out reaction between `Xe " and " PtF_(6)`. (ii) (i) `I_(2) lt F_(2) lt Br_(2) lt CI_(2)` (ii) `NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3) gt BiH_(3)` (c) (i) `2NaOH + CI_(2) to NaCI + NaOCI + H_(2)O` (ii) `2Fe^(3+) + SO_(2) + 2H_(2)O to 2Fe^(2+) + SO_(4)^(2+) + 4H^(+)` |
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| 74. |
Arrange the following groups of substances in the order of the property indicated against each group: (i) `NH_(3)PH_(3)AsH_(3)SbH_(3)` increasing order of boiling points (ii)O,S,Se,To increasing order of electron gain enthalphy with negative sign (iii) `F_(2),CI_(2),Br_(2),I_(2)` increasing order of bond dissociation enthalphy |
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Answer» (i) `PH_(3)lt` As `H_(3)ltNH_(3)ltSnH_(3)ltBiH_(3)` As we more form `PH_(3)` to `BiH_(3)` molecular mass in creases vander waals forces increases and hence B.P but B.P of `NH_(3)` is abnormally higher than those of `PH_(3)` and As`H_(3)` because of intermolecular H Bonding (ii) `SgtsegtTegtO` Oxygen has lowest due to small size and electron repulsion (iii) See Q 25 (a) Part set I (D.B)-2010 |
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| 75. |
Arrange the following compound in an increasing order of their reactivity in nucleophilic addition reactions : ethanal propanal, butanone, propanone. |
| Answer» Butanone `lt` Propanone `lt` Ethanol. | |
| 76. |
How would you account for the following ? (i) With the same d-orbital configuration `(d^(4)) Cr^(2+)` is reducting agent while `Mn^(3+)` is an oxidizing agent. (ii) The actionoids exhibits a larger numbe of oxidation states than the corresponding members in the lanthanoid series. (iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions. |
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Answer» (i) `Cr^(2+)` has the configuration `d^(4)` and easily changes to `Cr^(3+)` which has half `t_(2g)` configuration and hence more stable. Therefore `Cr^(2+)` is reducing. On the other hand, the change from `Mn^(3+)` to `Mn^(2+)` results in the half filled , `d^(5)` configuration which has extra stability. Therefore `Mn^(3+)` is oxidising. (ii) the actionoids exhibit a larger number of oxidation state than the correponding member in the lanthanoid series. This is due to very small energy gap between 5f, 6d and 7s. Orbital in the actinoid series. (iii) Most of the transmition metal ions exhibit characteristic in colours in aqueous solution. This is due to the particle absorpation of visible light. The absorbed light promotes the electron from one obrbital to another orbital to the same d subshell. Due to presence of unpaired electrons in d-orbitals and d -d- transition. |
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| 77. |
How would you account for the following :(i) `Cr^(2+)` is reducing in nature while with the same d-orbital configuration `(d^(4)),Mn^(2+)` is oxidising in nature. (ii) In the transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. |
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Answer» (i) `Cr^(2+)` has the configuration `d^(4)` and easily changes to `d^(3)` has filled orbitals and hence stable. Therefore, `Cr^(2+)` is reducing .On the other hand `Mn^(2+)` is more stable due to half filled `d^(5)` configuartion and `Mn^(3+)` easliy changes to `Mn^(2+)` and hence is oxiding . (ii) This is due to the large no.of unpaired in d-orbitals in the middle of the series. |
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| 78. |
What is the no. of atoms per unit cell (z) in a body -centred cubic strcuture ? |
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Answer» In a body- centred cubic structure : `z=1+1=2` Each of the eight corner atoms will contribute only one atom per unit cell. Cell atom contributes only one atom per unit cell. |
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| 79. |
What is the role of collectors in Froth Floatation process ? |
| Answer» Froth floatation process is used when ores has wettiability towards oil while impurties have wettiability towards water collectors are used in froth floatation process to collect ore particles and to remove them with froth. | |
| 80. |
Define the following terms: (i) Nucleotide (ii) Anomers (iii) Essential amino acids |
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Answer» (i) Nucleotide : It is the monomer unit of DNA which is formed by nitrogenous base of DNA which is formed by nitrogenous base Deoxyrobose sugar and Phosphoric acid (ii) Anomer: Anomers are cyclic monosaccaride which are differing from each other in the configureation of c -1 if they are aldose or in the configruation a c-2 if they are ektoses (iii) Essential amino acids: The amino acid can not synthesised by body and essential for |
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| 81. |
Anino acids may be acidic, alkaline or neutral, How does this happen? What are essential and non-essential amino acids? Name one of each type. |
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Answer» Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Equal number of amino and carboxyl groups makes it neutral, more number of amino than carboxyl groups makes it basic and more carboxyl as compared to amino groups makes it acidic. Essential amino acids : The amino acids which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids. Example : Valine, Leucine. Non-essential amino acids : The acids, which can be synthesised in the body are known as non-essential amino acids. Exmple : Alanine, Aspartic acid. |
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| 82. |
Explain the following terms with one example in each case : (i) Food preservatives (ii) Enzymes (iii) Detergents |
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Answer» (i) Food preservatives : Food preservatives are the substances used to prevent spoilage of food due to microbial growth during storage. The most common preservatives used are table salt, sugar, vegetable oils and sodium benzoate `(C_(6)H_(5)COONa)` (ii) Enzymes : The proteins which perform the role of biological catalysts in the body are called enzymes. Example : E. coli. (iii) Detergents : Detergents may be defined as ammonium, sulphate or hydrogen sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms. Example : The more common detergents are the sodium salts of long chain sulphonic acids. |
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| 83. |
Answer the following: (i) Haloalkanes easily dissolve in organic solvents, why? (ii) What is known as a racemic mixture ? Give an example. (iii) Of the two bromoderivatives, `C_6H_5CH (CH_3) Br and C_6H_5CH (C_6H_5)Br`, which one is more reactive in `S_(n^1)` substitution reaction and why? |
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Answer» (i) Haloalkanes are easily dissolve in organic solvents becouse the new forces of attraction set up between holoalkanes and solvent are of the same strength as the forces of attraction being broken between solvent molecules and haloalJltanes. Hence energy released. (ii) A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be canceiled by the rotation due to other isomer. Such a mixture is known as racemic mixture : for example(`pm`) butan-2-ol. (iii) The carbocation intermediate obtained from `C_6H_5CH (CH_5) Br [ i.e., C_6H_5CH (C_6H_5)^+]` is more stable than that obtained from `C_6H_5CH (CH_3) Br (i.e., C_6H_5 CH (CH_3)^+]` because it is stabilised bg two phenyl groups due to resonance. For `S_(N^1)` reaction : `C_6H_5CH (C_6H_5) Brgt C_6H_5CH (CH_3) Br` |
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| 84. |
Compare the following complexes with respect to structural shapes of units , magnetic behaviour and hybrid orbitals involved in units (i) `[Ni(CN)_(4)]^(2-)` (ii) `[NiCl_(4)]^(2-)` (iii) [At . Nos .: Ni =28 , Co =27] |
| Answer» `{:("Complexes behaviour","Magnetic","Hybridisation","Shape"),(("i")[Ni(CN)_(4)]^(2-),"Diamagnetic"," dsp"^(2),"square planar"),((ii)[NiCl_(4)]^(2-),"Paramagnetic"," sp"^(3),"tetrahedral"),((iii)[CoF_(6)]^(3-),"Paramagnetic"," sp"^(3)d^(2),"octahedral"):}` | |
| 85. |
Write the principles of the following methods: (i) Vapour phase refining (ii) Zone refining (iii) Chromatography |
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Answer» (i) Vapour phase refining: In this method, crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at lower temperature and then decomposing the volatile compound at some higher temperature to given the pure metal. (ii) Zone refining: It is based upon the principle that the impurities are more soluble in the molten state than in the solid state of the metal. (iii) Chromatography: It is based upon the principle that the different components of a mixture are adsorbed to different extents on an dsorbent |
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| 86. |
Define the following (i) Anionic detergents (ii) Limited spectrum antibiotics (iii) Tranquilizers |
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Answer» (i) Anionic detergents: See Q.20 (i), Delthi Board, Set -I, 2017 (ii) Limited spectrum antibiotics: Limited spectrum antibiotics are designed to kill a specific group of pathogens. Example: Pencillin (iii) Tranquilizers : Drugs which are used for the treatment of stress, fatigue, mild and sevre mental disease are called tranquilizers Example: Chlordiazepoxide, Iproniazid, equanil |
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| 87. |
Define the following (i) Cationic detergents (ii) Narrow spectrum antibiotics (iii) Disinfectants |
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Answer» (i) Cationic detergents: In these detergents, a major part of their molecules are cations and it is the cationic part of molecule which is invoved in their cleasing action. Example quaternary ammonium salts (chlorides, bromides, acetates etc.) `underset("(Cetyltrimethylammonium bromide)")([CH_(3) - (CH_(2))_(15) overset(o+)(-) underset(CH_(3))underset(|)overset(overset(CH_(3))(|))(N) - CH_(3)]Br^(-))` (ii) Narrow spectrum antibiotics: Antibiotics which are effective only against certain disease e.g., pencillin is narrow spectrum antibiotic which is effective only against disease caused by various cocci and some gramm positive bacteria. (iii) Disinfectants: They are the chemical substances which kill micro-organisms but are not safe to be applied to the living tissues. These are generally used to kill the micro organisms present in drains, toilets, floors etc. For example : 1% solution of phenol |
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| 88. |
Explain the following situations : (i) In the structure of `HNO_(3)` molecule , the N-O bond (121 pm) is shorter than N-OH bond (140pm). (ii) `SF_(4)` is easily hydrolysed whereas `SF_(6)` is not easily hydrolysed. (iii) `XeF_(2)` has a straight linear structure and not a bent angular structure. |
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Answer» Due to smaller size of N-O , than N-O bond length of `HNO_(3)` is smaller than N-H bond length . (ii) In `SF_(4)` , S is not sterically since it is surrounded by only four F atoms , so attack of `H_(2)O` molecules can take place easily and hence hydrolysis take place . On the other hand , S is sterically protected by six F atoms in `SF_(6)` and does not allow `H_(2)O` molecule to attack the S atoms. Thus `SF_(6)` does not undergo hydrolysis. (iii) As `XeF_(2)` has 5 pairs (10) electrons around Xe forming a `sp^(3)`d hybridisation. Therefore , its geometry is linear. |
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| 89. |
Determine the values of equilibrium constant `(K_C) and DeltaG^o` for the following reaction : |
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Answer» The cell reaction is `Ni(s)+2Ag^+(aq)toNi^(2+)(aq) +2Ag(s)`, `DeltaG^o=-nFE^o` `E^o=1.05V,n=2,f=96500C//"mol"`. `DeltaG^o=-(2mol)xx(96500C//mol)xx(1.05V)`. `=-202650 CV (1CV=1)` `=-202650J` Colculation of K in the from `DeltaG^o`, `DeltaG^o=-202650J,R=8.314J//mol//K`. `T=298k`. `logK=(DeltaG^o)/(2.303RT)=(-202650J)/((2.303)xx(8.314J//mol//K)xx(298K))` `=K=(-202650)/5705.84832=35.51` `or K=0.35xx10^2` |
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| 90. |
How would you account for the following : (i) `H_(2)S` is acidic than`H_(2)O`. (ii) The N-O bond in `NO_(2)^(-)` is shorter than the `N-O` bond in `NO_(3)^(-)`. (iii) Both `O_(2)` and `F_(2)` stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of flurine. |
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Answer» (i)Size of centre atom is `H_(2)S` is larege than `H_(2)O` Hence the S-H bond is weaker than O-H bond. Hence `H_(2)S` can dissociate to given `H+`ions (iii)` O_(2)` exceeds `F_(2)` in doing so due to ability of oxygn to from multiple bonds with metals. |
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| 91. |
State reasons for each of the following : (i) All the P-Cl bonds in `PCl_5` molecule are not equivalent. (ii) Sulphur has greater tendency for catenation than oxygen. |
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Answer» (i) The three equatorial P-Cl bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as corn pared to equatorial bond pairs. (ii) Catenation is the tendency of an atom to form bonds with identical atoms. In this group, only sulphur has a tendency for catenation. Due to small size, the long pairs of e-s on the oxrgen atoms repel the bond pair of O-O bond to a greater extend than that the long pairs of e-s on the sulphur atom in S-S bond. As a result, S-S bond is much stronger than · O-O bond. Hence S has a tendency for catenation than c. |
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| 92. |
(a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment ? (b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture `2 xx 10^(-4)m.` The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. |
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Answer» (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double slit experiment is modified by diffraction from each of two slits. (b) Given that : Wavelength of the light beam, `lambda_(1) =590 nm = 5.9 xx 10^(-7) m` Wavelength of another light beam, `lambda_(2) =596 nm = 5.96 xx 10^(-7) m` Distance of the slits from the screen = D = 1.5 m Distance between the two slits `= a = 2 xx 10^(-4) m` For the first secondry maxima, `sin theta=(3lambda_(1))/(2a)=(x_(1))/(D) " OR " x_(1)=(3lambda_(1)D)/(2a) and x_(2)=(3lambda_(2)D)/(2a)` `therefore ` Spacing between the positions of first secondary maxima of two sodium lines `x_(1)-x_(2)=(3D)/(2a) (lambda_(2)-lambda_(1))=6.75xx10^(-5)m.` |
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| 93. |
Ram is a student of class `X` in a village school. His uncle gifted him a bicycle with a dynomo fifted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. He asked this questions to his teacher. The teacher considered it an opprtunity to explain the working to the whole class. Answer the following questions : `(i)` State the principle and working of dynamo. `(ii)` Write two values each displayed by Ram and his school teacher. |
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Answer» `(i)` Principle : It works on the principle of electromagnetic induction. When a coil is rotated in a magnetic field, magnetic flux linked to it changes continuously giving rise to an alternating emf. Working : When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil. Considering the armature of be in vertical position and as it rotates in anticlock wise direction, the side ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the circuit, the current flows along `B_(1)`, `R_(1)`, `B_(2)`. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current it reversed and in external circuit it flows along `B_(2)R_(l)B_(1)`. Thus, the direction of induced emf and current changes in the external circuit after each half revoluation. `(ii)` Ram is very curious to know, quest for knowledge. Teacher is very helping and having good knowledge of science. |
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| 94. |
Many plant and animal species are on the verge of their extinction because of loss of forest land by indiscriminate use by the humans .As a biology student what method would you suggest along with its advantages that can protect such threatened species from getting extinct ? |
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Answer» As a biology student ,I would suggest the given method that can protect can protect these threatened spercies from becoming extinct. Ex Situ Convervation : The threatene species of palnts and animals are taken out of their habitats and are kept in special setting such as zoological parks, botanical gradens and wild life parks. The gametes of endangered species can be preserved by methods such as cryopreservation and can be fertilized in vitro and plants can be propageted through tissue culture methods similarly ,seeds can be preserved in seed banks. This method is called an off side conservation method . |
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| 95. |
Why do hermaphrodite angiosperms develop out breeding devices? Explain any two such devices with the help of examples. |
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Answer» Majority of flowering plants produce hermaphrodite flowers and pollen grains are likely to come in contact with the stigma of the same flower. Continued self-pollination result in inbreeding depression. Flowering plants have developed many devices to discourage self- pollination and to encourage cross-pollination. The out breeding devices are as follows : 1. In some plant species pollen release and stigma receptivity are not synchronized. 2. In some plant species the anther and stigma are placed at different positions so the the pollen cannot come in cont.act with the stigma of the same flower . . |
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| 96. |
Why do we add an inoculum of curd to milk for curdling it |
| Answer» The LAB (Lactic Acid Bacteria) produce acids that coagulate and partially digest the milk proteins. A small amount of curd added to the fresh milk as inoculum or starter contains millions of LAB, which at suitable temperatures multiply, thus converting milk to curd. | |
| 97. |
Which of the following substances are para-magnetic? Bi, Al, Cu, Pb, Ni |
| Answer» Para-magnetic substances are Aluminium (Al) and Calcium (Ca). | |
| 98. |
An aqueous solution of `2` per cent `(wt.//wt)` non-volatile solute exerts a pressure of `1.004` bar at the boiling point of the solvent. What is the molecular mass of the solute? |
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Answer» `P^(@)A=1atm=1.013"bar", Ps =1.004"bar"` `W_(B)=2g" "W_(s)=100g//mol " "W_(A)=98g` `(P^(@)A-PS)/(P^(@)A)=X_(B)" "implies" "(P^(@)A-PA)/(P^(@)A)~(nB)/(nA)` `(1.013-1.004)/(1.013)=(2)/(M_(B))xx18/98impliesM_(B)=(2xx18)/(98)xx(1.013)/(0.009)` ` thereforeM_(B)=41.35g//mol` |
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| 99. |
I-V graph for a metallie wire at two different tempearture, `T_(1)and T_(2)` is as shown in the figure. Which of the two temperatures is lower and why ? |
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Answer» V=IR For given voltage `1 alpha 1/R` Resistance increase with increase of temperature and temperature is inversely proportional to voltage ` T_(1) lt T_(2)` |
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| 100. |
Describe the role of the following : (i) NaCN in the extraction of silver ore. (ii) Cryolite in the extraction of aluminium from pure alumina. |
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Answer» (i) Na CN is used to form soluble complex of cyanide while impurities remain unaftected. (ii) To make laumina a good conductor of electricity and to low down the melting point of both from 2323 k to 1140 k |
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