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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.
| 1. |
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is `1s^(2),2s^(2)2p^(3)`, the simplest formula for this compound isA. `Mg_(2)X`B. `MgX_(2)`C. `Mg_(2)X_(3)`D. `Mg_(3)X_(2)` |
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Answer» Correct Answer - D Given, electronic configuration of `X=1s^(2)2s^(2)2p^(3)` `therefore` The valency of X will be 3. The valency of Mg is +2 `therefore` Magnesium reacts with element X to form an ionic compound with formula `Mg_(3)X_(2)`. |
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| 2. |
Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: a. 60 mL `(M)/(10) HCl + 40 mL (M)/(10)` NaOH b. 55 mL `(M)/(10) HCl + 45 mL (M)/(10)` NaOH c.75 mL `(M)/(5) HCl + 25 mL (M)/(5)` NaOH d. 100 mL `(M)/(10) HCl + 100 mL (M)/(10)` NaOH pH of which one of them will be equal to 1 ?A. IVB. IC. IID. III |
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Answer» Correct Answer - D `75 mL(M)/(5) HCl+25mL (M)/(5)NaOH` Milliequivalent of HCl `=75 mL " of" (M)/(5) HCl=(1)/(5)xx75=15` Milliequivalent of NaOH `=25 mL " of" (M)/(5) NaOH` `=(1)/(5)xx25=5` `therefore` Milliequivalent of HCl left unused =15-5=10 Volume of solution =100 mL `therefore` Molarity of `[H^(+)]` in the resulting mixture `=(10)/(100)=(1)/(10)` `therefore " " pH="log" (1)/([H^(+)])` =log (10)=1 |
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| 3. |
According to Hugo de Vries, the mechanism of evolution isA. phenotypic variationsB. saltationC. multiple step mutationsD. minor mutations |
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Answer» According to Hugo de Vries, the mechanism of evoluation is saltation. Hugo de Vries (1901) proposed mutation theory of evolutioin and stated that evolutoin is a jerky process in which new species are evolved due to discontinous sudden variations or saltation. These are the signle step large mutations occuring in population. |
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| 4. |
Which one of the following population interaction is widely used in medical science for the production of antibiotics ?A. ParasitismB. MutualismC. CommensalismD. Amensalism |
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Answer» Amensalism is widely used in medical science for the production of antibiotics. It involves, the secretion of chemicals called allochemics by one microbial group to harm other microbes, e.g., Penicillium secretes chemicals to inhibit the growth of Staphylococcus bacteria. These chemicals can be used in medical science for the production of antibiotics. On the other hand, no such chemicals are secreted in parasitism, mutualism and commertsalism. |
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| 5. |
Which one of the statements is incorrect?A. Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atomsB. Glycolysis occurs in cytosolC. Enzymes of TCA cycle are present in mitochondrial matrixD. Oxidative phosphorylation takes place in outer mitochindrial membrane |
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Answer» Oxidative phosphorylation is the process of ATP formation due to the transfer of electrons from NADH or `FADH_(2)` to oxygen molecule `(O_(2))` by a series of electron carriers . This process occurs in the inner mitochondrial membrane because of its less permeability , presence of ETC proteins and ATP synthase . The rest three statements are correct . |
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| 6. |
Select the incorrect mathc.A. `{:("Submetacentric chromosomes" ,-,"L-shaped chromosomes"):}`B. `{:("Allosomes" ,-,"Sex chromosomes"):}`C. `{:("Lampbrush chromosomes" ,-,"Diplotene bivalents"):}`D. `{:("Polytene chromosomes " ,-,"Oocytes of amphibians"):}` |
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Answer» Polytene chromosomes are glant chromosomes that are quite common in the salivary glands of insects therefore they are popularly called as salivary chromosomes . The Lampbrush chromosomes are highly elongated special kind of synapsed mid-prophase or diplotene chromosome that are bivalents . Sex chromosome are also called as allosomes . They determine the sex of an organism . Submetacentric chromosomes have a submedian centromere . They appear L-shaped during metaphase. Therefore , except option (d) , all are correctly matched. |
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| 7. |
Which of the following terms descibe humans dentition ?A. Pleurodont, Monophyodont, HomodontB. Thecodont, Diphyodent, HeterodontC. Thecodont, Diphyodont, HomodontD. Pleurodont, Diphyodont, Heterodont |
| Answer» The terms, thecodont , diphyodont and heterodont describe human dentition . In men , two types of teeth are found , milk or deciduous teeth and permanent teeth . Thus , they have diphyodont teeth . The teeth are thecodont , i.e., they remain embedded in the sockets of the jaw bones . Men have four types of teeth , incisors , canine , premolars and molars , i.e., heterodont teeth. | |
| 8. |
Which of the following features is used to identify a male cockroach from a female cockroach ?A. Forewings with darker tegminaB. Presence of caudal stylesC. Presence of a boat-shaped sternum on the 9th abdominal segmentD. Presence of anal cerci |
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Answer» In male cockroach, the `9^(th)` sternite bears a pair of sn~all and spine-like unjointed caudal or anal styles which are absent in female cockroach. The anal styles are believed to functiou as motion detector. Besides this, the other three characters, i.e. anal cerci, boat-shaped sternum on 9th abdominal segment and forewings with darker tegmina are found in both male and female cockroaches. |
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| 9. |
Which one of these animals is not a homeotherm ?A. CamelusB. CheloneC. MacropusD. Psittacula |
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Answer» Among the given animals Chelone is not a homeotherm. It is green sea turtle belonging to class-Reptilia which are ectotherms or cold-blooded and their internal body temperature varies according to the ambient environment. In contrast, Camelus and Macropus belonging to class- Mammalia and Psittacula belonging to class-Aves are homeotherms. They can maintain constant body temperature irrespective of surrounding temperature. |
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| 10. |
The increasing order of atomic radii of the following group 13 elements isA. `B lt Ga lt Al lt Tl lt In`B. `B lt Al lt Ga lt In lt Tl`C. `B lt Al lt In lt Ga lt Tl`D. `B lt Ga lt Al lt In lt Tl` |
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Answer» Correct Answer - D The atomic radii as well as ionic radii increases on moving down the group 13 elements because of the successive addition of one extra shell of electrons. However, there is an anomaly at gallium in case of atomic radii. Atomic radii of Ga is lesser as compared to Al. Gallium (Ga) with electronic configuration, `[Ar]_(18)3d^(10)4s^(2)4p^(1)` has an extra d-electrons which do not screen the nucleus effectively. Consequentl, electrons of Ga are more attracted by nucleus. Thus, the increasing order of atomic radii of the group 13 elements is B (85 pm) `lt` Ga (135 pm) `lt` Al (143 pm) `lt` In (167 pm) `lt` Tl (170 pm). |
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| 11. |
Which of the following is not true for halogens ?A. All but fluorine show positive oxidation statesB. All are oxidising agentsC. All form monobasic oxyacidsD. Chlorine has the highest electron-gain enthalpy |
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Answer» Correct Answer - A Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d-orbitals and therefore, can expand their octets and show `+1, +3, +5` and `+7` oxidation states. Thus, option (a) is incorrect. |
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| 12. |
The correct difference between first and second order reactions is thatA. a first-order reaction can be catalysed , a second-order reaction cannot be catalysedB. the half-life of a first-order reaction does not depend on `[A]_(0)`, the half-life of a second-order reaction does depend on `[A]_(0)`C. the rate of a first-order reaction does not depend on reactant concentrations, the rate of a second-order reaction does depend on reactant concentrationsD. the rate of a first-order reaction does depend on reactant concentrations, the rate of a second-order reaction does not depend on reactant concentrations |
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Answer» Correct Answer - B For first order reactions, the rate of reaction is proportional to the first power of the concentration of the first power of the concentration of the reactant. For, `A to B` Rate`=-(d[A])/(dt)=k[A] " " ["where,k=constant"]` Half-life `(t_(1//2)=(0.693)/(k))` `therefore` Rate of first order reaction depends upon reactant concentrations and half life does not depend upon initial concentration of reactant, `[A]_(0)`. For second order reactions, the rate of reaction is proportional to the second power of the concentration of the reactant. For, `2A to B` Rate `=k[A]^(2)` Half-life`(t_(1//2)=(1)/(k[A]_(0)))` `therefore` Rate of second order reaction depends upon reactant concentration and half life also does depend on `[A]_(0)`. |
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| 13. |
Which of the following is correct with respect to `-I` effect of the substitutes? `(R=alkyl)`A. `-NH_(2) gt -OR gt F`B. `-NR_(2) lt -OR lt -F`C. `-NH_(2) lt -OR lt -F`D. `-NH_(2) gt -OR gt -F` |
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Answer» Correct Answer - B::C I effect is related to the ability of substituent for the electron attraction capacity from the attached carbon atom. i.e. it is based on electronegativity of an atom. This effect increases with increase in the electronegativity of an atom. From above we can conclude that options (b) and (c ) are correct. `-NR_(2) lt -OR lt -F` (-I effect) `-NH_(2) lt - OR lt -F` (-I effect) Also, options(a) and (d) shows the order of +I effect. `-NH_(2) gt -OR gt -F` (+I effect) `-NR_(2) gt -OR gt -F` (+ I effect) |
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| 14. |
World Ozone Day is celebrated onA. 16th SeptemberB. 21st AprilC. 5th JuneD. 22nd April |
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Answer» Correct Answer - A World Ozone Day is celebrated on 16th September to control `O_(3)` depletion . Ozone layer is a fragile shield of gas that protects earth from harmful UV-rays . On 21st April the Civil Service Day and National Yellow Bat Day is celebrated . 5th June of every year is celebrated as annual event , celebrated on 22nd April of every year. |
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| 15. |
Regarding cross-linked or network polymers, which of the following statements is incorrect?A. Examples are bakelite and melamineB. They are formed from bi-and tri-functional monomersC. They contain covalent bonds between various linear polymer chainsD. They contain strong covalent bonds in their polymer chains |
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Answer» Correct Answer - D Cross-linked or network polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains. These are hard, rigid and brittle due to cross-links e.g. bakelite, melamine etc. Thus, option (d) is incorrect. |
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| 16. |
Pollen grains can be stored for several years in liquid nitrogen having a temperature ofA. `-196^(@)C`B. `-80^(@)C`C. `-120^(@)C`D. `-160^(@)C` |
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Answer» Correct Answer - A Pollen grains can be stored for several years in liquid nitrogen having a temperature of `-196^(@)C` Pollen grains can be later used in plant breeding programmes . |
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| 17. |
Select the correct match.A. `{:((a) , "Matthew Meselson and" , : , "Pisum sativum F. Stahl"):}`B. `{: ((b) , "Alfred Hershey and" , : , TMV "Martha Chase"):}`C. `{:((c) , "Alec Jeffreys" , : , "Streptococus pneumoniae"):}`D. `{:((d) , "Francois Jacob and Jacques Monod" , : , "Lac operon"):}` |
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Answer» Correct Answer - D Jacob and Monod (1916) discovered the lac operon . An operon is a part of genetic material or DNA which acts as a single regulated unit . It possesses one or more structural genes , an operator gene , a promoter gene , a regulator gene , a repressor gene and an inducer or corepressor . Matthew Meselson and F Stahl discovered the semi-conservative mode of DNA replication in E coli. Alfred Hershey and Martha Chase use `T_(2)` Bacteiriophage in their experiments to infect E.coli and proved that DNA is the genetic material . Alec Jeffreys (1984) invented the DNA fingerprinting technique . This technique determines nucleotide sequences of certain areas of DNA which are unique to each individual. |
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| 18. |
The Golgi complex participates inA. respiration in bacteriaB. formation of secretory vesiclesC. fatty acid breakdownD. activatio of amino acid |
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Answer» Golgi complex participates in the formation of secretory vesicles . It is a cytoplasmic structure found in eukaryotic cells . It is made up of four parts , cisternae , tubules , vesicles and vacuoles . The forming face or cisternae receives vesicles from endoplasmic reticulum . Their contents pass through various cisternae with the help of coated vesicles and intercisternal connectives . They ultimately reach the maturing face where they are budded off as , coated secretory or Golgian vesicles or vacuoles . In bacteria , respiration occurs with the help of mesosomes . The break down of fatty acid |
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| 19. |
Which one of the following pairs is wrongly matched ?A. XO type sex-determination - GrasshopperB. ABO blood grouping - CodominanceC. Strach synthesis in pea - Multiple allelesD. TH Morgan - Linkage |
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Answer» Correct Answer - C In the given pairs , option (c) is wrongly matched. Starch synthesis in pea is an example of pleiotropy . A pleiotropic gene is a single gene which produces many or multiple unrelated phenotypes . Rest of the pairs are correctly matched . Concept Enhancer The gene for starch synthesis in pea seeds has two alleles B and b . In BB genotype , large starch grains are produced . After maturation the seeds are round . In bb homozygous condition , smaller starch grains are produced and mature seeds are wrinkled . Bb heterozygotes form round seeds so that B seems to be dominant allele. However , Bb seeds have starch grains of intermediate size , showing incomplete dominance. |
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| 20. |
Select the correct matchA. TH Morgan - TransductionB. `F_(2) xx` Recessive parent - Dihybrid crossC. Ribozymes - Nucleic acidD. G Mendel - Transformation |
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Answer» Correct Answer - C Ribozymes are RNA molecules having enzymatic activity ,i.e., they are capable of catalysing specific biochemical reactions . Hence , they are nucleic acids with enzymatic function. TH Morgan is known as the Father of Experimental Genetics . He worked on linkage , crossing over , linkage maps . etc. In dihybrid cross , two allelic pairs are used for crossing . Mendal is considered as the Father of Genetics . He proposed the laws of inheritance. |
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| 21. |
Which of the following organsims are known as chief producers in the oceans ?A. CyanobacteriaB. DiatomsC. DinoflagellatesD. Euglenoids |
| Answer» Diatoms are chief producers in the oceans and they contribute 40% of marine primary productivity. They constitute a major group of unicellular eukaryotic microalgae and are among the most common types of phytoplanktons. The other given organisms also exhibit autotrophic mode of nutrition. | |
| 22. |
Select the wrong statement .A. Pseudopodia are locomotory and feeding structures in sporozoansB. Mushrooms belong to BasidiomycetesC. Cell wall is present in members of Fungi and PlantaeD. Mitochondria are the powerhouse of the cell in all kingdoms except Monera |
| Answer» Sporozoans are endoparasites . They lack locomotory organelles like cilia , flagella , pseudopodia , etc . , e.g. Plasmodium , Pseudopodia are found in amoeboid protozoans , e.g., Amoeba , Entamoeba , etc . Therefore , statement (a) is wrong while rest of the statements are correct . | |
| 23. |
All of the following are parts of an operon exceptA. an enhancerB. structural genesC. an operatorD. a promoter |
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Answer» Except enhancer, all the given components are parts of an operon, Enhancer sequences are present in eukaryotes that, when bound by specific proteins or transcription factors, enhance the trascription of an associated gene. On the other hand , operon is a regulatory unit of DNA containing a cluster of genes in prokaryotes. The promoter of operon is the site where RNA polymerase binds . The promoter of operon is the site where RNA polymerase binds . The operator acts as on-off switch to control transcription. The structural genes code for enzymes involved in metabolic pathway. |
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| 24. |
All of the following are included in ex-situ conservation exceptA. botanical gardensB. sacred grovesC. wildlife safari parksD. seed banks |
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Answer» Sacred groves is a mode of in situ conservation in which forest fragments of varying size are protected by religious communities . it helps to protect the biota of that area on site . On the other hand, botanical gardens, seed banks and wildlife safari parks are the examples of ex situ conservation in which the biota is protected outsideits natural habitat. |
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| 25. |
Match the items given in Column I with those in Column II and select the correct option given below A. `{:(1 , 2, 3), ("i" , "ii", "iii" ):}`B. `{:(1 , 2, 3), ("i" , "iii", "ii" ):}`C. `{:(1 , 2, 3), ("iii" , "i", "ii" ):}`D. `{:(1 , 2, 3), ("ii" , "i", "iii" ):}` |
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Answer» The atrioventricular opening between the left atrium and left ventricle is guarded by the bicuspid valve. It is also called as mitral valve . The right atrioventricular opening is guarded by the tricuspid valve . It has three flaps . Semilunar valve is found in right ventricle and pulmonary artery. Therefore , option (c) is correct . |
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| 26. |
In a growing population of a country .A. reproductive and pre-reproductive individuals are equal in numberB. reproductive individuals are less than the post-reproductive individualsC. pre-reproductive individuals are more than the reproductive individualsD. pre-reproductive individuals are less than the reproductive individuals |
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Answer» In a growing population younger population (or pre-reproductive individuals) size is larger than that of reproductive individuals . Such population is represented by a triangular-shaped age pyramid . Whereas , the equal number of reproductive and pre-reproductive individuals represents a stable population and the age pyramid is bell-shaped Less number of pre-reproductive individuals than reproductive individuals represents declining population and age pyramid appears urn-shaped . The similar case is seen when reprodutive individuals are less than the post - reproductive individuals . |
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| 27. |
AGGTATCGCAT is a sequence from the coding strand of a gene . What will be the corresponding sequence of the transcribed mRNA ?A. ACCUAUGCGAUB. UGGTUTCGCATC. AGGUAUCGCAUD. UCCAUGCGUA |
| Answer» Coding strand is the one that codes for mRNA. It has same nucleotide sequence as that of mRNA except thymic (T) is replaced by uracil (U) in mRNA. Hence, the corresponding sequence of transcribed mRNA by template or non-coding strand (complementary to RNA) is AGGUAUCGCAU. | |
| 28. |
In which of the following forms is iron absorbed by plantsA. Free elementB. FerrousC. FerricD. Both ferric acid and ferrous |
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Answer» Correct Answer - C According to NCERT , plants absorb iron mostly in the form of ferric `(Fe^(3+))` ions. However , plants in acidic soil can absorb iron in ferrous `(Fe^(2+))` as well as ferric `(Fe^(3+))` form . It is an important constituent of proteins involved in the transfer of electrons like oxidised from `Fe^(2+)` to `Fe^(3+)` during electron transfer . It activates catalase enzyme . It is essential for the formation of chlorophyll. |
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| 29. |
Use of bioresources by multinational companies and organisations with out authorisation from the concerned country and its people is calledA. biodegradationB. biopiracyC. bio-infringementD. bioexploitation |
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Answer» Correct Answer - B Biopiracy is referred to the use of bioresources by multinational companies and other organisations without proper authorisatioin from the countries and people concerned without compensatory payment . Bio-infringement is the commission of a prohibited act with respect to a patented invention without permission from the patent holder . Bio-exploitation means taking advantage of biological resources of other country without permission . Biodegradation is biological breakdown of organic material by bacteria , fungi , etc. |
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| 30. |
Which one of the following statements is correct ?A. Horsetails are gymnospermsB. Selaginella is heterosporous , while Salvinia is homosporousC. Ovules are not enclosed by ovary wall in gymnospermsD. Stems are usually unbranched in both Cycas and Cedrus. |
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Answer» In gymnosperm ovules are not enclosed by ovary wall . Seeds do not occur inside a fruit . They are naked . Horsetail is the common name of Equisetum . Pteridophytes like Selaginella and Salvinia are heterosporous and possess two types of spores , i.e. microspores and megaspores . Cycas has an unbranched columnar stem while Cedrus possess branched stem . Therefore , only statement (c) is correct. |
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| 31. |
The two functional groups characteristic of sugars areA. Carbonyl and phosphateB. Carbonyl and methylC. Hydroxyl and methylD. Carbonyl and hydroxyl |
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Answer» Sugars are chemically carbohydrates . They are polyhydroxy aldoses , ketoses and their condensatioin products . Aldoses bear a terminal aldehyde or -CHO group while ketoses have an internal ketone or -CO group. Thus they posses two functional groups , i.e., carbonyl and hydroxyl . |
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| 32. |
In stratosphere , which one of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen ?A. FeB. ClC. CarbonD. Oxygen |
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Answer» Correct Answer - B In stratosphere , Cl acts as a catalyst in the degradation of ozone and release of molecular oxygen . It is released by action of UV rays on a chlorofluorocarbon . Chlorine reacts with ozone in a series of chain reaction , converting it into oxygen . One active chlorine can destroy 5000 molecules of ozone in one month . `CFCl_(3) overset(UV-C)(to) CFCl_(2) + Cl` `CFCl_(2) overset(UV-C)(to) CFCl + Cl` `Cl + O_(3) to ClO + O_(2)` `ClO + O_(3) to Cl + 2O_(2)` Iron (Fe) , carbon (C) and oxygen (O) are not Ozone Depleting Substances (ODS). |
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| 33. |
Which of the following is an amino acid derived hormone ?A. EstradiolB. EcolysoneC. EpinephrineD. Estriol |
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Answer» Among the following, epinephrine is an amino acid derived hormone. It is a catecholamine which is produced in the chromaffin cells of adrenal medulla from amino acids tyrosine. On the other hand, estradiol and estriol are steroid hormone that are involved in the regulation of estrous and menstrual cycles. Ecdysone is also a steroid hormone that controls moulting in insects . |
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| 34. |
Which of the following structures or region is incorrectly paired with its function ?A. `{:((a) , "Hypothalamus" , "Production of releasing hormones and regulation of temperature hunger and thirst"):}`B. `{:((b) , "Limbic system", "Consists of fibre tracts that interconnect different regions of brain , controls movement"):}`C. `{:((c) , "Medulla oblongata", "Controls respiration and cardiovascular reflexes."):}`D. `{:((d) , "Corpus callosum", "Band of fibres connecting left and right cerebral hemispheres."):}` |
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Answer» Limbic system consists of four major components namely hippocampus, amygdala, septal nuclei and manunilary bodies. It controls the emotional behaviour. food habits and sex behaviour of an organism. It is not involved in controlling movements . The rest three options are correctly paired with their functions. |
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| 35. |
The difference between spermiogenesis and spermiation isA. In spermiogenesis , spermatozoa from Sertoli cells are released into the cavity of seminiferous tubules , while in spermiation spermatozoa are formedB. In spermiogenesis , spermatozoa are formed , while in spermiation spermatids are formedC. In spermiogenesis , spermatids are formed while in spermiation spermatids are formedD. In spermiogenesis , spermatozoa are formed while in spermiation spermatozoa are released from Sertoli cells into the cavity of seminiferous tubules |
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Answer» Spermiogenesis is the process of transformation of spermatids (n) into spermatozoa (n ) or sperms. It involves the differentiation phase in which onespermatid develops into one spermatozoan . Spermiation involves the release of sperms from seminiferous tubules through Sertoli cells. |
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| 36. |
Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?A. `N_(2)O`B. `NO_(2)`C. `N_(2)O_(5)`D. `NO` |
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Answer» Correct Answer - C Nitrous oxide `(N_(2)O)`, nitrogen dioxide `(NO_(2))` and nitric oxide (NO) are the common pollutant introduced into the atmosphere. `N_(2)O` occurs naturally in environment. NO and `NO_(2)` causes considerable amount of air pollution. They are given off in car exhaust fumes and when fossil fuels are burnt as well as produced during thunderstorms. In each case NO is formed first and then `NO_(2)`. |
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| 37. |
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will beA. 2.8B. `3.0`C. 1.4D. 4.4 |
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Answer» Correct Answer - A Key Concept firstly, write the reaction of formic acid and oxalic acid with conc. `H_(2)SO_(4)`, respectively. Then, find the gaseous products formed and identify the remaining gaseous product after passing through KOH. Finally, calculate the total number of moles of gaseous product. `underset("Formic acid")(HCO OH)overset(Conc. H_(2)SO_(4))to CO(g)+H_(2)O(l)` `{:("Initial moles",(2.3)/(46)=(1)/(20)mol,0),("Final moles"," "0,(1)/(20)):}` Similarly, `{:(CO OH),(|),(CO OH),("Oxalic acid"):}overset(Conc.H_(2)SO_(4))toCO(g)+CO_(2)(g)+H_(2)O(l)` `{:("Initial moles",(4.5)/(90)=(1)/(20)mol,0,0),("Final moles"," "0,(1)/(20),(1)/(20)):}` Now, `H_(2)O(l)` gets absorbed by conc. `H_(2)SO_(4)`. gaseous mixture CO and `CO_(2)` when passed through KOH, only `CO_(2)` gets absorbed. Thus, CO is the remaining gas. Total number of moles of CO formed in the above equation `=(1)/(20)+(1)/(20)=(1)/(10)` `because " " " Moles"=("Weight")/("Molar mass")` `therefore " Weight of CO formed" =(1)/(10)xx25=2.8 g` Thus, weight of the remaining product at STP will be 2.8 g. |
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| 38. |
Which one of following elements is unable to from `MF_(6)^(3-)` ion?A. BB. AlC. GaD. In |
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Answer» Correct Answer - A Boron belongs to 2nd period of the periodic table with electronic configuration `1s^(2), 2s^(2),2p^(1)`. It does not have vacant d-orbitals, thus cannot increase its covalency above four. Therefore, boron (B) cannot form `MF_(6)^(3-)` ion. In contrast, aluminium (Al), gallium (Ga), indium (In) have the vacant 3d-orbitals, thus can increase their covalence above four and form `MF_(6)^(3-)` ion. |
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| 39. |
The correct order of N-compounds in its decreasing order of oxidation states isA. `HNO_(3), NH_(4)Cl,NO,N_(2)`B. `HNO_(3),NO,NH_(4)Cl, N_(2)`C. `HNO_(3),NO,N_(2),NH_(4)Cl`D. `NH_(4)Cl,N_(2),NO,HNO_(3)` |
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Answer» Correct Answer - C Let the oxidation state of nitrogen in each of the given N-compounds be x. `(i) HNO_(3) : +1+x+3(-2)=0` x=+5 `therefore` Oxidation state of N in `HNO_(3)` is +5. (ii) `NO : x+1(-2)=0` x=+2 `therefore` Oxidation state of N in NO is +2 `(iii) NH_(4)Cl : x+4(+1)+1(-1)=0` x= -3 `therefore` Oxidation state of N in `NH_(4)Cl` is -3. `(iv) N_(2) : x=0 [ therefore N_(2)` is present in elemental state] `therefore` Oxidation state of N in `N_(2)` is 0. Thus, the correct decreasing order of oxidation states of given N - compounds will be `overset(+5)(HNO_(3)) gt overset(+2)(NO) gt overset(0)(N_(2)) gt overset(-3)(NH_(4)Cl)` |
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| 40. |
Which one of the following ions exhibits d-d transition and paramagnetism as well ?A. `MnO_(4)^(-)`B. `Cr_(2)O_(7)^(2-)`C. `CrO_(4)^(2-)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - D Key Concept In d-d transition, an electron in a d-orbital of the metal is exicted by a photon to another d-orbital of higher energy. Paramagnetism The complex compound which contains unpaired electrons shows paramagnetism while which contains paired electrons shows diamagnetism. The complex which contains unpaired electrons exhibit d-d transition and paramagnetism. (i) In `MnO_(4)^(-)`, The electronic configuration of `Mn^(7+)` is `[Ar]3d^(0)`. Number of unpaired electrons =0 Therefore, it will be diamagnetic and will not show d-d transition. (ii) In `Cr_(2)O_(7)^(2-)`, The electronic configuration of `Cr^(6+) " is" [Ar]3d^(0)`. Number of unpaired electrons=0 So, it will be diamagnetic and will not show d-d transition. (iii) In `CrO_(4)^(2-)`, The electronic configuration of `Cr^(6+) " is "[Ar]3d^(0)`. Number of unpaired electrons=0 Therefore, it is also diamagnetic and will not show d-d transition. (iv) In `MnO_(4)^(2-)`, The electronic configuration of `Mn^(6+)` is `[Ar]3d^(0)`. Number of unpaired electrons=1 Since, it contains one unpaired electron so it will exhibit both d-d transition and paramagnetism. |
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| 41. |
Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : A. `{:(1,2,3,4),(iv,I,ii,iii):}`B. `{:(1,2,3,4),(I,ii,iii,iv):}`C. `{:(1,2,3,4),(iv,v,ii,i):}`D. `{:(1,2,3,4),(iii,v,I,ii):}` |
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Answer» Correct Answer - C Key Concept Spin magnetic moment can be calculated as `mu=sqrt(n(n+2))BM` where, `mu`=magnetic moment BM=Bohr Magneton (unit of `mu`) n=number of upaired electrons in d-orbital. The electronic configuration of `CO^(3+)` is `[Ar]3d^(6)`. Here, n=4 `mu=sqrt(4(4+2))+sqrt(24)BM` The electronic configuration of `Cr^(3+) " is" [Ar]3d^(3)` Here, n=3 `mu=sqrt(3(3+2))=sqrt(15)BM` The electronic configuration of `Fe^(3+) " is " [Ar]3d^(5)`. Here, n=5 `mu=sqrt(5(5+2))=sqrt(35)BM` The electronic configuration of `Ni^(2+) " is" [Ar]3d^(8)`. Here, n=2 `mu=sqrt(2(2+2))=sqrt(8)BM` So, the correct option is ( c)`. |
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| 42. |
Calcium is important in skeletal muscle contraction because itA. detaches the myosi head from the actin filamentB. activates the myosin ATP ase by binding to itC. binds to troponin to remove the masking of active sites on actin for mysoinD. Prevents the formation of bonds between the mysoin cross bridges and the actin filament |
| Answer» Calcium plays a key regulatory role in muscle contraction . `Ca^(2+)` ions bind to troponin and changes its shape and position . This in turn , alter the shape and position of tropomyosin and hence , the active sites on F-actin are exposed . Due to this myosin cross-bridges are able to bind to these active sites and muscle contraction occurs. | |
| 43. |
Which of the following is an occupational respiratory disorder?A. BotulismB. SilicosisC. AnthracisD. Emphysema |
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Answer» Silicosis is an occupational respiratory disorder which is caused due to excessive inhalation of silica dust . It usually affects the workers of grinding or stone breaking industries . The long-term exposure can cause lung fibrosis (or stiffening ) , leading to breathing difficulties. Anthracis or Anthrax is a bacterial infection caused by Bacillus anthracis. Botullism is food poisoning infection caused by Clostridium botulinum . Its symptoms include diarrhoea , vomiting , abdominal distention, etc. Emphysema is a lung disease , that damages the air sacs and causes shortness or breathe , it may be caused by the smoking , deficiency of enzymes alpha-I antitrypsin and air pollution. |
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| 44. |
Match the items given in Column I with those in Column II and select the correct option given below A. `{:(1 , 2, 3,), ("i" , "iii", "ii",):}`B. `{:(1 , 2, 3,), ("i" , "ii", "iii",):}`C. `{:(1 , 2, 3,), ("iii" , "ii", "i",):}`D. `{:(1 , 2, 3,), ("ii" , "iii", "i",):}` |
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Answer» Fibrinogen is a soluble plasma protein that is stimulated by thrombin and gets converted into insoluble form fibrin . The latter helps in the formation of blood clot to seal the wound and stop bleeding . Globulins are simple proteins that form a large fraction of blood serum proteins involved in defence mechanism . There are four main types of globulins that are manufactured in liver , namely alpha-1 , alpha-2, beta and gamma. Albumin is a plasma protein that is manufactured by the liver . It helps in maintaining osmotic pressure which prevents the fluid-leakage out into the tissues from the blood stream. |
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| 45. |
Match the items given in Column I with those in Column II and select the correct option given below . A. `{:(1 , 2, 3), ("ii" , "iii", "i" ):}`B. `{:(1 , 2, 3), ("i" , "iii", "ii" ):}`C. `{:(1 , 2, 3), ("iii" , "ii", "i" ):}`D. `{:(1 , 2, 3), ("iii" , "i", "ii" ):}` |
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Answer» During proliferative phase. The follicles start growing in size under the influence of Follicle stimulating Horsome (FSH). Hence, this phase is also called follicular phase. During secreatory phase, corpus luteum secreates progesterone that helps to thicken the endometrial lining. Due to the persistance of corpus luteum, this phase is also called luteal phase. Menstruation or bleeding occurs due to the breakdown of endometrial lining in the absence of gregnancy. During this phase, corpus luteum regresses and progesterone level decreases. |
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| 46. |
Match the items given in Column I with those in Column II and select the correct option given below . A. `{:(1 , 2, 3, 4), ("iii" , "iv", "i" , "ii"):}`B. `{:(1 , 2, 3, 4), ("i" , "iii", "iv" , "ii"):}`C. `{:(1 , 2, 3, 4), ("ii" , "i", "i" , "iv"):}`D. `{:(1 , 2, 3, 4), ("i" , "ii", "iv" , "iii"):}` |
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Answer» Eutrophication is the nutrient enrichment of water bodies containing excessive population of phytoplanktons . Sanitary landfill is a method of solid waste disposal in which the waste material is burried in the pits dug on the ground and later they get covered by soil . Snow blindness is caused due to UV-B radiations exposure . These radiations can reach the earth surface due to the depletion of ozone layer . In Jhum cultivation land is cultivated temporarily and then abandoned so that , it can revert to its natural vegetation . it is a long term process and usually leads to deforestation. |
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| 47. |
Match the items given in Column I with those in Column II and select the correct option given below A. `{:(1 , 2, 3,4), ("i" , "iv", "ii","iii" ):}`B. `{:(1 , 2, 3,4), ("iii" , "i", "iv","ii" ):}`C. `{:(1 , 2, 3,4), ("iii" , "ii", "i","iv" ):}`D. `{:(1 , 2, 3,4), ("iv" , "iii", "ii","i" ):}` |
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Answer» Tidal Volume (TV) is the volume of air inspired or expired during normal breath . It is about 500-550 mL. Inspiratory Reserve Volume (IRV) is the extra amount of air that can be inspired directly after a normal inspiration . It is about 2500-3000 mL. Expiratory Reserve Volume(ERV) is the extra amount of air that can be expired forcibly after a normal expiration . It is about 1000- 1100 mL. Residual Volume(RV) is the volume of air which remains still in the lung after the most forceful expiration . It is about 1100-1200 mL . Therefore , option (b) is correct . |
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| 48. |
Match the items given in Column I with those in Column II and select the correct option given below A. `{:(1 , 2, 3, 4), ("ii" , "iv", "iii" , "i"):}`B. `{:(1 , 2, 3, 4), ("iii" , "ii", "i" , "iv"):}`C. `{:(1 , 2, 3, 4), ("i" , "iv", "iii" , "ii"):}`D. `{:(1 , 2, 3, 4), ("iii" , "iv", "i" , "ii"):}` |
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Answer» Herbarium is a place where dried and pressed plant specimens, mounted on sheets are kept systematically. It is a repository or store house for future use. Key is a booklet containing list of characters and their alternates which are helpful in identification of various taxa-class, order, family, genus and species. Museum is an institution where artistic and educational materials are exhibited to the public. The materials available for observation and study are called a collection. Catalogue is a list or register that enumerates methodically all the species found in a particular place. It often possesses brief description of species that aids in identification. Therefore option (d) is correct . |
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| 49. |
Which one is wrongly matchedA. Germma cups - MarchantiaB. Biflagellate zoospores - Brown algaeC. Uniflageliate gametes - PolysiphoniaD. Unicellular organism - Chlorella |
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Answer» Polysiplronia is a red algae. In it sexual reproduction is of oogamous type. The male sex organ, sperma tangium produces non-flagellate male gametes. In Brown algae, sexual reproduction varies from isogamy, anisogamy to oogamy. In isogamy and anisogamy both the gametes are motile while in oogamy only male gametes are motile. These motile gametes have two unequal laterally attached flagella. Chlorella is a unicellular organism. It is green algae belonging to class Chlorophyta. In Marchanthia , gemma cups are found on its dorsal surface . It contains gammae which help in vegetative propagation. |
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| 50. |
Which of the following organisms breeds only once in lifetime ?A. MangoB. JackfruitC. Bamboo speciesD. Papaya |
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Answer» Correct Answer - C Bamboo plants are perennial , monocarpic plants . They flower only once in their lifetime , usually after 50-100 years . They produce large number of fruits and die. Mango , Jackfruit and Papaya are polycarpic plants, i.e. they flower repeatedly at regular intervals every year. |
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