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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.
| 601. |
Identify and illustration in the following |
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Answer» (i) A-AATTC (Recognition Site of the restriction endonuclease) (ii) B-Pru II (Origin of replication) (b) Palindromic nucleotide sequence is depicted from A & C reads same on the strands when orientations of reading is kept the same. A - Coding strand B - Template strand C - PCR strand for polymerase chain reaction. PCR is used to amplify DNA segments to a large number with in a short span of time. |
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| 602. |
a) What is inbreeding depression? b) Explain the importance of "selection" during inbreeding in cattle. |
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Answer» a) Continued inbreeding, especially, close inbreeding, usually reduces fertility and even productivity. This is called inbreeding depression. b) Inbreeding increases homozygosity. Thus, inbreeding exposes harmful recessive genes that are eliminated by selection. Where, there is selection at each step, increase the productivity of inbred population. A superior female, in case of cattle, is the cow or buffalo that produces more milk per lactation. |
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| 603. |
Seccondary treatment of the sewage is also called Biological treatment Justify this statement and explain the process. |
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Answer» Secondary treatment or Biological treatment : In secondary treatment , the primary effluent is passed into large acration tanks where it is contantly agitated mechanically and air is pumped to it . This allows vigorous growth of useful acrobic microbes into flocs. While growing , these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD of the effluent. BOD is a measure of the organic matter present in the water . the greater the BOD of waste water , more is its polluting potential. BOD refers to the amount of the oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria. Once the BOD of sewage or waste water is reduced significantly the effluent is then passed into a settling tank where the bacteria , flocs, allowed to sediment . this sediment is called activated sludge .A small part of the activated sludge is pumped back into the acration tank to serve as the inooculum. The remaining major part of the slugde is pumped into lage tanks called anaerobic sludge digesters . Here other kinds of bacteria , which grow anaerobically , digest the bacteria and the fungi in the sludge . During this digestion , bacteria produce a mixture of gases such as methane , hydrogen sulphide and candon dioxide . these gases form biogas and can be used as source of energy as it s inflammable . the efflent from the secondary treatment plant is generally released into natural wator bodies like rivers and streams. Secondary treatment , of the sewage is also called biological treatment because in secondary treatment , biological factors/organisms play a major role in treatment . So it is called Bioligical treatment. |
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| 604. |
(i) Deficiency of which vitamin causes scurvy ? (ii) What type of linkage is responsible for the formation of proteins ? (iii) Wrtie the product formed when glucose is treated with HI. |
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Answer» (i) Deficiency of vitamin C causes scurvy. Bleeding gums is a sympton of scurvy. (ii) Peptide linkage is responsible for the formation of proteins. Protein is formed by the combination of the amino acids that contain amino `(-NH_(2))` and carboxyl(-COOH) as functional groups. (iii) When glucose is treated with HI, n-Hexane if formed as the final product. Reaction involved : `OHC-underset(("Glucose"))((COOH)_(4))-CH_(2)OH+ HI to CH_(3)-underset(("n - Hexane"))(CH_(3)-)CH_(2)-CH_(2)-CH2-CH_(2)-CH_(2)` |
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| 605. |
The short weve length limit for the Lyman series of the hydrogen spectrum is `913.4 Å` Calculate the short wevelength limit for Balmer series of the hydrogen spectrum. |
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Answer» Using `(1)/(lambda) = R ((1)/(n_(i)^(2)) - (1)/(n_(i)^(2)))` Here, `n_(i) = 1, n_(f) = 2 ` for short weve `(1)/(lambda) = R((1)/(1^(2)) - (1)/(2^(2)) ) rArr lambda = (4)/(3lambda) ` …(i) Also, for Balmer short weve length, `n_(1) = 2, n_(f) = 3` `(1)/(lambda) = R((1)/(1^(2)) - (1)/(2^(2)) ) = (R5)/(36) rArr lambda = (36)/(5R)` Using `lambda = (36)/(5xx4) xx3 xxlamda = (27xx913.4)/(5) = 4932.36Å`. |
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| 606. |
For what kinetic energy of a proton, will the associated de-Broglie wavelength be 16.5 nm? |
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Answer» `lambda=16.5 nm = 16.5 xx 10^(-9)m`. `m=1.6xx10^(-27)kg` Using de-Broglie equation, we have `lambda=(h)/(mv) rArr 16.5xx10^(-9)=(6.626xx10^(-34))/(1.6xx10^(-27)xxv)` `upsilon = (6.626xx10^(-34))/(16.5xx10^(-9)xx1.6xx10^(-27))=(6.626xx10^(-34))/(26.4xx10^(-36))` `=0.251xx10^(2)=25.1 m//s` Hence, kinetic energy `= (1)/(2) mv^(2) = (1)/(2) xx 1.6xx10^(-27)xx(25.1)^(2)` `=504.008xx10^(-27)=5.04xx10^(-25)J`. |
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| 607. |
(a) Write the principle of method used for the refining of germanium (b) Out of PbS and `PbCO_(3)` (ores of lead), which one is concentrated by froth floatation process preferably ? (c) What is the significance of leaching in the extraction of aluminium ? |
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Answer» (a) Zone refining method is used for the refining of germanium. Principle : This method is based upon the principle that the impurities are more soluble in the molten state than in the solid state of the metal (b) PbS is concentrated by froth floatation process (c) To remove the impurities like Silica `(SiO_(2))`, iron oxide `(Fe_(2)O_(3))`, leaching is carried out. During leaching, powdered bauxite ore is heated iwht conc. `NaOH` solution at `473 - 523` K. `underset("Alumina")underset("Bauxite")(Al_(2)O_(3)(s)) + 2NaOH + 3H_(2)O overset(473 - 523)rarr underset("Sodium meta- aluminate")(2Na[Al(OH)_(4)])` `underset("Silica")(SiO_(2)) + 2NaOH (aq) rarr underset("(Sodium Silicate)")(Na_(2)SiO _(3) + H_(2)O)` Leaching leaves behind `Fe_(2)O_(3), TiO_(2)` and other impurities |
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| 608. |
Write the IUPAC name of the given compound : |
| Answer» IUPAC name : 2-Phenylethanol. | |
| 609. |
Write the structure of an isomer of compound `C_(4)H_(9)Br` which is most reactive towards `S_(N)1` reaction. |
| Answer» `H_(3)C-overset(underset(|)(Br))underset(overset(|)(CH_(3)))(C)-CH_(3)` is most reactive isomer of `C_(4)H_(9)Br` towards `S_(N)1` reaction `3^(@)` carbocation is most stable. | |
| 610. |
Give an example each of a molecular solid and an ionic solid. |
| Answer» `"Molecular solid" -CH_(4), "lonic solid "-NaCI.` | |
| 611. |
`Pb(NO_(3))_(2)` on heating gives a brown gas which undergoes dimerization on cooling? Identify the gas. |
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Answer» Nitrogen dioxide gas `(NO_(2))` `2 Pb (NO_(3))_(2) overset(Delta)(to) 2PbO + 4NO_(2) + O_(2)` `2NO_(2)to N_(2)O_(4)` |
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| 612. |
What happpens when : (i) `CH_(3)-CI` is treated with aqueous KHO? (ii) `CH_(3)-CI` is treated with KCN ? (iii) `CH_(3)-Br` is treated with Mg in the presence of day ether ? |
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Answer» `(i) CH_(3)Cl+KOH (aqu)to underset("methanol")(CH_(3)OH)` (ii) `CH_(3)Cl+KCNtounderset("ethylnitrile")(CH_(3)CH)` (iii) `CH_(3)Br+Mgoverset("dry ehter")tounderset(("Grigard reagent"))(CH_(3)MgBr)` |
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| 613. |
The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. |
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Answer» `mu gt gt 1` for ferromagnetic. Ferromagnetic material :- (i) Pulls all the field lines inside the material. (ii) Moves towards the strongest field in a non-uniform field. |
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| 614. |
Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing . |
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Answer» Metal ring 1 is clockwise. Metal ring 2 is anticlockwise. |
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| 615. |
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily. |
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Answer» Metal ring 1 is clockwise. Metal ring 2 is anticlockwise. |
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| 616. |
Give reasons for the following (a) The presence of `-NO_2` group at ortho or para position increases the reactivity of haloarences towards nucleophilic substution reactions . ( b) P-dichlorobenzene has higher melting point than of ortho or mela isomer . ( c) Thionyl chloride method is preferred for preparing alkyl chloride from alcohols. |
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Answer» The presence of `-NO_2` group at ortho or para position increases the rectivity of haloarenes towards nucleophilic substitution reactions because nitro groups at ortho or para position withdraw the electron density from the benzene ring facilitating the attack of the nucleophile. ( b) Melting point of para isomer is quite higher than that of orhto or meta isomers.This is due to the fact that it has a summetrical structure and therefore ,its molecules can easily pack closely in the atmosphere leaving behind pure alkyl chlorides , |
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| 617. |
Complete the following reactions: `(a) MnO_2+KOH +O_2 rarr` (b)`I^-+MnO_4^-+H^+ rarr` ( c) `Cr_2O_7^(2-)+Sn^(2+)+H^+rarr` |
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Answer» See Q.23 (ii) (or) (ii) Paper - 2016 ,Delhi Board ,set-I [page 108] ( b)`2MnO_4^-+10I^-+16H^-rarr 2Mn^(2+)+5l_2+8H_2O` ( c) See Q.7 (or) (ii) Paper - 2016 Delhi Board ,Set-I[page 338]. |
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| 618. |
What are the transcriptional products of RNA polymerase III? |
| Answer» RNA- Polymerase III catalyses transcription of Trna, 5 Sr RNA, SrRNA. | |
| 619. |
Describe any two modes by which apomictic seeds and be produced. |
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Answer» Types of apomixis in flowering plants Non-recurrent apomixis : In this type "the megaspore mother cell undergoes the usual meiotic divisions and a haploid embryo sac meqagametophyte is formed. The new embryo may then arise either from the egg (haploid parthenogenesis) or from some other cell of the gametophyte (haploid apogamy)." The haploid plants have half as many chromosomes as the mother plant, and "the process is not repeated from one generation to another" (i) Recurrent apomixis, is now more often called gametophytic apomixis : In this type, the megagametophyte has the same number of chromosomes as the mother plant because meiosis was not completed. It generally arises either from an archesporial cell or from some other part of the nucellus. (ii) Adventive embryony, also called sporophytic apomixis, sporophytic budding, or nucellar embryony: Here there may be a megagametophyte in the ovule, but the embryos do not arise from the cells of the gametophyte, they arise from cells of nucells or the integument. Adventive embryony is important in several species of Citrus, in Garcinia, Euphorbia dulcis, Mangifera indica etc. (iii) Vegatative apomixis : In this type "the flowers are replaced by bulbils or other vegetative propagules which frequently germinate while still on the plant". (iv) Vegatative apomixis is important in Allium, Fragaria, Agave, and some grasses among others. |
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| 620. |
Identify and write the correct statement : (a) Drosophila male has one X and one Y chromosome. (b) Drosophila male has two X chromosomes. |
| Answer» (a) True. (Drosophila male has one X and one Y chromosome) | |
| 621. |
Why are seeds of some grasses called apomictic? Explain |
| Answer» Seeds of some grasses are called apomictic because they develop with out fertilisation | |
| 622. |
Give one example each of a plant that reproduces by : (a) runner (b) offset |
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Answer» (i) runner in Grasses. (ii) offset of water hyacinth |
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| 623. |
Explain how natural selection operates in natural by taking an example of white winged and dark winged moths of England. |
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Answer» In the nineteenth century it was noticed that in towns and cities it was actually the black form of the moth that was more common than the pale peppered form. Industrialisation and domestic coal fires had caused sooty air pollution which had killed off lichens and blackened urban tree trunks and walls. So, now it was the plae form of the moth that was more obvious to predators, while the melanic form was better camouflaged and more likely to survive and produce offspring. As a result, over successive generations, the black moths came to outnumber the pale forms in towns and cities. Since, moths are short-lived, this evolution by natural selection happened quite quickly. For example, the first black Peppered Moth was recorded in Manchester in 1848 and by 1895 98% of Peppered Moths in the city were black. |
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| 624. |
People are quite apprehensive to use GM crops. Give three arguments in support of GM crops so as to convince the people in favour of such crops. |
| Answer» See Q. 19, Set-I., Outside Delhi-2015. | |
| 625. |
Write the reactions involved in the following reactions: (i) Clemmensen reduction (ii) Cannizzaro reaction |
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Answer» (i) Clemmensen reduction: Reduction of an aldehyde or ketone to give alkanes. For eg: `underset("Acetaldehyde")(CH_(3) CHO + 4[H]) overset(Zn-Hg, HCl)rarr underset("Ethane")(CH_(3) CH_(3) + H_(2)O)` `underset("Propanone")(CH_(3)COCH_(3)) + 4[H] overset(Zn-Hg, HCl)rarr underset("Propane")(CH_(3)CH_(2)CH_(3)) + H_(2)O` (ii) Cannizzaro reaction: Aldehydes which do not contain `alpha-` hydrogen, when treated with conc. NaOH undergo self oxidation -reduction `underset("Formaldehyde")(2H - overset(O)overset(||)(C) - H) + underset((50%))(NaOH) rarr underset("Methylalcohol")(CH_(3) - OH) + H - underset("Sod. formate")(overset(O)overset(||)(C) - ONa)` |
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| 626. |
Write IUPAC name of the following compound `(CH_(3))_(2) N - CH_(2) CH_(3)` |
| Answer» `underset("N,N -Dimethylethanmine")(H_(3) C - underset(CH_(3))underset(|)(N) - CH_(2) CH_(3))` | |
| 627. |
The following data were obtained during the first order thermal decomposition of `N_(2)O_(5)` (g) at constant volume : `2N_(2) O_(5)(g)to 2N_(2)O_(4)(g) +O_(2)(g)` Calculate the rate constnat. |
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Answer» `2N_(2)O_(5)(g) to 2N_(2)O_(4)(g)+ O_(2)(g)` `{:("At " t=0, 0.5" atm", 0 " atm", 0 " atm"),("At time " t, 0.5-2x " atm" , 2x" atm", x" atm"):}` `p_(t) = p_(N_(2)O_(5)) + p_(N_(2)O_(4)) + p_(O_(2))` = (0.5 -2x) + 2x +x =0.5 +x `x = p_(t)-0.5` `p_(N_(2)O_(5))= 0.5-2x =0.5-2(p_(1)-0.5) =1.5-2p_(1)` `"At "t = 100s, p_(1) = 0.512 atm` `p_(N_(2)O_(5))= 1.5-2 xx 0.512 = 0.476 "atm"` `k = (2.303)/(t) "log" ("pi")/("pA")` `k = (2.303)/(100s) "log" (0.5 " atm")/(0.476" atm")` `k = (2.303)/(100s) xx 0.0216 = 4.98 xx 10^(-4)s^(-1)` |
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| 628. |
(a) Write symbolically the `beta`-decay process of `""_(15)^(32)P`. (b) Derive an expression for the average life of a radionuclide . Give its relationship with the half-life. |
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Answer» (b) Average or Mean Life : Average life is defined as the total life time of all the atoms of the element divided by the total number of atoms present in the sample of the element . By definition , average life time of the radioactive element is `tau = (int_(0)^(N_(0)) t dN)/(N_(0))` Since , `" " dN = - lambda N` and `N = N_(0) e^(-lambda t)` `therefore " " dN = - lambda N_(0) e^(-lambda t)` dt When N = 0 then `t = oo` and when N = `N_(0)` then t = 0 `tau= (int_(oo)^(0) t. - lambda N_(0) e^(-lambda t) dt)/( N_(0)) = lambda int_(oo)^(0) t e^(-lambda t) dt` `tau = lambda [ (t e^(-lambda t))/(lambda) - (e^(-lambda t))/(lambda^(2))]_(0)^(oo) " "` (Using integration by parts ) `tau = lambda [ (0-0) - (0 - (1)/(lambda^(2)))] = lambda xx (1)/(lambda^(2)) = (1)/(lambda)` Since , `" " lambda = (0.693)/(t_(1//2))` `therefore " " tau = (1)/((0.693)/(t_(1//2))) = (tau_(1//2))/(0.693) = 1.44 tau_(1//2)` Hence , the average life period of a radioactive element is 1.44 times the half life period of the element . |
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| 629. |
What are analgesic drugs ? How are they classified and when are they usually recommended for use ? |
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Answer» Analgesic drugs : The medicines which are used to relieve body pains are called analgesic drugs. For example : Aspirin and paracetamol. These drugs are classified as follows : (i) Non - narcotic (non-addictive) analgesics (ii) Narcotic drugs. Non- narotic drugs like aspirin is used in prevention of hert attacks and the narcotic drugs like as morphine is used for the relief of postoperative pain, cardiac pain and pains of terminal cancer. |
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| 630. |
Write a distinguishing feature of metallic solids. |
| Answer» Merallic solids are malleable and ductile and good conductors of electricity heat. | |
| 631. |
Write a distinguishing feature of a metallic solid compared to an ionic solid. |
| Answer» Ionic solids are electrical insulator while metals are good electrical conductors. | |
| 632. |
Giving one example in each of the following cases, discuss briefly the role of coordination compounds in (i) extraction metallurgy of metals (ii) analytical chemistry |
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Answer» (i) The extraction and purification of the valuable metals by hydrometallurgical processes palladium : trans`[Pd(NH_(3))_(2)Cl_(2)], ` platinum : `[PtCl_(6)]^(2-)` gold : `[Au(CN)_(2)]^(-)` and `[AnCl_(4)]^(-)`, uransium :`[UO_(2)(SO_(4))_(3)]^(4-)`, cobalt: `[Co(D_(2)EHPA)_(2)]`. The neutral cobalt `D_(2)` EHPA complex is unique in its being soluble in kerosene , a property that makes possible the industrial solvent extration process. (ii) Application of the coordination compound in the analytical chemistry covers both the aspects that is quantitative and qualitative. for example addition of thiocyanate ion to the solution containing `Fe^(3+)` ions give rise a red colour to the thiocyanide complex. `[Fe(OH)_(5)(SCN)]^(2+)` ion concentration by constructing a calibration curve from a solution of the known concentration of `Fe^(3+)` concentration . |
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| 633. |
Silver metal crystallises with a face centred cubic lattice. The length of the unit cell is found to be `3.0xx10^(-8)`cm. Calculate atomic radius and density of silver. Molar mass of Ag`=108 g mol^(-1), N_(A)=6.02xx10^(23)mol^(-1)`). |
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Answer» `a=3xx10^(-8)cm, M=108g//mol ` `f.c.c.z =4Na=6.02xx10^(23)mol^(-1)` For f.c.c. `a=2sqrt(2r)rArr r=(a)/(2sqrt(2))xx(sqrt(2))/(sqrt(2))` `r=(asqrt2)/(4)` `r=(3xx10^(-8) xx1.414)/(4)=1.06xx10^(-8) cm` `d=(ZM)/(a^3xxNa)` `d=(4xx108)/((3xx10^(-8))^(3)xx6.02xx10^(23))` `d=2.66xx10^(1)` `d=26.6g//cm^(3)` |
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| 634. |
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.0xx10^(4)N//C(Fig.a)` Calculate the time it takes to fall through this distance starting from rest. If the direction of the field is reversed (fig .b) keeping its magnitude unchanged , calculate the time taken by a proton to fall through this distance starting from rest. |
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Answer» The field is upward , so the negatively charged particle , electron experiences a downward froce of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae `=(eE)/(me)` Starting from rest , the time required by the electron to fall through a distance h `t_(e)=sqrt((2h)/(ae))=sqrt((2hme)/(eE))=sqrt((2xx1.5xx10^(-2)xx9.11xx10^(-31))/(1.6xx10^(-19)xx2.0xx10^(-4)))=2.9xx10^(-9)S`. The field is downward , and the positively charged proton experiences a downward force of magnitude eE . The acceleration of the proton is `aP=(eE)/(m_(p))` Hence , the time taken by the proton `t_(p)=sqrt((2h)/(a_(p)))=sqrt((2hm_(p))/(eE))=sqrt((2xx1.5xx10^(-2)xx1.67xx10^(-27))/(1.6xx10^(-19)xx2.0xx10^(-4)))=1.3xx10^(-7)sec`. |
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| 635. |
Define electric flux and write its SI unit . The electric field components in the figure shown are : `E_(x)=alphax,E_(y)=0,E_(z)=0` where `alpha=(100N)/(Cm)`. Calculate the charge within the cube , assuming a = 0.1 m . |
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Answer» Electric flux : Number of electric field lines passing through a given area normally is called electric flux . It is a scalar quantity . It is denoted by `phi_(E)`. `phi_(E)=int vecE.vec ds` where E is electric fields intensity and ds is area element. Its SI unit is `Nm^(2)c^(-1)`. Given : a = 0.1 M `Ex=alphax` " " `alpha=(100N)/(Cm)` Ey = 0 Ez = 0 Electric flux associated with surface ABCE `phi_(1)=vecE_(x).vecS=E_(x)Scos180^(@)=(alpha a)(a^(2))(-1)=-alpha a^(3)`. Electric flux associated with surface EFGH `phi_(2)=vecE_(x).vecS=E_(x)Scos0^(@)=alpha(2a)(a^(2))(1)=2alphaa^(3)`. Net flux associated with the cube `phi=phi_(1)+phi_(2)` `phi=-alpha a^(3)+2alphaa^(3)=alpha a^(3)=100 xx0.1xx0.1xx0.1=0.1NM^(2)C^(-1)` According to Gauss theorem `q=in_(0)phi=8.85xx10^(-12)xx0.1=8.85xx10^(-13)C`. |
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| 636. |
Two nuclei have mass numbers in the ratio 8 : 125. What is the ratio of their nuclear radii ? |
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Answer» Here, `A_(1) : A_(2)=8 : 125 implies A_(1)/A_(2)=8/125` Since, `R=R_(0)A^(1//3)` `:. R_(1)/R_(2)=A_(1)^(⅓)/A_(2)^(⅓)=(A_(1)/A_(2))^(⅓)=(8/125)^(⅓)=2/5` |
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| 637. |
Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of their nuclear radii ? |
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Answer» `R_(1)/R_(2)=[A_(1)/A_(2)]^(⅓)=[27/125]^(⅓)=3/5` `:.` The ratio of their nuclear radii `= 3 : 5` |
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| 638. |
Account for the following (i)Primary amines `(R-NH_(2))` have higher boiling point than tertiary amines `(R_(3)N)` (ii) Aniline does not undergo friedel - crafts reaction (iii) `(CH_(3))_(2)` NH is more basic than `(CH_(3))_(3)N` in an aquaous solution |
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Answer» (i) Due to maximum intermolecualr hydrogen bonding in primary amines (due to presence of more number of H atoms) primary aminal have high BP in comp[arison to tertiary amines (ii) Aniline does not undergo Fredel craft reaction due to acid base reaction betweenn basic compound salt is formed with AI `CI_(3)` As a result ,N of aniline acquires (+) ve charge hence act strong deactivating group (iii) In `(CH_(3))` N there is maximum steric hindrance and least solvation but in `(CH_(3))` NH the solvatin is more and the syeric hindrance is less than in `(CH_(3))` although + I effect is less since there are two methyl goup dimethly amine is stilell a stronger base than trimethl amine (i) `C_(6)H_(5)NO_(2)overset(Sn_HCI)rarr(A)` |
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| 639. |
(i) What is the relation between critical angle and refractive index of a material ? (ii) Does critical angle depend on the color of light ? Explain. |
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Answer» (i) `mu= sin i_(c) or n_(21)= sin i_(c)` Where `n_(21)` is the refractive index of rare medium 1 with respect to denser medium 2. (ii) As `mu` depends on wavelength, therefore critical angle for the same pair of media in contact wil be different for different colours. |
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| 640. |
Give the structers of A,B and C in the following reaction : (i) `C_(6)H_(5)NO_(2)overset(Sn+HCI)rarr` `Aoverset(NaNO_(2)+HCI)underset(273 K)rarr B overset(H_(2)O)rarrC` (ii) `CH_(3)CNoverset(H_(2)O //H^(+))rarr``Aoverset(NH_(3))underset(Delta)rarr` `Boverset(Br_2+KOH)rarrC` |
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Answer» `C_(6)H_(5)NH_(2)overset(NaNO_(2)+HCI)rarr(B)C_(6)H_(5)N_(2)CIoverset(H_(2)O)rarr(C )C_(6)H_(5)OH` (ii) `CH_(3)CNoverset(H_(2)O/H^(+))rarr` `CH_(3)-COOHoverset(NH_(3))rarr(B)CH_(3)-CONH_(2)overset(Br2+KOH)rarr(C )CH_(3)-NH_(2)` |
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| 641. |
Different animals respond to changes in their surroundings in different ways . Taking one example each explain some animals indergo aestivation while some others hibernation " .How do fungi respond to adverse climatic conditions ? |
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Answer» Some animals go into aestivation to avoid summer related problems (heat and dessications) . Example , snail and fish . Some animals go into hibernation to avoid winter related problems (extreme cold) . Example , Bear. Fungi form various kinds of thick walled spores which help them to survive unfavoaurable conditions. |
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| 642. |
How does smoking tabacco in human lead to oxygen deficiency in their body? |
| Answer» Smoking increases the carbon mono-oxide content in blood and reduces the concentration of oxygen bound to haem molecule of haemoglobin. | |
| 643. |
Different animals respond to changes in their surroundings in different ways . Taking one example each explain some animals undergo aestivation while some others hibernation " .How do fungi respond to adverse climatic conditions ? |
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Answer» Some animals go into aestivation to avoid summer related problems (heat and dessications) . Example , snail and fish . Some animals go into hibernation to avoid winter related problems (extreme cold) . Example , Bear. Fungi form various kinds of thick walled spores which help them to survive unfavoaurable conditions. |
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| 644. |
Mention the unique flowering phenomenon exhibited by Strobilanthus Kunthiana (neelakuranaji). |
| Answer» Strobilanthus Kunthiana (Neela Kuranaji) exhibits flowers once in 12 years. | |
| 645. |
Explain the significance of meiocytes in a dipoid organism . |
| Answer» Meiocytes are gamete mother cells which undergo meiosis ,It is through meiosis that only one set of chromosomes gets transferred to the gametes in diploid organisms . | |
| 646. |
Which of the following is a fibre? Nyon , Neoprene ,PVC |
| Answer» Correct Answer - Nylon | |
| 647. |
The conversion of primary aromatic amines into diazonium salts is known as |
| Answer» Diazotisation Reaction | |
| 648. |
What is the role of tertiary-butyl peroxide in the polymerisation alkene ? |
| Answer» Tert-butyl peroxide acts as a free radical generating initiator (catalyst.) | |
| 649. |
Write the structures of the cross-aldol products between ethanal and propanal. |
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Answer» (i) 2-Methylbut -2-enal `CH_(3)-CH=overset(underset(|)(CH_(3)))(C)H-CHO` (ii) Pent-2-enal `CH_(3)-CH_(2)-CH=CH-CHO` |
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| 650. |
Identify the reaction and write the IUPAC name of the product formed : (a) `CH_(3)-CH_(2)-COOH overset((i) Br_(2)//"Red phosphorous")(to )` (b) |
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Answer» (a) Reaction : Hell-Volhard -Zelinsky reaction. IUPAC : 2-Bromopropanoic acid. (b) Reaction : Rosenmund reduction reaction. IUPAC : Benzaldehyde. |
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