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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.
| 351. |
Explain the role of each of the following in the extraction of metals from their ores : (i) CO in the extraction of nickel. (ii) Zinc in the extraction of silver. (iii) Silica in the extraction of copper. |
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Answer» (i) Impure nickel reacts with Co to form volatile compound which on heating forms pure nickel. `Ni+4COoverset(330-350k)tounderset("Volatile") (Ni(CO)_4)overset(450-470k)tounderset(("Pure"))(Ni+4CO)`. (ii) Zinc in the extration of silver : Zn acts as reducing agent in the extraction of silver. `2[Ag(CN)_2]^(-) (aq)+Zn(s) to2Ag(s) +[Zn(CN)_4]^(2-)(aq)]` (iii) The ore is heated in a reverberatory furnance after mixing with silica. In the furnance, iron oxide slags of as iron silicate and copper is produced in the form of copper matte. This contains `Cu_2D and FeS`. `FeO+SiO_2tounderset(("Slag"))(FeSiO_3)` |
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| 352. |
Explain what is meant by (i) a peptide linkage (ii) a glycosidic linkage |
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Answer» Peptide linkage : Proteins are condensation polymers of `alpha`-amino acids in which the same or different `alpha`-amino acids are connected by peptide bonds . Chemically , a peptide bond is an amide linkage formed between -COOH group of one `alpha`-amino acid and `NH_(2)` group of the other `alpha`-amino acid by a molecular of water. (ii) Glycosidic linkage : The two monosaccharides units are joined together through an oxide linkage formed by loss of a molecule of `H_(2)O` . such a linkage between two monosaccharides units through oxygen atom is called glycosidic linkage . |
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| 353. |
Explain what is meant by (i) a peptide linkage, (ii) a glycosidic linkage. |
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Answer» (i) Proteins are the polymers of a-amion acids and they are connected to each other by peptide bond or peptide linkage . Chemically, peptide linkage is an amide formed between -COOH group and `-NH_2` group. Peptide linkage is an amide linkage `-overset(O)overset(|)C -NH-` (ii) Disaccharides on hydrolysis with dilute acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide Linkage formed by the loss of a water molecule. Such a linkage between two monoccharides units torough oxygen atom is called glycosidic linkage. |
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| 354. |
A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction . (Given : log 2 = `0.3010` , log 4 = `0.6021` , R = `8.314 J K^(-1) mol^(-1)`) . |
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Answer» `t_(1//2)` for first order reaction : `t_(1//2) = (0.693)/(K)` For 300 K , `" " K_(1) = (0.693)/(t_(1//2)) = (0.693)/(40) S^(-1)` For 320 K , `" " K_(2) = (0.693)/(t_(1//2)) = (0.693)/(20) S^(-1)` `log ((K_(2))/(K_(1))) = (E_(a))/(2.303R) ((1)/(T_(1)) - (1)/(T_(2)))` log `(((cancel0.693)/(20))/((cancel0.693)/(40))) = (E_(a))/(2.303 xx 8.314) [(1)/(300) - (1)/(200)]` `log(2) = (E_(a))/(2.303 xx 8.314)xx ((320 - 300)/(300 xx 320))` `0.3010 = (E_(a))/(2.303 xx 8.314) xx (cancel20)/(cancel200_(150) xx cancel320)` `E_(a) = 0.3010 xx 2.303 xx 8.314 xx 150 xx 32` `E_(a) = 27663, 8 J//"mol". |
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| 355. |
An equilateral glass prism has a refrective index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index `4sqrt2/5`. |
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Answer» `mu=(sin(A+delta_(m)))/(sin-A/2)` `mu=16/((4sqrt2//5))=(1.6xx5)/(4sqrt2), A=60^(@)` `(1.6xx5)/(4sqrt2)=(sin(60+delta_(m)/2))/(sin""60/2)` `(1.6xx5)/(4sqrt2)xx1/2=sin((60+delta_m)/2)` `sin((60+delta_(m))/2)=1/sqrt2` `(60+delta_(m))/2=45^(@)` `60+delta=90^(@)` `delta_(m)=30^(@)` |
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| 356. |
Explain the following giving one suitable example in each case (i) Elastomers (ii) Condensation polymers (iii) Addition polymers |
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Answer» (i) Elastomers : These are rubber-like solid with elastic properties . In these elastomeric polymers , the polymer chain are held together by the weakest intermolecular forces . These weak binding forces permit the polymer to be stretched. (ii) Condensation polymers : These are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units . Which results in the lose of small molecules like `H_(2)O`, HCl etc . (iii) Addition polymers : The addition polymer are formed by the repeated addition of monomer molecules processing double or triple bonds , e.g., the formation of polythene from ethane and polypropene from propene. |
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| 357. |
Giving one example of each of : (i) addition polymers (ii) condensation polymers (iii) copolymers. |
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Answer» (i) Addition polymers -Polythene, PVC (ii) Condensation polymers -Nylon-66,Nylon-6,Terylene (iii) Copolymers-Buna-S, Buna-N |
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| 358. |
The relation between angle of incidence `i`, angle of prism `A` and angle of minimum deviation for a triangular prism is. |
| Answer» The relation between the angle of incidence I, angle of prism A, and angle of minimum deviation, `Delta_(m)` for a triangular prism is given as in given by `i=(A+Delta_(m))/(2)`. | |
| 359. |
A capacitor has been charged by ad.c. source. What are the magnitudes of conduction and displacement currents, when it is fully charged? |
| Answer» Both conduction and displacement current will be zero. | |
| 360. |
(a) Define the term molar conductivity . How is it related to conductivity of the related solution ? (b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration . Its other half-cell consists of a zinc electrode dipping in `1.0`M solution of `Zn(NO_(3))_(2)` . A voltage of `1.48` V is measured for this cell . Use this information to calculate the concentration of silver nitrate solution used. `(E_(Zn^(2+)|Zn)^(@) = -0.76V , E_(Ag^(+)|Ag)^(@) = + 0.80V`). |
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Answer» (b) Half cell reactions : At Cathod : `" " 2Ag^(+) (aq) + 2e^(-) to 2Ag(s)` At Anode : `" " Zn(s) to Zn^(2+) (aq) + 2e^(-)` The cell can be represented as Zn(s) `|Zn^(2+) (aq) || Ag^(+) (aq) |Ag(s)` The cell reaction is Zn(s) + `2Ag^(+)(aq) to Zn^(2+) (aq) + 2 Ag(s)` `therefore " " E_("cell")^(@) = E_("cathod2")^(@) - E_("anode")^(@)` ` = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56V` `therefore " " E_("cell") = E_("cell")^(@) - (0.059)/(n) "log" ([Zn^(2+)])/([Ag^(+)]^(2))` `1.48 = 1.56 - (0.059)/(2) "log" (1)/([Ag^(+)]^(2))` `(0.059)/(2) "log" (1)/([Ag^(+)]^(2)) = 1.56 - 1.48` log `[Ag^(+)]^(-2) = (0.08 xx 2)/(0.059)` `-2 "log"[Ag^(+)] = (0.08 xx 2)/(0.059)` `therefore " " log [Ag^(+)] = -1.356 = -2 + 2 - 1.356` `[Ag^(+)] = "anti-log" (2644) = 4.406 xx 10^(-2) M` |
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| 361. |
A Voltaic cell is set up at `25^(@)C` with the following half cells ? |
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Answer» `K = (1)/(R) xx ((1)/(A))` `^^_(m) = (K xx 1000)/(M)` (`{:("at anode": , Al(s) to Al^(3+) (aq) + 3e^(-)"]" xx 2), ("at cathode" , Ni^(2+)(aq) + 2e^(-) to Ni(s) xx 3):}/(2Al(s) + 3Ni^(2+)(aq) to Al^(3+) (aq) + Ni(s)))/` `E_("cell") = E_("cell")^(0) - (0.0591)/(n) "log" ([Al^(3+)]^(2))/([Ni^(2+)]^(3))` `n = 6 , [Al]^(3+) = 0.001 M = 1 xx 10^(-5) M` `[Ni^(2+)] = 0.5 M E^(@) Ni^(2+)//Ni - E^(@) Al^(3+)//Al = -0.25 V - (-1.66V)` `E_("cell")^(@) = 1.41 V` `E_("cell")^(@) = 1.41 - (0.0591)/(6) "log" ((10^(-3))^(2))/((0.5)^(3)) = 1.41 - (0.0591)/(6)"log" (10^(-6))/(0.125)` `= 1.41 - (0.0591)/(6) "log" (10^(-6) xx 8) = 1.41 - (0.0591)/(6) ("log" 10^(-6) + "log" 2^(3))` `= 1.41 - (0.0591)/(6) (-6 "log" + 3 "log" 2)` `= 1.41 - (0.0591)/(6) (-6 + 3 xx 0.3010) = 1.41 - (0.0591)/(6) (-5.097)` `= 1.41 + (0.3012)/(6) = 1.46V` |
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| 362. |
Explain the following observations : (i) Many of the transition elements are known to form interstitial compounds . (ii) There is a general increase in density from titanium (Z =22) to copper (Z = 29). (iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. |
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Answer» (i) Many of the transition elements are known to form interstitial compounds , they are capable of entrapping small atoms like H , C or N in the interstitial sites in their crystals lattice. (ii) There is a general increase in density from titanium (Z =22) to copper ( z = 29) . On moving from titanium to copper in general atomic mass increases where as atomic size decreases , therefore density increases in general . (iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series because in actinoids 5f , 6 d and 7 s subshells are of comparable energies . |
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| 363. |
Where is acrosome present in humans? Write its function. |
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Answer» (a) Head of Human sperm. (b) It contains hydrolytic enzymes that help in dissolving the membrane of the ovum for fertilization. |
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| 364. |
Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruit-flies for his experiments ? State any three reasons. |
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Answer» Margan worked with Drosophila melanogaster a fruitfly. 1. Fruit flies are prolific breeders and have a generation time of two weeks. 2. Fruit flies have three pairs of autosomes and a pair of sex chromosomes (XX in females, XY in males). The sexes could thus be easily differentiated. 3. They could be grown on simple synthetic medium in laboratory. |
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| 365. |
What is meant by (i) petide linkage (ii) biocatalysts ? |
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Answer» (i) Peptide linkage : The amide linkage `(-overset(O)overset(||)(C )-NH-)` formed between two `alpha`-amino acid molecules with the loss of a water molecule in a polypeptide is called a peptide linkage. (ii) Bio catalysts : Biocatalyst is the use of natural catalysis, such as protein enzymes to perform chemical transformations or organic compound. Both enzymes that have been more or less isolated and enzymes still residinh linving cells are employed for this task. |
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| 366. |
Which one of the following compounds is more reactive towards `S_(N)2` reaction and why ? `CH_3CH(Cl) CH_2CH_3 or CH_3CH_2CH_2Cl` |
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Answer» `CH_3overset(Cl)overset(|)CH-underset((a)(2^@))CH_2CH_3` `underset((b)(1^@))(CH_3CH_2CH_2Cl` Primary alkyl halides are more reactive towards `SN_2` reaction because primary alkyl halides are less hindered by alkyl groups rather than `2^@ or 3^@` which are having one more bulky groups which create hindrance for halogen to get detached. |
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| 367. |
The conductivity of a `0.01 ` M solution of acetic acid at 298 k is `1.65xx10^(-4) S cm ^(-1)`. Calculate molar conductivity `(wedge_(m))` of the solution. |
| Answer» `wedge_(m)^(c)=(Kxx1,000)/(C)=(1.65xx10^(-4)xx1000)/(0.01) =1.65xx10^(1)=16.5Scm^(2)//"mol"`. | |
| 368. |
Write the IUP AC name of the following compound : `(CH_3)_3C CH_2Br` |
| Answer» 1-Bromo-2, 2-dimethylpropane (i) | |
| 369. |
Draw the structure of 2 bromopentane |
| Answer» `CH_(3)-CHBr-CH_(2)-CH_(2)-CH_(3)` | |
| 370. |
Write the IUP AC name of the following compound : `CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-underset(Cl)underset(|)CH-CH_(3)` |
| Answer» 3-Chloro, 2-dimethylbutane. | |
| 371. |
Write the IUPAC name of `PH-CH=CH-CHO`. |
| Answer» 3-Phenyl prop-2enal. | |
| 372. |
Rearrange the following compounds in the increasing order of their boiling points : `CH_(3)-CHO,CH_(3)-CH_(2)-OH,CH_(3)-CH_(2)-CH_(3)` |
| Answer» `CH_(3)-CH_(2)-CH_(3)ltCH_(3)-CHOltCH_(3)-CH_(2)-OH` | |
| 373. |
Arrange the following in increasing order of boiling points : `(CH_(3))_(3)N,C_(2)H_(5)OH,C_(2)H_(5)NH_(2)` |
| Answer» `(CH_(3))_(3)NltC_(2)H_(5)NH_(2)ltC_(2)H_(5)OH` | |
| 374. |
Write the IUPAC name of the following compound: `CH_(3)underset(CH_(3))underset(|)-CH-CHO` |
| Answer» Correct Answer - 2 methyl propanal | |
| 375. |
Out of `CH_(3)-NH_(2)` and `(CH_(3))_(3)N` which one has higher boiling point? |
| Answer» `CH_(3)NH_(2)` has higher B.P because it has 2 hydrogen to form hydrogen bonding hence higher boiling point | |
| 376. |
(a) For a reaction ,`A+B rarr Product` the rate law is given rate =`k[A]^(1)[B]^(2)` .What is the order of the reaction ? (b) write the unit of rate constant k for the first order reaction |
| Answer» (a) order =3 (b) unit of k =`s^(-1)` | |
| 377. |
How is the vapour pressure of a solvent affected when a non volatile solute is dissolved in it? |
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Answer» When solute is added to a solvent the vapor presure of the solvent (above the resulting solution) is lower than the vapor pressure above the pure solvent. When a non volatile solute is added to sovent it reduces the escaping tendency of solvent molecule in to vapour phase because some of the solute particle occupy the surface and thus lower the vap pressure of solvent .This behaviour is shown by rault law `P_(solvent)=X_(solvent)P_(solvent)^(0)` Where: `P_(solvent)`is the vapour pressure of the solvent above the solution `X_(solvent )` is the mole fraction of the solvent in the solution and `P_(solvent)^(0)` is the vapor pressure of the pure solvent |
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| 378. |
The graphs, drawn here, are for the phenomenon of photoelectric effect. (i) Identify which of the two characteristics (intensity/frequency) of incident light, is being kept constant in each case. (ii) Name the quantity, corresponding to the in eact case. (iii) Justify the existence of a threshold for a given photosenitive surface. |
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Answer» (i) Intensity is constant in graph I. Frequency current in graph II. (ii) Saturation current and stopping potential. (iii) `VltV^(0)` i.e., frequency of incident radiation is less than the threshold frequency the K.E. of photoetectrons becomes negative which has no physical meaning. So, there is a minimum frequency that is called threshold frequency at which photoelectric effect occure. |
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| 379. |
Obtain the relation `N=N_(0)e^(-lamdat)` for a sample of radio active material having decay constant `lamda` where N is the number of nuclei present at instant t. Hence, obtain the relation between decay constant `lamda` and half life `T_(1//2)` of the sample. |
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Answer» According to radion active law, the rate of decay at any instant is proportional to number of atoms present at that time. i.e., `-(dN)/(dt)propN" "implies" "(dN)/(dt)=-lamdaN" "....(i)` `(dN)/(N)=-lamdadt" "impliesint(dN)/(N)=-lamdaintdt" "....(ii)` `log_(e)N=-lamdat+C" at "t=N=N_(0)` Equation (ii) `implieslogN_(0)=C` Put in equ. (ii) `logN=-lamdat+logN_(0)logN-logN_(0)=-lamdat` `log""(N)/(N_(0))=-lamdat" "implies" "(N)/(N_(0))=e^(-lamdat)` `N=N_(0)e^(-lamda)t` Furthre `t=T_(1//2),N=(N_(0))/(2)` `N=N^(0)e^(-lamdat)or(N_(0))/(2)=N_(0)e^(-lamdaT_(1//2))` `implies(1)/(2)=e^(-lamdaT_(1//2))impliese^(-lamdaT_(1//2))=2` `lamdaT_(1//2)=elog2=.693impliesT_(1//2)=(0.693)/(lamda)`. |
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| 380. |
(a) Define the following terms: (i) Molarity (ii) Molal elevation constant `(k_(b))` (b) A solution containing 15 g (molar mass=60 g `mol^(-1))` per litre of solution water has the same osmotic pressure (isotonic ) as a solution of glucose (molar mass =180g `mol^(-1))` in water calculate the mass of glucose present in one litre of its solution. |
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Answer» (a)(i) Molarity (M) : Molrity can be defined as no of moles of solute dissolved per liitre of solutiohn lolarity M= vol of solutio (liter moles of solute) (ii) Molal elevation constant (kb) When 1 molal solution is prepared the elevation in boiling point is called as molal boling point elevation constant (b) For isotonic solution `pi_("urea")=("WBRT")/("MB"xxV)=(15 RT)/(60 V)` `pi_("glucose")=("WBRT")/(180V)` `pi_("urea")=pi_("glucose")` `(15"RT")/(60 V)=("WBRT")/(180V)` WB=`(15xx180)/(60)=45g` |
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| 381. |
(a) what type of deviation is shown by a mixture of ethanol and acetone? Give reason (b) A solution of glucose (molar mass =180 g`mol^(-1))` in water is labelled as 10% (by mass) what would be the molality and molarity of the solution ? (Density of solution =1.2 `gL^(-1))` |
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Answer» (a) Ethanol and acetone shows + ve deviation beacuase both are non polar compunds and after mixing force of attraction decreases Like particel force of attractio `gt` unike particle force of attraction (b) Molarity `=("WB"xx1000)/("MB"xx"WA")=(10xx1000)/(180xx90)=0.617 M` Vs =`("MS")/("ds")rarr(100)/(1.2 g//ml)` Molarity `=("WB"xx1000)/("MB"xx"Vs")rarr(10xx1000)/(180xx100)xx1.2=0.667` m |
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| 382. |
(a) Complete the following equation : (i) `CrO_(4)^(2-)+2OH^(-)rarr` (ii) `MnO_(4)^(-)+4H^(+)+3e^(-)rarr` (b) Account for the following : (i)Zn is not considered as transition element (ii) Transition metals form a large number of complexes (iii) The `E^(@)` value for the `(Mn^(3+))/(Mn^(2+))` couple is much more positive than for `(Cr^(3+))/(Cr^(2+))`couple. |
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Answer» (a)(i)`Cr_(2)O_(7)^(-2)+2OHrarr2crO_(4)^(-2)+H_(2)O` (ii)`MnO_(4)+4H+3e^(-)rarrrMnO_(2)+2H_(2)O` (b) (i) Because Zn has complelety filled d orbital (ii) Smaller sizes of metal ion high ionic changes vacant d orbital (iii) `Mn^(3+)` less stable than `Mn^(2+)`(3ds) which is stabel half filled configuration `Cr^(3+)` has stable `t_(22)g^(3)` configureation `Cr^(3+)` cannot reduce to `Cr^(2+)` (i) with reference to structural variabliltiy and chemical reactivity write the differenence between lanthanoids and actinoids (ii) Name a memeber of the lanthanoid series which is well known to exhibit + 4 oxidation state (iii) Complelte the following equation : `MnO_(4)^(-)+8H^(+)+5e^(-)rarr` (iv) out of `Mn^(+)` and `Cr^(3+)` which is more paramagnetic and why ? Atomic nos : Mn =25 Cr =24 |
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| 383. |
Write the coordination isomer of `[Cu(NH_3)_4][PtCl_4]`. |
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Answer» `[Cu(NH_3)_4][PtCl_4]` Coordination isomer is `[CuCl_4][Pt(NH_3)_4]` |
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| 384. |
For the reaction `AtoB`, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction ? |
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Answer» `A to B` Rate (R) `=K[A]^(alpha)` New Rate law `3R=K[9A]^(alpha)` From (1) and (2) `(1)/(3)=(1/9)^(alpha)rArr(1)/(2)=(1/3)^(2alpha)` `rArr1=2alpharArralpha=(1)/(2)` |
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| 385. |
A current of `1.50` A was passed through an electrolytic cell containing `AgNO_3` solution with inert electrodes. The weight of silver deposited was `1.50g`. How long did the current flow ? (Molar mass of `Ag=108 g " mol "^(-1), 1F=96500C " mol" ^(-1)`). |
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Answer» Current (F)` =1.50A` `W=1.50g` `t=?` Molar mass (M)`=108//mol` `1F=96500C//mol` `W=ZIT=("Equiy weight")/(96500)xxIxxT" "[Eq. weight =(108)/(I)]` `1.50=(108)/(96500)xx1.50xxt` `t=(96500)/(108)=893.52Sec`. |
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| 386. |
(a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given : `E^(@)Zn_(2+1//Zn) = -0.76V, E^(@)A_(g+//)A_(g) = +0.80V` What is the effect of increase in concentration of `Zn^(2+) " on the " E_(cell)` ? (b) Write the products of electrolysis of aqueous solution of NaCI with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: `"Ni(s)"//"Ni"^(2+)(0.01M)////"Cu"^(2+)(0.1M)//"Cu(s)"` `["Given"E_(Ni2+//Ni)^(@) = -0.025 V E_(Cu2+//Cu)^(@) = +0.34V]` Write the overall cell reaction. |
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Answer» (a) `E_(cell) " decreases."` (b) `"Anode" : CI_(2)uarr` `"Cathode" : H_(2)uarr` (c) `Cu^(2+)(aq) + Ni(s) to Ni^(2+)(aq) + Cu(s)` `E_(cell)^(@) = E_("cathode")^(@) - E_("anode")^(@)` `E_(cell)^(@) = 0.34-(-0.25)` `E_(cell)^(@) =0.59V` `E_(cell)^(@) =E_(cell)^(@)-(2.303RT)/(nF) "log" ([Ni^(2+)])/([Cu^(2+)])` `E_(cell)^(@) = 0.59 - (059)/(2)"log" ([0.01])/([0.1])` `E_(cell) =0.6195V` |
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| 387. |
(a) What type of a battery is the lead storage battery ? Write the anode and the cathode reactions and the overall occuring in a lead storage battery when current is drawn from it. (b) In the buttom cell, widely used in watches the following reaction take place `Zn_((s)) + Ag_(2)O_((l)) rarr Zn^(2+)(aq) + 2Ag_((s)) + 2OH_((aq))^(-)`. Determine `E^(@)` and `DeltaG^(@)` for the reaction. (given : `E_(Ag^(+)//Ag)^(@) = +0.80 V , E_(Zn^(2+)//Zn)^(@) = - 0.76 V)` |
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Answer» (a) It is a secondary cell. `{:("Anode":" "Pb+SO_(4)^(2-)rarrPbSO_(4)2e^(-)),("Cathode":" "PbO_(2)+PBO_(2)+4H^(+)+2SO_(4)^(2-)rarrPbSO_(4)+2H_(2)O):}` Overall reaction : `Pb+PBO_(2)+4H^(+)+2SO_(4)^(2-)rarrPbSO_(4)+2H_(2)O` (b) Zn is oxidised and `Ag_(2)O` is reduced. `E_("cell")^(0)=E_(Ag_(2)O),Ag_("reduction")-E_(Zn//Zn_(("oxidation"))^(2+)` `= 0.8 - (-0.76) = 1.564 V` `DeltaG^(@)= - nFF_("cell")^(0) = 2 xx 96500 xx 1.56 = 301080 J//"Mol"` |
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| 388. |
A bar magnet of magnetic moment 6 `J//T` is aligned at `60^(@)` with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii). |
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Answer» Formula of work done is given by . `W = MIB (cos theta f - cos thetai)` M = magnetic moment `betaf` = magnetic field `thetaf` = Final angle-substanded by bar magnet. `thetai` = Initial angle substanded by bar magnet. (a) (i) Normal to the magnetic field `theta f = 90^(@)` `thetai = 60^(@)` `M = 6J//T and B = 0.44T` So, `W =- 6xx 0.44 [ cos 90^(@) - cos 60^(@)]` `=- 2.46[ 0-1//2] = 1.32J` (ii) Opposite to the magnetic filed. `thetaf = 180^(@)" "thetai = 60^(@)` `W =- 6 xx 0.44 [ cos 180^(@) - cos 60^(@)] = - 3.64 xx (-1-1//2) = 3.96J` (b) `T = MB sin theta` `=- 6 xx 0.44 sin (180^(@)) =0` |
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| 389. |
Name the scientist who disproved spontaneous generation theory. |
| Answer» Louis Pasteur disproved spontaneous generation theory. According to this theory living organisms arose from decaying and rotting matter. | |
| 390. |
Waste Disposal and Waste Management poses a major problemin present times. Generation of garbage and its disposal is a major threat and consequently leads to servers environmental isssue. The problem is not with biogradable and recycled wastes. We realies that the need is to reduce non-biogradable wastes. (a) Why is there a great concern of managing non-biodegrable waste in comparison to biodegradable waste ? Explain. (b) As a member of eco club of your school, suggest any two ways that you will discuss with your fellow members to organise for a 'Zero garbage day' once in a month in the school. |
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Answer» (a) Non-biodegradable wastes is a type of waste that cannot be broken down into its base by micro-organisms, air moisture or soil. `*` It causes land and water pollution. Herbours pathogens and vectors and promotes the spread of disease such as cholera and typhoid. `*` Burning of polystyrene polymers. `*` Death of animals they ingest the plastic and it entraps in the digestive system. That is why, it is necessary ti manage non-niodegradable waste in comparison to biodegradable waste. (b) Zero Garbage Day : (1) Use of Garbege bags and say no to plastic : Whenever you see solid waste, disposed on roads of near by your house, collect all the solid waste and throw it in garbage. * Use dustbin in place. Do not use plastic materials as they are toxic to nature, humans and animals. * separately collect the waste which can be recycled and reused. Dumping station or landfills should be constructed so thet no wast should be left open on roads. (2) More use Eo-friendly materials : As they are for our enviroment and they do not emit any harmful substance. Go Green and Plant more trees. |
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| 391. |
List two advantages of the use of unleaded petrol in automobiles as fuel. |
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Answer» Unleaded petrol does not release lead compounds from exhaust fumes into the atomsphere and caues less pollution . Unleaded petrol does not emit harmful componds ,It helps in preventing health diseases like bronchits ,asthma and lung diseases . |
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| 392. |
With the help of any two suitable examples explain the effect of anthropogenic actions on organic evolution |
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Answer» Anthropogenic activities have caused increased rate of evolution. (1) The excess use of herbicides and pesticides in the agricultural field by the humans to kill pests and insects has resulted in selection of resistant variety of pests and insects over a short time span. The change favoured resistant pests and insects which lead to their evolution. (2) The excess use of antibiotics has caused selection of drug resistant microbes, the microbes sensitive to antibiotics died but few variants having resistance against it survived. This led to evolution of more fatal microbes. |
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| 393. |
Aflower of brinjal plant following the process of sexual reprodution prroduces viable seeds. Answer thr following questions giving reasons: How many ovules are minimaly involved ? ( b) How many megaspore mother cells are involed ? (c ) What is the minimum number of pollen grains that must land on stigma for pollination ? ( d) How many male gametes in the above cese ? |
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Answer» 360 one ovule after fertilisation forms on seed . (b) 360 each MMC forms four megaspores out of which only one remains functional. (c ) 360 one pollen grain particiates in fertillisation in fertillisation of one ovlue . 720 each pollen grain carries two male gametes ( e) 90 , each micropore mother cells divides to from four pollen grains. |
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| 394. |
DNA being hydrophilic cannot pass through the cell membranes of a hot cell. Explain how does recombinant DNA get introduced into the host cell to transfrom the latter. |
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Answer» Introduction of rDNA into host cell. (i) Heat shock method- In this method, DNA is treated with a specific concentration of a divalent cation, such as calcium which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. The rDNA is forced into the cell by incubating the cell with rDNA on ice followed by placing them briefly at `42^(@)`C (heat shock) and then patting them back on ice ? (ii) Micro Injection- In this method, the rDNA is directly injected into the nucleus of an animal cell. (iii) Gene `Gun// Biolistics- In this method, cells are bombarded with hifh velocity microparticles of gold or tungsten coated with DNA. |
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| 395. |
Name an organsim where cell division in itself is a mode of reproduction. |
| Answer» Amoeba and Paramoecium | |
| 396. |
State the role of transposons in silencing of mRNA in eukaryotic cells. |
| Answer» Transposons or mobile genetic elements in viruses are the sources of the complementary dsRNA, which in turn bind to specfic mRNA and cause RNA interference of the parasite. | |
| 397. |
When does a human body elicit an anamnestic response? |
| Answer» When the immune system encounters a pathogen for the first time, a primary response is produced which is of low intensity. An anamnestic response is produced which is of very high intensity when the immune system encounters the pathogen again | |
| 398. |
A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? |
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Answer» Let the concetration of the reactant , [A] =a Order of reaction =2 so that Rate fo reactioin will Rate =`k[A]^(2)" "…(i)` Plug the values we get Rate `=ka^(2)` (i) Given that concentration of the reactant of the reactant is double So that `[A]=2a,` Plug the value in equation (1) we get New Rate of reaction, `R_(1)=k(2a)^(2)=4ka^(2)` We have already calculate that initial rate `R=Ka^(2)` Plud the value we get `R_(1)=4R` Hence rate of reaction will increased to 4 times (ii) Given that concentration of the reactant is reduced to half So that [A]=(1/2)a Plug the values in equation (1) we get New Rate of reaction , `R_(2)=k((1//2)a)^(2)` `=(1//4)ka^(2)` We have already calculate that initial rate `R=ka^(2)` Plug the value we get `R_(1)=(1//4)R` Hence rate of reaction will reduced to 1/4 |
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| 399. |
Calculate the freezing point of a solution containing 8.1 g of HBr in 100g of water, assuming the acid to be 90% ionized. [Given : Molar mass Br = 80 g/mol, `K_(f)` water = 1.86 K kg/mol]. |
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Answer» `HBr to H^(+) + Br^(-)` `i = 1-alpha+mx` n=2 `i=1+alpha` `DeltaT_(f) = iK_(f)m` `DeltaT_(f) = (1-alpha)1.86 xx (8.1)/(81) xx (1000)/(100)` `rArr DeltaT_(f) = (1+0.9) xx 1.86 xx (8.1)/(81) xx 10` `DeltaT_(f) = 3.53` `T_(f)^(@) = 0^(@)C` `DeltaT_(f) = T_(f)^(@) - T_(f)` `T_(f) = -3.534^(@)C` |
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| 400. |
mention any two human diseases caused by round worms. Name their causative agents and their mode of transmission into the human body. |
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Answer» Ascariasis (round worms). 1. It is caused by an intestinal endoparasite of humans, Ascaris lumbnco1des, commonly called round worm. 2. Eggs of parasite are excreted along with faces of infected person, which contaminate water, soil and plants. 3. Infection takes place through contai:ninated vegetables, fruits and water. 4. Abdominal pain, indigestion, fever, muscular pain, internal bleeding, nausea and headache are symptoms of round worm infection. |
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