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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.
| 751. |
When do the oogenesis and the spermatogenesis initiate in human females and males respectively? |
| Answer» Oogenesis is initiated during the embryonic development, Spermatogenesis begins at puberty. | |
| 752. |
State the significance of the study of fossils in evolution. |
| Answer» Study of fossils indicates the geological period in which various life form arose. The calculation of geological period can be done via radioactive dating. | |
| 753. |
Why does the bluish colour predominate in a clear sky? |
| Answer» A blue colour has a shorter wavelength than red, therefore, blue colour is scattered much more strongly. Hence the sky looks blue. | |
| 754. |
Write the IUPAC name of the following compound: `H_(3)C-underset(CH_(3))underset(|)(C)=underset(Br)underset(|)(C)-CH_(2)-OH` |
| Answer» IUPAC name, 2-Bromo-3-methyl but-2-en-1-ol. | |
| 755. |
Mention the property of plant cells that has helped them to grow into a new plant in in-vitro conditions. Explain the advantages of micropropagation. |
| Answer» The advantages of micropropagation are: (i) BY this method, it is possible to achieve propagation of a large number of plants in very short .durations. Plants like tomato, banana, apple, etc., have been produced on commercial scale. (ii) Healthy plants can be recovered from diseased plants by micropropagation. This is done by removing the meristem, which is disease-free and growing it in vitro. This has been done in banana, sugar cane, potato, etc" | |
| 756. |
What type of colloid is formed when a solid is dispersed in a liquid? Given an example. |
| Answer» Sol is formed when solid is dispersed in liquid medium. Example, Mudy water and Paint. | |
| 757. |
A solution of glycerol `(C_(3)H_(8)O_(3))` in water was prepared by dissolving some glycerol in `500g` of water. This solution has a boiling point of `100.42^(@)C` while pure water boils at `100^(@)C`. What mass of glycerol was dissolved to make the solution ? (`K_(b)` for water `= 0.512 K kg mol^(-1)`) |
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Answer» `W_(H_(2)O) = 500 g = 0.5 kg, T_(b)^(0) = 100^(0)CW_("glycerol")` `:. T_(b) = 100.42 .^(@)C`. `DeltaT_(b) = 0.42.^(@)C` `DeltaT_(b) = K_(bm)` `DeltaT_(b) = (W_("glycerol"))/(W_(H_(2)O) xx M_("glycerol"))` `0.42 = 0.512 xx (W_("glycerol"))/(0.5 xx 92)` `W_("glycerol") = (0.42 xx 0.5 xx 92)/(0.512) = 37.73 g`. |
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| 758. |
(a) Define the following terms : `{:((i)"Ideal solution",(ii)"Osmitic pressure"):}` (b) Calculate the boiling point elevation for a solution prepared by adding 10g `CaCl_(2)` to 200g of water, assuming that `CaCl_(2) `is completely dissociated. `(K_(b)"for water"=0.512Kg mol^(-1),"Mole mass of "`CaCl_(2)=111g mol^(-1))` |
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Answer» (a) (i)Ideal solution : An ideal solution of the components A and B is defined as the solution in which the intermolecular interactions between A-B are of the same magnitude as theintermolecular interactions found in the pure coponent (A-A and B-B). Example: Benzene + Toluene (ii) Osmotic Pressure : The minimum execess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure. (b) `DeltaT_(b)=K_(b)m` `T_(S)-T_(A)^(@)K_(b)xx(wB)/(M_(B)xxwA(kg))` `T_(A)^(@)=100^(@)C` `{:(w_(B)=10g,M_(B)=111g//mol),(w_(A)=10g,K_(b)=0.512kg//mol):}` As, `CaCl toCa^(2+)+2Cl^(-)` i = 3 [Completely dissociates] `T_(S)-T_(A)^(@)=ixx(Kfxxw_(B))/(M_(B)xxw_(A)(kg))` `T_(S)100=(3xx0.512xx10^(5)xx1000)/(111xx200)=0.692` `T_(S)=100.692^(@)C` |
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| 759. |
How have human activities caused desertification ? Explain. |
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Answer» Human activities contribute to desertification- 1. Deforestation- Humans cut down trees to serve their own purpose like construction of houses and roads which is the main cause of desertification. 2. Improper farming practices-If same crop is grown continuously, it makes the soil deficient of nutrients resulting in the loss of fertility of soil. 3. Soil Erosion-That is by different human activities like construction of houses and industrialization. Or An algal bloom is the phenomenon of excessive growth of planktonic forms in a nutrient rich water body. As the planktonic species multiply on the surface they form a layer that eventually covers the entire surface of the water body. They block sunlight, which does not reached submerged aquatic plants that may have a role in supplying necessary nutrients to other aquatic life forms and keeping the water clean. Some algal species can release toxic substances. Due to the high respiratory needs of such a huge concentration of biomass on the surface, the biological oxygen demand (BOD) of the water body increases, causing many of the life forms to die, their carcasses further contribute to the deterioration of the quality of the water content. |
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| 760. |
Write two factors justifying the need of modulation for transmission of a signal. |
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Answer» Need for modulation : The sound waves cannot be transmitted from a radio transmitter by converting them into electrical wave directly for the following reasons : (i) For efficient transmission and reception, the transmitting and receiving antennas must have a length equal to quarter wavelength of the audio signal. To set up a vertical antenna of this size is paractically impossible. (ii) The energy radiated from an antenna is practically zero, when the frequency of the signal to be transmitted is below 15 kHz. It also makes the direct transmission of audio signal as impracticable. |
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| 761. |
Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons. |
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Answer» The use of A.C. voltage is preferred over the use of D.C. voltage because of the following reasons : (i) The loss energy in transmitting the A.C. voltage over long distances with the help of step up transformers is negligible as compared to D.C. voltage. (ii) A.C. voltage can be stepped up and stepped down as per the requirement using a transformer. |
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| 762. |
Why is it found experimentally difficult to detect neutrinos in nuclear `beta`-decay? |
| Answer» Neutrinos are difficult to detect experimentally in a decay because they are uncharged particles with almost no mass. Also, neutritions interact very weakly with matter, so they are very difficult to detect. | |
| 763. |
Write the name of the cel which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. |
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Answer» Mercury cell is used in hearing aids Reaction: At Anode: `Zn(s)+2OH^(-)toZnO(s)+H_(2)O(l)+2e^(-)` At Cathode: `underline(" "HgO(s)+H_(2)O(l)+2e^(-)toHg(l)+2OH^(-)" ")` `underline(" "Zn(s)+HgO(s)toZnO(s)+Hg(l)" ")` |
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| 764. |
Write the IUPAC names fo the following compounds : (i) `CH_(2) =ChCH_(2)Br` (ii) `("CC"l_(3))_(3)"CC"l` |
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Answer» (i) Cu(I) are unstable in aqueous soluton and undergo disproportionation . `2Cu^(+) to Cu^(2+) + Cu CY^(2+)` is stable due to more (-) ve `Deltayd` H of `Cu^(2+)` (aq) that compensates the 2nd I.E. of Cu. (ii) Thus is due to the tendency of oxygen to form multiple bonds. |
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| 765. |
Write the IUPAC name of the following compound : `CH_(2) = CHCH_(2) Br` |
| Answer» Correct Answer - 3-Bromol-1-Propene. | |
| 766. |
Draw the structure of 4-choloropentan-2-one. |
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Answer» 4-chloropentan - 2-one . `overset(1)CH_(3)-overset(2)CO-overset(3)CH-underset(Cl)underset(|)overset(4)CH-overset(5)CH_(3)` |
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| 767. |
How will you convert ethanol to ethene? Write chemical equation |
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Answer» Ethonal to ethene. `CH_(3)underset("Ethanol")(CH_(2))OH underset(443k)overset("Conc."H_(2)SO_(4))toCH_(2)= CH_(2)` |
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| 768. |
How would you account for the following : (i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur. (ii) Phosphorus shows greater tandency for catenation than nitrogen. (iii) Fluorine never acts as the central atom in polyatomic interhalogen compounds. |
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Answer» (i) Because of the compact nature of oxygen atom oxygen has less negative electron gain enthalpy than sulphur. (ii) It is due to weaker N-N bond than P-P bond due to more repulsion between valence `e^(-)s` of nitrogen (smaller in size) than phosphorus. (iii) It is becuase `F` does not have vacand d-orbitals and cannot show positive and higher oxidalion state. |
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| 769. |
Give the prepartion and uses of PVC ( Polyvinyl Chloride) |
| Answer» `underset("Vinyl chloride")(underset(Cl)underset(|)nCH = CH_(2))underset("Polyvinyl chloride")([CH_(2)-underset(Cl)underset(|)CH])` | |
| 770. |
Which aerosol depletes ozone layer? |
| Answer» CFC (Chloroflor-Carbon) | |
| 771. |
Write the characteristics of Ramapithecus , Dryopithecus and Neanderthal man. |
| Answer» Ramapithecus about 15 mya , primates called Ramapitheus were existing . They were hairy and walked like gorillas and chimpanzees Ramapithecus was more man like. Dryopithecus about 15 mya primates called Dryopitheus existing was more ape like. The Neanderthal man with a brain size of 1400 c c lived in near east and central Asia between 40,000 -1,00.000 years back . they used hides to protect their body and burried their dead. | |
| 772. |
How does the resistivity of a consuctor depend upon temperature electrical consuctivity ? |
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Answer» (i) The resistivity of a conductor increases with increase in temperature `:. rho_(T)=rho_(o) [1+alpha (T-T_(o))]` (ii) The resistivity of a consuctor is the reciprocal of electrical conductivity `:. rho=1/sigma` |
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| 773. |
Write the function of (i) Transducer and (ii) Repeater in the context of communication system. |
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Answer» (i) Transducer : A transducer is a device, which converts one form of energy into another. (ii) Repeater : A repeater is a combination of receiver and transmitter placed along the path of signal so as to extent the range of the communication system. |
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| 774. |
An object is kept in front of a concave mirror of focal length 15 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror. |
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Answer» `{:(f=-15 cm),(m=-3):}` `m=(-v)/(u)=-3` `rArr" "v=3u` Mirror equation `(1)/(f)=(1)/(v)+(1)/(u)` `(1)/(-15)=(1)/(3u)+(1)/(u)` `(1)/(-15)=(1+3)/(3u)=(4)/(3u)` `u=(4xx(-15))/(3)=-20 cm` `u=-20cm` |
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| 775. |
An electric dipole is held in a uniform electric field. (i) Show that the net force acting on it is zero. (ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle fo `180^(@)`. |
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Answer» (i) Force acting on charge - q and +q of a dipole is equal and in opposite direction. So, net force is zero but their line of action may be different so the dipole may experience a torque. (ii) `W=pE(cos q2- cos q1)` `=pE (cos 180^(@)-cos 0^(@))=pE(-1-1)=-2pE` |
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| 776. |
What type of ores can be concentrated by magnetic separation method ? |
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Answer» These are four types: 1. Hydraulic washing 2. Magentic separation 3. Froth floatation method 4. Leaching |
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| 777. |
For a decomposition reaction the values of rate constant k at two different temperatures are given below : `K_(1)=2.15xx10^(-8)L"mol"^(-1)s^(-1)"at " 650K` `K_(2)=2.39xx10^(-7)L"mol"^(-1)s^(-1)"at " 700K` Calclate the value of activation energy for this reaction. `(R=8.314JK^(-1)"mol"^(-1))` |
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Answer» Here `K_(1)=2.5xx10^(-8)L "mol"^(-1)S^(-1),K_(2)=2.39xx10^(-7)L" mol"^(-1)S^(-1)` `T_(1)=650K,T_(2)=700 K and R=8.314JK^(-1)" mol"^(-1)`. Using the formula, `"log"(K_(2))/(K_(1))=(Ea)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `"log" (2.39xx10^(-7))/(2.15xx10^(-8))=(Ea)/(2.303xx8.314)[(700-650)/(650xx700)]` `"log" 1.11xx10=(Ea)/(19.147)xx(50)/(455000)` `1.0457=(Ea)/(19.147)xx(1)/(9100)` `:.Ea=1.0457xx19.147xx9100` `=182200.36J` |
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| 778. |
Identify the compound that on hydrogenation produces an optically active compound from the following compounds : |
| Answer» Correct Answer - B | |
| 779. |
(a) For a reaction `ABrarrP` , the rate law is given by, `r=k[A]^(1//2)[B]^(2).` What is the order of this reaction? (b) A first order reaction is found to have a rate constant `k=5.5xx10^(-14)s^(-1)`, Find the half life of the reaction. |
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Answer» (a) `0.5+2=2.5` (b) For the first order reaction, `t_(1//2)=(0.693)/(K)=(0.693)/(5.5xx10^(-14)S^(-1))=1.26xx10^(13)S.` |
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| 780. |
Calculate the equilibrium constant for the reaction at 298 K `Zn(s)+Cu^(2+)(aq)harr Zn^(2+)(aq)+Cu(s)` Given` " " E_(Zn^(2+)//Zn)^(@)=-0.76 V` and `E_(Cu^(2+)//Cu)^(@)=+0.34 V` |
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Answer» `DeltaG^(@)=-"RT In Kc "=-2.303" RT log Kc"` `therefore" "-212300=-2.303xx8.314xx298xxlog Kc` `"or "log Kc =(212300)/(2.303xx8.314xx298)=37.2074` `Kc" = Antilog "37.2074=1.6xx10^(37)` |
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| 781. |
A sugarcane has been affected by virus. How can a virus free cane be developed from it ? Explain the procedure. |
| Answer» The virus free sugarcane can be developed by the process of tissue culture where in the whole plant is regenerated from explants. i.e., any part of a plant grown in a test tube under sterile conditions in special nutrient media. The technique makes use of the property of totipotency and the fact that the meristem (apical and axillary) of the sugarcane plant is free of virus. Hence, the meristem can be removed and grown In-Vitro to get disease free plant. | |
| 782. |
Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes : `[CoCl_(4)]^(2-),[Ni(CN_(4)]^(2-), [Cr(H_(2)O)_(2)(C_(2)O_(4))_ (2)]^(-)` ` ("At.No.": Co= 27, Ni = 28, Cr = 24)` |
| Answer» `{:("Complexes-dization","Hybridisation"," Shape","Magnetic behaviour"," IUPAC Name"),((i) [CoCl_(4)]^(2-)," "sp^(3),"tetrahedral"," paramagnetic","tetrachlorido cobaltate (ii) ion"),((ii) [Ni(CN_(4)]^(2-)]," "dsp^(2),"square planar"," diamagnetic","tetracyanonickelate (ii) ion "),((iii)[Cr(H_(2)O)_(2)(C_(2)O_(4))_(2)]^(-1)," "d^(2)sp^(3),"octahedral"," paramagnetic","diaquadioxalato chromate (iii) ion"):}` | |
| 783. |
Write any two factors on which internal resistance of a cell depends .The reading on a high resistance voltmer,where a cell is connected across it , is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 `Omega` as shown in the circuit , the voltmeter reading drops to 1.8 V .Find the interval resistance of the cell. |
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Answer» Internal resistance of a cell depends upon the following factors : (i) Distance between the electrodes . (ii) The nature of the electrolyte . (iii) The nature of electrodes . Area of the electrodes , immersed in the electrolyte . If area increases , internal resistance decreases . (b) Given : emf of cell , `epsi = 2.2 V , R = 5 Omega , V = 1 . 8 V` `r = ((epsi)/(V) - 1) R implies ((2.2)/(1.8) - 1) xx 5 = ((2.2 - 1.8)/(1.8)) xx 5` `therefore " " r = (0.4)/(1.8) xx 5 = (2)/(1.8) = 1.1 Omega`. |
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| 784. |
(i) What are disinfectants ? Given an example. (ii) Given two example of macro-molecules that are chosen as drug targets. (iii) What are anionic detergents ? Give an example . |
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Answer» (i) Disinfectants : A disinfectant is used primarily to sanitise objects. If acts as an antimicrobial that kills the microbes that cause infection. Disingectants usually come as liquids or sprays . Example : Chlorine in the concentration of 0.2 to 0.4 ppm, in an aquenous solution, acts as a disingectan. (ii) Enzymes and receptors are the macromoleculars that are chosen as drug targets. (iii) Aniomic datergents : The detergents are sodium salts of sulphonated long chain alcohols or hydrocarbone eg. sodium lauryl sulphate. Anionic detergents are of two types : 1. Sodium alkyl sulphates : Sodium lauryl sulphate `(C_(11)H_(23)CH_(2)OSO^(-3) Na^+)` and sodium stearyl sulphate `(C_(17)H_(35)CH_(2)OSO^(-3)Na^+)`. 2. Sodium alkyl benzenesulphonates : Sodium `4-(1-dodecy) benzenesulphonate (SDS). |
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| 785. |
Study the diagram showing replication of HIV in humans and answer the following questions accordingly : (i) Write the chemical nature of the coat A . (ii) Name the enzyme B acting on X to produce molecule 0 . N ame 0 . (iii) Mention the riame of the host cell D the HIV attacks first when it enters into the human body. (iv) Name the two different cells the new viruses E suibsequently attack. |
| Answer» (i) Coat A is made up of protein. (ii) The enzymeV B is reverse transcriptase, C is viral DNA. (iii) The host cell D is macrophage. (iv) The new viruses E subsequently attack macrophages and helper T-lymphocytes | |
| 786. |
Differentiate between primary and secondary succession. Provide one example of each. |
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Answer» Primary succession : 1. The invading species are either lichens (on rocks) or phyto planktons (in water) 2. Rate of succession is slow 3. Climax is slow to reach Secondary succession : 1. The invading species depend on the condition of soil, availability of water and type of seeds 2. Rate of succession is faster 3. Climax is reached faster. |
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| 787. |
Complete the following chemical reaction equations : (i) `XeF_(2)+H_(2)Oto` (ii) `PH_(3)+HgCl_(2)to` |
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Answer» (i) `2XeF_(2)+2H_(2)Oto2Xe+4HF+O_(2)` (ii) `2PH_(3)+3HgCl_(2)toHg_(3)P_(2)+6HCl` |
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| 788. |
A large number married couples in the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless. (a)Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as biology student can promote to check this social evil. (b) State any two reasons responsible for the cause of infertility. (c) Suggest a technique that can help the couple to have a child where the problem is with the male partner. |
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Answer» In India the female partner is often blamed for the childlessness of a couple because of the male-dominated nature of our society. As a student of Biology I can promote the following values to overcome this evil. 1. A childless couple should try to known the problems scientifically and not swarsed by social mores. They should approach a health care facility to seek help for infertility if possible. 2. The couple should look for ART (Assisted reproductive technologies) to have children if the cause of infertility cannot be corrected. (b) The causes of infertility are : 1. Diseases 2. Psychological (c) Artificial insemination is the technique which can help overcome infertility when the problem lies with the male partner. In this technique semen is collected from the husband or a healthy donor and introduced into the vagina artificially of the female artificially. |
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| 789. |
How did Hershey and Chase prove that DNA is the hereditarymaterial ? Explain their experiment with suitable diagrams. Observation and Conclusions |
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Answer» Procedure : Hershey and Chase grew some bacteriophage virus on a medium th at contained radioactive phosphorus (`""^32P`) and some in another medium with radioactive sulphur (`""^32S`). Viruses grown in the presence of radioactive phosphorus (`""^32P`) contained radioactive DNA. Similar viruses grown in presence of radioactive sulphur (`""^35S`) contained radioactive protein. Both the radioactive virus types were allowed to infect E.coli separately. Soon after infection,the bacterial cells were gently agitated in blender to remove viral coats from the bacteria. The culture was also centrifuged to separate the viral particle from the bacterial cell. Only radioactive (`""^32P`) was found to be associated with the bacterial cell, whereas radioactive (`""^35S`) was only found in surrounding medium and not in the bacterial cell. This indicates that only DNA and not protein For Figure See Q. 30., Delhi Board, Set-II, 2008 |
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| 790. |
What is a detritus food chain made up of? How do they meet their energy and nutritional requirements? |
| Answer» A detritus food chain is made of dead organic matter (detritus) and decomposers or heterotrophs such as bacteria and fungi. Decomposers secrete enzymes that break down dead and waste material into simple inorganic material which is later absorbed by them. | |
| 791. |
Explain with the help of a suitable exmaple, the naming of a restriction endonuclease. |
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Answer» The naming of a restriction endonuclease is done in the following way: i) Ist letter represents the genus of the organism from which the enzyme is derived. ii) Iind and IIIrd letters represent the species of the organism. iii) Ivth letter represents the name of the strain. iv) Roman number represents order of isolation. In Eco RI "Derived from E.coli, Strain R. It is the Ist to be discovered. |
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| 792. |
(a) List any three ways of measuring population density of a habitat. (b) Mention the essential information that can be obtained by studying the population density of an organism. |
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Answer» (a) Three ways of measuring population density of a habitat are : (i) By physical counting. (ii) By percent cover or total biomass. (iii) Counting pug marks and counting faecal pellets. (b) (i) The status of habitat. (ii) Whether competition for survival exists or not. (iii) Whether population is increasing or declining. |
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| 793. |
a) Differentiate between dominance and co-dominance. b) Explain co-dominance taking an exmaple of human blood groups in the population. |
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Answer» a) 1. Dominance- It is condition in which one gene show its effect and the other is marked by it. The gene that express its character is dominant while the other is recessive. 2) Co-dominance- It is a condition in which both the genes are equally dominant and so the character of both genes is well expressed in next generations. b) ABO blood groups are controlled by gene I. Gene I has three alleles `I^(A), I^(B)` and `I^(0)//i.I^(A)` and `I^(B)` produce RBC surface antigens A and B, respectively, whereas i does not produce any antigen. `I^(A)` and `I^(B)` are the dominant alleles whereas i is the allele. When `I^(A)` and `I^(B)` are present together, both express equally and produce the surface antigens A and B, as they are co-dominant. |
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| 794. |
a) Write the importance of measuring the size of a population in a habitat or an ecosystem. b) Explain with the help of an exmaple how the percentage cover is a more meaningful measure of population size than mere numbers. |
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Answer» a) The size of the population tells us a lot about its status in the habitat.Whatever ecological processes are to be investigated in a population, be the outcomes of completion with another species, the impact of a predator, it is evaluated in terms of change in the population size. b) Population size more technically called population density. Population size of a species is the number of individuals of a species per unit area volume. `=("Number of individuals in a region "N)/("Unit area in a region"(S))` `rArr P_(Delta) = N/S` A population at any given time is compound of individuals of different ages. When the age of distribution (per cent individuals of a given age or age group) is plotted for the populations, the resulting structure is called age pyramid. |
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| 795. |
Write the stpes you would suggest to be undertaken to obtain a foreign-gene-product. |
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Answer» Step-I: When you inset a piece of DNA into cloning vector and transfer into a bacterial plant or animal cell, the alien DNA gets multiplied. Step II: After having cloned the gene of interest and having optimised the conditions to induce the expression of the target protien, one has to consider producing it on a large scale. Step III. The cultures may be ued for extracting the desired protein and then purifying it by using different separation techniques. iv) The cells can also multiplied in a continuous culture system. The used medium is drained out from one side while fresh medium is added from the other to maintain the cells in their physiologically most active log exponential phase. |
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| 796. |
Why do lepidopterans die when they feed on Bt cotton plant? Explain how does it happen. |
| Answer» Some strains of Bacillus thuringiensis produce protien that kill certain insects each as lepidopterons (tobacco budworms, armyworm, bettles and flies). B. thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxin insecticidal protein. The activated toxin binds to the surface of midget epithelial cells and create pores that causes cell swelling and lysis. The toxin is coded by a gene named cry. | |
| 797. |
The density of lead is `11.35 g cm^(-3)` and the metal crystallizes with fee unit cell. Estimate the radius of lead atom. (At. Mass of lead `= 207 g mol^(-1) and NA = 6.02xx10^23 mol^(-1))` |
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Answer» Density of Lead `=11.35g//cm^3` Atomic mass of lead `=270g//mol` `N_A+6.02xx10^23mol^(-1)," Edges"=a" PM"=axx10^(-10)` Valume `=a^3xx10^(-30)cm^3` `"Density of the unit cell"=("Mass of unit cell")/(a^3xxN_Axx10^(-30))g//cm^3` `11.35g//cm^3=(4xx207g//mol)/(a^3xx6.02xx10^23mol)g//cm^3` `a^3=(4xx207cm^3)/(11.35xx6.02xx10^(-3))` `a^3=121.18xx10^(-24)cm^3` `a=4.948xx10^(-8)cm=494.8p m`. `"In" Fcc,r=(sqrt2a)/4=(1.414xx494.8)/4=175 p m` |
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| 798. |
Complete the following chemical equations : (i) `P_(4)`+ `SOCl_(2)` `rarr` (ii)` F_(2) "(Excess)" + Cl_(2) overset(300^(@)C)to` |
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Answer» (i) `P_(4)+ 8 SoCl_(2) to 4 PCl_(3) + 4SO_(2) + 2S_(2)Cl_(2)` (ii) `3F_(2) + Cl_(2) to 2ClF_(3)` |
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| 799. |
Assign reasons for the following : (i) Transition metals and many fo their compounds act as good catalyst. (ii) Transition metals generally from coloured compounds. |
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Answer» (i) These metal shows variable oxidation states and form unstable intermediate which redily change into product . Hence these are good catalyst. (ii) Because they have unpaired electron in d-orbitals and under go d-d transition by absorbing light from visible region and radiate complementary colour. |
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| 800. |
How can crop varieties be made disease resistant to overcome food crisis in India ? Explain. Name one disease resistant variety in India of : Wheat to leaf and stripe rust (b) Brassica to White rust |
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Answer» Method of breeding for disease resistance: The conventional method of breedm for disease resistance is that of hybridisation and selection. Its steps are essentially identical to those for breeding for any other agronomic characters such as high yield Some crops varieties breed by hybridisation and selection, for disease resistance to fungi bacteria and viral diseases are released. Wheat to leaf and stripe rust Wheat, variety Himgiri and the resistance to diseases is leaf and stripe rust, hill bunt. Brassica to White rust assica variety is Pusa Swamim (Karan Rai), and resistance to diseases is white rust. |
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