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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.
| 851. |
Name the stange of cell division where segregation of an independent pair of chromosomes occurs. |
| Answer» The segragation of an independent pair of chromosomes occurs during Anaphase of Meiosis - I. | |
| 852. |
Two types to aquatic organisms in a lake show specific growth patterns as shown below, in a brief period of time. The lake is adjacent to an agricultural land extensively with fertilizers. Answer the question based on the facts given above : (i) Name the organisms depicting the patterns A and B. (ii) State the reason for the growth pattern seen in A. (iii) Write the effects of the growth patterns seen above. |
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Answer» (i) A Dissolved Oxygen - algae B Biochemical oxygen demend - fish/aquatic animal. (ii) Due to excessive loading nutrients of fertilizers from adjacent agriculture land resuting in increase in nutrients. (iii) Decrease in dissolved, in increase in nutrients. (iii) Decrease in dissolved oxygen, increase in BOD, fish mortality, unpleasant odour of Eutrophication. |
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| 853. |
Two types to aquatic organisms in a lake show specific growth patterns as shown below, in a brief period of time. The lake is adjacent to an agricultural land extensively with fertilisers. Answer the question based on the facts given above : (i) Name the organisms depicting the patterns A and B. (ii) State the reason for the growth pattern seen in A. (iii) Write the effects of the growth patterns seen above. |
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Answer» (i) A Dissolved Oxygen - algae B Biochemical oxygen demend - fish/aquatic animal. (ii) Due to excessive loading nutrients of fertilizers from adjacent agriculture land resuting in increase in nutrients. (iii) Decrease in dissolved, in increase in nutrients. (iii) Decrease in dissolved oxygen, increase in BOD, fish mortality, unpleasant odour of Eutrophication. |
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| 854. |
Explain , giving three reasons, why tropics show greatest levels of species diversity. |
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Answer» (i) Tropical latitude have remained relatively undisturbed, have a long evolutionary time for species diversification. 2. Less seasonal variations, constant and predictable environmant condition, promote niche specialization for greater species diversity. 3. More availability of solar energy, contributes to higher productivity. |
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| 855. |
Name the physical quantity which has its unit joule `"coulomb"^(-1)`. Is it a scalar or vector quantity? |
| Answer» Physical quantity is Electric potential . Yes , it is a scalar quantity . | |
| 856. |
How are dominance, codominance and incomplete dominance patterns of inheritance different from each other ? |
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Answer» Dominance In diploid organism, there are two copies of each gene, i, e. a pair of alleles. Now, these two alleles need not always be identical, as in a heterozygote. One of them may be different due to some changes that it has undergone which modifies the information that particular allele contains. Co-dominance The `F_1` resembled either of the two parents, was in between. But, in the case of co-dominance the `F_1` generation resembles both parents. A good example is red blood cells that determine ABO bllood grouping in human beings. ABO blood group are controlled by the gene I In coinplete dominance When experiments on peas were repeatedu sing other traits in other plants, it was found that sometimes the `F_1` had a phenotype that did not resemble either of the parents and was in between the two. The inheritance of flower colour in the dog flower is a good example to understand incomplete dominance. |
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| 857. |
A genticist intersted in studing variations and patterns of inheritance in living being prefers to choose organisms for experiments with shorter life cycel Provide a reason. |
| Answer» A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms with short life cycles for experiments because it enables the geneticist to study many generations of those organisn1s in their short life spans | |
| 858. |
Certain species wasps are seen to frequently visit flowering fig trees . What type of interaction is seen between then and why ? |
| Answer» The interaction is Mutualism . The female wasps use the fruit not noly as an oviposition (egg-laying) site but use the developing seeds within the fruit for nourishing their larvae . In return the wasp acts as a pollinator for the fig plant. | |
| 859. |
State what is apomixis. Write its significance. How can it be commercially used ? |
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Answer» Apomixis is a form of asexual reproduction in which embryo is formed without meiosis and fusion of haploid gametes. It can occur through diploid saprophytic cells of ovule or from an unfertilised egg itself . Significance - Certain species of plants belong to Asteraceae and grasses exclussively use the phenomenon of apomixes for producing seeds. Commercial applications of apomixes 1. By apomixis, hybrid varieties of seeds can be produced, which will provide higher and better yield 2. Apomixis prevents the loss of specific characteristics in the.hybrid plants. 3.Apomixis is a cost-effective method of producing seeds. |
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| 860. |
How are the following formed and involved in DNA packaging in a nucleus of a cell? (i) Histone octomer (ii) Nucleosome (iii) chromatin |
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Answer» (i) Histone octamer : Histones are positively charged basic proteins Histones are rich in basic amino acids lysine and orginine both of which carry positive charge in their side chains , the Histone proteins are arranged to form a unit of eight molecules called the histone octamer. (ii) Nucleosome : the negatively charged DNA molecule is wrapped around a positively charged histone octamer to form a nuclesome A usual nuclesome contains 200 bp of DNA holix. (iii) chromatin : Nucleosome consitutes the repeating unit found in the necledelous called chromatin , chromation appers as thread- like stained bodies bodies and nucleosomes in chromatin appea as a bead -on- string structure when viewed under the electroin microsccopre. |
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| 861. |
Given below is a part of the template strand of a structural gene `bar("TAC CAT TAG GAT")` (a) Write its transcribed mRNA strand with its polarity. (b) Explain the mechanism involved in initiation of the transcription of this strand. |
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Answer» Transcription of the DNAtemplate strand produces and RNAtranscript with the same sequence as the non-template or coding strand of the DNA. However, there would be uracil opposite to adenine inplace of thymine. The sequence of mRNA-51- AUG GUAAUC CUA-3’ (b) Template strand pf DNA acts as a template for the synthesis of mRNA during transcription. It runs from 3’ to 5’ where as, the coding strand is a sequence of DNA that has the same base sequence as that of mRNA. It runs from 5’ to 3’., (c) The process of transcription is initiated with the help of enzyme called DNA dependant RNA polymerase in only one direction, which is 5’ 3’ direction there by initiating the process of synthesis of RNA called transcription. |
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| 862. |
Study the given pedigree chart and answer the queation that follow. (a) In the trait recessive or dominant ? (b) Is the trait sex-linked or autosomal ? (c) Give the genotypes of the parants in generation I and of their third and fourth child in generation. |
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Answer» (a) the trait is recessive. (b) the trait is Autosomal recessive trait. (c) This is an autosome linked recessive trait that can be transmitted from parents to the of offspring when both the partners are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele, `Hb^(A) and Hb^(S)` . Out of the three possible genotypes only homozygous individuals for `Hb^(S) (Hb^(S) Hb^(S))` show the diseased phenotype Heterozygous (Hb Hbs) individuals appear unaffected but they are carrier of the disease as there is 50% probability of transmission of the mutant gene to the progeny. Genotype I generation - Aa and Aa Genotype of generation II of third child is aa and fourth child is Aa. |
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| 863. |
List the charges that occur when an ovule matures into seed. |
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Answer» `**" "`Integuments of ovules harden as tough protective seed coats. `**" "`The micropyle remains as a small pore in the seed coat. `**" "` As the seed matures, its water content is reduced. `**" "`Seeds become relatively dry (10-15 per cent mosisture by mass). `**" "`The general metabolic activity of the embryo slows down. `**" "`The embryo may enter a state of inactivity called dormancy. |
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| 864. |
a) Why does endosperm development precede development in angiosperm seeds? State the role of endosperm in mature albuminous seeds. b) Describe with the help of three labelled diagrams the different embryonic stages that include mature embryo of dicot plants. |
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Answer» a) The development of endopserm preceds that the embryo in angiosperm seeds, because the endosperm cells provide nutrition to the develoment embryo. In mature albuminous seeds, endosperm exists as strong as strong tissue. It stores starch and fat. b) Development of embryo: The enzymes starts dividing and gives rise to proembryo. This proembryo further divides forming a globular, heart-shaped and mature embryo. Following are the steps that occur during the development of embryo: 1) The embryo develops at the microplyar end of the embryo sac where the zygote is situated. 1) The embryo deveops at the micropylar end of the embryo sac where the zygote is sitauated. 2) First, the zygote give rise to proembryo and, then, to the globular, heart-shaped, mature embryo. 3) A typical dicot embryo consists of an embryonal axis and two cotyledons. 4) The portion of the embryonal axis above the level of cotyledons is called epicotyl. It contains the plumule (shoot tip). The portioin below the axis is called hypocotyl. It contains the radicle (root tip). The root tip is covered by the root cap. |
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| 865. |
A youth in his twenties met with an accident and succumbed to the injuries. His parents agreed to donate his organs. List any two essential clinical steps to be undertaken before any organ transplant. Why is the transplant rejected sometimes ? What views would you share with your health club members to promote organ donation ? |
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Answer» 1. Blood group matching of donor. 2. Tissue Matching. 2. The grafts are rejected some times because of the cell mediated immunity involving the T-lymphocytes. The T-lymphocytes are responsible for the body to differentiate self and non-self leading to graft rejection. The views to be shared with health club members regarding organ donation are : (i) Organ donation is a healthy practice and is life saving for those in need. (ii) With the availability of immune suppressants like cyclosporine A the acceptability of donor organs in recipients has increased. |
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| 866. |
Name of the cells and the event they undergo to produce pollen grains. |
| Answer» See Q.28, Set-I, Outside Delhi-2010. | |
| 867. |
Comment upon the mode of pollination in Vallisneria and Eichhornia which have emergent flowers. |
| Answer» In Eichhornia the flowers emerge above the level of water and are pollinated by insects or wind. In Vallisneria, the female flower reches the surface of water by the long stalk and the male flowers or pollen grains are released on to the surface of water. They are carried passively by water current, some of them eventually reach the female flowers and the stigma. | |
| 868. |
The energy levels of an atom are as shown in figure . Which one of those transition will result in the emission of a photon of wavelength `275 nm`? |
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Answer» (a) For element A Ground state energy, `E_(1) = -2eV` Excited state energy, `E_(2) = 0 eV` Energy of photon emitted, `E = E_(2) - E_(1)` `= 0 - (-2) = 2 eV` `:.` Wavelength of photon emitted, `lamda = (hc)/(E) = (6.626 xx 10^(-34) xx 3xx 10^(8))/(2 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(3.2) = 6.211 xx 10^(-7) m = 621 : 1nm` For element B `E_(1) =4.5 eV, E_(2) = 0eV` `E = 0 - (-4.5) = 4.5 eV` `:. lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(4.5 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(7.2) = 2.760 xx 10^(-7) = 276 nm` For element C `E_(1) = -4.5 eV, E_(2) = -2 eV` `E = -2- (-4.5) = -2 + 4.5 = 2.5 eV` `:. " " lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(2.5 xx 1.6 xx 10^(-19))` `= (19.878 xx 10^(-7))/(4) = 4.969 xx 10^(-7) m = 496.9 nm` For element D `E_(1) = -10 eV, E_(2) = -2eV` `E = -2-(-10) = 8 eV` `:. lamda (6.626 xx 10^(-34) xx 3 xx 10^(8))/(8xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(1.28) = 1.552 xx 10^(-7) m = 155.2 nm` `:. Element B has a photon of wavelength 275 nm (b) Element A has radiation of maximum wavelength 621 nm |
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| 869. |
Three identical capacitors `C_(1) , C_(2)` and `C_(3)` of capacitance `6 mu F` each are connected to a `12 V` battery as shown in Fig. Find (i) charge on each capcitor (ii) equivalent capacitance of the network. (iii) energy stored in the network of capacitors. |
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Answer» (i) Here, `V = 12V` and `C_(1) = C_(2) = C_(3) = 6mu F = 6 xx 10^(-6)F` Charge on capacitor `C_(3)` is `q_(3) = C_(3)V = 6 xx 10^(-6) xx 12 = 72 xx 10^(-6) = 72 mu C` Since, capacitor `C_(1) and C_(3)` are in series `:.` Equivalent capacitance `(1)/(C_(S)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(C_(S)) = (1)/(6) + (1)/(6) = (2)/(6) = (1)/(3)` `:. C_(S) = 3 mu F` Charge on capacitor `C_(1) and C_(2)` is `q = C_(S) V = 3 xx 10^(-6) xx 12` `= 36 xx 10^(-6) = 36 mu C` `:.` Charge on each capacitor `C_(1) and C_(2) " is " 36 mu C` (ii) Since, `C_(1) and C_(2)` is in series `:.` Equivalent capacitance `C_(S) = 3mu F` Now `C_(3) and C_(S)` are in parallel `:.` Equivalent capacitance `C = C_(3) + C_(S)` `= 6 + 3 = 9mu F` (iii) Energy stored `= (1)/(2) CV^(2) = (1)/(2) xx 9 xx 10^(-6) xx (12)^(2) = (1)/(2) xx 9 xx 10^(-6) xx 144` `= 648 xx 10^(-6) = 6.48 xx 10^(-4) J` |
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| 870. |
Calculate the shortest wavelength of light emitted in the Paschen series of hydrogen spectrum. Which part of the electromagnetic spectrum, does it belong ? ( Given : Rydberg constant , R = `1.1 xx 10^(7)m^(-1)` |
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Answer» The wave number of the spectral lines forming Paschen series is given by `vec(v)=1/lambda=R(1/3^(2)-1/n_(i)^(2)).n_(i)=4,5,6,...` The shortest wavelength of the spectral lines (series limit) of Paschen series is given by `1/(lambda Pa_(min)) =R(1/3^(2)-1/(oo^(2)))= R//9 rArr lambda Pa_min=9/R` `9//_1.1xx 10^(7)=8199A^(@)` The series lies in infrared region of em spectrum. |
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| 871. |
When are two object just resolved ? Explain . How can the resolving power of a compound microscope be increased ? Use relevant formula to support your answer . |
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Answer» When two objects placed nearby and their separate and distinct clean image is formed by an instrument the objects are resolved. Here, central maxima of diffraction pattern of one lies on `1^(st)` secondary minima of the other. Resolving power of a microscope `= 1/d = 2mu (sin theta)/(lambda)`. Increasing `mu` and `theta` increases resolving power and decreasing `lambda`. |
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| 872. |
In the study of a photoelectric effect the graph between the stopping potential V and frequency `nu` of the incident radiation on two different metals P and Q is shown in fig. (i) which one of two metals have higher threshold frequency (ii) Determine the work function of the metal which has greater value (iii) Find the maximum kinetic energy of electron emitted by light of frequency `8xx10^(14) Hz` for this metal. |
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Answer» (i) Threshold frequency of metal Q is higher than P. (ii) Also from, `W = hv_(0)` work function of Q is `6 xx 10^(14) xx 6.63 xx 10^(-34)` `W = 3.9 xx 10^(-19)` unit `W_(Q) gt W_(P)` (iii) `E_("max") = hv - varphi_(0) = hv - hv_(0) = h(v-v_(0))` `= 6.63 (8 xx 10^(14) - 6 xx 10^(14)) J xx 10^(-34) J= 1.33xx 10^(-19) J`. |
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| 873. |
Two identical capacitors of `12 pF` each are connected in a series across a battery of 50V. How much electrostatic energy is stored in the combination ? If these were connected in parallel across the same battery , how much energy will be stored in combintion now ? Also find the charge drawn from the battery in each case. |
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Answer» In series combination . `1/(C_(eq)) = (1/12+1/12) Pf rArr C_(eq) = 6xx 10^(-12)F` Electrostatic energy stored is `1/2 C_(eq)V^(2) = 1/2 xx 6 xx 10^(-12) xx 50^(2) = 7.5 xx 10^(-9) J` Charge drawn from the battery `=Q = 6 xx 10^(-12) xx 50 = 3 xx 10^(-9) J` In parallel `C_(eq) = C_(1) xx C_(2) = 12 + 12 = 24 pF` Energy stored `= 1/2C_(eq) V^(2) = 1/2 xx 24 xx 10^(-12) xx 50^(2) = 3 xx 10^(-8) J` Charge drawned from battery `Q = C_(eq) V = 24 xx 10^(-12) xx 50 = 1.2 xx 10^(-9) C`. |
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| 874. |
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ? If another capacitor of 6 pF is connected across the combination, find the charge stored and potential difference across each capacitor. |
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Answer» Energy stored = `1/2CV^(2) = 1/2 12 xx 10^(-12) xx 50^(2) J` `= 1.5 xx 10^(-8) J` If, `1/(C_(eq)) = (1/12+ 1/6) pF` `1/(C_(eq)) = 1/4 pF , C_(eq) = 4 xx 10^(-12)F` `Q = C_(eq) V = 4 xx 10^(-12)C xx 50= 200 xx 10^(-12)C` Voltage across 12 pF capacitor `= Q/C = (200 xx 10^(-12))/(12 xx 10^(-12)) =16.67V` Voltage across `16pF` capacitor `= (200 xx 10^(-12))/(16 xx 10^(-12)) = 33.33 V`. |
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| 875. |
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vaccum. |
| Answer» Velocity/Speed of light remains the same. | |
| 876. |
An electric lamp having coil of negligible inductance connected in series with a capacitor and an a.c. source is glowing with certain brightness, Fig. How does the brightness of the lamp change on reducing (i) capacitance (ii) frequency |
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Answer» When AC source is connected , the capacitor offers capacitative reactance ` X_C-underset(omegaC)(1)=underset(2pinuC)(1)`. The current flows in the circuit and the lamp glows. (i) On reducing capacitance, `C,X_C` increases. Therefore, the brightness of the bulb reduces. (ii) On reducing frequency `v,X_C` increases. Therefore, the brightness of the bulb reduces. |
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| 877. |
The radius of curvature of the faces of a double convex lens are `10 cm and 15 cm`. If focal length of lens of lens is `12 cm`, find the refractive index of the material of the lens. |
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Answer» Hence, `R_1=10 cm, R_2= -15 cm , f = 12 cm , mu = ? ` Using lens formula, we have ` (1)/( f) =(mu-1) ((1)/(R_1)-(1)/(R_2))rArr (1)/(12) =(mu-1)((1)/(10)-(1)/(15))` ` rArr (1)/(12) = (mu-1) ((1)/(10)+(10)/(15))` ` rArr (1)/(12)=(mu-1)((3+2)/(3))rArr (1)/(12)=(mu-1)((5)/(3))` `rArr (mu-1)=(1)/(12)xx(30)/(5)rArr mu-1=0.5rArr mu=1+0.5therefore mu=1.5`. |
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| 878. |
write the following radiations in ascending order in respect of their frequencies: X-rays, microwaves, UV rays and radio waves. |
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Answer» The ascending order of the frequencies of the radiation are: Radio waves, microwaves, UV-rays and X-rays |
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| 879. |
Arrange the following electro-magnetic radiations in ascending order of their frequencies : (i) Microwave (ii) Radiowave (iii) X-rays (iv) Gamma rays. Write two uses of any one of these. |
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Answer» Given electromagnetic radiations in ascending order of their frequencies are Radiowave, Microwave , X-rays, Gamma rays. Uses of Microwave : (i) Microwaves are used in Radar system for aircraft navigation. (ii) Microwaves are used in microwave ovens for cooking purpose. |
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| 880. |
How do organisms manage with stressful conditions existing in their habitat for short duration? Explain with the help of one example each. |
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Answer» If the stressful conditions remains for short duration the organisms manage either by-1. Migration 2. Suspend. (I) Migration-The temporary movement of oi-ganism from the stressful habitat to a mom hpspitable area and return when stressful period is over is called migration. For example-Migratory birds from Siberia come to Keolado National Park (Bharatpur) every winter. (II) Suspend-Thoge animals who fail to migrate might avoid the stress by escaping in time Hibernation of bears during winter or aestivation of snails and iish to avoid in summer related prOblemsheat and dessication are examPles of this phenomenon. |
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| 881. |
How do organisms cope with stressful external environmental conditions whichare localised or of short duration ? |
| Answer» The following methods are employed by organisms to cope with stressful conditions: (i) Migrate temporarily from the stressful habitat to a hospitable area, (ii) Suspend activities, (iii) form thick walled spores, (iv) form dormant seeds (v) hibernate during winter, (vi) aestivate during summer, (vii) Planktons undergo diapause. | |
| 882. |
(a) Draw a diagram of the adult human female reproductive system and label the different : (i) parts of fallopian tube (ii) layers of uterus wall (b) Explain the events during fertilization of an ovum in humans. |
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Answer» (a) diagram of female reproductive system. (b) The process of fusion of a sperm with an ovum is called fertilization. During fertilization, a sperm comes in contant with zona pellucids layer of the ovum and induces changes in the membrane that block the entry of additional sperms. The secrections of the acrosome help the sperm enter into hte cytoplasm of the ovum through the zona pellucida and the plasma membrane . This induces the completion of hte meiotic division of the secondary socyte. The secondary meiotic division is also unequal and results in the formation of a second polar body and haploid ovum. the haploid nucleus of the ovum fuse together to form a diploid zygote. |
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| 883. |
(a) The graph given below represents the organisms response to temperature as an environmental condition. (i) Which one of the two lines represents conformers and why ? (ii) What does the other line in the graph represent and why ? (b) Mention the different adaptations the parasites have evolved with, to be able to successfully complete their life cycles in their hosts. |
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Answer» (i) Conformers : Aquatic animals and plants in which the osmotic concentration of body fluids changes according to the ambient conditions of water are called conformers. (ii) Regulators : Some organisms are able to maintain homeostasis by physiological means which ensures constant body temperature, constant osmotic concentration etc. |
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| 884. |
Suggest and explain assisted reproductive techniques which will help a couple to have children, where the female had a blockage in the fallopian tube and the male partner had a low sperm count. |
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Answer» With special techniques called assisted repoductive technologies, infertile could be assisted to have children. ART includes : (i) In-Vitro Fertilisation (IVF) : It is transfer of an ovum collected from a donor into the fallopian tube of another female for fertilisation and further development. (ii) Zygote Intrafallopian Transfer (ZIFT) : Ova from wife/donor (female) and sperms from husband/donor (male) are collected and are induced to form zygote under simulated conditions in laboratory. The zygote with upto 8 blastomeres is then transferred into the fallopian tube. (iii) Intra Cytoplasmic Sperm Injection (ICSI) : Here the sperm is directly injected into the ovum. |
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| 885. |
(a) Draw a diagram of the adult human female reproductive system and label the different : (i) parts of fallopian tube lt,brgt (ii) layers of uterus wall (b) Explain the events during fertilization of an ovum in humans. |
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Answer» (a) diagram of female reproductive system. (b) The process of fusion of a sperm with an ovum is called fertilization. During fertilization, a sperm comes in contant with zona pellucids layer of the ovum and induces changes in the membrane that block the entry of additional sperms. The secrections of the acrosome help the sperm enter into hte cytoplasm of the ovum through the zona pellucida and the plasma membrane . This induces the completion of hte meiotic division of the secondary socyte. The secondary meiotic division is also unequal and results in the formation of a second polar body and haploid ovum. the haploid nucleus of the ovum fuse together to form a diploid zygote. |
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| 886. |
(a) Where do the signals for parturition originate from in humans? (b) Why is it important to feed the newborn babies on colostrum? |
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Answer» (a) Parturition signals originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex. (b) The milk that comes out of the mammary glands during initial days of lactation is called colostrum. It contains several antibodies and nutrients for the baby. |
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| 887. |
Both nucellus and endosperm have abundant reserve food materials. How is their food reservoir utilised in angiosperms ? |
| Answer» In angiosperms, the seed is the final product of sexual reproduction. The food reservoir utilised in angiosperms through integuments of ovules. The micropyle remains as a small pore in the seed coat. This facilitates entry of oxygen and water into the seed. | |
| 888. |
Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium. (Atomic mass = 93u). |
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Answer» `r = (sqrt(3))/(4)a` `143.1 = (sqrt(3))/(4)a` a = 330.4 pm `rho = (zM)/(a^(3)N_(A))` `rho = (2 xx 93)/((330.4 xx 10^(-10)) xx 6.023 xx 10^(23))` `p = 8.58 g//cm^(3)`. |
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| 889. |
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymeri -zation reaction ? |
| Answer» The condensation polymerization reaction takes place between two different bi -functional or tri - functional monometric units. | |
| 890. |
(a) How will you bring about the following conversions : (i) Ethanol to 3- hydroxybutanal (ii) Benzaldehyde to Benzophenone (b) An organic compound A has the molecular formula `C_(8)H_(16)O_(2)`. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B . C on dehydration reaction gives but - l - ene . write equations for the reactions involved. |
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Answer» (a) (i) Ethanol to 3- hydroxybutanal `underset("Ethanal 2molecules")(CH_(3)CHO+H-CH_(2)-CHO)overset("Dil NaOH")underset(("Aldol condensation"))tounderset("3- hydroxybutanal")(CH_(3)-overset(OH)overset(|)CH-CH_(2)-CHO)` (ii) `underset("Benzaldehyde")(C_(6)H_(5)CHO)overset(K_(2)Cr_(2)O_(7))underset(H_(2)SO_(4))toC_(6)H_(5)COOHoverset(CaCO_(3))tounderset(("Calcium Benzoate"))((C_(6)H_(5)COO)_(2)Ca)overset("Dry")underset("distill")toC_(6)H_(5)CO-C_(6)H_(6)` (b) `underset(A(C_(8)H_(16)O_(2)))underset(("Butyl butanoate"))(CH_(3)CH_(2)CH_(2)overset(O)overset(||)COCH_(2)CH_(2)CH_(2)CH_(3))overset("dil" H_(2)SO_(4))tounderset((B))underset(("Butanoic acid"))(CH_(3)CH_(2)CH_(2)COOH)+underset((C ))underset("Butan -1-al")(CH_(3)CH_(2)CH_(2)OH)` |
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| 891. |
What is the effect of catalyst on: (i) Gibbs energy `(DeltaG)` and (ii) activation energy of a rection ? |
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Answer» (i) A caralyst does not alter gibbs energy `(DeltaG)` of a reaction. (ii) A catalyst decreases the activation energy of a reaction |
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| 892. |
What is the geometrical shape of equipotential surfaces due to a single isolated charge ? |
| Answer» For an isolated charge the equipotential surfaces are co-centric spherical shells and the distance between the shells increases with the decreases in electric field. | |
| 893. |
Explain briefly the following terms used in communication system : (i) Transducer (ii) Repeater (iii) Amplification |
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Answer» (i) Transducer : Any device that converts one form of energy into another can be termed as a transducer. An electrical transducer may be defined as a device that converts some physical variable in the electrical signals. (ii) Repeater : A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receivers sometimes with a change in carrier frequency. (iii) Amplification : It is the process of increasing the amplitude of a signal using an electronic circuit called the amplifier. Amplification is done at a place between the source and the destination. |
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| 894. |
`15.0 g` of an unknown molecular material was dissolved in `450 g` of water. The resulting solution was found to freeze at `-0.34 ^(@)C`. What is the the molar mass of this material. (`K_(f)` for water `= 1.86 K kg mol^(-1)`) |
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Answer» Here ` w^(2) = 15 g, w_(1) = 450 g, K_(f) = 1.86" K kg "mol^(-1)` ` M_(2) `=? `DeltaT_(f) = 0 - (-0.34) = 0.34^(@)C or 0.34" K "` Using the formula, `Delta T_(f) = K_(f) m = K_(f) xx (w_(2) xx 1000)/ (M_(2) xx w_(1))` ` 0.34 = (1.86 xx 15 xx 1000)/(M_(2) xx 450)` ` :. " " M_(2) = (1.86 xx 15xx 1000)/(0.34 xx 450) = (186 xx 100)/(34 xx 3)` ` = (18600)/102 = 182.35" g "mol^(-1)` |
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| 895. |
(a) Define the following terms : (i) Mole fraction (ii) Ideal solution (b) `15.0`g of an unknown molecular material is dissolved in 450g of water . The resulting solution freezes at `-0.34^(@)C` . What is the molar mass of the material ? `(K_(f)` for water = `1.86` K kg `mol^(-1))` |
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Answer» (b) `Delta`mix H = 0 `Delta`mix V = 0 `DeltaT_(f) = K_(f)X` molality `DeltaT_(f) = (1.86 xx 15.0 xx 1000)/(M xx 450)` `M = (1.86 xx 150 xx 1000)/(0.34 xx 450)` `M = 182.3` gm `mol^(-1)`. |
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| 896. |
State the reason for the following : (i) Cu (I) ion is not stable in an aqueous solution. (ii) Unlike `Cr^(3+),Mn^(2+),Fe^(3+)` and the subsequent other `M^(2+)` ions of the 3d series, the 4d and 5d series metals generally do not form stable oxidation states. |
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Answer» (ii) `Cr^(2+)` is stronger reducing agent than `Fe^(2+)` Reason : `d^(4) to d^(3)` occurs in case of `Cr^(2+) ` to `Cr^(3+)` But `d^(6) to d^(5)` occurs in case of `Fe^(2+)` to `Fe^(3+)` In a medium (like water) `d^(3)` is more stable as compared to `d^(5)`. |
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| 897. |
`15.0 g` of an unknown molecular material was dissolved in `450 g` of water. The reusulting solution was found to freeze at `-0.34 .^(@)C`. What is the the molar mass of this material. (`K_(f)` for water `= 1.86 K kg mol^(-1)`) |
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Answer» `W_(A) = 15.0 gm` `W_(B) = 450 gm` `KF_(2) = 1.86 K kg//"mol"` For water, ` DeltaT_(f) = K_(f) xx "molality"` `0 - (0.34) = 1.86 K kg mol^(-1) xx (15 g xx 1000)/(M xx 450)` `M = (1.86 K kg mol^(-1) xx 15 gm xx 1000)/(0.34 K xx 450 kg) = 182.35 gm//mol`. |
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| 898. |
Of `PH_(3)` and `H_(2)S` which is more acidic and why ? |
| Answer» `H_(2)S` is acidic because S is the more electronegative than P. This results in the S-H bond oecoming more polar than a p-H bond, and thus easier to remove. | |
| 899. |
(a) Given chemical tests to distinguish between the following : (i) Benzoic and ethyl benzoate. (ii) Benzaldehy and acetophanone. (b) Complete each synthesis by giving missing regents or products in following. (i) (ii) `C_(6)H_(3)CHO overset(H_(2)NCONHNH_(2)) to ` (iii) |
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Answer» (a) (i) Benzoic acid and ethyl benzoate: These two componunds can be distinguished by the following tests: `to NaHCO_(3)` test: Benzoic acid being an acids prouduces brisk effervescene with `NaHCO_(3)` solution while ethylbenzote does not. `underset("Benzoic acid")(C_(6)H_(5)COOC_(2)H_(5))+ NaHCO_(3) to underset("Sod. formate")(C_(6)H_(5)COONa + CO_(2) uarr H_(2))` `to `Indoform test |
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| 900. |
plants that inhabit a rain - forest are not found in a wetland ,Explain . |
| Answer» plants in rain forest are adapted to high density of rainfall and sunlight as against those fond in wetlannds which are which are suited to waterlogged and saline conditions . | |