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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.
| 801. |
Mention the role of cyanobacteria as a hiofertiliser. |
| Answer» They fix atmospheric nitrogen and increase the organic matter of the soil through their photosynthetic activity, e.g. Nostoc, Anabaena, oscillatoria etc. | |
| 802. |
Comment of the similarity between the wing of a cockroach and the wing oi a bird. What do you infer from the above, with reference to evolution? |
| Answer» Analogous organs (convergent evolution) Analogues-organs as they have dissimilar structural design but perform same function. | |
| 803. |
a) While cloning vectors, which of the two will be preferred by biotechnologists bacteriophases or plasmids, Justify with reason. b) Name the first transgenic cow developed and state the improvement in the quality of the product produced by it. |
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Answer» (a) Biotechnologist will prefer bacteriophage as the preferred cloning vector. Reason : Bacteriophage have very high copy numbers of their genome within the bacterial cell whereas, plasmids have variable copy number per cell, some may have only one or two copies per cell and others may have 15-100 copies per cell. (b) Rosie is the first transgenic cow developed which produces human protein entriched milk (human `alpha`-albumin 2.4 g/L). |
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| 804. |
(a) Name the causative agent of typhoid in humans. (b) Name the test administered to confirm the disease. (c) How does the pathogen gain entry into the human body ? Write the diagnostic symptoms and mention the body organ that gets affected in severe cases. |
| Answer» Salmonella typhi is a pathogenic bacterium which causes typhoid fever in human beings. (b) Typhoid fever can be confirmed by widel test. (c) These Pathogens generally enter the small intestine through food and water contaminated with them and migrate to other organs through blood. Symptoms : Sustained high fever (`39 ^@ a 40^@`C) weakness, stomach pain, constipation headache and loss of appetite are some of the common symptoms of this disease. Intestinal perforation and death may occur in severe cases. In severe cases the disease affects the Alimentary canal and brain. | |
| 805. |
In the meter bridge experiment, balance point was observed at J with AJ=I. (i) The values of R and X were doubled and then interchanged. What would be the new position of balance point (ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected? |
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Answer» (i) The value of R and X were doubled and then interchanged. Hence, the new position of balance point will become 100-l. (ii) From the principle of Wheatstones Bridge. `(R)/(X)=(l)/(100-l)X=R((100-l)/(l))` Hence the galvonometer and cell are interchanged, the condition for a balance bridge is still satified. Therefore, the galvanometer will not show any deflection. |
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| 806. |
In the given block diagram of a receiver, Figure, identify the boxes labelled as X and Y and write their functions. |
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Answer» X `rarr" IF stage "rarr` Intermediate frequency stage to change the carrier frequency to lower frequency. `Yrarr" Amplifier "rarr` It amplifies the signal i.e. it increases the amplitude of the detected signal to compensate the attenuation of the signal. The detected signal may not be strong enaough to use. |
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| 807. |
An alternating voltage given by `E = 280 sin 50 pi t` is connected across a pure resistor of 40 ohm. Find frequency of source and rms current through the bulb. |
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Answer» (i) `V_(0)=280 V,` `omega = 50 pi rArr 2pi v = 50 pi rArr v=25 Hz` (ii) `I_(rms)=(V_(rms))/(R)=(V_(0)//sqrt(2))/(R)=(280)/(sqrt(2)xx40)=3.5xx1.414=4.95 A.` |
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| 808. |
A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate (i) the resistance of the bulb, (ii) the rms current through the bulb. |
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Answer» (i) `P=(V^(2))/(R ) rArr R=(V^(2))/(P)=(220xx220)/(10)=484 Omega.` (ii) `I_(rms)=(V_(rms))/(R ) =(220)/(484)=0.45` ampere. |
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| 809. |
A light bulb is rated 200 W for 220 V supply at 50 Hz. Calculate resistance of the bulb and rms current through the bulb. |
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Answer» (i) Resistance of the bulb `R=(E^(2))/(P)=((220)^(2))/(200)=242 Omega.` (ii) Peak voltage `V_(0)=Vsqrt(2)=200 sqrt(2)=311 V.` rms current through the bulb, `I=(P)/(E )=(311)/(220)=0.450 A.` |
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| 810. |
Given reasons for the following : (i) High reverse voltage do not appear across a LED. (ii) Sunlight is not always required for the working of a solar cell. (iii) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5 V. |
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Answer» (i) Because of its heavy doping as reverse voltage is applied minority carriers increases near junction and current starts flowing. (ii) Because energy produced by solar cell can be stored in batteries for later use, when sunlight is not available. (iii) As the depletion width si very small `(lt10^(_7))` even a small voltage say 5V can set up a very high electric field as `E=(V)/(d)`. |
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| 811. |
In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? |
| Answer» The angular width of central diffraction band `=(2lamda)/(q)alpha(1)/(a),` i.e., when slit width is doubled the width of central band becomes half of the initial value. As intensity of central maximum `alpha a^(2),` if slit width is doubled, the intensity becomes four times. | |
| 812. |
How does the angular separation between frings in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled ? |
| Answer» The angular width of central diffraction band `=(2lamda)/(a)alpha=(1)/(a).` When slit width is doubled the width of central band becomes half of the initial value. As intensity of central maximum `alphaa^(2)`, if slit width is doubled, the intensity become four times. | |
| 813. |
The figure, drawn here, shows the geometry of path differences for diffraction by a single slit of width a. Given appropriate reasoning to explain why the intensity of light is (i) maximum of the central point C on the screen. (ii) (nearly) zero for point P on the screen when `theta=lamda//a`. Hence, write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. |
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Answer» At center of screen C all wavelets : (i) From slit meet in the same phase so they add constructively to produce a central maxima i.e., bright firnge. (ii) When `theta=lamda//a`, path difference from two halves of slit will be `lamda//2` i.e., wavelets from two halves reach at point P in opposite phase, giving a dark or minima. (iii) `beta_(0)=(2Dlamda)/(a)`. |
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| 814. |
The temperature coefficient of resistance for two material A and B are `0.0031^(@)C` and `0.0068^(@)C^(-1)` respectively .Two resistance `R_(1) and R_(2)` made from material A and B respectively . Have resistance of `200 Omega` and `100 Omega` at `0^(@)C`. Show as a diagram the colour cube of a carbon resistance that would have a resistance equal to the series combination of `r_(1) and R_(2)` at a temperature of `100^(@)C` (Neglect the ring corresponding to the tolerance of the carbon resistor) |
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Answer» `R_(1)("at "100^(@)C)=R_(1)(0^@C)[1+alpha.t]` `=200(1+.0031xx100)` `=200(1.31)=262Omega` `R_(2)("at "100^(@)C)=R_(2)(0^(@)C)[+alpha.t]` `=100.[1+.0068xx100]` `=100(1.68)=168Omega` `R_(1)+R_(2)=262+168=430Omega` Yellow Orange Brown. |
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| 815. |
A circualr coil, having 100 tures of wire, of readius (nearly) 20cm each, lies in the XY plane with its centre at the origin of co-ordinates. Find the magnetic field, at the point `(0,0,20sqrt3Cm)`, when this coil carries a current of `((2)/(pi))A`. |
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Answer» `B=(mu_(0))/(4pi)(2pinIa^(2))/((a^(2)+x^(2))^(2//3))` `=(10^(-7)xx2pixx100xx(2)/(pi)xx(20xx10^(-2))^(2))/[(400+1200)xx10^(-4)]^(3//2)=(10^(-7)xx200xx400xx10^(-4))/(64000xx10^(-6))` `=(8xx10^(-7))/(64xx10^(-3))=(1)/(8)xx10^(-4)=0.125xx10^(-4)=1.25xx10^(-5)T`. |
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| 816. |
a) Explain "birth rate" in a population by taking a suitable example. b) Write the other two characteristics which only a population shows but an individual cannot. |
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Answer» a) Birth rate is expressed as the number of births to existing individuals of a populations in given period of time. Eg. If in a pond of 20 lotus 5 new lotus plants are added taking the population by 25, the birth rate is `5/20=0.25` b) i) Sex ratio: An individual is either a male or a female but a population has sex ratio. Eg: 60 percent of the population are males and 40 percent are females. ii)Age Pyramid: A population at a given time is composed of individuals of different age groups. If the age distribution is plotted for the populations, the resulting structure is called age pyramid. |
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| 817. |
Name the host and the site where the following occur in the life cycle of a malarial parasite : (a) Formation of gametocytes. (b) Fusion of gametocytes. |
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Answer» (a) Human Host `to` RBC (b) Mosquito Host `to` Female Anopheles. |
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| 818. |
(a) Why is it that the father never passes on the gene for haemophilia to his sons? Explain. (b) State the functions of the following in a prokaryote : (i) tRNA, (ii) rRNA |
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Answer» (a)Haemophilia and colour blindness are x-chromosome linked recessive disorder. Males have only one copy of x-chromosome and one y-chromosome. The genes on these chromommee do not have corresponding alleles. Therefore, the presence of only single defective x-chromoeome will result in these disorders. Women can develop these disorders only when they have both the defective copies of x-chromosomes, which is rare. Since, the father passes on only the ychromosome to the son a son is not haemophiliac. (b) (i) (a) tRNA-It acts as an adaptor molecule that picks up a particular amino acid from cellular pool and takes the same over to A site of mRNA for incorporation in the polypeptide chain. (b) It recognises the codon on mRNA by its anticodon. (ii) (a) rRNA-It constitutes the ribosomal structure. (b) It helps to form peptide bond. |
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| 819. |
Explain how do the following act as contraceptives : (a) CuT (b) “Saheli” |
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Answer» (a) Cu-ions released suppress sperm motility, lowers the fertilising capacity of sperms. . (b) Inhibit ovulation, implantation, as well as alter the quality of cervical mucus to prevent or retard the entry of sperm. |
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| 820. |
How does the study of different parts of a flower help in identifying wind as its pollinating agent? |
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Answer» 1. When pollination is effected by wind, it is called anemophily. 2. Pollen grains are light in weight, non-sticky, dry and winged. 3. Well-exposed stamens for easy dispersal of pollen grains in the wind. 4. The stigma is sticky, large, have feather to trap pollen grains floating in the air. 5. Numerous flowers are packed together to form inflorescene. |
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| 821. |
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers. |
| Answer» The genotype of a pea plant bearing violet flowers is cross. | |
| 822. |
Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel. |
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Answer» 1. Flower colour is purple of white. 2. Flower position is axil or terminal. 3. Stem height is long or short. 4. Seed Shape is round of winked. |
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| 823. |
ON adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with `Cu^(2+)` ions. Identify the gas. |
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Answer» A colourless gas with pungent odour is ammonia. The following reactions are carried out. `(NH_(4))_(2)SO_(4)+2NaOH overset(Delta)rarr 2NH_(3)+2H_(2)O+Na_(2)SO_(4)` `underset((aq))(4NH_(3))+underset((aq))(Cu^(2+))rarr underset("Blue coloured complex")([Cu(NH_(3))_(4)]^(2+))` |
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| 824. |
State one reason for adding blue-green algae to the agricultural soil. |
| Answer» Blue green algae fix atmospheric nitrogen into organic forms and act as biofertilesers. Therefore, these are added to agricultural soil. Eg : Anabaena, Nostor. etc. | |
| 825. |
Calculate the emf of the following cell at `298K`: Fe(s) abs(Fe^(2+)(0.001M))abs(H^+(1M))H_2(g)(1"bar"),Pt(s) ("Give`E_("Cell")^@=+0.44V)` |
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Answer» `"At anode":FetoFe^(2+)+2e^(-)" Atcathode ",2H^(+)+2e^(-)toH_2` So, total number of electrons (n) transferred = 2 `"Given: " E_("Cell")^@=+0.44" Volt"` `" Temperature " (T) =298K` `E_("Cell")=E_("Cell")^@-((2.303RT)/(nF))log.(a_("oxi"))/(a_("red"))` `=E_("Cell")^@-((0.05916V)/n)log. (a_"oxidation")/a_("reduction")` `0.44-(0.0591V)/2log. (0.001)/1` `" Therefore," E_("Cell")=0.44(-0.02955xx(-3))` `=0.44+0.08865=0.53"Volt"` |
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| 826. |
Name the blank spaces a, b, c and d in the table given below |
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Answer» a = Penicillium notatum b =Acetic acid c = mugus d = Saccharomyces cerevisiae. |
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| 827. |
Name the blank spaces a, b, c and d given in the following table |
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Answer» (a) Streptococcus (b)Yeast (fungus) (c) Cyclosporin-A (d) Clostridium butylicum |
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| 828. |
Name the blank spaces a, b, c and d from the table given below : |
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Answer» a = Lactobacillus b = Trichoderma Polysporum c = Fungus (Yeast) d = Penicillin |
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| 829. |
Name the blank spaces a, b, c and (1 given in the following table : |
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Answer» (a) Lactobacillus. (b)Trichodenna Polysporum (c)Yeast(fungus) (d)Penicillim |
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| 830. |
Suggest any two techniques which can help in early detection of becterial`//`viral infections much before the symptoms appear in the body. |
| Answer» The symptoms of the disease can be detected by amplifications of Nucleic acid by PCR and using Autoradiography. | |
| 831. |
Write the location and functions of the following in human testes : (a) Sertoli cells (b) Leydig cells |
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Answer» (a) Sertoid cells : Each seminiferous tubule is lined on its inside by two types of cells called male germ cells and sertoli cells provide nutrition to the germ cells. (b) Leydig cells : The regions outside the seminiferous tubules called interstitial spaces, contain small blood vessels and interstitial cells of Leydig cells. Leydig cells synthesise and secrete testicular hormones called Androgens. |
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| 832. |
State the difference between the structural genes in a Transcription Unit of Prokarytes and Eukaryotes. |
| Answer» In Eukaryotes, the structural gene is monocistronic and split with exons (the coding sequence) and introns (the intervening sequnces that do not appear in mature RNA). On the other hand the Structural gene is polycistronic in prokaryotes. | |
| 833. |
A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in Fig. (i) Calculate the potential difference between A and C (ii) At what point [of A and C] is the electric potential more and why? |
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Answer» P.D. does not depend upon the path alongwith the test charge q moves `therefore" "E=(-dv)/(dr)=-((V_(C)-V_(A))/(d))-(V_(A)-V_(C))/(d)` `d_(AC)=4" So, "V_(A)-V_(C)-Exx4-4^^(E )` At point C, electric potential will be more since as potential decreases in the direction of electric field `therefore" "V_(A)-E_(0)=E_(0)xxd.` |
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| 834. |
State the significance of biochemical similarities amongst diverse organism in evolution. |
| Answer» Biochemical-similarities point to the same shared ancestry as structural similarities among diverse organisms. | |
| 835. |
in certain seasons we sweet profusely while in some other season we shiver . |
| Answer» we maintain a constant body temperature of `37^(@)C` in summer when the outside temperature is more than our body temperature we sweet to bring down the body temperature in winters when the outside temperature is lower than body temperature of `37^(@)`C we shiver to produce heat and raise the body temperature this is done to regulate the body temperature . | |
| 836. |
Write the names of two semi-dwarf and high yielding rice varieties developed in India after 1966. |
| Answer» Correct Answer - Jaya/Ratna. | |
| 837. |
Mention the difference between spermiogenesis and spermiation. |
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Answer» Spermiogenesis-Transformation of spermatids into sperms. Spermiation-The process of release of spermatozoa from Sertoli cells into the cavity of sperminiferous tubules. |
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| 838. |
What is an interaction called when an orchid grows on a mango plant? |
| Answer» This interaction is called commensalism. | |
| 839. |
Coconut palm is monoecious while date palm is dioecious. Why are they called so? |
| Answer» In some flowering plants, both male and female flowers may be present on the same individual, (monoecious) or in separate indiviuals (dieocious). Some example of monoecious plant are cucurbits and coconuts and of dioecious plants are papaya and date palm. | |
| 840. |
How does the floral pattern on Mediterranean orchid Ophrys guarantee cross pollination ? |
| Answer» The orchid bears flowers which resembles the female wasp in colour, smell as well as appearance. The male pollinators mistake them as their female counterparts. In the process of their pseudocopulating, they pollinate the flower. | |
| 841. |
The composition of a sample of wustite is `Fe_(0.93)O_(1.00)` What percentage of iron is present in the form of `Fe(III)`? |
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Answer» `Fe_(0.93)O_(1.0)` is non - stoichiometric and a mixture of `Fe^(2+) and Fe^(3+)` ions. Let x atoms of `Fe^(3+)` ions are present in the compound. This means that x ` Fe^(3+)` ions have been replaced by `Fe^(2+)` ions. No of `Fe^(2+) ions = 0.93 - x` For electrical neutrality, Positive charge on the compound = Negative charge on the compound. `2(0.93 - x) + 3x = 2` ` 1.86 + x = 2` ` x = 0.14` `% "of " Fe^(3+)" ions present" = (0.14)/(0.93) xx 100 = 15.05 %` |
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| 842. |
(i) Can two equaipotential surfaces intersect each other ? Give reasons (ii) Two charges `-q and +q` are located at point A (0, 0 -a) and B(0, 0, +a) respectively. How much work is done in moving a test charge from point `P (7, 0, 0) " to " Q(-3, 0, 0)` ? |
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Answer» (i) No, if two equipotential surfaces intersect then at the point of intersection two tangents can be drawn and there will be two directions of electric field intensity which is not possible. (ii) Since both the points are in the equatorial line of the dipole and V = 0 at every point on it, work done will be zero. Also the force on any charge is perpendicular to the equatorial line, so work done is zero. |
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| 843. |
Two nuclei have mass number in the ratio 1:2. What is the ratio of their nuclear densities? |
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Answer» Since, the nuclear density is same for all nuclei. The ratio of their densities are `1 : 1` i.e., `rho_(1) : rho_(2) = 1 : 1` |
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| 844. |
An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g `cm^(-3)`. Calculate the number of atoms in 108 g of the element. |
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Answer» For f.c.c structure z =4 `a=300 "pm"=300x10^(-10)cm,d=10.8cm^(-3)` `d=(zxxM)/(a^(3)xxNA)` `M=(dxxa^(3)xxNA)/z=(10.8xx(300xx10^(-10)^(3)xx6.022xx10^(23)))/(4)=43.9 gmol^(-1)` `because` 43.9 g of element contains = `6.022xx10^(23)` atoms 108 g of the elements contains `=(6.022xx10^(23))/43.9xx108=14.81xx10^(23)` atoms |
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| 845. |
A 4% solution (w/w) of sucrose (M 342 g `mol^(-1)`) in water has a freezing point of 271.15K Calculate the freezing point of 5% glucose (M= 180 g `mol^(-1)`) in water. (Given: Freezing point of pure water 273.15 K) |
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Answer» Mass of sucrose `(W_(B))4` g Mass of water `(W_(A))=100-4= 96 g` `M_(B)=342` `DeltaT_(f)=273.15-271.5=2K` `K_(f)=(DeltaT_(f)xxM_(B)xxW_(A))/(W_(B)xx1000)` `K_(f)=(2xx342xx96)/(4xx1000)=16.416` Now, for solution of glcose Mass of glucose (WB)= 5g Mass of water (WA) =100 -5 =95 `M_(B)=180gmol^(-1)` `DeltaT_(f)=(K_(f)xxW_(B)xx1000)/(M_(B)xxW_(A))=(16.416xx5xx1000)/(180xx95)=4.8 K` `DeltaT_(f)=T_(f)^(@)-T_(f)=4.8=273.15-Tf` `T_(f)=273.15-4.8=268.35K` |
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| 846. |
(i) Name the method of refining to obtain silicon of high purity. (ii) What is the role of SiO_2 in the extraction of copper? (iii) What is the role of depressants in froth floatation process ? |
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Answer» (i) Zone refining to obtain silicon of high purity. Zone melting is a group of similar methods of purifying crystals, in which a narrow region of a crystal is molten, and this molten zone is moved along the crystal and it is based on the priniciple that the impurities are more soluble in molten state than in the sqlid state of metal. (ii) The sulphide ores of copper are heated in reverberatory furnace. If the ore contains iron, it is mixed with silica before heating. Iron oxide "slags of" as iron silicate and copper is produced in the form 9f copper matte which contains `Cu_(2)S` anmd FeS. `FeO+SiO_(2)rarrunderset("(Slag)")(FeSiO_(3))` It is utilised for manufacturing `H_(2)SO_(4)`. (iii) Sometimes, it is possible to separate two sulphide ores by adjusting proportion of oil to water or by using" depressants". It selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. |
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| 847. |
(a) Name the method of refining which is (i) used to obtain semiconductor of high purity (ii) used to obtain low boiling metal. (b) Write chemical reactions taking place in the extraction of copper from `Cu_(2)S`. |
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Answer» (i) Zone refining. (ii) Distillation. (b) It is first heated in air to get copper oxide. `2Cu_(2)S+3O_(2)to2Cu_(2)O+2SO_(2)` It is then reduced to the corresponding copper metal by the following reaction `2Cu_(2)O+Cu_(2)Sto6Cu+SO_(2)` |
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| 848. |
Give reasons for the following: (i) Transition elements and their compounds act as catalysts (ii) `E^(@)` value for `(Mn^(2+) | Mn)` is negative whereas for `(Cu^(2+) | Cu)` is positive. (iii) Actinoids show irregularities in their electronic configuration. |
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Answer» (i) Refer Q. 25 (i), Paper-2009, Outside Delhi, Set-1, [Page 54]. (i) `E_(M^(2+)//M)^(@)` for any metal is related to the sum of enthalpy of atomisation, ionization and hydration Copper has high enthalpy of atomisation and low enthalpy of hydration. Hence `E_(Cu^(++)//Cu)^(@)` is positive. (iii) This happens because the energy difference between 5f, 6d and 7s subshell of the actinides is very small and hence, electrons can be accommodated in any of them. |
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| 849. |
Enlist the steps involved in the inbreeding of cattle. Suggest two disadvantages of this practice . |
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Answer» Superior female It is the cow or buffalo that produces more milk per lactation superior )females - It is the bull that gives rise to superior progeny as compared to other males. The steps involved in the inbreeding of cattle are as follows : Superior males and superior females of the same breed are identified and are made to mate in pairs. The progeny obtained is evaluated and superior males and females among them are identified for further mating. |
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| 850. |
A team of students are perparing to participate in the inter school sports meet.During a practice session you find some vials with labels of certain cannabionoids . (a) Will you report to the authorities ? Why ? ( b) Name a plant from which such chemicals are obtained . ( c) Write the effect of there chemicals on human body . |
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Answer» Yes ,I would report the matter to the authorites because cannabionids are classified under drugs of abuse . ( b) Cannabionoids can be obtained from a plant called Cannobis Sativa . ( c) The cannabionoids bind to cannobinoid receptors in the brain and affect the cardiovascular system. |
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