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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.
| 651. |
Mole Fraction |
| Answer» It is the ratio of number of moles of one componen to the total number of moles present in the solution. | |
| 652. |
Calculate the molality of ethanol solution in which the mole fraction of water is 0.88. |
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Answer» Mole fraction of water `chi_(H_(2)O) = 0.88` Mole fraction of ethanol, `chi_(C_(2)H_(5)OH) = 1-0.88 = 0.12` `chi_(C_(2)H_(5)OH) = (n_(2))/(n_(1) + n_(2)) " " ...(1)` `n_(2) = "number of moles of ethanol"` `n_(1) = "number of moles of water"` Molality of ethanol means the number of moles of ethanol present in 1000 g of water. `n_(1) = (1000)/(18) = 55.5 "moles"` Substituting the value of `n_(1)` in equation (1) `(n_(2))/(55.5+n_(2)) = 0.12` `n_(2) = 7.57 "moles"` Molality of ethanol `(C_(2)H_(5)OH) = 7.57 m` Alternatively, Mole fraction of water = 0.88 Mole fraction of ethanol = 1-0.88 = 0.12 Therefore, 0.12 moles of ethanol are present in 0.88 moles of water. Mass of water `=0.88 xx 18 = 15.87`g of water Molality = Number of moles of solute (ethanol) present in 1000 g of solvent (water) `=12 xx 1000//15.84 = 7.57m` Molality of ethanol `(C_(2)H_(5)OH) = 7.57m` |
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| 653. |
Mention the uses of cloning vector in biotechnology. |
| Answer» Cloning vector are used for large DNA fragments and can be easily detected at thetime of cloning experiments. | |
| 654. |
Explain primary productivity the factors that influence it. Describe how do oxygen and chemical composition of detritus control decomposition. What is El Nino effect ? Explain how it accounts for biodiversity loss ) Explain any three measures that you as an n individual would take, to reduce environmental pollution. |
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Answer» Primary productivity has two aspects (i) Gross pramiry productivity : It is the rate of production of organic matter during photosynthesis. (ii) Net primary productivity: It is the amount of energy left in the producers alter utilization of some energy for respiration, i.e., GPP R = N PP, it is the amount of energy available in the producers for the consumption of herbivores. Primary productivity depends on a number of environmental factors, availability of nutrients and photosynthetic capacity of plants. The annual net primary productivity of the whole biosphere is approximately 170 billion tons of organic matter. (b) Chemical composition of detritus: Decomposition is faster when detritus is rich in nitrogen and water - soluble substances like sugars. (i) Decomposition is slow when detritus is rich in lignin and chitin. (ii) Docomposition is an oxygen consuming process. Anaerobic conditions inhibit decomposition Nlno effect: El Nino i- . severe atmospheric and oceanic disturbance in thePacific ocean that occur. ovary seven to fourteen yearn. It is called El Nino, Meaning The Chnld" because it in usually appears near the Christmas season. Warm surface water Meaning the central Pacific towards eastern pacific. El Nino are thought to be potentially more damaging on a global scale, as they may cause floods and Mudalidel in Latin America, but also month-long droughts in Southeut Asia and Australia. (b) Thou arc the three measures that we are an individual would take, to reduce environmental pollution. (I ) Uses of renewable onergy resources. (ii) Uses of television and other instrument at low pitch. (iii) Minimum use of fossil fuel. (iv) No tobacco smoking. (v) Planting more trees |
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| 655. |
How does Penicillium reproduce asexually |
| Answer» Penicillium reproduce asexually by spore formation | |
| 656. |
Using IUPAC norms write the formulae for the following: (i) Hexaamminecobalt (III) sulphate. (ii) Potassium trioralatochromate (III). |
| Answer» (i) `[Co(NH_(3))_(6)]SO_(4)` (ii) `K_(3)[Cr(C_(2)O_(4))_(3)]` | |
| 657. |
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get a affected when its radius is increased ? |
| Answer» It is independent of radius of Gaussian surface . | |
| 658. |
A point charge Q is placed at the point O as shown in Fig. Is the potential difference `(V)_(A) - V_(B))` positive, negative or zero if Q is (i) possible (ii) negative ? |
| Answer» `V_(A) -V_(B) = ive` | |
| 659. |
A point charge Q is placed at the point O as shown in Fig. Is the potential difference `(V)_(A) - V_(B))` positive, negative or zero if Q is (i) possible (ii) negative ? |
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Answer» (i) If q is positive charge, `V_A-V_B`= positive. (ii) If q is negative charge, `V_A-V_B`= Negative. |
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| 660. |
Define electric dipole moment. Write its SI unit ? |
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Answer» Electric dipole moment Dipole moment `overset(rightarrow)P` is a measure of strength of electric dipole. It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and the distance between them. SI unit of dipole moment is coulomb-metre (cm). |
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| 661. |
How are radio waves produced? |
| Answer» By rapid acceleration and deacceleration of electrons in aerials. | |
| 662. |
The angles of dip at two places are respectively `0^@` and `90^@`. Where are these values on earth? |
| Answer» At North and South Pole | |
| 663. |
A hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V.What is the potential at the centre of the sphere? |
| Answer» Potential inside the charged sphere is constant and equal to potential on the surface of the conductor. So, potential at the centre of the sphere is 12V. | |
| 664. |
Why the electric field at the outer surface of a hollow charged conductor is normal to the surface? |
| Answer» The electrostatic field be normal to the surface at every point of a charge conductor. Electric field inside the conductor is always zero. The electric lines of force exert lateral pressure on each other leads to explain repulsion between like charges. | |
| 665. |
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere? |
| Answer» Potential inside the charged sphere is constant and equal to potential on the surface of the Conductor. So, potential at the centre of the sphere is 5V. | |
| 666. |
Prove that an ideal capacitor in an a.c. circuit does not dissipate power. |
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Answer» In a circuit containing inductor L, current I lags behind the voltage E be a phase angle of `90^(@) or (pi)/(2)` `therefore " "E=E_(0) sin omega"t then I"=I_(0)sin(omegat-(pi)/(2))` `therefore " "I=-I_(0)cos omegat` Work done in one complete cycle is `W=underset(0)overset(T)int EI dt=underset(0)overset(T)int E_(0) sin omegat(-I_(0) cos omegat) dt=E_(0)I_(0) underset(0)overset(T)int sin omegat cos omegat` `=-E_(0)I_(0) underset(0)overset(T)intsin (sin 2omegat)/(2)=-(E_(0)I_(0))/(2)[-(cos2 omegat)/(2omega)]_(0)^(T)=-(E_(0)I_(0))/(2)[(cos 2 omegaT)/(2omega)-(cos 0)/(2omega)]=(E_(0)I_(0))/(2)[(cos 2.(2pi)/(T).T)/(2omega)-(1)/(2omega)] ` `=(E_(0)I_(0))/(2)[-(cos4pi)/(2 omega)-(1)/(2omega)]=(E_(0)I_(0))/(2)[(1)/(2omega)-(1)/(2omega)]=0` Average power `=(W)/(T)=(0)/(T)=0`. Hence, average power supplied to an ideal capacitor by the source over a complete cycle of a,c cycle is zero. |
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| 667. |
A resistor ` R_(1)` consumes electrical power `P_(1)` when connected to an `emf epsilon`. When resistor `R_(2)` is connected to the same `emf` , it consumes electrical power `P_(2)` . In terms of `P_(1) and P_(2)`, what is the total electrical power consumed when they are both connected to this emf source (a) in parallel (b) in series |
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Answer» Two resistance are `R_(1) a R_(2)`. Both consumed the Power `P_(1) and P_(2)` respectively. `P=(V^(2))/(R)therefore P_(1)=(V^(2))/(R_(1)) and P_(2)=(V^(2))/(R_(2))` `"(i) ""Power in series"=(1)/(P_(1))+(1)/(R_(2))=(R_(1))/(V^(2))+(R_(2))/(V^(2))=(R_(1)+R_(2))/(V^(2))` `"(ii) ""Power in parallel"=P_(1)+P_(2)` `" "=(V^(2))/(R_(1))+(V^(2))/(R_(2))=(V^(2)(R_(1)+R_(2)))/(R_(1)R_(2))` |
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| 668. |
Explain how a hereditary disease can be corrected. Give an example of first successful attempt made towards correction of such diseases. |
| Answer» A hereditary disease can be corrected by gene therapy. In this method, genes are inserted into a person s cells and tissues to treat a disease. The first successful attempt was gene therapy for adenosine deaminase (ADA) deficiency | |
| 669. |
Study the graph given below. Explain how is oxygen concentration affected in the river when sewage is discharged into it. |
| Answer» When sewage is discharged into river, the oxygen concentration declines sharply because a large amount of oxygen is consumed by aerobic micro organisms in river to decompose the organic matter in river. When the amount of organic matter reduces, the amount of dissolved oxygen again increases. | |
| 670. |
Identify A, D, E and F in the diagram of an antibody molecule given below : |
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Answer» Antigen binding site D : Light chain E : Heavy chain F : Disuflde bond |
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| 671. |
Explain the events that occur, upto fertilisation, when the compatible pollen grain lands on the stigma. |
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Answer» When compatible pollen grain land on stigma, the pollen grains germinate and produce a pollen tube which passes through the tissue of stigma and style and finally reaches the ovary. In the pollen tube generative cell mitotically divides to divides to produce two male gametes during the growth of pollen tube in the stigma. In some plants pollen grains shed at the three celled stage. In such plants pollen tubes already carry two male gametes. Pollen tube after reaching ovary, enter the ovule through the micropyle and them enters one of the synergids through the filiform apparatus. One of the male gametes moves towards the egg cell and fuses with tis nucleus. The fusion of male and female gametes is called syngamy and results in the formation of zygote. |
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| 672. |
What does an interaction between pollen grains and its compatible stigma result in after pollination ? List two steps in sequence that follow after the process. |
| Answer» See Q.29, (b) Set-III., Delhi Board-2013. | |
| 673. |
State any three functions of placenta in human female. |
| Answer» See Q.24, (or) Set-I., Delhi Board -2015. | |
| 674. |
Differentiate between out-crossing and cross breeding |
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Answer» Out crossing - 1. It is mating of animals within the same breed but having no common ancestors upto 4-6 generations. 2. It is the best method to overcome in breeding depression and to increase milk productivity in milch cattle. 2. Cross breeding - 1. It is breeding between two individuals of different species. 2. It helps to combine desirable qualities of two different breeds. |
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| 675. |
Why are microbes like Spirulina being produced on a commercial scale ? Mention its two advantages. |
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Answer» Because they can be grown on waste materials and can serve as rich source of proteins. Its advantages : (i) It reduces pollution. (ii) It is powerful antioxidant, which can protect against oxidative damage. |
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| 676. |
Give the IUPAC name of the following compound. `CH_(2)=C-underset(CH_(3))underset("| ")("C ")H_(2)Br` |
| Answer» 3-Bromo-2-methyl propene. | |
| 677. |
(a) How many mole of mercury will be produced by electrolysing 1.0 MHg `(NO_(3))_(2)` solution with a current of 2.00 A for 3 hours? [Hg`(NO_(3))_(2) = 200.6 g mol^(-1)`]. (b) A voltaic cell is set up at `25^(@)`C with the following half-cells `Alk^(3+)` (0.001M) and `Ni^(2+)` (0.50M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. (Given : `E_(Ni^(2+)//Ni)^(2)=- 0.25 V,E_(Al^(3+)//Al)^(@)= - 1.66V`) |
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Answer» Time = 3 hours `=3xx 60xx60= 10800 Sec`. Current = 2A Charge = Current `xx` time `= 2 xx 10800 = 21600 C` According to the questions Solution of `Hg(NO_(3))_(2) = 1 mol`. Given `Hg(NO_(3))_(2) = 200.g//mol` We required 2F or `2 xx 96487C`to deposite 1 mole 63 g. of Hg For 21600 C, the mass of mercury is `=((200.6g//mol)xx (21600^(@)C))/((2xx 96487C//mol)) = (4332960 g)/(192975)=22.45g` Moles `= (22.45)/(200.6)= 0.112 mol` (b) Given `E_(Ni^(2+)//Ni)^(2) = - 0.25V`, `E_(Al^(3+)//Al)^(o) = - 1.66V` Half cell equations are `Al to Al^(3+) + 3e^(-)" ""(at anode)"` `Ni^(2) + 2e^(-) to Ni" ""(at Cathod)"` `2Al + 3Ni^(2+) to 2Al^(3+) + 3Ni" ""over all recation"` The cell may be represented as `Al|Al^(3+)|| Ni^(2+) | Ni` `E_("Cell")^(o) = E_("right")^(o) - E_("left")^(o)` `=(-0.25)-(1.66)` `= -0.25 + 1.66 rArr E_("cell")^(o) = 1.41 V`. |
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| 678. |
How and why is bacterium Thermus aquaticus employed in recomb technology ? Explain. |
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Answer» Becterium thermus aquaticus produces an enzyme called Taq polymerase (DNA polymerase) which is resistant to denaturation by heat tre atment. This enzyme is used to amplify a specific DNA fragment m polymerase chain reaction (PCR) technique, The enzyme utilises building blocks [dNTPs] to extend the perimer. In the first step, the double stranded DNAmoleculeB are heated to a high temperature. So, that the two strands separate to a single stranded DNA molecule. This process is called denaturation. Then this Ss DNA molecule is used as a template strand for the synthesis of a new strand by the DNA polymerase enzyme by the process of annealing and extension which results in the duplication of the original DNA molecule. This process is repeated over several cycles to obtain multiple copies of the DNA fragment. |
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| 679. |
Explain the various steps involved in the production of artificial insulin |
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Answer» Insulin used for diabetes was earlier exlier extracted from pancreas of slaughtered cattle and pigs. Insulin consists of two short polypeptide chains A and chain B that are linked together by disulphide bridges . In mammals including humans , insulin is synthesised as aprohormone (like a pro - enzyme the pro - hormone ) also need to be processed before it becomes a fully mature and functional hormone. In America , a company prepared two DNA sequences corresponding to A and B chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains . Chains A and B were produced separately , extracted and combined by creating disulfide bonds to form human insulin. |
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| 680. |
Why the pyramid of energy is always upright? Explain. |
| Answer» Pyramid of energy is never inverted because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat at each step. Each bar in the energy pyramid indicates the amount of energy present at each trophic level in a given time. | |
| 681. |
Name the disorder caused due to the absence of one of the X-chromosomes in a human female. |
| Answer» Turners syndrome is caused due to the absence of one of the X-chromosomes in a human female. | |
| 682. |
In a botanical garden of a city there is a huge banyan tree growing on which hundreds of birds and thousands of insects live. Draw the pyramids of numbers and also biomass represented by this community. Comment giving reasons on the two different pyramids drawn. |
| Answer» See Q.28., Set-II., Delhi Board-2012. | |
| 683. |
After wathicng a programme on TV about te presence of carcinogens (cancer causing agents) Potassium bromate and Potassium iodate in bread and other bakery products, Rupali a class XII student decided to make other awarre about the adverse effects of these carcinogens in food. she consulted the school principle and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers and other bakery products to the students. The principal took an immediate action ad instructed the canteen contractor to replace the bakery products with some protein and vitamin rich food like fruits, salads, sprouts, etc. The decision was welcomed by the parents and the students. After reading the above passage, answer the following questions: (a) What are the values (at least two) displayed by Rupali? (b) which polysaccharide component of carbohydrates is commonly present in bread? ltBrgt (c) Write the two types of secondary structures of proteins (d) Give two examples of water soluble vitamin. |
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Answer» (i) Ritu is having scientific attitude and she is having critical thinking, social responsibility (ii) Starch (iii) Secondary structure of proteins are of two types: (i) `alpha-`helix (ii) `beta`-pleated structure (iv) Water soluble vitamins are vitamin B-complex and vitamin C |
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| 684. |
List the specific symptoms of amoebiasis. Name the causative organism |
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Answer» The symptoms of amoebiasis are Abdominal pain and cramps. Stools with excess mucus and blood clots. Constipation, Causative organism-Entamoeba histolytica |
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| 685. |
Plenty of algal bloom is observed in a pond in your locality. a) Write what has caused this bloom and how does it affect the quality of water. b) Suggest a preventive measure. |
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Answer» a) Presence of large amounts of nutrients in water also cause growth of planktonic algae cause algal bloom. It effects on quality of water in many ways. 1) Deterioration of water quality. 2) Increased fish mortality rate. c) Some bloom forming algae are toxic to human and animals. 4) Accelerated eutorphication. 1) Avoid excess use of fertilisers and manures on agricultural land. 2) Protecting soil from erosion. 3) Treating sewage to remove the nutrients, nitrogen and phosphorous. 4) Artificial destratification. 5) Biomanipulation |
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| 686. |
A crane had DDT level as 5 ppm in its body. What would happen population of such birds ? Explain giving reasons |
| Answer» The high concentration of DDT (5ppm) in cranes disturb their calcium metabolism which causes thinning of egg shell and their premature breaking Thereby causing decline in thebird populations. | |
| 687. |
Distinguish between the roles of flocks and anaerobic sludge digesters in sewage treatments. |
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Answer» Flocks : 1. these are masses of bacteria associated with fungal filaments to form a mesh. 2. these help to reduce the Bioogical oxygen Demand (BOD) of the primary effiuent. Anaerobic sludge digesters 1. Here the activated sludge consisting of bacteria fungi is digested by anaerbic bacteria . 2. these help in the producation of biogas consisting of methane , hydrogen sulphide and carbon dixide . |
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| 688. |
Conditions Of Total Internal Reflection |
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Answer» Essential conditions for total internal reflection: (i) Light should travel from a denser medium to a rarer medium. (ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact. |
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| 689. |
In a typical unclear reaction, e.g. `""_(1)^(2)H+""_(1)^(2)H to ""_(2)^(3)He + n +3.27 Me V`, although number of nucleons is conserved is conserved, yet energy is released. How ? Explain. (b) Show that nuclear dendity in a given nucleus is independent of mass number A. |
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Answer» (a) In a nuclear reaction, the sum of the masses of the target nucleus `(""_(1)^(2)H)`, and the bombarding particle `(""_(1)^(2)H)` may be greater or less than the sum of the masses of the product nucleus `(""_(1)^(3)He)` and the outgoing particle `(""_(0)^(1)n)`. So, from the law of conservation of mass energy, some energy (3.27 MeV) is evolved or involved in a nuclear reaction. This energy is called Q-value of the nuclear reaction. (b) Density of the nucleus `=("mass of nucleus")/("volume of nucleus")` Mass of the nucleus = a amu `= A xx 1.66 xx 10^(-27) `kg Volume of the nucleus `=(4)/(3) pi R^(3)` `=(4)/(3) pi (R_(0)A^(1//3))^(3) =(4)/(3) pi R_(0)^(3) A` Thus, ` " density"= (Axx1.66xx 10^(-27))/(((4)/(3)piR_(0)^(3))A)=(1.66 xx 10^(-27))/(((4)/(3)piR_(0)^(3)))` which shows that the density is independent of mass number A. Using `R_(0)=1.1 xx 10^(-15)m` and density ` =2.97 xx 10^(17) kg//m^(3)` |
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| 690. |
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 `mu`C. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 `mu`C. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100V? |
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Answer» (i) Intial voltage, `V_(1)=V` volts, Charge stored, `Q_(1)=300 mu C` `Q_(1)=Cv_(1) " " ` …(i) Charged potential, `V_(2)=V-100V.` `Q_(2)=100 mu C` `Q_(2)=CV_(2) " " ` ...(ii) By dividing (ii) from (i), we get `(Q_(1))/(Q_(2))=(CV_(1))/(CV_(2))rArr (Q_(1))/(Q_(2))=(V_(1))/(V_(2))` `(300)/(100)=(V)/(V-100)` `V=150` volts. `C=(Q_(1))/(V_(1))=(300xx10^(-6))/(150)=2xx10^(-6)F=2 mu F` (ii) If the voltage applied had increased by 120V, then `V_(3)=150 +100=250V` Hence, charge stored in the capacitor, `Q_(3)=CV_(3)=2xx10^(-6)xx250 =500 mu C` |
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| 691. |
An optical instrument uses eye-lens of power 16 D and objective lens of power 50D and has a tube length of 16.25 cm. Name the optical instrument and calculate its magnifying power if forms the final image at infinity. |
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Answer» `P_(e) = 16 D rArrf_(e) = 1/P_(e)=1/16 m = 100/16 cm =25/4cm` `P_(0)=50 D rArr f_(0) = 1/ P_(d)=1/50 m=100/50 cm=2 cm` `f_(e) = 25/4 cm and f_(0) = 2 cm` length of tube = l= 16.25 cm Optical instrument is compound microscope. `M= L/(f_(0)) xx D/(f_(e))" " =16.25/2 xx cancel (25)/cancel (25)_(4)=16.25 xx 2` M=32.5 |
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| 692. |
Calculate the capacity of unknown capacitance is connected acrosss a battery of V volts. The charge stored in it is `360 muC`. When potential across the capacitor is reduced by 120V, the charge stored in it becomes `120 muC`. Calculate (i) the potential V and unknown capacitance C. (ii) What will be the charge stored in the capacitor. If the voltage applied had increased by 120 V |
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Answer» From the relation `Q=CV=360 mu C` When potential is reduced to 120V `120 mu C= C( V-120) = 120 C` `rArr" " 120 mu C = 360 mu C- 120 C rArr " "120=240 mu C` `rArr C=(cancel (240)muC)/cancel (120)=2 mu F` Potential `V= (360 mu C)/(2 mu F) = 180 V` Charged stored when voltage applied is increased by `120V Q ^(1)=2 mu F xx (180V + 120 V) = 600 mu C` |
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| 693. |
(a)For the reaction : `2AgCl (s) + H_(2)(g) ` (1 atm) ` to 2 Ag(s) + 2H^(+) (0.1M) + 2Cl^(-) (0.1M) , DeltaG^(@) = -43600J` at `25^(@)`C. Calculate the e.m.f. of the cell . [log `10^(-n) = -n`]. (b) Define fuel cell and write its two advantages. |
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Answer» (a) `2AgCl(s) + H_(2)(g)` (1 atm) `to 2Ag(s) + 2H^(+) (0.1M) + 2Cl^(-) (0.1M)` `DeltaG^(@) = -43600 J` at `25^(@)C` `Delta G^(@) = - n F E^(@)` cell `" "` [n =2] `-43600 = -2 xx 96500 E^(@)` cell `E_("cell")^(@) = (cancel43600^(218))/(cancel2 xx cancel 96500)` `E_("cell")^(@) = 0.225V` `E_("cell")^(@) = E_("cell")^(@) - (0.0591)/(2) "log"([H^(+)]^(2)[Cl^(-)]^(2))/(1) = 0.025 - (0.0591)/(2) "log" 0.1 xx 0.1 xx 0.1 xx 0.1` `= 0.225 - 0.02955 "log" 10^(-4) = 0.225 - 0.02955 xx (-4) = 0.225 + 0.1182` `E_("cell")^(@) = 0.3432V` (b) Fuel cell : A device in which the heat produced as a result of the combustion of fuel (like `H_(2) - O_(2)`) is converted into electricity is called fuel cells. Advantages : (i) Because of the continous supply , fuel cell never become dead . (ii) They do not cause any pollution problem unlike thermal plants which burn fossils fuels like coal , gas oil etc. (iii) They give an efficiency of 60-70% which is superior to thermal plants. |
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| 694. |
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K : Sn(s) `|Sn^(2+) (0.004M)||H^(+) (0.020M) |H_(2)(g)` (1 bar )| Pt(s) (Given `E_(Sn^(2+)//Sn)^(@) = -0.14V`). (b) Give reasons : (i) On the basis of `E^(@)` values , `O_(2)` gas should be liberated at anode but `Cl_(2)` gas which is liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of `CH_(3)COOH` decreases on dilution . |
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Answer» (a) `Sn(s) | Sn^(2+) (0.004M) ||H^(+) (0.020M) | H_(2)(g) ` (1 bar ) |Pt(s) Cell reaction : `Sn + 2H^(+) to Sn^(2+) + H_(2)` `E_("cell") = overset(n=2)(E_("cell")^(@)) - (0.0591)/(chi)"log" ([Sn^(2+)])/([H^(+)]^(2)) = 0-(-0.14) - (0.0591)/(2) "log" (0.004)/(0.020xx 0.020)` `= 0.14 - (0.059)/(2) "log" 10 = 0.14 - (cancel0.0591)/(cancel2) = 0.14 - 0.02955 = 0.111V.` (b) (i) `E_("OX")^(@)` value for `O_(2)` is more than that of `E^(@)` for `Cl_(2)` but `Cl_(2)` is produced not `O_(2)` . This unexpected result is explained on the basis of the concept of 'over voltage' , i.e., water needs greater voltage for oxidation to `O_(2)` as it is kinetically slow process than that needed for oxidation of `Cl^(-)` ions to `Cl_(2)` .This extra voltage required is called over voltage . `{:(2Cl^(-)(aq) to Cl_(2) (g) + 2e^(-) " " E_(OX)^(@) = -1.36V) , (H_(2)O(l) to (1)/(2)O_(2)(g) + 2H^(+) (aq)+ 2e^(-) E_(OX)^(@) = -1.23V):}` (ii) On dilution , although `CH_(3)COOH` is completely dissociated but concentration of ions per unit volume is so low that the conductivity decreases with dilution . |
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| 695. |
Which basic mode of communcations is used in satellite communication ? What type of wave propagation is used this mode ? Write, giving reason, the frequency range used in this mode of propagation. |
| Answer» Satellite communications uses broadcast mode in which space wave mode of propagation is used. For frequency range `30MHz- 1000MHz`, `lambda` falls in the range `0.3m` to `10m`. In this range, diffraction of wave is high and also looses directional property so range of frequency is `5.925 - 6.425 GHz`is satellite communication. | |
| 696. |
(i) Find equivalent capacitance between A and B in the combination give below . Each capacitor is a of `2 muF` capacitance . (ii) If a dc source of 7V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network ? |
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Answer» (i) `C_(eq)` in `PRST = 2+2+2= 6 muF`, Remaining capacitors between A and B are in series . `:. 1/C =1/2 + 1/6 + 1/2 = (3+1+3)/(6) = 7/6` `rArr C = 6/7muF` (ii) `Q = CV` `= 6/7 xx 10^(-6) xx 7= 6 xx 10^(-6) C` `E= 1/2 CV^(2) = 1/2 xx (6xx10^(-6))/(7) xx 49 = 21 xx 10^(-6)J`. |
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| 697. |
When primary coil P is moved towards secondary coil S, Fig. the galvanometer shows memetary deflection in the can be done to have larger deflection in the galvanometer with same battery ? State the related law ? |
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Answer» (i) To get the larger deflection in the galvanometer with the same battery, the coil P should be moved faster. (ii) The related law is electromagnetic induction. It is the process in which an emf is induced in a circuit placed in a magnetic field when the magnetic flux linked with the circuit changes. |
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| 698. |
An alternating voltage given by `V=70 sin 100 pi t` is connected across a pure resistor of `25 Omega.` Find (i) the frequency of the source. (ii) the rms current through the resistor. |
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Answer» `V_(0)=70, omega =100 pi` (i) `2piV=100 pi therefore V=50 Hz` (ii) `I_(rms)=(V_(rms))/(R ) =(V_(0)//sqrt(2))/(R )=(70)/(sqrt(2xx25))` `=(14)/(5xx1.414)=1.98` ampere. |
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| 699. |
Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 cm away, find the height of the final image when it is formed 25 cm away from the eye piece. |
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Answer» Magnifying power of an astronomical telescope is defined as the ratio of the angle subtended at the eye by the final image of the angle subtended at the eye, by the object directly. The expression for magnifying power : `m=(f_(0))/(f_(e ))(1+(f_(e ))/(d))` (a) In normal adjustment `m= -(f_(0))/(f_(e ))` (b) When final image is formed at infinity `m=(f_(0))/(f_(e ))(1+(f_(e ))/(d)), f_(0)=150 cm implies f_(e )=5 cm`. Using `(1)/(f )=(1)/(v)-(1)/(u)` for objective. `(1)/(150)=(1)/(v)-(1)/(-3000) implies (1)/(v)=(1)/(150)-(1)/(3000)=(20-1)/(3000) implies v=(3000)/(19)` `m_(0)=(3000)/(19)xx(1)/(-3000)=-(1)/(19)` Using `(1)/(f)=(1)/(v)-(1)/(u)` for eye piece, `(1)/(5)=(1)/(-25)-(1)/(u) implies (1)/(u) =(1)/(-25)-(1)/(5) implies u=(1+5)/(-25)` `u=(-25)/(6) cm implies m_(e )=(v)/(u)` `m_(e )=(-25)/(-25)xx6=6` `therefore` Total magnification `=m_(0)xxm_(e )= -0.315` `h_(1)=0.315xx100=31.578`. |
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| 700. |
An electromagnetic wave `Y_1`, has a wavelength 1 cm while other electromagnetic wave `Y_2` has a frequency of `10^15Hz`. Name these two types of waves and write one useful application for each. |
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Answer» `Y_1` is microwave, use in radar communication. `Y_2`: ultraviolet, ioize atom in atomsphere, |
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