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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.
| 1101. |
(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so ? (ii)Which transition metal of 3d series ha positive `E^(@)(M^(2+)//M)` value and why ? (iii) Out of `Cr^(3+) and Mn^(3+)`, which is a stronger oxidizing agent and why ? (iv) Name a member of the lanthanoid series which is well known to exhibit + 2 oxidation state. (v) Complete the following equation : `MnO_(4)^(-) + 8H^(+) + 5e^(-) to ` . |
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Answer» In 3d -series of transition metals, manganese has an atomic number of 25 that gives tha electronic configuration as `[Ar] 3d 4s2,` where we see that the maximum number of ltbr. Unpaired electrons is found in manganese atom, so, it can show a maximum oxidation state upto + 7. (ii) Copper is the transition metal of 3d series that exhibits positive `MnO_(4)^(-) + 8H^(+) + 5e^(-) to ` value of a metal depends on the energy changes involved in the following : 1. Sublimation energy : The energy required for converting one mole of an atom from Solid state to the gaseous state. `M_(g) to underset("Sublimation energy")(M_((g)) Delta s H = high` 2. Ionisation energy : The energy required to take out electons from one mole of atom in the isolated gaseous state. `M_((g)) to M_((g)) DeltaH = high` ("Ionisation energy") 3. Hydration energy : The energy released when one mole of ions are hydrated . `M_((g)) to M_((g)) Delta s H ("low energy") ("Hydration energy") ` Since, copper has a high energy of atomisation and low hydration energy, the `E(@) (M^(2+)//M)` value for copper is positive. (iii) Out of `Cr^(3+) and Mn^(3+), Mn^(3+)` is a stronger oxidising agent because it has 4 electrons in its valence shell and when it gains one electron to form `Mn^(2+)`, it results in the half-filled `(d^(5))` configuration that has extra stabillity. (iv) Europium (Eu) is well-known to echibit + 2 oxidation state due to its half filled forbital in + 2 oxidation state. (v) `MnO_(4)^(-)+8H^(+)+ 5e to Mn^(2+) + 4H_(2)O` |
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| 1102. |
Write the structure of 4 chloropentan 2 one |
| Answer» `H_(3)C-overset(Cl)overset(|)(C )H-CH_(2)-overset(O)overset(||)(C )-CH_(3)` | |
| 1103. |
What are the products of hydrolysis of maltose? |
| Answer» Alpha D Glucose Disaccharides | |
| 1104. |
List the crteria a molecule that can act as genetic material must fulfill. Which one of the critria are best fulfilled by DNA or by RNA thus making one of them a better gentic material than can other ? Explain . |
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Answer» A molecule the can act genetic must fulfill the following critra. 1. It should be able to replicate . 2. It should be chemically and structurally stable . 3. It should provide chance for mutation that is required for evolution 4. it should be able to express itself in the form of genes . DNA is more stable than RNA and better suited for storge of genetic information while RNA is more suitable for the transmission of gentic information because , 1. Both DNA and RNA can replicaate . 2. presence of thymine in place of Uracil makes DNA more stable 3. Presence 2, -OH group in RNA makes it more labile RNA is also catalytic and more reactive ,DNA is chemically and structurally more stable . 4. RNA mutates faster than DNA since it is unstable. 5. RNA can directly code for protein synrthesis while DNA is dependant on RNA for protein sythesis this shows that protein sythesis has evolved around RNA . |
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| 1105. |
Differentiate between analogy and homology giving one example one each of plant and animal respectively . ( B) How are they considered as an evidence in support of example of evolution ? |
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Answer» Homology : this refers to simlar anatomy but different functioning . example : thorn and tendril of bouganivlllea and cucurbita and forelimbs of whale and bats . Analogy : this refers to similar functioning different anatomy Example : Eye of octopus and mammals and sweet potao ( root modification ) and potato ( stem modification ) these structures are considered as an evidence of evolution because : Homology points to development of a similar structure in different directions due to different needs indicating divergent evolution . 2. Analogy point to different structures evolving for the same function indicating convergent evolution . |
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| 1106. |
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current ? |
| Answer» Ideal capacitor offers infinite resistance for dc. When an ac source is used, the current flows continuously, but we know that the capacitor has dielectric between its plates. So, ideally there is no current, and circuit would be incomplete. In real capacitor is charged due to contribution of varying electric field. The current between the capacitor plates is given by displacement current. | |
| 1107. |
Name an opioid drug and its source plant. How does the drug affect the human body? |
| Answer» Heroin is an opioid drug obtained from the poppy plant Papaver Somniferum. It is a depressant and slows down the body functions. | |
| 1108. |
Essential elements of a communication system are |
| Answer» Transmitter, channel and receiver. | |
| 1109. |
(a) (i) Write the disproportionation reaction of `H_(3)PO_(3)` (ii) Draw the structure of `XeF_(4)` (b) Account for the following: (i) Although Fluorine has less negative electron gain enthalpy yet `F_(2)` is strong oxidizing agent. (ii) Acidic character decreases from `N_(2)O_(3)` to `Bi_(2)O_(3)`, in group 15. (c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. |
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Answer» (i) See Q. 14 (ii), Paper-2013, Outside Delhi, Set-I, IPage 202]. (ii) See Q 11 (i), Paper-2009, Delhi Board, Set-I, [Page 28] (b) (i) F has although low electron affinity but low dissociation energy and have high hydration energy of its ion therefore, fluorine is strongest oxidizing agent. (ii) On moving down the group, the atomic size increases, electronegativity decrease5 and metallic character increases and hence, acidic character decreases. (c) Aqueous acidified solution of potassium dichromate is orange in colour. When sulphur dioxide gas is bubbled through the solution the colour changes to green. This is because sulphur dioxide reduces the dichromate `CrO_(7)^(2-)` which contains chromium in +6 oxidation state to `Cr^(3+)`. `K_(2)Cr_(2)O_(7)(aq)+3SO_(2)(g)+H_(2)SO_(4)(aq)toCr_(2)(SO_(4))_(3)+K_(2)SO_(4)(aq)+H_(2)O(l)` |
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| 1110. |
(a) Give reasons for the following: (i) Sulphurin vapour state shows paramagnetic behaviour. (ii) N-N bond is weaker than P-P bond. (iii) Ozone is thermodynamically less stable than oxygen. (b) Write the name of gas released when Cu is added to: (i) dilute `HNO_(3)` and, (ii) conc. `HNO_(3)` |
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Answer» (a) (i) See Q.29 (i), Paper-2008, Outside Delhi, Set-I, [Page 22]. (ii) See Q. 24 (ii), Paper-2014, Delhi Board, Set-II, [Page 235]. (iii) Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat. Ozone consists of three molecules of oxygen and thus in this it is in unstable state so in order to get stable it gives up its one molecule of oxygen to restore diatomic state. (b) (i) dilute HNO:See Q. 26 (or) (ii) (a), Paper-2018 (C), [Page 484]. (ii) conc. `HNO_(3): NO_(2)` gas released when Cu is added to con. `HNO_(3)`. `Cu+4HNO_(3)(conc.)toCu(NO_(3))_(2)+2NO_(2)+2H_(2)O`. |
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| 1111. |
Explain glucose. What is the role of glucose (cane sugar) from Sucrose. |
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Answer» Glucose occurs in nature in free as well as in the combined form. It is present in sweet fruits and honey. In the combined form glucose occurs in abundance in cane sugar and polysaccharides such as starch and cellulose. When sucrose is boiled with dilute HCl or `H_(2)SO_(4)` in alcoholic solution, glucose and fructose are obtained in equal amounts. `underset("Sucrose ")(C_(12)H_(22)O_(11))+H_(2)Ooverset(H^(+))rarrunderset("Glucose")(C_(6)H_(12)O_(6))+underset("Fructose")(C_(6)H_(12)O_(6))` |
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| 1112. |
Define the term "threshold freq uence", in the context of photolelectric emission. |
| Answer» Threshold frequency is defined as the minimum frequency of incident light which can cause photo electric emission. | |
| 1113. |
Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a valocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for (i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ. |
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Answer» Let length `PQ =l` and width of rod = dx and let area be A. Velocity = v, resistance = r. Current in the loop (i) `i = (epsilon)/(r)` `i=(Blv)/(r)` (ii) Force in the loop `F=ilB` `F=(B^(2)l^(2)v)/(r)` (iii) `P=Fv=(B^(2)l^(2)v^(2))/(r)` There is a decrease in area Hence emf is `epsilon=(-dphi)/(dt)` `epsilon =(-d)/(dt)(BA) , epsilon = -B(dA)/(dt)` `epsilon = +(Bldx)/(dt) , epsilon = Blv` |
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| 1114. |
The galvanometer, in each of the two given circuits deos not show any deflection. Find the ratio of the resistors `R_(1)` and `R_(2)` used in these two circuits. |
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Answer» Current sensitivity of a galvanometer is defined as the deflection in galvanometer per unit current. Its SI unit is Radians/Ampere. For balanced Wheatstone bridge, there will be no deflection in the galvanometer. `(4)/(R_(1))=(6)/(9)" "R_(1)=(4 xx 9)/(6) = 6 Omega`. For the equivalent circuit, when the Wheatstone. bridge is balanced, there will be no deflection in the galvanometer. `(12)/(8)=(6)/(R_(2)) rArr R_(2)=(6 xx 8)/(12) = 4 Omega` `(R_(1))/(R_(2))=(6)/(4)=(3)/(2) rArr 3 : 2`. |
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| 1115. |
Distinguish between Sky wave and Space wave modes of propagation in a communication system. |
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Answer» In Sky wave propagation, radio waves are emitted from the transmitter antenna antenna and reach the receiving antenna after reflection by the ionosphere. Frequency ranges from 1.7 mHz to 40 mHz and is used by short wave broad cast service. In space wave communication, we send, receive and process the information through space by using tall antenna Here, loss of power called attenuation of ground wave is felt due to induced current of the Earth. |
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| 1116. |
(a) write three characteristic properties of nuclear force. (b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph. |
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Answer» (a) Characteristic properties of nuclear forces are : (i) The nuclear forces are neither gravitational nor electrostatic as the former is too weak to hold nucleus together while the latter will blow it apart. (ii) They become negligible for distances greater than gf( 1 fermi = `10^(-15)` m (iii) They do not follow inverse square law. They are short range, strong attractive forces. |
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| 1117. |
Show that the electric field at the surface of a charged conductor is given by `vecE = (sigma)/(epsi_(0)) hatn` , where `sigma` is the surface charge density and `hatn` is a unit vector normal to the surface in the outward direction . |
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Answer» Electric field at a point on the surface of charged conductor , `E = (1)/(4 pi epsi_(0)) (Q)/(R^(2))` . Let , consider charged conductor as a sphere of radius R. if `sigma` is in surface density , then `Q = 4 pi R^(2) sigma` `E = (1)/(4 pi epsi_(0)) = (4 pi R^(2) sigma)/(R^(2)) = (sigma)/(epsi_(0))` `vecE = (sigma)/(mu_(0)) hatn` ... where n is a unit vector normal to the surface in the outward direction . |
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| 1118. |
A proton and deuteron are accelerated by same potential difference.Find the ratio of their de-Broglie wavelengths. |
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Answer» The de Broglie wavelength `lambda` associated with same potential V is `lambda = (h)/(sqrt(2 meV))` `therefore " " lambda alpha (1)/(sqrtm)` [For same potential] Since alpha particle is `""_(2)^(4)He` and proton is `_(1)^(1)H` `therefore` mass of alpha particle = 4 times mass of proton `therefore " " alpha _("proton") alpha (1)/(sqrtM)` and `lambda_("alpha") alpha (1)/(sqrt4m)` `therefore (lambda_("proton"))/(lambda_("alpha")) = (1)/(sqrtm) xx (sqrt(4m))/(1) = (2)/(1) therefore lambda_("proton") : alpha_("alpha") = 2 : 1` |
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| 1119. |
Two identical loops , one of copper and the other of aluminium , are rotated with the same angular speed in the magnetic field . Compare (i) the induced emf and (ii) the current produced in the two coils . Justify your answer . |
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Answer» Since induced emf of a coil rotating in magnetic field B with angular velocity w is given by `E_(0) = N` B A `omega` and induced current `I_(0) = (E_(0))/(R)` (i) The ratio of induced emf = ` 1: 1` (ii) The ratio of induced current = `1 :1` Since , the induced emf does not depend upon the nature of coil . |
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| 1120. |
Silver crystallizes in face-centered cubic unit cell. Each side of this unit cell has a length of 400 Pm· Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four comer atoms.) |
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Answer» Here, a = 400 pm Radius for face-centred cubic unit cell is `r=a/(2sqrt2)=400/(2sqrt2)=400/(2xx1.41)=400/2.82=141.84` |
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| 1121. |
Arrange the following polymers in increasing order of their intermolecular forces : (i) Nylon 6 6, Buna-S, Polythene. (ii) Nylon 6, Neoprene, Polyvinyl chloride. |
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Answer» In order of increasing intermolecular forces : (i) Buna-S, Polythene, Nylon 6, 6, (ii) Neoprene, Polyvinyl chloride, Nylon 6 |
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| 1122. |
Draw the structures of the monomers of the following polymers: (i) Polythene (ii) PVC (iii) Teflon |
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Answer» (i) Etheve `CH_(2) = CH_(2)` (ii) Vinyl Choride `underset(Cl)underset(|)(CH_(2) - CH)` (iii) Tetra fluoro ethene `CF_(2) = CF_(2)` |
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| 1123. |
Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation. (b) For the cell reaction `Ni(s)|Ni^(2+)(aq)"||"Ag^(+)(aq)|Ag(s)` Calculate the equilibrium constant at `25^@C`. How maximum work would be obtained by operation of this cell ? `E_((Ni^(2+)//Ni))^(@)=-0.25V and E_((Ag^(+)//Ag))^(@)=0.80V` |
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Answer» Dissociation is a process in which molecules. Separate or split into smaller particles such as atoms. Degreed dissociation is the fraction of mole of reactant that underwent dissociation. `alpha=sqrt((K)/(C))` degree of dissociation of a weak electrolyte is propotional to the inverse square root of the concentration or the square root of the dilution. the concentration of any one ionic species is given by the root of the product of the dissociation constant and the concentration of the electrolyte. (b) `At Anode to Nito Ni^(2+)+2e^(-)` At Cathode `to [Ag^(+)+e^(-)to Ag]xx2` Cell reaction `-Ni+2Ag^(+) to Ni^(2+)+2Ag` `E_("Cell")^(@)=E_("Cathode")^(@)-E_("Anode ")^(@)=E_(Ag//Ag)^(@)-E_(Ni^(2+)//Ni^+)^(@) = 80V -(-0.250)` `E_("cell")^(@)=1.05V` `E_("cell")=E_("cell")^(@)-(0.059)/(n) log.([Ni^(2+)])/[Ag^+]^2` `n=2, E_("cell")^(@) =1.05V`, `[Ni^(2+)]=0.1M`, `[Ag^(+)]=1.0M` `^E.cell=1.06V-(0.059)/(2)log.((0.1))/((1^2))` `=1.05+0.0295V=1.0795V` `E_("cell")^(@)(0.0591)/(2)log KC` `1.05=(0.0591)/(2)logKC` `log KC = (1.05xx2)/(0.0591)=35.553` `KC=Anti (35.533) =3.412xx10^(35)` `DeltaG=-nFE^@=-2xx96500xx1.05=-202650J//Mol`. |
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| 1124. |
What type of stoichiometric defecit is shoen by ZnS ? |
| Answer» Frenkel defect is shown by ZnS, | |
| 1125. |
A modulating signal is a square wave as shown in figure. The carrier wave is given by `c(t)=2 sin(8pit)"volt"`. The modulation index is |
| Answer» `mu=(A_(m))/(A_(c))=(1)/(2)=0.5.` | |
| 1126. |
State the reason, why GaAs is most commonly used in making of a solar cell. |
| Answer» Ga As (gallium arsenide) is most commonly used in making of a solar cell because it absorbs relatively more energy from the incident solar radiations being of relatively higher absorption coefficient. | |
| 1127. |
Give reasons for the following : (a) Leather gets hardened after tannig ( b) `Fe Cl_3` is preferred over KCl in case of a cut leading to bleeding . |
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Answer» Animal skin is colloida in nature and has positively charged partices ,whereas taninng consits of negatively charged colloidal particles .When animal skin is soaked in tanning mutual coagulation takes palce and as a result,leather gets hardened . ( b) This is because the valency of iron in ferric chloric is 3+ whereas that of potassium in KCL is 1+ The higer the better is the coagulation efficieny of the compond .Also ferric chloride is more compatible with skin than KCl. ( c) At high pressure `1/n=0` ,hence extent of adsorption becomes of pressure . |
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| 1128. |
Chromium metal can be plated out from an acidic solution containing `CrO_(3)` according to the following equation `:` `CrO_(3)(aq)+6H^(o+)+6H^(o+)(aq)+6e^(-) rarr Cr(s)+3H_(2)O` `a.` How many grams of chromium will be plated out by `24000C` ? `b.` How long will take to plate out `1.5g` of chromium by using `12.5 A` current ? |
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Answer» `6XX96500` coulombs deposite `Cr=(52xx24000)/(6xx 96500 )g = 2.1553 g ` (ii) 52 g of Cr is deposit by electricity`=6xx 96500 C` 1.5 g required electricity `=(6xx 96500)/(52)xx1.5 C = 16071 C` Time for which the current is required to be passed`=(16071.9C)/(12.5 A)=1336 S` |
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| 1129. |
Calculate `Delta ,G^@` and log `K_c` for the following reation: `Cd^(+2)(aq)+Zn^(2+)(aq)+Cd(s)` Given : `E_(cd^(2+)//cd)^(0)=0.403 V` `E_(Zn^2+//Zn)^(0)=0.763 V` |
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Answer» `E^@ "cell"= E^@C-E^@ a ` `=-0.403-(0.763)=0.360 V ` `log K_c=(nE_("cell"^(@))/(0.059))=((2xx 0.360)/(0.059))=((0.720)/(0.059))=12.20` (ii) `Delta G^@=-nFE_"cell"^(@)=-2xx96500xx0.360=-69480` J |
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| 1130. |
Chromium crystallises in bcc structure .If its edge length is 300 p m , find its density .Atomic mass of chromium is 52u .`[N_A=6.022 xx 10^(23) cm ,M = 52 _u ` |
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Answer» `a= 300 pm = 300 xx 10^(-10) cm ,M = 53 _u` For bcc Z= 2 `d=(ZM)/(a^3Na)=(2xx52)/((300xx10^(-10))^3xx6.023 xx 10^(23)) =(104)/(2.7 xx 6.022 )=6.396 g//cm^(3)` |
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| 1131. |
Give the stuctures of monomers of the following polymes : ( a) Nylon -6,6 (b) Buna -s |
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Answer» (a) Nyslon-6,6 : See Q .25 ( iii) Paper - 2013 Delhi Board ,Set-I [ page 188] (b) Buna - S: See Q. 25 ( i) ,Paper - 2013 Delhi Board ,Set-I [ page 188] |
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| 1132. |
Use the data to answer the following and also justify giving rason: `{:(," "Cr," "Mn," "Fe," "Co),(E_(M^(2+)//M)^0,-0.91,-1.18,-0.44,-0.28),(E_(M^(2+)//M^(2+))^0,-0.41,+1.57,+0.77,+1.97):}` ( a) Which is a stronger redusing agent in aqueous medium `Cr^(2) or Fe^(2+)` and why ? ( b) Which is the most stable ion in +2 oxidartion and why ? |
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Answer» `Cr^2` is stronger reducting agent then `fe^(2+)` because `E^(@)_(Cx^(2+)//Cx^(2+)` is -ve (-0.14 V) whereas `E^(@)_(Fe^3+//Fe^(2+))` is + Ve (+0.77 V). Thus `Cr^2+` is easily oxidized to `Fe^(3+)` but `Fe^(2+)` cannot be easily oxidized to `Fe^(3+)` .Hence `Cr^2` is stronger reducting agent than `Fe^(2+)` ( b ) `E^@` value for Mn(-118) is more negative than expected from gerneral trend due to extra stability of half filled `Mn^(++)`ion |
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| 1133. |
(a) Under what conditions is the phenomenon of total internal reflection between the critical angle of incidence and the refractive index of the medium. (b) Three lenses of focal lengths+10 cm, -10 cm and + 30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination. |
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Answer» `"For Lens "L_(1)` `1/f=1/v-1/u,` `1/v=(30-10)/300=20/300=1/15` `For Lens"L_(2)` `u=(+15-5)=10cm,` `1/v=0,` `"For Lens"L_(3)` `u=oo` `f=+30cm,` `1/v=1/30,` Final Image will be +30 cm from `L_(3)`. `u=-30xm` `110=1/v-1/(-30),` `v=15 cm` `f=-10cm,` `v=oo` `1/f=1/v-1/u,` `v=+30 cm` f=+10cm `1/10-1/30-1/v` 1/f=1/v-1/10` `1/(+30)=1/v-1/oo` |
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| 1134. |
The velocity of a certain monochromatic light, in a given transparent medium is `2.25xx 10^(8)m//s`. What is the (a) critical angle of incidence, (b) polarising angle for this medium? |
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Answer» `"Here "v=2.25xx10^(8)m//s` Speed of light in vaccum `c=3xx10^(8)m//s` `" "mu=(c)/(v)=(3xx10^(8))/(2.55xx10^(8))underset(8x to 0)lim=1.33` (a) If C is the critical angle, then `mu=(1)/(Sin C)` `" "1.33=(1)/(Sin C)" ""Sin C"=(1)/(1.33)=0.75` (b) If `beta` is the polarising angle then, `tan i_(beta)=mu` `tan i_(beta)=1.33=tan53.06^(@) Rightarrow i_(beta)=53.06^(@)` |
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| 1135. |
The energy level diagram of an element is given below. Identify, by doing necessary calculations , which transition corresponds of the emission of a spectral line of wavelength 102.7 nm. |
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Answer» `lambda=102.7 nm = 102.7xx10^(-9)m` The energy of the emitted photons is, `E=(hc)/(lambda)=(6.626xx10^(-34)xx3xx10^(8))/(102.7xx10^(-9))=(19.878xx10^(-26))/(102.7xx10^(-9))=1.9355xx10^(-18)J` `:.` Energy corresponds `=(1.9355xx10^(-18))/(1.6xx10^(-19))` `eV=12.097 ev ~ 12.1 eV`. |
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| 1136. |
An inductor 200 mH, capacitor `500 mu F`, resistor 10 ohm are connected in series with a 100 V, variable frequency a.c. source. Calculate (i) frequency at which power factor of the circuit is unity (ii) current amplitude at this frequency (iii) Q factor. |
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Answer» Here `L=200mH = 200xx10^(-3)=0.2H` `C=500 mu F = 500 xx 10^(-6) = 5xx10^(-4)F` `R=10Omega and F_(V)=100V` (i) Power factor, `cos phi =1 ` (given) `(R)/(Z)=1 " " Z=R` `sqrt(R^(2)+(X_(L)-K_(C))^(2))=R` `R^(2)+(X_(L)-X_(C))^(2)=R^(2) rArr X_(L)-X_(C)=0` `rArr X_(L)=X_(C) rArr 2 pi v L = (1)/(2pi v C)` `rArr 4pi^(2)v^(2)LC=1` `:. v=(1)/(2pi sqrt(LC))=(1)/(2xx3.14sqrt(0.2xx5xx10^(-4)))=(1)/(2xx3.14xx10^(-2))=(100)/(6.28)=15.92 ~~ 16Hz`. (ii) Current amplitude at resonance `I_(0)=(E_(0))/(Z)=(sqrt(2 in_(rms)))/(R)=(1.414xx100)/(10)=14.14A` (iii) Q-factor `=(1)/(R) sqrt((L)/(C))=(1)/(10)sqrt((0.2)/(5xx10^(-4)))` `=(1)/(10)sqrt((2)/(50xx10^(-4)))=(1)/(10) sqrt((1)/(25xx10^(-4)))=(1)/(10xx5xx10^(-2))=(10^(2))/(50)=(100)/(50)=2`. |
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| 1137. |
An air cored solenoid is of length 0.3 m, area of cross section `1.2 xx 10^(-3) m^(2)` and has 2500 turns. Around its central section, a coil of 350 turns is wound. The solenoid and the coil are electrically insulated from eachother. Calculate the e.m.f. induced in the coil if the initial current of 3 A in the solenoid is reversed in 0.25 s. |
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Answer» `N_(1)=2500, " " N_(2)=350 " " A=1.2xx10^(-3)m^(2)` `1=0.3m. " " dI=3-(-3)=3+3=6A " " dt=0.25s` Since mutual inductance, `M=(mu_(0)N_(1)N_(2)A)/(l)=(4pi xx10^(-7)xx2500xx350xx1.2xx10^(-3))/(0.3)` `=(4xx3.14xx1.05xx10^(-4))/(0.3)=43.96xx10^(-4)=4.39xx10^(-3)H` Induced emf `|E|=M(dI)/(dt)=(4.39xx10^(-3)xx6)/(0.25)=105.36xx10^(-3)=0.10536 V`. |
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| 1138. |
A battery of emf `12V` and internal resistance `2 Omega` is connected two a `4 Omega` resistor. Show that the a voltmeter when placed across cell and across the resistor in turn given the same reading |
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Answer» Total resistance `=2+4=6Omega` `(a) 1=(E)/(R )=(12)/(6)=2A` `V=E-Ir=12-2xx2=*"Volt"` `"Voltmeter reading"=8V` `"Ammeter reading"=2A` Now when voltmeter is put parallel to `4Omega` resistance, potential drop across `4Omega` resistance is `4xx2=8V`. So, voltmeter will show `8V`. `(b)` Voltmeter is placed parallel and ammeter in series in the circuit because ammeter is a very low resistance device and voltmeter has very high resistance. |
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| 1139. |
Given a uniform electric field `vecE=5xx10^(3)hatiN//C`, find the flux of this field through a square of 10cm on a side whose plane is parallel to the y - z plane. What would be the flux through the same square if the plane makes a `30^(@)` angle with the x-axis? |
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Answer» When the plane is parallel to the y-z plane : Electric flux, `phi=EA` Here `vecE=5xx10^(3) hati N//CA=10cm^(2) hati=10^(-2) hati m^(2),phi=5xx10^(3)hati. 10^(-2)hati rArr phi=50` weber. When the plane makes a `30^(@)` angle with the x-axis, the area vector makes `60^(@)` with the x-axis, `phi=E.A rArr phi=E A cos theta` `rArr" "phi=5xx10^(3).10^(-2)cos 60^(@)` `rArr" "phi=50xx(1)/(2)rArr phi=25Wb.` |
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| 1140. |
What happens to the width of depletion layer of a p-n junction when it is (i) forward biased, (ii) reverse biased? |
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Answer» (i) In forward biased: It decreases. (ii) In reverse biased: It increases. |
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| 1141. |
The radius of innermost electron orbit of a hydrogen atom is `5.3xx10^(-11)m`. What is the radius of orbit in second excited state? |
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Answer» Radius of nth orbit of hydrogen atom `r=n^2xx 5.3xx 10^-11m` `therefore` Radius of second state ` n=3)` is orbit of hydrogen atom ` r=(3)^2xx5.3xx10 ^-11 m` `9xx5.3 xx10^-11 m= 4.77 xx10^-11 m`. |
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| 1142. |
On what factors, does the maximum range of ground wave propagation depend? |
| Answer» When the waves propagates near to the surface, the waves glide over the surface of the carth, they are called ground. Waves. The maximum range of coverage depends on the transmitted power and frequency. | |
| 1143. |
Given below is a list of six micro- organisms . State their usefulness to humans. (a) Nuclelolyedrovirus ( b) Sachharomycess cerevisiae ( c) Monascus purpureus ( d) Trichoerma polysporum ( E) Penicllium notatum ( F) Prpopionibacterium sharmanii |
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Answer» Monascus purpures : used for producing statins for lowering blood chlestrol ( D) Trichoderma polysporum : used for producing cyclosporin A used as an immunosppessive agent in organ transplant patients . ( E) Penicillium notatum : Used for prducing antibiotic penicillin . |
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| 1144. |
Name a free-living and symbiotic bacterium that serve as bio-fertilizer. Why are they so called ? |
| Answer» A free living bacterium is Azotobacter and symbiotic bacterium is Rhizobuim. They are called biofertilisers because these organisms increase the nutrient quantity of soil. | |
| 1145. |
(i) Name a substance which can be used as an antiseptic as well as disinfectant. (ii) name an artificial sweetener whose use in limited to cold foods and drinks. (iii) What are contionic detergents ? |
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Answer» (i) `0.2%` solution of phenol acts as antiseptic whereas `1%` solution acts as an disinfectant. (ii) Aspartame is artificial sweetner whose use is limited to cold food and drinks. (iii) Cationic detergents : Those detergents in which major part of their moleculaes are cations and cationic part involves in their cleansing action. Ex. Quaternary ammonium salts of chloride, bromides, acetates etc. `[{:(" "CH_(3)),(" "+),(" "|),(CH_(3)(CH_(2))_(15)-N-CH_(3)),(" "|),(" "CH_(3)):}]Br^(-)` Cetyl trimethyl ammonium bromide |
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| 1146. |
Write the IUPAC name of the following compound: |
| Answer» 3 Phenylprop-2-en-1-ol. | |
| 1147. |
Write the formula of the compound of sulphur which is obtained when conc. `HNO_(3)` oxidises `S_(8)`. |
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Answer» Sulphuric acid `(H_(2)SO_(4))` is formed `S_(8)+48HNO_(3)to8H_(2)SO_(4)+48NO_(2)+16H_(2)O` |
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| 1148. |
Write the formula of the compound of phosphorus which is obtained when conc. `HNO_(3)` oxidises `P_(4)`. |
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Answer» Orthophosphoric acid `(H_(3)PO_(4))` `underset((conc.))(P_(4)+2OHNO_(3))to2ONO_(2)+4H_(3)PO_(4)+4H_(2)O`. |
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| 1149. |
A solution prepared by dissolving 8.95 mg of a given fragment in 35.0 mL of has an osmotic pressure of 0.335 torr at `25^(@)C`. Assuming that the given fragment is non-electolyty. Calculate its molar mass. |
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Answer» `omega = 8.95 mg = 8.95xx 10^(-4) = 0.0895g`. `T = 25^(@)C = 25 + 273 = 289K` `V = 35.0mL = (35)/(1000)L,` `pi = 0.335 "Torr" = (0.335)/(760)"atm"` Molar Mass `M = (omegaRT)/(piV)` `= (8.95 xx 10^(-3) xx 0.0821 xx 298)/((0.335)/(760)xx(35)/(1000))=14193 g//mol^(-1)` |
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| 1150. |
Community service department of your school plans a visit to a slum area near the school with an objective to educate the slum dwellers with respect to health and hygiene. (a) why is there a need to organise such visit ? (b) Write steps you will higlight, as a member of this department in your interaction with them to enable them to lead a healthy life. |
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Answer» (a) There is a need to improve the health condition of the residents so that they can combat fatal diseases. (b) the following steps should be taken to improve the hygienic condition of the area. (i) We will highlight the need for safe disposal of garbage. (ii) will provide some basic tips regarding hygiene. (iii) We will advise not to litter the road and to check over flow of sewage. |
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