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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.
| 1. |
With which halogen the reactions of alkanes are explosive ?A. FluorineB. ChlorineC. BromineD. Iodine |
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Answer» Correct Answer - A Reaction of alkanes with fluorine are explosive. This is an explosive peocess which is difficult to control, so carried out in dark. In this process, replacement of one or more hydrogen of alkane by halogen takes place. For this reaction the order of reactivity of halogen is `l_(2)ltBr_(2)ltCl_(2)ltF_(2)`. |
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| 2. |
Excess of ammonia with sodium hypochlorite solution in the presense of gule or gelatin givesA. `NaBH_(2)`B. `NH_(2)NH_(2)`C. `N_(2)`D. `NH_(4)Cl` |
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Answer» Correct Answer - B Excess of ammonia `(NH_(3))` when reacts with sodium hypochlorite solution (NaOCl) in presence of glue or gelatin gives hydrazine `(NH_(2)*NH_(2))`. This process is known as Reschig process `2NH_(3)+NaOC l toN_(2)H_(4)+NaCl+H_(2)O` Addition of gelatin or glue suppress the side reaction by complexing with the metal ions and yield of hydrazine is increased. |
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| 3. |
Which of the following polymers is used to manufacture clothes for firefighters ?A. ThiokolB. KevlarC. NomexD. Dynel |
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Answer» Correct Answer - C Nomex is a flame - resistant polymer and it is used to manufacture clothes for fire fighters. It is prepared by condensation of m-phenylenediamine and isopthalamide monomers. `*` Thiokol is obtained by the condensation copolymerisation of ethylene dichloride and sodium tetrasulphide `(Na_(2)S_(4))`. It is used as binder in solid rocket fuel. `*` Kevlar is obtained by condensation copolymerisation of terephthalic acid and diaminobenzene. It is used for making bullet proof vests. `*` Dynel is a copolymer of acrylonitrile and vinyl chloride monomers. it is used for making synthetic hair wigs. |
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| 4. |
What is the density of solution of sulphuric acid used as an electrolyte in lead accumulator ?A. `1.5gmL^(-1)`B. `1.2gmL^(-1)`C. `1.8gmL^(-1)`D. `2.0gmL^(-1)` |
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Answer» Correct Answer - B An aqueous solution of sulphuric acid i.e. `38%` by weight or having a density of `1.30gmL^(-1)` is used as an electrolyte in lead accumulators. It is consumed during discharging and when the concentration of `H_(2)SO_(4)` falls to `1.20gmL^(-1)`, it requires recgarging. Thus, the most suitable option is (b). |
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| 5. |
The most basic hydroxide from following isA. `Pr(OH)_(3)(Z=59)`B. `Sm(OH)_(3)(Z=62)`C. `Ho(OH)_(3)(Z=67)`D. `La(OH)_(3)(Z=57)` |
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Answer» Correct Answer - D Due to lanthanoid contraction, the basic character of hydroxides decreases from `La^(3+)` to `Lu^(3+)` As a result, covalent nature of Ln - OH bond also increases. Thus the most basic hydroxide among the gicen option is `La(OH)_(3)`. |
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| 6. |
Identify the oxidation states of titinium (Z=22) and copper (Z=29) in their colourless compounds.A. `Ti^(3+),Cu^(2+)`B. `Ti^(2+),Cu^(2+)`C. `Ti^(4+),Cu^(+)`D. `Ti^(4+),Cu^(2+)` |
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Answer» Correct Answer - C Electronic configuration of titanium (Ti) (Z=22) is `[Ar]_(18)3d^(2)4s^(2)`. In case of `Ti^(4+)`, the electronic configuration will be `[Ar]_(18)3d^(0)4s^(0)` Similarly, electronic configuration of copper (Cu) (Z=29) is `[Ar]_(18)3d^(10)4s^(1)`. On loosing one electron it acquires stable electronic configuration, i.e `[Ar]_(18)3d^(10)`. In both `Ti^(4+)` and `Cu^(+)` no unpaired electron is present. As a result, no d-d transition occur and therefore these are colourless in nature. |
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| 7. |
What is the catalyst used for oxidation of `SO_(2)` to `SO_(3)` in lead chamber process for manufacture of sulphuric acid ?A. Nitric oxideB. Nitrous oxideC. Potassium iodideD. Dilute HCl |
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Answer» Correct Answer - A In lead chambar process, a mixture containning `SO_(2)`, air NO is treated with steam `(H_(2)O)`. In this reaction, NO (nitric oxide) act as a catalyst. `2SO_(2)^(+)underset("(From air)")(O_(2))+2H_(2)O+underset("Catalyst")([NO])to2H_(2)SO_(4)+underset("Catalyst")([NO])` |
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| 8. |
In which substance does nitrogen exhibit the lowest oxidation state ?A. Nitrogen gasB. AmmoniaC. Nitrous oxideD. Nitric oxide |
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Answer» Correct Answer - B Oxidation state of nitrogen gas `(N_(2))` is zero, as both the nitrogen atom possess same electronegativity as well as same number of electrons. The oxidation states of ammonia. Nitrous oxide and nitric oxide are given below Ammonia `(NH_(3))` `x=3(1)=0` `x=-3` Nitrous oxide `(N_(2)O)` `2(x)+1(-2)=0` `2x=2` `x=+1` Nitric oxide (NO) `x+(-2)=0` `x=+2` Thus, nitrogen exhibits lowest oxidation state in case of ammonia |
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| 9. |
Which of the following compounds reacts immediately with Lucas reagent ?A. `CH_(3)CH_(2)OH`B. `CH_(3)CH_(2)CH_(2)OH`C. `CH_(3)-underset(OH)underset(|)(CH)-CH_(3)`D. `CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)(C)-CH_(3)` |
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Answer» Correct Answer - D Lucas reagent is used to distinguish primary secondary or tertiary alcohols. Alcohols are soluble in Lucas reagent while their halides are immiscible and produces tubidity in solution. In case of tertiary alcohols, turbidity is produced immediately as they from the halides easily Primary alcohols do not produce turbidity at room temperature. (c) `underset((2^(@)))underset()(CH_(3)-overset(OH)overset(|)(CH)-CH_(3))underset(anhy.ZnCl_(2))overset(Conc.HCl+)tounderset("Turbidity appears in 5 min")(H_(3)C-overset(Cl)overset(|)(CH)-CH_(3))` (d)`underset((3^(@)))(H_(3)C-underset(O)underset(|)overset(CH_(3))overset(|)(C)-CH_(3))underset("anhy.ZnCl"_(2) "(Lucas reagent)")overset("Conc.HCl") to underset("immediately")underset("Turbidity appears")(H_(3)C-underset(Cl)underset(|)overset(CH_(3))overset(|)(C)-CH_(3))+H_(2)O` Thus, option(d) is correct |
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| 10. |
Which among the following metals crystallise as a simple cube ?A. PoloniumB. IronC. CopperD. Gold |
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Answer» Correct Answer - A Polonium is a radiactive element having two metallic allotropic forms , `alpha-" form "andbeta-` form. Out of these two forms, `alpha` - Po form crystallises as simple cubic crystal. Copper and gold crystallises in face centred cubic crystal and iron crystallises in body centred cubic crystal and iron crystallises in body centred cubic crystal . |
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| 11. |
What is the oxidation number of gold in the complex `[AuCl_(4)]^(1-)` ?A. `+4`B. `+3`C. `+2`D. `+1` |
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Answer» Correct Answer - B Let x be the oxidation number of gold (Au) in the complex `[AuCl_(4)]^(-1)` `:." "x+4(-1)=-1` `rArr" "x=-1+4` `rArr" "x=+3` Hence, the oxidation number of gold is `+3` |
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| 12. |
Which among the following oxoacids of phosphorus shows a tendency of disproportaonation ?A. Phosphinic acid `(H_(3)PO_(2))`B. Orthophosphoric acid `(H_(3)PO_(4))`C. Phosphonic acid `(H_(3)PO_(3))`D. Pyrophosphoric acid `(H_(4)P_(2)O_(7))` |
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Answer» Correct Answer - C Phosphonic acid `(H_(3)PO_(3))`, oxoacid of Phosphorus shows a tendency of disproportionation. On heating `H_(3)PO_(3)` disproportionates to give orthophosphoric acid and phoosphene gas. `4H_(3)PO_(3)overset(Delta)to3H_(3)PO_(4)+PH_(3)` |
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| 13. |
A transistor is used as a common emitter amplifier with load resistance `2kOmega`. The changed by `20muA` which results in a change in collector current by 1.5 mA. The voltage gain of the amplifier isA. 900B. 1000C. 1100D. 1200 |
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Answer» Correct Answer - B Given, load resistance `(R_(e))=2kOmega=2xx10^(3)Omega` Input resistance `(R_(i))=150Omega` Change in collector current `(Deltal_(c))=1.5mA` `=1.5xx10^(-3)A` Change in base current `(Deltal_(B))=20xx10^(-6)A` We know that, voltage gain `=(V_(o))/(v_(i))=(R_(o)xxDeltal_(c))/(R_(i)xxDeltal_(B))` `=(2xx10^(-3)xx1.5xx10^(-3))/(150xx20xx10^(-6))=(3xx10^(6))/(150xx20)` `=1000` |
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| 14. |
A mass attached to one end of a string crosses top - most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be (where, g=gravitational acceleration)A. glancing angleB. 3gC. 4gD. 6g |
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Answer» Correct Answer - B We know that, velocity of particle at top - most point on vertical circle, `(v_(top))=sqrt(3rg)` . . . (i) But centripetal acceleration `(a_(c))=(v^(2))/(r)` `:." "a_(c)(sqrt(3rg)^(2))/(r)" "` [from Eq. (i)] `a_(c)=(3rg)/(r)` `a_(c)=3g` |
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| 15. |
The line `5x+y-1=0` coincides with one of the lines given by `5x^(2)+xy-kx-2y+2=0`, then the value of k isA. `-11`B. 31C. 11D. `-31` |
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Answer» Correct Answer - C Given, 5x+y-1=0 coincides with one of the lines given by `5x^(2)+xy-kx-2y+2=0` On putting the value of `y=-5x+1` in the above equation, we get `5x^(2)+x(-5x+1)-(kx-2(-5x+1)+2=0` ltbrlt`rArr" "5x^(2)-5x^(2)+x-kx+10x-2+2=0` `rArrx-kx+10x=0` `rArr11x-kx=0` `rArr11-k=0` `k=11` |
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| 16. |
The angle made by incident ray of light with the reflecting surface is calledA. glancing angleB. angle of incidenceC. angle of deviationD. angle of refraction |
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Answer» Correct Answer - A The angle made by incident ray of light with the reflecting surface is called grancing angle. |
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| 17. |
In a non - uniform circular motaion the ratio of tangential to radial acceleration is (where, r= radius of circle, v= speed of the particle, `alpha=` angular acceleration)A. `(alpha^(2)r^(2))/(v)`B. `(alpha^(2)r)/(v^(2))`C. `(alphar^(2))/(v^(2))`D. `(v^(2))/(r^(2)alpha)` |
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Answer» Correct Answer - C Given, radius of circle = r Speed of particel = v Angular acceleration = `alpha` We known that, tangential acceleration `=alphar` . .. (i) Radial acceleration `(v^(2))/(r)` . . .(ii) On dividing Eq. (i) by Eq. (ii), we get `("Tangential acceleration")/("Radial acceleration")=(alphar)/(v^(2))xxr(alphar^(2))/(v^(2))` |
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| 18. |
In amplitude modulationA. amplitude remains constant but frequency changesB. both amplitude and frequency do not changeC. both amplitude and frequency changeD. amplitude of the carrier wave changes according to the information single |
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Answer» Correct Answer - D In amplitute modulation, amplitude of the carrier wave changes according to the information signal. |
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| 19. |
If numerical aperture of a microscope is increased, then itsA. resolving power remanis constantB. resolving power becomes zeroC. limit of resolution is decreasedD. limit of resolution is increased |
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Answer» Correct Answer - C Resolving limit of microscope is given by `RL=(lamda)/(2musinalpha)` where, `musinalpha` is called numerical aperture [NA] and `lamda` = wavelenghth of light. `:.RL=(lamda)/(2(NA))` Hence, if numerical aperture of a microscope is increased, then its limit of resolution is decreased. |
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| 20. |
Let `x={:[(a^(2)b^(2))/(c)]:}` be the physical quantity. If the percentage error in the measurement of physical quantities a,b, and c is 2,3 and 4 per cent respectively, then percentage error in the measurement of x isA. `7%`B. `14%`C. `21%`D. `28%` |
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Answer» Correct Answer - B Given `x=(a^(2)b^(2))/(c)` `(Deltax)/(x)=2(Deltaa)/(a)+2(Deltab)/(b)+(Deltac)/(c)` `(Deltax)/(x)xx100=2((Deltaa)/(a)xx100)+2((Deltab)/(b)xxb)+((Deltac)/(c)xxc)` Here, percentage error in a=2 Percentage error in b=3 Percentage error in c=4 `((Deltax)/(x)xx100)=2xx2+2xx3+4` `=4+6+4=14%` |
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| 21. |
In compound microscope, the focal ,ength and aperture of the objective used isrespectivelyA. large and largeB. large and smallC. short and largeD. short and small |
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Answer» Correct Answer - D In compound microscope, the focal length and aperture of objective used respectively, short and small. |
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| 22. |
When source of sound moves towards a stationary observer, the wavelength of sound received by himA. decreases while frequency increasesB. remains the same, whereas frequency increaseC. increase and frequency also increasesD. decreases while frequency remains the same |
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Answer» Correct Answer - A When source of sound moving towards to stationary observer. Apparent frequency `(n_(a))=n[(v)/(v-v_(s))]` Hence, the wavelength of sound received by him decreases while frequency increases. |
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| 23. |
A certain reaction ocuurs in two steps as (I) `2SO_(2(g))+2NO_(2(g)) to 2SO_(3(g))+2NO_((g))` `(ii) 2NO_((g) ) +O_(2) (g) to 2NO_(2(g))` In the reaction ,______.A. `NO_(2)(g)` is intermediateB. `NO(g)` is intermediateC. `NO(g)` is cataystD. `O_(2)(g)` is intermediate |
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Answer» Correct Answer - B Given Step I : `2SO_(2)(g)+2NO_(2)(g)to2SO_(3)(g)[email protected](g)` Step II : `2NO(g)+O_(2)(g)to2NO_(2)(g)` An intermediate formed in the reaction is not a part of overall reaction. Net reaction : `2SO_(2)(g)+O_(2)(g)to2SO_(3)(g)` As, No(g) formed in first elementary step is neither a reactant nor product in the net reaction. It is formed in one elmentary step and consumed in the next step. Thus NO,=(g) act as an intermediate. |
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| 24. |
Identify the element that forms amphoteric oxide.A. CarbonB. ZincC. CalciumD. Sulphur |
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Answer» Correct Answer - B Zn (zinc) element of forms amphoteric oxide i.e. ZnO. This oxide reacts with both acids and bases, exhibiting a dual behaviour. They show characteristics of both acidic as well as basic oxides. Following reactions are given by ZnO `ZnO+2HCl to ZnCl_(2)+H_(2)O` `ZnO+2NaOHtoNa_(2)ZnO_(2)+H_(2)O` Note : Among oxides, Co is neutral, CaO is basic and `SO_(2)` is acidic in nature. |
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| 25. |
DUring galvanisation of iron, which metal is used for coating iron surface ?A. CopperB. ZincC. NickelD. Tin |
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Answer» Correct Answer - B The process of coating iron objects with a thin layer of zinc is called galvanisation. It is done by dipping the object in molten zinc. This methods is used to prevent corrosion on iron objects |
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| 26. |
An alternating electric field, of frequency v, is applied across the dees (radius=R) of a cyclotron that is being used to accelerate protons (mass=m) the operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by:A. `(2pimv)/(e),2pi^(2)mv^(2)R^(2)`B. `(2pi^(2)mv)/(e^(2)),4pi^(2)mv^(2)R^(2)`C. `(pimv)/(e),pi^(2)mv^(2)R^(2)`D. `(2pi^(2)m^(2)v^(2))/(e),2pi^(2)mv^(2)R^(2)` |
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Answer» Correct Answer - A In cyclotron frequency, `v=(eB)/(2pim)` or `B=(2pir*v)/(e)` Also, relation for radius (R) is `R=(mv)/(eB)orv=(eBR)/(m)` `:." "KE=(1)/(2)mv^(2)=(1)/(2)mxx(e^(2)B^(2)R^(2))/(m^(2))=(e^(2)B^(2)R^(2))/(2m)` `(e^(2)R^(2))/(2m)xx(4pi^(2)m^(2)v^(2))/(e^(2))=2pi^(2)mv^(2)R^(2)` |
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| 27. |
Two periodic waves of intensities `I_(1)` and `I_(2)` pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:A. `(l_(1)+l_(2))`B. `2(l_(1)+l_(2))`C. `(sqrt(l_(1))+sqrt(l_(2)))`D. `(sqrt(l_(1))-sqrt(l_(2)))` |
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Answer» Correct Answer - B We know that, `l_(max)=(a_(1)+a_(2))^(2)` . . .(i) `l_(min)=(a_(1)-a_(2))^(2)` . . .(ii) On adding Eqs. (i) and (ii) we, get `l_(max)+l_(min)=(a_(1)+a_(2))^(2)+(a_(1)-a_(2))^(2)` `l_(max)+l_(min)=a_(a)^(2)+a_(2)^(2)+2a_(a)a_(2)+a_(a)^(2)+a_(2)^(2)-2a_(1)a_(2)` `l_(max)+l_(min)=2(a_(1)^(2)+a_(2)^(2))` But `lpropa^(2)` Therefore, `l_(max)+l_(min)=2(l_(1)+l_(2))` |
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| 28. |
A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails, then P(X=1)=A. `(1)/(2)`B. `(2)/(3)`C. `(1)/(6)`D. `(3)/(4)` |
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Answer» Correct Answer - D Given, A coin is tossed three times and X - absolute difference between the number of heads and number of tails. Now, X=1 when exactly two head or two tail comes `:.P(X=1)=(3)/(8)+(3)/(8)=(6)/(8)=(3)/(4)` `{:[(because" probability of exacty two head "=(3)/(8)),("and probability of exactly two tails"=(3)/(8))]:}` |
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| 29. |
`int(1)/(sinx*cos^(2)x)dx=`A. `secx+log|secx+tanx|+C`B. `secx+log|secx+tanx|+C`C. `secx+log|secx-tanx|+C`D. `secx+log|"cosec"x-cotx|+C` |
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Answer» Correct Answer - D We have, `l=int(dx)/(sinx*cos^(2)x)` `rArr" "l=int(sin^(2)x+cos^(2)x)/(sinxcos^(2)x)dx` `rArrl=int(sin^(2)x)/(sinxcos^(2)x)dx+int(cos^(2)x)/(sinxcos^(2)x)dx` `rArrl=intsecxtanxdx+int"cosec"xdx` `rArrl=secx+log|"cosec"x-cotx|+C` |
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| 30. |
If `y=(tan^(-1)x)^2`, then `(x^2+1)^2 (d^2y)/(dx^2) + 2x(x^2+1) dy/dx=`A. 4B. 2C. 1D. 0 |
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Answer» Correct Answer - A We have, `y=(tan^(-1)x)^(2)` On differentiating w.r.t.x, we get `(dy)/(dx)=(2tan^(-1)x)/(1+x^(2))` `rArr(1+x^(2))(dy)/(dx)=2tan^(-1)x` On squaring both sides, we get `(1+x^(2))^(2)((dy)/(dx))^(2)=4(tan^(-1)x)^(2)` `rArr(1+x^(2))^(2)((dy)/(dx))^(2)=4y" "[becausey=(tan^(-1)x)^(2)]` Again, differentiating w.r.t.x, we get `(1+x^(2))^(2)(2(dy)/(dx)*(d^(2)y)/(dx^(2)))+2(1+x^(2))(2x)((dy)/(dx))^(2)=4(dy)/(dx)` On dividing both sides by `2(dy)/(dx)` we get `(1+x^(2))^(2)((d^(2)y)/(dx^(2)))+2x(1+x^(2))(dy)/(dx)=4` |
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| 31. |
If `log_(10)((x^(3)-y^(3))/(x^(3)+y^(3)))=2`, then `(dy)/(dx)=`A. `(x)/(y)`B. `-(y)/(x)`C. `-(x)/(y)`D. `(y)/(x)` |
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Answer» Correct Answer - D Given, `log_(10)((x^(3)-y^(3))/(x^(3)+y^(3)))=2` `rArr" "(x^(3)-y^(3))/(x^(3)+y^(3))=10^(2)=100` `rArrx^(3)-y^(3)=100(x^(3)+y^(3))` `rArr101y^(3)=-99x^(3)` On differentiating both sides w.r.t.x, we get `101xx3y^(2)(dy)/(dx)=-99*(3x^(2))` `rArr101y^(2)(dy)/(dx)=-99x^(2)` On multiplying by x both sides, we get `rArr101xy^(2)(dy)/(dx)=-99x^(3)` `rArr(dy)/(dx)=(-99x^(3))/(101xy^(2))` `rArr(dy)/(dx)=(101y^(3))/(101xy^(2))" "[because-99x^(3)=101y^(3)]` `rArr(dy)/(dx)=(y)/(x)` |
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| 32. |
If `f(x)={{:(x^(2)+alpha," for "xge0),(2sqrt(x^(2)+1)+beta," for "xlt0):}` is continuous at x=0 and `f((1)/(2))=2`, then `alpha^(2)+beta^(2)` isA. 3B. `(8)/(25)`C. `(25)/(8)`D. `(1)/(3)` |
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Answer» Correct Answer - C We have, `f(x)={(x^(2)+alpha",",xge0),(2sqrt(x^(2)+1)+beta",",xlt0):}` is continuous at x=0. `:.underset(xto0^(-))(lim)f(x)=underset(xto0^(+))(lim)f(x)=f(0)` `rArrunderset(xto0^(-))(lim)x^(2)+alpha=underset(xto0^(+))(lim)2sqrt(x^(2)+1)+beta=alpha" "[becausef(0)=alpha]` `rArr" "alpha=2+beta` `rArralpha-beta=2` . . . (i) Given, `f((1)/(2))=2` `f((1)/(2))=((1)/(2))^(2)+alpha" "[because(1)/(2)lt0]` `rArr2=(1)/(4)+alpha` `rArralpha=(7)/(4)` On putting the value of in Eq. (i), we get `beta=(7)/(4)-2=-(1)/(4)` Hence `alpha^(2)+beta^(2)=((7)/(4))^(2)+((-1)/(4))^(2)` `=(49+1)/(16)=(50)/(16)=(25)/(8)` |
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| 33. |
If `f:R-{2}toR` is a function defined by `f(x)=(x^(2)-4)/(x-2)`, then its range isA. RB. `R-{2}`C. `R-{4}`D. `R-{-2,2}` |
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Answer» Correct Answer - C We have, `f(x)=(x^(2)-4)/(x-2),xne2` `f(x)=((x+2)(x-2))/(x-2),xne2` `f(x)=(x+2),xne2` `:.` Range of `f(x)=R={4}` |
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| 34. |
The molarity of urea (molar mass 60 g `mol^(-1)`) solution by dissolving 15 g of urea in 500 `cm^(3)` of water isA. `2"mol dm"^(-3)`B. `0.5"mol dm"^(-3)`C. `0.125"mol dm"^(-3)`D. `0.0005"mol dm"^(-3)` |
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Answer» Correct Answer - B Given, mass of urea = 15 g Molar mass urea `=60gmol^(-1)` volume `=500cm^(3)=500mL=0.5L` `{:[(because1cm^(3)=1mL),(1L=1000mL)]:}` `:.` Molarity `=("Mass of solute in grams")/("Molar mass of solute "(M)xx"Volume of solution V(L)")` Molarity of urea `=(15g)/(60gmol^(-1)xx0.5L)` `rArr" "0.5mol//L=0.5mol//dm^(3)(because1L=1dm^(3))` |
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| 35. |
A series combination of `N_(1)` capacitors (each of capacity `C_(1)`) is charged to potential difference 3V. Another parallel combination of `N_(2)` capacitors (each of capacity `C_(2)`) is charged to potential difference V. The total energy stored in both the combinations is same, The value of `C_(1)` in terms of `C_(2)` isA. `(C_(2)N_(1)N_(2))/(9)`B. `(C_(2)N_(1)^(2)N_(2)^(2))/(9)`C. `(C_(2)N_(1))/(9N_(2))`D. `(C_(2)N_(2))/(9N_(1))` |
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Answer» Correct Answer - A In the first condition, `C_(eq)=(C_(1))/(N_(1))` Potential difference `(V)=3V` `:.` Energy stored `(E_(1))=(1)/(2)CV^(2)` `=(1)/(2)((C_(1))/(N_(1)))(3V)^(2)` `=(9C_(1))/(2N_(1))` . . . (i) In the second condition, `C_(eq)=N_(2)C_(2)`, potential difference =V energy stored `(E_(2))=(1)/(2)CV^(2)` `=(1)/(2)N_(2)C_(2)V^(2)` . . . (ii) According to the question, `E_(1)=E_(2)` From Eqs. (i) and (ii), we get `(9)/(2)(C_(1))/(N_(1))V^(2)(1)/(2)N_(2)C_(2)V^(2)` `C_(1)=C_(2)(N_(2)N_(1))/(9)` |
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| 36. |
The molar specific heat of an ideal gas at constant pressure and constant volume is `C_(p)` and `C_(v)` respectively. If R is the universal gas constant and the ratio of `C_(p)` to `C_(v)` is `gamma`, then `C_(v)`.A. `(1-gamma)/(1+gamma)`B. `(1+gamma)/(1-gamma)`C. `(gamma-1)/(R)`D. `(R)/(gamma-1)` |
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Answer» Correct Answer - D According to Mayer formula, `C_(p)-C_(v)=R` . . .(i) where, `C_(p)` = specific heat at constant pressure, `C_(v)` = specific heat at constant volume and R = gas constant Now , `gamma=(C_(p))/(C_(v))` `rArr" "C_(p)=gammaC_(v)` . . . .(ii) From Eqs. (i) and (ii), we get `rArr" "gammaC_(v)-C_(v)=RrArrC_(v)(gamma-1)=R` `rArr" "C_(v)=(R)/(gamma-1)` |
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| 37. |
Magnetic susceptibility for a paramagnetic and diamagnetic materials is respectively,A. small, positive and small, positiveB. large, positive and small, negativeC. small, positive and small, negativeD. large, negative and large, positive |
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Answer» Correct Answer - C The magnetic susceptibility of a magnetic substance is defined as the ratio of intensity of magnetisation (l) to magnetic intensity (H), i.e. `x=(l)/(H)` For paramagnetic material, x is positive and small `(0ltxltepsilon)`, where is smaal positive number. For diamagnetic meterial, x is negative and small `(-1lexle0)`. |
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| 38. |
A unit vector is represented as `(0.8hat(i)+bhat(j)+04hat(k))`. Hence, the value of b must beA. 0.4B. `sqrt(0.6)`C. 0.2D. `sqrt(0.2)` |
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Answer» Correct Answer - D Given, unit vector `=0.8hat(i)+bhat(j)+0.4hat(k)` Magnitude of vector `(Asqrt(x^(2)+y^(2)+z^(2)))` Here, `x=0.8,y=b,z=0.4` Now, `=sqrt((0.8)^(2)+(b)^(2)+(0.4)^(2))` `rArr1=sqrt(0.64+b^(2)+0.16)` `rArr1=sqrt(0.64+b^(2)+0.16)` `rArr1=0.64+b^(2)+0.16` `rArr1=0.80+b^(2)` `rArrb^(2)=1-0.80` `rArrb^(2)=0.2` `rArrb=sqrt(0.2)` |
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| 39. |
A mass is suspended from a vertica spring which is executing SHM of frequency 5 Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (take, acceleration due to gravity, `g=10m//s^(2)` )A. `2pi m//s`B. `pi m//s`C. `(1)/(2pi)m//s`D. `(1)/(pi)m//s` |
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Answer» Correct Answer - D Given, frequency of SHM (n) =5 Hz Acceleration due to gravity (g) `=10m//s^(2)` WE know that, `T=2pisqrt((m)/(k))` But frequency, `n=(1)/(T)` `or" "n=(1)/(2pi)sqrt((k)/(m))or5=(1)/(2pi)sqrt((k)/(m))` On taking square both sides, we get `25=(1)/(4pi^(2))(k)/(m)` `k=100pi^(2)m` . . . (i) But `kA=mg` `rArrA=(mg)/(k)` . . . (ii) Now, `V_(max)=omegaA` `=(2pi)/(T)xx(mg)/(k)" "` [from Eq. (ii)] `or" "v_(max)=2pinxx(mg)/(k)" "(becausen=(1)/(T))` `v_(max)=2pinxx(mg)/(100pi^(2)m)` [from Eq. (i)] `v_(max)=(nxxg)/(50pi)` `v_(max)=(5xx10)/(50pi)` `v_(max)=(1)/(pi)m//s` |
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| 40. |
A bomb at rest explodes into 3 parts of same mass. The momentum of two parts is `-3phat(i)and2phat(j)`. Respectively. The magnitude of momentum f the third part isA. pB. `sqrt(5)p`C. `sqrt(11)p`D. `sqrt(13)p` |
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Answer» Correct Answer - D According to the question, we get can draw the following diagram. Now, the magnitude of momentum of the third part is given by `p_(3)=sqrt((p_(1)^(2))+(p_(2))^(2))` `p_(3)=sqrt((-3p)^(2)+(2p)^(2))` `p_(3)=sqrt(9p^(2)+4p^(2))` `p_(3)=sqrt(13p^(2))` `:." "p_(3)=sqrt(13)p` |
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| 41. |
If `vec(a),vec(b),vec(c)` are mutually perpendicular vectors having magnitudes 1,2,3 respectively, then `[vec(a)+vec(b)+vec(c)" "vec(b)-vec(a)vec(c)]=` |
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Answer» Correct Answer - C We have, `vec(a),vec(b)` and `vec(c)` are mutually perpendicular vectors. `:.vec(a).vec(b)=vec(b).vec(c)=vec(a).vec(c)=0` and `|vec(a)|=1,|vec(b)|=2, |vec(c)|=3` Now, `[vec(a)+vec(b)+vec(c) vec(b)-vec(a)vec(c)]` `=(vec(a)+vec(b)+vec(c))*((vec(b)-vec(a))xx vec(c))` `=(vec(a)+vec(b)+vec(c))*[(vec(b)xx vec(c))-(vec(a)xx vec(c))]` `=vec(a)*( vec(b) xx vec(c))-vec(a).(vec(a)xx vec(c))+ vec(b)* (vec(b)xx vec(c))-vec(b)*(vec(a)xx vec(c))+ vec(c)* (vec(b)xx vec(c))-vec(c)* (vec(a)xx vec(c))` `=[vec(a) vec(b) vec(c)]- [vec(a) vec(a) vec(c)] + [ vec(b)vec(b)vec(c)]- [ vec(b) vec(a) vec(c)] + [ vec(c) vec(b) vec(c)]- [ vec(c)vec(a)vec(c)]` `= [ vec(a)vec(b)vec(c)]-0+0+[vec(a)vec(b)vec(c)]+0-0` `=2 [ vec(a) vec(b) vec(c)]` `= 2vec(a)* ( vec(b)xx vec(c))=2vec(a)*(|vec(b)||vec(c)| sin. (pi)/(2) n)` `=2vec(a) *(2xx3xx1xx hat(n))=12vec(a)* hat(n)` `= 12 | vec(a)||hat(n)| cos 0^(@)=12xx1xx1xx1=12` Hence, option (c) is correct. |
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| 42. |
The moment of inertia of a ring about an axis passing though the centre and perpendicular to its plane is l. It is rotating ring is gently placed `omega` Another identical ring is gently placed on it, so that their centres coincide. If both the rings are rotating about the same axis, then loss in kinetic energy isA. `(lomega^(2))/(2)`B. `(lomega^(2))/(4)`C. `(lomega^(2))/(6)`D. `(lomega^(2))/(8)` |
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Answer» Correct Answer - B According to the law of conservation of angular momentum, `lomega=` constant Now, according to the question `l_(1)omega_(1)=l_(2)omega_(2)orlomega=(2l)omega_(2)` `omega_(2)=(omega)/(2)` New kinetic energy `=[(1)/(2)l_(2)omega_(2)^(2)]=(1)/(2)(2l)xx((omega)/(2))^(2)=(lomega^(2))/(4)` Loss in kinetic energy `(K_(L))=K_(i)-K_(j)` `=(1)/(2)lomega^(2)-(lomega^(2))/(4)=(lomega^(2))/(4)` |
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| 43. |
The equation of line passing through `(3,-1,2)` and perpendicular to the lines `vec(r)=(hat(i)+hat(j)-hat(k))+lamda(2hat(i)-2hat(j)+hat(k))` and`vec(r)=(2hat(i)+hat(j)-3hat(k))+mu(hat(i)-2hat(j)+2hat(k))` isA. `(x+3)/(2)=(y+1)/(3)=(z-2)/(2)`B. `(x-3)/(3)=(y+1)/(2)=(z-2)/(2)`C. `(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`D. `(x-2)/(2)=(y+1)/(2)=(z-2)/(3)` |
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Answer» Correct Answer - C Direction ration of the line perpendicular to the lines `vec(r)=vec(a_(1))+lamdavec(b_(1))andvec(r)=vec(a_(2))+muvec(b_(2))` is `alpha(vec(b_(1))xxvec(b_(2)))` `:.` Direction ratio of line perpendicular to the lines `vec(r)=(hat(i)+hat(j)-hat(k))+lamda(2hat(i)-2hat(j)+hat(k))` `andvec(r)=(2hat(i)+hat(j)-3hat(k))+mu(hat(i)-2hat(j)+2hat(k))` is `alpha|{:(hat(i),hat(j),hat(k)),(2,-2,1),(1,-2,2):}|` `alpha[(-4+2)hat(i)-(4-1)hat(i)+(-4+2)hat(k)]` `=alpha[-2hat(i)-3hat(j)-2hat(k)]` Now, equation of line passing through (3,-1,2) and parallel to `-2hat(i)-3hat(j)-2hat(k)` is `vec(r)=3hat(i)-hat(j)+2hat(k)+beta(2hat(i)+3hat(j)+2hat(k))` Hence, cartesian from of the above equation is `(x-3)/(2)=(y+1)/(3)=(z-2)/(2)` |
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| 44. |
Letters in the word HULULULU are rearranged. The probability of all three L being together isA. `(3)/(20)`B. `(2)/(5)`C. `(3)/(28)`D. `(5)/(23)` |
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Answer» Correct Answer - C Given words, HULULULU `:.` Total number of samle space `=(8!)/(4!3!)` and total number of ways all three L being together `=(6!)/(4!)` `:.` Required probability `=((6!)/(4!))/((8!)/(4!3!))` `=(6!xx3!)/(8!)=(3)/(28)` |
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| 45. |
A pipe closed at one end has length 83 cm. The number of possible natural oscilations of air column whose frequencies lie below 1000 Hz are (take, velocity of sound in air 332 m/s)A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C Fundamental frequency of one closed pipe, `V=(nv)/(4L)` Here, `n=1` `or" "v_(1)=(332)/(4xx83xx10^(-2))` `or" "v_(1)=100` As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are `3v_(1)=300Hz,5v_(2)=500Hz`, `7v_(2)=700Hz,9v_(2)=900Hz` Hence, the number of possible natural oscillation of air colume is 5. |
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| 46. |
A satelite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. The relation between h and R isA. h=2RB. h=3RC. h=5RD. h=7R |
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Answer» Correct Answer - D According to the question, `v_(c)=(1)/(4)v_(e)` `sqrt((GM)/((R+h)))=(1)/(4)sqrt((2GM)/(R))` On taking square root both sides, we get `rArr" "(GM)/(R+h)=(1)/(16)(2GM)/(R)` `rArr" "(GM)/(R+h)=(GM)/(8R)` `rArr" "R+h=8R` `rArr" "h=8R-R` `rArr" "h=7R` |
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| 47. |
If A,B,C are the angles off `DeltaABC`, then `cotA*cotB+cotB*cotC+cotC*cotA`= |
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Answer» Correct Answer - B Given, `A+B+C=pi` `rArrA+B+pi-C` `rArrcot(A+B)=cot(pi-C)` `rArr(cotAcotB-1)/(cotA+cotB)=-cotC` `rArrcotAcotB-1=-cotAcotC-cotBcotC` `rArrcotAcotB+cotBcotC+cotCcotA=1` |
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| 48. |
An ideal transformer converts 220 V AC to 3.3 kV AC to transmit a power of 4.4 kW. If primary coil has 600 turns, then alternating current in secondary coil isA. `(1)/(3)A`B. `(4)/(3)A`C. `(5)/(3)A`D. `(7)/(3)A` |
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Answer» Correct Answer - B Given, input voltage `(V_(P))=220V` Output voltage `(V_(S))=3.3xx10^(3)V` Power `(P)=4.4kW=4.4xx10^(3)W` Number of turns in primary coil = 600 We know that, For an ideal transformer, `(V_(S))/(V_(P))=(I_(P))/(I_(S))` `I_(S)=I_(P)xx(V_(P))/(V_(S))` `I_(S)=(20xx220)/(3.3xx10^(3))` `I_(S)=(44)/(33)=(4)/(3)A` |
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