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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.
| 1. |
Advantage of optical fibreA. high band width and EM interferenceB. low bnad width and EM interferenceC. high band width low transmission capacity and no EM interferenceD. high band width, high data transmission capacity and no. EM interference |
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Answer» Correct Answer - D Few advantages of optical fibres are that the number of signals carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it. |
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| 2. |
A solid reflects incident light and its electrical conductivity decreases with temperature. The binding in this solidsA. ionicB. covalentC. metallicD. molecular |
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Answer» Correct Answer - C Metals reflect incident light by the vibrations of free electrons under the influence of electric field of incident wave. The conductivity of metals decreases with increase of temperature dut to increases in random motion of free electrons. The bonding is therefore metallic. |
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| 3. |
The equation `12x^(2)+7xy+ay^(2)+13x-y+3=0` represents a pair of perpendicular lines. Then, the value of a isA. `(7)/(2)`B. `-19`C. `-12`D. 12 |
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Answer» Correct Answer - C Comapring the given equation with standard eqution, we get `a=12 and b=a` , for perpendicular lines coefficient of `x^(2)+` coefficient of `y^(2)=0` `:. 12+a=0` `rArr a=-12` |
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| 4. |
Which of the following is the weakest base ?A. Ethyl amineB. AmmoniaC. Dimethyl amineD. Methyl amine |
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Answer» Correct Answer - B Alkyl group (an electron relasing(+I) group) increases electron density, at N-atom hence, basic nature is increases. In ammonia, no alkyl group in present, so it is least basic. |
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| 5. |
An isobar of `._(20)Ca^(40)` isA. `""_(18)Ar^(40)`B. `""_(20)Ca^(38)`C. `""_(20)Ca^(42)`D. `""_(18)Ar^(38)` |
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Answer» Correct Answer - A Isobars have same atomic mass but different atomic number. Thus, the isobar of `""_(20)C^(40)` is `""_(18)Ar^(40)` . |
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| 6. |
Solubility of `Ca(OH)_(2)` is s mol `L^(-1).` The solubility product `(K_(sp))` under the same condition isA. `4s^(3)`B. `3s^(4)`C. `4s^(2)`D. `s^(3)` |
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Answer» Correct Answer - A `Ca(OH) hArr Ca^(2+)+2OH^(-)` `K_(sp)=[Ca^(2+)][OH^(-)]^(2)` `=(s)(2s)^((2)=4s^(2)` |
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| 7. |
A radioactive isotope having `t_(1//2)` =3 days was read after 12 days . If 3 g of the isotope is now left in the container, the initial weight of isotope wasA. 12 gB. 24 gC. 36 gD. 48 g |
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Answer» Correct Answer - D `T=nxxt_(1//2)` `12=nxx3` `n=4` `N=N_(0)xx((1)/(2))^(n)` `3=N_(0)xx((1)/(2))^(4)` `N_(0)=16xx3=48g` |
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| 8. |
A transformer is used to light a `100 W` and `110 V` lamp from a `220V` mains. If the main current is `0.5 A`, the Efficiency of the transformer is approximately:A. `30%`B. `50%`C. `90%`D. `10%` |
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Answer» Correct Answer - C The efficiency of transformer `=("Energy obtained from the secondary coil")/("Energy given to the primary coil")` or `eta=("Output power")/("Input power")` or `eta=(V_(s)I_(s))/(V_(p)I_(p))` Given, `V_(s)I_(s)=100W,V_(p)=220V,I_(p)=0.5A` Hence, `eta=(100)/(220xx0.5)=0.90=90%` Note The efficiency of an ideal transformer is 1 (or 100%). But in practice due to loss in energy, the efficiency of trensformer is always less than 1 (or less than 100%) |
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| 9. |
The pH of a 0.1 M solution of `NH_(4)OH` (having `K_(b)=1.0xx10^(-5))` is equal toA. 10B. 6C. 11D. 12 |
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Answer» Correct Answer - C `[OH^(-)]=sqrt(K_(b)xxC)` `=sqrt(1xx10^(-5)xx10^(-1))` `sqrt(10^(-6))=10^(3)` `K_(w)=[H^(+)][OH^(-)]` `10^(-14)=[H^(+)][OH^(-3)]` `[H^(+)]=10^(-11)` Hence, `pH=- logH^(+)` `=-log(1xx10^(11))=11` |
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| 10. |
The purest zinc is made byA. electrolytic refiningB. zone refiningC. the van Arkel methodD. the Mond process |
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Answer» Correct Answer - B The purest zince is made by zone refinig. |
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| 11. |
In mass spectrometer used for measuring the masses of ions, the ions are initaily accerlerated by an electric potential `V` and then made to describe semicircular paths of radius `R` using a magnetic field `B`.if `V` and `B` are kept constant, the ratio `(("charg e on the ion")/("mass of the ion"))` will be propertional to:A. `1/R`B. `(1)/(R^(2))`C. `R^(2)`D. R |
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Answer» Correct Answer - B The radiue of the orbit in which ions moving is determinet by the relation as given below `(mv^(2))/(R)=qvB` where m is the mass, v is velocity, q is charge of ion and B is the flux density of the magnetic field, so that qvB is the magnetic force acting on the ion, and `(mv^(2))/(R)` is the centripetal force on the ion moving in a curved path of radius R. about the verical field B is given by `omega=(v)/(R)=(qB)/(m)=2piv` where v is frequency. Energy of ion is given by `E=(1)/(2)mv^(2)=(1)/(2)m(Romega)^(2)` `=(1)/(2)mR^(2)B^(2)(q^(2))/(m^(2))` or `E=(1)/(2)(R^(2)B^(2)q^(2))/(m)" "......(i)` Form Eqs. (i) and (ii), we get `qV=(1)/(2)(R^(2)B^(2)q^(2))/(m)` or `(q)/(m)=(2V)/(R^(2)B^(2))` If V and B are kept constant, then `(q)/(m)prop(1)/(R^(2))` |
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| 12. |
Tranquilisers are also known asA. psychosomatic durgsB. psychoterapeutic durgsC. psychosystolic drugsD. None of the above |
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Answer» Correct Answer - B Tranquilisers are the drugs which used to releive mental aliment. These are also known as psychoterapeutic drugs as they act on the central nervous system. |
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| 13. |
The correct order of boiling point for primary `(1^(@)),` secondary `(2^(@))` and tertiary `(3^(@))` alcohols isA. `1^(@)gt2^(@)gt3^(@)`B. `3^(@)gt2^(@)gt1^(@)`C. `2^(@)gt1^(@)gt3^(@)`D. `2^(@)gt3^(@)gt1^(@)` |
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Answer» Correct Answer - A Alcohols with same m olecular weight are expected to have almost boiling point however two more factors other than the molecule weight are important, they are namely H- bonding and surface area of molecule. Both theses factors are least in `3^(@)` alcohol and maximum in `1^(@)` alcohols. Hence, `3^(@)` alcohols have least boiling point while `1^(@)` alcohols have maximum biling point. |
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| 14. |
In the reaction, `2A+` dry silver oxide `overset(Delta)to` ether + 2Ag X A isa/anA. primary alcoholB. acidC. alkyl halideD. alcohol |
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Answer» Correct Answer - C An alkyl halid on heating withd dry silver oxide gives ether. `underset("alky halide")(2R-X)+Ag_(2)Ounderset(dry)overset(Delta)to R-underset("ether")O-R+2AgX` |
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| 15. |
Iodoform test is not given byA. 2-pentanoneB. ethanolC. ethanalD. 3-pentanone |
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Answer» Correct Answer - D Idoform test is given by compounds which have `(CH_(3)-CO-)"group or" (CH_(3)-underset(OH)underset(|)overset(H)overset(|)C-)` group. Hence 2=-pentanone, `CH_(3)CHO and C_(2)H_(5)OH` give this test. But 3- pentanone `(CH_(3)CH_(2)COCH_(2)CH_(3))` does give idoform test. `underset("2-pentanone")(CH_(3)COCH_(2)CH_(2)CH_(3))+3I_(2)+4NaOH to underset("iodoform (yello ppt)")(CHI_(3))darr` `+CH_(3)CH_(2)CH_(2)COONa+3NAI+2H_(2)O` `underset("ethanal")(CH_(3)-CHO)+3I_(2)+4NaOHrarrunderset(("yellow ppt."))underset("iodoform")(CHI_(3))darr+HCOONa+3NaI+3H_(2)O` `underset("ethanal")(C_(2)H_(5)OH)+4I_(2)+6NOaOH to underset("iodoform(yellow ppt)")(CHI_(3)darr)+HCOONa+5NaI+5H_(2)O` `CH_(3)-CH_(2)-underset(3"pentanone")(CO)-CH_(2)-CH_(3)+I_(2)+NaOH to No "reaction"` |
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| 16. |
Which of the following solution will have highest boiling point ?A. 0.1 M `FeCl_(3)`B. `0.1M BaCl_(2)`C. `0.1MNaCl`D. 0.1 M urea `(NH_(2)CONH_(2))` |
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Answer» Correct Answer - A Elevation in boiling point is a colligative propert, ie, depends only on nump or ions 0.1 `FeCI_(3)` gives maximum number of ions, thus has highest bioling point |
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| 17. |
End product of the following reaction is `CH_(3)CH_(2)COOH overset(Cl_(2))underset(red P)to overset("alcoholic KOH")to`A. `CH_(3)underset(OH)underset(|)(CH)COOH`B. `underset(CH)underset(|)(CH_(2))CH_(2)COOH`C. `CH_(2)=CHCOOH`D. `underset(Cl)underset(|)(CH_(2))underset(OH)underset(|)(C)HCOOH` |
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Answer» Correct Answer - C `CH_(3)CH_(2)COOH underset(( "red pHVZ reaction"))overset(CI_(2)) to CH_(3)underset(CI)underset(|)(CHCOOH) underset(("elimination"))overset("Alcoholic KOH") to CH_(2)=underset("acrylic acid")(CHCOOH)` |
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| 18. |
For a reaction, `A+2B to C,` rate is given by `+(d[C])/(dt)=k[A][B],` hence, the order of the reaction isA. 1B. 2C. 1D. 0 |
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Answer» Correct Answer - B Rate `((+dc)/(dt))=K[A][B]` Thus, the order of reaction w.r.t A=1 the order of reaction w.r.t.B=1 Total order of reaction `=1+1=2` |
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| 19. |
If `Delta` E is the heat of reactin for `C_(2)H_(5)OH(l)+3O_(2)(g) to 2CO_(2)(g)+3H_(2)O(l)` at constant volume the `DeltaH` (heat of reaction at constant pressure), at constant temperature isA. `DeltaH=DeltaE+RT`B. `DeltaH=DeltaE-RT`C. `DeltaH=DeltaE-RT`D. `DeltaH=DeltaE-2RT` |
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Answer» Correct Answer - B We know, half, `Delta H= Delta E+ Delta nRT` where, `Deltan` = number of moles of gasesous products - number of moles of gaseous reactants `=2-3=-1` So, `Delta H= Delta E-RT` |
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| 20. |
The IUPAC name of `H_(3)C-underset(OC_(3)H_(7))underset(|)(CH)-C_(3)H_(7)`A. 4-propoxy pentaneB. pentyl-propyl etherC. 2-propoxy pentaneD. 2-pentoxy propane |
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Answer» Correct Answer - C `H_(3)overset(1)C-overset(H)overset(|)underset(2"propoxy pentane")underset(O-CH_(2)-CH_(2)-CH_(3))underset(|)(""^(2)C)-overset(3)(CH_(2))-overset(4)(CH_(2))-overset(5)(CH_(3))` |
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| 21. |
What is the general electronic configuration of transition elementsA. `(n-1)d^(10),(n+1)s^(2)`B. `(n-1)d^(1-10),(n+1)s^(1-2)`C. `(n-1)d^(1-10), np^(6),ns^(2)`D. `(n-1)d^(1-10), ns ^(1-2)` |
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Answer» Correct Answer - D Generally, d-block elements are called transton elements as they contain nner partially filled d-subshell. Thus their general e electronic configuration is `(n-1)d^(1-10), ns^(1-2)` |
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| 22. |
Solution A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 `MBaCl_(2)` and 0.1 M `AlCl_(3).` Which one of the following paris is istonic ?A. A and BB. B and CC. A and DD. A and C |
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Answer» Correct Answer - A Isotoic solutiosn have same molar concentration of solute in slution. Molar concentration of particles in solutiona are `0.1` in glucose, `2xx0.05 M ` in NacI`3xx0.05` in `BaCI_(2) and 4xx0.1` in `AICI_(3)`. Theirfore, `0.1 M` glucose and `0.05` M NaCI solutions are isotonic. |
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| 23. |
What is an ideal gas ? Explain its main characteristics.A. One that consists of moleculesB. A gas satisfying the assumptions of kinetic theoryC. A gas having Maxwellian distribution of speedD. A gas consisting of massless particles |
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Answer» Correct Answer - B An ideal gas is a gas which satisfying the assumptions of the kinetic energy. |
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| 24. |
Chloramine-R is aA. disinfectantB. antiseptic qC. analgesicD. antipyretics |
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Answer» Correct Answer - B Antiseptics drugs cause destruction of micro organism, which produce septic diseases, eg, dettol, savlon acriflavin, boric acid, phenol, iodoform, `KMnO_(4)` and some dyes such as chloramine T methylene blue. |
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| 25. |
What is the value of inductance `L` for which the current is a maximum in series `LCR` circuit with `C=10 muF` and `omega=1000(rad)/s`?A. 100 mHB. 1 mHC. Cannot be calculated unless R is unknowD. 10 mH |
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Answer» Correct Answer - A Current in LCR series circuit, `i=(V)/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))` where V is rms valus of current, R is resstance, `X_(L)` is inductive reactance and `X_(C)` is capacitive reactance. For current to be maximum, denominator should be minimum which can be done, if `X_(L)=X_(C)` This happens in resonance state of the circuit ie, `omegaL=(1)/(omegaC)` or `L=(1)/(omega^(2)C)" "......(i)` Given, `omega=1000s^(-1),C=10muF` `=10xx10^(-6)F` Hence, `L=(1)/((1000)^(2)xx10xx10^(-6))` `=0.1H` `=100mH` |
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| 26. |
The velocity of particle at time t is given by the relation `v=6t-(t^(2))/(6)`. The distance traveled in 3 s is, if s=0 at t=0A. `(39)/(2)`B. `(57)/(2)`C. `(51)/(2)`D. `(33)/(2)` |
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Answer» Correct Answer - C Given `v=(ds)/(dt)=6t-(t^(2))/(6)` On intergrating both side, we get `s=3t^(2)-(t^(3))/(18)+` constant Now, put at t=0 , we get constant =0 `:. S=3t^(2)-(t^(3))/(18)` Now, distance traveled in `3s=3(3)^(2)-((3)^(3))/(18)` `=27-(27)/(18)=(51)/(2)` |
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| 27. |
If `x=(1-t^(2))/(1+t^(2))` and `y=(2t)/(1+t^(2))`, then `(dy)/(dx)` is equal toA. `(a(1-t^2))/(2t)`B. `(a(t^(2)-1))/(2t)`C. `(a(t^(2)+1))/(2t)`D. `(a(t^(2)-1))/(t)` |
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Answer» Correct Answer - B Given, `x=(1-t^(2))/(1+t^(2)) and y=(2at)/(1+t^(2))` On differentiating w.r.t. respectively, we get `(dx)/(dt)=((1+t^(2))(0-2t)-(1-t^(2))(0+2t))/((1+t^(2))^(2))` `=(-4t)/((1+t^(2))^(2))` and `(dy)/(dt)=((1+t^(2))2a-2at(2t))/((1+t^(2))^(2))=(2a(1-t^(2)))/((1+t^(2))^(2))` `:. (dy)/(dx)=(dy//dt)/(dx//dt)=(a(1-t^(2)))/(-2t)` `rArr (dy)/(dx)=(a(t^(2)-1))/(2t)` |
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| 28. |
The solution of the differential equation `(dy)/(dx)=(x-y+3)/(2x(x-y)+5)` isA. `2(x-y)+log(x-y)=x+c`B. `2(x-y)-log(x-y+2)=x+c`C. `2(x-y)+log(x-y+2)=x+c`D. None of the above |
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Answer» Correct Answer - C Given differential equation is `(dy)/(dx)=(x-y+3)/(2(x-y)+5)` Let `x-y= v rArr (dy)/(dx)=1-(dv)/(dx)` `(dv)/(dx)=(v+2)/(2v+5)` `rArr int (2v+5)/(v+2) dv =int dx` `rArr int (2+(1)/(v+2))dv= int dx` `rArr 2v + log (v+2) =x+c` `rArr (2x-y)+log (x-y+2)=x+c` |
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| 29. |
If `y=x^(n)logx+x(logx)^(n)," then "(dy)/(dx)` is equal toA. `x^(n-1)(1+nlogx)+(logx)^(n+1)[n+logx]`B. `x^(n-2)(1+nlogx)+(logx)^(n-1)[n+logx]`C. `n^(n-1)(1+nlogx)+(logx)^(n-1)[n-logx]`D. None of the above |
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Answer» Correct Answer - A Given `y=x^(n) log x+(log x)^(n)` `(dy)/(dx)=nx^(n-1) log x+x^(n)*(1)/(x)+xn (log x)^(n-1)((1)/(x))+1*( log x)^(n)` `=x^(n-1)(1+n log x)+(log x)^(n-1)[ n+log x]` |
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| 30. |
The solutiion of `(x,y,z)` the equation `[(-1,0,1),(-1,1,0),(0,-1,1)][(x),(y),(z)]=[(1),(1),(2)]` is `(x,y,z)`A. (1,1,1)B. (0,-1,2)C. (-1,2,2)D. (-1, 0, 2) |
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Answer» Correct Answer - D `[{:(1,0,1),(-1,1,0),(0,-1,1):}][{:(x),(y),(z):}]=[{:(1),(1),(2):}]` `[{:(x,+0y,z),(-x,1,0z),(0x,-y,+z):}]=[{:(1),(1),(2):}]` `rArr x+z=1` `-x+y=1` and `-y+z=2` On solving these equations, we get `x=-1,y=0,z=2` |
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| 31. |
Two nuclei have their mass numbers in the ratio of `1:3`. The ratio of their nuclear densities would beA. `1:3`B. `3:1`C. `(3)^(1//3):1`D. `1:1` |
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Answer» Correct Answer - D Density of nuclear matter is independent of mass number,so the required ratio is 1:1. Alternative `A_(1):A_(2):=1:3` Their radii will be in the ratio `R_(0)A_(1)^(1//3):R_(0)A_(2)^(1//3)=1:3^(1//3)` Density `=(A)/((4)/(3)piR^(3))` `:.P_(A_(1)):P_(A_2)=(1)/((4)/(3)piR_(0)^(3).1^(3))=(3)/((4)/(3)piR_(0)^(3)(3^(1//3))^(3))` Their nuclear densities will be the same. |
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| 32. |
The velocity of electromagnetic radiatior in a medium of permittivity `epsilon_(0)` and permeability `mu_(0)` is given byA. `sqrt((epsi_(0))/(mu_(0)))`B. `sqrt(mu_(0)epsi_(0))`C. `(1)/(sqrt(mu_(0)epsi_(0)))`D. `sqrt((mu_(0))/(epsi_(0)))` |
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Answer» Correct Answer - C Velocity of electromagnetic radiation is the velocity of light (c), ie `c=(1)/(sqrt(mu_(0)epsi_(0)))` where `mu_(0)` is the permeability and `epsi_(0)` is the permittivity of free space. |
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| 33. |
For the reaction : `H_(2)+I_(2)to 2HI,` the differential rate law isA. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=2(d[HI])/(dt)`B. `-(d[H_(2)])/(dt)=-2(d[I_(2)])/(dt)=(d[HI])/(dt)`C. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=(d[HI])/(dt)`D. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=(d[HI])/(dt)` |
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Answer» Correct Answer - B `H_(2)+I_(2) to 2HI` Rate of reaction `=-(d[H_(2)])/(dt)=(-d[I_(2)])/(dt)=(1)/(2)(d[HI])/(dt)` or `(-2d[H_(2)])/(dt)=(-2d[I_(2)])/(dt)=(d[HI])/(dt)` |
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| 34. |
In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t` `i=I_(0) sin (omegat-phi)` The average power in the circuit over one cycle of `AC` isA. `(E_(0)I_(0))/(2)`B. `(E_(0)I_(0))/(2)sinphi`C. `(E_(0)I_(0))/(2)cos phi`D. `E_(0)I_(0)` |
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Answer» Correct Answer - C Power= rate of work done in one complete cycle. or `P_(av)=(W)/(T)` or `P_(av)=((E_(0)I_(0)cosphi)T//2)/(T)` or `P_(av)=(E_(0)I_(0)cosphi)/(2)` where `cosphi` is called the power factor of an AC circuit. |
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| 35. |
A travelling wave in a stretched string is described by the equation `y = A sin (kx - omegat)` the maximum particle velocity isA. `Aomega`B. `omega//k`C. `domega//dk`D. `x//l` |
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Answer» Correct Answer - A Given that, the displacement of a particle is `y=Asin(omegat-kx)" "......(i)` The particle velocity `v_(p)=(dy)/(dt)" "......(ii)` Now, on differentiating Eq. (i) w.r.t.t, `(dy)/(dt)=Acos(omegat-kx).omega` `implies(dy)/(dx)=Aomegacos(omegat-kx)` From Eq. (ii) `impliesv_(p)=Aomegacos(omegat-kx)` For maximum particle velocity, `cos(omegat-dx)=1` So, `v_(p)=Aomegaxx1` `impliesv_(p)=Aomega` |
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| 36. |
Absorption co-efficient of an open window is...A. zeroB. 0.5C. 1D. 0.25 |
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Answer» Correct Answer - C Open window behaves like a perfectly black body. |
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| 37. |
A charge of 8.0 mA in the emitter current brings a charge of 7.9 mA in the collector current. The values of `alpha and beta` areA. `0.99, 90`B. `0.96,79`C. `0.97,99`D. `0.99,79` |
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Answer» Correct Answer - D Given that, change in emitter current, `DeltaI_(E)=8mA` and change in collector current, `DeltaI_(C)=7.9mA` We know that, `alpha=(DeltaI_(C))/(DeltaI_(E))impliesalpha=(7.9)/(8)impliesalpha~-0.99` Also we know that `beta=(alpha)/(1-alpha)` `impliesbeta=((7.9)/(8))/(1-(7.9)/(8))=(7.9)/(8-7.9)` or `beta=(7.9)/(0.1)=79` Hence, the required answer is `alpha=0.99andbeta=79` |
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| 38. |
The energy required to charge a parallel plate condenser of plate separtion `d` and plate area of cross-section `A` such that the unifom field between the plates is `E` isA. `1/2epsi_(0)E^(2)//Ad`B. `epsi_(0)E^(2)//Ad`C. `epsi_(0)E^(2)Ad`D. `1/2epsi_(0)E^(2)Ad` |
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Answer» Correct Answer - C Energy given by the cell `E=CV^(2)` Here, C = capacitance of condenser `=(Aepsi_(0))/(d)` V= potential difference across the plates =Ed Therefore, `E=((Aepsi_(0))/(d))(Ed)^(2)=Aepsi_(0)E^(2)d` |
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| 39. |
A `p-n` photodiode is made of a material with a band gap of `2.0 eV`. The minimum frequency of the radiation that can be absorbed by the material is nearlyA. `10xx100^(14)Hz`B. `5xx10^(14)Hz`C. `1xx10^(14)Hz`D. `20xx10^(14)Hz` |
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Answer» Correct Answer - B p-n photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage. Energy of radiation = band gap energy ie, `hv=2.0eV` or `v=(2.0xx1.6xx10^(-19))/(6.6xx10^(-34))~~5xx10^(14)Hz` |
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| 40. |
The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy isA. 3.4 eVB. 6.8 eVC. 10.2 eVD. zero |
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Answer» Correct Answer - C Given ground state energy of hydrogen atom `E_(1)=-13.6eV` Energy of electron in first excited state (ie, n=2) `E_(2)=-(13.6)/((2)^(2))eV` Therefore, excitation energy `DeltaE=E_(2)-E_(1)` `=-(13.6)/(4)-(13.6)` `=-3.4+13.6=10.2eV` |
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| 41. |
The chances to fail in Physis are 20% and the chances to fail in Mathematics are 10%. What are the chances to fail in atleast one subject ?A. 0.28B. 0.38C. 0.72D. 0.82 |
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Answer» Correct Answer - A Let P `(A)=(20)/(100)=(1)/(5),P(B)=(10)/(100)=(1)/(10)` Since, event are independent and we have to find `P(AuuB)=P(A)+P(B)-P(A)*P(B)` `=(1)/(5)+(1)/(10)-(1)/(5)*(1)/(10)` `=(3)/(10)-(1)/(50)=(14)/(50)=(14)/(50)xx100=28%` |
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| 42. |
A wholesale merchant wants to start the business of cereal with ₹ 24000. Wheat is ₹ 400 per quintal and rice is ₹ 600 per quintal. He has capacity to store 200 quintal cereal. He earns the profit ₹ 25 per quintal on wheat and 40 per quintal on nee. If he stores x quintal rice and y quintal wheat, then for maximum profit the objective function isA. `25x+40y`B. `40x+25y`C. `400x+600y`D. `(400)/(40)x+(600)/(25)y` |
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Answer» Correct Answer - B For maximum profit `z=40x+25y` |
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| 43. |
Negation of the conditional 'If it rains, I shall go to school' isA. It rains and I shall go to schoolB. It rains and I shall not go to schoolC. It does not ranis and I shall go to schoolD. None of the above |
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Answer» Correct Answer - B Let P : It rains, q: I shall go to school Thus, we have `p rArr q` Its negation is `~(p rArr q)` is `pvv~q` ie, it rains and I shall not go to school. |
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| 44. |
The value of `intxsinxsec^(3)xdx` isA. `(1)/(2)[sec^(2)x-tanx]+c`B. `(1)/(2)[xsec^(2)x-tanx]+c`C. `(1)/(2)[xsec^(2)x+tanx]+c`D. `(1)/(2)[sec^(2)x+tanx]+c` |
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Answer» Correct Answer - B `int x sin x sec^(2) x dx = int x sin x(1)/(cos^(3))dx` `= int x tan x* sec^(2)x dx` Put `tan x=t rArr sec^(2) x dx=dt` and `x= tan^(-1)t` Then, it reduces to `int tan^(-1) t* t dt =(t^(2))/(2) tan^(-1) t- int (t^(2))/(2(1+t^(2))dt` `= (x tan^(2) x)/(2)-(1)/(2)t+(1)/(2) tan^(-1)t+c` `=(x (sec^(2)x-1))/(2)-(1)/(2) tan x+(1)/(2)x+c` `=(1)/(2)[ xsec^(2)x- tan x] +c` |
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| 45. |
For a first order reaction, the half-life period isA. dependednt on the square of the initial concentrationB. dependent on first power of initial concentrationC. dependent on the square root of initial concentrationD. indipendent on initial concentration |
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Answer» Correct Answer - D `t_(1//2)=(1)/((a)^(n-1))` where, n= order o reaction = a initial concentration For first order reaction, `n=1` `t_(1//2)=(1)/(a^(1-1))` `a=(1)/(a^(0))=0` thus, for a first order reaction, `t_(1//2)` is independent on intial concentration. |
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| 46. |
The vapour pressure of pure benzene at a certain temperature is `640 mm` of `Hg`. A non-volatile non-electrolyte solid weighing `2.175 g` added `39.0 g` of benzene. The vapour pressure of the solution is `600 mm` of `Hg`. What is the molecular weight of solid substance?A. `49.50`B. `59.60`C. `69.60`D. `79.82` |
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Answer» Correct Answer - C Given , vapour pressure of benzene, `P^(@)=640` mm Hg Vapour pressure of solution, `P=600` mm Hg Weight of solurte `w=2.175g` Weighof benzene, `W=39.08g` Molecular weight of benzene, `M=78g` Molecular weight of solute m=? According to law, `(P^(@)-p)/(p^(@))=(wxxM)/(mxxW)` `(640-600)/(640) =(2.175xx78)/(mxx39.08)` `(40)/(640)=(2.175xx78)/(mxx39.08)` `m=(16xx2.175xx78)/(39.08)` `m=69.60` |
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| 47. |
A frame made of metalic wire enclosing a surface area A is covered with a soap film. If the area of the frame of metallic wire is reduced by `50%` the energy of the soap film will be changed by:A. `100%`B. `75%`C. `50%`D. `25%` |
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Answer» Correct Answer - C Surface energy = surface tension `xx` surface area `E=Txx2A` New surface energy, `E_(1)=Txx2((A)/(2))=TxxA` % decrease in surface energy `=(E-E_(1))/(E)xx100` `=(2TA-TA)/(2TA)xx100=50%` |
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