q, r, s r s q Since Q has taken 2 days sick leave, he has worked only 5 days on the end of seventh day Ratio of their work

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1 . Ans. A. Before superlaive aricle he has o be used. one of he expression should ae plural noun and so opion C and D can be he answer.. Ans. B. lose is verb. 3. Ans. A. effeciveness is noun and prescribed is verb. These words are ap and befiing wih he word medicine. 4. Ans. A. 5. Ans. C. 6. Ans. C. P (a person infeced bu does no show sympoms) =.5 x.7 =.35 The percenage is 35% 7. Ans. B. The words was confiden ha hey would reciprocae and las wee proved him wrong lead o saemens iii and iv as logically valid inferences. 8. Ans. D. From given daa, he following arrangemen is possible Shiva Leela Pavihra Among four alernaives, opion D is TRUE. 9. Ans. C. a b c q, r, s r s q a b c q r, r s, s q r q ( s ) s a c a ac b ac b abc s r ( s ) s abc. Ans. C. Q's one hour wor = 5 R's one hour wor = 5 Since Q has aen days sic leave, he has wored only 5 days on he end of sevenh day. Wor compleed by Q on 7 h day= (5x) 5 Wor compleed by R on 7 h day=(7x8) 5 Raio of heir wor = 5 7 / : 5 5. Ans. C. 4 4K 4 M. M M. M M. M I. M M A is no correc 4 4K M. M M. M M. M I. M M B is no correc 43 4K M M M M M M I I I C is correc M M M M M I M D is no correc. Ans. A. We now ha if X is parameer of poison s disribuion Then, Firs momen Second momen Given ha ( ) or Firs momen P a g e

2 3. Ans. B. We now ha every differeniable funcion is coninuous bu converse need no be rue 4. Ans. A. 5. Ans. C. Since he inegraion of an odd funcion is even in his logic A and B canno be he answer as hey are odd funcions. However boh C and D are even funcions bu he inegraion of a linear curve has o be parabolic in naure and i canno be a consan funcion. Based on his Opion C is correc. 6. Ans. C. If he inpu o a sysem is is eigen signal, he response has he same form as he eigen signal 7. Ans. B. Consider x () = cos w If x () is sampled wih a sampling period T, s xn cos n is obained Here, T s m NTs T T N T s m Which mus be a raional number T 6 Thus, lives a periodic signal afer sampling. T 5 s 8. Ans. B. n a u n ; ROC : z a R n b u n ; ROC : z b R n a u n bnu n ; ROC : R R z b 9. Ans. D. A wo por newor is reciprocal in ransmission parameers if AD BC = i.e. Deerminan(T) =. Ans. A. Sinusoidal signal frequency = 33 Impulse rain frequency = 46 Resulan signal conains specral frequencies ±33, ±3, ±7.9, ±59ec, Thus if i is passed hrough ideal LPF of cuoff frequency 3Hz only ±3Hz frequency is filered ou. Oupu signal fundamenal frequency = 3 Hz.. Ans. A. New energy level is near o conducion band, so i is penavalen aoms o form n-ype semiconducor.. Ans. C. I sol IOFF L VDS L rd II c W ( V V V ) sol I ON D n ox us DS If he channel lengh reduces, hen hreshold volage also changes L I ON So opion C. is maching. 3. Ans. B. Volage a (+) erminal V V CC Vref Volage a emier of PNP BJT VE VCC Vref The curren I I I E E C I E hrough R V ( V V R Vref R I CC CC ref ) Vre f IC IE ( ) I E IC I( ) R 4. Ans. A. The volage a (-) erminaed of he OP-Amp V ( ) V 8 The oupu of he op-amp goes o +V cc whenever V. i V( ) i e Vi V maes BJT urn ON. So, he LED glows. The secions of he wave form become more han V for he range [a o b, (c o d) and (e o f)] So LED glows 3 imes. 5. Ans. A. Sol. When he oupu volage is posiive he diode D i is urned on maing resisor o become parallel o.. So he gain is reduced. When he oupu volage becomes negaive he diode D is urned on hereby again resisor o become parallel o.. So he gain is reduced. Wih he use of diodes, he non-ideal OP-Amp is made sable o produce seady P a g e

3 6. Ans. A. 9. Ans. D. 7. Ans. C. 3. Ans. A. x y z 7 3,6 3,6,7 m m m ABC ABC ABC ABC AB B( AC B) B( A c) 8. Ans. C. The ransfer characerisics of he CMOS inverer is as follows 3. Ans. B. 3. Ans. B. 33. Ans. B. 5 6 s cos cos8 5cos Am Ac [ cos ]cos(8 ) A A c c Ac Since he inverer is conneced in feedbac loop formed by connecing OXQ resisor beween he oupu and inpu, he oupu goes and says a he middle of he characerisics VIR VIH Va V Swiching hreshold of inverer a 34. Ans. A. Consider a Gaussian surface a sphere of radius m To ensure D a radius m, he oal charge enclosed by Gaussian surface is zero Q enc Ps 8 P.5 nc / m s 3 P a g e

4 35. Ans. B. D ( x y ) dxdy r(cos sin ) rdrd r (Changing ino polar coordinaes by x rcos ) 37. Ans. A. 36. Ans. B. Mehod-I:- 4x ( x y ) dxdy x y 4x 4 x ( x ) dx 4x ( x )( y) dx ( 4 4 x x x dx 4 x dx x 4 x 4 x sin Mehod II:- 38. Ans. A. (by Cauchy inegral formula) 4 P a g e

5 39. Ans. A. Volume= 5 / /8 3 dddz Ans. B. Sol. Laplace ransiion of one cycle of f r e S s / Laplace ransiion of causal periodic square wave given in f() is, ST / () s e Fs ST ( e ) ST / S e ST / ST / ST / e e S e 4. Ans. A. The propery of any LTI sysem or newor is if he exciaion conains 'n' number of differen frequency hen he response also conains exacly n number of differen frequency erm and he oupu frequency and inpu frequency mus be same however depending on componens here is a possible change in ampliude and phase bu never he frequency. If he source has 3 frequency erms as given 3 a cs o b and e T e Y s Y X R Y P Z Q T / T / K K K K3 s ( s ) s s s K u( ) K u( ) K e / 3 s / y( ) e / e X ( s) hen any volage or any curren of any elemen should have also 3 erms based on his opion B. and D. are eliminaed. If we ae opion C.. I has 3 frequency erm bu i also sugges here is a phase change so 4 bu ampliude mus be same as inpu as a is presen which may no be rue always. So opion A. is correc, as i sugges frequency erm of oupu and inpus are same wih possible change in ampliude and phase, because we have ( b and ). 4. Ans. C. In general he firs order, L.P.F filer ransfer funcion is G s because G() = and G (co) =, if we s ae his ransfer funcion as reference and give differen inpu such as s().r().u() if inpu is s() Y s X ( s) s s / y() e if inpu is u() Y s / s K e if inpu is r() Y Y s X R Y P Z Q K K K K3 s ( s ) s s s K u( ) K u( ) K e 43. Ans. A. / Ans. C. To find maximum power ransferred o load we need o obain Thevenin equivalen of he circui Obaining V oc 5 P a g e

6 V 5 V VC V 6V 4 5 Obaining I SC 46. Ans. A. V 5 V 3 V ISC ma VC 6 Rm 8 I SC So he newor is For MPTR = 8 P mcr V n 6.8 wa. 4 R (48) 3 n 47. Ans. A. IDs = I g DS ds W NCOX VGS VT V L di DS W = = NCO VGS VT dv L DS s 48. Ans. B. DS 45. Ans. A. Consider 6 P a g e

7 49. Ans. A. ni.5 Pn D 6 N 5 P G 3.5 / po cc P() Pe / p 4.5 / 6.5. P( ) P( ) P P( ) no 3 P() P.5 / cc P ( 3) 7.47 / 5. Ans. A. cc cc 5. Ans. A. As we now in a decoder w.r. any binary inpu combinaion he corresponding oupu pin is high and remaining low. Similarly o he encoder one inpu is high among all and is equivalen binary combinaion is available a oupu. In his case o idenify he funcionaliy, le give some arbirary binary inpu and observe he oupu. Le [X X X ] is [ ] respecively hen OP 5 = = IP? Then [Y Y Y ] is [ ] If [X X X ] is [ ] hen [OP 7 = IP 5 = ] so [Y Y Y = ] [X X X ] is [ ] hen [OP 4 = IP 6 = ] so [Y Y Y = ] From he above we can say ha If inpu hen oupu is So inpu binary and oupu gray. 53. Ans. B. Decoder inpus will behaves as MUX selec lines and when he oupu of decoder is high hen only corresponding buffer will be enable and passed he inpus (P,Q,R,S) o he oupu line, so i will wor as 4-o- muliplexer. 54. Ans. C. In push operaion 3 cycles involved: 6T+3T+3T = 7 POP operaion 3 cycles involved: 4T+3T+3T = 7 So in he opcode fech cycle T saes are exra in case of push compared o POP and his is needed o decremen he SP. 5. Ans. D. 55. Ans. B. Sol. In his firs we need o find he brea poin by d ds finding he roo of and hen by using magniude condiion value of can be obained. K Gs s 5s5 ( s 5s 5) d ds s 5 s.5 7 P a g e

8 Applying magniude condiion Gs s K 5s5 s (.5 ) 5 x(.5) 5 We can proceed here by aing his polynomial as characerisic equaion and conclusion can be draw by using RH crierion. As we are ineresed o now how many roos are lying on righ half of s plane. 56. Ans. A. K =.86 Pea over shoo % e. In. In In..34 The characerisic equaion of above ransfer funcion Comparing wih sandard equaion s is s s n n The number of roos i.e. he number of zeros in his case in righ half of plane is number of sign changes Number of sign changes = Ans. A. i i i log () log i i i i i i i d i ia aia a ( a ) da i i i d i a / a a bis da i ( a) / Ans. A. Cross over Probabiliy P=. X = number of errors P( x ) P( x ) sen sen (.) (.9) (.) (.9) 7 8 3(.)(.9) (.3) Ans. A. 8 P a g e

9 6. Ans. A. r S n s n r s u h u du n r u h u du Y s SNR = E [ Y s ] N h u du s u du Es SNRop N No Es SNRop if En Es N o s u h udu By CS in equaliy ifh u CS u 6. Ans. A. 63. Ans. A. Given Lossless horn anennas ηt = ηr = 6. Ans. D. 9 P a g e

10 64. Ans. C. 65. Ans. B. / P r ad / Wradr sin d d C r cos r / cos..56 sindd C C C 5 5 max( Wrad ) 4 Max. direciviy 4 r P.56 r ad Max. Direciviy in db log db. *** P a g e