CSIR Engineering Sciences Solved December 2012

8/10/2019 CSIR Engineering Sciences Solved December 2012

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S 7 5 P OK/12 6 A E 1 8

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P A R T A

1 . W hat i s th e nu m b e r o f d is t i nc t a r ran g em e n ts of he l e t ter s of h e w or d UG CCS IR s o that U and I can n ot co m e toge t h er?

1 252 0 2. 720 3 152 0 4. 180 0

2. Supp o se t h e sum of th e se v en p osi t iv e n u m be rs i s 21 Wh at is the m inim um p oss i b levalu e o f t h e av e rage of t h e sq a ures o f h e se n u m be r s?

1 63 2. 21 3 9 4 7

3 Let

W hic h o f t h e fol lowin g is t n ie?

1 B C A 2 . A < B < C 3. B A C 4. C A B

4. W h ic h of the fo llow ing n umb e rs is the la rges t?34 43 24 42 23 32

2 2 3 3 4 4 .

5 T h e cub e B C DE F GH i n th e fig u re h a s ~ ch e d ge e q ua l to a T he a rea o f th e tria n glew i th ve r tices at A , C an d F is

J 32 1 a

4

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6 If th e s u m of the n ext tw o te r m s o f th e ser ie s be l o w is X w hat is the v a lue o f lo g2 ?

2, - 4, 8, -16, 3 2, -6 4, 12 8,

1 128 2. 1 0 3 . 25 6 4 8

7

A co n ical vesse l wi t h sem i-ver tical a ngle 30 an d h eight 10.5 em h as a thin l id. A sphe re

k ept i n side i t to uc hes the l id. T he ra d ius o f th e sphe r e in em is

1 3 .5 2 5 3. 6.5 4 7

8 A c ircle of ra d ius 5 un its in the X Y pl a ne h as it s cent re in the f ir st qu adra n t , to u ch e s th e

x axis a nd h a s a c h o rd of le n gth 6 un i ts on they axis . T h e c oord inate s of i ts cen tre a re

1 4 ,6 ) 2 3 ,5) 3 . 5, 4 ) 4. 4,5)

9 A wire of l e ngth 6m is us e d to m ak e a t e trah e dron o f e a ch e dge 1m, usin g on l y on e

s t rand o f w ire o ~ eac h e dg e . T h e m i n im u m nu mb e r of ti m es th e w i r e has to b e cut is

1 2 3 3. 4 0

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10 . W ha t d o e s t h is d i ag ram d em on s tr a te ?

/

I 1 1 / 1 j

v

1. 1 + 2 + 3 + + n = n .o....n_ _1- -)2

/

2. 12 + 22 +32 n 2 = n (n + 1 ) ( 2 n + 1 ) 6

3. 1 + 3 + 2n = - 1 ) = n

5

2 2 ( ) 2 2 n ( n + 1 ( 2 n + 1 4. 2 + 4 + + 2 n = -- - - - -- - -- - - - -- -3 .

J4S _

11 . S upp os e th ere ar e s o ck s o f di f fe ren t co lo r s in a b o x . If y ou tak e o ut o ne so c k a t a tim e, w ha t is th e m ax im u m n u m b e r of soc ks th a t y ou h a ve to ta k e o ut b e f o r e a m atc h in g p ir is f ou n d? A ss u m e t h a t is an ev e n n um be r.

l N 2 . N 1 3 . N 1 4 . N /2

i2 . A t w ha t t i m e aft e r 4 0 c lo c k , the h o u r an d th e m in u te ha n d s w il l lie op p os it e to eac hot h er ?

1. 4 - 5 0 ' - 3 1 , 2. 4 - 5 2 ' -5 1 3. 4 - 5 3 '- 23

13 . W h ich o f th e fo l lo w in g c u rv e s j u st to u ch e s the x a x is ?

1. y = x

- x 1

3. y = x

- 1 Ox 52 . y x

- 2 x + 2

4. y = x

- 7 x 12

4. 4 - 5 4 ' -- 3 3

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1 4 S uppose we mak e ide n tical sm aller s p heres f rom a b ig sphe re. The total su rface ar e a

of he sm a ller sph eres is times the tota l surfac e area o f he b ig sphere , where is

l J N 2 1 3 N 3 4.

15 W hat is the nex t numb e r in the sequen ce 24, 3 0 33 3 9 51,-- ----?

1 57 2 69 3 54 4 81

16 Fou r lines a re draw n on a plane w ith no two par a llel and no thr ee conc u rrent. Lines

ar e drawn joining the po ints of intersec tion of the prev ious fou r lines. The nu mber o f

ne w lines obtaine d this w ay is

17

1 3 2. 5 3 12 4 2

If AB is paral led to C D and A 0=20D then the area of trian gle OA B is big ger tha n thear e a o f r i n g l ~O CD by a factor of

1 2 2 3 3 4 4 8

18 A mar, A kbar an d Antho ny are three fr iends, o ne of w hom is a doct o r, ano ther is a n

enginee r and th e third is a pr o fessor. Amar is not an engin e er. Ak b ar is th e short est.

The ta llest pe rson is a doct o r. The engine er s hei ght is the geo m etric m ean o f theheights ofthe o ther tw o. Then which o f the fo llowing is true?

1 A ma r is a docto r and he is the t allest2. Akb ar is a p rofesso r and he is the t a llest3. An thony is an en g ineer ari d he is shortest

4. A nthony is a do c tor and he is th e tallest

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19 . If 100 cats ca tch 100 m i e in 1 00 minutes then how long will i t take for 7 ca t s to ca tch7 m i c e?

20 .

1 100 7 m inutes

3 49/1 0 0 m i n u t e s

2 . 100 minu tes

4 7 m inutes

A semi-circula r a rch of radius R ha s a vert ica l pole put o n th e ground toge th er wi th oneo f its legs. An a n t on the top of the a rch f inds the angula r he i g h t of the tip of the po leto be 45. The he igh t of he p o le is

1 fiR 2 J3R 3. J4R 4 J5R

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PA R t B )

IMATHEMATICS

I21 T h ~value of

zacosh 1l . 1ll m [sin 8 + h) - si n 8 ]h->o Sh 6 6

is

(1).1 (2) ~2 . (3) ./3

2

(4) .../3

22. The number of distinct real roots of the equation 13x 13 - e - x - 1 = 0 is

(1) 0 2) 1 (3) 2 (4) 13

23 The minimum value of the function f: IR ..Im. defined by f x) = ~ : x etet 2 dt is

1) f O) (2) f l ) (3) f 2) (4) f - 2 )

24 cos n cos n ( . 1 )Let an= - 2- and n = - 2- s 2 fo rn EN. Thenn n n(1) L ~ = lan converges and ~ = l n diverges

(2) L ~ = ln converges and L ~ = lan diverges

25 LetS be the sphere x2 + y2 + z2 = l The value dfthe surface integralffs xsin y cos 2 x, z z sin y x, y z) da ~

(1) ~3

(2) 27r3

(3)3

(4) Btr3

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2 6 T he val ue o f t h e in teg ral

27

2 8

i o s z d z

~ ~ ~ i s z ~ z z 2 ) z - 4 )

(1 ) r r i4

(2) r ri 2

(3 ) r ri (4 ) 2 rr i

Le t A b e a 2 x 2 m a tr ix . S u p p o s e , 1

) and ~) a r e e i g en v e c t o r s cd r r e s p o n d in g to t h e e ig e n v a l u e s 1 a n d 2 o f A r e s p e c t iv e ly . T h e n A is

(2 ) ~ ~] (3 ) ~ ~] (4 ) ~ ~ ]

C 'd h d d ffi . I . 4 + y2

I f ( 1) 2 h ( 2 ) .ns1 e r t e o r m ar y 1 e re n t1 a e q u a t io n y = y = t e n y IS l + x ( 1 ) 1 (2) 5 (3) 1 4 (4 ) 21

2 9 L et D = { x x 2 , x 3 x 4 : x 1 = 0 o r 1 fo r i = 1 ,2 ,3 ,4 }. A v e c t o r is c h o s e n a t r a n d o m f ro m t h e s e t D . T h e p r o b a b i l i t y th a t th e p ro d u c t o f h e f i r s t a n d th i rd c o o rd i n a t e i s 0 , is

3 (2 ) -

4 ( 3 ) . :

2 (4 ) ~

8

3 0 . T h e d if f e r e n t ia l e q u a t i o n f o r w h ic h x, x ln x a n d x 2 a r e l i n e a r l y in d e p e n d e n t s o lu ti o n s i s

(1 ) x 3 y ' + x 2 y -3 x y ' + 3 y = O (3 ) x 3 y ' - x 2 y + 2 x y ' - 2 y = 0

I E N G I N E E R I N G A P T I T U D E I

(2 ) x 3 y ' - 2 x 2 y + 3 x y '- 6 y = 6

(4 ) y - y + 2 y - 3 = 0

3 1 W h i c h o f th e fo l lo w in g s ta te m e nt s is tr u e r e g a r d i n g t h e e ff e c t o f e a r t h o n t h e c a p a c i ta n c e o f s i n g l e - p h a s e a n d th r e e - p h a s e t r a n s m i s s io n l i n e s ?

1 ) Id e a l ly th e p r e s e n c e o f t h e e a rt h d o e s n o t a f f e c t t h e c a p a c i t a n c e . (2 ) Id e a l ly th e e ar th in c r e a s e s th e c a p a c i ta n c e b ut fo r p r a c t i c a l v a lu e s o f h e i g h t s o f

c o n d u c t o r s ab o ve th e g r o u n d , t h i s e n h a n c e m e n t c a n b e n e g l e c t e d . (3 ) Id e a lly , t h e e a rt h r e d u c e s t h e c a p a c i ta n c e b u t fo r p ra c t ic a l v a lu e s o f h e ig h ts o f

c o n d u c t o r s a bo v e th e g ro u n d , t h is r e d u c t io n c a n be n e g l e c t e d .(4 ) Id e a lly , t h e p re s e n c e o f . e a r t h e n h a n c es t h e c a p a c i t a n c e s ig n i f ic a n t ly a n d fo r

p ra c t ic a l v al u e o f h e i g h ts o f c o n d u c to r s ab o v e th e g r o u n d , th is e n h a n c e m e n t c a n n o t b e n e g l e c t e d .

J .

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32 . C o n s id e r a u n ifo rm ly c h a rg e d sp h e re o f radiu s , R. T h e ch a rg e p e r u n i t v o lu m e is p .

T h e m a g n itu d e o ft h e e le c tr ic field a t a d is tan c e o r fro m th e c en tre o f t h e sph e re r

R ), is g iv e n by : Eo is th e p e rm ittiv ity f free space)

1) E = pr/3Eo

3) E = p R3 47t ~: :

2) E = p r2 /3Eo

4) E = pr/47tEo

33. A long c u r r e n t ca rry in g c o n d u c to r h a v in g rad iu s R , is c a r ry in g a u n i fo rm ly d is tr ib u ted

cu rren t th ro u g h its c ro ss -s ec tio n . T h e m a g n e tic fie ld in ten s ity in s id e t h e c o n d u c to r is

Hi w h ile th e m ag n e tic fie ld in ten s ity o u ts id e the c o n d u c to r is Ho. W h ich o f the

fo llo w in g s ta tem en ts is tru e ?

1) B o th Hi and H 0 a re in v erse ly p ro p o rt io n a l to th e d istan ce f ro m th e cen tre o f th e

c o n d u c to r .

2) B o th Hi and H o a re p ro p o rtio n a l t o th e d is tan ce from th e c e n t re o f th e c o n d u c to r

3) Hi is p ro p o rtio n a l to th e d is tan ce fro m th e cen tre o f th e c o n d u c to r a n d H 0 v a r ie s

in v e rs e ly w ith the d is ta n ce from th e c en tre o f th e c o n d u c to r

4) Hi v a r ie s in v erse ly w i th the d is ta n c e from the c e n t re o f the c o n d u c to r and H 0 is

p ro p o rt io n a l to th e d is ta n ce from th e cen tre o f th e c o n du c to r .

34. T he p ro p e rt ie s o f M o n o c l in i c c ry sta l s y ste m are

1) a:fb :fc , a:fP:fy o n e 2 -fo ld ro ta tio n a l a x is

2) a:fb :fc, a= p = 9 0:fy , tw o 2 - fo ld ro ta tio n a l ax is

3) a :fb :fc , a= y= 90:f P , o n e 2 -fo ld ro ta tio n a l ax is

4) a :f b :fc , a= y= 90 :fp , tw o 2 -fo ld ro ta t io n a l ax is

35 . In an x - ra y d iffrac tio n pa tte rn , the f irs t o rd e r re f lec tio n from o n e o f th e a to m ic p la n es

in n ick e l cry s ta l a p p e a rs a t 62 .13 w h e n x -ray s o f wav e len g th o f 0 .15 42 n m w a s

used . W h a t are the in d ice s o f t h e p la n e i ft h e la ttice p a ram e te r o f n ic k e l is 0 .3 5 2 n m ?

1) 111 2 ) 2 0 0 3) 2 2 0 4 ) 4 0 0

36. T h e n u m b e r ave rage m o le cu la r w e ig h t o f p oly v in y l ch lo r id e P V C ) , w h e n its d e g re e

o f .p o ly m eriza t io n is 3 4 0 , is

a to m ic w e ig h ts o fC H a n d C l a r e 1 2 .0 1 , 1.01 an d 3 5 .45 g /m ol, re s p e c tiv e ly )

1) 2 0 5 6 3 2 ) 2 1 2 5 0

37. F or a f lo w to be in co m p ress ib le

1 d e n s i ty m ust be co n s tan t.

2) v e lo c ity field m u s t be d iv e rg e n c e free .

3) v o r tic ity v ec to r m u s t be a nu ll v ec to r .

3) 3 2 9 5 9

4) s tream lin e , s tre ak lin e and p a th lin e a re id en tica l.

4 ) n o n e o f th e s e

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3 8. A sm al l m a ss m is t ie d t o a s tr in g an d s w u n g in th e a n t ic lo c k w is e d ire c ti on ab o ut the ce n te r P o f a ci rc le o f r a d i u s r .a t a c on s ta n t s p e e d s (se e F ig u re ) . W h e n it is a t p o in t A ,th e s tr in g s n a p s o ff. A s s u m e th a t the a i r o f fe r s no d rag o n th e m ass . N eg lec t th e e ff ec to f g rav it y . W h a t a re t he co o rd in a te s x,y) o f h e m ass t s e co n d s af te r th e s tr in g sn aps ?

v

~ ~ ~ ~ o ~ . x

m s2

2

) m s2

2

) (3 ) r s t ) 1 ) r z ; - . s t (2 ) r - - ; : - , 0 (4 ) m s z tz s t ) 2 r '

39. A c y li nd e r (r ad ius r , le n gt h L an d m ass m ) is k ep t s ta ti on a ry a t p o in t A o n a n in c lin edplan e . If it is re le a se d t o ro l l do w n th e s l o p e ( w it h ou t s l ip p in g ) by a h ei g h t be tw een p o in ts A a n d B , w h a t w o u l d b e it s ve loc it y w h e n it re a ch e s p o i n t B ?

h

. (1 ) ~ g h (2 ) ioh (3 ) fiJ i

(4) ~oh

. 40. N it r o ge n g a s is s to re d i n a c y l in d e r o f 5 li tr e s v o lu m e a t a p re s su re o f 2 0 0 b ar and a

tem pe rat ur e o f2 7 C . A c om p res s ib il it y fa c to r o f 1 1 fo r n i t r o g e n m a y b e a ss u m e d at th es e c on d it io n s. S p ec if ic g a s c on s ta n t o f n it ro g en is 0 .2 9 7 kJ /k g K . T h e ac tu a l m ass o f n it ro g e n in th e c y li n d e r is a pp ro x im a te ly

(1 ) 1 p g ram s le ss t ha n tha t o bt ai n ed f ro m id eal g as e q u at io n o f sta te (2) 1 0 0 g ra m s m o re t h an t h a t o b ta in e d fr o m ide a l ga s e q u a t i o n o f s ta te (3) 1 130 g ra m s le s s th an t h a t o b ta in e d fr o m ide a l ga s e q u a ti o n o f s ta te (4 ) 11 30 g ra m s m o re th an t h a t o b ta in ed fr o m ide a l g as e q u a ti o n o f s ta te

4 1 . T h e n e t c h a n g e in en tro p y a t t h e en d o f a p ro c es s u n de rgo n e b y a c lo s e d s y st em tu rns o u t to b e ze ro . T h e a bo v e p ro c es s

- (1) m u s t b e r e v e rs ib le

(3) m u s t be i rr e ve rs ib le an d a d ia b at ic

(2) m u st be r e ve r s ib l e an d ad ia ba ti c:

(4 ) m a y b e ir re v er si b le , in vol vi n a ne the a t t ra n s fe r

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42 . W h e n m o is t a ir is o o l e ~ s e n s ib ly in a c o o l ing c o il

1 ) B o th w e t b u lb tem p e ra tu re a n d re la ti v e h u m id i ty dec re a se

2 ) W e t bu lb t e m p e ra tu re d e c re a s e s an d re la t iv e h u m id i ty in c re a s e s

3 ) W e t bu lb te m p e ra tu re d e c re a s e s a n d r e la t iv e hu m id ity r e m

a in s c o n s ta n t

4 ) B o th w e t b u lb t e m p e ra tu re a n d re la tiv e h u m id i ty rem ain c o n s t a n t

tl3 . A c u r re n t o f 1 A /c m2 is fo rced th ro u g h a n n - t y p e Si b ar w ith a re :;is tiv ity o f 1 0 - c

m at

a te m p e ra tu re o f 3 0 0 K. H o w lo n g d o es it ta k e fo r an e le c tro n to d rift th ro u g h th e Si

b a r o f l m m lo n g i f th e m o b il i ty o f e lec tro n a t 3 0 0 K is 1450 c m

/v o lt-se c ?

1 ) 6 9 lS 2 ) 145 lS 3 ) 4 9 ns 4 ) 1 m s

44. T h e e le c tr ic f ie ld w ith in th e s p a c e ch a rg e re gio n o f a PN j u n c t io n is ~ o wn b e lo w .

E = E l e c t r i c f i e l d

= d i s t a n c e

T h e bu ilt-in . p o te n t ia l o f he P N ju n c t io n is 0 .7 5 V. Is th e d io d e fo rw ard o r re v e rse

b ia s e d and w i th h o w m u ch v o l ta g e ?

1 ) F o rw ard b ia se d , 1.75 V

3 ) F o rw ard b ia s e d , 0 .45 V

2 ) R ev e rse b ia s e d , 0 .75 V

4 ) R ev e rse b ia s e d , 8 .25 V

45. T h e h a lf-w av e rec tif ie r s h o w n . b e lo w is o p e ra t in g a t a f re q u e n c y o f 5 0 H z an d the

R M S value o f t ra n s fo rm e r o u tp u t v o ltag e is 6 .3 V. W h at is t h e a p p ro x im a te v a lu e o f

d e o u tp u t v o l ta g e an d w h a t is th e m in im u m v a lu e o f c a p a c i ta n c e C ) req u ired to

m a in ta in th e r ip p le v o ltag e w i th in 0 .25 V ? A ssu m e: R = 5 k O an d d io d e v o ltage

d r o p = 1 V.

c _

{ i ) - 7 . 9 V , 12 6 1 1F 2 ) 7 .9 V , 126 1 1F 3 ) 5.3 V, 631-lF 4 ) - 5 .3 V , 631-lF

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P A R T C )

C O M P U T E R S C IE N C E IN F O R M A T IO N T E C H N O L O G Y

46 . L e t A d e n o te t h s e t o f p o s itiv e in teg e rs an d R b e a re la tio n d e f in e d by aR b i f a n d o n ly i fb = a 1. T h e t ra n s it iv e c lo s u re T o f R is

( 1 ) a T b i f a n d o n ly i f a , b a re in te g e rs a nd ~ 0, b 0.2 ) a T b i f an d o n ly i f a a nd b a re in tege rs a n d a > 0, b > 0. 3) a T b i f a n d o n l y i f a a nd b a re p o s it iv e i n te g e r s a n d a > b. 4 ) a T b if an d o n ly if a and b a r e p o s itiv e i n te g e r s an d b > a .

4 7 . In d e le ti o n b y m e rg in g in a B in a ry S ea rch T re e , w h en a n o d e With tw o c h i ld re n is d e le te d , th e ri g h t s u b -tr ee o f th e d e le ted node m a y be a tta c h ed t o th e n o de c o n ta in in g la rg es t k ey in th e le f t su b -tree . I f th e node 15 is d e le te d from th e tree in F ig u re 1, w e o b ta in a s tru c tu re a s s h o w n in F ig u re 2.

T h e v a lu es in th e n o d e s m ark ed X a n d Y in F ig u re 2 are

F I G U R E 1

l ) X = 5 , Y = l 0 . 3 ) X = 1 0 , Y = 2 0 .

F I G U R E 2

2) X = 11, Y = 3 0 . 4 ) X = 2 0 , Y = 3 0 .

4 8 . T h e p r io r it y q u e u e A D T , w it h o p e ra tio n s In ser t a n d D e le te M in o v e r a se t o f in teg e rs , is im p le m en ted u s in g a lin ked li s t an d heap d a ta s tru c tu re s . L e t the co m p lex i ty o f o p e ra t io n s in l in k ed li s t for I n se rt an d D ele te M in be LI, L D ) w h i le t h e c o m p lex ity o fth e o p e ra t io n s in H e a p be H I, H D ). W h ich o f h e fo ll o w in g is T R U E ?

1) L I, L D ) = 0 1 ), 0 1)),

H I, H D ) = 0 I ) , 0 lo g n )).3 ) L l , L D ) = 0 1 ), 0 n )) ,

H I, H D ) = O Iog n ), 0 lo g n )).

2 ) L I , L D ) = 0 1 ), 0 (n )),

H I, H D ) = 0 1 ) , 0 lo g n)).

4 ) L I , L D ) = 0 n ) , 0 n )),

H I, H D ) = 0 1 ), 0 l )).

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49 . S u ppose a cc e ss to a n in s tr uc ti o n in th ecach e is 2 0 ti m es fa s ter t h a n a ccess to a n

in st ruc tio n in the m a in m em o ry . A s su m e th at a re q u e st e d in s tr uc tio n is fo und in th e

cache w ith p ro ba b il ity 0 .8 4 , a nd a ls o a ss u m e t hat i a n in st ruc tio n is n o t f ound in th e

cache , it m u s t firs t b e fet ch e d f rom th e m ai n m em o ry to the c a c h e a n d then f e tc h e d

from th e c a ch e t o b e e x ec u te d . T h e sp ee d u p fa ctor re su lti n g fr om th e p rese nc e o f th e

cac he is :

1 ) 2 (2) 1 5 .987 (3) 6 .8 2 (4 ) 4 .7 62

50 . I f an 8 -w a y se t-a s so c ia tiv e cache is m a de up o f 32 b i t w ords, 4 w o rd s p er li ne a n d

409 6 se ts , ho w big is th e c ac h e in b y te s ?

(1 ) 12 b y te s (2) 512 k il o by te s (3) 2 m e g ab y te s ( 4 ) 10 k il o b y te s

51 . T ra n sa c ti o n p ro c ess in g i s a sso c ia te d w ith every th in g b e lo w e x c e p t

(1) p ro d u c in g deta il, s u m m a ry , o r e x c ep t io n r epor ts.

(2 ) r e c o ~ d i n g a b u sin e ss a ctiv ity .

(3 ) c o n f ir m in g an a c ti o n o r tr ig g er in g a response .

(4 ) m a in ta in in g dat a.

52 . W hich o f h e f o ll o w ing is c o rre c t?

(1 ) 2n < n < n3 < 3 n

(3 ) n3 log n < n3 < 3n < 2 2n

(2 ) n log n < n lo g2n < n2

< 2

(4 ) n < n2

< n l og n < n2 lo g n

5 3 . C o n s id er an e xe cu ti o n o f Q U I C K S O R T w it h the f irs t i te m o f an a rr a y se g m e n t act ing

as p iv o t o r sp li tt er. A f te r 2 p asses o f runn in g q u ic k sort o n th e a rr a y

2, 8, 7, 1, 3, 5, 6, 4

le t the resu lti ng se q u e n c e b e

a ~ az, a3, a., as, ~ . a7, as.

T hen , w h ic h one o f h e fo ll ow ing is T R U E ?

l ) a 3 = 6 , ~ = 5

(3 ) a3 = 2, = 7

(2 ) a3 = 5, = 6

(4 ) a3 = 7, = 2

54 . L et P be a B o o le a n functi on an d P = A B B A. T h e n th e fu nc tio n

PEB A E BPE B A EB PEB B EB PE BA ev a lu a te s to

l )A E B B ( 2 ) A , (3 ) B (4 ) 0

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55 . O r th o g o n a l arr ay s a l l o w m ax im u m t es t c o v e r a g e f ro m m mt m u m se t of t e s t p ro c ed u r e s . S u p p o s e th e re ar e th re e p a r am et e rs A, B , C e a c h o f w hi ch h a s o n e o ft h re e p o s s ib l e v a lue s 1, 2 or 3. T h e s t r en g th of th e o r t h o g o n a l arr ay is t he n u m be r of p a ra m e te r s w ho s e co m b in a t i o n o f v a lue s oc cu r o n l y o n c e .

T h e num b er o f te s t c a s e s r eq u ir e d fo r an o r th o g o n a l a r ra y o f s tr e ng th 2 f o r A, B a n d C a re

1 ) 6 2 ) 9 3 ) 8 4 ) 27

I E L E C T R IC A L S C I E N C E I

5 6 . T h e b a ck em f o f a s ep a r a te l y e xc it e d d e m o t or is e s t im a t e d an d is f o u n d t o b e 18 0 V w he n it i s r u n n in g a t a s p e e d o f 10 00 rp m an d w h i le ra t ed v o lt a ge is ap p l ied to it s fi e ld w in d in g . N eg l ec t in g ro t a ti o na l lo s s e s , t h e to r q u e d ev el o pe d b y th is m a ch in e w he n it is dr aw in g I OA o f a r m a tu re c u r re n t a n d w h il e h a lf th e r a te d v o l t ag e is a p p li ed t o i ts fi e ld w i nd in g is

5 4 1 - N m rr

2 72 ) - N m rr

3 ) ~ Nm rr

4 )0

9

N m rr

5 7 A 4 0 0 V , th r ee p h as e, 50 H z 1 42 0 rpm w ou nd ro to r i n d u c t i on m o t o r is ha v in g p er p h a se r o to r r es is ta nc e r e f e rr e d to th e s t a to r a s 0 .2 5 n a n d is h av in g p er p h a s es ta n ds t il l ro to r re ac t an c e r ef e rr ed to t h e s t a to r a 0 .6 0 W h i l e d ri v in g a ce r ta in loa d a t1425 rp m , th e p er p h as e r o t o r cu r re n t re f er r ed to t h e s t a to r is fo u n d to be 5 A .N e g l ec t in g r o ta ti o n a l l o ss e s , th e t o rq ue d e v e l o p ed by th e m ac h i n e is

5 8

3 75 .(1) N m

S O rr

1 2 5 2 ) - N m

SO rr

A di o de b r id g e re c ti f ie r co n n ec t ed to a 230 V, 5 0 H z si n gl e p h se si n u s oi d al ac su p p l y is fe e d i n g an R - L loa d a ss h o w n in t he f ig u re . T h e in d u c ti v eco m p o n e n t o f th e lo a d , L c an b eco ns i de re d to b e v er y h ig h s o th a t t h e lo ad c u rr e n t , iL c an be as su m e d to b e a p er f ec t d e c ur ren t _of m a gn i tu d e 10 A .T h e r m s v a lu e o f the so u r c e c u rre n t, sIS

I) 3 X 1 0 2 rr

2 ) 6 X 1 0 A rr

7 5 3) N m

S O rr

3 ) 3 X 10 A rr

254) N m

s or r

4) l O A

R

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5 9

6 0

t - :

16

The s p eed o f a 400V , 50 Hz, 1400 rp m , 3 -p hase in d u c tio n m o to r h avin g neg lig ib le

st a to r se rie s . im pedance, is co n tro lle d b y V / f m ethod o f speed c o n tro l to d riv e a ra te d

load a t 700 rpm . W h ic h o f h e fo llow in g st a tem ent is t ru e f or th is c a se ?

-(1) S p eed o f t h e ro ta tin g m ag neti c fie ld is 700 rp m a n d slip sp

eed is 50 rp m

2.) S p eed o f he ro ta tin g m agn etic f ie ld is 1500 rp m an d sli p sp eed is8 0 0 rpm

(3) S p e e d o f he ro ta ti n g m ag n etic f ie ld is 800 rpm a n d slip speed is 100 rpm

(4 ), S p ee d o f he r o ta ti n g m ag net ic fie ld is 750 rpm an d sl ip speed is 50 rpm .

The sw it ch , S o f th e d e to de b u ck co n v e rte r sho w n in th e figure is op era ted w it h a

duty c y c le o f 0.5 a t a sw it ch in g freq u en cy o f 5 kH z w h il e it is fee d in g a cer ta in lo ad .

W hile the in put v o lta g e o f the c o n v e rte r is m ain ta in ed a t 100 V d e, th e stea dy s ta te

avera ge v o ltage a t th e o u tp u t te rm in a ls o f the co n v e rt e r, V 0 is fo u nd to be 70 V .

W hich o f the fo llo w in g sta te m en ts is tru e fo r t h is c o n v e r te r? L

V

( 1) T h e co nv e rte r is o p era tin g u n d e r con tin u ou s m o d e o f co n d u ction

(2) T h e co nv e rt e r is o p er a ti n g und er d is co ntin uo us m o d e o f c o n d u c ti o n

(3) T h e con v e rte r is o pera tin g on th e bou ndary o f con tin uo u s and d is co n tin uo u s

m o d e o f co n d u c tio n

( 4) T h e co n v ert e r is o p era ti n g in u n sta b le m ode o f o p era tio n

61 . In th e n e tw o rk sh o w n in the figure, th e paralle l co m b in atio n o f R and L co n s ti tu tes

the in te rnal im pedance o f the so u rce w hereas th e se ri es com p in a tio n o f R 1 an d C

co n s tit u te s th e im p ed an ce o f th e load . l f t h e load is d raw in g m a x im u m pow er, Rt and

C can be exp re ssed in te rm s o f R an d L a s R

v sin t L

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(1) R = R(roL)z . C = Rz+((oL)z1 R2+(roL)2 m2LR 2

(2)R2 +(roL) 2

R = C =R(mL) 2

(3) R. = RmL2 . C = R2 +(roL) 2

1 R2+(mL)2 m2LR 24) R = R(OlL)2 c =

1 R2+(wL)2

62 . The state-space model of a dynamic system is given as;X1 = a X 2Xz 2ux2 = 3x 1 a 4 x 2 3u

2LR 2

R2 + o1 L) 2

R2+(oLro2LR2

What should be the relation e t w e e n ~and a 1 so that this system is controllable?

63 . In the single line diagram o f a three phase radial distribution system shown in thefigure, the line-to-line rated voltage is 11 kV. The per-phase real (P) and reactive (Q)

power drawn by the load at the receiving end is 400 kW and 100 kVAR respectively.Considering the receiving end voltage as reference, the power factor and total realpower in the three phase system at the sending end are

+ Sending

(1) 0.9 and 1250 kW

(3) 0.966 and 1225 kW

endP Q

Receivinbentl

(2)0.92and 1270kW

(4) 0.82 and 1300 kW

64 . In a two-terminal HVDC link, there are two following modes of operation:Mode 1: Converter 1 acts as a rectifier and converter 2 acts as an inverter.Mode 2: Converter 1 acts as an inverter and converter 2 acts as a rectifier.Under these conditions, which one of the following statements is true under normal,steady state operating conditions,?

(1) Conver ter 1 will always have constant current (CC) control (both in mode 1 andmode 2) while converter 2 will always have constant extinction angle (CEA)control (both in mode 1 and mode 2).

(2) Converter 1 will always have constant extinction angle (CEA) control (both inmode 1 and mode 2) while converter 2 will always have constant current (CC)control (both in mode 1 and mode 2).

S/75 POK/12-6 AE 2A

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(3) Th e conv erter, wh ich act s as a rec tifier, wi l l a lwa ys hav e constan t curre nt (CC)

contro l wh i l e the other one (wh ich act s as an in verter) wi l l a lwa ys hav e constan t

exti nct ion an gle (CE A) contro l.(4 ) The co nverter, w hich a cts as a r ectifier , will alw ays hav e cons tan t extin ction ang le

(C E A) contr o l wh i le the oth e r one (which a cts as an inver t er) will a lways have

constant cu r ren t (CC) con trol.

65 . T he term perc en tage in a per centage different ial rela y deno tes the fa ct ~ h t

(1) The diffe rential operat ing curren t is alway s a fix ed percen tage o fthe l ine curren tat the pr imar y s ide of h e CT s

(2) The ave r age res training c u r ren t is always a fixed percenta ge of the line cu rrent a tthe pr imary s ide ofthe C Ts

3) Th e numb er o f tu r ns in t he operat ing co il is alwa ys a f ixed perc entage of thenumb er of urn s in th e restraini ng coil

(4) Th e diffe rential o peratin g current is alw ays a fix ed per centage of the a veragerestr aining current fo r give n values of n u m ber of tu r ns in t he operat ing an d the

re straining c oil.

I M ATERIAL SC I ENCE I

66 . S ui tably co nnect the impro vemen t of prope rty wit h the enli s ted m icrostructu ral

fe a tures.

1. Creep2. Fati gue3 . Fractu re4. H ardnes s

( 1 1- b, 2-c, 3-a , 4 -d

(3) 1- a, 2-b, 3 -c , 4-d

a. low inclusio n vo lum e fracti onb. smooth surfac e finishc. larg e grain s i zed. finely distrib uted preci pitates

(2) 1-c, 2- b, 3-a, 4-d

(4) 1 -d, 2-b , 3-a , 4-c

67 . A zi n c rod im mersed in an ac id solutio n loses 2 5 m g d u ring 12 h o u r of exposu re. Theeq uivalen t current flowing dur ing c orrosio n is(a tomi c weight o f zinc is 65 .4 g / mol)

(1) 0.85 rnA (2) 1.7 rnA (3) 2.5 rnA (4) 3.4 rnA

68 . The diffu sion coef f ic ient o f s i lver in copp er at 650 C and 900C is 5.5 x6

m2/s

and 1.3 x 10 13

m2/s, r especti vely. The activa tion energ y for d i ffusion in J /mo l is

(1) 4 9.2 (2) 85.4 3) 98.6 (4 ) 196.7

S /75 P O K/12 6 A E 2 8

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69 . M atch the f ollow ing p roper t ies w ith th e too ls use d t o m easu re the m

1. elec trical cond uctiv i ty2. im p a ct en e rgy3. therm al ex p ansi o n4. s pecif ic hea t

(1) 1-b, 2 -c, 3 -d, 4- a

(3) 1-d, 2 -c, 3 -b, 4- a

a. cal or ime tryb. fou r prob e tec h niqu e c. Izod tes td . dial atom e try

(2 ) 1-c, 2-a, 3 -d, 4 -b

(4 ) 1-b, 2-d, 3 -c, 4 -a

70 . M atch the follow ing t y pe o f phas e dia g rams with the r e actio ns in v olve d

I. E utect ic2. Eu tecto id3. Per itecti c 4 . Mon otect ic

(1) 1-a, l -c 3 -d, 4-b (3 ) 1-d, 2 -c, 3 -b, 4- a

a. S - ~ S2 S3

b. Ll- -+L2+ S 1c . L--+ S 1+ S2 d . L S 1~ S2

(2) 1-c , 2-a, 3-d, 4 -b(4 ) 1-c, 2-b, 3 -d, 4 -a

71 . The yield stre n gth o f a po lycry stalli n e m a terial incre ases from 120 M Pa to 220 M P awhen its g rain s ize is decre ased from 0.04. m m to 0 .0 1 mm . The y ield s treng th of hesame mater ial : h en it s grai n size is 0. 0 01 m m is

(1 ) 628 M Ya (2 ) 652 M Pa (3) 6 78 M Pa (4 ) 69 6 M P a

72 M atch the fo llowi ng ty p e o f heat t reatm ent w ith th e pha ses o b taine d

1. Q uenc hing2. Mara g ing3. T empe ring4. A u s temp ering

(1 ) 1 -a , 2 -c, 3 -d, 4- b

(3 ) 1-d, 2'-C, 3 -b, 4 -a

a. interm etall ic com poun dsb. mart ensite c. bainit e d. epsil o n ca rbide

(2) 1- b, 2-a , 3-d, 4-c

(4 ) 1-b, 2-d, 3 -a, 4 -b

73 . Ma tch th e fol lowin g m at erials with their natur e

1. N b3Sn2. G aAs

3. Fe-S i allo y4. B aTi0 3

1) 1- c, 2-b , 3-a, 4-b

(3) 1 -c, 2-a , 3-d , 4-b

a. se m icon d ucto r b. di e lectr ic

c. s u perc o nduc tord. so ft ma g netic

(2) 1-b, 2 -a, 3 -d, 4- c

(4) 1-b, 2 -c, 3 -a, 4- d

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74 . An electrochemical cell consists of a nickel electrode in a solution ofNi 2+ ions on oneside, while a cadmium electrode immersed in a solution of Cd 2+ ions on the otherside. If the concentrations o f Ni 2+ and Cd 2+ ions are 0.002 and 0.6 M, respectively,and assuming Ni is oxidized, the change in the cell potential at 300K is close to(The half cell potentials ofNi and Cd are -0.250 and -0.403 V, respectively, R is

8.314 J/mol.K, F = 96500 Columb/mol)

( l ) 0.073 v (2) -0.08 v (3)-0.153V (4) -0.226 v

75 . What is the vacancy concentration in copper at 727C? Copper has density of 8.4glee and atomic weight of 63.5 g/mol.(The activation energy for vacancy nucleation is 0.9 eV/atom, Boltzmann constant is8.62 x 10-5 eV/K and Avogadro number is 6.023 x 1023 atoms/mol)

( l ) 1.7 x 10 24 vacancies/m 3

(3) 2.3 x 10 24 vacancies/m 3

I LUID MECHANICS I

(2) 2.0 x 10 24 vacancies/m 3

(4) 2.5 x 1024 vacancies/m 3

76 . Consider an incompressible flow over a flat plate with a free stream velocity of U.The flow is in the x direction and y is perpendicular to x. The similarity variables are

(u/U) and ll The variable is expressed as , where v refers to kinematic~ v x u .

viscosity of the fluid. The boundary layer thickness, o t t ~ i n sits maximum value of5.0 as

YJ

oo, i.e. at =tmax

= 5. The variable u/U is a function of dimensionlessstream function, f(ll) and given as {u/U) = f (11). It is known that at = 5, f(ll) =3.2833 and f ( l l ) = 0.9915. What is the ratio of displacement thickness to boundarylayer thickness at = llmax?

1) 0.51 (2) 2.12 (3) 0.343 (4) 0.25

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77 . T he ga p b e tw e e n th e tw o p la te s , as sh ow n inth e fig ur e, is fil l ed w it h a f l u i d o f co n st an td e n s it y p I 0 0 0 k g /m 3 T h e to p p la te m ov e sdo w n w ith a v e lo ci ty o I m m /s. T h ev el oc it y di st ri bu ti on a t th e ex it is gi ve n by

u y ) = Uo {h~t h~t )) 2} W h er e h t) is the g a p he ig h t a s a fu n ct io n o f t im e , y is th e c o o rd in a te a s sh o w n in th e f ig ur e, a n d U 0 is a fu nc ti o n o f tim e .

G i ve n h a t t 0 ) ho = I m m , L = I m m .W h ic h on e o f th e fo ll o w in g o p t io n s is tr u e ?

( I) A tt = O .I s , h = 0 . 9 m m , U 0 = 11 .1 1 m m /s

2) A t t = 0 1 s, h = 0. ~ m m , U 0 = II . I I m m /s

3 ) A t t = 0 1 s, h = 0. 1 m m , U 0 = 8. 33 m m /s

4 ) A t t = 0 1 s h = 0 .9 m m ,. U 0 = 8. 33 m m /s

7 8 . M o o d y s c h ar t pr es e nt s a f\ Jn ct io na l re la ti o n b e tw e en f ri ct io n fa ct o r a g ai ns t R ey n ol ds n um b e r, fo r d if fe re nt p ip e r ou g hn e ss v al ue s . O n e o f the f o l lo w in g is N O T a fe a tu re o f t h is d ia gr am .

( I ) Th e p lo t is p re se n te d on a log - lo g ax es .2) In a fu ll y t u r b u l ~ n t re g im e , th e p ip e r ou g h n e ss is ind e pe n de n t o f R ey n ol ds

n u m be r. 3 ) F o r la m in a r f lo w s, th e f ric ti on fa ct or is d ir ec t ly p ro po r ti o n al t o Re y no ld s n um b e r.

4 ) In th e tu rb ul en t re g im e fo r th e s am e v a lu e o f R ey n ol d s nu m b er , as th e p ip e ro u g h n es s in c re as es , th e va lu e o f fr ict io n fa c to r in cr ea se s .

79 . A s i m pl if ie d m od e l o f an in d us tr ia l c o ni ca l ta n k o f h e ig ht I 0 m , w hi ch is fu ll y fi ll ed w it h a pr o ce ss f lu id is sh ow n in F ig ur e . Th e to p a nd b o tt om su rf a ce s ar e ci rc ul ar w ith ra d ii 1 .0 m a n d 0 1 m r es p ec ti ve ly a n d to ta l co n e an g le is 6 0 . T h e ra te o f c ha n ge o f s u rf ac e hei ght ~~) at a h ei gh t (h ) = 3 .3 3 m is g iv e n as

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1 - 2 3:7 8 5 m /s 2 - 2 .3 7 8 5 m /s 3 ) - 0 .1 1 3 3 m / s 4 - 0 .0 2 1 8 m /s

80 . T he v elo city fie ld and th e d e n s ity field in a d if fu se r is g iv en b y u = u 0 e-x1L

p = p0e - 2x L w h ere L is th e le n g th o f th e d if fu se r. T h e to ta l ra te o f c h a n g e o f density

at x = L is g iv en by

(1 ) 0 3 ) - 2 p o u o e-3 L

81 . M atch th e fo ll o w ing n o n -d im en s io n a l g ro u p s w hich a re ca te g o r ized in to tw o se ts A

and B. ~ ~ B

1 Froude n u m b e r

2 W eber n u m b er 3 E ule r n u m b er 4 G ra s h o fn u m b e r

(1 ) 1 - T ; 2 - Q ; 3 - P ; 4 - S

3) 1 - Q ; 2 - T ; 3 P; 4 S

P L o cal a cce le ra tio n t o co n v ec tiv e accel erat io n

Q ) In e rtial fo rc e to g rav it y fo rce

R) In er ti a l fo rce to p ressu re fo rce

S) T h er m a l b u o y an cy force t o v isc o u s f orce

T ) In e rt ia l force to su rfa ce ten s io n fo rc e

2) 1 - Q ; T; 3 - R; 4 P

4 ) 1 - T ; Q ; 3 S; 4 P

82 . A v en tu r im e te r is u sed to m eas ure flow ra te in a p ip e o f 3 2 0 m m diam eter. T h e

d iam ete r o f th e th roat o f th e v en turim e te r is 160 m m . T h e d iffe ren tia l m ano m eter

that co n n ec ts the p ip e and th e th ro at s h o w s a readin g o f 1 m a nd sp ec ific g ravity o f

gauge f lu id is 1.4. I f th e co e ff ic ie n t o f d ischarge o f th e v en tu r im ete r is 0 .9 8, and the

value o f g is 9.8 1 m /s2

, w h a t is t he flow ra te?

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83 . C o n si d e r a tw o -di m en s ion a l in c om p re s s ib l e flo w o ver a fl a t p la te w ith x a s th e flo w d ire c tio n . T he b ou n da r y la y er g row s in y d ir ec t ion . U s u al l y a b ou n d ar y lay er w ou l d g ro w wi th i n cr e a sin g x . H ere un if orm su c tio n , v 0 is app l ie d to tqe p la t e on th e d o wn s tre a m in o rde r to k eep th e bo u nd ary la ye r th i c kn ess con s tan t . T h e ki n em atic vi sco s ity is v A full y de v elo ped si t u atio n e v o l v~ s far d o wn stre a m . T h e far fi e ldve locit y is o o ~Wh at is th e vel o ci t y di s t ri b u tio n u y ) in this reg io n?

1) u y) = Uco [ 1 - e ~ y]

3) u y ) = Uco

2) U y ) = Uco [ e ~Y]4 ) u y) =

84 . A u nifo rm je t o f w ater is i s suin g o u t o f a noz z le o f di a m e ter d = 0 .0 I m. Th e j et sp e ed w he n it e xi t s fro m t h e n o zz le is 2 .0 m /s . Th e je t im p in g es o n a ho r iz on tal f latp la te and flo w s ra d ial ly o u tw a r ds in th e fo rm o f a flat s hee t a s s ho w n in fig ure . If thedi s tan c e b e tw e e n t he n o zz le an d fl at p la te is 1.5 m wh at is th e for ce re q ui re d t o ho ldth e pla te in pla c e? Neg lect th e w e ij h t o ft h e p late and the w ei g ht o f th e w a ter she e t.A ss u m e w ate r de nsity a s I 00 0 kg lm .

I) 0. 31 41 { 2 ) 0. 9 08 N 3) 1.81 6 N 4) 6 28 N

8 5. V elo c ity co m po n en t s in a flo wfi e ld are g iv e n by u = 0 v = a l - z 2 ) , w 2a y z .Ex p re s s ion s fo r the stre am fun c tion \fl) an d ve lo ci t y po ten t ial ), res p e ctiv ely , ar e

az 3I) \f = a y 2 z - - + co nsta nt; I>d oe s no t ex ist

3

az3 3 .2 ) \f = ay 2 z - 3 + co ns t a nt I> = -a yz

2 + ~ + c o ns t an t az3

3) \f = -a y z 2 + 3 + con st a n t; I> d o es no t exi s taz3 2 ay34) \f a y 2 z + 3 + c o n s ta n t ; I> = ay z - 3 + c on s ta n t

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O L ID M E C H N I C S

86 . T he th in so l id s h af t o f le ng t h a nd r ad iu s r (sh o wn in th e F igu r e ) is fi x e d a t on e

e n d . A r ig i d ro d is a t tac h ed t o th is s h a f t a t t h e o t he r e nd . Th e ri g id r o d i s su b j ec t e d

to a se t of e qua l op posi te f o rce s P a t it s en d s p e rp e n d i c u la r to t h e a x is o f th e sh af t

a s sh o w n . Th e r ig id i ty m od u lus o f th e m a t er ia l of th e sha ft is g iv en b y G.

D e te r m in e th e m a xim um ten s i le s tr e s s a t po i n t A o n t he su r face of th e sh a f t a s

sh ow n in the F ig u r e .

P h

) n r3

Sh af t

4P h

2 ) n r3

p

. 3

) B P L n r3

87 . A s im p l y -s u p p o r te d co n c re t e b r id g e gi rd er h as a sp a n o f m. U nd er su n li g h t, th e to p

su r fac e o f th e g irde r h e a ts up to +4 0C , bu t t h e bo t tom su r f ace rem a in s a t + 25 C se e

F ig u re ). T h e cr o ss s e c t i o n o f th e g i r d e r is r e c ta n g u la r a n d o f si z e 6 m m x 8 m m .

T h e M o d ulu s o f E la s t ic i t y E an d the co e ffic ien t o f th erm al e xpan sio n a of c o n c r e te

a r e 25 ,0 00 M P a an d J O xJ 0-6 ICC r esp e cti v ely . A ss um e th a t a ) th e gir d er r e m a in s u n

c rack ed un d e r th e d iffe r en t ia l lo ad in g d ue t o t e m p e ra t u re , and b ) th e g ra d ie n t of t em p e ra t u re is u n i f o rm ac ross the d e p th of th e g ir d er. W ha t is t h e m ax i m u m

a d d i t io n a l fl e xu ra l n o rm a l s t r ess in t h e g i rd e r?

4 0 o

2 5 o

~ ~ ~ B m ~ 4

1 ) 0 MP a 2 ) 1 .5 M Pa 3 ) 3 . 0 M P a 4) 8.0 M P a

A

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8 8 . A tw o -b a r p ro p is u se d to su p po rt an a rt if ac t in a M u se u m see F igu re) ; th e bars a re co nn ected to ea ch o th e r by fr ic ti o n les s h in ges. B oth th e bars are m ad e of the sa m e m ate ri a l p ro p er ti es E a n d v and h av e n eg lig ib le m ass . T he area s o f cro ss-sec tio n o f the tw o b ar s are A an d fiA as sh o w n in th e F igu re . I f the w eigh t o f the ar ti fa c tsu p po rte d is W w hat is th e elast ic a x ia l st ra in en erg y s to red in th e tw o bars tog e th e rdue to ap p li ca ti o n o f w eig h t W?

1 ) w z L 2EA

2 ) . w z L EA

w

3) W2

L

2EA

89 . O bta in th e e la st ic f le xural defle ct io n .:: \ a t t he free e nd o f th e c ant il ev e r beam show n in the F ig u re sub jected a co ncen tra te d loa d at a d is ta n ce o f L from th e su ppo rt. T h ebeam cr o ss -s ec tio n h as a m om en t o f iner ti a o f L T h e m ate ri a l o f th e beam has M od ul us o f E las ti ci ty E

1 ) J. = P L

4 8 E l

S /7 5 P O K / 1 2 6 A E 3 A

P L 2 ) = 2 E l

f

3 ) J. = S 3

6 E l 4 ) = SP L

3

3 E J

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90. A b eam w ith a c an til ev e r o v er h a n g is loa d e d a s sho w n in t h e F ig ure . W h at a re th e

S h ea r F o rc e a n d B e n d in g M o m e n t D ia g ra m s ?

1 )

F or c e

B end in g M o m en t

O +o- ~m oi o~ol . l. l

3 )

S h e a r F or ce

I

w

l ******l l ~

:2 l_ L ___ _ _ _ _ _ _ +

2 )

S h e a r

F or c e

B en d ing M om e n t

o ~ m w L L I

4)

S h ea r

F o rc e

B en d in g , .. , ~mTi rT TTT' IT' IT 'ImrTTTTT TT TI'

ml - 1 M o m e nt 1

I

w e 2

B e n di n g M o m e nt

91. A fix ed-b ase p o rt al fr am e h as u n if o rm c ros s -s e ctio n o f b rea d th b an d de p th h A

se c ti on o f the p o rt al fr am e is su b ject ed to s h e ar fo rc e V an d b en d in g m o m e n t

= Vh . W hat is th e fle xura l n o rm a l str ess a t a he igh t o f h 4 fro m th e to p fib re o f th e

se c ti o n?

v I -

b h

2 v ) 2 b h

v 3) 3 b h

v 4 ) 4 b h

S /7 5 P O K / 1 2 6 A E 3 8

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9 2 . A n e la s t ic sp rin g o s tif fn e ss k is lo a d ed w ith a W eight W at ts free e n d (S ee F ig u re ) . T w o o p t io n s a re av a ila b le fo r th e e th o d o ap p l ica tio n lo ad in g :

9 3 .

O p t io n A : S lo w ly ( a lm o s t s ta tic a lly ) O p tio n B : S u d d en ly (d y n a m ic a l ly )

A s s u m e th e re is n o d am p in g in t h sp r in g -m as s sy s te m . W h at is th e ra tio o th e m a x im u m d e flec tio n a t th e free nd o th e sp r in g g e n e ra te d in O p tio n B to th a t in O p t io n A ?

(1 ) 0.5 (2 ) 1 .0 3) 1.5

A n o b je c t m ad e o .e la s tic m a te r ia l (M o d u lu s o E las tic ity a n d P o i s s o n s R a tio p is re s tra in ed fro m m o v in g in th e x -d i rec tio n b y tw o r ig id w a l ls as s h o w n in t h e F ig u re ; it is f re e to m ove in t h e o th e r t w o d irec tio n s . A s tre ss o cr is ap p lie d o n it in th e d irec tio n a s sh o w n . F in d th e re la tio n sh ip b e tw een th e ap p lied c o m p re ss iv e s t r e s s Y an d th e co m p res s iv e stra in E in t h e y -d i re c t io n .

(1 ) a = (2 ) J

1 tt (3 ) a = .

1 ~ t

k

w

(4 ) 2 .0

y j

l X

j

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2 8

94 . A h a n d r o l le r o f m a s s m a n d d ia m e te r 5 0 0 m m s r eq u i r e d t o be m o v e d to a h ig h e r

g ro u n d le v e l o f s t e p 3 0 m m a s sh o w n in th e F ig u re . w o p o s s ib i l i ti e s e x is t t o ta k e it

to t h e h ig h e r g ro u n d le v e l, n a m e ly a) it c a n b e p u sh e ov e r t o th e h ig h e r le v el, o r b )

i t c a n b e p u l l e t o the h ig he r le v e l S e e F ig u r e ) . T h e c o e f f i c i e n t o f f r ict io n b e tw e e n

t h e r o lle r a n d th e g ro u n d i s 0 2 D ra w th e f re e b o d y d ia g r a m o f th e r o ll e r a lo n e a t th e

in s ta n t o f ro l l o v e r o r p u s h o v e r fo r th e tw o c a s e s .

F

1 )

2)

3)

4)

C a s e a ) C a s e b )

F

F

F

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H E R M O D Y N M IC S

96 . A w e l l- in s u la ted r ig id ta n k is co n n e c ted th ro u gh a v al v e to a m a in s u p p ly li ne w hic h

ca rr ie s a ir a t 0 .8 M P a an d 30 C . T he t a n k is in it i al ly co m p le t e ly ev ac u a ted . T h e valve

is now o p e n e d an d a ir is a l lo w e d to f lo w in to th e ta n k u n ti l th e tan k p re s su re reach es

0.8 M Pa, at w h ic h p o in t t h e va lv e is c los e d . K in etic an d p o ten ti a l en er g y e ff ec ts a re

neg li g ib ly s m al l . T h e fi na l t em p er at u re o f a i r in th e t an k w il l b e

( 1) less t h an 30 C (2 ) e q u a l to 30 C (3 ) gr ea te r t h a n 30 C

(4 ) a fu nc ti o n o f

ta n k v ol um e

97. A sea le d ri g id v es se l h a v in g a v o lu m e o f m3 co n ta in s a s a t u r a t ed m ix tu re o f liq u id

an d v ap o u r of ref ri g e ran t R -1 34 a at l0 C . T h e vess e is n o w he at ed to 50 C , an d th e

p ro cess st op s w h e n th e l iq u id ph as e ju s t d isa pp e a rs c o m p le te ly .

R l3 4a P ro p er ti e s : at 50 C , sa tu ra ti o n p re ss u re, Psat = 1 3 1 8 kPa ; sp ec if ic vol um e o f

sa tu ra ted v a p o ur , g = 0. 0 1 5 m3 /kg. .

A t l0 C , sa tu ra ti o n p re ssu re , Psat=4 1 4 .9 k P a; sp ec if ic v o l um e o f s a tu ra ted l iquid , Vr =

0 .00 0 8 m3/k g; s p e ci f ic v o lu m e o f sa tu ra ted v ap o ur , g = 0 .0 5 m

3 /kg .

T h e in iti a l m as s o f th e li q u id (i n kg) w a s

(1) 11. 54 (2 ) 2 8 .4 6 (3 ) 3 8. 48 (4) 9 4 .8 5

98 . A c o a l-f ir ed p o w e r p la n t, o p er at in g on a si m p le ide l R a n k in e cy cl e , has a b o il e r

pr es sure o f 3 2 b a r an d a co n d e ns e r p re s su re o f 0. 75 bar . T h e ex it st a te f ro m th e st eam

tu rb in e is d ry s a tur at ed v ap o u r . G iven sa tu ra ted w a te r p ro p e rtie s:

Pr es su re (ba r)

0. 75 32.0

vrm-3 /k g )

hr (k J /k g )

hg(kJ /k g )

0 .0 0 0 3 7 3 84 .3 6 2 6 6 2. 96

0 .0 0 1 2 2 4 0 2 5 280 3

S f

(k J/ k g K )

1.2129

2 .678

Sg

(k J /kg K )

7 .4 5 6 3

6 .1 6 0

S u p e rh ea t s te am pr op er ti e s : a t 32 bar 592 C , hg = 36 63 k J /k g , S g = 7 .4 5 63 k J/kg K .

T he s p ec if ic h ea t a d di tio n in the b o iler ( in kJ /k g) an d t h e s p e ci f ic p u m p w o rk inpu t

(in k J/ kg) w o u ld be

I ) 1778 a n d 3 . 2 4

(3) 263 8 an d 6 4 1 .6 4

(2 ) 24 15 .3 a n d 6 .4 8

4 ) 3 2 7 5. 4 a n d 3. 2 4

9 9. A fr ict io n le ss p i s to n -c yl in d e r d evi ce c o n ta in s I kg o f a g as , h a v in g asp ec if ic gas

co ns ta n t o f 0 .5 kJ /k gK . A s s u m e 1 5 to be th e sp ec ifi c h e at ra ti o o f th e ga s. T he pis to n

is free to m ov e . H eat is a d d e d to th e g as un til th e te m p e ra tu re r is es b y 10C. T he

am ou n t o f h e a t a dd ed ( in k J) is

I) 1 2 ) 10 3) 15 4 ) 28 3 .

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100. Con side r atm os p he r ic ai r wh ich is a n ai r-wa ter v ap o ur m ix t u re (con side red to be anide a l g as m ixtu re) a t 100 kPa, a t a dry bu lb te m pe ratu re o f 15 C and 40 rel ativ e h u m id ity. W a ter s atur atio n pr e ss u re t 15C = 1.7 05 kPa. (s p eci fic g as c on s tant) of w at e r v a p ou r = 0 .461 kJ/k.gK an d o f air== 0 .28 7 kJ/k.gK. T h e hu m id ity ratio in gmof w ate r/kg of d ry a ir) is

(1 )4.2 75 (2) 6 .867

10 I. An in ven tor cl a im s to ha v e de v elop ed a cy clic he a t c onv e rsio nsy stem (H E) w hi c h s im u lta n e ou s lypr o du c es pow er, refr iger atio n a n dh e atin g o utpu t' b y in tera ctin g w ithth re e ther mal re s e rv o irs as sh own inthe f igure . Th e sy stem p r o po s ed i s

(3) 10 ,688

(1) no t f ea sible as it vio late s fir st la w o f th e rm o d yn a m ic s(2 ) no t fea sible as it vi o late s se con d law o f ther m od yn a m ic s (3) f easi ble and it is a re ver s ible sys t em(4) feas ible and it is an irrev ersi b le s yste m

(4 ) 17 .168

I 0 2. A n ev ac u a ted rigi d ta n k is fill ed u p to ce r tain pres sure from a pres suri zed gas supp lylin e. C on s ider tw o case s bo th f ollo w in g the i e ~ l gas modeL In th e firs t ca s e (c ase I)th e ta n k is ins ulat ed a nd in th e se cond ca se (c ase 2) h eat is lo st to the su rroun din gsfrom th e gas in th e tank . Ig nore kin etic and po tent ial e nerg y e f fect s. W hic h of the foll o win g st atem en t s is c or r ect a bou t the two ca ses r egar ding fin al m ass o g as in the ta nk?

I) F ina l ma ss in ca s e I is eq ual to th at in ca s e 2(2) F ina l m a ss in ca s e I is les s th a n th at in ca se 2(3 ) Fi nal m ass in c as e I is grea ter tha n that in c ase 2(4) C om pari son betw ee n fina l m a ss in ca se a n d c ase 2 is not p oss ible from giv en d ata

103. A stea d y f low ad i abat ic t u rb i n e o pe r a tes wit h ai r (a ssum ed .to be a n i dea l gas) ent erin g at 500 kP a , 425 K a nd 15 0 m /s an d le avin g at I 00 kP a, 3 25 K a n d 50 m/s.A ssu m e o (s u rrou ndin g te m pe ratu re ) = 27 C. A ls o ass ume , fo r air, sp e c ific he a t cp = I .0 kJ/k.gK an d s pec i fic g as c on s tant R = 0. 3 kJ /kgK . T he m ax im u m w or k (inkJ / kg) that can be p rod uced by this turb ine is

(1 ) 11 0 .0 (2) 1 64.0 (3) 1 74.4 (3) 348 .8

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1 04 .

32

L en oi r c yc le , w h ic h a p p ro xi m a tes th e o p e ra t io n

o f a p u lse je t , is sh o w n in th e f ig u re o n th e

pr es su r e -s p e c if ic v o l u m e p la n e. Th e p ro ce s se s a re

as fo llo w s: p

i 1 -2, is o ch ori c, h e a t a dd it i o n o f2 3 0 k J/k g ;

ii 2 -3 , a di a b a ti c e xp a n s io n , ch a n g e in in te r n a l e ne rg y = 75 kJ /kg ;

iii) 3- 1 , is o ba r ic co m p re ss ion , h ea t r ej e ct i on o f

1 9 0 k J /k g.

3

v

T h e n e t w o rk ou tpu t k J/ kg ) o f th e c yc le an d c ha n g e in in te rna l e n e r gy k J/k g) du r in g

pr o ce ss 3-1 a re

1 )- 4 0 an d -15 5 2 ) 4 0 a nd - 15 5 3) 4 0 a n d 15 5 4) -4 0 an d 155

1 0 5 . T h e s lop e o f sat u ra t io n p re s su re - te m p e ra tur e c u r v e P t v s. T o f am m o n i a a t 50 C is

50 k N / m 2K . Th e s pe ci f ic vo lum e o f sa tu r a te d a m m o n i a v a p o ur a t 50 C is 0 .06 34

m3

/k g, w h ile th e sp e ci f ic vo lu m e of sa tur a te d l iq u id is ne g li g ib le c o m p a re d to tha t of

th e v ap ou r . T h e e n t h a lp y o f v ap or iza t io n in kJ /k g ) o f am m on ia at 50 C , u si n g

Cl a p e yr o n eq u a t io n , i s

1) 15 .85 2 ) 15 7. 5 3) 10 23 .9 4) 3 9 68 0

I E L E T R O N I S J

10 6 . A s su m in g ide a l O P A M P s , the o u tp u t v ol tag e Vo in the ci rcu it s ho w n b e lo w w il l b e:

lien 2 kn

2kn

1) 2 V 2) 4 V 3) 6V 4) 8V

1 0 7 . A n a m p li f ie r h as a m id -b a n d g a in of 1 00, a low er 3 -dB f req u e n cy o f 0.1 H z an d anup pe r 3 -dB f r eq ue n c y o f 120 H z. W h a t is th e r a ng e o f f re q u e nc y for w h ic h t h e ga in

w ill b e at lea st 9 9 ?

1 ) 0 .3 - 8 H z 2) 0 .7 - 1 6 .9 7H z 3) 1 - 1 0 H z 4 ) 0 .6 - 1 2 H z

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108 . In the circuit shown below, what will be the value of R for Yo to become zero?Assume the following:

Diode conduction drop Vn) = 0.7 V

Transistor: VBr: = 0.7 V and hn, is very large.

1) 2.06 k 2) 1.56 k 3) 3.6 kn 4) 2.63 k

109 . Consider a fast-food restaurant in which a customer is nine times as likely to order achicken burger as a fish sandwich. Suppose that each customer orders either one

chicken burger or one fish sandwich. If the restaurant serves an average of eightcustomers per minute, the information rate of food order is

1) 0.0625 bit/sec 2) 0.55 bit/sec 3) 3.25 bit/sec 4 1 bit/sec

110 . The ratio of total average power to the average carrier power of sinusoidal amplitudemodulation is 1.18. The rms value of antenna current for such t ransmission is 10.86ampere. The corresponding unmodulated current in ampere is given by

l ) 15 2) 8 3) 10 4) 20

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3 4

111 . A n L T I s ys te m is ch a ra ct eri ze d b y t h e d if fe r en ce e qu a ti o n

y n ] 0 .2 y [ n - 1] + 0 .0 1 y [ n 2] = x [n ]

T he im pu ls e r e sp o n se of t he sys te m is

1) n(0 .1 ) u(n )

3) n( 0.1 ) + 1 u (n + 1)

(2) (n+ 1) (0 .1 ) +1u( n)

(4 ) (n +1 )(0 . 1 ) u( n+ 1 )

11 2 Th e in ve rs e z-t ra n s fo rm of z - 2 lo g 1 - 2z -1

) w it h R O C = lzl > 121 ) is

2 n .

(1) n 2 ) u n - 2)

(3 ) ~ 2 nu n 2)n - 2 )

2n (2) - u n )

n

113 . A r an d o m si gn al is gen e ra te d bls p a ss i n g w h it e no is e si gn al t h r ou g h an ide al L T I

sy st em w ith t ra n s fer fu n c tio n e 5

I f th e au toc or re la ti o n o f i np u t r an do m s ig na l is

R -r) = 3 e- 21TI th e p ow e r s pe c tr a l de ns ity of t h e o u tp u t s i gn al is

(1) 1 2e - js w (2 ) 1 2 e- jl O w

1 1 4. A B o o l ea n al g e br ai c e x pre s s io n F is g iv en a s:

F = A + B) A + A + B )C + A B + C ) + A B + AB C

w h er e A B a nd Ca re lo g ic a l va ri ab le s. T he m o st s i m pl if i ed e x p re ss io n fo r F is:

1 ) C A + B ) + B + C

(3 ) C A + A B

(2 ) (C + A B) A+ B C )

(4 ) C A B+ AB + BC

115 . T h e an a lo g ou tp u t vo lt a ge o f a R -2 R la d der ne tw o rk b a sed 6 - b it D AC w it h a

re fe re nc e vo lt ag e of 5 v o lt s a n d fo r a d ig ita l i n pu t of 01 11 0 0 is

1) 1 .0 9 3 7 v (2 ) 2.1 8 75 v 3) 0 .5 4 68 v (4 ) 4.37 50 v

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- m f R O U G H W O R K

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7

~ ~ R O U G H W O R

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8

~ ~ R O UG H W OR

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ttfJ q;p;iJROUGH W O R

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C fi P l/R O U G H W O R

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I IE N G I N E E R I N G

S I E N E S N S W E R S H E E T i

LS_e_r_ia_I_N_o__________ _ __Roll Number Booklet Code Medium Subject Code Centre Code

IF:I :: J = ~I l lFll [ ] [] [] m ~ _____ _ , _ __ , _____, , _____,CD CDCDCDCDCD 0 0 0 0 0 0

0 0 0 0 0 0

ipj ji

ANSWERQ.N o. CD 0 1 0 0 0 0 0 03 0 0

0

4 0 0 05 0 0 06 0 0 0 0 0 08 0 0 0 0 0 01 0 0 00 0 012 0 0 0 3 0 0 014 0 0 015

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0 0 0

27 0 0 028 0 0 0 29 0 0 03 0 0 031 0 0 032 0 0 033 0 0 0

CD

0

CDCD 0 0 0 0

A N S W E R C O L U MN S

ANSWERQ.No. CD 35 0 0 036 0 037 0

0

38 0 0 9 0 04 0 0 41 0 0 042 0 043 0 0 44 0 0 0 45 0 0~ m iD46 0 0 047 0 ~ 046 0 0

9 0 0 05 0 05 0 0 2 0 053 0 05 4 0 055 0 056 0 057 0 056 0 0 0 59 0 0

0

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0

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ANSWERO .No. CD @ @ 069 0 0 07 0 0 0

7 0 0 0

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75 0 0 0

76 0 0 0

77 0 0 078 0 0 0

79 0 0 0

8 0 0 ~ 081 0 0 0

82 0 0 0

84 0 0 085 0 .

86 o o o67 0 0 0

88 0 0 tf 0

89 0 0 09 0 0 09 0 0 0

92 0 0 0

93 0 0 0

94 0 0 95

96 097 0 98 099 01

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ill

0

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0 0

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INS T RUCTION FOR INVI GILA TO R

Check Roll Number Booklet CodeSu bject Code Centre Cod e Signature of

Candidate before sig ning.

Signature of lnvigilator

ANSWERO . No. CD 01 4 0 0 01 5 0 0 01 6 0 0 01 7 0 0 01 8 0 0 01 9 0 0 011 0 0 ~ 0

0 0 0 12 0 0 0113 0 0 0114 0 0 0 115 0 0 0

I I

o zCD o ~ aro Jc..o: ; =

Z T C

CD30 0OCD X _ - .CDC/

n

i.2:CD

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CSIR Engineering Sciences Solved December 2012