SCTO Sampldjfjde Paper

8/18/2019 SCTO Sampldjfjde Paper

1/14


2/14

[A–2]

10. Contour lines on a map join places of:

(1) Equal mean atmospheric pressure

(2) Equal average temperature

(3) Equal rainfall

(4) Equal height above sea level

10. ekufp= ij leksPp js[kk,a fdu LFkkuks dks tksM+rh gS%

(1) leku ek/; okrkoj.kh; ncko(2) leku vkSlr rkiØe(3) leku o’kkZ (4) leqæ Lrj ds Åij leku Å¡pkbZ

11. A writ, which is in the nature of command issued by

the court asking a public authority to perform a publicduty which it is bound to perform or to refrain from

performing a particular act which it should not

perform, is called:

(1) Certiorari (2) Habeas corpus

(3) Mandamus (4) Quo warranto

11. U;k;ky; }kjk fuxZfer ,d lekns”k tks ,d daekM dh rjg fdlh

lkoZtfud çkf/kdkjh dks lkoZtfud dk;Z djus tks oks djus dks ck/; gS

rFkk fdlh dk;Z dks ugha djus dks tks mls ugha djuk pkfg,] dgk

tkrk gS%

(1) lfVZ;ksjkjh (2) gS fc;l dkiZl(3) eUMel (4) Doks okjUVks

12. The capillary action phenomenon of water climbing up

a narrow tube dipped in w ater is due to:

(1) Surface friction

(2) Surface tension

(3) Atmospheric pressure

(4) Differential temperature

12. ty es Mwch ,d ldjh uyh es ikuh ds Åij p


3/14

[A–3]

18. The average age of 10 students is 15 years. When 5

new students joined, the average age rose by one

year. The average age in years of the new students is:

(1) 18 (2) 16

(3) 15 (4) 17

18. 10 fo|kfFkZ;ks dh vkSlr vk;q 15 o’kZ gSA tc 5 u, fo|kFkhZ “kkfey

fd, tkrs gS rks vkSlr vk;q 1 o’kZ ls c


4/14

[A–4]

26. Three wheels make 60, 36 and 24 revolutions per

minute. Each has a red spot on its rim, which is at the

lowest position at time zero. The red spot will all be at

this position again after:

(1) 2 seconds (2) 5 seconds

(3) 4 seconds (4) None of these

26. rhu ifg;s 60, 36 ,oa 24 pôj çfrfeuV yxkrs gSA çR;sd ds fje

ij ,d yky nkx gSA tks fd “kw U; dky ij fuEure fLFkfr ij gSA

yky nkx iqu% mlh fLFkfr ij gksxk fdrus le; i”pkr%

(1) 2 ls ds .M (2) 5 ls ds .M(3) 4 ls ds .M (4) bues a ls dks bZ ugha

27. If 2 logx (x–2) = log

x 4, then the value of x is:

(1) 1 (2) 3

(3) 4 (4) 2

27.

;fn

2 logx (x–2) = log x 4,

rks

X

dk eku D;k gS%(1) 1 (2) 3

(3) 4 (4) 2

28. Which of the following in the largest number?

(1) 31/3

(2) 51/5

(3) 21/2

(4) 71/7

28. fuEufyf[kr es ls dkSu&lh lokZf/kd cM+h la[;k gS\

(1) 31/3

(2) 51/5

(3) 21/2

(4) 71/7

29. In a throw of two dice, the probability of getting a sum

of 9 or 11 is:

(1) 2/9 (2) 7/9

(3) 5/9 (4) None of these

29. nks ik¡lks dh ,d Qsd es vadks dk ;ksx 9 vFkok 11 vkus dh

çkf;drk gksxh%

(1) 2/9 (2) 7/9(3) 5/9 (4) bues a ls dks bZ ugha

30. The triangle joining the points (2, 7), (4, –1), (–2, 6) is:

(1) Equilteral

(2) Right angled

(3) Isosceles

(4) None of these

30. fcUnqvks (2, 7), (4, –1), (–2, 6) dks feykus ls cuk gqvk f=Hkqt

gksxk%

(1) leckgq (2) ledks .k(3) lef}ckgq (4) bues a ls dks bZ ugha

DIRECTIONS: For answering question, study diagram which is

represented as follows and select the appropriate choice.

ABDE represents Artists

FGHI represents Scientists

JKLM represents Administrators

OPQN represents healthy people

31. Scientists who are also artists but in not good state of

health belong to the area:

(1) 4 (2) 5

(3) 9 (4) 10

funs Z”k% fuEufyf[kr ç”ua ds mÙkj ns us ds fy, js[kkfp= dk v/;;u djus ds i”pkr mfpr fodYi dk pquko djs aA

ABDE iz nf”kZr djrk gS dykdkjks a dks FGHI çnf”kZr djrk gS oS Kkfudks a dksJKLM çnf”kZ r djrk gS ç”kkldks a dks OPQN çnf”kZ r djrk gS LoLFk O;fä dks

31. oSKkfud tks dykdkj Hkh gS ysfdu LoLFk ugha gS] fdl {ks= ls

lEcfU/kr gS%

(1) 4 (2) 5

(3) 9 (4) 10

DIRECTION: Select the suitable alternative to satisfy the

relationship in the following question.

32. Buffalo : Leather : : Sheep:

(1) Wool (2) Cotton

(3) Fur (4) Silk

funs Z”k% fuEufyf[kr ç”u ls muds lEcU/kks a dks lUrq ’V djus okys lgh fodYi dks pqfu,A

32. HkSl % peM+k % % HksM+%

(1) Åu (2) dikl(3) Qj (4) js”ke


5/14

[A–5]

DIRECTION: Select the choice out of the given choices which

gives the given words in the correct alphabetical order.

33. Nasal, New, Nine, Noble:

(1) Noble, New, Nasal, Nine

(2) Nine, Noble, New, Nasal

(3) Nasal, New, Nine, Noble

(4) New, Nasal, Nine, Noble

funs Z”k% fuEufyf[kr fn;s gq , “kCnksa dks fn;s x;s fodYiks a es a ls vaxzs th o.kZekyk ds vuq lkj lgh Øe esa vkus okys fodYi dk pquko djs aA

33. Nasal, New, Nine, Noble%

(1) Noble, New, Nasal, Nine

(2) Nine, Noble, New, Nasal

(3) Nasal, New, Nine, Noble

(4) New, Nasal, Nine, Noble34. Five students participated in an examination and each

scored different marks. Naina scored higher than

Meena. Kamla scored lower than praveen but higher

than Naina. Anuj’s score was between Meena and

Naina. Which of the following pairs represents the

highest and the l owest scores respectively?

(1) Praveen, Naina (2) Naina, Praveen

(3) Praveen, Anuj (4) Praveen, Meena

34. ik¡p fo|kfFkZ;ks us ,d ijh{kk es Hkkx fy;k vkSj çR;sd us fHké vad

vftZr fd;kA uSuk us ehuk ls T;knk vad vftZr fd;kA deyk us

çohu ls de ijUrq uSuk ls T;knk vad vftZr fd;kA vuqt dk vad

ehuk vkSj uSuk ds chp es FkkA fuEufyf[kr es ls dkSu&lk tksM+k

Øe”k% mPpre ,oa fuEure vad n”kkZrk gS%

(1) çohu] uS uk (2) uS uk] çohu(3) çohu] vuq t (4) çohu] ehuk

INSTRUCTIONS: In this question a piece of paper is folded and then cut as shown below. The dotted lines shown are the portion which

have been folded. The curve arrow shows the directions of folding. And the number of scissors beneath the figure show the number of

portions cut. From the given responses, indicate how it will appear when opened. The opening is in the same order as folding.

funsZ”k% fuEufyf[kr ç”u esa dkxt ds Vq dM+s dks eks M+ dj dkVk x;k gS] tSlk fd uhps n”kkZ ;k x;k gSA fcUnq js [kk,a eks M+ s gq, Hkkx dks çnf”kZr djrh gSaA oØkRed rhj eq M+ s Hkkx dh fn”kk dks n”kkZ rk gSA fp= ds uhps nh xbZ dS afp;k¡ dVs gq , Hkkx dh la[;kvks dks n”kkZ rh gSaA mÙkj fodYiksa esa ls ;s crk,a fd bl çfØ;k ds vuqlkj bls [kksyus ij ;g dS lk fn[ksxkA eks M+ us ds Øekuqlkj gh [kks yus dk Øe gks xkA

35. Question figure ¼ç”u vkÑfr½ Answer figure ¼mÙkj vkÑfr½

35. (1) (2) (3) (4)

DIRECTIONS: At a public meeting there were 8 speakers A, B,

C, D, E, F, G and H. Each spoke for some time according to the

following scheme–

I. ‘A’ spoke after ‘F’ and took more time than ‘B’

II. ‘C’ spoke before ‘G’ and after ‘B’ and took less time

than E.

III. ‘D’ spoke after ‘H’ and before ‘B’ and took less time

than ‘H’, but more time than ‘E’.

IV. ‘H’ spoke after ‘A’ and took less time than ‘B’

36. Who spoke for the longest time?

(1) A (2) B

(3) C (4) D

funs Z”k% ,d lkoZ tfud lHkk es a A, B, C, D, E, F, G, ,oa H 8 oäk FksA fuEufyf[kr ;ks tuk ds vuq lkj çR;s d us dq N le; cks yk&

I. ̂A* ^F* ds ckn cks yk vkSj ^B* ls T;knk le; fy;kII. ^C* ^G* ls igys cksyk ys fdu ^B* ds ckn vkS j ^E* ls de

le; fy;kIII. ^D* ^H* ds ckn cks yk vkS j ^B* ls igys ,oa ^H* ls de le;

fy;k ysfdu ^E* ls T;knkIV. ^H* ̂A* ds ckn cks yk vkSj ^B* ls de le; fy;k

36. dkSu lcls T;knk le; cksyk%

(1) A (2) B

(3) C (4) D

DIRECTION: Some words have been coded but no specificcode is indicated. Use your judgement to pick the coded word

which represents the given word the best.

37. RECOMMEND:

(1) 1 5 4 5 9 5 2 4 7

(2) 1 5 6 7 2 2 5 4 3

(3) 2 2 7 9 1 9 2 8 1

(4) 5 6 6 6 7 1 2 8 1

funs Z”k% dq N “kCnks a dks dw V Hkk’kk esa fy[kk x;k gS ysfdu mlds fy;s dks bZ fuf”pr dw V fufnZ’V ugh gSA vki vius fu.kZ; {kerk dk ç;ks x djrs gq, dw V esa fy[ks gq, la [;k dks crk, tks ç”u es a fn;s gq , “kCn dks lokZf/kdvPNh rjg ls iznf”kZ r djrk gS%

37. RECOMMEND%

(1) 1 5 4 5 9 5 2 4 7

(2) 1 5 6 7 2 2 5 4 3

(3) 2 2 7 9 1 9 2 8 1

(4) 5 6 6 6 7 1 2 8 1


6/14

[A–6]

38. If BAT = 69 and BOOK = 172, then PEN = ?

(1) 66 (2) 105

(3) 144 (4) 183

38. ;fn BAT = 69 vkSj BOOK = 172, rks PEN = \

(1) 66 (2) 105

(3) 144 (4) 183

39. In a certain code language ‘bring the white board’ is

written as ‘ka na di pa’ and ‘White and black board’ is

written as ‘na di sa ra’. How is ‘the’ written in that

code?

(1) ka (2) pa

(3) ka or pa (4) ra

39. fdlh fuf”pr dksM Hkk’kk es ^bring the white board* dks ^ ka na

di pa* ,oa ^white and black board* dks ^ na di sa ra* fy[kk

tkrk gS rks ^ the* dks ml dks M es dSls fy[kk tk;sxk\

(1) ka (2) pa(3) ka or pa (4) ra

40. If M+N means M is brother of N, M/N means M is father

of N and M×N means M is sister of N. Which of the

following means A is uncle of B?

(1) A / C x B

(2) C x B / A

(3) A + D / E / B

(4) A + G / H x B

40. vxj M+N ek;us M HkkbZ gS N dk] M/N ek;us M firk gS N dk

rFkk MxN ek;us M cgu gS N dkA fuEufyf[kr es ls fdldk ek;us

B dk pkpk A gS \

(1) A / C x B

(2) C x B / A

(3) A + D / E / B

(4) A + G / H x B

41. JE, LH, OL, SQ, ____

(1) WV(2) WX

(3) VW

(4) XW

41. JE, LH, OL, SQ, -----

(1) WV(2) WX

(3) VW

(4) XW

42. ‘B’, the son of ‘A’ was wedded to ‘C’ where as ‘D’ was

married to ‘E’. If E is the brother of ‘B’, how is ‘D’

related to ‘A’?

(1) Daughter-in-law

(2) Daughter

(3) Sister

(4) Cousin

42. ^B*] ^ A* dk iq= gS ftldk ^C* ls fookg gqvk gS tcfd ^D* dk

fookg ^E* ls gqvk FkkA ;fn ^E* ^B* dk HkkbZ gS rks ^ D* dk ^ A* ls

D;k laca/k gS\

(1) cgw(2) iq =h(3) cgu(4) ppsjh cgu@HkkbZ

43. Sonu starts from his home towards the South. Afterwalking for 60 m, he turns right and goes for 40 m. He

turns right again and walks for 80 m before turning

left. He then walks for 30 m and reaches his school.

How far is his school from his home and in which

direction?

(1) 10√43 m, North East

(2) 10√13 m, North West

(3) 10√53 m, North West

(4) None of these

43.lksuw vius ?kj ls nf{k.k dh vksj pyuk çkjEHk djrk gSA

60 eh-

pyus ds ckn og nk,¡ eqM+dj 40 eh- pyrk gSA og iqu% nk,¡ eqM+rk

gS ,oa ck,¡ eqM+us ls igys 80 eh- pyrk gSA fQj og 30 eh- pydj

vius fo|ky; igqprk gSA mldk fo|ky; mlds ?kj ls fdruh nwjh

ij ,oa fdl fn”kk es gS\

(1) 10√43 eh-] mÙkj&iw oZ (2) 10√13 eh-] mÙkj&if”pe(3) 10

√

53 eh-] mÙkj&if”pe(4) bues a ls dks bZ ugha

DIRECTIONS: In the following number series only one number

is wrong. Find out the wrong number.

44. 4, 9, 21, 49, 101

(1) 21

(2) 49

(3) 101

(4) None of these

funs Z”k% fuEufyf[kr la[;k Ja `[kyk esa ,d la [;k xyr gSA xyr la [;k dks Kkr djs aA

44. 4, 9, 21, 49, 101

(1) 21

(2) 49(3) 101 (4) bues a ls dks bZ ugha

45. Pick the odd one out.

(1) Copper (2) Tin

(3) Zinc (4) Brass

45. fuEufyf[kr es ls vlaxr dks pqfu,%

(1) rk¡ck (2) Vhu(3) tLrk (4) ihry


7/14

[A–7]

TECHNICAL APTITUDE

46. The value of gravitational constant between two

bodies in vacuum is 6.67×10 –11

Nm2 kg

–2. Its value in a

denser medium of density 10 gm cm –3

will be:

(1) 6.67×10 –10

Nm2 kg

–2

(2) 6.67×10 –21

Nm2 kg

–2

(3) 6.67×10 –11

Nm2 kg

–2

(4) 6.67×

10 –15 Nm2 kg –2

46. fuokZr es nks oLrqvks ds chp xq:Rokd’kZ.k fu;rkad dk eku 6.67×10 –

11 Nm

2 kg

–2 gSA 10 gm cm

–3 ?kuRo ds l?ku ek/;e es bldk

eku gksxk%

(1) 6.67×10 –10

Nm2 kg

–2

(2) 6.67×10 –21

Nm2 kg

–2

(3) 6.67×10 –11

Nm2 kg

–2

(4) 6.67×

10 –15 Nm2 kg –2

47. The drift velocity of free electrons in a conductor is vd

and a current i is flowing is it. If both the radius and

current are doubled, the drift velocity will be:

(1) vd/8 (2) vd/4

(3) vd/2 (4) vd

47. ,d pkyd e eä byDVkuk dk vuxeu ox vd g tc fd ble i

/kkjk çokfgr gk jgh gA ;fn pkyd dh f=T;k rFkk mle cgu okyh

/kkjk nxuh dj nh tk, rk vuxeu ox gk tk;xk%

(1) vd/8 (2) vd/4

(3) vd/2 (4) vd

48. The pitch of sound note depends upon:

(1) Its amplitude (2) Its frequency

(3) Its wavelength (4) Its velocity

48. /ofu Loj dk ¼fip½ rkjRo fuHkZj djrk gS%

(1) mlds vk;ke ij (2) mldh vko`fÙk ij(3) mldh rja x nS/;Z ij (4) mlds os x ij

49. Two waves of equal amplitudes and equal wavelengths

superimpose in different phases. The amplitude ofresultant wave will be maximum when the phase

difference between the waves is:

(1) Zero (2) π/2(3) π (4) 3π/2

49. leku vk;ke o leku rjaxnS/;Z dh nks rjaxs fofHké dykvks es

v/;kjksfir dh tkrh gSA ifj.kkeh rjax dk vk;ke vf/kdre gksxk tc

muds chp dykUrj gS%

(1) “kw U; (2) π/2(3) π (4) 3π/2

50. A tank is filled with water upto height H. When a hole

is made at a distance h below the level of water, the

horizontal range of water jet will be:

(1) h) –(Hh2 (2) h)(Hh4

(3) h) –(Hh4 (4) h)(Hh2

50. ,d VSd es H Å¡pkbZ rd ty Hkjk gSA tc ikuh ds Åijh ry ls h

xgjkbZ ij Nsn dj fn;k tk, rks fxjus okys ikuh dh {kSfrt ijkl

gksxh%

(1) h) –(Hh2 (2) h)(Hh4

(3) h) –(Hh4 (4) h)(Hh2

51. In an ideal L–C circu it, the capacitor is charged by

connecting it to a d.c. source which is then

disconnected. The current in the circuit:

(1) Becomes zero instantaneously

(2) Grows monotonically

(3) Decays monotonically

(4) Oscillates instantaneously

51. ,d vkn”kZ L–C ifjiFk es la?kkfj= dks fn’V /kkjk lzks r ls tksM+dj

vosf”kr fd;k tkrk gS rFkk vkosf”kr gksus ds ckn lz ksr dks gVk fy;k

tkrk gS] ifjiFk es /kkjk%

(1) rq jUr “kwU; gks tk;s xh(2) ,d leku :i ls c


8/14

[A–8]

55. A convex lens is in contact wi th a concave lens. The

ratio of their focal length is 2/3. The focal length of

combination is 30 cm. What are their individual focal

lengths:

(1) –75, 50 (2) –10, 15

(3) 75, 50 (4) –15, 10

55. ,d mÙky ySl ,d vory ySl ds lEidZ es j[kk x;k gSA mudh

Qksdl nwfj;ks dk vuqikr 2/3 gSA mudh la;qä Qksdl nw jh 30 lseh-

gSA mudh vyx&vyx Qksdl nwfj;k¡ D;k gS%

(1) –75, 50 (2) –10, 15

(3) 75, 50 (4) –15, 10

56. The amplitude of a particle executing simple harmoni c

motion is 4 cm. At what displacement from the mean

position its half energy will be in the form of kinetic

energy and half energy will be in the form of potential

energy:

(1) √2 cm (2) 2√2 cm

(3) 1 cm (4) 2 cm

56.

ljy vkorZxfr djus okys fdlh d.k dk vk;ke

4 cm

gSA lkE;fLFkfr ls fdrus foLFkkiu ij mldh ÅtkZ vk/kh xfrt o vk/kh

fLFkfrt ÅtkZ gksxh%

(1) √2 cm (2) 2√2 cm

(3) 1 cm (4) 2 cm

57. A current is flowing in a wire of non-uniform cross-

section. Which one of the following is constant

throughout the length of the wire:

(1) Current only

(2) Current and drift velocity

(3) Drift velocity only

(4) Current, electric field and drift velocity only

57. fdlh vleku ifjPNsn okys rkj es /kkjk çokfgr gks jgh gSA rkj dh

iwjh yEckbZ es fuEufyf[kr es ls D;k vpj gksxk%

(1) ds oy /kkjk(2) /kkjk ,oa vuq xeu osx(3) ds oy vuq xeu os x(4)

/kkjk] fo|qr {ks = ,oa vuq xeu os x

58. Two thin, long parallel wires separated by a distance b

are carrying a current of i amp. each. The magnitude of

the force per unit length exerted by one wire on the

other is:

(1)2

2

b

io (2)

b2

io 2

π

µ

(3)b2

io

π

µ

(4)2b2

io

π

µ

58. nks yEcs irys lekUrj rkj ijLij b ehVj nw jh ij fLFkr gSA çR;sd

es i amp. /kkjk cg jgh gSA ,d rkj }kjk nwljs ij vkjksfir çfr

,dkad yEckbZ ij yxus okys cy dk ifjek.k gS%

(1)2

2

b

io (2)

b2

io 2

π

µ

(3)b2

io

π

µ

(4)2b2

io

π

µ

59. The ratio of the coefficient of thermal conductivity of

two different material is 5:3. If the thermal resistance

of the rods of same thickness of these materials is

same, then the ratio of the length of these rods w ill be:

(1) 3:5 (2) 5:3

(3) 3:4 (4) 3:2

59. nks inkFkks dh m’ek pkydrk xq.kkad 5:3 ds vuqikr es gSA ;fn bu

NM+ks dk m’eh; çfrjks/k ,d leku eksVkbZ ds fy;s cjkcj gks rks bu

NM+ks dh yEckbZ;ks dk vuqikr gksxk%

(1) 3:5

(2) 5:3

(3) 3:4

(4) 3:2

60. When electron jumps from n=4 to n=2 orbi t, we get:

(1) Second line of Lyman series

(2) Second line of Balmer series

(3) Second line of Paschen series

(4) An absorption line of Balmer series

60. tc bysDVªkWu n=4 ls n=2 es dwnrk gS rc ges çkIr gksxh%

(1) ykbeu Js.kh dh f}rh; js [kk(2) ckej Js .kh dh f}rh; js [kk(3) ik”pu Js.kh dh f}rh; js [kk(4) ckej Js .kh dh vo”kks ’k.k js [kk

61. An α particle of energy 5 MeV is scattered through

180°

by a fixed uranium nucleus. The distance of

closest approach is of the order of:

(1) 10 –8

cm (2) 10 –10

cm

(3) 10 –12

cm (4) 10 –15

cm

61. 5 MeV ÅtkZ dk ,d α d.k fLFkj ;w jsfu;e ukfHkd ls 180° ds

dks.k ij çdhf.kZr gksrk gS]

α

d.k dh ukfHkd ds fudVre igqp dh

nwjh dk dksfVeku gS%

(1) 10 –8

cm (2) 10 –10

cm

(3) 10 –12

cm (4) 10 –15

cm

62. Fast neutrons can easily be slowed down by:

(1) The use of lead shielding

(2) Passing them through heavy water

(3) Elastic collisions with heavy nuclei

(4) Applying a strong electric field

62. rhozxkeh U;wVªkWuks dks ljyrk ls eUn eUn djus ds fy;s%

(1) lhls ds dpo dk mi;ks x djuk pkfg;s (2) mudks Hkkjh ty esa ls xq tkjuk pkfg;s (3) mudk Hkkjh ukfHkdks a ls çR;kLFk la ?kê djk;k tkuk pkfg;s(4) rhoz oS|q r {ks= yxk;k tkuk pkfg;s


9/14

[A–9]

63. Of which physical quantity the unit is parsec:

(1) Length (2) Mass

(3) Time (4) Electric field

63. ikjlsd fdl HkkSfrd jkf”k dk ek=d gS%

(1) yEckbZ (2) æO;eku(3) le; (4) fo|q r {ks=

64. The focal length of convex lens, of a simple

microscope, is 5 cm. For least distance of distinct

vision, its magnifying power will be:

(1) 2 (2) 4(3) 5 (4) 6

64. ,d ljy lw{en”kÊ ds mÙky ySl dh Qksdl nw jh 5 cm gS ] Li’V

n`f’V dh U;wure nwjh ds fy;s bldh vko/kZu {kerk gksxh%

(1) 2 (2) 4

(3) 5 (4) 6

65. Boiling water is changing into steam. At this stage the

specific heat of water is:

(1) 0 (2) ∞ (3) 1 (4) 0.5

65. mcyrk gqvk ty Hkki es cny jgk gSA bl fLFkfr es ikuh dh fof”k’V

Å’ek gS%

(1) 0 (2) ∞ (3) 1 (4) 0.5

66. The path difference between the two monochromatic

waves of wavelength λ for constructive interference

must be:

(1) (2n – 1)4

λ

(2) (2n – 1)2

λ

(3) nλ

(4) (2n + 1)2λ

66. rjax nS/;Z λ dh nks ,do.kÊ çdk”k rja xks ds e/; laiks’kh O;frdj.k

gsrq iFkkUrj gksuk pkfg,%

(1) (2n – 1)4

λ

(2) (2n – 1)2

λ

(3) nλ (4) (2n + 1)

2

λ

67. A small ball of radius r is falling in a viscous liquid. Its

terminal velocity is proportional to:

(1) r (2) r 2

(3) r 3

(4) r –1

67. r f=T;k dh ,d NksVh xksyh “;ku æo es fxj jgh gS ] mldk lhekUr

osx vuqØekuqikrh gS%

(1) r (2) r 2

(3) r 3

(4) r –1

68. Which of the following has maximum unpaired d-

electrons?

(1) Zn+2

(2) Fe+2

(3) Ni+2

(4) Cu+

68. fuEu es fdles lcls vf/kd v;qfXer d&bysDVªkWUl gS\

(1) Zn+2

(2) Fe+2

(3) Ni+2

(4) Cu+

69. The ion that is isoelectronic with CO is:

(1) –2O (2) 2N

(3) CN

(4)2

O

69. og vk;u tks CO dk vkblks&bysDVªkWfud gS%

(1) –2O (2) 2N

(3) CN

(4) 2O

70. Which enzymes is used as catalyst to convert glucose

into ethanol?

(1) Zymase (2) Invertase

(3) Maltase (4) Diastase

70. ,UtkbZe tks Xywdkst dks ,FkSkuky es ifjofrZr djus ds fy, mRçsfjr

djrk gS\

(1) tkbes l (2) bUoVsZ t(3) ekYVs l (4) Mk;LVs l

71. The value of ∆H for the reaction

N2 + 3H2 2NH3 is:

(1) Negative (2) Positive

(3) Zero (4) None of these

71. vfHkfØ;k N2 + 3H2 2NH3 ds fy;s ∆H dk eku gS%

(1) _.kkRed (2) /kukRed(3) “kw U; (4) bues a ls dks bZ ugha

72. The oxidation state of Cr in K2 Cr 2 07 is:

(1) +7 (2) +6

(3) +3 (4) +2

72. K2 Cr 2 07 es Cr dh vkWDlhdj.k la[;k gks rh gS%

(1) +7 (2) +6

(3) +3 (4) +2

73. The outer most electronic configuration of most

electronegative element is:

(1) ns2 np

3(2) ns

2 np

4

(3) ns2 np

5(4) ns

2 np

6

73. vf/kdka”k fo|qr _.kkRed rRoks ls okº;re dks”k dk bysDVªkWfud

foU;kl gksrk gS%

(1) ns2 np

3(2) ns

2 np

4

(3) ns2 np

5(4) ns

2 np

6


10/14

[A–10]

74. For the reaction, C(5) + CO2(g) 2CO(g), the partial

pressure of CO2 and CO are 2.0 and 4.0 atm.

respectively at equilibrium. The Kp for the reaction is:

(1) 0.5 (2) 4.0

(3) 8.0 (4) 32.0

74. vfHkfØ;k C(5) + CO2(g) 2CO(g) es CO2 o CO dk

vkaf”kd nkc lkE;koLFkk es Øe”k% 2.0 o 4.0 atm gSA fØ;k ds fy;s

Kp dk eku gS%

(1) 0.5 (2) 4.0

(3) 8.0 (4) 32.0

75. The vapour pressure of two pure liquids A & B are 100

and 80 torr respectively. The total pressure of solutionobtained by mixing 2 m of A and 3 mole of B would be:

(1) 120 torr (2) 36 torr

(3) 88 torr (4) 180 torr

75. nks “kq) æoks A o B dk ok’i nkc Øe”k% 100 o 80 torr gS] A

ds

2 m

B

ds

3

eksy feykus ds i”pkr dqy nkc fdruk gksxk\

(1) 120 torr (2) 36 torr

(3) 88 torr (4) 180 torr

76. Which of the following is used as an anti knocking

material?

(1) TEL (2) C2H5OH

(3) Glycol (4) Freon

76. fuEu es dkSu ,sUVh ukSfdax inkFkZ ds :i es ç;ksx fd;k tkrk gS%

(1) Vs y (2) bFkkby ,Ydksgy(3) Xykbdksy (4) fÝ;kW u

77. What volume of 0.25N acid would be required to react

with 0.500gm of Ca(OH)2

[Mol.Wt.:Ca-40,0-16,H-1]:

(1) 54.0ml (2) 108ml(3) 100ml (4) 27ml

77. 0.500gm Ca(OH)2

ls vfHkfØ;k ds fy;s 0.25 N vEy dk fdruk

vk;ru yxsxk% [Mol.Wt.:Ca-40,0-16,H-1] %

(1) 54.0ml (2) 108ml

(3) 100ml (4) 27ml

78. The precipitate of CaF2(Ksp=1.7×10 –10

) is obtained when

equal volumes of the following are mixed:

(1) 10 –4

M Ca2+

+10 –4

MF –

(2) 10 –2

M Ca2+

+10 –3

MF –

(3) 10 –5

M Ca2+

+10 –3

MF –

(4) 10 –3

M Ca2+

+10 –5

MF –

78. fuEufyf[kr es ls fdlds leku vk;ru feykus ij

CaF2(Ksp=1.7×10 –10

) ds vo{ksi çkIr gksrs gS%

(1) 10 –4

M Ca2+

+10 –4

MF –

(2) 10 –2

M Ca2+

+10 –3

MF –

(3) 10 –5

M Ca2+

+10 –3

MF –

(4) 10 –3

M Ca2+

+10 –5

MF –

79. In the process of catalysis, catalyst change the rate of

reaction because it:

(1) Lowers the energy of activation

(2) Enhances the energy of activation(3) Gets consumed after the reaction

(4) Can be regenerated after reaction

79. mRçsjdÙo dh çfØ;k es] mRçsjd vfHkfØ;k dk osx cny nsrk gS D;ksfd

;g%

(1) ,fDVoslu dh ÅtkZ dks fuEu dj ns rk gS

(2) ,fDVoslu dh ÅtkZ dks c


11/14


12/14

[A–12]

92. Compilers can detect:

(1) Syntax errors

(2) Semantic errors

(3) Logical errors

(4) All of these

92. dEikbYlZ irk yxk ldrk gS%

(1) okD; jpuk =q fV;ks a dks (2) vFkZ fc’k;d =qfV;ks a dks (3) rdZ la xr =q fV;ks a dks (4) mi;q Zä lHkh

93. Internet uses:

(1) Packet switching (2) Circuit switching(3) Telephone switching (4) Telex switching

93. baVjusV fuEu dk iz;ksx djrk gS%

(1) iS ds V fLofpax (2) ifjiFk fLofpa x(3) Vs yhQksu fLofpa x (4) Vs ys Dl fLofpa x

94. 220

is referred to as:

(1) 1 KB (2) 1 MB

(3) 1GB (4) 1TB

94. 220 dks lanfHkZr fd;k tkrk gS%

(1) 1 KB (2) 1 MB

(3) 1GB (4) 1TB

95. ASCII stands for:

(1) American storage code for information interchange

(2) American standard code for information interchange

(3) American standard code for incoming information

(4) None of these

95. ASCII dk rkRi;Z gS%

(1) ves fjdu LVksjs t dks M QkWj bUQks jes'ku bUVjpsa t(2) ves fjdu LVS UMMZ dksM QkWj bUQks jes'ku bUVjpsa t(3) ves fjdu LVS UMMZ dks M QkW j budfea x bUQksjes'ku(4) bues a ls dks bZ ugha

96. Pipelining improves CPU performance due to;(1) Reduced memory access time

(2) Increased clock speed

(3) The introduction of parallellism

(4) Additional functional units

96.

ikbZiykbZfuax

CPU

dh fu iknu dks lq/kkjrh gS%(1) ?kVh gq bZ Le`fr vfHkxe dky ds dkj.k(2) c0, is:

(1)ds

)s(dFk – (2)

ds

)s(dF –

(3)ds

)k/s(dFk – 1 (4)

ds

)k/s(dFk – 2

97. ;fn ƒ(t) dk ykIykl :ikUrj F(s) gS rks tƒ (kt), k>0 dk ykIykl

:ikUrj gksxk%

(1)ds

)s(dFk – (2)

ds

)s(dF –

(3)ds

)k/s(dFk – 1 (4)

ds

)k/s(dFk – 2

98. If∫

=

t

0

3tduu) –(t(u)f then f(t) is:

(1) t2/3 (2) 6t

(3) 3t (4) t4

98.;fn ∫ =

t

0

3tduu) –(t(u)f rks

f(t) gksxk%

(1) t2/3 (2) 6t(3) 3t (4) t4

99. If the equations 2x – y + 3z = α, x + 3y–6z = α2 and 3x –

5y + 12z = 4 have a solution then the value of α is:

(1) 0 (2) ± i

(3) 3i1± (4) 2

99. ;fn lehdj.k 2x – y + 3z = α, x + 3y–6z = α2 vkSj 3x – 5y

+ 12z = 4 dk ,d gy gS rks α dk eku gS %

(1) 0 (2) ± i(3) 3i1± (4) 2

100. If λ is an eigen value of a non-singular matrix A then

the eigen value of the matrix (Adj A) is:

(1) |A| / λ (2) 1 / λ (3) λ (4) λ / |A|

100. ;fn λ ,d O;qRØe.kh; vkO;w g A dk vfHky{kf.kd eku gS rks vkO;w g

(Adj A) dk vfHky{kf.kd eku gksxk%

(1) |A| / λ (2) 1 / λ (3) λ (4) λ / |A|

101. The particular integral of the differential equation x2 y"

+ xy' + y = cos(log x) is:

(1) sin(log x) (2) x)sin(log2

xlog –

(3)x)sin(log

2

xlog

(4) x)sin(log

2

x –

101. vody lehdj.k x2 y" + xy' + y = cos(log x) dk fof'k V

lekdyu gS%

(1) sin(log x) (2) x)sin(log2

xlog –

(3)x)sin(log

2

xlog

(4) x)sin(log

2

x –


13/14

[A–13]

102. The function f(z) =z |z| is:

(1) Analytic for every z

(2) Analytic only at z = 0

(3) Analytic everywhere except at z = 0

(4) Analytic for no point z

102. Qyu f(z) =z |z| gS%

(1) çR;sd z ds fy, fo'ys "kd(2) ds oy z = 0 dk fo'ys "kd(3) z = 0 dks NksM+dj gj txg fo'ys "kd(4) No Iokb± V z ds fy, fo'ys "kd

103. The value of∫

C

3

)2 –z)(3 –z(

dz[ taken around the circle

C:|z–2|=2

1 in the positive direction is:

(1) π i (2) 2 π i

(3) 4 π i (4) 8 π i

103. /kukRed fn'kk es o`Ùk C:|z–2|=2

1 ds pkjks vksj fy, x,

∫

C

3)2 –z)(3 –z(

dz[ dk eku gS %

(1) π i (2) 2 π i(3) 4 π i (4) 8 π i

104. The residue of4

2

z

zcos –1(z)f = at z = 0 is:

(1) 1/2 (2) 2

(3) 4 (4) 0

104. z = 0 ij4

2

z

zcos –1(z)f = dk vof'k V gS%

(1) 1/2 (2) 2(3) 4 (4) 0

105. On eliminating the arbi trary function ƒ from the

relation z = ƒ(y2 /x), the partial differential equation

obtained is:

(1) 2px + qy = 0

(2) px + qy = 0

(3) 2py + qx = 0

(4) py + qx = 0

105. lecU/k z = ƒ(y2 /x) ls LosPN Qyu ƒ dks foyqIr djus ij vka f'kd

vody lehdj.k çkIr fd;k tkrk gS%

(1) 2px + qy = 0(2) px + qy = 0(3) 2py + qx = 0(4) py + qx = 0

106. tan (cos –1

x) is equal to:

(1) x/x – 1 2 (2)2x1

x

+

(3) x/x1 2+ (4) x 2x – 1

106. tan (cos –1

x) cjkcj gksxk%

(1) x/x – 1 2 (2)2x1

x

+

(3) x/x1 2+ (4) x 2x – 1

107. If A is a square matrix of order 3 and determinant of A

has value 5 then the value of determinant of the matrix(2A) is:

(1) 5 (2) 3

(3) 15 (4) None of these

107. ;fn A ,d 3 Øe dk oxZ vkO;wg gS rFkk lkjf.kd A dk eku 5 gSA

rks vkO;wg ¼

2A½ ds lkjf.kd dk eku gksxk%

(1) 5 (2) 3

(3) 15 (4) bues a ls dks bZ ugha

108. The function f(x)=|x| is:

(1) Differantiable at the origin

(2) Continuous but not differentiable at the origin

(3) Not continuous at the origin

(4) None of these

108. Qyu f(x)=|x| gksxk%

(1) ew y fcUnq ij vodyuh;(2) ew y fcUnq ij lrr vkS j vodyuh; ugha (3) ew y fcUnq ij lrr ugha(4) bues a ls dks bZ ugha

109. The line 1b

y

a

x= touches the curve y=be

–x/a at the

point:(1) (a, b/a)

(2) (–a, b/a)

(3) (a, a/b)

(4) None of these

109. js[kk 1b

y

a

x=

oØ y=be –x/a

dks fdl fcUnq ij Li”kZ djrh gS %

(1) (a, b/a)(2) (–a, b/a)

(3) (a, a/b)

(4) bues a ls dks bZ ugha

110. The area of the ellipse 1b

y

a

x2

2

2

2

= is:

(1) π ab (2) π (a+b)(3) π (a2+b2) (4) None of these

110. nh?kZo`Ùk 1b

y

a

x2

2

2

2

=

dk {¨=Qy gksxk%

(1) π ab (2) π (a+b)(3) π (a2+b2) (4) bues a ls dks bZ ugha


14/14

[A–14]

111. Solution of the differential equationdx

dy sin 2x –y=tan x

is:

(1) x)(tancxtany (2) x–y sin x = c

(3) xy tan x = c (4) None of these

111. vody lehdj.kdx

dy sin 2x –y=tan x dk gy gksxk%

(1) x)(tancxtany (2) x–y sin x = c

(3) xy tan x = c (4) bues ls dksbZ ugha

112. A particle is thrown vertically upwards with a velocity

of 400 cm per second. It will return to this positionafter:

(1) 1 second (2) 0.5 second

(3) 2 second (4) None of these

112. 400 cm çfr lsds.M dh xfr ls ,d d.k Åij dh vksj Qsdk x;k

gSA og bl fLFkfr es nqckjk gksxk%

(1) 1 ls ds .M ckn (2) 0.5 ls ds .M ckn(3) 2 ls ds .M ckn (4) bues a ls dks bZ ugha

113. Let∧

kp jia and∧

k jib then |b||a||ba|

holds for:

(1) All real p

(2) No real p

(3) p = –1

(4) p = +1

113. ;fn∧

kp jia vkSj∧

k jib rks |b||a||ba| gksus

ds fy;s p dk eku gksxk%

(1) lHkh okLrfod p (2) fdlh Hkh okLrfod p ds fy;s ugha (3) p = –1

(4) p = +1

114. If in a triang le ABC, a=2, b=3, c=5 then ∠ C=:

(1) π/6 (2) π/3

(3) π/2 (4) None of these

114. ;fn fdlh f=Hkqt ABC es a=2, b=3, c=5 rks ∠ C=%

(1) π/6 (2) π/3

(3) π/2 (4) bues a ls dks bZ ugha

115. If x is real , then the least value of the expression

1x2x

5x6 –x2

2

is:

(1) –1 (2) –1/3


115. ;fn x okLrfod gks rks in1x2x

5x6 –x

2

2

dk de ls de eku gksxk%

(1) –1 (2) –1/3


116. The number of normals that can be drawn from a given

point to the parabola is:

(1) 3 (2) 4

(3) 2 (4) 5

116. fdlh fn;s x;s fcUnq ls fdlh ijoy; ij [khaps x;s vfHkyEcks dh

la[;k gksxh%

(1) 3 (2) 4

(3) 2 (4) 5

117. ∫ (tan x + cot x)2 dx =:

(1) tan x + cot x + c

(2) cot x – tan x + c

(3) tan x – cot x + c

(4) sec x + c

117. ∫ (tan x + cot x)2 dx =%

(1) tan x + cot x + c

(2) cot x – tan x + c

(3) tan x – cot x + c

(4) sec x + c

118. Equation o f the curve passing through (1, 1) and

satisfying the differential equation

0)y0,(x;x

y2

dx

dy> is:

(1) x2=y (2) x=y2

(3) x=2y (4) y=2x

118. ¼1, 1½ ls xqtjus okyk rFkk vody lehdj.k

0)y0,(x;x

y2

dx

dy> dks larq’V djus okyk oØ dk lehdj.k

gksxk%

(1) x2=y (2) x=y2

(3) x=2y (4) y=2x

119. The smallest positi ve integer n for which 1i –1

i1n

=

⎠

⎞

⎜

⎝

⎛

is:

(1) 8 (2) 4


119. 1i –1

i1n

=

⎠

⎞

⎜

⎝

⎛

ds fy, U;wure /kukRed iw.kk±d n gS%

(1) 8 (2) 4


120. The centre of gravity of the surface of a hollow cone

lies on the axis and divides it in the ratio:

(1) 1:2 (2) 1:3

(3) 2:3 (4) None of these

120. ,d [kks[kys “kadq dk xq:Ro dsUæ “ka dq dh v{k ij gksrk gS rFkk mls

vuqikr es foHkkftr djrk gS%

(1) 1:2 (2) 1:3

(3) 2:3 (4) bues a ls dks bZ ugha

SCTO Sampldjfjde Paper

Documents

Transcript of SCTO Sampldjfjde Paper