CAREER INSTITUTE Pat S KOTA (RAJASTHAN) … · -2/32 SPACE FOR ROUGH WORK / jQ dk;Z ds fy;s txg ALL INDIA OPEN TEST/JEE (Min)/22-03-2015 00E314002 3. 20 / g mpty conveyo b ovi horizontall

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1. Immediately fill in the form number on this page of the Test Bookletwith Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2. The candidates should not write their Form Number anywhere else(except in the specified space) on the Test Booklet/Answer Sheet.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are360.

5. There are three parts in the question paper A,B,C consisting ofPhysics, Chemistry and Mathematics having 30 questions in eachpart of equal weightage. Each question is allotted 4 (four) marks forcorrect response.

6. One Fourth mark will be deducted for indicated incorrect responseof each question. No deduction from the total score will be madeif no response is indicated for an item in the Answer Sheet.

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bl ijh{kk iqfLrdk dks rc rd u [kk sysa tc rd dgk u tk,A

1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys@dkys ckWy ikbaV isuls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA

2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkji= ij dgha vkSj u fy[ksaA

3. ijh{kk dh vof/k 3 ?k aVs gSA

4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 360 gSaA

5. bl ijh{kk iqfLrdk es a rhu Hkkx A, B, C gSa] ftlds izR;sd Hkkx esa

HkkSfrd foKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj lHkh iz'uksa ds vad

leku gSaA izR;sd iz'u ds lgh mÙkj ds fy, 4 (pkj)vad fuèkkZfjr fd;s x;s gSaA

6. izR;sd xyr mÙkj ds fy, ml iz'u ds dqy vad dk ,d pkSFkkbZ vad dkVktk;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls½.kkRed vadu ugha gksxkA

7. mÙkj i= ds i`"B&1 ,oa i`"B&2 ij okafNr fooj.k ,oa mÙkj vafdr djus gsrqdsoy uhys@dkys ck Wy ikbaV isu dk gh iz;ksx djsaAisfUly dk iz;ksx loZFkk oftZr gSA

8. ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkhizdkj dh ikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa]eksckby Qksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU;izdkj dh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA

9. jQ dk;Z ijh{kk iqfLrdk esa dsoy fu/kkZfjr txg ij gh dhft;sA

10. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{k fujh{kddks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iqfLrdk dks ys tkldrs gSaA

11. mÙkj i= dks u eksM+sa ,oa u gh ml ij vU; fu'kku yxk,saA

IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

ALL INDIA OPEN TEST # 02 DATE : 22 - 03 - 2015TARGET : JEE (Main) 2015

ENTHUSIAST & LEADER COURSE

Test Type : Major Test Pattern : JEE (Main)

FORM NUMBER

PAPER CODE 0 0 C E 3 1 4 0 0 2Hindi

(ACADEMIC SESSION 2014-2015)CLASSROOM CONTACT PROGRAMME

H-1/32

SPACE FOR ROUGH WORK / jQ dk;Z ds fy;s txg

ALL INDIA OPEN TEST/JEE (Main)/22-03-2015

00CE314002

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

HAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

1. It is estimated that per minute each cm2 of earth

receives about 2 cal (1 cal = 4.18J) of heat energy

from the sun. This is called Solar constant. In SI

units the value is :-

(1) 1060 (2) 178.4

(3) 17.84 (4) 1393.33

2. The potential energy for a force field Fr

is

given by U(x,y) = cos (x + y). The force acting

on a part icle at position g iven by

coordinates 0,4pæ ö

ç ÷è ø

is :-

(1) ( )1 ˆ î j2

- + (2) ( )1 ˆ î j2

+

(3) 1 3ˆ î j2 2

æ ö+ç ÷

è ø(4)

1 3ˆ î j2 2

æ ö-ç ÷

è ø

1. ;g ifjdfyr fd;k x;k gS fd lw;Z ls i`Foh ds izR;sd

oxZ lseh {ks= esa izfr fefuV 2 dSyksjh (1 cal = 4.18 J)

Å"ek ÅtkZ izkIr gksrh gSA ;g lkSj fu;rkad dgykrk

gSA bls SI bdkbZ esa O;Dr dhft;sA

(1) 1060 (2) 178.4

(3) 17.84 (4) 1393.33

2. fdlh cy {k s= Fr

ds fy;s fLFkfrt ÅtkZ dk eku

U(x,y) = cos (x+y) gSA funsZ'kkad 0,4pæ ö

ç ÷è ø

okyh

fLFkfr ij d.k ij dk;Zjr cy dk eku gS%&

(1) ( )1 ˆ î j2

- + (2) ( )1 ˆ î j2

+

(3) 1 3ˆ î j2 2

æ ö+ç ÷

è ø(4)

1 3ˆ î j2 2

æ ö-ç ÷

è ø

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00CE314002

3. Sand is deposited at a uniform rate of 20 kg/s

and with negligible kinetic energy on an empty

conveyor belt moving horizontally at a constant

speed of 10 meters per minute. Find the force

required to maintain constant velocity of the

conveyor belt.

(1) 10/3 N (2) 10/4 N

(3) 1/4 N (4) 5/4 N

4. The radiation of intensity I is falling obliquely

on a surface. Area of surface exposed to radiation

is A. Select correct alternative

3. 10m izfr feuV dh fu;r pky ls {kSfrt :i ls xfr

dj jgh [kkyh dUos;j csYV ij ux.; xfrt ÅtkZ

okyh jsr 20 kg/s dh ,dleku nj ls Mkyh tkrh gSA

csYV dk osx fu;r cuk;s j[kus ds fy, vko';d cy

gksxk%&

(1) 10/3 N (2) 10/4 N

(3) 1/4 N (4) 5/4 N

4. fdlh lrg ij I rhozrk dh fofdj.k fr;Zd :i ls

vkifrr gSA fofdfjr gksus okyh lrg dk {ks=Qy A

gSA lgh dFku pqfu;s

H-3/32



00CE314002

(1) If surface is perfectly reflecting surface thenforce experienced due to radiation is

2IAcosc

q

(2) If surface is perfectly absorbing surface thenforce experienced due to radiation is

2IAcosc

q

(3) If surface is perfectly absorbing surface thenforce experienced due to radiation is

IA cosc

q

(4) If surface is partially reflecting surface thenforce experienced due to radiation is

2IA cosc

b q

where b is a constant and depends upon natureof surface.

5. A simple pendulum is suspended in a lift whichis going up with an acceleration 5 m/s2.An electric field of magnitude 5 N/C anddirected vertically upward is also present inthe lift. The charge of the bob is 1 mC andmass is 1 mg. Taking g = p2 and length of thesimple pendulum 1m, the time period of thesimple pendulum is :(1) 1 s (2) 2 s(3) 0.5 s (4) None of these

(1) ;fn lrg] iw.kZ ijkorZd lrg gS rks fofdj.k ds

dkj.k mRiUu cy 2IAcos

cq

gksxkA

(2) ;fn lrg] iw.kZ vo'kks"kd lrg gS rks fofdj.k ds

dkj.k mRiUu cy 2IAcos

cq

gksxkA

(3) ;fn lrg] iw.kZ vo'kks"kd lrg gS rks fofdj.k ds

dkj.k mRiUu cy IA cos

cq

gksxkA

(4) ;fn lrg] vkaf'kd :i ls ijkorZd gS rks fofdj.k

ds dkj.k mRiUu cy 2IA cos

cb q

gS]

tgka b lrg dh izÏfr ij fuHkZj ,d fu;rkad gSA

5. ,d ljy yksyd dks fy¶V esa yVdk;k x;k gS tks fd

5 m/s2 ds Roj.k ls Åij tk jgh gSA fy¶V esa 5 N/C

ifjek.k okyk rFkk m/okZ/kj Åij dh vksj fn'kk esa ,d

fo|qr {ks= Hkh fo|eku gSA xksyd dk vkos'k 1 mC rFkk

nzO;eku 1mg gSA ljy yksyd dh yEckbZ 1m rFkk

g = p2 yhft,A ljy yksyd dk vkorZdky gksxk&

(1) 1 s (2) 2 s

(3) 0.5 s (4) buesa ls dksbZ ugha

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00CE314002

6. A mixture of plane polarised and unpolarisedlight falls normally on a polarizing sheet. Onrotating the polarizing sheet about the directionof the incident beam, the transmitted intensityvaries by a factor of 4. Find the ratio of theintensities I

P and I

0, respectively of the polarized

and unpolarized components in the incidentbeam.

(1) 3

2(2)

5

2

(3) 1

2(4)

3

4

7. A silicon specimen is made into a p-type

semiconductor by doping, on an average, one

indium atom per 5 × 107 silicon atoms. If the

number density of atoms in the silicon specimen

is 5 × 1028 atom m–3, then the number of acceptor

atoms in silicon per cubic centimeter will be

(1) 2.5 × 1030 atom cm–3

(2) 2.5 × 1035 atom cm–1

(3) 1 × 1013 atom cm–3

(4) 1 × 1015 atom cm–3

6. lery /kzqfor rFkk v/kzqfor izdk'k dk ,d feJ.k èkzqfor

'khV ij yEcor~ vkifrr gksrk gSA bl /kzqfor 'khV dks

vkifrr iqat dh fn'kk ds lkis{k ?kqekus ij ikjxfer

rhozrk pkj ds xq.kd ls ifjofrZr gks tkrh gSA vkifrr

iqat esa /kzqfor rFkk v/kzqfor ?kVdksa dh rhozrkvksa Øe'k%

IP rFkk I

0 dk vuqikr gksxk %&

(1) 3

2(2)

5

2

(3) 1

2(4)

3

4

7. ,d flfydkWu izfrn'kZ dks vifeJ.k }kjk ,d p- izdkj

ds v¼Zpkyd esa ifjofrZr fd;k tkrk gSA blds fy;s

5 × 107 flfydkWu ijek.kqvksa esa ,d bf.M;e ijek.kq

feyk;k tkrk gSA ;fn flfydkWu izfrn'kZ esa ijek.kqvksa

dk la[;k ?kuRo 5 × 1028 atom m–3 gks rks flfydkWu esa

izfr?ku ls.VhehVj xzkgh ijek.kqvksa dh la[;k gksxh%&

(1) 2.5 × 1030 atom cm–3

(2) 2.5 × 1035 atom cm–1

(3) 1 × 1013 atom cm–3

(4) 1 × 1015 atom cm–3

H-5/32



00CE314002

8. A plane electromagnetic wave travelling alongthe X-direction has a wavelength of 3 mm. Thevariation in the electric field occurs in theY-direction with an amplitude 66 Vm–1. Theequations for the electric and magnetic fields asa function of x and t are respectively :-

(1) Ey = 33 cos p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 1.1 × 10–7 cos p × 1011

xt

cæ ö-ç ÷è ø

(2) Ey = 11 cos 2p × 1011 x

tc

æ ö-ç ÷è ø

,

Bz = 11 × 10–7 cos 2p × 1011

xt

cæ ö-ç ÷è ø

(3) Ex = 33 cos p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 11 × 10–7 cos p × 1011

xt

cæ ö-ç ÷è ø

(4) Ey = 66 cos 2p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 2.2 × 10–7 cos 2p × 1011

xt

cæ ö-ç ÷è ø

8. X fn'kk ds vuqfn'k xfr'khy ,d lery fo|qr pqEcdh;

rjax dh rjaxnS/; Z 3 mm gSA Y fn'kk esa 66 Vm–1

vk;ke ds lkFk fo|qr {ks= esa ifjorZu laiUu gksrs gSaA x

rFkk t ds Qyu ds :i esa fo|qr rFkk pqEcdh; {ks=ksa dh

lehdj.ksa Øe'k% gksxha%&

(1) Ey = 33 cos p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 1.1 × 10–7 cos p × 1011

xt

cæ ö-ç ÷è ø

(2) Ey = 11 cos 2p × 1011 x

tc

æ ö-ç ÷è ø

,

Bz = 11 × 10–7 cos 2p × 1011

xt

cæ ö-ç ÷è ø

(3) Ex = 33 cos p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 11 × 10–7 cos p × 1011

xt

cæ ö-ç ÷è ø

(4) Ey = 66 cos 2p × 1011

xt

cæ ö-ç ÷è ø

,

Bz = 2.2 × 10–7 cos 2p × 1011

xt

cæ ö-ç ÷è ø

H-6/32



00CE314002

9. If the B-H curves of two samples of P and Q ofiron are as shown below, then which one of thefollowing statements is CORRECT ?

H H

B B

Sample P Sample Q

(1) Both P and Q are suitable for makingpermanent magnet

(2) P is suitable for making permanent magnetand Q for making electromagnet

(3) P is suitable for making electromagnet andQ is suitable for permanent magnet

(4) Both P and Q are suitable for makingelectromagnets

10. Two standing waves are established. One istransverse wave in taut string and an another islongitudinal sound wave in organ pipe. Both thewaves are having equal power and of samewavelength in same mode of vibration.Statement 1: In order to establish these waves,the power of the source would be more forlongitudinal wave in organ pipe.andStatement 2: The standing waves does notrequire any source to establish as it can't transferany energy.(1) Statement–1 is True, Statement–2 is True ;

Statement–2 is a correct explanation forStatement–1.

(2) Statement–1 is True, Statement–2 is True ;Statement–2 is not a correct explanation forStatement–1.

(3) Statement–1 is True, Statement–2 is False.(4) Statement–1 is False, Statement–2 is True.

9. ;fn yksgs ds nks izfrn'kks ± P o Q ds fy, B-H oØfp=kuqlkj izkIr gksrs gks rks lgh dFku pqfu, %&

H H

B B

Sample P Sample Q(1) P o Q nksuksa gh LFkk;h pqEcd ds fuekZ.k ds fy,

mi;qä gSaA(2) LFkk;h pqEcd ds fuekZ.k ds fy, P rFkk fo|qr

pqEcd ds fuekZ.k ds fy, Q mi;qä gSA(3) fo|qr pqEcd ds fuekZ.k ds fy, P rFkk LFkk;h

pqEcd ds fuekZ.k ds fy, Q mi;qä gSA(4) P o Q nksuksa gh fo|qr pqEcd ds fuekZ.k ds fy,

mi;qä gSaA10. ,d dlh gqbZ jLlh esa ,d vuqizLFk rjax rFkk vkxZu

ikbi esa vuqnS/; Z /ofu rjax tSlh nks vizxkeh rjaxsa LFkkfirdh tkrh gSA nksuksa rjaxksa dh 'kfDr leku gS rFkk dEiUudh leku fo/kk esa rjaxnS/; Z leku gSAoDrO;–1: bu rjaxksa dks LFkkfir djus ds fy;s vkxZuikbi esa vuqnS/;Z rjax ds fy;s lzksr dh 'kfDr vf/kdgksxhAvkSjoDrO;–2 : vizxkeh rjaxsa izkIr djus ds fy;s fdlhlzksr dh vko';drk ugha gksrh gS D;ksafd bueas dksbZ ÅtkZLFkkukUrfjr ugha gks ldrh gSA(1) oäO;&1 lR; gS] oäO;&2 lR; gS] oäO;&2]

oäO;&1 dk lgh Li"Vhdj.k gSA(2) oäO;&1 lR; gS] oäO;&2 lR; gS ; oäO;&2]

oäO;&1 dk lgh Li"Vhdj.k ugha gSA(3) oäO;&1 lR; gS, oäO;&2 vlR; gSA(4) oäO;&1 vlR; gS] oäO;&2 lR; gSA

H-7/32



00CE314002

11. A horizontal plane supports a stationary verticalcylinder of radius R = 1 m and a disc A attachedto the cylinder by a horizontal thread AB oflength l

0 = 2m (seen in figure, top view). An

intial velocity (v0 = 1m/s) is imparted AB to

the disc as shown in figure. How long will itmove along the plane until it strikes against thecylinder? (All surface are assumed to besmooth)

v0

R

B Al0

(1) 1 sec (2) 2 sec (3) 5 sec. (4) 3 sec.12. Consider two containers A and B containing

identical gases at the same pressure, volume andtemperature. The gas in container A iscompressed to half of its original volumeisothermally while the gas in container B is

compressed to half of its original valueadiabatically. The ratio of final pressure of gasin B to that of gas in A is :-

(1) 12g - (2) 1

1

2

g -æ öç ÷è ø

(3) 2

1

1 gæ öç ÷è - ø

(4) 2

1

1gæ öç ÷è - ø

11. ,d {kSfrt ry ij f=T;k R = 1 m okyk fLFkj Å/okZ/kj

csyu j[kk gqvk gS rFkk bl csyu ls ,d pdrh A dks

l0 = 2m yEckbZ okyh {kSfrt jLlh AB dh lgk;rk ls

tksM+k tkrk gSA bl O;oLFkk dk Åij ls fy;k x;k n';fp= esa n'kkZ;k x;k gSA izkjfEHkd osx (v

0 = 1m/s) pdrh

AB dks fp=kuqlkj iznku djrs gSA csyu ls Vdjkus ls

iwoZ pdrh ry ij fdruh nsj rd xfr djsxh\ lHkh

lrgsa fpduh ekfu;s%&

v0

R

B Al0

(1) 1 sec (2) 2 sec (3) 5 sec. (4) 3 sec.

12. nks ik=ksa A rFkk B esa leku nkc] vk;ru rFkk rkieku ij

,dtSlh xSlsa Hkjh gqbZ gSA ik= A esa Hkjh xSl dks lerkih;

:i ls blds izkjfEHkd vk;ru ls vk/ks vk;ru rd

lEihfM+r fd;k tkrk gS tcfd ik= B esa Hkjh xSl dks

:}ks"e :i ls mlds izkjfEHkd eku ls vkèks rd laihfM+r

fd;k tkrk gSA B rFkk A esa Hkjh xSl ds vfUre nkc dk

vuqikr gksxk %&

(1) 12g - (2) 1

1

2

g -æ öç ÷è ø

(3) 2

1

1 gæ öç ÷è - ø

(4) 2

1

1gæ öç ÷è - ø

H-8/32



00CE314002

13. Two electric lamps A and B radiate the samepower. Their filaments have the samedimensions and have emissivities eA and eB.Their surface temperatures are TA and TB. The

ratio A

B

TT

will be equal to :–

(1)

1/ 4

B

A

ee

æ öç ÷è ø

(2)

1/ 2

B

A

ee

æ öç ÷è ø

(3)

1/ 2

A

B

ee

æ öç ÷è ø

(4) 1/ 4

A

B

ee

æ öç ÷è ø

14. The relative density of a metal may be found byhanging a block of the metal from a springbalance and noting that in air the balance reads(5 ± 0.05) N while in water it reads (4 ± 0.05)N.The relative density would be quoted as:

(1) (5 ± 0.05) (2) 5 ± 11%

(3) (5 ± 0.10) (4) 5 ± 6%

15. A parallel plate capacitor with air between theplates has a capacitance of 9 pF. The separationbetween its plates is 'd'. The space between theplates is now filled with two dielectrics. One ofthe dielectric has dielectric constant K

1 = 6 and

thickness d

3 while the other one has dielectric

constant K2 = 12 and thickness

2d

3. Capacitance

of the capacitor is now

(1) 18 pF (2) 25 pF (3) 81 pF (4) 20 pF

13. nks fo|qr cYc A o B leku 'kfDr fofdfjr djrs gSaAmuds rUrqvksa dh foek,a ,d leku gS rFkk mRltZdrk,aeA o eB gSA budh lrgksa ds rki TA rFkk TB gSA vuqikr

A

B

TT dk eku gksxk&

(1)

1/ 4

B

A

ee

æ öç ÷è ø

(2)

1/ 2

B

A

ee

æ öç ÷è ø

(3)

1/ 2

A

B

ee

æ öç ÷è ø

(4) 1/ 4

A

B

ee

æ öç ÷è ø

14. fdlh /kkfRod CykWd dks fLizax rqyk ls yVdkdj /kkrq

dk lkisf{kd ?kuRo Kkr fd;k tk ldrk gSA ekuk ok;q

esa fLizax rqyk dk ikB~;kad (5 ± 0.05) N o ty esa

(4 ± 0.05)N izkIr gksrk gS rks bldk lkisf{kd ?kuRo gksxk&

(1) (5 ± 0.05) (2) 5 ± 11%(3) (5 ± 0.10) (4) 5 ± 6%

15. ,d lekUrj iê la/kkfj= dh IysVksa ds e/; ok;q Hkjh gS

rFkk bldh /kkfjrk 9 pF gSA bldh IysVksa ds e/; nwjh d

gS rFkk IysVksa ds e/; ds LFkku dks vc nks ijkoS|qrksa }kjk

Hkj fn;k tkrk gS ftlesa ls ,d dk ijkoS|qrkad K1 = 6

rFkk eksVkbZ d

3 gS tcfd nwljs dk ijkoS|qrkad K

2 = 12

rFkk eksVkbZ 2d

3 gSA la/kkfj= dh /kkfjrk vc D;k gksxh %&

(1) 18 pF (2) 25 pF (3) 81 pF (4) 20 pF

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00CE314002

16. The mean lives of a radioactive sample are

30 years and 60 years for a-emission and

b–emission respectively. If the sample decays

both by a–emission and b–emission

simultaneously, the time after which, only

one-fourth of the sample remain is :-

(1) 10 years

(2) 20 years

(3) 40 years

(4) 45 years

17. n drops of a liquid, each with surface energy E,

joining to form a single drop

(1) some energy will be released in the process

(2) some energy will be absorbed in the process

(3) the energy released or absorbed will be

E

n (n–n2/3)

(4) the energy released or absorbed will be

nE (22/3–1)

16. a-mRltZu o b–mRltZu ds fy;s jsfM;kslfØ; izfrn'kZ

dh v/kZ vk;q Øe'k% 30 o"kZ rFkk 60 o"kZ gSA ;fn

izfrn'kZ a–mRltZu o b–mRltZu nksuksa }kjk ,d lkFk

fo?kfVr gksrk gS rks fdrus le; i'pkr~ izfrn'kZ dsoy

,d pkSFkkbZ jg tk,xk\

(1) 10 o"kZ

(2) 20 o"kZ

(3) 40 o"kZ

(4) 45 o"kZ

17. æo dh n cwanksa ftuesa izR;sd dh i`"B ÅtkZ E gS] dks

tksM+dj ,d cwan cukbZ tkrh gS %&

(1) bl izfØ;k esa dqN ÅtkZ eqDr gksxhA

(2) bl izfØ;k esa dqN ÅtkZ vo'kksf"kr gksxhA

(3) eqDr ;k vo'kksf"kr gksus okyh ÅtkZ E

n(n–n2/3)

gksxhA

(4) eqDr ;k vo'kksf"kr gksus okyh ÅtkZ nE (22/3–1) gksxhA

H-10/32



00CE314002

18. The Pitot tube shown in the figure is used tomeasure fluid flow velocity in a pipe of crosssectional area S. It was invented by a Frenchengineer Henri Pitot in the early 18th century.The volume of the gas flowing across the sectionof the pipe per unit time is (The difference inthe liquid columns is Dh, r

0 and r are the

densities of liquid and the gas respectively) :-

OUTIN

(1) Q = 0h g2S

D rr

(2) Q = 02 h gS

D rr

(3) Q = 0hgS

D rr

(4) Q = 0

2 h gS

D rr

19. The plot given below is of the average powerdelivered to an LRC circuit versus frequency.The quality factor of the circuit is :

3 5 7 9 11frequency (kHz)

aver

age

pow

er (m

icro

wat

ts)

0.0

0.5

1.0

(1) 5 (2) 2.4 (3) 2.8 (4) 1.4

18. fp= esa iznf'kZr Pitot uyh dk mi;ksx S vuqizLFk dkV{ks=Qy okys ikbi esa æO; izokg ds osx dks Kkr djus dsfy, fd;k tkrk gSA bl uyh dk vfo"dkj 18th 'krkCnhds izkjEHk esa ,d Ýsap baftfu;j Henri Pitot us fd;kFkkA iznf'kZr ikbi ds vuqizLFkdkV {ks= ls izfr bdkbZle; esa izokfgr gksus okyh xSl dk vk;ru gksxk(æo LrEHkksa esa vUrj Dh gS rFkk æo ,oa xSl ds ?kuRoØe'k% r

0 o r gSA) :-

OUTIN

(1) Q = 0h g2S

D rr

(2) Q = 02 h gS

D rr

(3) Q = 0hgS

D rr

(4) Q = 0

2 h gS

D rr

19. iznf'kZr fp= esa ,d LRC ifjiFk dks nh xbZ vkSlr

'kfä rFkk vko`fÙk ds e/; vkjs[k n'kkZ;k x;k gSA ifjiFk

dk fo'ks"krk xq.kkad gksxk %&

3 5 7 9 11frequency (kHz)

aver

age

pow

er (m

icro

wat

ts)

0.0

0.5

1.0

(1) 5 (2) 2.4 (3) 2.8 (4) 1.4

H-11/32



00CE314002

20. A shell is fired at a speed of 200 m/s at an angleof 37° above horizontal from top of a tower 80mhigh. At the same instant another shell was firedfrom a jeep travelling away from the tower at aspeed of 10 m/s as shown. The velocity of thisshell relative to jeep is 250 m/s at an angle of53° with horizontal. Find the time (in sec) takenby the two shells to come closest.

sh e l l 1

sh e ll 2

jeep

(1) 1 (2) 2 (3) 3 (4) 4

21. In the given figure linear acceleration of solid

cylinder of mass m2 is a

2. Then angular

acceleration a2is (given that there is no slipping).

\\\\\\\\\\\\\\\\\\\\\\\\\\\

m ,R1

m ,R2

(1) 2aR (2) 2(a g)

R+

(3) 22(a g)R

+(4) None of these

20. ,d xksyh dks 80 m Åaps ehukj ds 'kh"kZ ls {kSfrt ls

Åij 37° dks.k ij 200 m/s dh pky ls nkxrs gSaA mlh

{k.k ,d vU; xksyh dks fp=kuqlkj 10 m/s dh pky ls

ehukj ls nwj tk jgh thi ls nkxk x;k FkkA bl xksyh dk

thi ds lkis{k osx] {kSfrt ls 53° dks.k ij 250 m/s gSA

nksuksa xksfy;ka s ds e/; nwjh U;wure gksus esa yxk le;

(lsd.M esa) Kkr djsaAsh e l l 1

sh e ll 2

jeep

(1) 1 (2) 2 (3) 3 (4) 4

21. iznf'kZr fp= esa m2 nzO;eku ds ,d Bksl csyu dk

j s[kh; Roj.k a2 gS] rc dk s.kh; Roj.k a

2gk sx k

(;gk¡ dksbZ fQlyu ugha gSA)\\\\\\\\\\\\\\\\\\\\\\\\\\\

m ,R1

m ,R2

(1) 2aR (2) 2(a g)

R+

(3) 22(a g)R

+(4) buesa ls dksbZ ugh

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00CE314002

22. A billiard ball is hit by a cue at a point distance

h above the centre. It acquires a linear velocity

v0. Let m be the mass and r be the radius of the

ball. The angular velocity acquired by the

ball is

(1) 20

r2h5v

(2) 20

r3h2v

(3) 20

r5h2v

(4) 20

rh2v

23. A sphere of mass M and radius r slips on a rough

horizontal plane. At some instant it has

translational velocity v0 and rotational velocity

about the centre 0V2r . The translational velocity

when the sphere starts pure rolling motion is

(1) 0V5 (2) 02V

7

(3) 0V3 (4) 0V6

7

22. ,d fcfy;MZ xsan dks dsUæ ls h Å¡pkbZ ij fLFkr fcUnq ij

NM+ }kjk fgV fd;k tkrk gSA ;g js[kh; osx v0 izkIr dj

ysrh gSA ekuk xsan dk æO;eku m o f=T;k r gSA xsan }kjk

izkIr dks.kh; osx gksxk%&

(1) 20

r2h5v

(2) 20

r3h2v

(3) 20

r5h2v

(4) 20

rh2v

23. M nzO;eku rFkk r f=T;k dk ,d xksyk [kqjnjs {kSfrt

ry ij fQlyrk gSA fdlh {k.k ij bldk LFkkukUrjh;

osx v0 rFkk dsUnz ds lkis{k ?kw.kZu osx 0V

2r gSA tc xksyk

'kq¼ yksVuh xfr izkjEHk djrk gS rks bldk LFkkukUrjh;

osx gksxk%&

(1) 0V5 (2) 02V

7

(3) 0V3 (4) 0V6

7

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00CE314002

24. A small block slides down from rest at point Aon the surface of a smooth cylinder, as shown.At point B, the block falls off (leaves) thecylinder. The equation relating the angles q

1 and

q2 is given by

A

B

q1

q2

(1) 22 13sin sinq = q (2) 3

2 12sin sinq = q

(3) 22 13cos cosq = q (4) 3

2 12cos cosq = q

25. Charge is uniformly distributed on the surface

of a hollow hemisphere. Let O and A be two

points on the base of the hemisphere and Vo and

VA be the electric potentials at O and A

respectively. Then,

O AO A

(1) VA = V

o(2) V

A < V

o

(3) VA > V

o(4) insufficient data

24. ,d NksVk CykWd fp=kuqlkj ,d fpdus csyu dh lrg

ij fcUnq A ls fojkekLFkk ls uhps dh vksj xfr djrk gSA

fcUnq B ij CykWd csyu ls fxj tkrk gSA dks.k q1 o q

2 ds

e/; lehdj.k gksxh%&

A

B

q1

q2

(1) 22 13sin sinq = q (2) 3

2 12sin sinq = q

(3) 22 13cos cosq = q (4) 3

2 12cos cosq = q

25. ,d [kks[kys v¼Zxksys dh lrg ij vkos'k ,dleku :i

ls forjhr gSA ekuk v¼Zxksys ds vk/kkj ij fLFkr fcUnq O

rFkk A ij fo|qr foHko Øe'k% Vo rFkk V

A gSA

rc %&

O AO A

(1) VA = V

o(2) V

A < V

o

(3) VA > V

o(4) vk¡dM+s vi;kZIr gS

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00CE314002

26. Four charges q, 2q, –4q and 2q are placed in

order at the four corners of a square of side b.

The net field at the centre of the square is

(1) 20b2

qpe from +q to –4q

(2) 20b2q5

pe from +q to –4q

(3) 20b2q10

pe from +q to –4q

(4) 20b2q20

pe from –4q to +q

27. A glass rod has ends as shown in figure.

The refractive index of glass is m. The object O

is at a distance 2R from the surface of larger

radius of curvature. The distance between apexes

of ends is 3R. The range of m for which the

image is real is given by

R

G la ss ro d

3 R

O2 R

R /2

air

(1) 2–2.5

(2) 2 – 2.25

(3) for any value of m>1

(4) for any value of m<2.25

26. pkj vkos'k q, 2q, –4q o 2q dks b Hkqtk okys oxZ ds

pkjksa dksuks ij Øe esa j[kk tkrk gSA oxZ ds dsUæ ij dqy

{ks= gksxk%&

(1) 20b2

qpe ; +q ls –4q dh vksj

(2) 20b2q5

pe ; +q ls –4q dh vksj

(3) 20b2q10

pe ; +q ls –4q dh vksj

(4) 20b2q20

pe ; –4q ls +q dh vksj

27. ,d dkap dh NM + ds fljs fp=kuqlkj gSaA dkap dk

viorZukad m gSA fcEc O vf/kd oØrk f=T;k okyh

lrg ls 2R nwjh ij gSA fljksa ds 'kh"kks ± ds chp dh nwjh

3R gSA m ds ekuksa dh fdl ijkl ds fy;s izfrfcEc

okLrfod gksxk

R

G la ss ro d

3 R

O2 R

R /2

air

(1) 2–2.5

(2) 2–2.25

(3) m>1 ds fdlh Hkh eku ds fy,

(4) m<2.25 ds fdlh Hkh eku ds fy,

H-15/32



00CE314002

28. In a uniform electric field, a cube of side 1 cm isplaced. The total energy stored in the cube is8.85µJ. The electric field is parallel to four ofthe faces of the cube. The electric flux throughany one of the remaining two faces is.

(1) 1

V / m5 2

(2) 100 2V / m

(3) 5 2V / m (4) 10 2V / m29. A charged particle oscillates about its mean

equilibrium position with a frequency of109 Hz. The electromagnetic waves produced:(1) will have frequency of 3 × 109 Hz.(2) will have frequency of 2 × 109 Hz.(3) will have a wavelength of 0.8 m.(4) fall in the region of radiowaves.

30. Suppose that two heat engines are connected inseries, such that the heat exhaust of the firstengine is used as the heat input of the secondengine as shown in figure. The efficiencies ofthe engines are h

1 and h

2, respectively. The net

efficiency of the combination is given by

(1) hnet

= h2 + (1 – h

1)h

2 Th

w1

Tm

w2

Tc

Q2

Q2

Q1

Q3

(2) ( )1

net1 21

hh =

- h h

(3) hnet

= h1 + (1 – h

1)h

2

(4) ( )1

net2 2

1

1

- hh =

- h h

28. fdlh le:i fo|qr {ks= esa 1 cm Hkqtk okyk ,d ?ku

j[kk gSA ?ku esa lafpr dqy ÅtkZ 8.85µJ gSA ;gk¡ fo|qr

{ks= ?ku ds pkj Qydksa ds lekUrj gSA 'ks"k cps nksuksaQydksa esa ls fdlh ,d ls fuxZr fo|qr ¶yDl gksxk

(1) 1

V / m5 2

(2) 100 2V / m

(3) 5 2V / m (4) 10 2V / m29. dksbZ vkosf'kr d.k viuh ekè; larqyu fLFkfr ds nksuksa

vksj 109 Hz vko`fÙk ds nksyu djrk gSA blls mRiUuoS|qrpqEcdh; rjaxksa(1) dh vko`fÙk 3 × 109 Hz gksxh(2) dh vko`fÙk 2 × 109 Hz gksxh(3) dh rjaxnS?; Z 0.8 m gksxh(4) ds fofdj.k jsfM;ks rjaxksa ds {ks= esa gksaxs

30. nks Å"ek baftuksa dks Js.khØe esa bl izdkj tksM+k tkrk gSfd izFke batu ls mRlftZr Å"ek nwljs batu ds fy;s

fuos'kh Å"ek dk dk;Z djrh gS] fp= ns[ksaA ;fn bu

baftuksa dh n{krk,sa Øe'k% h1 o h

2 gks rks bl fudk; dh

dqy n{krk gksxh%&

(1) hnet

= h2 + (1 – h

1)h

2 Th

w1

Tm

w2

Tc

Q2

Q2

Q1

Q3

(2) ( )1

net1 21

hh =

- h h

(3) hnet

= h1 + (1 – h

1)h

2

(4) ( )1

net2 2

1

1

- hh =

- h h

H-16/32


00CE314002


PART B - CHEMISTRY31. Work done in isothermal reversible process for

ideal gas is-

(1) –nRT ln 2

1

VV

æ öç ÷è ø

(2) – P(V2 – V1)

(3) 2 2 1 1P V P V

1æ ö-ç ÷g -è ø

(4) 0

32. Osmotic pressure of a 4.44% solution ofanhydrous CaCl

2 was found to be 16.42 atm at

27°C. What is the degree of dissociation ofCaCl

2 :

(1) 100% (2) 33%

(3) 66% (4) 50%

33. The standard reduction potentials of Cu2+, Zn2+,Sn2+ and Ag+ are 0.34, –0.76, – 0.14 and 0.80 Vrespectively. The storage that is possible withoutany reaction is for (under standard condition) :

(1) CuSO4 solution in a zinc vessel

(2) AgNO3 solution in a zinc vessel

(3) AgNO3 solution in a tin vessel

(4) CuSO4 solution in a silver vessel

34. 50 ml of 0.02M NaHSO4 is mixed with 50 mlof 0.02M Na2SO4. Calculate pH of the

resulting solution.

[pKa2 (H2SO4) = 2]

(1) 2 (2) 2–log ( 2 )–1

(3) 2 + log ( 2 )+1 (4) 1.7

31. vkn'kZ xSl ds fy, lerkih; mRØe.kh; izØe esa fd;kx;k dk;Z gS-

(1) –nRT ln 2

1

VV

æ öç ÷è ø

(2) – P(V2 – V1)

(3) 2 2 1 1P V P V

1æ ö-ç ÷g -è ø

(4) 0

32. futZyh; CaCl2 ds 4.44% foy;u dk 27°C ij

ijklj.k nkc 16.42 atm ik;k x;kA CaCl2 ds fo;kstu

dh ek=k D;k gksxh(1) 100% (2) 33%

(3) 66% (4) 50%

33. Cu2+, Zn2+, Sn2+ rFkk Ag+ ds ekud vip;u foHkoØe'k% 0.34, –0.76, – 0.14 rFkk 0.80 V gSaA fdlhHkh vfHkfØ;k ds fcuk fuEu esa ls fdls laxzg.k djuklEHko gSA (ekud ifjfLFkfr;ksa esa) :(1) ftad ik= esa CuSO

4 dk foy;u

(2) ftad ik= esa AgNO3 dk foy;u

(3) fVu ik= esa AgNO3 dk foy;u

(4) flYoj ik= esa CuSO4 dk foy;u

34. 50 ml 0.02M NaHSO4 dks 50 ml 0.02M Na2SO4

ds lkFk fefJr fd;k x;kA ifj.kkeh foy;u dh pH

Kkr dhft,A [pKa2 (H2SO4) = 2]

(1) 2 (2) 2–log ( 2 )–1

(3) 2 + log ( 2 )+1 (4) 1.7

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00CE314002


35. Phosphine [PH3(g)] decomposes to produce

vapours of phosphorus (P4) and H

2 gas. What

will be the change in volume when 100 mL of

phosphine is decomposed ?

(1) +50 mL (2) 500 mL

(3) +75 mL (4) –500 mL

36. In NaCl type structure shortest distance between

two nearest neighbours is 100 pm, then distance

between two next nearest neighbours in the

same unit cell will be

(1) 100

2 pm (2) 100 2 pm

(3) 200 pm (4) 200 2 pm

37. A solution of Na2S

2O

3 is standardized

iodometrically against 0.167 g of KBrO3 where

BrO3– changes to Br–. This process requires

45 mL of the Na2S

2O

3 solution. What is the

strength of the Na2S

2O

3 ?

[Mw. of KBrO3 = 167]

(1) 2

N15

(2) 2

N30

(3) 1

N30

(4) 1

N60

35. QkWLQhu [PH3(g)] fo;ksftr gksdj QkWLQksjl (P

4) rFkk

H2 xSl dh ok"iksa dk fuekZ.k djrh gS tc 100 mL

QkWLQhu fo;ksftr gksrh gS rc vk;ru esa D;k ifjorZu

gksxk ?

(1) +50 mL (2) 500 mL

(3) +75 mL (4) –500 mL

36. NaCl izdkj dh lajpuk esa nks fudVre iM+ksfl;ks dse/; U;wure nwjh 100 pm gS ] rc leku bdkbZ lSy esavxys nks fudVre iMksfl;ksa ds e/; nwjh gksxh

(1) 100

2pm (2) 100 2 pm

(3) 200 pm (4) 200 2 pm

37. Na2S

2O

3 foy;u dks 0.167 g KBrO

3 ds fo:¼

vk;k sMks ferh; :i ls ekudhd`r fd;k tkrk g S]

tgk¡ BrO3

– dk Br– esa ifjorZu gksrk gSA bl izØe ds

fy, Na2S

2O

3 foy;u ds 45 mL dh vko';drk gksrh

gSA Na2S

2O

3 dh lkeF;Z D;k gksrh gS ? [KBrO

3 dk

v.kqHkkj = 167]

(1) 2

N15

(2) 2

N30

(3) 1

N30

(4) 1

N60

H-18/32


00CE314002


38. 35X88 is an unstable isotope, decays in two suc-

cessive steps to produce stable isotope 32Z84as

35X88 I¾¾® Y II¾¾® 32Z

84

The correct statement is (possible emission area, b–,positron, neutron and 1H

1 and in each steponly one particle is involved)

(1) I may involve a b–emission.

(2) II may involve a neutron emission

(3) Y and Z may be isodiaphers

(4) X and Z may be isotone

39. For a liquid normal boiling point is –173ºC then at2 atm pressure it's boiling point should be nearly(DHvap = 200cal/mole, R = 2 cal/mol-K,ln2 = 0.7)

(1) –73ºC (2) 333ºC

(3) 60ºC (4) 103ºC

40. Enthalpy of dilution of 4M HCl to 2M HCl is–2.5 kJ/mol. Find enthalpy change when500 ml of HCl is diluted from 4M to 2M

(1) –2.5 kJ/mol (2) –5 kJ/mol

(3) –10 kJ/mol (4) –1.25 kJ/mol

41. Which of the following pair will produce all

product same on reaction with water :

(1) Na2O, Na (2) CO2, K2O

(3) Na, NaH (4) None of these

38. 35X88 ,d vLFkk;h leLFkkuhd gS] ;g nks Øekxr inksa

ds {k; gksdj LFkk;h leLFkkfud 32Z84 fuEu izdkj

mRiknhr djrs gS 35X88 I¾¾® Y II¾¾® 32Z

84

lgh dFku gSa (lEHkkfor mRltZu a, b–, ikWthVªkWu] U;wVªkWu]

1H1 rFkk izR;sd in esa dsoy ,d d.k lfEefyr gS)

(1) I esa b– mRltZu lEehfyr gks ldrk gS

(2) II esa U;wVªkWu mRltZu lEehyhr gks ldrk gS

(3) Y rFkk Z vkblksMkbZQj gks ldrs gS

(4) X rFkk Z leU;wVªkWfud gks ldrs gS

39. ,d æo ds fy, lkekU; DoFkukad –173ºC gS rks 2 atm

nkc ij bldk DoFkukad yxHkx gksuk pkfg, %

(DHvap = 200 cal/mole, R = 2 cal/mol-K,

ln2 = 0.7)

(1) –73ºC (2) 333ºC

(3) 60ºC (4) 103ºC

40. 4M HCl dk 2M HCl esa ruqdj.k dh , sUFk sYih–2.5 kJ/mol gSA tc 500 ml HCl dks 4M ls 2Mesa ruq fd;k tkrk gS] rc ,sUFkSYih ifjorZu Kkr dhft,A(1) –2.5 kJ/mol (2) –5 kJ/mol

(3) –10 kJ/mol (4) –1.25 kJ/mol

41. fuEu esa ls dkSulk ; qXe ty ds lkFk fØ;k djkus ij

lHkh mRikn] leku nsxk :(1) Na2O, Na (2) CO2, K2O

(3) Na, NaH (4) buesa ls dksbZ ugha

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00CE314002


42. Which of the following complex compound is

optically inactive :

(1) [M(AA)2cd] (2) [M(AB)(CD)(EF)]

(3) [M(AA)2] (4) Mabcd(tetrahedral)

43. Total number of compounds / ions which have

atleast two bond lengths same out of

CO3

2–, HCOOH, S2O3

2–, NO

1

2

(1) 2 (2) 4 (3) 3 (4) 1

44. Three complex compounds I, II and III have

Blue, Red and Green colour respectively then

which of the following order of labs (absorbed

wavelength) is CORRECT :(1) I > II > III (2) III > II > I

(3) II > I > III (4) III > I > II

45. Which of the following statement is

CORRECT :

(1) Out of four quantum number, azimuthal

quantum number may have highest

numerical value for a particular subshell

(2) On increasing electronegativity,

ionisation energy decreases generally

(3) If intermolecular axis is x-axis then dxy

will form pie (p) bond with px orbital

(4) None of these

42. fuEu esa ls dkSulk ladqy ;kSfxd izdkf'kd vfØ; gS

(1) [M(AA)2cd] (2) [M(AB)(CD)(EF)]

(3) [M(AA)2] (4) Mabcd(prq"Qydh;)

43. ;kSfxdksa @ vk;uksa dh dqy la[;k ftuesa de ls de

nks ca/k yEckbZ;k¡ leku gSa] gS

CO3

2–, HCOOH, S2O3

2–, NO

1

2

(1) 2 (2) 4 (3) 3 (4) 1

44. rhu ladqy ;kSfxd I, II rFkk III Øe'k% uhyk] yky rFkk

gjs jax ds gSa rks labs (vo'kksf"kr rajxnS/; Z) ds fy;s fuEu

esa ls dkSulk Øe lgh gS :

(1) I > II > III (2) III > II > I

(3) II > I > III (4) III > I > II

45. fuEu esa ls dkSulk dFku lgh gS :

(1) ,d fo'ks"k midks'k ds fy;s] pkjksa DokWUVe la[;kvksa

esa ls fnxa'kh DokWUVe la[;k dk vkfdad eku vf/kdre

gks ldrk gS

(2) lkekU;r% fo|qr½.krk c<+us ds lkFk vk;uu ÅtkZ

de gksrh gS

(3) ;fn vUrjvkf.od v{k] x-v{k gS rks dxy d{kd]

px d{kd ds lkFk (p) ca/k cuk;sxk

(4) buesa ls dksbZ ugha

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00CE314002


46. Which of the following will be dissolved in

excess of NaOH :

(1) Cu(OH)2 (2) Zn(OH)2

(3) Bi(OH)3 (4) None of these

47. 'A' salt which gives green flame with bunsen

burner, on reaction with dil. H2SO4 produces gas

(2) and white turbidity (Z). The saturated

solution of gas (2) on reaction with H2S gas

produces 'Z' then select CORRECT statement

(1) Gas 'B' is CO2

(2) Gas 'B' is SO2

(3) White turbidity is of elemental sulphur

(4) Both 'B' and 'C' are correct

48. Refining of nickel is done by:

(1) liquation (2) poling

(3) cupellation (4) Mond's process49. A metal (M) which exists in oxide form and its

oxide is not possible to reduce by carbon up to800ºC but it is possible to reduce by Al evenbelow 800ºC. Select correct Ellingham diagramfor metal (M)

(1) DG0

T

Al

C CO® 2

M

C CO

® (2) D 0G

T

AlM C

CO®

C CO® 2

(3) D 0G

T

AlM

C CO® 2

C CO

® (4) D 0G

T

Al

MC CO® 2

C CO

®

46. fuEu es a ls dk Su NaOH ds vkf/kD; esa foys;

gksxk :(1) Cu(OH)2 (2) Zn(OH)2

(3) Bi(OH)3 (4) buesa ls dksbZ ugha

47. ,d yo.k 'A' tks cqUlu cuZj dks gjh Tokyk iznku djrkgS] ruq H2SO4 ds lkFk fØ;k djkus ij xSl (2) rFkk 'osr/kqa/kykiu (Z) nsrk gSA xSl (2) dk larÌr foy;u H2S

xSl ds lkFk fØ;k djkus ij 'Z' nsrk gS rks lgh dFkupqfu;sa:

(1) xSl 'B', CO2 gS(2) xSl 'B', SO2 gS

(3) 'osr /kq a/kykiu lYQj rRo dk gS(4) 'B' rFkk 'C' nksuksa lgh gSa

48. fudy dk ifj'kks/ku fdlds }kjk fd;k tkrk gS&

(1) æo.k (2) ikWfyax

(3) D;wisyhdj.k (4) ekWUM izØe49. ,d /kkrq (M) tks blds vkWDlkbM :i esa vfLrRo j[krh

gS vkSj blds vkWDlkbM dk 800ºC rd dkcZu }kjkvip;u lEHko ugha gksrk gS] ijUrq Al }kjk 800ºC lsde rki ij Hkh vipf;r fd;k tk ldrk gSA /kkrq (M)ds fy, lgh ,fyxae oØ dk p;u dhft,

(1) DG0

T

Al

C CO® 2

M

C CO

® (2) D 0G

T

AlM C

CO®

C CO® 2

(3) D 0G

T

AlM

C CO® 2

C CO

® (4) D 0G

T

Al

MC CO® 2

C CO

®

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00CE314002


50. Which of the following metal produces N2O on

reaction with 20% HNO3 -

(1) Sn (2) Pt

(3) Zn (4) Pb

51. An alcohol with molecular formula C5H

12O

does not change the colour of acidic dichromate

solution. Identify that alcohol from the

following-

(1) OH (2)

OH

(3) OH

(4) OH

52. The alcohol which show turbidity immediate

when reacted with (HCl + ZnCl2)

(1) OH

(2) OH

(3) OH (4)

OH

50. fuEu esa ls dkSulh /kkrq 20% HNO3 ds lkFk fØ;k

djkus ij N2O cukrh gS

(1) Sn (2) Pt

(3) Zn (4) Pb

51. C5H

12O v.kqlw= dk ,d ,YdksgkWy vEyh; MkbØksesV

ds foy;u dk jax ifjo£rr ugha djrk gSA fuEu esa ls

og ,YdksgkWy igpkfu, -

(1) OH (2)

OH

(3) OH

(4) OH

52. ,slk ,YdksgkWy tks (HCl + ZnCl2) ds lkFk fØ;k djds

rqjUr /kqa/kykiu n'kkZrk gS&

(1) OH

(2) OH

(3) OH (4)

OH

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00CE314002


53.O

Anhydrous HI

conc.HI

A

B

A & B are respectively:

(1)OH

I

& I

OH

(2) OH

I

&

I

OH

(3) Both I

OH(4) Both

OHI

54. NBS P1

MgD.E. P

2

CH –CH3

H /Å D

O

P3

(Major product).

Identify P3

(1) (2)

(3) OH

(4)

55. Find out the number of 1-2-shifts during the

conversion of

B /BH

- D-¾¾¾®

(1) 2 (2) 3

(3) 4 (4) 5

53. O

futZyh; HI

lkUnz HI

A

B

A rFkk B Øe'k% gS&

(1)OH

I

rFkk I

OH

(2)OH

I

rFkk

I

OH

(3) nksuksa I

OH(4) nksuksa

OHI

54. NBS P1

MgD.E. P

2

CH –CH3

H /Å D

O

P3

(eq[; mRikn)

P3 igpkfu,s&

(1) (2)

(3) OH

(4)

55. fuEu :ikUrj.k ds nkSjku 1-2-f'k¶V dh la[;k Kkr

dhft,A

B /BH

- D-¾¾¾®

(1) 2 (2) 3

(3) 4 (4) 5

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00CE314002


56. Which of the following has highest heat of

hydrogenation.

(1) (2)

(3) (4)

57. Which amino acid is achiral ?

(1) Alamine

(2) Valine

(3) Proline

(4) Glycine

58. Which of the following is not correct about

product -

OH OHconc.H SO2 4

D

(1) It is a spiro compound

(2) It is a Ketone

(3) It can show tautomerism

(4) Its double bond equivalent is 4

56. fuEu esa ls fdldh gkbMªkstuhdj.k dh Å"ek vf/kdre

gksrh gS&

(1) (2)

(3) (4)

57. dkSulk vehuks vEy vfdjSy gS&

(1) ,sykfeu

(2) osyhu

(3) izksfyu

(4) Xykbflu

58. mRikn ds fy;s fuEu esa dkSulk lgh ugha gS&

OH OHlkUnz H SO2 4

D

(1) ;s Likbjks ;kSfxd gS

(2) ;s dhVksu gS

(3) ;s pyko;ork iznf'kZr dj ldrk gS

(4) bldk f}cU/k rqY;kad 4 gS

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00CE314002


59. In which of the following pairs first one is having

more resonance energy than the second one-

(1) ,

(2) Å ,

Å

(3)NH

, NH2

(4) Å ,

Å

60. Find out correct stability order in the following

carbocations-

(I)Å

(II)

Å

(III)

Å

(IV) Å

OH

(1) IV > I > III > II

(2) IV > III > I > II

(3) I > IV > III > II

(4) I > III > IV > II

59. fuEu esa ls fdl ;qXe esa nwljs ;kSfxd dh rqyuk esa izFkedh vuqukn ÅtkZ vf/kd gksrh gS&

(1) ,

(2) Å ,

Å

(3)NH

, NH2

(4) Å ,

Å

60. fuEu dkcZ/kuk;uksa esa LFkkf;Ro dk lgh Øe Kkr dhft;s&

(I)Å

(II)

Å

(III)

Å

(IV) Å

OH

(1) IV > I > III > II

(2) IV > III > I > II

(3) I > IV > III > II

(4) I > III > IV > II

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00CE314002


PART C - MATHEMATICS61. If z1, z2, z3 Î C such that |z1| = |z2| = |z3| = 2,

then greatest value of expression|z1 – z2|.|z2 – z3| + |z3 – z1|.|z1 – z2| + |z2 – z3||z3 – z1| is

(1) 18 (2) 36 (3) 9 (4) 72

62. Let z1, z2 Î C such that 1 2| z z | 3+ = and

|z1| = |z2| = 1, then the value of |z1 – z2| is -

(1) 12 (2) 2 (3) 1 (4) 4

63. Two distinct numbers a and b are chosenrandomly from the set {2, 22, 23, ........ 225}.What is the probability the logab is an integer ?

(1) 31

300 (2) 31

150(3)

750 (4)

225

64. Figure shows DABC with AB = 3, AC = 4 &BC = 5. Three circles S1, S2 & S3 have theircentres on A, B & C respectively and theyexternally touches each other. The sum of areasof three circles is -

B C

A

S2

S3

S1

(1) 11p (2) 12p (3) 13p (4) 14p

61. ;fn z1, z2, z3 Î C bl izdkj gS fd |z1| = |z2| = |z3| = 2gS] rks O;atd |z1 – z2|.|z2 – z3| + |z3 – z1|.|z1 – z2|+ |z2 – z3||z3 – z1| dk egÙke eku gksxk

(1) 18 (2) 36 (3) 9 (4) 72

62. ekuk z1, z2 Î C bl izdkj gS fd 1 2| z z | 3+ = rFkk

|z1| = |z2| = 1 gks] rks |z1 – z2| dk eku gksxk -

(1) 12 (2) 2 (3) 1 (4) 4

63. leqPp; {2, 22, 23, ........ 225} ls nks fHkUu la[;kvksaa rFkk b dk ;kn`PN;k p;u djrs gSA logab ds iw.kk±dgksus dh izkf;drk D;k gksxh ?

(1) 31

300 (2) 31

150(3)

750 (4)

225

64. fp=kuqlkj f=Hkqt ABC ftlesa AB = 3, AC = 4rFkk BC = 5 gSA rhu o`Ùk S1, S2 ,oa S3 ftudsdsUnz Øe'k% A, B ,oa C ij gS rFkk ;s o`Ùk ,d nwljs dkscká Li'kZ djrs gSA rhuksa o`Ùkksa ds {ks=Qyksa dk ;ksxQygksxk-

B C

A

S2

S3

S1

(1) 11p (2) 12p (3) 13p (4) 14p

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65. The range of function ƒ : R ® R, ƒ(x) = 4

4

(x 1)x 1++

is -(1) [0, ¥) (2) [0, 16](3) [0, 8] (4) [0, 32]

66. Twenty persons arrive in a town having 3 hotelsx, y and z. If each person randomly chooses oneof these hotels, then what is the probability thatatleast 2 of them goes in hotel x, atleast 1 inhotel y and atleast 1 in hotel z ? (each hotel hascapacity for more than 20 guests)

(1) 18

222

2

C

C (2) 20 18 17 16

2 1 120

C . C . C .33

(3) 20

99

C3

(4) 20 20

20

3 13.2 43

3

- +

67. If

21 2 3

1 2 3

1 2 3

3 1 2 1 3 3 3 8 1 a a a8 9 5 3 2 7 1 9 1 b b b1 1 3 3 7 9 2 5 3 c c c

ì üæ öæ öæ ö æ öï ïç ÷ç ÷ç ÷ ç ÷=í ýç ÷ç ÷ç ÷ ç ÷ï ïç ÷ç ÷ç ÷ ç ÷è øè øè ø è øî þ

,

then the value of |a2 – b1| + |a3 – c1| + |b3 – c2| is-(1) 0 (2) 1 (3) 2 (4) 3

68. The minimum value ofƒ(x) = |x – 1| + |2x – 1| + |3x – 1| + ...... + |119x – 1|occurs at x. Then x is equal to-

(1) 1

84 (2) 151 (3)

180 (4)

194

65. Qyu ƒ : R ® R, ƒ(x) = 4

4

(x 1)x 1++

dk ifjlj

gksxk -(1) [0, ¥) (2) [0, 16](3) [0, 8] (4) [0, 32]

66. chl O;fDr ,d dLcs esa igq¡prs gS ftlesa 3 gksVy x, yrFkk z gSA ;fn izR;sd O;fDr ;kn`PN;k buesa ls fdlh,d gksVy dk p;u djrk gS] rks D;k izkf;drk gS rkfdmuesa ls de ls de 2 O;fDr gksVy x esa] de ls de1 O;fDr gksVy y esa rFkk de ls de 1 O;fDr gksVyz esa tk;s ? (izR;sd gksVy dh {kerk 20 vfrfFk;ksa lsvf/kd gS)

(1) 18

222

2

C

C (2) 20 18 17 16

2 1 120

C . C . C .33

(3) 20

99

C3

(4) 20 20

20

3 13.2 43

3

- +

67. ;fn

21 2 3

1 2 3

1 2 3

3 1 2 1 3 3 3 8 1 a a a8 9 5 3 2 7 1 9 1 b b b1 1 3 3 7 9 2 5 3 c c c

ì üæ öæ öæ ö æ öï ïç ÷ç ÷ç ÷ ç ÷=í ýç ÷ç ÷ç ÷ ç ÷ï ïç ÷ç ÷ç ÷ ç ÷è øè øè ø è øî þ

gS] rks |a2 – b1| + |a3 – c1| + |b3 – c2| dk eku gksxk(1) 0 (2) 1 (3) 2 (4) 3

68. ƒ(x) = |x – 1| + |2x – 1| + |3x – 1| + ...... + |119x – 1|dk U;wure eku ftl x ij izkIr gksxk] ml x dk ekugksxk&

(1) 1

84 (2) 151 (3)

180 (4)

194

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00CE314002


69. The value of

xloge x e

x e

é ùæ öç ÷ê úè øê ú" >-ê ú

ê úë û

is equal to

(where [.] denotes greatest integer function)(1) 1(2) 0(3) 2(4) does not take unique value

70. If ƒ(n) = 1n n 1

e 1tan n Ne e

-- +

-æ ö" Îç ÷+è ø,

then n 1

ƒ(n)¥

=å is equal to -

(1) 1 1cote

- æ öç ÷è ø

(2) cot–1(1)

(3) 1 2tane

- (4) 1 1tane

-

71./ 2 2

0

4x sin x x cos x dx2 sin x

p +ò is equal to -

(1) 2p

(2) 2

4p

(3) 4p

(4) 2

16p

69.

xloge x e

x e

é ùæ öç ÷ê úè øê ú" >-ê ú

ê úë û

dk eku gksxk (tgk¡ [.] egÙke

iw.kk±d Qyu dks n'kkZrk gS)(1) 1(2) 0(3) 2(4) vf}rh; eku xzg.k ugha djsxkA

70. ;fn ƒ(n) = 1n n 1

e 1tan n Ne e

-- +

-æ ö" Îç ÷+è ø gks]

rks n 1

ƒ(n)¥

=å cjkcj gksxk -

(1) 1 1cote

- æ öç ÷è ø

(2) cot–1(1)

(3) 1 2tane

- (4) 1 1tane

-

71./ 2 2

0

4x sin x x cos x dx2 sin x

p +ò cjkcj gksxk -

(1) 2p

(2) 2

4p

(3) 4p

(4) 2

16p

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00CE314002


72. Let ƒ : (0, ¥) ® R & g : (0, ¥) ® R be two

functions where g(x) =1xx

+ .

If 1 < ƒ(x).g(x) < 10 " x > 0, then xlim ƒ(x)®¥

is -

(1) 1 (2) 10(3) 0 (4) can not be determined

73. The area bounded by curves y = 1 – cos(px),

y = – x2 and the line 1x2

= and 1x2

= - is -

(1) 13 212

-p

(2) 12 113

-p

(3) 12 113 2

-p

(4) 13 112

-p

74. The number of 5 digit numbers that can beformed with digits 1,2,3,4,5,6 if the number mustinclude 1 and 2 is-(1) 65 – 2.55 + 45 (2) 10.54

(3) 20.54 (4) 5.45

75. The vertex of parabola is at (1,2) and its axis isparallel to y-axis. If parabola passess through(0,6), then its latus rectum is-

(1) 1

2(2) 2 (3)

1

4(4) 4

76. The varience of data 1001, 1003, 1006, 1007,1009, 1010 is -(1) 10 (2) 15 (3) 20 (4) 50

72. ekuk nks Qyu ƒ : (0, ¥) ® R rFkk g : (0, ¥) ® R

tgk¡ g(x) =1xx

+ gSA ;fn 1< ƒ(x).g(x) <10 " x > 0

gks] rks xlim ƒ(x)®¥

gksxk -

(1) 1 (2) 10(3) 0 (4) Kkr ugha fd;k tk ldrk

73. oØ y = 1 – cos(px), y = – x2 ,oa js[kk

1x2

= ,oa 1x2

= - }kjk ifjc¼ {ks=Qy gksxk&

(1) 13 212

-p

(2) 12 113

-p

(3) 12 113 2

-p

(4) 13 112

-p

74. vadksa 1,2,3,4,5,6 ds iz;ksx ls fufeZr dh tk ldusokyh 5 vadksa dh la[;kvksa dh la[;k] ;fn la[;k esa 1,oa 2 vko';d :i ls gks] gksxh-(1) 65 – 2.55 + 45 (2) 10.54

(3) 20.54 (4) 5.45

75. ijoy; dk 'kh"kZ (1,2) gS rFkk bldk v{k y-v{k dslekUrj gSA ;fn ijoy; (0,6) ls xqtjrk gS] rks bldsukfHkyEc dh yEckbZ gksxh-

(1) 1

2(2) 2 (3)

1

4(4) 4

76. vkadM+ksa 1001, 1003, 1006, 1007, 1009, 1010 dkizlj.k gksxk -(1) 10 (2) 15 (3) 20 (4) 50

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77. ( ) ( ) ( )p ~ q ~ r ~ p q ~ r ~ p ~ q rÙ Ù Ú Ù Ù Ú Ù Ù is

equivalent to-

(1) ( ) ( ) ( )( )~ p q q r r pÙ Ú Ù Ú Ù

(2) p q rÚ Ú

(3) ( ) ( ) ( ) ( )( )p q q r r p p q rÙ Ú Ù Ú Ù Ù Ú Ú

(4) ( ) ( ) ( )( ) ( )( )~ p q q r r p p q rÙ Ú Ù Ú Ù Ù Ú Ú

78. The shaded region in venn-diagram can berepresented by which of the following ?

A B

C

U

(1) ( ) ( ) ( ) ( )C C C C C CA C A B A C B CÈ Ç È È È È È

(2) ( ) ( ) ( ) ( )C C C C C CA C A B A C B CÈ Ç È Ç È Ç È

(3) ( ) ( ) ( ) ( ) ( )C C C C C CA C A B A C B C A B CÈ Ç È Ç È Ç È È Ç Ç

(4) ( ) ( ) ( ) ( ) ( )C C C C C CA C A B A C B C A B CÈ Ç È Ç È Ç È Ç Ç Ç

79. From the point C(0,l) two tangents are drawnto ellipse x2 + 2y2 = 4 cutting major axis at Aand B. If area of DABC is minimum, then valueof l is-

(1) 2 (2) 2 (3) 2 2 (4) 8

77. ( ) ( ) ( )p ~ q ~ r ~ p q ~ r ~ p ~ q rÙ Ù Ú Ù Ù Ú Ù Ù

lerqY; gksxk -

(1) ( ) ( ) ( )( )~ p q q r r pÙ Ú Ù Ú Ù

(2) p q rÚ Ú

(3) ( ) ( ) ( ) ( )( )p q q r r p p q rÙ Ú Ù Ú Ù Ù Ú Ú

(4) ( ) ( ) ( )( ) ( )( )~ p q q r r p p q rÙ Ú Ù Ú Ù Ù Ú Ú

78. osu&vkjs[k esa Nka;kfdr {ks= dks fuEu esa ls blds }kjkiznf'kZr dj ldrs gSa&

A B

C

U

(1) ( ) ( ) ( ) ( )C C C C C CA C A B A C B CÈ Ç È È È È È

(2) ( ) ( ) ( ) ( )C C C C C CA C A B A C B CÈ Ç È Ç È Ç È

(3) ( ) ( ) ( ) ( ) ( )C C C C C CA C A B A C B C A B CÈ Ç È Ç È Ç È È Ç Ç

(4) ( ) ( ) ( ) ( ) ( )C C C C C CA C A B A C B C A B CÈ Ç È Ç È Ç È Ç Ç Ç

79. fcUnq C(0,l) ls nh?kZoÙk x2 + 2y2 = 4 ij [khaph xbZ nksLi'kZ js[kk;sa nh?kZv{k dks A rFkk B ij dkVrh gSA ;fnf=Hkqt ABC dk {ks=Qy U;wure gks] rks l dk ekugksxk&

(1) 2 (2) 2 (3) 2 2 (4) 8

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00CE314002


80. If 1 n 1

n 1

0

x 1I dx

x 1

+

+-

=+ò , then the value of

I10 + I11 + 2log2 is-

(1) 1 (2) 1

9(3)

1

10(4)

1

11

81. Least integer in the range of

( ) ( ) ( ) 2ƒ x x 4 1 x log x= + - - is-

(1) –2 (2) –1 (3) 0 (4) 1

82. Let A º (l + 2, 1 – 2l, l + 2) and

B º (2k + 1, k, k +1) and l, k Î R. Thenminimum distance between A and B is -

(1) 0 (2) 1

35(3)

3

35(4)

3

35

83. A circle has centre C on axes of parabola & ittouches the parabola at point P. CP makes anangle of 120º with axis of parabola. If radius ofcircle is 2, then latus rectum of parabola is-

(1) 2 (2) 4 (3) 8 (4) 16

84. Let ƒ be derivable funciton ƒ : R ® R satisfying

the equation ( ) ( ) ( )x2

20

ƒ tƒ x 1 x 1 dt x R

1 t

é ù= + + " Îê ú+ë û

ò ,

then ƒ(1) is-

(1) 1

e(2) e (3) 2e (4) 4e

80. ;fn 1 n 1

n 1

0

x 1I dx

x 1

+

+-

=+ò gks] rks I10 + I11 + 2log2

cjkcj gksxk -

(1) 1 (2) 1

9(3)

1

10(4)

1

11

81. ( ) ( ) ( ) 2ƒ x x 4 1 x log x= + - - ds ifjlj es a

U;wure iw.kk±d gksxk -

(1) –2 (2) –1 (3) 0 (4) 1

82. ekuk A º (l + 2, 1 – 2l, l + 2),

B º (2k + 1, k, k +1) rFkk l, k Î R gks] rks A ,oa Bds e/; U;wure nwjh gksxh -

(1) 0 (2) 1

35(3)

3

35(4)

3

35

83. ,d o`Ùk ftldk dsUnz C ijoy; ds v{k ij fLFkr gSrFkk ;g ijoy; dks fcUnq P ij Li'kZ djrk gSA CPijoy; ds v{k ds lkFk dks.k 120º cukrk gSA ;fn o`Ùkdh f=T;k 2 gks] rks ijoy; dk ukfHkyEc gksxk-

(1) 2 (2) 4 (3) 8 (4) 1684. ekuk ƒ : R ® R esa vodyuh; Qyu gS] tks lehdj.k

( ) ( ) ( )x2

20

ƒ tƒ x 1 x 1 dt x R

1 t

é ù= + + " Îê ú+ë û

ò dks larq"V djrk gS]

rks ƒ(1) gksxk&

(1) 1

e(2) e (3) 2e (4) 4e

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00CE314002


85. The number of solutions of the equation3 3 3sin 1 sin sin 1 0q- + q + q+ = in [0,4p]is-(1) 2 (2) 4 (3) 5 (4) 6

86. If the function ( )-

-

ì <= í

+ l ³î

1

1

tan x;x 1ƒ x

sec x ;x 1

has local minima at x = 1, then range of l is-

(1) 0,4

pæ ùç úè û

(2) 0,4

pé ö÷êë ø

(3) ,4

pæ ù-¥ç úè û (4) ,

4

pæ ö-¥ç ÷è ø

87. The value of ( ) ( )ll

x 2de n x

d nx at x = e, is -

(1) ( )ee e 2+ (2) e 1e +

(3) 2ee+1 (4) ee(e + 1)

88.2xe

x 0

d e elim

dx x®

æ ö-ç ÷ç ÷è ø

is -

(1) 0 (2) –e (3) e (4) e2

89. Let a, b, crr r be three unit vectors such that

3a.b b.c a.c

2+ - =r rr r r r

. Then the value of

a.b b.c c.a+ +r rr r r r

is-

(1) 1

2(2) 1 (3)

3

2(4)

1

2-

90. If the equation x8 – kx2 + 3 = 0 has a realsolution, then least integral value of k is-(1) 3 (2) 2 (3) 1 (4) 4

85. vUrjky [0,4p] esa lehdj.k3 3 3sin 1 sin sin 1 0q- + q + q+ = ds gyksa dhla[;k gksxh -(1) 2 (2) 4 (3) 5 (4) 6

86. ;fn Qyu ( )-

-

ì <= í

+ l ³î

1

1

tan x;x 1ƒ x

sec x ;x 1dk x = 1 ij

LFkkuh; fufEu"B eku gks] rks l dk ifjlj gksxk-

(1) 0,4

pæ ùç úè û

(2) 0,4

pé ö÷êë ø

(3) ,4

pæ ù-¥ç úè û (4) ,

4

pæ ö-¥ç ÷è ø

87. x = e ij ( ) ( )ll

x 2de n x

d nx gksxk -

(1) ( )ee e 2+ (2) e 1e +

(3) 2ee+1 (4) ee(e + 1)

88.2xe

x 0

d e elim

dx x®

æ ö-ç ÷ç ÷è ø

gksxk -

(1) 0 (2) –e (3) e (4) e2

89. ekuk a, b, crr r rhu bdkb Z lfn'k bl izdkj gS fd

3a.b b.c a.c

2+ - =r rr r r r gSA rc a.b b.c c.a+ +

r rr r r r dk eku

gksxk&

(1) 1

2(2) 1 (3)

3

2(4)

1

2-

90. ;fn lehdj.k x8 – kx2 + 3 = 0 dk okLrfod gy gks]rks k dk U;wure iw.kk±d eku gksxk&(1) 3 (2) 2 (3) 1 (4) 4

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CAREER INSTITUTE Pat S KOTA (RAJASTHAN) … · -2/32 SPACE FOR ROUGH WORK / jQ dk;Z ds fy;s txg ALL INDIA OPEN TEST/JEE (Min)/22-03-2015 00E314002 3. 20 / g mpty conveyo b ovi horizontall