allen paper code 01ct3140 69 · paper code 01ct3140 69 p nstructio ully. y ll 5 in specifical . ......

CLASSROOM CONTACT PROGRAMME

FORM NUMBER

PAPER CODE 0 1 C T 3 1 4 0 6 9

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.Ïi;k bu funsZ'kks a dks /;ku ls i<+ sA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA

Do not open this Test Booklet until you are asked to do so.

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 marksare 360.

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.

7. Use Blue/Black Ball Point Pen only for writting particulars/markingresponses on Side–1 and Side–2 of the Answer Sheet.

Use of pencil is strictly prohibited.

8. No candidate is allowed to carry any textual material, printed or written,bits of papers, pager, mobile phone any electronic device etc, exceptthe Identity Card inside the examination hall/room.

9. Rough work is to be done on the space provided for this purpose inthe Test Booklet only.

10. On completion of the test, the candidate must hand over the AnswerSheet to the invigilator on duty in the Room/Hall. However, thecandidate are allowed to take away this Test Booklet with them.

11. Do not fold or make any stray marks on the Answer Sheet.

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 @ mÙkj

i= 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 dkVk

tk;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls

½.kkRed vadu ugha gksxkA

7. mÙkj i= ds i`"B&1 ,oa i`"B&2 ij okafNr fooj.k ,oa mÙkj vafdr djus gsrq

dsoy uhys@dkys ckWy ikbaV isu dk gh iz;ksx djsaA

isfUly 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] istj]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{kd

dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iqfLrdk dks ys tk

ldrs 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 Mek/;e % fgUnh

Your Target is to secure Good Rank in JEE 2015

Corporate OfficeALLEN CAREER INSTITUTE

“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2436001 [email protected]

www.allen.ac.in

(ACADEMIC SESSION 2014-2015)

DATE : 01 - 04 - 2015ALLEN JEE (Main) TEST

TARGET : JEE (Main) 2015 LEADER & ENTHUSIAST COURSE : SCORE

Leader & Enthusiast Course/Score/01-04-2015

H-1/32Kota/01CT314069

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

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

1. Diffusion current in a p-n junction is greater

than the drift current in magnitude :-

(1) If the junction is forward-biased

(2) If the junction is reverse-biased

(3) If the junction is unbiased

(4) In no case.

2. Curie temperature is the temperature above

which :

(1) a ferromagnetic material becomes para

magnetic

(2) a paramagnetic material becomes diamagnetic

(3) a ferromagnetic material becomes diamagnetic

(4) a paramagnetic material becomes ferromagnetic

3. The average value of electric energy density

in an electromagnetic wave is :

(1) 2

0

1E

2e (2)

2

0

E

2e

(3) e0E2 (4)

20

1E

4e

1. fdlh p-n laf/k esa folj.k /kkjk dk eku viogu èkkjk ls

(ifjek.k esa) vf/kd gksrk gS :-

(1) ;fn laf/k vxz ck;flr gks

(2) ;fn laf/k i'p ck;flr gks

(3) ;fn laf/k ck;flr uk gks

(4) fdlh Hkh fLFkfr esa ugha

2. D;wjh rki og rki gS] ftlls vf/kd rki ij%&

(1) ykSg pqEcdh; inkFkZ] vuqpqEcdh; cu tkrk gS

(2) vuqpqEcdh; inkFkZ] izfrpqEcdh; cu tkrk gS

(3) ykSg pqEcdh; inkFkZ] izfrpqEcdh; cu tkrk gS

(4) vuqpqEcdh; inkFkZ] ykSg pqEcdh; cu tkrk gS

3. fo|qr pqEcdh; rjax esa fo|qr ÅtkZ ?kuRo dk vkSlr eku

gksrk gS %&

(1) 2

0

1E

2e (2)

2

0

E

2e

(3) e0E2 (4)

20

1E

4e

Kota/01CT314069H-2/32

Target : JEE (Main) 2015/01-04-2015


4. In amplitude modulation, the modulation index

m, is kept less than or equal to 1 because

(1) m > 1, will result in interference between

carrier frequency and message frequency,

resulting into distortion.

(2) m > 1 will result in overlapping of both side

bands resulting into loss of information.

(3) m > 1 will result in change in phase between

carrier signal and message signal.

(4) m > 1 indicates amplitude of carrier signal

is greater than amplitude of message signal

resulting into distortion.

5. A surveyor's 30-m steel tape is correct at some

temperutre. On a hot day the tape has expanded

to 30.01 m. On that day, the tape indicates a

distance of 15.52 m between two points. The

true distance between these points is :-

(1) 15.515 m

(2) 15.520 m

(3) 15.525 m

(4) 15

4. vk;ke ekWMwyu esa ekWMwyu lwpdkad m dk eku 1 ls de

;k blds cjkcj j[kk tkrk gS] D;kasfd %&

(1) m > 1 gksus ij okgd vko`fr rFkk lans'k vkofr ds

e/; O;frdj.k gksus ls fo:i.k mRiUu gksxkA

(2) m > 1 gksus ij nksuksa ik'oZ cSaMks dk vfrO;kiu gksxk

ftlds QyLo:i lans'k esa âkl gks tkrk gSA

(3) m > 1 gksus ij okgd ladsr rFkk lans'k ladsr ds

eè; dyk esa vUrj vk tkrk gSA

(4) m > 1 dk vFkZ gS okgd ladsr dk vk;ke] lans'k

ladsr ds vk;ke ls vf/kd gS ftlds QyLo:i

fo:i.k gksrk gSA

5. ,d LVhy dk Qhrk 30 m yEck gS tks fdlh rki ij lgh

eki n'kkZrk gSA fdlh xeZ fnu ;g 30.01 m rd foLrkfjr

gks tkrk gSA bl fnu bl Qhrs }kjk nks fcUnqvksa ds chp dh

nwjh 15.52 m ekih tkrh gSA bu fcUnqvksa ds e/; lgh

nwjh gksxh %&

(1) 15.515 m

(2) 15.520 m

(3) 15.525 m

(4) 15


H-3/32Kota/01CT314069


6. A given mass m of a hypothetical solid issupplied with heat continuously at a constantrate and the graph shown in the adjacent figureis plotted. If L

f and L

v are latent heats of fusion

and latent heats of vaporization and Sl and S

s

are specific heats of l iquid and solidrespectively. It can be concluded that :-

10 90 100 120

temperature

200

3010

t(sec)

melting

Boiling

(1) ,f v SL L S S> =l

(2) ,f v SL L S S< >l

(3) ,f v SL L S S> <l

(4) ,f v SL L S S= >l

7. A large cylindrical tank of cross-sectional area1m2 is filled with water. It has a small hole at aheight of 1m from the bottom. A movablepiston of mass 5 kg is fitted on the top of thetank such that it can slide in the tank freelywithout friction. A load of 45 kg is applied onthe top of water by piston, as shown in figure.The value of v when piston is 7m above thebottom is (g = 10 m/s2) :-

45kg

v

(1) 120 m/s (2) 10 m/s

(3) 1 m/s (4) 11 m/s

6. fdlh dkYifud Bksl ds m nzO;eku dks fu;r nj ijyxkrkj Å"ek nh tkrh gS rFkk blds laxr vkjs[k [khapktkrk gS ftls fp= esa n'kkZ;k x;k gSA ;fn L

f o L

v Øe'k%

laxyu rFkk ok"iu dh xqIr Å"ek,sa gks o Sl o S

s Øe'k%

nzo o Bksl dh fof'k"V Å"ek,sa gks rks lgh fodYi pqfu;sA

10 90 100 120

temperature

200

3010

t(sec)

melting

Boiling

(1) ,f v SL L S S> =l

(2) ,f v SL L S S< >l

(3) ,f v SL L S S> <l

(4) ,f v SL L S S= >l

7. vuqizLFk dkV {ks=Qy 1m2 okys ,d cM+s csyukdkj Vsaddks ikuh ls Hkjk x;k gSA blds iSans ls 1m dh ÅapkbZ ij,d NksVk fNnz gSA ,d 5 kg nzO;eku ds pyk;eku fiLVudks Vsad ds 'kh"kZ ij bl izdkj ls dlk x;k gS fd ;g Vsadesa fcuk ?k"kZ.k eqDr :i ls xfr dj ldrk gSA fiLVu }kjkikuh ds 'kh"kZ ij ,d 45 kg dk Hkkj fp=kuqlkj yxk;ktkrk gSA tc fiLVu iSans ls 7m Åij gS rks v dk eku gksxk(g = 10 m/s2) :-

45kg

v

(1) 120 m/s (2) 10 m/s

(3) 1 m/s (4) 11 m/s

Kota/01CT314069H-4/32

Target : JEE (Main) 2015/01-04-2015


8. If y1 = 5 (mm) sinpt is equation of oscillation

of source S1 and y

2 = 5 (mm) sin(pt + p/6) be

that of S2 and it takes 1 sec and ½ sec for the

transverse waves to reach point A from sources

S1 and S

2 respectively then the resulting

amplitude at point A, is :-

S 1 S 2

A

(1) 5 2 3+ mm (2) 5 3 /2 mm

(3) 5 mm (4) 5 2 mm

9. A bob of mass 10 M is suspended through an

inextensible string of length l. When the bob is

at rest at the equilibrium position, two particles

of mass m each moving with velocity u making

an angle 60° with the string strike and get

simultaneously attached to the bob. What is the

value of impulsive tension (l) in the string

during the impact ?

(1) 0 (2) 2 mu (3) mu (4) 12 mu

8. ;fn y1 = 5 (mm) sinpt lzksr S

1 ds nksyu dh lehdj.k

rFkk y2 = 5 (mm) sin(pt + p/6) lzksr S

2 ds nksyu dh

lehdj.k gS rFkk vuqizLFk rjaxksa dks lzksrksa S1 rFkk S

2 ls

fcUnq A rd igqapus esa Øe'k% 1 lsd.M rFkk ½ lsd.M

dk le; yxrk gS rks fcUnq A ij ifj.kkeh vk;ke gksxk%&

S 1 S 2

A

(1) 5 2 3+ mm (2) 5 3 /2 mm

(3) 5 mm (4) 5 2 mm

9. ,d 10 M nzO;eku dk xk syd] l yEckb Z dh

vforkU; jLlh }kjk yVdk gqvk gSA tc xksyd

lkE;koLFkk esa fojkekoLFkk esa gS rks m nzO;eku okys

nks d.k jLlh ls 60° dk dks.k cukrs gq, u osx ls

xksyd ls Vdjkrs gSa rFkk ,d lkFk blls fpid tkrs

gSaA VDdj ds nkSjku jLlh esa mRiUu vkosxh; ruko l

dk eku D;k gksxk\

(1) 0 (2) 2 mu (3) mu (4) 12 mu


H-5/32Kota/01CT314069


10. A ball of mass m strikes the inclined face ofthe wedge normally with speed v

0. The wedge

is at rest on a rough horizontal surface beforecollision. The conservation of momentum isapplicable for the event of collision for

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\a

9 0 °

m

M

Y 'Y

X '

X

(i) m as system, along Y'

(ii) M as system, along Y'

(iii) (M + m) as system, along X

(iv) (M + m) as system, along Y

Which of the following is correct?

(1) (i) only (2) (i) and (ii) only

(3) (iii) only (4) (iii) and (iv) only11. A small disk can slide in a circular path on a

frictionless inclined plane inclined at an angleof 30° with the help of a thread as shown. Massof the disk is m and acceleration due to gravityis g. If the disk is released, when the thread ishorizontal, expression for the tension in thethread at the lowest point is :-

30°

(1) 1

2mg (2)

3

2mg (3) 2 mg (4) 3 mg

10. ,d m nzO;eku dh xsan ost ds > qds gq, Qyd ls v0

pky ls yEcor~ :i ls Vdjkrh gSA ;g ost VDdj lsigys [kqjnjh {kSfrt lrg ij fojkekoLFkk esa j[kk gSAVDdj ds fy, laosx laj{k.k fu;e ykxw gksrk gS

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\a

9 0 °

m

M

Y 'Y

X '

X

(i) ,d fudk; ds :i esa m ds fy,] Y' ds vuqfn'k

(ii) ,d fudk; ds :i esa M ds fy,] Y' ds vuqfn'k

(iii) ,d fudk; ds :i esa (M+m) ds fy,] X ds vuqfn'k

(iv) ,d fudk; ds :i esa (M+m) ds fy,] Y ds vuqfn'k

fuEu esa ls dkSulk fodYi lR; gS\

(1) dsoy (i) (2) dsoy (i) o (ii)

(3) dsoy (iii) (4) dsoy (iii) o (iv)

11. ,d NksVh pdrh] fdlh 30° dks.k okys ?k"kZ.kjfgr ur

ry ij ,d Mksjh dh lgk;rk ls fp=kuqlkj o`Ùkkdkj iFk

esa xfr djrh gSaA pdrh dk nzO;eku m o xq:Roh; Roj.k

g gSA tc Mksjh {kSfrt gS ml le; pdrh dks fojkekoLFkk

ls NksM+us ij fuEure fcUnq ij Mksjh esa ruko gksxk %&

30°

(1) 1

2mg (2)

3

2mg (3) 2 mg (4) 3 mg

Kota/01CT314069H-6/32

Target : JEE (Main) 2015/01-04-2015


12. A small 100 g sleeve B can slide on a smooth,circular and rigid wire frame A of radius 5 mplaced in vertical place. The wire frame isrotating about its vertical diameter at 2 rad/s.When the sleeve is brought at a particularangular position other than the bottom and thetop of the ring, the sleeve will not slide on thewire frame. What is force of interactionbetween the sleeve and the wire frame at thisposition?

A

B

(1) 2 N (2) 3 N (3) 4 N (4) 5 N13. A metal wire of length L

1 and area of cross

section A is attached to a rigid support. Anothermetal wire of length L

2 and of the same cross

sectional area is attached to the free end of thefirst wire. A body of mass M is then suspendedfrom the free end of the second wire. If Y

1 and

Y2 are the Young’s moduli of the wires

respectively, the effective force constant of thesystem of two wires is :

(1) ( )

( )1 2

1 2 2 1

Y Y A

2 Y L Y L

é ùë ûé ù+ë û

(2) ( )

( )1 2

1/ 2

1 2

Y Y A

L L

é ùë û+

(3) ( )

( )1 2

1 2 2 1

Y Y A

Y L Y L

é ùë û+ (4)

( )( )

1/ 2

1 2

1/ 2

1 2

Y Y A

L L+

12. ,d 100 g dh NksVh oy; B] Å/okZ/kj ry esa fLFkr ,d

fpdus oÙkkdkj n`<+ rkj Ýse A, ftldh f=T;k 5 m gS]

ij fQly ldrh gSA ;g rkj Ýse Å/oZ O;kl ds lkis{k

2 rad/s dh dks.kh; pky ls ?kw.kZu dj jgk gSA oy; Bdks rkj Ýse ds U;wure rFkk mPpre fcUnq ds vfrfjDr

fdlh ,d dks.kh; fLFkfr ij ykdj NksM+us ij ;g ugha

fQlyrh gSA bl fLFkfr esa oy; B rFkk rkj Ýse ds chp

vU;ksU; cy dk ifjek.k fdruk gksxk\

A

B

(1) 2 N (2) 3 N (3) 4 N (4) 5 N

13. yEckbZ L1 rFkk vuqizLFk dkV {ks=Qy A okys ,d èkkfRod

rkj dks n<+ vk/kkj ls tksM+ fn;k tkrk gSA bl rkj ds eqäfljs ls L

2 yEckbZ rFkk leku vuqizLFk dkV {ks=Qy okys

,d vU; /kkfRod rkj dks tksM+ nsrs gSaA vc bl nwljs rkjds eqä fljs ls M æO;eku dk ,d fi.M yVdk fn;ktkrk gSA ;fn bu rkjksa ds ; ax izR;kLFkrk xq.kkad Øe'k% Y

1

o Y2 gks rks nksuksa rkjksa ls cus fudk; ds izHkkoh cy fu;rkad

dk eku gksxk %&

(1) ( )

( )1 2

1 2 2 1

Y Y A

2 Y L Y L

é ùë ûé ù+ë û

(2) ( )

( )1 2

1/ 2

1 2

Y Y A

L L

é ùë û+

(3) ( )

( )1 2

1 2 2 1

Y Y A

Y L Y L

é ùë û+ (4)

( )( )

1/ 2

1 2

1/ 2

1 2

Y Y A

L L+


H-7/32Kota/01CT314069


14. The fundamental frequency of a sonometerwire increases by 6 Hz if its tension is increasedby 44% keeping the length constant. Thechange in the fundamental frequency of thesonometer wire in Hz when the length of thewire is increased by 20%, keeping the originaltension in the wire will be :-

(1) 2 (2) 4 (3) 5 (4) 10

15. A ring of mass m can freely slide along the

massless curved rod as shown. At the lower

most point, the curved path becomes vertical.

If whole system is released from rest, velocity

of ring (v) at lowermost point just before

touching the block M (all surfaces are smooth)

is :-

M

HRing

curved

m

(1) v = 2gH (2) v < 2gH

(3) v > 2gH (4) Data insufficient

16. In an H2 gas process, PV2 = constant. The ratio

of work done by gas to change in its internal

energy is

(1) 2/3 (2) 0.4 (3) – 0.4 (4) 1.5

14. ;fn lksuksehVj rkj dh yEckbZ dks fu;r j[krs gq, blds

ruko esa 44% of¼ dh tkrh gS rks bl rkj dh ewy vkofÙk

esa 6Hz dh o`f¼ gksrh gSA rkj ds ewy ruko dks fu;r

j[krs gq, tc rkj dh yEckbZ esa 20% o`f¼ dh tkrh gS rks

lksuksehVj rkj dh ewy vko`fr esa ifjorZu (Hz esa)

gksxk

(1) 2 (2) 4 (3) 5 (4) 10

15. nzO;eku m okyh oy; fp=kuqlkj nzO;ekughu oØh;

NM+ ds vuqfn'k eqDr :i ls fQly ldrh gSA fuEure

fcUnq ij oØh; iFk] ÅèokZèkj gks tkrk gSA ;fn laiw.kZ

fudk; dks fojkekoLFkk ls NksM+k tk;s rks CykWd M dks

Li'kZ djus ls Bhd igys fuEure fcUnq ij oy; dk osx

(v) gksxk (lHkh lrgsa fpduh gS)

M

HRing

curved

m

(1) v = 2gH (2) v < 2gH

(3) v > 2gH (4) vkadM+s vi;kZIr gSa

16. H2 xSl ds fy, fd, x, izØe PV2 = fu;rkad esa xSl

}kjk fd, x,s dk;Z rFkk bldh vkUrfjd ÅtkZ esa ifjorZu

dk vuqikr gksxk %&(1) 2/3 (2) 0.4 (3) – 0.4 (4) 1.5

Kota/01CT314069H-8/32

Target : JEE (Main) 2015/01-04-2015


17. On a TP diagram, two moles of ideal gas

perform process AB and CD. If the work done

by the gas in the process AB is two times the

work done in the process CD then what is the

value of T1/T2?

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

18. The electric field at A due to dipole p is

perpendicular to p. The angle q is :-

Ap

q

(1) 0° (2) 90°

(3) tan–12 (4) tan–1 2

19. Uniform electric field of magnitude 100 V/m

in space is directed along the line y = 3 + x.

Find the potential difference between point A

(3, 1) & B (1, 3)

(1) 100 V (2) 200 2 V

(3) 200 V (4) 0

17. fdlh vkn'kZ xSl ds nks eksyksa ij fd;s x;s izØeksa AB o

CD dks TP vkjs[k ij n'kkZ;k x;k gSA ;fn izØe AB esa

xSl }kjk fd;k x;k dk;Z izØe CD dh rqyuk esa nks xquk

gks rks T1/T2 dk eku gksxk %&

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

18. fp= esa f}/kzqo p ds dkj.k fcUnq A ij fo|qr {ks= p ds

yEcor~ gSA dks.k q gksxk :-

Ap

q

(1) 0° (2) 90°

(3) tan–12 (4) tan–1 2

19. lef"V esa 100 V/m ifjek.k dk le:i fo|qr {ks= js[kk

y = 3 + x ds vuqfn'k fo|eku gSA fcUnq A (3, 1) o B

(1, 3) ds e/; foHkokUrj gksxk %&

(1) 100 V (2) 200 2 V

(3) 200 V (4) 0


H-9/32Kota/01CT314069


20. A galvanometer G deflects full scale when a

potential difference of 0.50 V is applied. The

internal resistance of the galvanometer rg is

25 ohms. An ammeter is constructed by

incorporating the galvanometer and an

additional resistance RS. The ammeter deflects

full scale when a measurement of 2.0 A is made.

The resistance RS is closest to :

(1) 0.25 W (2) 2.5 W

(3) 0.45 W (4) 0.1 W

21. In the circuit shown in the figure K1 is open.

The charge on capacitor C in steady state is q1.

Now key is closed and at steady state charge

on C is q2. The ratio of charges q1/q2 is

C

K1R =21 W

R =32 WE

(1) 5/3 (2) 3/5

(3) 1 (4) 2/3

20. tc 0.50 V dk foHkokUrj yxk;k tkrk gS rks ,d

xsYosuksehVj G iw.kZ fo{ksi n'kkZrk gSA xsYosuksehVj dk

vkarfjd izfrjks/k rg dk eku 25 ohms gSA xsYosuksehVj

rFkk ,d vfrfjä izfrjks/k RS dh lgk;rk ls ,d vehVj

dk fuekZ.k fd;k tkrk gSA ;g vehVj 2.0 A ds izs{k.k

ysus ij iw.kZ fo{ksi n'kkZrk gSA izfrjksèk RS dk eku yxHkx

gS %&

(1) 0.25 W (2) 2.5 W

(3) 0.45 W (4) 0.1 W

21. fp= esa iznf'kZr ifjiFk esa dqath K1 [kqyh gSA LFkk;h voLFkk

esa la/kkfj= C ij vkos'k q1 gSA vc dqath dks can dj nsrs

gSa rFkk C ij LFkk;h voLFkk vkos'k dk eku q2 izkIr gksrk

gSA vkos'kksa dk vuqikr q1/q2 gksxk %&

C

K1R =21 W

R =32 WE

(1) 5/3 (2) 3/5

(3) 1 (4) 2/3

Kota/01CT314069H-10/32

Target : JEE (Main) 2015/01-04-2015


22. A circular loop of radius R carries a current I.

Another circular loop of radius r(<<R) carries

a current i and is placed at the centre of the

larger loop. The planes of the two circles are at

right angle to each other. Find the torque acting

on the smaller loop.

(1) 2

02 iIr

R

m p(2)

20iIr

2 R

mp

(3) 2

0 iIr

2R

m p(4) 0iIr

2 R

mp

23. A student peddles a stationary bicyle.

The pedals of the bicycle are attached to a

100 turn coil of area 0.10 m2. The coil rotates

at half a revolution per second and it is placed

in a uniform megnetic field of 0.01 T

perpendicular to the axis of rotation of the coil.

What is the maximum voltage generated in the

coil ?

(1) 1.314 V (2) 1.214 V

(3) 2.314 V (4) 0.314 V

22. f=T;k R okys ,d oÙkkdkj ywi esa I /kkjk izokfgr gSA

f=T;k r(<<R) okyk ,d vU; o`Ùkkdkj ywi cM+s ywi ds

dsUæ ij fLFkr gS rFkk blesa i /kkjk izokfgr gks jgh gSA nksuksa

o`Ùkksa ds ry ,d&nwljs ds yEcor~ gSA NksVs ywi ij dk;Zjr

cyk?kw.kZ gksxk %&

(1) 2

02 iIr

R

m p(2)

20iIr

2 R

mp

(3) 2

0 iIr

2R

m p(4) 0iIr

2 R

mp

23. ,d fo|kFkhZ ,d fLFkj lkbfdy ds iSMy dks ?kqekrk gSA

iSMy dk lacaèk 100 Qsjksa rFkk 0.10 m2 {ks=Qy okyh

,d dq.Myh ls gSA dq.Myh izfr lsd.M vkèkk ifjØe.k

(pDdj) dj ikrh gS rFkk ;g ,d 0.01 T rhozrk okys

,dleku paqcdh; {ks= eas] tks dq.Myh ds ?kw.kZu v{k ds

yacor gS] j[kh gSA dq.Myh esa mRiUu gksus okyh vfèkdre

oksYVrk D;k gksxh\

(1) 1.314 V (2) 1.214 V

(3) 2.314 V (4) 0.314 V


H-11/32Kota/01CT314069


24. The mean and rms value of an alternatingvoltage for half cycle as shown in figure arerespectively:-

V

t3T2

T2

T0

V0

–V0

(1) V0, V0 (2) 00

V, V

2

(3) 0 03V V,

2 2(4) 0 0V V

,4 2

25. When a metallic surface is illuminated withmonochromatic light of wavelength l, thestopping potential is 5 V0. When the samesurface is illuminated with light ofwavelength 3l, the stopping potential is V0.Then the work function of the metallicsurface is :

(1) hc

6l(2)

hc

5l

(3) hc

4l(4)

2hc

4l

24. fp= esa iznf'kZr izR;korhZ oksYVrk dk vk/ks pØ ds fy;s

ek/; rFkk oxZ ek/; ewy eku Øe'k% gksxk%&

V

t3T2

T2

T0

V0

–V0

(1) V0, V0 (2) 00

V, V

2

(3) 0 03V V,

2 2(4) 0 0V V

,4 2

25. tc ,d /kkfRod i`"B dks l rjaxnS/; Z okys ,do.khZ;

izdk'k ls izdkf'kr fd;k tkrk gS rks fujks/kh foHko dk

eku 5 V0 gksrk gSA blh i`"B dks 3l rjaxnS/; Z okys

izdk'k ls izdkf'kr djus ij fujks/kh foHko V0 izkIr

gksrk gSA /kkfRod i`"B dk dk;ZQyu gksxk %&

(1) hc

6l(2)

hc

5l

(3) hc

4l(4)

2hc

4l

Kota/01CT314069H-12/32

Target : JEE (Main) 2015/01-04-2015


26. A sample of hydrogen atom is in excited state

(all the atoms). The photons emitted from this

sample are made to pass through a filter through

which light having wavelength greater than

800 nm can only pass. Only one type of photons

are found to pass the filter. The sample's excited

state initially is : [Take hc = 1240 eV -nm,

ground state energy of hydrogen

atom = –13.6 eV.]

(1) 5th excited state (2) 4th excited state

(3) 3rd excited state (4) 2nd excited state27. Three identical rods, each of length l, are

joined to form a rigid equilateral triangle. Itsradius of gyration about an axis passingthrough a corner and perpendicular to theplane of the triangle is

(1) l (2) 2

l

(3) 2

l(4)

3

l

28. A thin rod AB is sliding

A D

BC

w

between two fixed rightangled surfaces. Atsome instant its angularvelocity is w. If I

x

represent moment ofinertia of the rod about an axis perpendicularto the plane and passing through the point X(A, B, C or D), the kinetic energy of the rod is

(1) 2

A

1

2I w (2)

2B

1

2I w (3)

2C

1

2I w (4)

2D

1

2I w

26. gkbMªkstu ijek.kq dk ,d uewuk (lHkh ijek.kq) mÙksftrvoLFkk esa gSA bl uewus ls mRlftZr QkWVksukas dks ,d , slsfQYVj ls xqtkjk tkrk gS ftlls 800 nm ls vfèkdrjaxnS/; Z okyk izdk'k gh xqtj ldrk gSA ;gk¡ dsoy ,dizdkj ds QkWVksu gh bl fQYVj ls xqtj ikrs gSaA bl uewusdh izkjfEHkd mRrsftr voLFkk gS%&[hc = 1240 eV -nm] gkbMªkstu ijek.kq dh ewy voLFkkdh ÅtkZ = –13.6 eV.]

(1) 5th mRrsftr voLFkk (2) 4th mRrsftr voLFkk

(3) 3rd mRrsftr voLFkk (4) 2nd mRrsftr voLFkk

27. izR;sd l yEckbZ okyh rhu ,d tSlh NM+ksa dks tksM+dj,d n`<+ leckgq f=Hkqt cuk;k tkrk gSA bl f=Hkqt dsry ds yEcor~ rFkk blds fdlh ,d fljs ls xqtjusokyh v{k ds lkis{k bldh ifjHkze.k f=T;k gksxh %&

(1) l (2) 2

l

(3) 2

l(4)

3

l

28. ,d iryh NM+ AB nks fLFkj

A D

BC

w

yEcor ~ lrgk s a d s e/;fQly jgh gSA fdlh {k.kbldk dks.kh; osx w gSA ;fnI

x blds ry ds yEcor~ rFkk

fcUnq X (A, B, C vFkok D) ls xqtjus okyh v{k ds

lkis{k NM+ ds tM+Ro vk?kw.kZ dks n'kkZrk gks rks NM+ dhxfrt ÅtkZ gksxh %&

(1) 2

A

1

2I w (2)

2B

1

2I w (3)

2C

1

2I w (4)

2D

1

2I w


H-13/32Kota/01CT314069


29. A uniform bar of length '6l' and mass '8m' lies

on a smooth horizontal table. Two point masses

m and 2m moving in the same horizontal plane

with speed 2v and v respectively, strike the bar

(as shown in the fig.) and stick to the bar after

collision. Total energy after collision (about the

center of mass, c) will be :-

2ll

(1) 22mv

5(2)

2mv

5(3)

23mv

5 (4) mv2

30. A force F is applied on the top of a cube asshown in the figure. The coefficient of frictionbetween the cube and ground is µ. If F isgradually increased the cube will topple beforesliding then range of µ is:-

F

(1) µ > 1 (2) µ < 1

2(3) µ >

1

2(4) µ < 1

29. ,d ,dleku NM+ ftldh yEckbZ '6l' vkSj nzO;eku

'8m' gS] ,d fpduh {kSfrt Vscy ij j[kh gqbZ gSA m rFkk

2m nzO;eku ds nks d.k leku {kSfrt ry esa Øe'k% 2v

o v pkyksa ls fp=kuqlkj NM + ls yEcor~ Vdjkrs gSa vkSj

NM + ls fpid tkrs gSaA VDdj ds i'pkr~ nzO;eku dsUnz c

ds lkis{k dqy ÅtkZ gksxh %&

2ll

(1) 22mv

5(2)

2mv

5(3)

23mv

5 (4) mv2

30. ,d ?ku ds 'kh"kZ ij cy F dks fp=kuqlkj yxk;k tkrk

gSA bl ?ku rFkk /kjkry ds e/; ?k"kZ.k xq.kkad µ gSA ;fn

F dk eku /khjs /khjs c<+k;k tk;s rks f[kldus ls igys ;g

?ku iyV tk;sxk rc µ dh ijkl gS :-

F

(1) µ > 1 (2) µ < 1

2(3) µ >

1

2(4) µ < 1

Kota/01CT314069H-14/32

Target : JEE (Main) 2015/01-04-2015


PART B - CHEMISTRY31. A cubic unit cell contains maganese ions at the

corners and fluoride ions at the centres of eachedge. The coordination number of maganeseion is-(1) 8 (2) 2 (3) 4 (4) 6

32. Calculate DGº (in kcal/mole) for decomposition

of Cl2(g) �� 2Cl(g), if chlorine molecules

are 50% dissociated at 1000 K at a pressure of

15 atm at equilibrium (ln 20 = 2.99)

(1) –2.6 (2) –5.99 (3) –2.01 (4) –4.56

33. The moles of Ag+ which must be added to

decrease the concentrat ion of Cl– from

4 × 10–5 M to 10–5M in 100 ml solution, if Ksp

for AgCl is 10–10M2 at 25ºC

(1) 4 × 10–5 mole (2) 2 × 10–5 mole

(3) 3 × 10–6 mole (4) 4 × 10–6 mole

34. Choose the incorrect statement -(1) Activation energy of reaction decreases

on decreasing temperature(2) Order of reaction may change with

change in temperature(3) When slowest step is the first step in a

mechanism, then the rate law of overallreaction is the same as the rate law for thisstep

(4) Rate of photochemical reaction is directlyproportional to intensity of photons.

31. ,d ?kuh; bdkbZ dksf"Bdk esa eSaxuht vk;u dksuksa ij

rFkk ¶yksjkbM vk;u izR;sd fdukjs ds dsUnz ij fLFkr gSA

eSaxuht vk;u dh leUo; la[;k D;k gksxh\(1) 8 (2) 2 (3) 4 (4) 6

32. Cl2(g) �� 2Cl(g) ds fo;k stu ds fy,

DGº (kcal/mole esa) dh x.kuk dhft;s] ;fn lkE; ij

1000K rFkk 15atm ds nkc ij Dyk sjhu v.k q

50% fo;ksftr gksrs gksa (ln 20 = 2.99)

(1) –2.6 (2) –5.99 (3) –2.01 (4) –4.56

33. ;fn 25ºC ij AgCl ds fy, Ksp

= 10–10M2 gS] rksAg+ ds eksy] tks 100 ml foy;u esa mifLFkr Cl– dhlkUnzrk dks 4×10–5M ls 10–5M rd ?kVkus ds fy,]feyk;k tkuk pkgh;sA(1) 4 × 10–5 mole (2) 2 × 10–5 mole(3) 3 × 10–6 mole (4) 4 × 10–6 mole

34. xyr dFku dk p;u dhft;sA

(1) vfHkfØ;k dh lfØ;.k ÅtkZ rki ?kVkus ij ?kVrh gS

(2) rki esa ifjorZu ds lkFk vfHkfØ;k dh dksfV esa

ifjorZu gks ldrk gS

(3) tc ,d fØ;kfof/k esa izFke in lcls /khek in

gS] rc lEiw.kZ vfHkfØ;k dk nj fu;e] bl in

ds fy, nj fu;e ds leku gksrk gS

(4) izdk'k jklk;fud vfHkfØ;kvksa dh nj QksVksuks dh

rhozrk ds lh/ks lekuqikrh gksrh gS


H-15/32Kota/01CT314069


35. The position of equilibrium lies to the right in

each of these reactions :

N2H

5+ + NH

3 �� NH

4+ + N

2H

4

NH3 + HBr �� NH

4+ + Br

–

N2H

4 + HBr �� N

2H

5+ + Br

–

Based on this information, what is the order ofacidic strength -(1) HBr > N

2H

5+ > NH

4+

(2) N2H

5+ > N

2H

4 > NH

4+

(3) NH3 > N

2H

4 > Br

–

(4) N2H

5+ > HBr > NH

4+

36. Boiling point of equimolar ideal solution of

hexane, C6H

14 and heptane , C

7H

16 is 80ºC.

What is the boiling point of mixture of hexane

and heptane containing equal mass of both.

Boiling points of hexane and heptane are 69ºC

& 98ºC respectively -

(1) Greater than 98ºC

(2) Greater than 80ºC

(3) Between 69ºC and 80ºC

(4) Below 69ºC

37. Gadolinium – 153, which is used to detect

osteoporosis (porous bones), has a half life of

242 days. Which value is closest to the

percentage of Gd– 153 left in a patient's system

after 2 years -

(1) 33 % (2) 26 %

(3) 12.0 % (4) 6.25 %

35. fuEu vfHkfØ;kvkas esa ls izR;sd esa nk;ha vksj lkE;fLFkr gS&N

2H

5+ + NH

3 �� NH

4+ + N

2H

4

NH3 + HBr �� NH

4+ + Br

–

N2H

4 + HBr �� N

2H

5+ + Br

–

bl lqpuk ds vk/kkj ij] vEyh; lkeF;Z dk Øe D;k gS-

(1) HBr > N2H

5+ > NH

4+

(2) N2H

5+ > N

2H

4 > NH

4+

(3) NH3 > N

2H

4 > Br

–

(4) N2H

5+ > HBr > NH

4+

36. gSDlsu] C6H

14, gSIVsu rFkk C

7H

16 ds leeksyj vkn'kZ

foy;u dk DoFkukad 80ºC gSA gSDlsu rFkk gSIVsu nksuksa

ds leku nzO;eku okys feJ.k dk DoFkukad D;k gksxkA

gSDlsu rFkk gSIVsu ds DoFkukad Øe'k%69ºC rFkk 98ºC gS

(1) 98ºC ls vf/kd

(2) 80ºC ls vf/kd

(3) 69ºC rFkk 80ºC ds e/;

(4) 69ºC ls de gS

37. xSMksyhuh;e – 153, ftldk mi;ksx vksLVh;ksQksjksfll

(fNfnzr gfM~M;ksa) nks"k ds fy, gksrk gS] fd v/kZ vk;q

242 fnu gSA 2 o"kksZ ds i'pkr~ jksxh ds 'kjhj esa 'ks"k

Gd– 153 dk fudVre izfr'kr eku gS -

(1) 33 % (2) 26 %

(3) 12.0 % (4) 6.25 %

Kota/01CT314069H-16/32

Target : JEE (Main) 2015/01-04-2015


38. The change in electrode potential of

Cr3+/Cr2O

7– – electrode at 25ºC due to change

in pH of its electrolytic solution from 1 to 3 is

(Assume : [Cr2O

7– –] & [Cr3+] remains constant)

Use : 2.303RT 0.06

F=

(1) 0.21 V (2) 0.28 V

(3) 0.14 V (4) No change

39. Which among the following electron transition

(n2 ® n

1) in H-atom, minimum change in its

orbit radius (r2 – r

1) is observed-

(1) 4 ® 1 (2) 5 ® 4

(3) 6 ® 5 (4) 4 ® 2

40. The density of a gas 'A' at 1 atm and 750K is

0.3 gm/litre. If the mol.wt of 'A' is 27 then

choose the correct statement -

(1) 'A' behaves ideally

(2)'A' behaves non-ideally with +ve deviation

(3 'A' behaves non-ideally with –ve deviation

(4) 'A' can be liquified by applying pressure

at 750K.

38. 25ºC ij Cr3+/Cr2O

7– – bySDVªksM ds oS|qr vi?kV~;

foy;u dh pH esa 1 ls 3 rd ifjorZu ds dkj.k bySDVªksM

foHko esa fdruk ifjorZu gksxk

(eku fyft;s: [Cr2O

7– –] & [Cr3+] fu;r jgrs gS)

fn;k gS : 2.303RT 0.06

F=

(1) 0.21 V (2) 0.28 V

(3) 0.14 V (4) dksbZ ifjorZu ugha

39. H- ijek.kq esa fuEu bySDVªkWu laØe.k (n2 ® n

1) esa ls

dkSulk] bldh d{kk dh f=T;k (r2 – r

1) esa U;wure

ifjorZu izs{khr gksxk-

(1) 4 ® 1 (2) 5 ® 4

(3) 6 ® 5 (4) 4 ® 2

40. ,d x Sl 'A' dk ?kuRo 1 atm rFkk 750K ij

0.3 gm/litre gS ;fn 'A' dk vkf.od Hkkj 27 gS] rks

lgh dFku dk p;u dhft,

(1) 'A' vkn'kZ O;ogkj iznf'kZr djrk gS(2) 'A' /kukRed fopyu ds lkFk vu&vkn'kZ O;ogkj

iznf'kZr djrk gS(3) 'A' ½.kkRed fopyu ds lkFk vu&vkn'kZ O;ogkj

iznf'kZr djrk gS(4) 'A' dks 750K ij nkc yxkdj nzohr fd;k tk

ldrk gS


H-17/32Kota/01CT314069


41. Which of the following statement is

CORRECT :(I) Where orbitals are available in degenerate

sets maximum spin multiplicity is

observed

(II) Where two electrons occupy the same

shell they may have same spins

(III) All noble gases do not have same valence

shell electronic configuration

(1) Only I and III (2) Only II and III

(3) Only I and II (4) I, II and III

42. A + 2Cl¯ ® B

Where A is PCl+

4 while B is PCl

6

Which of the following statement is

INCORRECT :(I) B to A involvement of number of

d-orbital in hybridization increases

(II) A to B precentage of p-character in P–Cl

bond increases

(III) Number of covalent bond in A and B are

same

(IV) B is product of Lewis acid-base

interaction which is associated with super

octet molecule only

(1) Only I, II and III (2) Only II and III

(3) Only III and IV (4) I, III and IV

41. fuEu esa ls dkSulk dFku lgh gS :

(I) tc leHkaz'k leqPp; esa d{kdksa dh miyC/krk gks

rks vf/kdre pØ.k cgqyrk izsf{kr dh tkrh gS

(II) tc nks bysDVªkWu leku dks'k esa mifLFkr gSa rks mudk

pØ.k leku gks ldrk gS

(III) lHkh mRd`"V xSlksa ds la;ksth dks'k dk bysDVªkWfu;

foU;kl leku ugha gks ldrk gS

(1) dsoy I rFkk III (2) dsoy II rFkk III

(3) dsoy I rFkk II (4) I, II rFkk III

42. A + 2Cl¯ ® B

tgk¡ A, PCl+

4 gS rFkk B, PCl

6 gS

fuEu esa ls dkSulk dFku xyr gS :

(I) B ls A esa ifjorZu ij] ladj.k esa lfEefyr

d-d{kdksa dh la[;k c<+rh gS

(II) A ls B esa ifjorZu ij] P–Cl ca/k esa p-y{k.k

dh izfr'kr~rk c<+rh gS

(III) A rFkk B esa lgla;kstd ca/kksa dh la[;k leku

gS

(IV) B, yqbZl vEy&{kkj vU;ksU; fØ;k dk mRikn

gS tks dsoy lqij v"Vd v.kq ls lEcfU/kr gS

(1) dsoy I, II rFkk III (2) dsoy II rFkk III

(3) dsoy III rFkk IV (4) I, III rFkk IV

Kota/01CT314069H-18/32

Target : JEE (Main) 2015/01-04-2015


43.

O2

O (excess)2

'X'

'Y'P4

Define the order of dP–O

(Single bond length)

in given compound X and Y is :

(1) X < Y (2) X > Y

(3) X = Y (4) Can not be predicted

44. Complex which have maximum number of

unpaired electron in set of eg orbital is :

(1) [Cr(CN6)]–4 (2) [Ni(en)

3] Cl

2

(3) [Sc(H2O)

6]3+ (4) [V(H

2O)

6]2+

45. (NH4)

2Cr

2O

7 on heating gives :

(1) N2O (2) NH

3

(3) NO2

(4) N2

46. HgI2

ppt.

D HgI2Cooling HgI2

'X''Y' 'Z'

On rubbing

Sublimation

(1) 'X' ® Yellow, 'Y' ® Red, 'Z' ® Yellow

(2) 'X' ® Red, 'Y' ® Yellow, 'Z' ® Red

(3) 'X' ® Red, 'Y' ® Yellow, 'Z' ® Yellow

(4) 'X' ® Yellow, 'Y' ® Red, 'Z' ® Red

43.

O2

O ( )2 vkf/kD;

'X'

'Y'P4

fn;s x;s ;kSfxd X rFkk Y esa dP–O

(,dy ca/k yEckbZ)

dk Øe gS :

(1) X < Y (2) X > Y

(3) X = Y (4) vuqeku ugha yxk ldrs

44. ladqy ftlds] eg d{kd ds leqPp; esa v;qfXer bysDVªkWuksa

dh vf/kdre la[;k mifLFkr gS] gS :

(1) [Cr(CN6)]–4 (2) [Ni(en)

3] Cl

2

(3) [Sc(H2O)

6]3+ (4) [V(H

2O)

6]2+

45. (NH4)

2Cr

2O

7 xeZ djus ij nsrk gS :

(1) N2O (2) NH

3

(3) NO2

(4) N2

46. HgI2

vo{ksi

D HgI2B.Mk

djus ijHgI2

'X''Y' 'Z'

jxM+us ij

Å/oZikru

(1) 'X' ® ihyk, 'Y' ® yky, 'Z' ® ihyk

(2) 'X' ® yky, 'Y' ® ihyk, 'Z' ® yky

(3) 'X' ® yky, 'Y' ® ihyk, 'Z' ® ihyk

(4) 'X' ® ihyk, 'Y' ® yky, 'Z' ® yky


H-19/32Kota/01CT314069


47. 2Na2O

400 ºC�� A + 2Na

A + SO2 ® Na

2SO

4

A + CO ® Na2CO

3

In above reaction A is :

(1) NaO2

(2) Na2O

2

(3) NaOH (4) Na2O

3

48. Which of the following match is

INCORRECT :

(1) Hg2I

2 ® Green

(2) Ag2CrO

4 ® Brick red

(3) BiI3 ® Orange

(4) BaCrO4 ® Yellow

49. In the metallurgy of iron, when lime stone is

added to the blast furnance the calcium ions

ends up in :

(1) Slag (2) Gangue

(3) Metallic Ca (4) CaO

50. During the electrolytic refining of silver anode

mud contains :

(1) Zn, Cu, Ag, Au (2) Zn, Au

(3) Cu, Au (4) Au

47. 2Na2O

400 ºC�� A + 2Na

A + SO2 ® Na

2SO

4

A + CO ® Na2CO

3

mijksDr vfHkfØ;k esa A gS :

(1) NaO2

(2) Na2O

2

(3) NaOH (4) Na2O

3

48. fuEu esa ls dkSulk feyku xyr gS :

(1) Hg2I

2 ® gjk

(2) Ag2CrO

4 ® bZV tSlk yky

(3) BiI3 ® ukjaxh

(4) BaCrO4 ® ihyk

49. vk;ju ds /kkrqdehZ; izØe esa] tc okR;k HkV~Vh esa ykbe

LVksu feyk;k tkrk gS rks dSfY'k;e vk;u cukrs gSa\

(1) /kkrqey (2) xSax (Gangue)

(3) /kkfRod Ca (4) CaO

50. flYoj ds oS|qr vi?kVuh; 'kqf¼dj.k ds nkSjku ,uksM

eM+ esa mifLFkr gksrs gSa :

(1) Zn, Cu, Ag, Au (2) Zn, Au

(3) Cu, Au (4) Au

Kota/01CT314069H-20/32

Target : JEE (Main) 2015/01-04-2015


51.

OEt

(i) Cl

O

(ii) KMnO4

(iii) MeOH/H /+

/ AlCl3 Product ;

Product is :

(1)

OEt

OMe

(2)

OEtCOOH

(3)

CO Me2

OEt

(4)

COOH

OEt

52. On another planet jupiter, gauche & anti form

of 1, 2-dichloroethane freezes and single bond

rotation stops. Under these conditions these two

forms can be considered as :

(1) Enantiomers (2) Diastereomers

(3) Meso compounds (4) Chain isomers

51.

OEt

(i) Cl

O

(ii) KMnO4

(iii) MeOH/H /+

/ AlCl3

mRikn ;

mRikn gS&

(1)

OEt

OMe

(2)

OEtCOOH

(3)

CO Me2

OEt

(4)

COOH

OEt

52. vU; xzg twfiVj ij] 1, 2-MkbZDyksjks,sFksu ds xkWp rFkk

izfr :i te tkrs gSa vkSj ,dy cU/k ?kw.kZu :d tkrk

gSA bu ifjfLFkfr;ksa esa bu nksuksa :iksa dks fdl izdkj ekuk

tk ldrk gS&

(1) izfrfcEc:i leko;oh (2)foofje leko;oh

(3) ehlks ;kSfxd (4) J`a[kyk leko;oh


H-21/32Kota/01CT314069


53.

Br

ClCl

Above conversion can be brought about by ?

(1) (i) HBr ; (ii) Cl2

(2) (i) Cl2 ; (ii) NBS

(3) (i) NBS ; (ii) Cl2

(4) (i) Cl2 ; (ii) BrCl / CCl4

54.

NO2

BrF

F

CH SH3 (A)

Major product for the above reaction is :

(1)

NO2

SCH3F

F(2)

NO2

SCH3

BrF

(3)

NO2

CH3S BrF

(4)

NO2

F BrF

SCH3

53.

Br

ClCl

mijksDr :ikUrj.k dks fuEu esa ls fdlds }kjk djk;k tk

ldrk gS&

(1) (i) HBr ; (ii) Cl2

(2) (i) Cl2 ; (ii) NBS

(3) (i) NBS ; (ii) Cl2

(4) (i) Cl2 ; (ii) BrCl / CCl4

54.

NO2

BrF

F

CH SH3 (A)

mijksDr vfHkfØ;k ds fy;s eq[; mRikn gS&

(1)

NO2

SCH3F

F(2)

NO2

SCH3

BrF

(3)

NO2

CH3S BrF

(4)

NO2

F BrF

SCH3

Kota/01CT314069H-22/32

Target : JEE (Main) 2015/01-04-2015


55.OH

OH

HIO4 (A) NaOH (B)

Product "B" is :

(1)

OH(2)

OH

O

(3) C H–O

(4) O

56. The correct statement(s) about the followingsugar X and Y is :

CH OH2

O O O

O

HO

HHOH H

H

O

CH OH2H

CH OH2

HO

HCH OH2

HOH H

OHHOH H

H OHH

OHOH HO

H

OH

(X)(Y)

CH OH2

OH HH HO

(1) Both (X) & (Y) are non-reducing sugar

(2) (X) is non reducing sugar & (Y) is a

reducing sugar

(3) (X) is reducing sugar & (Y) is non

reducing sugar

(4) Both (X) & (Y) are reducing sugar

55.OH

OH

HIO4 (A) NaOH (B)

mRikn "B" gS&

(1)

OH(2)

OH

O

(3) C H–O

(4) O

56. fuEu 'kdZjk X rFkk Y ds ckjs esa lgh dFku gS&

CH OH2

O O O

O

HO

HHOH H

H

O

CH OH2H

CH OH2

HO

HCH OH2

HOH H

OHHOH H

H OHH

OHOH HO

H

OH

(X)(Y)

CH OH2

OH HH HO

(1) (X) rFkk (Y) nksuksa vuvipk;d 'kdZjk gS

(2) (X) vuvipk;d 'kdZjk rFkk (Y) vipk;d

'kdZjk gS

(3) (X) vipk;d 'kdZjk rFkk (Y) vuvipk;d

'kdZjk gS

(4) (X) rFkk (Y) nksuksa vipk;d 'kdZjk gS


H-23/32Kota/01CT314069


57. H

H

(i) O3(ii) (CH ) S3 2

(A) (Ph) P=CH3 2 (B)

Product B may be :

(1)

HCH2 (2) CH2

H

(3) (4) Both (1) & (2)

58. Which of the following statement correctly

describes the migration aptitude of aspartic acid

during electrophoresis ?

(pK1 = 2 ; pK2 = 3.90 ; pK3 = 10.0)

HOOC–CH —CH—COOH2

NH2

(1) at pH = 1 ; aspartic acid migrate towards

(+) electrode

(2) at pH = 2.45 aspartic acid show no net

migration towards any electrode

(3) at pH = 7.0 ; aspartic acid show no net

migration toward any electrode

(4) at pH = 9.0 ; aspartic acid show a net

migration towards (–) electrode

57. H

H

(i) O3(ii) (CH ) S3 2

(A) (Ph) P=CH3 2 (B)

mRikn B gksuk pkfg;s&

(1)

HCH2 (2) CH2

H

(3) (4) (1) rFkk (2) nksuksa

58. fuEu esa ls dkSulk dFku] oS|qr d.k lapyu ds nkSjku

,LikfVZd vEy ds vfHkxeu O;ogkj dks lgh :i ls

iznf'kZr djrk gS&

(pK1 = 2 ; pK2 = 3.90 ; pK3 = 10.0)

HOOC–CH —CH—COOH2

NH2

(1) pH = 1 ij ; ,LikfVZd vEy (+) bysDVªkWM dh rjQ

vfHkxeu djrk gS

(2) pH = 2.45 ij] ,LikfVZd vEy fdlh Hkh bysDVªkWM

dh rjQ dqy vfHkxeu ugha n'kkZrk gS

(3) pH = 7.0 ij ; ,LikfVZd vEy fdlh Hkh bysDVªkWM

dh rjQ dqy vfHkxeu ugha n'kkZrk gS

(4) pH = 9.0 ij ; ,LikfVZD vEy (–) bysDVªkWM dh

rjQ dqy vfHkxeu n'kkZrk gS

Kota/01CT314069H-24/32

Target : JEE (Main) 2015/01-04-2015


59. Radical chain reaction of ethane with chlorine

may proceed according to the following

mechanism. Which of the following chain

propagation step is most likely ?

(1) (i) Cl + CH3–CH3 ¾® CH3–CH2 + HCl

(ii)CH3–CH2+Cl2�¾® CH3–CH2Cl+ Cl

(2) (i) Cl + CH3–CH3 ¾® CH3–Cl + CH3

(ii) CH3 + Cl2 ¾® CH3–Cl + Cl

(3) (i) Cl + CH3–CH3 ® CH3–CH2–Cl+ Cl

(ii) H + Cl2 ¾® HCl + Cl

(4) (i) CH3–CH2 +Cl–Cl ¾® CH3–Cl + Cl

(ii) Cl–CH3+CH3–CH3¾®CH3–CH2 + Cl60. If butadiene is polymerised by a free radical

synthesis the possible product contains whichrepeatation units ?

(i) C CCH2

H

H C2

H n

(ii) C CCH2

HH C2

H n

(iii) CH2–CH2–CH2–CH n

(1) (i) & (ii) (2) (iii) only

(3) (i) only (4) (i), (ii) & (iii)

59. ,sFksu dh Dyksfju ds lkFk ewyd J`a[kyk vfHkfØ;k fuEu

fØ;kfof/k ds vuqlkj gksrh gSA fuEu esa ls dkSulk Ja`[kyk

lapj.k in lokZf/kd mi;qDr gS&

(1) (i) Cl + CH3–CH3 ¾® CH3–CH2 + HCl

(ii)CH3–CH2+Cl2�¾® CH3–CH2Cl+ Cl

(2) (i) Cl + CH3–CH3 ¾® CH3–Cl + CH3

(ii) CH3 + Cl2 ¾® CH3–Cl + Cl

(3) (i) Cl + CH3–CH3 ® CH3–CH2–Cl+ Cl

(ii) H + Cl2 ¾® HCl + Cl

(4) (i) CH3–CH2 +Cl–Cl ¾® CH3–Cl + Cl

(ii) Cl–CH3+CH3–CH3¾®CH3–CH2 + Cl

60. ;fn C;qVkMkbZbu eqDr ewyd la'ys"k.k }kjk cgqydhd`r

gksrh gS lEHkkfor mRikn] ftlesa iqujkofr bdkbZ;k¡ gksrh gS&

(i) C CCH2

H

H C2

H n

(ii) C CCH2

HH C2

H n

(iii) CH2–CH2–CH2–CH n

(1) (i) rFkk (ii) (2) dsoy (iii)(3) dsoy (i) (4) (i), (ii) rFkk (iii)


H-25/32Kota/01CT314069


PART C - MATHEMATICS61. If 12a + 5b = 9, where a, b Î R, then minimum

value of a2 + b2 is -

(1) 319 (2)

16981 (3)

81169 (4)

913

62. x

sec x(1 tan x)dx ƒ(x) C(e sec x)-

+= +

+ò ,

where ƒ(0) = ln2, then ƒ4pæ ö

ç ÷è ø

is -

(1) 4n 1 e 2pæ ö

+ç ÷è ø

l (2) n( 2)l

(3) ( )n 2 2l (4) 4en 12

pæ öç ÷+ç ÷ç ÷è ø

l

63. If the roots of the equation

x3 – 9x2 + ax – 15 = 0 are in A.P., then a is -

(1) 0 (2) 20 (3) 21 (4) 23

64. The value of the expression

3(1!) – 4(2!) + 5(3!) – 6(4!) ...... – 2008(2006)!+ (2007)! is -

(1) –2007 (2) –1 (3) 1 (4) 2007

65. Let A, B, C are three sets such that n(A Ç B)= n(B Ç C) = n(C Ç A) = n(A Ç B Ç C) = 2,then n((A × B) Ç (B × C)) is equal to -

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

61. ;fn 12a + 5b = 9, tgk¡ a, b Î R gks] rks a2 + b2 dkU;wure eku gksxk -

(1) 319 (2)

16981 (3)

81169 (4)

913

62. x

sec x(1 tan x)dx ƒ(x) C(e sec x)-

+= +

+ò ,

tgk¡ ƒ(0) = ln2 gks] rks ƒ4pæ ö

ç ÷è ø

gksxk -

(1) 4n 1 e 2pæ ö

+ç ÷è ø

l (2) n( 2)l

(3) ( )n 2 2l (4) 4en 12

pæ öç ÷+ç ÷ç ÷è ø

l

63. ;fn lehdj.k x3 – 9x2 + ax – 15 = 0 ds ewy lekUrjJs.kh esa gks] rks a gksxk -

(1) 0 (2) 20 (3) 21 (4) 23

64. O;atd 3(1!) – 4(2!) + 5(3!) – 6(4!) ......

– 2008(2006)! + (2007)! dk eku gksxk -

(1) –2007 (2) –1 (3) 1 (4) 2007

65. ekuk A,B,C rhu leqPp; bl izdkj gS fd n(A Ç B)= n(B Ç C) = n(C Ç A)= n(A Ç B Ç C) = 2 gks]rks n((A × B) Ç (B × C)) cjkcj gksxk -

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

Kota/01CT314069H-26/32

Target : JEE (Main) 2015/01-04-2015


66. If

4

2

xt

9x

ƒ(x) 5 dt= ò , then

h 0

ƒ(3 h) ƒ(3 h)limh®

+ - - is equal to -

(1) 0 (2) 108(59) (3) 55 (4) 54(58)

67. Let a1, a

2 are two values of a for which the

system 2ax + y = 5, x – 6y = a and x + y = 2 is

consistent, then |2(a1 + a

2)| is -

(1) 21 (2) 23 (3) 25 (4) 27

68. Let ƒ(x) = ex – e–x + cosx, then ƒ(x) is -

(1) always increasing

(2) always decreasing

(3) non differentiable at x = 0

(4) local maxima at x = 1.

69. If a, b are two real numbers satisfying

a2 + b2 = 5 and 3(a5 + b5) = 11(a3 + b3), then

ab is -

(1) 2 (2) 1 (3) 7 (4) 9

70. Let (1 + x)(1 + x + x2)(1 + x + x2 + x3) ......

(1 + x + x2 + ..... + x30) = a0 + a

1x + a

2x2 .....

+ a465

x465, then sum of a0 + a

2 + a

4 + ......... +

is -

(1) (31)! (2) (31)!

2 (3) (30)! (4) (60)!

2

66. ;fn 4

2

xt

9x

ƒ(x) 5 dt= ò gks] rks h 0

ƒ(3 h) ƒ(3 h)limh®

+ - -

cjkcj gksxk -

(1) 0 (2) 108(59) (3) 55 (4) 54(58)

67. ekuk a ds nk s eku a1, a

2 ftlds fy, fudk;

2ax + y = 5, x – 6y = a rFkk x + y = 2 laxr gks] rks|2(a

1 + a

2)| dk eku gksxk -

(1) 21 (2) 23 (3) 25 (4) 27

68. ekuk ƒ(x) = ex – e–x + cosx gks] rks ƒ(x) gksxk -

(1) lnSo o/kZeku

(2) lnSo Îkleku

(3) x = 0 ij vodyuh; ugha

(4) x = 1 ij LFkkuh; mfPp"B

69. ;fn a, b nks okLrfod la[;k;sa gS] tks a2 + b2 = 5 rFkk3(a5 + b5) = 11(a3 + b3) dks lUrq"V djrh gS] rks abdk eku gksxk -

(1) 2 (2) 1 (3) 7 (4) 9

70. ekuk (1 + x)(1 + x + x2)(1 + x + x2 + x3) ......

(1 + x + x2 + ..... + x30) = a0 + a

1x + a

2x2 .....

+ a465

x465 gks] rks a0 + a

2 + a

4 + ...... + dk ;ksxQy

gksxk -

(1) (31)! (2) (31)!

2 (3) (30)! (4) (60)!

2


H-27/32Kota/01CT314069


71. The solution of differential equationxdy (sin y e )

dx ( ny x cos y)+

=-l

is -

(1) y((lny) – 1)) = ex + xsiny + C

(2) lny = xsiny + C

(3) y(lny + 1) = ex – xsiny + C

(4) xlny = ex – xsiny + C

72.cos

sin

ƒ(x tan )dxq

q

qò is (where n , n I2p

q ¹ Î )

(1) tan

1

cos ƒ(x sin )dxq

- q qò

(2) cos

sin

tan ƒ(x)dxq

q

- q ò

(3) tan

0

sin ƒ(x cos )dxq

q qò

(4) sin tan

sin

cot ƒ(x)dxq q

q

q ò

73. If µx

100xx

(e 5)lim(e 7)®¥

++

exists, then sum of all

possible positive integral values of µ is -

(1) 5051 (2) 50

(3) 4950 (4) 5050

74. 80C40

is not divisible by -

(1) 7 (2) 23 (3) 11 (4) 29

71. vody lehdj.kxdy (sin y e )

dx ( ny x cos y)+

=-l

dk gy gksxk -

(1) y((lny) – 1)) = ex + xsiny + C

(2) lny = xsiny + C

(3) y(lny + 1) = ex – xsiny + C

(4) xlny = ex – xsiny + C

72.cos

sin

ƒ(x tan )dxq

q

qò gksxk (tgk¡ n , n I2p

q ¹ Î )

(1) tan

1

cos ƒ(x sin )dxq

- q qò

(2) cos

sin

tan ƒ(x)dxq

q

- q ò

(3) tan

0

sin ƒ(x cos )dxq

q qò

(4) sin tan

sin

cot ƒ(x)dxq q

q

q ò

73. ;fn µx

100xx

(e 5)lim(e 7)®¥

++

fo|eku gks] rks µ ds lHkh lEHko

/kukRed iw.kk±d ekuksa dk ;ksxQy gksxk -(1) 5051 (2) 50

(3) 4950 (4) 5050

74. 80C40

fuEu ls foHkkftr ugha gksxk -

(1) 7 (2) 23 (3) 11 (4) 29

Kota/01CT314069H-28/32

Target : JEE (Main) 2015/01-04-2015


75. If z1, z

2, z

3 are the roots of the equation

z3 – z2(4 + 3i) + z(3 + 8i) – 5i = 0, then

Re(z1) + Re(z

2) + Re(z

3) is -

(1) 0 (2) –1 (3) 3 (4) 4

76. Which of the following statement is false

(where A & B are two non empty sets)

(1) A – B = A Ç B'

(2) A – B = A – (A Ç B)

(3) A – B = A – B'

(4) A – B = (A È B) – B

77. Two numbers x & y are chosen at random(without replacement) from the set

{1, 2, 3, ......, 1000}. Then the probability that|x4 – y4| is divisible by 5, is -

(1) 113999 (2)

400999 (3)

679999 (4)

1999

78. Let 2ƒ(x) + ƒ(–x) =1 1sin xx x

æ ö-ç ÷è ø

, then value

of e

1/ e

ƒ(x)dxò is -

(1) 0 (2) (e + p) (3) 1ee

+ (4) 2e

79. X,Y,Z are sets of all positive divisors of 1060,

2050 and 3040 respectively n(X È Y È Z) is -

(1) 70301 (2) 30701

(3) 73001 (4) 70031

75. ;fn z1, z

2, z

3 lehdj.k

z3 – z2(4 + 3i) + z(3 + 8i) – 5i = 0 ds ewy gks]

rks Re(z1) + Re(z

2) + Re(z

3) dk eku gksxk -

(1) 0 (2) –1 (3) 3 (4) 4

76. fuEu esa ls dkSulk dFku vlR; gksxk(tgk¡ A rFkk B nks vfjDr leqPp; gSa)(1) A – B = A Ç B'

(2) A – B = A – (A Ç B)

(3) A – B = A – B'

(4) A – B = (A È B) – B

77. nks la[;kvksa x rFkk y dk leqPp; {1, 2, 3, ......, 1000}

ls (fcuk iqujkofÙk ds) ;kn`PN;k p;u djrs gS] rc

|x4 – y4| ds 5 ls foHkkftr gksus dh izkf;drk gksxh -

(1) 113999 (2)

400999 (3)

679999 (4)

1999

78. ekuk 2ƒ(x) + ƒ(–x) =1 1sin xx x

æ ö-ç ÷è ø

gk s] rk s

e

1/ e

ƒ(x)dxò gksxk -

(1) 0 (2) (e + p) (3) 1ee

+ (4) 2e

79. 1060, 2050 rFkk 3040 ds lHkh /kukRed Hkktdksa dsleqPp; Øe'k% X,Y,Z gks] rks n(X È Y È Z) gksxk -(1) 70301 (2) 30701

(3) 73001 (4) 70031


H-29/32Kota/01CT314069


80. AB is a line segment of length 24 cm and C is

its middle point. On AB, AC and CB semcircles

are described in same side. Then radius of the

circle which touches all the three semicircles is-

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

81. The reflection of the complex number (3 + 2i)

in the straight line z iz= - is-

(1) (–2 –3i) (2) (2 – 3i)

(3) (2 + 3i) (4) (i + 5)

82. Let ƒ(q) is distance of the line

( ) ( )sin x cos y 1 0q + q + = from origin.

Then the range of ƒ(q) is -

(1) 1/ 4

1,

2é ö¥÷êë ø

(2) 1, 2é ùë û

(3) [1,¥) (4) 1/ 4

1,1

2é ùê úë û

83.( )( )

x21

0

x 0

tan t dt

limsin x x

-

® -

ò is-

(1) 0 (2) –2

(3) 2 (4) 1

2

80. js[kk[k.M AB ftldh yEckbZ 24 cm gS rFkk C bldk

eè; fcUnq gSA Hkqtk AB, AC rFkk CB ij ,d gh vksj

v¼Zo`Ùk cukrs gSA rc ml o`Ùk dh f=T;k tks lHkh rhuksa

v¼Zo`Ùkksa dks Li'kZ djrk gS] gksxh -

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

81. lfEeJ la[;k (3 + 2i) dk ljy js[kk z iz= - esa

ijkorZu gksxk-

(1) (–2 –3i) (2) (2 – 3i)

(3) (2 + 3i) (4) (i + 5)

82. ekuk ƒ(q) js[kk ( ) ( )sin x cos y 1 0q + q + = dh

ewyfcUnq ls nwjh gSA rc ƒ(q) dk ifjlj gksxk-

(1) 1/ 4

1,

2é ö¥÷êë ø

(2) 1, 2é ùë û

(3) [1,¥) (4) 1/ 4

1,1

2é ùê úë û

83.( )( )

x21

0

x 0

tan t dt

limsin x x

-

® -

ò gksxk-

(1) 0 (2) –2

(3) 2 (4) 1

2

Kota/01CT314069H-30/32

Target : JEE (Main) 2015/01-04-2015


84. The distance between the focii of the ellipse

(3x – 9)2 + 9y2 = ( )2

2x y 1+ + is-

(1) ( )3 2 1- (2) ( )+3 2 1

3

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

4 3

+

85. A line makes the same angle q with each of thex and z axes. If the angle b which it makeswith y-axis is such that sin2b = 3sin2q, thencos2q is-

(1) 2

5(2)

1

5(3)

3

5(4)

2

3

86. The circumradius of an isosceles triangle ABCis four times as that of inradius of the triangle,if A = B. Then -(1) 8 cos2A – 8cosA + 1 = 0(2) 4 cos2A – 10cosA + 1 = 0(3) cos2A – cosA – 3 = 0(4) cos2A – cosA – 8 = 0

87. Let A, B, C are three angles such thatsinA + sinB + sinC = 0, then

( )sin Asin Bsin C

sin 3A sin 3B sin 3C+ +

(wherever definied) is -

(1) 12 (2) –12 (3) 1

12- (4)

1

12

84. nh?kZo`Ùk (3x – 9)2 + 9y2 = ( )2

2x y 1+ + dh ukfHk;ksa

ds e/; nwjh gksxh -

(1) ( )3 2 1- (2) ( )+3 2 1

3

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

4 3

+

85. ,d js[kk x rFkk z v{k ds lkFk leku dks.k q cukrh

gSA ;fn y-v{k ds lkFk dks.k b bl izdkj gS fd

sin2b = 3sin2q gks] rks cos2q gksxk-

(1) 2

5(2)

1

5(3)

3

5(4)

2

386. lef}ckgq f=Hkqt ABC dh ifjf=T;k bl f=Hkqt dh

vUr%f=T;k dh pkj xquk gS] ;fn A = B gSA rc -

(1) 8 cos2A – 8cosA + 1 = 0

(2) 4 cos2A – 10cosA + 1 = 0

(3) cos2A – cosA – 3 = 0

(4) cos2A – cosA – 8 = 0

87. ekuk A, B, C rhu dks.k bl izdkj gS fdsinA + sinB + sinC = 0 gks] rks

( )sin Asin Bsin C

sin 3A sin 3B sin 3C+ +

(tgk¡ dgha Hkh ifjHkkf"kr) gksxk -

(1) 12 (2) –12 (3) 1

12- (4)

1

12


H-31/32Kota/01CT314069


88. 12

r 1

3tan

r r 9

¥-

=

æ öç ÷- +è ø

å is -

(1) 3

p(2)

6

p

(3) 2

p(4)

12

p

89. Let ( ) ( )( ) ( )( ) ( )3 2

5 ƒ x ƒ xh x 2ƒ x 100

3 2= + + + .

Where ƒ(x) is a differentiable function. Then

which one of the following is correct-

(1) h(x) always increases

(2) h(x) always decreases

(3) h(x) increases as ƒ(x) increases

(4) h(x) increases as ƒ(x) decreases

90. Let three points A(2,3,4) B(3,4,2) and C(4,2,3)

in space are given. A point D in space is such

that it is at a distance of 6 units from 3 given

points. Then volume of tetrahedron ABCD is -

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

88. 12

r 1

3tan

r r 9

¥-

=

æ öç ÷- +è ø

å gksxk -

(1) 3

p(2)

6

p

(3) 2

p(4)

12

p

89. ekuk ( ) ( )( ) ( )( ) ( )3 2

5 ƒ x ƒ xh x 2ƒ x 100

3 2= + + +

gSA tgk¡ ƒ(x) vodyuh; Qyu gSA rc fuEu esa ls dkSulk

lgh gksxk -

(1) h(x) lnSo o/kZeku

(2) h(x) lnSo Îkleku

(3) ƒ(x) ds Îkleku gksus ij h(x) o/kZeku

(4) h(x) ds Îkleku gksus ij ƒ(x) o/kZeku

90. ekuk lef"V esa rhu fcUnq A(2,3,4) B(3,4,2) rFkk

C(4,2,3) fn, x, gSA ,d fcUnq D lef"V esa bl izdkj

gS fd bldh fn, x, rhuksa fcUnqvksa ls nwjh 6 bdkbZ gS]

rc prq"Qyd ABCD dk vk;ru gksxk -

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

Kota/01CT314069H-32/32

Target : JEE (Main) 2015/01-04-2015



allen paper code 01ct3140 69 · paper code 01ct3140 69 p nstructio ully. y ll 5 in specifical . ......

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