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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

SCORE – I DATE : 16 - 03 - 2015

0 1 C E 3 1 4 0 6 1

PHASE – ELC, ELD, ELP

Corporate OfficeALLEN CAREER INSTITUTE

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

+91-744-2436001 [email protected]

CLASSROOM CONTACT PROGRAMME

www.allen.ac.in

JEE (Main) : LEADER COURSE

MAJOR TEST Test Pattern : JEE (Main)

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 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.

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, except

the 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 @ 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 esaHkk Sfrd foKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj lHkh iz'uksa dsvad 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] 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{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

1/27Kota/01CE314061

SPACE FOR ROUGH WORK

Leader Course/Phase-ELC, ELD & ELP/Score-I/16-03-2015

1. The mean radius of the earth's orbit round the sunis 1.5 × 1011. The mean radius of the orbit ofmercury round the sun is 6×1010m. The mercurywill rotate around the sun in :-(1) A year (2) Nearly 4 years

(3) Nearly 14

(4) 2.5 years

2. A clock S is based on oscillation of a spring and aclock P is based on pendulum motion. Both clocksrun at the same rate on earth. On a planet havingthe same density as earth but twice the radius :-(1) S will run faster than P(2) P will run faster than S(3) They will both run at the same rate as on the

earth(4) None of these

3. Two inductor coils of self inductance 3H and 6Hrespectively are connected with a resistance 10Wand a battery 10 V as shown in figure. The ratio oftotal energy stored at steady state in the inductorsto that of heat developed in resistance in 10 secondsat the steady state is (neglect mutual inductancebetween L1 and L2) :-

(1)1

10(2)

1100

L =3H1

L =6H2

10V 10W(3) 1

1000(4) 1

1. i`Foh dh lw;Z ds pkjks vksj ifjØe.k d{k dh ek/; f=T;k1.5 × 1011 ehVj gSA cq/k dh lw;Z ds pkjksa vksj ifjØe.kd{k dh ek/; f=T;k 6×1010ehVj gSA cq/k lw;Z dk pDdjyxk;sxk :-(1) ,d o"kZ esa (2) yxHkx pkj o"kZ esa

(3) yxHkx 14

o"kZ esa (4) 2.5 o"kZ esa

2. fLizax nksyu ij vk/kkfjr ,d ?kM+h S gS rFkk yksYkd xfrij vk/kkfjr ,d ?kM+h P gSA nksuksa ?kfM +;k¡ i`Foh ij lekuj¶rkj ls pyrh gSaA i`Foh ds leku ?kuRo ijUrq nksxquhf=T;k okys ,d xzg ij :-(1) S ls P rst pysxh(2) P ls S rst pysxh(3) nksuksa mlh j¶rkj ls pysaxh tSls i`Fih ij(4) mijksDr esa ls dksbZ ugha

3. fp= esa fn[kk;s vuqlkj Øe'k% 3 gsujh o 6 gsujh Lo&izsdRodh nks izsjd dq.Myh 10 W ds izfrjks/k ,oa 10V dh cSVjh

ls tqM+h gSA LFkk;hvoLFkk ij 10 sec esa iz sjd dq.Mfy;ksa

esa lafpr dqy ÅtkZ rFkk izfrjks/k esa mRiUu Å"ek dk vuqikr

gksxk (L1 o L2 ds e/; vU;ksU; izsj.kk dks ux.; ekurs

gq,):-

(1)1

10(2)

1100

L =3H1

L =6H2

10V 10W(3) 1

1000(4) 1

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART A - PHYSICS

Kota/01CE3140612/27


Target : JEE(Main) 2015/16-03-2015

4. A non conducting ring of radius R and mass mhaving charge q uniformly distributed over itscircumference is placed on a rough horizontalsurface. A vertical time varying uniform magneticfield B = 4t2 is switched on at time t=0. Thecoefficient of friction between the ring and the table,if the ring starts rotating at t = 2 sec, is :-

(1) 4qmR

g (2) 2qmR

g (3) 8qRmg (4)

qR2mg

5. The radius of the circular conducting loop shownin figure is R. Magnetic field is decreasing at aconstant rate a. Resistance per unit length of theloop is r. Then current in wire AB is (AB is one ofthe diameters) :-

A B

(1) R2

ar from A to B (2)

R2

ar from B to A

(3) 2Ra

r from A to B (4) Zero

4. ,d R f=T;k rFkk m nzO;eku dh vpkyd oy; ftl ijq vkos'k ,d leku #i ls forfjr gS dks [kqngjh {kSfrtlrg ij j[kk tkrk gSA ,d m/okZ/kj le; ds lkFk ifjofrZrpqEcdh; {ks= B = 4t2, t = 0 ij yxk;k tkrk gSA lrgrFkk oy; ds chp ?k"kZ.k xq.kkad gksxkA ;fn t = 2 lSd.Mij oy; ?k weuk izkjEHk dj nsa :-

(1) 4qmR

g (2) 2qmR

g (3) 8qRmg (4)

qR2mg

5. fp= esa R f=T;k dk o`Ùkkdkj pkyd oy; iznf'kZr gSApqEcdh; {ks= fu;r nj a ls ?kV jgk gSA ywi dh , adkdyEckbZ dk izfrjks/k r gS rks AB (AB rkj dk O;kl gS ) rkjesa /kkjk gksxh :-

A B

(1) R2

ar A ls B (2)

R2

ar B ls A

(3) 2Ra

r A ls B (4) 'kwU;

izR;sd iz'u dks vtqZu cudj djksA

3/27Kota/01CE314061



6. Some magnetic flux is changed in a coil ofresistance 10 ohm. As a result an induced currentis developed in it, which varies with time as shownin figure. The magnitude of change in flux throughthe coil in Webers is (Neglect self inductance ofthe coil):-

4

0.1t(s)

i (amp)

(1) 2 (2) 4(3) 6 (4) 8

7. A bar magnet is released from rest coaxially alongthe axis of a very long, vertical copper tube. Aftersome time the magnet :-(1) Will mover with an acceleratin g(2) Will move with almost constant speed(3) Will stop in the tube(4) Will oscillate

8. Four identical bulbs each rated 100 watt, 220 voltsare connected across a battery as shown. The totalelectric power consumed by the bulbs is :-

220V

(1) 75 watt (2) 400 watt(3) 300 watt (4) 400/3 watt

6. 10 ohm izfrjks/k dh ,d dq.Myh ls dqN pqEcdh; ¶yDl

ifjofrZr gksrk gSA ftlds ifj.kke#o#i blesa fo|qr /kkjk

izsfjr gksrh gS tks le; ds lkFk fp=kuqlkj ifjofrZr gksrh

gS rks dq.Myh ls ¶yDl esa ifjorZu dk ifjek.k oscj esa

gksxk (dq.Myh ds LoizsjdRo dks ux.; ekusa):-

4

0.1t(s)

i (amp)

(1) 2 (2) 4(3) 6 (4) 8

7. rkacs dh ,d cgqr yEch uyh dh v{k ds vuqfn'k lek{kh;,d NM + pqEcd dks fxjk;k tkrk gSA dqN le; i'pkr~pqEc :-(1) g Roj.k ds lkFk xfr djsxkA(2) yxHkx fu;r pky ls xfr djsxkA(3) uyh esa #d tk;sxkA(4) nksyu djsxkA

8. 100 okV rFkk 220 oksYV ds pkj le#i cYcksa dks n'kkZ;svuqlkj ,d cSVjh ds lkFk tksM+rs gSA cYcksa }kjk miHkksx esayh xbZ dqy 'kfDr gS :-

220V

(1) 75 watt (2) 400 watt(3) 300 watt (4) 400/3 watt

Kota/01CE3140614/27


Target : JEE(Main) 2015/16-03-2015

9. A 50 W bulb is in series with a room heater and thecombination is connected across the mains. To getmax. heater output the 50 W bulb should bereplaced by :-(1) 25 W (2) 10 W(3) 100W (4) 200W

10. 12 cells each having the same emf are connectedin series and are kept in a closed box. Some of thecells are wrongly connected. This battery isconnected in series with an ammeter and two cellsidentical with each other and also identical withthe previous cells. The current is 3 A when theexternal cells aid this battery and is 2A when thecells oppose the battery. How many cells in thebattery are wrongly connected :-(1) one (2) two(3) three (4) none

11. A potentiometer wire of length 100 cm has aresistance of 10 ohm. It is connected in series witha resistance and an accumulator of emf 2V and ofnegligible internal resistance. A source of emf10 mV is balanced against a length of 40 cm of thepotentiometer wire. What is the value of externalresistance :-(1) 890 W (2) 600 W(3) 640 W (4) 790 W

9. ,d 50 W dk cYc ,d dejs ds ghVj ds lkFk Js.khØe

esa gS rFkk ;g la;kstu L=ksr ls tqM+k gSA ghVj ls egÙke

fuxZeu ds fy, 50W ds cYc dks fuEu cYc ls cnyuk

pkfg, :-

(1) 25 W (2) 10 W

(3) 100W (4) 200W10. leku foñokñcy ds 12 lSyksa dks Js.khØe esa tksM+dj ,d

cUn cDls esa j[kk tkrk gSA dqN lsy xyr tqM + x;s gSaA bl

cSVjh ds lkFk nks le#i lsyksa (tks igys okys lsyksa ds

lkeu gS) o ,d vehVj dks Js.khØe esa tksM+k x;k gSA

tc ; s lsy cSVjh dks lg;ksx djrs gS] rks ifjiFk esa /kkjk

3A gS rFkk tc ; s fojks/k djrs gS rks /kkjk 2A gSA cSVjh esa

tqM+s xyr lsy gS :-(1) ,d (2) nks

(3) rhu (4) dksbZ ugha

11. ,d 100 cm yEcs foHkoekih rkj dk izfrjks/k 10 vkse

gSA ;g ux.; vkUrfjd izfrjks/k o fo-ok-cy 2V dh ,d

cSVjh rFkk ,d izfrjks/k ds lkFk Js.khØe esa tksM+k tkrk gSA

10 mV fo-ok-cy dk ,d L=ksr foHkoekih rkj dh

40 cm yEckbZ ij lUrqfyr gSA cká izfrjks/k dk eku gS :-

(1) 890 W (2) 600 W

(3) 640 W (4) 790 W

dksbZ Hkh iz'u Key Filling ls xyr ugha gksuk pkfg,A

5/27Kota/01CE314061



12. Five capacitors together with their capacitances areshown in the adjoining figure. The potentialdifference between the points A and B is 60 volt.The equivalent capacitance between the point Aand B and charge on capacitor 5mF will berespectively :-

12 Fm 10 Fm 5mF 9mF 8mFA

B(1) 44mF, 30mC (2) 16mF, 150mC(3) 15mF, 200mC (4) 4mF, 50mC

13. A capacitor of capacitance 10 mF is connected to abattery of emf 2V. It is found that it takes 50 msfor the charge on the capacitor to become 12.6 mC.Then the resistance of the circuit is :(Take 1/e = 0.37):-(1) 4 kW (2) 5 kW (3) 6 kW (4) 7 kW

14. Four capacitors of capacitance 10mF and a batteryof 200V are arranged as shown. How much chargewill flow through AB after the switch S is closed :-

S

B A(1) 6000 mC (2) 4500 mC(3) 3000 mC (4) 4000 mC

12. ik¡p la/kkfj=ks a dks mudh /kkfjrk ds lkFk fp=kuqlkj

O;ofLFkr fd;k x;k gSA fcUnqvksa A o B ds chp foHkokUrj

60 oksYV gSA A o B ds chp rqY; /kkfjrk rFkk 5mF laèkkfj=

ij vkos'k Øe'k% gksaxs :-

12 Fm 10 Fm 5mF 9mF 8mFA

B

(1) 44mF, 30mC (2) 16mF, 150mC(3) 15mF, 200mC (4) 4mF, 50mC

13. 10 mF /kkfjrk ds la/kkfj= dks 2V fo-ok-cy cSVjh lstksM+us ij ;g ik;k tkrk gS fd la/kkfj= 12.6mC rdvkosf'kr gksus esa 50 ms le; ysrk gS rks ifjiFk dk izfrjksèkgS (1/e = 0.37):-(1) 4 kW (2) 5 kW(3) 6 kW (4) 7 kW

14. 10mF /kkfjrk ds pkj la/kkfj= o 200V dh ,d cSVjhfp= esa n'kkZ;s vuqlkj O;ofLFkr gSA dqath S dks can djusij fdruk vkos'k AB ls izokfgrk gksxk :-

S

B A(1) 6000 mC (2) 4500 mC(3) 3000 mC (4) 4000 mC

Kota/01CE3140616/27


Target : JEE(Main) 2015/16-03-2015

15. The lower plate of a parallel plate capacitor issupported on a rigid rod. The upper plate issuspended from one end of a balance. The twoplates are joined together by a thin wire andsubsequently disconnected. The balance is thencounterpoised. Now a voltage V = 5000 volt isapplied between the plates. The distance betweenthe plates is d =5 mm and the area of each plate isA = 100 cm2. Then find out the additional massplaced to maintain balance. [All the elements otherthan plates are massless and nonconducting] :-

Rigid rod

(1) 44 g (2) 4.4 g(3) 0.44 g (4) 440 g

16. A circular loop is kept in that vertical plane whichcontains the north-south direction. It carries acurrent that is towards south at the topmost point.Let A be a point on axis of the circle to the east ofit and B a point on this axis to the west of it. Themagnetic field due to the loop :-

(1) is towards east at A and towards west at B

(2) is towards west at A and towards east at B

(3) is towards east at both A and B

(4) is towards west at both A and B

15. lekUrj iê la/kkfj= dh fupyh IysV ,d n`<+ NM + dslgkjs fLFkr gSA Åijh IysV ,d rqyk ds ,d Nksj ls yVdh

gSA nksuks a IysVsa ,d irys rkj }kjk tksM+h tkrh gS ,oa fQj

vyx dj nh tkrh gSA bl le; larqfyr gSA vc ,d

foHkokUrj V = 5000 oksYV dk foHkokUrj IysVks a ij

vkjksfir fd;k tkrk gS IysVksa ds e/; nwjh d = 5mm vkSj

izR;sd IysV dk {ks=Qy A = 100 cm2 gSa rks rqyk dkslUrqfyr j[kus ds fy;s vko';d vfrfjDr nzO;eku Kkrdhft,A IysVk sa ds vfrfjDr lHkh rRo nzO;ekughu o

vpkyd gS :-

(n<+ NM+)

(1) 44 g (2) 4.4 g(3) 0.44 g (4) 440 g

16. ,d o`Ùkkdkj ywi mÙkj&nf{k.k fn'kk esa fLFkr Å/okZ/kjry esa j[kk gqvk gSA blds mPpre fcUnq ij /kkjk dh fn'kk

nf{k.k dh vksj gSA ekufld bldh v{k ij fcUnq A iwoZ

dh vksj rFkk fcUnq B if'pe dh vksj gSA ywi ds dkj.k

pqEcdh; {ks= :-

(1) A ij iwoZ dh vksj rFkk B ij if'pe dh vksj gSA

(2) A ij if'pe dh vksj rFkk B ij iwoZ dh vksj gSA

(3) A rFkk B nksuksa ij iwoZ dh vksj gSA

(4) A rFkk B nksuksa ij if'pe dh vksj gSA

7/27Kota/01CE314061



17. A dipole having dipole moment p is placed in frontof a solid uncharged conducting sphere as shownin the diagram. The net potential at point A lyingon the surface of the sphere is :-

R rf

A

p

(1) 2kpcos

rf

(2) 2

2kpcos

rf

(3) zero (4) 2

22kpcos

rf

18. An infinitely long wire carrying current I is along

Y axis such that its one end is at point A(0, b) while

the wire extends upto + ¥. The magnitude of

magnetic field strength at point (a, 0) :-

¥

IA

(0,0) (a,0)

(1)0

2 2

I b14 a a b

æ öm+ç ÷

p +è ø(2)

0

2 2

I b14 a a b

æ öm-ç ÷

p +è ø

(3)0

2 2

I b4 a a b

æ ömç ÷

p +è ø(4) None of these

17. ,d f}/kz qo (f}/kz qo vk?kw.kZ p) dks Bksl vukosf'kr pkydxksys ds lkeus fp=kuqlkj j[kk tkrk gSA xksys ds i`"B ds

fcUnq A ij foHko gksxk :-

R rf

A

p

(1) 2kpcos

rf

(2) 2

2kpcos

rf

(3) 'kwU; (4) 2

22kpcos

rf

18. ,d vUkUr yEckbZ ds rkj esa /kkjk Y- v{k ds vuqfn'k cg

jgh gS rFkk ftldk ,d fljk fcUnq A(0, b) ij gS tc fd

nwljk fljk + ¥ rd vxzlj gSA fcUnq (a, 0) ij pqEcdh;

{ks= dk ifjek.k gS :-

¥

IA

(0,0) (a,0)

(1)0

2 2

I b14 a a b

æ öm+ç ÷

p +è ø(2)

0

2 2

I b14 a a b

æ öm-ç ÷

p +è ø

(3)0

2 2

I b4 a a b

æ ömç ÷

p +è ø(4) buesa ls dksbZ ugha

Kota/01CE3140618/27


Target : JEE(Main) 2015/16-03-201519. There exists a uniform magnetic and electric field

of magnitude 1 T and 1 V/m respectively along

positive y-axis. A charged particle of mass 1kg and

of charge 1 C is having velocity 1 m/sec along x-

axis and is at origin at t = 0. Then the co-ordinates

of particle at time p seconds will be :-

(1) (0, 1, 2) (2) (0, – p2/2,–2)

(3) (2, p2/2,2) (4) (0, p2/2, 2)

20. A horizontal metallic rod of mass 'm' and length 'l'

is supported by two vertical identical springs ofspring of spring constant 'K' each and natural lengthl0. A current 'i' is flowing in the rod in the directionshown. If the rod is in equilibrium then the lengthof each spring in this state is :-

K B Kl

i m

(1) 0i B mg

K-

+l

l (2) 0i B mg

2K-

+l

l

(3) 0mg i B

2K-

+l

l (4) 0mg i B

K-

+l

l

19. /kukRed y-v{k ds vuqfn'k 1 VSlyk rFkk 1 oksYV@eh- ds

leku leku pqEcdh; rFkk fo|qr {ks= vfLrRo esa gSaA

1kg nzO;eku rFkk 1C vkos'k ds vkosf'kr d.k dk x-v{k

ds vuqfn'k osx 1 eh-@ls- gS rFkk t = 0 ij ;g ewyfcUnq

ij gSA rc d.k ds le; p lsd.M ij funsZ'kkad gksaxs :-

(1) (0, 1, 2) (2) (0, – p2/2,–2)

(3) (2, p2/2,2) (4) (0, p2/2, 2)

20. 'm' nzO;eku rFkk 'l' yEckbZ dh {kSfrt /kkfRod NM + 'K'

fLizax fu;rkad okyh nks le#i m/okZ/kj fLizaxks a ij fLFkr

gSA fLiz ax viuh okLrfod yEckbZ l0 esa gSA fp=kuqlkj NM +

esa izokfgr /kkjk 'i' iznfr'kZr gSA ;fn NM + lkE;koLFkk esa gS

rks bl fLFkfr esa izR;sd fLaiz dh yEckbZ gS :-

K B Kl

i m

(1) 0i B mg

K-

+l

l (2) 0i B mg

2K-

+l

l

(3) 0mg i B

2K-

+l

l (4) 0mg i B

K-

+l

l

Use stop, look and go method in reading the question

9/27Kota/01CE314061



21. For the circuit shown in figure, the direction andmagnitude of the force on PQR is :-

Q

i

P R

B60°

(1) No resultant force act on the loop(2) ILB out of the page

(3)12

ILB into the page

(4) ILB into the page

22. A uniform, constant magnetic field Bur

is directedat an angle of 45° to the x-axis in the xy-plane,PQRS is a rigid square wire frame carrying a steadycurrent I0, with its centre at the origin O. At timet = 0, the frame is at rest in the position shown inthe figure, with its sides parallel to the x and y axes.Each side of the frame is of mass M and Length L

Sy

R

I0I0 x

O

P Q

(1)2

0BI L ˆ ˆ( i j)2

t = - +r (2)

20BI L ˆ ˆ(i j)2

t = -r

(3) 2

0BI L ˆ ˆ(i j)2

t = +r (4)

20BI L ˆ ˆ( i j)2

t = - -r

21. fp= esa iznf'kZr ifjiFk esa PQR cy dh fn'kk rFkk ifjek.kgS :-

Q

i

P R

B60°

(1) ywi ij dksbZ ifj.kkeh cy dk;Z ugha djsxk(2) ILB dkxt ls ckgj dh vksj

(3)12

ILB dkxt esa vanj dh vksj

(4) ILB dkxt esa vanj dh vksj

22. ,d ,dleku o fu;r pqEcdh; {ks= Bur

, XY-v{k ls45° dks.k ij dk;Zjr gSA PQRS rkj dk n`<+ ,oe~ oxkZdkjÝse gS] ftlesa LFkk;h /kkjk I0 izokfgr gks jgh gS rFkk Ýsedk dsUnz fcUnq O ij gSA le; t = 0 ij] Ýse fp= esanf'kZr LFkku voLFkk esa gS rFkk bldh Hkqtk;sa X o Y-v{kks ads lekUrj gSaA Ýse dh izR;sd Hkqtk dk nzO;eku M rFkkyEckb L gS :-

Sy

R

I0I0 x

O

P Q

(1)2

0BI L ˆ ˆ( i j)2

t = - +r (2)

20BI L ˆ ˆ(i j)2

t = -r

(3) 2

0BI L ˆ ˆ(i j)2

t = +r (4)

20BI L ˆ ˆ( i j)2

t = - -r

Kota/01CE31406110/27


Target : JEE(Main) 2015/16-03-2015

23. A circular coil of radius R and a current I, whichcan rotate about a fixed axis passing through itsdiameter is initially placed such that its plane liesalong magnetic field B. Kinetic energy of loopwhen it rotates through an angle 90° is: (Assumethat I remains constant) :-

(1) pR2BI (2) 2R BI2

p(3) 2pR2BI (4) 23 R I

2p

24. A rod of length l having uniformly distributedcharge Q is rotated about one end with constantfrequency 'f'. Its magnetic moment :-

(1) pfQl2 (2) 2fQ

3p l

(3) 22 fQ

3p l

(4) 2pfQl2

25. If the frequency of the source e.m.f. in an ac circuitis n, the power varies with a frequency :-(1) n (2) 2n (3) n/2 (4) zero

26. An LCR series circuit with 100 W resistance isconnected to an AC source of 200 V and angularfrequency 300 radians per second. When only thecapacitance is removed, the current leads thevoltage by 60°. Then the current and powerdissipated in LCR circuit are respectively :-(1) 1A, 200 watt (2) 1A, 400 watt(3) 2A, 200 watt (4) 2A, 400 watt

27. A series LCR circuit containing a resistance of120 ohm has angular resonance frequency4 × 103 rad s-1. At resonance, the voltage acrossresistance and inductance are 60V and 40Vrespectively. The values of L and C are respectively(1) 20 mH, 25/8 mF (2) 2mH, 1/35mF(3) 20 mH, 1/40 mF (4) 2mH, 25/8 nF

23. ,d o`Ùkkdkj dq.Myh dh f=T;k R rFkk blesa /kkjk I gS]lg blds O;kl ls xqtjus okys fLFkj v{k ds lkis{k ?kw.kZudj ldrh gS] izkjEHk esa bl izdkj j[kh gqbZ gS fd bldkry pqEcdh; {ks= B ds vuqfn'k gSA dq.Myh ;k ywi dhxfrt mtkZ D;k gksxh tc ;g 90° dks.k ls ?kwe tk,A(I dks fu;r ekurs gq,) :-

(1) pR2BI (2) 2R BI2

p(3) 2pR2BI (4) 23 R I

2p

24. 'l' yEckbZ dh ,d NM + dks] tks fd ,d leku#i ls Qvkos'k ls vkosf'kr gSA ,f fljs ls fu;r vko`fr ls ?kqek;ktkrk gS bldk pqEchd; vk?kq.kZ gksxk :-

(1) pfQl2 (2) 2fQ

3p l

(3) 22 fQ

3p l

(4) 2pfQl2

25. ,d izR;korhZ /kkjk ifjiFk esa L=ksr fo-ok-cy dh vko`frn gS rks 'kfDr fdl vko`fr ls ifjofrZr gksxh :-(1) n (2) 2n (3) n/2 (4) zero

26. ,d LCR Js.kh ifjiFk 100 W izfrjks/k ds lkFk 200 VrFkk 300 jsfM;u izfr lSd.M dh dks.kh; vko`fÙk okysizR;korhZ /kkjk L=ksr tqM+k gSA tc ifjiFk ls dsoy laèkkfjrgVk;k tkrk gS rks /kkjk foHko ls 60° i'pxkeh gks tkrh gSAtc dsoy izsjdRo gVk;k tkrk gS rks /kkjk foHko ls 60°vxzxkeh gks tkrh gS rks LCR ifjiFk esa /kkjk rFkk 'kfDrO;; Øe'k% gS :-(1) 1A, 200 watt (2) 1A, 400 watt(3) 2A, 200 watt (4) 2A, 400 watt

27. ,d Js.kh LCR ifjiFk esa 120 vkse dk izfrjks/k tqM+k gSrFkk bldh dks.kh; vuquknh vko`fÙk 4 × 103 jsfM;u@lSñ

gSA vuqukn ij izfrjks/k rFkk iz sjdRo ds fljksa ij foHko

Øe'k% 60V rFkk 40V gS rks L rFkk C ds eku Øe'k% gS

(1) 20 mH, 25/8 mF (2) 2mH, 1/35mF(3) 20 mH, 1/40 mF (4) 2mH, 25/8 nF

11/27Kota/01CE314061



28. A transformer is used to light a 140 watt, 24 voltlamp from 240 V AC mains. The current in themain cable is 0.7 amp. The efficiency of thetransformer is :-(1) 48% (2) 63.8%(3) 83.3% (4) 90%

29. Find the rms value for the saw-tooth voltage of peakvalue V0 from t = 0 to t = 2T as shown in figure :-+V0

–V0

V0 – –

+ +T2

32

T t2T

(1) V0 (2) 0V2

(3) 0V2

(4) 0V3

30. For an infinite line of charge having charge densityl lying along x-axis, the work required in movingcharge q from C to A along arc CA is :-

B

a

Aa

C

++++++++++++ X

(1) eq log 2

0

lpe (2) e

q log 24 0

lpe

(3) eq log 2

4 0

lpe (4) e

q 1log2 20

lpe

28. 140 watt, 24 volt ds ySEi dks izdkf'kr djus ds fy,VªkalQkeZj dk mi;ksx djrs gq, 240 V ds eq[; izR;korhZ/kkjk L=ksr ls tksM+k tkrk gSA eq[; dscy (rkj) esa /kkjk0.7 amp gS rks VªkalQkeZj dh n{krk gS :-(1) 48% (2) 63.8%(3) 83.3% (4) 90%

29. fp=kuqlkj 'kh"kZ eku V0 okys f=Hkqtkdkj foHko dkoxZekè; ewy t = 0 ls t = 2T ds fy, Kkr djks :-

+V0

–V0

V0 – –

+ +T2

32

T t2T

(1) V0 (2) 0V2

(3) 0V2

(4) 0V3

30. x-v{k ds vuqfn'k vkos'kksa dh vUkUr yEch J`a[kyk dkvkos'k ?kuRo l gSA pki CA ds vuqfn'k vkos'k q dks Cls A rd ys tkus ds fym vko';d dk;Z gksxk :-

B

a

Aa

C

++++++++++++ X

(1) eq log 2

0

lpe (2) e

q log 24 0

lpe

(3) eq log 2

4 0

lpe (4) e

q 1log2 20

lpe

Kota/01CE31406112/27


Target : JEE(Main) 2015/16-03-2015

31. Which of the following can not give carbyl aminereaction -

(1) CH3NH2 (2) CH –C–NH3 2

O(3) PhNH2 (4) EtNH2

32. Formic acid and Acetic acid can be differentiatedby(1) Esterification (2) Tollen's test(3) 2,4-DNP test (4) NaHCO3 test

33. Which of the following is most reactive towardsnucleophilic attack

(1) CH –C–OC H3 2 5

O(2)

CH –C–Cl3

O

(3)CH –C–O–C–CH3 3

O O(4)

CH –C–NMe3 2

O34. Which compound will not undergo decarboxylation

on heating ?

(1) CH2

COOH

COOH(2) Ph–C–CH –COOH2

O(3) CH –C–CH –COOH3 2

NH

(4) CH –COOH2

CH –COOH2

31. fuEu esa ls dkSu dkfcZy ,ehu vfHkfØ;k ugh ns ldrk -

(1) CH3NH2 (2) CH –C–NH3 2

O

(3) PhNH2 (4) EtNH2

32. QkfeZd vEy rFkk ,lhfVd vEy es foHksn fd;k tk ldrkgS :-(1) ,LVjhdj.k (2) Vkysu ifj{k.k(3) 2,4-DNP ifj{k.k (4) NaHCO3 ifj{k.k

33. fuEu es ls dkSu ukfHkdLusgh vkØe.k ds fy, lokZf/kdfØ;k'khy gS :-

(1) CH –C–OC H3 2 5

O(2)

CH –C–Cl3

O

(3)CH –C–O–C–CH3 3

O O(4)

CH –C–NMe3 2

O34. fuEu esa ls dkSulk ;kSfxd xeZ djus ij fodkckZsfDlyhdj.k

ugh nsxk ?

(1) CH2

COOH

COOH

(2) Ph–C–CH –COOH2

O(3) CH –C–CH –COOH3 2

NH

(4) CH –COOH2

CH –COOH2

PART B - CHEMISTRY

Take it Easy and Make it Easy

13/27Kota/01CE314061



35. An optically active compound X has molecularformula C4H8O3. It evolves CO2 with NaHCO3. 'X'Reacts with LiAlH4 to give an achiral compound.'X' is

(1) CH –CH –CH–COOH3 2

OH

(2) CH –CH–COOH3

Me


CH –OH2

(4) CH –CH–CH –COOH3 2

OH

36. (x) C4H7 OCl 3NH¾¾¾® C4H9ON 2BrKOH¾¾¾®

CH3–CH2–CH2–NH2 compound x is

(1)

O

Cl

(2) C Cl

O

(3) OHCl

(4) CHOCl

35. ,d izdkf'kd lfØ; ;kSfxd X ftldk v.kqlw= C4H8O3gS ;g NaHCO3 ds lkFk CO2 xSl nsrk gSA 'X' dks LiAlH4ls vfHkd`r djus ij ,d vfdjsy ;kSfxd cukrk gS rks 'X'gSA

(1) CH –CH –CH–COOH3 2

OH


Me


CH –OH2

(4) CH –CH–CH –COOH3 2

OH

36. (x) C4H7 OCl 3NH¾¾¾® C4H9ON 2BrKOH¾¾¾®

CH3–CH2–CH2–NH2 ;kSfxd x gSA

(1)

O

Cl

(2) C Cl

O

(3) OHCl

(4) CHOCl

Kota/01CE31406114/27


Target : JEE(Main) 2015/16-03-2015

37. CCH3Ph

N–OHH SO2 4

D (B)

Product (A) & (B) respectively in the above reactionare :-

(1) Ph–C–NH–CH3

O,Ph–C–NH–CH3

O

(2) CH –C–NH–Ph3

O, CH –C–NH–Ph3

O



O

(4) CH –C–NH–Ph, Ph–C–NH–CH3 3

O O

38. The reaction of chloroform with alcoholic KOHand p-toludine forms

(1) CNH C3

(2) NCH C3

(3) N Cl2H C3

(4) NHCHCl2H C3

37. CCH3Ph

N–OHH SO2 4

D (B)

mijksDr vfHkfØ;k es (A) rFkk (B) gS :-


O,Ph–C–NH–CH3

O

(2) CH –C–NH–Ph3


O



O

(4) CH –C–NH–Ph, Ph–C–NH–CH3 3

O O

38. p-VkyqMhu ds lkFk DyksjksQkeZ rFkk ,Ydksgksfyd KOH dhvfHkfØ;k djus ij curk gSA

(1) CNH C3

(2) NCH C3

(3) N Cl2H C3

(4) NHCHCl2H C3

fdlh iz'u ij nsj rd :dks ugha A

15/27Kota/01CE314061



39.NH2

2NaNOHCl¾¾¾® (A) product of this

reaction is

(1)

(2)

(3)

(4)

40. In which of the following reaction cyanide will beobtained as is a major product

(1) Ph C CH3

O(i) LiAlH4

(i) H O3+

(2) Ph–C–NH2

ONaOHBr2

(3) Ph–C–NH2

OP O4 10

D

(4)

39.NH2

2NaNOHCl¾¾¾® (A) bl vfHkfØ;k dk

mRikn gSA

(1)

(2)

(3)

(4)

40. fuEu esa ls dkSulh vfHkfØ;k esa lkbukbM eq[; mRikn ds :iesa izkIr gksxk

(1) Ph C CH3

O(i) LiAlH4

(i) H O3+

(2) Ph–C–NH2

ONaOHBr2

(3) Ph–C–NH2

OP O4 10

D

(4)

Kota/01CE31406116/27


Target : JEE(Main) 2015/16-03-2015

41.D DOH

Compound D is :-

(1) (2)

(3)

CH –OH2

CH –OH2

(4)

42. NaOBr¾¾¾® (A) major

product of the reaction is :-

(1) (2)

(3) (4)

41.D DOH

;kSfxd D gS :-

(1) (2)

(3)

CH –OH2

CH –OH2

(4)

42. NaOBr¾¾¾® (A) vfHkfØ;k dk

mRikn gS :-

(1) (2)

(3) (4)

LoLFk jgks] eLr jgks rFkk i<+kbZ esa O;Lr jgks A

17/27Kota/01CE314061



43. A weak acid of dissociation constant 10–5 is beingtitrated with aqueous NaOH solution. The pH at thepoint of one-third neutralization of the acid will be:-(1) 5 + log 2 – log 3 (2) 5 – log 2(3) 5 – log 3 (4) 5 – log 6

44. What is the conjugate acid of HPO42– ?

(1) H3PO4 (2) H2PO4–

(3) H3O+ (4) PO43–

45. For the reversible reaction in equilibrium

N2(g) + O2(g) 1

2

k

k�� 2NO(g)

C0 = 32.1 10 tCe-- ´ for the forward reaction and

C'0 = 44.2 10 tC'e

-- ´ for the backward reaction, henceKc for the above equilibrium is :-(1) 5.0 (2) 2.0(3) 0.5 (4) 2

46. At a certain temperature the equilibrium constantKc is 0.25 for the reaction

A2(g) + B2(g) � C2(g) + D2(g)If we take 1 mole of each of the four gases in a 10litre container, what would be equilibriumconcentration of A2(g) ?(1) 0.331 M (2) 0.033 M(3) 0.133 M (4) 1.33 M

47. In which reaction will an increase in the volumeof the container favour the formation of products?(1) C(s) + H2O(g) � CO(g) + H2(g)(2) H2(g) + I2(g) � 2HI(g)(3) 4NH3(g) + 5O2(g) � 4NO(g) + 6H2O(l)(4) 3O2(g) � 2O3(g)

43. fo;kstu fLFkjkad 10–5 dk ,d nqcZy vEy] tyh; NaOHfoy;u ds lkFk vuqekfir gksrk gSA vEy ds ,d&frgkbZmnklhuhdj.k ds fcUnq ij pH gksxk :-(1) 5 + log 2 – log 3 (2) 5 – log 2(3) 5 – log 3 (4) 5 – log 6

44. HPO42– dk la;qXeh vEy D;k gS\

(1) H3PO4 (2) H2PO4–

(3) H3O+ (4) PO43–

45. lkE; esa mRØe.kh; vfHkfØ;k

N2(g) + O2(g) 1

2

k

k�� 2NO(g)

ds fy,] vxz vfHkfØ;k ds fy, C0 = 32.1 10 tCe-- ´ rFkk

izrhi vfHkfØ;k ds fy , C'0 = 44.2 10 tC'e

-- ´ gSA vr%mijksDr lkE; ds fy, Kc gS :-(1) 5.0 (2) 2.0(3) 0.5 (4) 2

46. vfHkfØ;k A2(g) + B2(g) � C2(g) + D2(g)ds fy, ,d fuf'pr rki ij lkE ; fLFkjkad Kc 0.25 gSA;fn ge ,d 10 yhVj ik= esa pkjksa xSlksa ds izR;sd ds 1 eksyysrs gS] rks A2(g) dh lkE; lkanzrk D;k gksxh\(1) 0.331 M(2) 0.033 M(3) 0.133 M(4) 1.33 M

47. fdl vfHkfØ;k esa ik=k ds vk;ru esa o`f¼] mRiknksa ds fuekZ.kds vuqdwy gS\(1) C(s) + H2O(g) � CO(g) + H2(g)(2) H2(g) + I2(g) � 2HI(g)(3) 4NH3(g) + 5O2(g) � 4NO(g) + 6H2O(l)(4) 3O2(g) � 2O3(g)

Kota/01CE31406118/27


Target : JEE(Main) 2015/16-03-2015

48. Hydrazine reacts with KIO3 in presence of HCl as;N2H4 + IO3

– + 2H+ + Cl– ¾® ICl + N2 + 3H2OThe equivalent masses of N2H4 and KIO3respectively are :-(1) 8, 87 (2) 8, 35.6 (3) 16, 53.5 (4) 8, 53.5

49. In which of the following reactions, hydrogen isacting as an oxidising agent ?(1) With iodine to give hydrogen iodide(2) With lithium to give lithium hydride(3) With nitrogen to give ammonia(4) With sulphur to give hydrogen sulphide

50. Consider the heterogeneous equilibrium in a closedcontainer

NH4HS (s) � NH3 (g) + H2S(g)If more NH4HS is added to the equilibrium(1) Partial pressure of NH3 increases(2) Partial pressure of H2S increases(3) Total pressure in the container increases(4) No effect on partial pressure of NH3 and H2S

51. A sample of 100 ml of 0.10 M acid HA(Ka = 1 × 10–5) is titrated with standard 0.2 M KOH.How many ml of KOH will have to be added whenthe pH in the titration flask will be 5.00 ?(1) 0 (2) 10 (3) 100 (4) 50

52. For preparing a buffer solution of pH 5 by mixingsodium acetate and acetic acid, the ratio of theconcentration of salt and acid should be (Ka = 10–5):-(1) 1 : 10 (2) 1 : 1 (3) 10 : 1 (4) 1 : 100

53. The compound which does not showparamagnetism is :-(1) [Cu(NH3)4]Cl2 (2) [Ag(NH3)2]Cl(3) NO (4) NO2

48. gkbMªkthu HCl dh mifLFkfr esa KIO3 ds lkFk fuEu izdkjls vfHkfØ;k djrk gS;N2H4 + IO3

– + 2H+ + Cl– ¾® ICl + N2 + 3H2ON2H4 o KIO3 dk rqY;kadh Hkkj Øe'k% gksrk gS :-(1) 8, 87 (2) 8, 35.6 (3) 16, 53.5 (4) 8, 53.5

49. fuEu esa ls fdl vfHkfØ;k esa gkbMªkstu vkWDlhdkjh ds lekudk;Z djrk gS\(1) vk;ksMhu ds lkFk gkbMªkstu vk;ksMkbM nsrk gS(2) yhfFk;e ds lkFk fyfFk;e gkbMªkbM nsrk gS(3) ukbVªkstu ds lkFk veksfu;k nsrk gS(4) lYQj ds lkFk] gkbMªkstu lYQkbM nsrk gS

50. ,d cUn ik= esa fo"kekaxh lkE; ij fopkj djrs gSNH4HS (s) � NH3 (g) + H2S(g)

lkE; ij NH4HS vksj feykus ij(1) NH3 dk vkaf'kd nkc c<+rk gS(2) H2S dk vkaf'kd nkc c<+rk gS(3) ik= esa dqy nkc c<+rk gS(4) NH3 rFkk H2S dk vkaf'kd nkc vizHkkfor jgrk gS

51. 100 ml 0.10 M vEy HA (Ka = 1 × 10–5) ds ,duewus dks ekud 0.2 M KOH ds lkFk vuqekfir fd;k tkrkgSA tc vuqekiu ykLd esa pH 5.00 gksxh rks KOH ds fdrusml feyk;s tk;sxsa \(1) 0 (2) 10 (3) 100 (4) 50

52. pH 5 ;qDr cQj foy;u cukus ds fy, lksfM;e ,lhVsVrFkk ,flfVd vEy feykus ij] yo.k rFkk vEy dh lkUnzrkdk vuqikr gksxk (Ka = 10–5)(1) 1 : 10 (2) 1 : 1 (3) 10 : 1 (4) 1 : 100

53. fuEufyf[kr essa ls dkSulk ;kSfxd vuqpqEcdRo iznf'kZr ughadjrk gS :-(1) [Cu(NH3)4]Cl2 (2) [Ag(NH3)2]Cl(3) NO (4) NO2

19/27Kota/01CE314061



54. When MnO2 is fused with KOH, a colouredcompound is formed. The product and itscolours is ?(1) K2MnO4, Green (2) KMnO4, purple(3) Mn2O3, brown (4) Mn3O4 black

55. [Cu(CN)4]2– is more stable that [Cu(NH3)4]2+

because :-(1) stability constant of [Cu(CN)4]2– is greater than

[Cu(NH3)4]2+

(2) CN– is stronger ligand than NH3(3) both (1) & (2)(4) none

56. Which of the following compounds is not coloured?(1) Na2[CuCl4] (2) Na2[CdCl2](3) K4[Fe(CN)6] (4) K3[Fe(CN)6]

57. In [Cr(C2O4)3]2–, the isomerism shown is :-(1) ligand (2) optical(3) geometrical (4) ionisation

58. Among the following elements, the one having thehighest ionisation energy is :-(1) (Ne) 3s2 3p3 (2) (Ne) 3s2 3p4

(3) (Ne) 3s2p5 (4) (Ar) 3d10 4s2 4p2

59. In which of the following process, energy isliberated :-(1) Cl ¾® Cl+ + e– (2) HCl ¾® H+ + Cl–

(3) Cl + e ¾® Cl– (4) O– + e ¾® O2–

60. The solubility of silver bromide in hypo solutionis due to the formation of :(1) Ag2SO3 (2) Ag2S2O3(3) [Ag(S2O3)]– (4) [Ag(S2O3)2]3–

54. tc MnO2 dks KOH ds lkFk laxfyr fd;k tkrk gS rks cuusokyk mRikn ,oa mldk jax D;k gksrk gS ?

(1) K2MnO4, Green (2) KMnO4, purple

(3) Mn2O3, brown (4) Mn3O4 black55. [Cu(CN)4]2– dk LFkkf;Ro [Cu(NH3)4]2+ ls T;knk

gS D;ksafd :-(1) [Cu(CN)4]2– dk LFkkf;Ro fLFkjkad [Cu(NH3)4]2+ ls

T;knk gS(2) CN– ] NH3 ls vf/kd izcy {ks= ligand gS(3) (1) ,oa (2) nksuksa(4) dksbZ ugha

56. fuEufyf[kr esa ls dkSu lk ;kSfxd jaxhu ugha gS ?(1) Na2[CuCl4] (2) Na2[CdCl2](3) K4[Fe(CN)6] (4) K3[Fe(CN)6]

57. [Cr(C2O4)3]2– esa fdl izdkj dh leko;ork iznf'kZr gksrh gS:-(1) fyxS.M (2) izdkf'kd(3) T;kferh; (4) vk;uhdkjd

58. fuEufyf[kr esa ls fdl vfHkfoU;kl dh vk;uu ÅtkZ mPpregSa :-(1) (Ne) 3s2 3p3 (2) (Ne) 3s2 3p4

(3) (Ne) 3s2p5 (4) (Ar) 3d10 4s2 4p2

59. fuEufyf[kr esa ls fdl izfØ;k esa ÅtkZ mRlftZr gksrhgS :-(1) Cl ¾® Cl+ + e– (2) HCl ¾® H+ + Cl–

(3) Cl + e ¾® Cl– (4) O– + e ¾® O2–

60. flYoj czksekbM] gkbiks foy;u esa ?kqy tkrk gS D;ksafd blladqy dk fuekZ.k gksrk gS :(1) Ag2SO3 (2) Ag2S2O3(3) [Ag(S2O3)]– (4) [Ag(S2O3)2]3–

Kota/01CE31406120/27


Target : JEE(Main) 2015/16-03-2015

PART C - MATHEMATICS

61. Let f be a function satisfying f(x)f(xy)y

= for all

positive real numbers x and y. If f(30) = 20, thenthe value of f(40) is-(1) 15 (2) 20 (3) 40 (4) 60

62. Let f (x) = sin2x + cos4x + 2 and g (x) = cos(cos x) + cos (sin x). Also let period of f (x) andg (x) be T1 and T2 respectively then(1) T1 = 2T2 (2) 2T1 = T2(3) T1 = T2 (4) T1 = 4T2

63. Let f : X ® Y be a function such that

f(x) x 2 4 x= - + - , then the set of X and Y forwhich f(x) is both injective as well as surjective,is-

(1) [2,4] and 2,2é ùë û (2) [3,4] and 2,2é ù

ë û(3) [2,4] and [1,2] (4) [2,3] and [1,2]

64. Let cos–1(x) + cos–1(2x) + cos–1(3x) = p. If xsatisfies the cubic ax3 + bx2 + cx – 1 = 0, then(a + b + c) has the value equal to-(1) 24 (2) 25(3) 26 (4) 27

65. The value of 1 1

n 1

n n 1tan tann 2 n 1

¥- -

=

æ ö-æ ö æ ö-ç ÷ç ÷ ç ÷+ +è ø è øè øå is

equal to-

(1) 4p

(2) 3p

(3) 2p

(4) 34p

61. ekuk Qyu f, lHkh /kukRed okLrfod la[;kvksa x rFkk

y ds fy, f(x)f(xy)

y= dks larq"V djrk gSA ;fn

f(30) = 20 gks] rks f(40) dk eku gksxk&

(1) 15 (2) 20 (3) 40 (4) 60

62. ekuk f (x) = sin2x + cos4x + 2 rFkk g (x) = cos (cosx) + cos (sin x) gSA ekuk f (x) rFkk g (x) ds vkorZdkyØe'k% T1 rFkk T2 gS] rks&(1) T1 = 2T2 (2) 2T1 = T2(3) T1 = T2 (4) T1 = 4T2

63. ekuk f : X ® Y ,d Qyu bl i zdkj gS fd

f(x) x 2 4 x= - + - gS] rks X rFkk Y dk leqPp;

ftlds fy, f(x) ,dSdh rFkk vkPNknd nksuksa gks] gksxk-

(1) [2,4] rFkk 2,2é ùë û (2) [3,4] rFkk 2,2é ù

ë û(3) [2,4] rFkk [1,2] (4) [2,3] rFkk [1,2]

64. ekuk cos–1(x) + cos–1(2x) + cos–1(3x) = p gSA ;fnx ?ku lehdj.k ax3 + bx2 + cx – 1 = 0 dks larq"V djrkgS] rks (a + b + c) dk eku gksxk -(1) 24 (2) 25(3) 26 ( 4) 27

65. 1 1

n 1

n n 1tan tann 2 n 1

¥- -

=

æ ö-æ ö æ ö-ç ÷ç ÷ ç ÷+ +è ø è øè øå dk eku gksxk -

(1) 4p

(2) 3p

(3) 2p

(4) 34p

J ges'kk eqLdjkrs jgs a A

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66. If ( )1 11 4x sin 2 tan 2 ,y sin tan2 3

- -æ ö= = ç ÷è ø, then-

(1) x = 1 – y(2) x2 = 1 – y(3) x2 = 1 + y(4) y2 = 1 – x

67. Let ƒ : R ® R be defined as ƒ(x) = 3–|x| – 3x + sgn(e–x) + 2(whre sgn x denotes signum function of x). Thenwhich one of the following is correct ?(1) ƒ is injective but not surjective(2) ƒ is surjective but not injective(3) ƒ is injective as well as surjective(4) ƒ is neither injective nor surjective

68.( )1

xx1

cot x 1 xLimit

2x 1secx 1

-

®¥-

+ -

ì ü+ï ïæ öí ýç ÷-è øï ïî þ

is equal to-

(1) 1 (2) 0(3) p/2 (4) non existent

69. Let f(x) be continuous and differentiable functionfor all reals.

f(x + y) = f(x) – 3xy + f(y). If h 0

f(h)Lim 7h®

= then

the value of f '(x) is-(1) – 3x (2) 7(3) –3x + 7 (4) 2f(x) + 7

66. ;fn ( )1 11 4x sin 2 tan 2 ,y sin tan2 3

- -æ ö= = ç ÷è ø gks] rks

(1) x = 1 – y(2) x2 = 1 – y(3) x2 = 1 + y(4) y2 = 1 – x

67. ekuk ƒ : R ® R, ƒ(x) = 3–|x| – 3x + sgn (e–x) + 2 }kjkifjHkkf"kr gS(tgk¡ sgn x, x ds flXue Qyu dks n'kkZrk) gSA rc fuEuesa ls dkSulk dFku lgh gksxk ?(1) ƒ ,dSdh ijUrq vkPNknd ugha gksxkA(2) ƒ vkPNknd ijUrq ,dSdh ugha gksxkA(3) ƒ ,dSdh rFkk vkPNknd nksuksa gksxkA

(4) ƒ uk rks ,dSdh uk gh vkPNknd gksxkA

68.( )1

xx1

cot x 1 xLimit

2x 1secx 1

-

®¥-

+ -

ì ü+ï ïæ öí ýç ÷-è øï ïî þ

dk eku gksxk -

(1) 1 (2) 0(3) p/2 (4) fo|eku ugha

69. ekuk f(x) lHkh okLrfod ds fy, larr~ rFkk vodyuh;Qyu g SA f(x + y) = f(x) – 3xy + f(y) ;fn

h 0

f(h)Lim 7h®

= gks] rks f '(x) dk eku gksxk&

(1) – 3x (2) 7(3) –3x + 7 (4) 2f(x) + 7

Kota/01CE31406122/27


Target : JEE(Main) 2015/16-03-201570. Let a = min · {x2 + 2x + 3, x Î R} and

b = x xx 0

sin x cosxLime e-® -

. Then the value of n

r n r

r 0a b -

=å

is

(1) n 1

n2 13 · 2

+ +(2)

n 1

n2 13 · 2

+ -

(3) n

n2 13 · 2

-(4)

n 1

n4 13 · 2

+ -

71. The value of 2cosec bx

x 0Lim(cosax)

® is-

(1) æ ö-ç ÷ç ÷è ø

2

28bae (2)

æ öç ÷ç ÷è ø

2

2–8a

be (3) æ öç ÷ç ÷è ø

2

2–a2be (4)


2

2– b2ae

72. If f (x) = 2x 1 2

2e (1 4x)

ln(1 x )- +

- for x ¹ 0, then f has

(1) an irremovable discontinuity at x = 0(2) a removable discontinuity at x = 0 and f (0) = – 4(3) a removable discontinuity at x = 0 and

f (0) = – 1/4(4) a removable discontinuity at x = 0 and f (0) = 4

73. Let tan(2p | sin q | ) = cot (2p | cos q | ), whereq Î R and f (x) = ( | sin q | + | cos q | )x. The value

of úû

ùêë

é¥® )x(

2Limx f

equals (Here [ ] represents

greatest integer function)(1) –2 (2) –1 (3) 0 (4) 1

70. ekuk a = U;wure {x2 + 2x + 3, x Î R} rFkk

b = x xx 0

sin x cosxLime e-® -

gSA rc n

r n r

r 0a b -

=å dk eku gksxk&

(1) n 1

n2 13 · 2

+ +(2)

n 1

n2 13 · 2

+ -

(3) n

n2 13 · 2

-(4)

n 1

n4 13 · 2

+ -

71.2cosec bx

x 0Lim(cosax)

® dk eku gksxk -

(1) æ ö-ç ÷ç ÷è ø

2

28bae (2)


2

2–8a

be (3) æ öç ÷ç ÷è ø

2

2–a2be (4)


2

2– b2ae

72. ;fn x ¹ 0 ds fy, f(x) = 2x 1 2

2e (1 4x)

ln(1 x )- +

- gks] rks f dh

(1) x = 0 ij ,d vfoLFkkiuh; vlar~rk gksxhA

(2) x = 0 ij foLFkkiuh; vlarr~k rFkk f (0) = – 4 gSA(3) x = 0 ij foLFkkiuh; vlarr~k rFkk f (0) = – 1/4 gSA

(4) x = 0 ij foLFkkiuh; vlarr~k rFkk f (0) = 4 gSA73. ekuk tan(2p | sin q | ) = cot (2p | cos q | ) tgk¡

q Î R rFkk f (x) = ( | sin q | + | cos q | )x gSA

úû

ùêë

é¥® )x(

2Limx f

dk eku gksxk (;gk¡ [ ] egÙke iw.kk±d Qyu

dks n'kkZrk gS) :-(1) –2 (2) –1 (3) 0 (4) 1

viuh {kerk dks iwjk olwyus dk iz;kl djs a A

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74. Let

1 b x 2acot , x 04 3

f(x) 2 , x 0n(1 cx) 2, 0 x

x 3

-é + -æ ö < <ç ÷ê è øêê= =ê -ê < <êë

l

If the

function f(x) is differentiable at x = 0, then find thevalue of (b2 – 2a + c6).(1) 18 (2) 38(3) 0 (4) 48

75. If ( )( )( ) ( )( )2x 2x 3y 1

x 1 x 2 x 3 x 2 x 3 x 3= + + +

- - - - - -,

then xy 'y is equal to (where y' =

dydx

) -

(1) 1 1 1

1 x 2 x 3 x+ +

- - -

(2) x x x

1 x 2 x 3 x+ +

- - -

(3) 1 2 3

1 x 2 x 3 x+ +

- - -

(4) 1 2 3

x 1 x 2 x 3+ +

- - -

74. ekuk

1 b x 2acot , x 04 3

f(x) 2 , x 0n(1 cx) 2, 0 x

x 3

-é + -æ ö < <ç ÷ê è øêê= =ê -ê < <êë

l

gSA ;fn

f(x), x = 0 ij vodyuh; gS] rks (b2 – 2a + c6) dk eku

Kkr dhft,A(1) 18 (2) 38(3) 0 (4) 48

75. ;fn ( )( )( ) ( )( )2x 2x 3y 1

x 1 x 2 x 3 x 2 x 3 x 3= + + +

- - - - - -

gks] rks xy 'y dk eku gksxk (tgk¡ y' =

dydx

) -

(1) 1 1 1

1 x 2 x 3 x+ +

- - -

(2) x x x

1 x 2 x 3 x+ +

- - -

(3) 1 2 3

1 x 2 x 3 x+ +

- - -

(4) 1 2 3

x 1 x 2 x 3+ +

- - -

Kota/01CE31406124/27


Target : JEE(Main) 2015/16-03-2015

76. If = + + + ¥y sec x sec x sec x ....... , then

value of ( )p æ ö-ç ÷

è øò/ 3

0

dy2y 1 dxdx

is equal to

(secx > 0)-

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

77. If xy sin 2xxe y e= + , then at x = 0, dydx

is

equal to-(1) –1 (2) 1 (3) 0 (4) 2

78. If the function ƒ(x) = 1 x

24e-æ ö

ç ÷è ø- + 1 + x + 2x

2 +

3x3

and

g(x) = ƒ–1(x); then the value of g¢76

æ ö-ç ÷è ø equals

(1) 15 (2)

15

- (3) 67 (4) 6

7-

79. Side of an equilateral triangle expands at the rateof 2 cm/s. The rate of increase of its area when eachside is 10 cm, is

(1) 10 2 cm2/sec (2) 10 3 cm2/sec(3) 10 cm2/sec (4) 5 cm2/sec

80. If at any point on a curve the subtangent andsub-normal are equal, then the length of the normalis equal to

(1) 2 ordinate (2) ordinate

(3) 2 abscissa (4) abscissa

76. ;fn = + + + ¥y sec x sec x sec x ....... gS] rks

( )p æ ö-ç ÷

è øò/ 3

0

dy2y 1 dxdx

, (secx > 0) dk eku

gksxk -

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

77. ;fn xy sin 2xxe y e= + gks] rks x = 0 ij dydx

dk eku

gksxk&(1) –1 (2) 1 (3) 0 (4) 2

78. ;fn Qyu ƒ(x) = 1 x

24e-æ ö

ç ÷è ø- + 1 + x + 2x

2 +

3x3

rFkk

g(x) = ƒ–1(x); rks g¢76

æ ö-ç ÷è ø dk eku cjkcj gksxk

(1) 15 (2)

15

- (3) 67 (4) 6

7-

79. ,d leckgq f=Hkqt dh Hkqtk 2 lseh/lSd.M dh nj ls c<+jgh gSA tc bldh Hkqtk dh yEckbZ 10 lseh gks] rks f=Hkqtdk {ks=Qy fdl nj ls c<+ jgk gS

(1) 10 2 lseh2/ls- (2) 10 3 lseh2/ls-(3) 10 lseh2/ls- (4) 5 lseh2/ls-

80. ;fn ,d oØ ds fdlh fcUnq ij v/k% Li'khZ rFkk v/kksyEccjkcj gS rc vfHkyEc dh yEckbZ cjkcj gS

(1) 2 dksfV (2) dksfV

(3) 2 Hkqt (4) Hkqt

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81. Let f '(x) > 0 and g'(x) < 0 for all xÎR then(1) f{g(x)} > f{g(x + 1)}(2) f{g(x – 1)} < f{g(x + 1)}(3) g{f(x – 1)} < g{f(x + 1)}(4) g{f(x)} > g{f(x – 1)}

82. The sum of the squares of intercepts on axesmade by a tangent at any point on the curvex2/3 + y2/3 = a2/3 is-(1) a (2) 2a (3) a2 (4) 2a2

83. If length of subnormal at any point of the curve

yn = an–1x is constant, then n equals-(1) 1 (2) 3 (3) 2 (4) 0

84. Let f(x) = ò xe (x –1)(x – 2) dx , Then f decreases in

the interval-(1) (–¥, –2) (2) (–2, –1)

(3) (1, 2) (4) (2, +¥)

85. In which interval f(x) = 2x2 – ln|x|, (x ¹ 0) ismonotonically decreasing-(1) )(–1/2, 1/2)(2) (–¥, –1/2)(3) (–¥, –1/2) È (0, 1/2)(4) (–¥, –1/2) È (1/2, ¥)

86. Let f(x) = (x – 4) (x – 5) (x – 6) (x – 7) then-(1) f¢(x) = 0 has four roots(2) Three roots of f¢(x) = 0 lie in (4, 5) È (5, 6) È (6, 7)(3) The equation f¢(x) = 0 has only one root(4) Three roots of f¢(x) = 0 lie in (3, 4) È (4, 5) È (5, 6)

81. ekuk fd f '(x) > 0 rFkk g'(x) < 0 rFkk xÎR ds fy,] rks(1) f{g(x)} > f{g(x + 1)}(2) f{g(x – 1)} < f{g(x + 1)}(3) g{f(x – 1)} < g{f(x + 1)}(4) g{f(x)} > g{f(x – 1)}

82. oØ x2/3 + y2/3 = a2/3 ds fdlh fcUnq ij [khaph xbZ Li'kZ js[kk}kjk funsZ'khZ v{kksa ij dkVs x, vUr% [k.Mksa dh yEckbZ;ksa dsoxk±s dk ;ksx gS&(1) a (2) 2a (3) a2 (4) 2a2

83. ;fn oØ yn = an–1x ds fdlh fcUnq ij v/kksyEc dh yEckbZvpj gS] rc n cjkcj gksxk-(1) 1 (2) 3 (3) 2 (4) 0

84. ekuk f(x) = ò xe (x –1)(x – 2) dx rc f fdl vUrjky esa

Îkleku gksxk-(1) (–¥, –2) (2) (–2, –1)(3) (1, 2) (4) (2, +¥)

85. fdl vUrjky esa f(x) = 2x2 – ln|x|, (x ¹ 0) ,dfn"VÎkleku gksxk-(1) )(–1/2, 1/2)(2) (–¥, –1/2)(3) (–¥, –1/2) È (0, 1/2)(4) (–¥, –1/2) È (1/2, ¥)

86. ekuk f(x) = (x – 4) (x – 5) (x – 6) (x – 7) rc-(1) f¢(x) = 0 pkj ewy j[krk gSA(2) f¢(x)=0 rhu ewy (4, 5) È (5, 6) È (6, 7) esa fo|eku

gSA(3) lehdj.k f¢(x) = 0 ,d ewy j[krk gSA(4) f¢(x) = 0 ds rhu ewy (3, 4) È (4, 5) È (5, 6) esa

fo|eku gSA

Kota/01CE31406126/27


Target : JEE(Main) 2015/16-03-2015

87. The maximum value of (1/x)x is-(1) e (2) (e)1/e

(3) (1/e)e (4) ee

88. If f(x) = x2 + 2bx + 2c2 and g(x) = –x2 – 2cx + b2

such that min. f(x) > max.g(x), then relation betweenb and c is(1) nor real value of b and c

(2) 0 < c < b 2

(3) |c| < |b| 2

(4) |c| > |b| 289. If the equation anxn + an–1 xn–1 + ..... + a1x = 0, a1

¹ 0, n ³ 2, has a positive root x = a, then the equationnanxn–1 + (n – 1) an–1xn–2 + ..... + a1 = 0has a positive root which is(1) Equal to a(2) Greater than or equal to a(3) Smaller than a(4) Greater than a

90. If f(x) = x3 – x2 + 100x + 1001; then(1) f(2010) > f(2011)(2) f(3x – 5) > f(3x)(3) f(x + 1) < f(x – 1)

(4) f1

1999æ öç ÷è ø > f

12000

æ öç ÷è ø

87. (1/x)x dk mfPp"B eku gS&(1) e (2) (e)1/e

(3) (1/e)e (4) ee

88. ;fn f(x) = x2 + 2bx + 2c2 rFkk g(x) = –x2 – 2cx + b2 ,sls

gSa fd min. f(x) > max.g(x) rks

(1) b, c ds dksbZ okLrfod eku fo|eku ugha gSA

(2) 0 < c < b 2

(3) |c| < |b| 2

(4) |c| > |b| 289. ;fn lehdj.k anxn + an–1 xn–1 + ..... + a1x = 0,

a1 ¹ 0, n ³ 2 dk ,d x = a /kukRed ewy gS] rks lehdj.knanxn–1 + (n – 1) an–1xn–2 + ..... + a1 = 0dk ,d /kukRed ewy fo|eku gksxk tks(1) a ds cjkcj gS(2) a ds cjkcj ;k cM+k gS(3) a ls NksVk gS(4) a ls cM+k gS

90. ;fn f(x) = x3 – x2 + 100x + 1001 gksa] rks(1) f(2010) > f(2011)(2) f(3x – 5) > f(3x)(3) f(x + 1) < f(x – 1)

(4) f1

1999æ öç ÷è ø > f

12000

æ öç ÷è ø

Your moral dutyis to prove that ALLEN is ALLEN

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

CAREER INSTITUTE Pat S KOTA (RAJASTHAN) CLASSROOM … · JEE (Main) : LEADER COURSE MAJOR TEST Test...

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Transcript of CAREER INSTITUTE Pat S KOTA (RAJASTHAN) CLASSROOM … · JEE (Main) : LEADER COURSE MAJOR TEST Test...