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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

SCORE – I DATE : 19 - 12 - 2014

0 1 C T 2 1 4 0 6 4

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) : ENTHUSIAST COURSE

TEST # 04 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, 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 esaHkkSfrd foKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj lHkh iz'uksa ds vadleku 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;sx kA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls½.kkRed vadu ugha gksxkA

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

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

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

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

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

IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

Enthusiast Course/Score-I/19-12-2014

1/30Kota/01CT214064

SPACE FOR ROUGH WORK

1. Two concentric conducting thin spherical shellshave radii a and b (a < b). If they are charged to+Q and –2Q, the E-r graph is :

(1)

E

r

b

a (2)

E

rba

(3)

E

rba (4)

E

rba

2. In a meter bridge the point D is neutral pointas shown in the figure.

BR S

CG 100 l1l1

D

( )(1) The meter bridge can have no other neutral

point for this set of resistances(2) When the jockey contacts a point on meter

wire left of D, current flows to B from thewire

(3) When the jockey contacts a point on themeter wire to the right of D, current flowsfrom B to the wire through galvanometer

(4) When R is increased, the neutral point shiftsto left

1. nks ladsUnzh; pkyd irys xksykdkj dks'kksa dh f=T;k a ob (a < b) gSA ;fn bUgsa +Q rFkk –2Q rd vkosf'krfd;k tk;s rks E-r vkjs[k gksxk%&

(1)

E

r

b

a (2)

E

rba

(3)

E

rba (4)

E

rba

2. iznf'kZr fp= esa ,d ehVj lsrq eas fcUnq D mnklhufcUnq gSA

BR S

CG 100 l1l1

D

( )

(1) izfrjks/kksa ds bl ; qXe ds fy;s ehVj lsrq dk vU; dksbZmnklhu fcUnq ugh gksxkA

(2) tc tkWdh ehVj rkj ij D ds cka;h vksj fdlh fcUnq ijLi'kZ djrh gS rks /kkjk rkj ls B dh vksj izokfgr gksrh gSA

(3) tc tkWdh ehVj rkj ij D ds nka;h vksj fdlh fcUnqij Li'kZ djrh gS rks /kkjk B ls xsYosuksehVj ls gksdjrkj dh vksj izokfgr gksrh gSA

(4) R dks c<+k;s tkus ij mnklhu fcUnq cka;h vksj foLFkkfirgksrk gSA

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

Kota/01CT2140642/30

Target : JEE (Main) 2015/19-12-2014


3. Which of the following graphs represent thevariation of power loss in the external load withexternal resistance R?

r

R

(1)

P

r

(2)

P

r

(3)

P

r

(4)

P

r

4. The maximum number of emission lines foratomic hydrogen that you would expect to seewith naked eye if the only electronic levelsinvolved are those shown in the figure, is

n=7n=6n=5

n=4

n=3

n=2

n=1

(1) 6 (2) 5 (3) 21 (4) ¥

3. fuEu esa ls dkSulk vkjs[k ckg~; izfrjks/k R ds lkFk ckg~;yksM esa 'kfDr âkl esa ifjorZu dks n'kkZrk gS\

r

R

(1)

P

r

(2)

P

r

(3)

P

r

(4)

P

r

4. ;fn fdlh ijekf.od gkbMªkstu ds mRltZu LisDVªe esadsoy fp= esa iznf'kZr ÅtkZ Lrj gh Hkkx ysa rks vki viuhvka[kks a ls fcuk fdlh vU; midj.k dh lgk;rk lsvfèkdre fdruh mRltZu js[kk,a ns[k ldrs gSa\

n=7n=6n=5

n=4

n=3

n=2

n=1

(1) 6 (2) 5 (3) 21 (4) ¥


3/30Kota/01CT214064


5. The radionuclide 116 C dacays by b+ emission.

Given that m( 116 C ) = 11.011434 u

m( 115 B ) = 11.009305 u

me = 0.000548 u, 1u = 931.5 MeV/c2

The Q-value of this decay process is :-

(1) 0.962 MeV

(2) 0.962 × 103 MeV

(3) 0.962 eV

(4) 06. A charged capacitor discharges through a

resistance R with time constant t. The two arenow placed in series across an AC source of

angular frequency 1w =

t. The impedance of the

circuit will be-

(1) R

2(2) R (3) 2R (4) 2R

7. A condenser of capacity 6 µF is fully chargedusing a 6-volt battery. The battery is removedand a resistanceless 0.2 mH inductor isconnected across the condenser. The currentwhich is flowing through the inductor whenone-third of the total energy is in the magneticfield of the inductor is :-

(1) 0.1 A (2) 0.2 A (3) 0.4 A (4) 0.6 A

5. fofdj.kd U;wDykbM 116 C dk {k; b+ ds mRltZu }kjk

gksrk gSA eku yks m( 116 C ) = 11.011434 u

m( 116 B ) = 11.009305 u

me = 0.000548 u, 1u = 931.5 MeV/c2

bl {k; izØe ds fy, Q-eku gksxk %&

(1) 0.962 MeV

(2) 0.962 × 103 MeV

(3) 0.962 eV

(4) 0

6. ,d vkosf'kr la/kkfj= dks ,d izfrjks/k R }kjk fujkosf'kr

fd;k tkrk gS] le; fLFkjkad t gSA vc nksuksa dks ,d izR;korhZ

lzksr ds lkFk Js.khØe esa tksM+ fn;k tkrk gS] lzksr dh dks.kh;

vko`fÙk 1

w =t

gSA ifjiFk dh izfrckèkk gksxh&

(1) R

2(2) R (3) 2R (4) 2R

7. ,d 6 oksYV dh cSVjh ds iz;ksx }kjk 6 µF /kkfjrk ds ,d

la/kkfj= dks iw.kZ vkosf'kr fd;k x;k gSA cSVjh dks gVkdj

0.2 mH ds izfrjks/k&jfgr izsjdRo dks la/kkfj= ds fljksa

ij tksM+k x;k gSA izsjdRo esa ls fdruh /kkjk izokfgr gks

jgh gksxh tcfd iwjh ÅtkZ dk ,d&frgkbZ Hkkx izsjdRo

ds pqEcdh; {ks= esa gksxk\

(1) 0.1 A (2) 0.2 A (3) 0.4 A (4) 0.6 A

Kota/01CT2140644/30

Target : JEE (Main) 2015/19-12-2014


8. The magnetic force between wires as shownin figure is :-

li

I

x

(1) 2

0iI xn

2 2xm +æ ö

ç ÷p è ø

ll (2)

20iI 2x

n2 2x

m +æ öç ÷p è ø

ll

(3) 0iI x

n2 x


ll (4) None of these

9. The figure shows an apparatus suggested byFaraday to generate electric current from aflowing river. Two identical conducting platesof length a and width b are placed parallelfacing one another on opposite sides of the riverfollowing with velocity u at a distance d apart.Now both the plates are connected by a loadresistance R. Then the current through the loadR is :- (Consider vertical component of themagnetic field produced by earth is B

v and the

resistivity of river water is r.)R

Bv

db

a

u

(1) vB ubR

(2) vB ubd

Rabr+

(3) vB udd

Rabr+

(4) None

8. fp= esa n'kkZ;s x;s rkjksa ds chp pqEcdh; cy gksxk%&

li

I

x

(1) 2

0iI xn

2 2xm +æ ö

ç ÷p è ø

ll (2)

20iI 2x

n2 2x


ll

(3) 0iI x

n2 x


ll (4) buesa ls dksbZ ugha

9. fp= esa QSjkMs }kjk izLrkfor ,d midj.k n'kkZ;k x;k gS]tks cgrh gqbZ unh ls fo|qr /kkjk mRiUu djus ds fy,iz;qä fd;k tkrk gSA ;g unh u osx ls cg jgh gSA yEckbZa rFkk pkSM+kbZ b okyh nks ,d tSlh pkyd IysVsa bl dpkSM+kbZ okyh unh ds nksuksa vksj ,d&nwljs ds lEeq[k rFkklekUrj j[kh gqbZ gSaA vc nksuksa IysVksa dks yksM izfrjks/k R}kjk vkil esa tksM+ nsrs gSaA ;fn i`Foh }kjk mRiUu pqEcdh;{k s= dk Å/ok Z/kj ?kVd B

v rFkk ty /k kjk dh

izfrjks/kdrk r gks rks yksM R esa ls izokfgr /kkjk gksxh %&R

Bv

db

a

u

(1) vB ubR

(2) vB ubd

Rabr+

(3) vB udd

Rabr+

(4)dksbZ ugh


5/30Kota/01CT214064


10. Apply Bohr’s atomic model to a lithium atom.Assuming that its two K-shell electrons are tooclose to nucleus such that nucleus and K-shellelectron act as a nucleus of effective positivecharge equivalent to electron. The ionizationenergy of its outermost electron is:-

(1) 30.6 eV (2) 3.4 eV

(3) 32.4 eV (4) 13.6 eV

11. A nucleus of mass M + Dm is at rest and decays

into two daughter nuclei of mass M 3M

&4 4

each. Speed of light is c. The speed of daughter

nucleus of mass M4

is:-

(1) cm

M mD+ D

(2) cm

M mD+ D

(3) c2 m

MD

(4) c6 m

MD

12. The half life of a radioactive substance is20 minutes. The approximate time interval(t2 – t1) between the time t2 when 3/4 of it hasdecayed and time t1 when 1/4 of it had decayed is

(1) 20

minn2l

(2) 20 n3

minn2l

l

(3) 20 min (4) 20 ln 2 min

10. cksgj ds ijek.kq izfr:i dks ,d yhfFk;e ijek.kq ds fy,iz;qDr dhft;sA ekukfd blds nks K-dks'k ds bysDVªkWuukfHkd ds brus lehi gSa fd ukfHkd rFkk ; s K-dks'k dsbysDVªkWu ,d ,sls ukfHkd dh Hkk¡fr O;ogkj djrs gSa] ftldkizHkkoh /kukRed vkos'k bysDVªkWu ds rqY; gSA blds lclsckáre bysDVªkWu dh vk;uu ÅtkZ gksxh %&(1) 30.6 eV (2) 3.4 eV(3) 32.4 eV (4) 13.6 eV

11. æO;eku M + Dm dk ,d ukfHkd fojke voLFkk esa gS

rFkk ;g æO;eku M 3M4 4

rFkk ds nks {k;tkr ukfHkdksa esa

{kf;r gksrk gSA izdk'k dh pky c gSA M4

æO;eku okys

{k;tkr ukfHkd dh pky gS :-

(1) cm

M mD+ D

(2) cm

M mD+ D

(3) c2 m

MD

(4) c6 m

MD

12. ,d jsfM;ks lfØ; inkFkZ dh v¼Z&vk;q 20 feuV gSA

blds 3/4 {kf;r gksus ds le; t2 vkSj 1/4 {kf;r gksus ds

le; t1 esa vUrj (t2 – t1) dk eku yxHkx gS%&

(1) 20

minn2l

(2) 20 n3

minn2l

l

(3) 20 min (4) 20 ln 2 min

Kota/01CT2140646/30

Target : JEE (Main) 2015/19-12-2014


13. After absorbing a slowly moving neutron ofmass mN (momentum ~0) a nucleus of mass Mbreaks into two nuclei of masses m1 and3m1(4m1 = M + mN), respectively. If the deBroglie wavelength of the nucleus with massm1 is l, then de Broglie wavelength of the othernucleus will be:-

(1) 9 l (2) 3 l (3) 3l

(4) l

14. The energy spectrum of b-particles [numberN(E) as a function of b-energy E] emitted froma radioactive source is :-

(1) N(E)

E0E

(2) N(E)

E0E

(3) N(E)

E0E

(4) N(E)

E0E

15. An a-particle of energy 4 MeV is scattered

through 180° by a fixed uranium nucleus. The

distance of the closest approach is of the order of

(1) 1 Å (2) 10–10 cm

(3) 10–12 cm (4) 10–15 cm

13. ,d /kheh xfr ls xfr'khy mN æO;eku ds U;wVªkWu

(laosx~0) dk vo'kks"k.k dj æO;eku M dk ,d ukfHkd

æO;eku Øe'k% m1 ,oa 3m1 (4m1 = M + mN) ds nks

ukfHkdksa esa VwVrk gS A ;fn æO;eku m1 okys ukfHkd dh

Mh&czkXyh rjaxnS/;Z l gS] rc nwljs ukfHkd dh Mh&czkXyh

rjaxnSè;Z gksxh :-

(1) 9 l (2) 3 l (3) 3l

(4) l

14. ,d jsfM;kslfØ; òksr ls mRlftZr b-d.kksa dk ÅtkZ

LisDVªe [la[;k N(E), b-ÅtkZ E ds Qyu ds :i esa]

gS%&

(1) N(E)

E0E

(2) N(E)

E0E

(3) N(E)

E0E

(4) N(E)

E0E

15. 4 MeV ÅtkZ dk dksbZ a-d.k fdlh fLFkj ;wjsfu;e ukfHkd

ls 180° }kjk izdhf.kZr gks tkrk gSA vR;Ur lehi mixeu

dh nwjh dh dksfV gS-

(1) 1 Å (2) 10–10 cm

(3) 10–12 cm (4) 10–15 cm


7/30Kota/01CT214064


16. The diagram shows the energy levels for an

electron in a certain atom. Which transition

shown represents the emission of a photon with

maximum wavelength?

n = 4n = 3

n = 2

n = 1I

II

III

IV

(1) III (2) IV (3) I (4) II

17. Three charged capacitors, C1 = 17µF,

C2 = 34µF, C

3 = 41µF and two open switches,

S1 and S

2 are assembled into a network with

initial voltages and polarities, as shown. Final

status of the network is attained when the two

switches, S1 and S

2 are closed. In the figure,

the final charge on capacitor C3 in mC, is closet to:

S2

S1

60V + + – –– – + +

+ +– –

90VC1

50VC3

C2

(1) Zero (2) 410 (3) 1200 (4) 3300

16. fdlh fuf'pr ijek.kq esa ,d bysDVªkWu ds ÅtkZ Lrj

fuEu vkjs[k esa n'kkZ, x, gSaA buesa ls dkSulk laØe.k

vfèkdre rjaxnS/; Z ds QksVkWu ds mRltZu dks fu#fir

djrk gS ?

n = 4n = 3

n = 2

n = 1I

II

III

IV

(1) III (2) IV (3) I (4) II

17. fp= es a iznf'k Zr tky es a rhu vkosf'kr la/kkfj =

C1 = 17µF, C

2 = 34µF, C

3 = 41µF rFkk nks [kqys

fLop S1 o S

2 yxs gSaA izkjfEHkd oksYVrk,sa rFkk /kzqo.krk,sa

fp= esa n'kkZ;h xbZ gSaA tc S1 o S

2 dks cUn dj nsrs gSa rks

bl tky dh vfUre voLFkk izkIr gksrh gSA laèkkfj= C3

ij vfUre vkos'k (mC esa) yxHkx gksxk %&

S2

S1

60V + + – –– – + +

+ +– –

90VC1

50VC3

C2

(1) 'kwU; (2) 410

(3) 1200 (4) 3300

Kota/01CT2140648/30

Target : JEE (Main) 2015/19-12-2014


18. A galvanometer G deflects full scale when apotential difference of 0.50 V is applied. Theinternal resistance of the galvanometer r

g is

25 ohms. An ammeter is constructed byincorporating the galvanometer and anadditional resistance R

S. The ammeter deflects

full scale when a measurement of 2.0 A is made.The resistance R

S is closest to :

(1) 0.25 W (2) 2.5 W

(3) 0.45 W (4) 0.1 W

19. In a potentiometer (see figure) a balance is

obtained at a length of 400 mm when using a

known battery of emf 1.6 volts. After removing

this battery, another battery of unknown emf is

used and balance is obtained at a length of

650 mm. The emf of unknown battery is

G

l

(1) 2.6 volt (2) 1.6 volt

(3) 3.4 volt (4) 4.7 volt

18. 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

19. iznf'kZr foHkoekih esa tc 1.6 volts fo|qr okgd cy

okyh Kkr cSVjh iz;qä dh tkrh gS rks lUrqyu yEckbZ

400 mm izkIr gksrh gSA vc bl cSVjh dks gVkdj vKkr

fo|qr okgd cy okyh ,d vU; cSVjh dks yxkus ij ;g

lUrqyu yEckbZ 650 mm izkIr gks rks bl vKkr cSVjh

dk fo|qr okgd cy gS %&

G

l

(1) 2.6 volt (2) 1.6 volt(3) 3.4 volt (4) 4.7 volt


9/30Kota/01CT214064


20. Statement-1: When ultraviolet light isincident on a photocell, its stopping potentialis V

0 and the maximum kinetic energy of the

photoelectrons is Kmax

. When the ultravioletlight is replaced by X-rays, both V

0 and K

max

increases.Statement-2: Photoelectrons are emittedwith speeds ranging from zero to a maximumvalue because of the range of frequenciespresent in the incident light.(1) Statement-1 is true, statement-2 is true and

statement-2 is correct explanation forstatement-1.

(2) Statement-1 is true, statement-2 is true andstatement-2 is NOT the correct explanationfor statement-1.

(3) Statement-1 is true, statement-2 is false.(4) Statement-1 is false, statement-2 is true.

21. A Young's double slit experiment uses amonochromatic source. The shape of theinterference fringes formed on a screen is :-

(1) circle (2) hyperbola(3) parabola (4) straight line

22. In a Young's double slit experiment the intensityat a point where the path difference is

6l (l being the wavelength of the light used) is

I. If I0 denotes the maximum intensity, I0/I isequal to:-

(1) 2 (2) 43

(3) 2 (4) 2

3

20. dFku-1: tc fdlh QksVks lsy ij ijkcSaxuh izdk'kvkifrr gksrk gS rks bldk fujks/kh foHko V

0 rFkk QksVks

bysDVªkWuksa dh vf/kdre xfrt ÅtkZ Kmax

gksrh gSA tcbl ijkcSaxuh izdk'k ds LFkku ij X-fdj.k iz;qä dhtkrh gS rks V

0 o K

max nksuksa c<+ tkrs gSaA

dFku-2: izdk'k bysDVªkWu 'kwU; ls vf/kdre eku ijklokyh pky ds lkFk mRlftZr gksrs gSaA , slk vkifrr izdk'kesa fo|eku fofHkUu ijkl dh vkofÙk;ksa ds dkj.k gksrkgSA(1) dFku–1 lR; gS] dFku–2 lR; gS; dFku–2

dFku–1 dh lgh O;k[;k djrk gSA(2) dFku–1 lR; gS] dFku–2 lR; gS; dFku–2

dFku–1 dh lgh O;k[;k ugha djrk gS(3) dFku–1 lR; gS, dFku–2 vlR; gSA(4) dFku–1 vlR; gS] dFku–2 lR; gSA

21. ;ax ds fdlh f}&f>jh iz;ksx esa ,do.khZ izdk'k òksr dkmi;ksx fd;k tkrk gSA insZ ij cuh O;fDrdj.k fÝUtksa dhvkÏfr gS-(1) oÙk (2) vfrijoy;(3) ijoy; (4) ljy js[kk

22. ;ax ds f}&fLyV iz;ksx e sa ,d fcUnq ij rhozrk tgk¡ iFkkUrj

6l (l izdk'k dh rjaxnS/; Z gS) gS] I gSA ;fn I0 vf/kdre

rhozrk gS] rc I0/I cjkcj gS %&

(1) 2 (2) 43

(3) 2 (4) 2

3

Kota/01CT21406410/30

Target : JEE (Main) 2015/19-12-2014


23. A transparent solid cylindrical rod has a

refractive index of 4

3. It is surrounded by a

medium of refractive index 2. A light ray is

incident at the mid-point of one end of the rod

as shown in the figure.

q

The incident angle q for which the light ray

grazes along the wall of the rod is :-

(1) sin–1 æ öç ÷è ø

12 (2) sin–1

æ öç ÷è ø

1

3

(3) sin–1æ öç ÷è ø

2

3(4) sin–1

æ öç ÷ç ÷è ø

32

24. A moving coil galvanometer has 100 equal

divisions. Its current sensitivity is 10 divisions

per milli ampere and voltage sensitivity is 2

divisions per milli volt. In order that each

division reads 1 V, the resistance in Ohm's

needed to be connected in series with the coil

will be:-

(1) 103 (2) 105

(3) 99995 (4) 9995

23. ,d ikjn'k Zd Bksl csyukdkj NM + dk viorZuk¡d

4

3 gSA ;g pkjksa rjQ 2 viorZukad okys ek/;e ls

f?kjh gSA NM + ds ,d fljs ds e/; fcUnq ij ,d izdk'k

dh fdj.k vkifrr gS] tSlk fd fp= esa fn[kk;k x;k gSA

q

og vkiru dks.k q ftlds fy, izdk'k dh fdj.k NM +

dh nhokj ds i`"BLi'khZZ gS] gS:-

(1) sin–1 æ öç ÷è ø

12 (2) sin–1

æ öç ÷è ø

1

3

(3) sin–1æ öç ÷è ø

2

3(4) sin–1

æ öç ÷ç ÷è ø

32

24. fdlh py dq.Myh /kkjkekih esa 100 cjkcj Hkkx gSaA

bldh èkkjk lqxzkfgrk 10 Hkkx izfr feyh,sfEi;j rFkk

oksYVrk lqxzkfgrk 2 Hkkx izfr feyhoksYV gSA bldk izR;sd

Hkkx 1 oksYV ikB~;kad i<+s] blds fy, bldh dq.Myh

ds lkFk Js.khØe esa la;ksftr vko';d izfrjks/k dk

vkse esa D;k eku gksxk%&

(1) 103 (2) 105

(3) 99995 (4) 9995


11/30Kota/01CT214064


25. A rectangular loop of wire, supporting a massm, hangs with one end in a uniform magnetic

field Br

pointing into the plane of the paper.A clockwise current is set up such thati > mg/Ba, where a is the width of the loop.Then :

× × × × × × × ××××

×××

×××

×××

×××

×××

×××

×××

ia

mg RS

y

x

P Q

(1) The weight decends due to a vertical forcecaused by the magnetic field

(2) The weight moves towards rightward(3) The weight rises due to a vertical force

caused by the magnetic field(4) None of these

26. Three alternating voltage sources V1 = 3 sinwt

volt, V2= 5 sin(wt + f

1) volt and

V3 = 5 sin(wt – f

2) volt connected across a

resistance 73

R = W as shown in the figure

(where f1 and f

2 corresponds to 30° and 127°

respectively). Find the peak current (in Amp)through the resistor.

V2

V3

V1

Ö7/3W

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

25. rkj ds ,d vk;rkdkj ywi ls æO;eku m yVdk gqvk gSrFkk bls dkxt ds ry esa vUnj dh vksj funsZf'kr le:i

pqEcdh; {ks= Br

esa ,d fljs ls yVdk;k tkrk gSA ;gk¡,d nf{k.kkorhZ /kkjk bl izdkj LFkkfir dh tkrh gS fdi > mg/Ba gks] tgk¡ a ywi dh pkSM+kbZ gS rc%&

× × × × × × × ××××

×××

×××

×××

×××

×××

×××

×××

ia

mg RS

y

x

P Q

(1) ;g Hkkj pqEcdh; {ks= ds dkj.k yxus okys ÅèokZèkjcy ds dkj.k uhps dh vksj tkrk gSA

(2) ;g Hkkj nka;h vksj xfr djrk gSA(3) ;g Hkkj pqEcdh; {ks= ds dkj.k yxus okys ÅèokZèkj

cy ds dkj.k Åij dh vksj tkrk gSA(4) bueas ls dksbZ ugha

26. rhu izR;korhZ oksYVrk lzksr V1 = 3sinwt oksYV,

V2=5sin(wt + f

1) oksYV rFkk V

3 =5sin(wt–f

2) volt

dks fp=kuqlkj 73

R = W izfrjks/k ds fljksa ij tksM+k x;k

gSA (;gka f1 rFkk f

2 ds eku Øe'k % 30° o 127° gS)

izfrjks/kd ls izokfgr f'k[kj /kkjk (Amp esa) Kkr dhft,A

V2

V3

V1

Ö7/3W

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

Kota/01CT21406412/30

Target : JEE (Main) 2015/19-12-2014


27. The shape of image formed of an object AB

due to the concave mirror shown in the figure

is best represented by (assume point A is at the

centre of curvature of the mirror) :-

45ºf

A

2f

B

(1) (2)

(3) (4)

28. Parallel rays are incident on a thick

plano-convex lens having radius of curvature

R, refractive index µ and thickness t. When rays

are incident on plane surface they converge at

a distance x from plane surface. When rays are

incident on curved surface then rays converge

at y distance from curved surface. Then

(1) x = y (2) x < y

(3) x > y (4) data insufficient

27. fp= esa iznf'kZr vory niZ.k }kjk cus fcEc AB ds

izfrfcEc dh vkdfr dks lokZf/kd lgh rjhds ls n'kkZus

okyk fodYi gS (ekuk fcUnq A niZ.k ds oØrk dsUæ

ij gS):-

45ºf

A

2f

B

(1) (2)

(3) (4)

28. ,d eksVs leryksÙky ysUl dh oØrk f=T;k R, viorZukad

µ o eksVkbZ t gSA bl ysUl ij lekUrj fdj.ksa vkifrr

gksrh gSA tc ;s lery lrg ij vkifrr gksrh gS rks

lery lrg ls x nwjh ij vfHklfjr gks tkrh gSA tc ; s

oØh; lrg ij fxjrh gS rks oØh; lrg ls y nwjh ij

vfHklfjr gksrh gSA rc %&

(1) x = y (2) x < y

(3) x > y (4) vkadM+s vi;kZIr gSA


13/30Kota/01CT214064


29. An electron (mass m) with an initial velocity

v = n0 i is in an electric field E = E

0 j . If

l0 = h/mn

0, it's de Breogile wavelength at time

t is given by

(1) l0

(2) 2 2 2

00 2 2

0

e E t1

ml +

n

(3)

0

2 2 20

2 20

e E t1

m

l

+n

(4)

02 2 2

02 2

0

e E t1

m

læ ö

+ç ÷nè ø

30. Two charged particles traverse identical helical

paths in a completely opposite sense in a

uniform magnetic field B = B0 k

(1) They have equal z-components of momenta

(2) They must have equal charges

(3) They necessarily represent a particle-

antiparticle pair

(4) The charge to mass ratio satisfy :

1 2

e e0

m mæ ö æ ö+ =ç ÷ ç ÷è ø è ø

29. æO;eku m okyk ,d bysDVªkWu izkjfEHkd osx v = n0 i ds

lkFk fo|qr {ks= E = E0 j esa gSA ;fn l

0 = h/mn

0 gks rks

le; t ij Mh&czkXyh rjaxnS/;Z gksxh%&

(1) l0

(2) 2 2 2

00 2 2

0

e E t1

ml +

n

(3)

0

2 2 20

2 20

e E t1

m

l

+n

(4)

02 2 2

02 2

0

e E t1

m

læ ö

+ç ÷nè ø

30. nks vkosf'kr d.k le:i pqEcdh; {ks= B = B0 k esa

iw.kZr;k foijhr fn'kk esa ,dtSls dq.Myhuqek iFk dk

vuqlj.k djrs gS%&

(1) buds fy;s laosx ds z-?kVd leku gksxsaA

(2) bu ij fuf'pr :i ls leku vkos'k gSA

(3) ;s vko';d :i ls ,d d.k&izfrd.k ; qXe cukrs gSA

(4) budk vkos'k&æO;eku vuqikr 1 2

e e0

m mæ ö æ ö+ =ç ÷ ç ÷è ø è ø

gSA

Kota/01CT21406414/30

Target : JEE (Main) 2015/19-12-2014


31. KMnO4 + HCl ® MnCl2 + Cl2 + KCl + H2O

In the above reaction how many moles of H2O

would be formed for each mole Cl2 liberated

(1) 5/4 (2) 4/5

(3) 5/8 (4) 8/5

32. An ideal gas is subjected to cyclic process

involving four thermodynamic states, the

amounts of heat (Q) and work (W) involved in

each of these process are -

Q1 = 6000 J, Q2 = - 5500 J;

Q3 = - 3000 J; Q4 = 3500 J

W1 = 2500 J; W2 = -1000 J;

W3 = -1200 J; W4 = x J.

The ratio of the net work done by the gas to

the total heat absorbed by the gas is h . The

values of |x| and h respectively are

(1) 500; 7.5% (2) 1300; 10.5%

(3) 1000; 21% (4) None

33. What is heat of atomisation of P4O6(s)

Given heat of sublimation of P4O6 is x kJ/mol

& P-O bond energy is y kJ/mol.

(1) x + 6y (2) x + y

(3) x + 8y (4) x + 12y

31. KMnO4 + HCl ® MnCl2 + Cl2 + KCl + H2O

mijksDr vfHkfØ;k esa izR;sd eksy Cl2mRltZu ds fy,

fdrus eksy H2O dk fuekZ.k gksxk

(1) 5/4 (2) 4/5

(3) 5/8 (4) 8/5

32. ,d vkn'kZ xSl dks ,d pØh; izØe esa fy;k x;k ftlesa

pkj Å"ekxfrd voLFkk,sa lfEefyr gSA bu izR;sd izØeksa

eas lfEefyr Å"ek (Q) rFkk dk;Z (W) dk eku fuEu

izdkj gS-

Q1 = 6000 J, Q2 = - 5500 J;

Q3 = - 3000 J; Q4 = 3500 J

W1 = 2500 J; W2 = -1000 J;

W3 = -1200 J; W4 = x J.

xSl }kjk fd;s x, dk;Z ls xSl }kjk vo'kksf"kr dqy Å"ek

dk vuqikr h gSA |x| rFkk h ds eku Øe'k% gS

(1) 500; 7.5% (2) 1300; 10.5%

(3) 1000; 21% (4) dksbZ ugha

33. P4O6(s) ds ijekf.o;dj.k dh Å"ek D;k gk sxh

fn;k gS P4O6 ds m/oZikru dh Å"ek x kJ/mol rFkk

P-O cU/k ÅtkZ y kJ/mol gS

(1) x + 6y (2) x + y

(3) x + 8y (4) x + 12y

PART B - CHEMISTRY


15/30Kota/01CT214064


34. The Henry law constant for dissolution of agas in aqueous medium is 3 × 102 atm. At whatpartial pressure of the gas, the molality of gas

in aqueous solution will be 5

m.9

(1) 1 (2) 4

(3) 3 (4) 68

35. In electrolysis of aq.NiI2 solution using Ni

electrodes on passing 1 equivalent charge, mass

of cathode

Given [Ni = 59, I = 127]

(1) Increase by 29.5 gm

(2) Increase by 59 gm

(3) Decreases by 127 gm

(4) None of these

36. For a cell,

4B(s) + 3O2(g) ® 2B2O3(g) ; Eºcell = 1.433 volt

What is molar entropy (in J/K) of oxygen gas

Given :2 3f B O

kJ( Hº ) (g) 840mol

D = -

2 3ºm B O(S ) (g) 280J / K mol= -

ºm B(S ) (s) 10 J / K mol= -

(1) 0.1963 (2) 1.963(3) 15.03 (4) 150.3

34. tyh; ek/;e esa ,d xSl ds foyk;du ds fy, gSujhfu;e dk fu;rkad 3 × 102 atm gSA xSl ds fdlvkaf'kd nkc ij] tyh; foy;u esa xSl dh eksyyrk

5m

9gksxh

(1) 1 (2) 4

(3) 3 (4) 68

35. 1 rqY;kad vkos'k dks izokfgr dj Ni bySDVªksMks ds mi;ksx

ls tyh;.NiI2 foy;u ds oS|qr vi?kVu esa] dSFkksM+ dk

nzO;eku &

fn;k gS [Ni = 59, I = 127]

(1) 29.5 gm ls c<+rk gS

(2) 59 gm ls c<+rk gS

(3) 127 gm ls ?kVrk gS

(4) buesa dksbZ ugha

36. ,d lSy ds fy,]

4B(s) + 3O2(g) ® 2B2O3(g) ; Eºcell = 1.433 volt

vkWDlhtu xSl dh eksyj ,.VªkWih (J/K esa) D;k gksxh

fn;k gS :2 3f B O

kJ( Hº ) (g) 840mol

D = -

2 3ºm B O(S ) (g) 280J / K mol= -

ºm B(S ) (s) 10 J / K mol= -

(1) 0.1963 (2) 1.963

(3) 15.03 (4) 150.3

Kota/01CT21406416/30

Target : JEE (Main) 2015/19-12-2014


37. What may be the first ionisation energy of Naif it has effective nuclear charge of 1.84assuming Bohr model of H-atom to be effectiveon it -

(1) 13.6 eV (2) 46 eV

(3) 1.51V (4) 5.1 eV

38. For strong electrolyte Lm = Lmº 10 C- . Thencalculate Lm at 0.01M if its value at 0.16Mis 200 Scm2 mol–1

(1) 203 Scm2 mol–1 (2) 193 Scm2 mol–1

(3) 207 Scm2 mol–1 (4) 197 Scm2 mol–1

39. Expansion of 1 mole of ideal gas is taking placefrom 2 litre to 8 litre at 300K against 1 atmpressure. Calculate DStotal in JK–1mol–1

(given R=8.3J

mol K-,1lit-atm=100 J,ln2 = 0.693)

(1) 11.5 (2) 13.5

(3) 9.5 (4) 22.5

40.M0.1M

+M0.01M

+M(s) M(s)

Initial EMF of above cell at 298K is

(1) –0.059 V (2) 0.059 V

(3) 0.59 V (4) 0.0059 V

37. Na dh izFke vk;uu ÅtkZ D;k gks ldrh gS ;fn bl

ij gkbMªkstu ijek.kq ds cksgj eksMy dks ekurs gq, bldk

izHkkoh ukfHkdh; vkos'k 1.84 gks -

(1) 13.6 eV (2) 46 eV

(3) 1.51V (4) 5.1 eV

38. izcy vi?kV~; ds fy, Lm = Lmº 10 C- gS rks0.01M ij Lm dh x.kuk dhft, ;fn 0.16M ij bldkeku 200 Scm2 mol–1 gks

(1) 203 Scm2 mol–1 (2) 193 Scm2 mol–1

(3) 207 Scm2 mol–1 (4) 197 Scm2 mol–1

39. 300K ij 1 eksy vkn'kZ xSl dk 1atm nkc ds fo:¼2 yhVj l s 8 yhVj rd i zlkj gk s jgk g SAJK–1mol–1 esa DStotal dh x.kuk dhft,A

(fn;k gS R=8.3J

mol K-,1lit-atm=100J, ln2 = 0.693)

(1) 11.5 (2) 13.5(3) 9.5 (4) 22.5

40.M0.1M

+M0.01M

+M(s) M(s)

298K ij mijksDr lSy dk izkjfEHkd EMF gS

(1) –0.059 V (2) 0.059 V

(3) 0.59 V (4) 0.0059 V


17/30Kota/01CT214064


41. What is the molecular shape of BrF3

(1) Bent T shape (2) Sea saw

(3) Square pyramid (4) Pyramidal

42. Select the correct statement about[Mn(CO)

4NO] which is diamagnetic -

(1) It is diamagnetic because Mn metal isdiamagnetic in free state

(2) It is diamagnetic because Mn is in +1

oxidation state in this complex

(3) NO is present as positive ligand

(4) All of these

43. How many geometrical isomers are possible forcomplex [Mab(AB)

2]±n

(1) 5 (2) 4 (3) 3 (4) 644. Which of the following reaction form N

2O :

(1) NH4NO

3 D¾¾®

(2) NH3(excess) + Cl

2 D¾¾®

(3) NH4NO

2 D¾¾®

(4) FeSO4 + HNO

3 ¾¾®

45. CO can be absorbed by

(1) Ammonical Cu2 Cl

2 solution

(2) Aqueous FeSO4 solution

(3) Aqueous H2NCONH

2 solution

(4) Aqueous KI solution

41. BrF3 dh vkf.od vkd`fr D;k gS

(1) eqM+h gqbZ T vkdfr (2) <sadqyh

(3) oxkZdkj fijkfeMh; (4) fijkfeMh;

42. [Mn(CO)4NO] tks izfrpqEcdh; gS] ds lUnHkZ esa lgh

dFku pqfu;sa -

(1) ;g izfrpqEcdh; gS D;ksafd eqDr voLFkk esa Mn

/kkrq izfrpqEcdh; gS

(2) ;g izfrpqEcdh; gS D;ksafd bl ladqy esa Mn, +1

vkWDlhdj.k voLFkk esa gSa

(3) NO, /kukRed fyxs.M ds :i esa mifLFkr gS

(4) mijksDr lHkh

43. ladqy [Mab(AB)2]±n ds fy;s fdrus T;kfefr;

leko;oh lEHko gSa(1) 5 (2) 4 (3) 3 (4) 6

44. fuEu esa ls dkSulh vfHkfØ;k esa N2O fufeZr gksrk gS:

(1) NH4NO

3 D¾¾®

(2) NH3(vkf/kD;) + Cl

2 D¾¾®

(3) NH4NO

2 D¾¾®

(4) FeSO4 + HNO

3 ¾¾®

45. CO dks vo'kksf"kr fd;k tk ldrk gS

(1) veksfu;ke; Cu2 Cl

2 foy;u }kjk

(2) tyh; FeSO4 foy;u }kjk

(3) tyh; H2NCONH

2 foy;u }kjk

(4) tyh; KI }kjk

Kota/01CT21406418/30

Target : JEE (Main) 2015/19-12-2014


46. ACl2 (excess) + BCl

2 ® ACl

4 + B ¯

BO 400ºCD

>¾¾¾¾®12 O

2 + B. If A & B are metal

then ore of B would be

(1) Siderite (2) Cinnabar

(3) Malachite (4) Horn silver

47. Which of the following reaction is not correctly

matched with its products :

(1) P4 + 10SO

2Cl

2 ¾¾® 4PCl

5 + 10SO

2

(2) P4 + 8SOCl

2 ¾¾® 4PCl

3 + 4SO

2 + 2S

2Cl

2

(3) 3CH3COOH+PCl

3 ¾¾®3CH3COCl+H

3PO

3

(4) Sn + PCl5 ¾¾® SnCl

2 + PCl

3

48. Which of the following statement is not

correct :

(1) Combustion of B2H

6 is exothermic

reaction

(2) B2H

6 reacts with CO give BH

3.CO

(3) K4[Fe(CN)

6] reacts with conc. H

2SO

4

gives CO

(4) syn gas is CO + N2

46. ACl2 (vkf/kD;) + BCl

2 ® ACl

4 + B ¯

BO 400ºCD

>¾¾¾¾®12 O

2 + B, ;fn A rFkk B /kkrq gS] rks

B dk v;Ld gksxk

(1) flMjkbV (2) flusckj

(3) esysdkbV (4) gkWuZ flYoj

47. fuEu esa ls dkSulh vfHkfØ;k dk blds mRiknksa ds lkFk

feyku] lgh ugha gS :

(1) P4 + 10SO

2Cl

2 ¾¾® 4PCl

5 + 10SO

2

(2) P4 + 8SOCl

2 ¾¾® 4PCl

3 + 4SO

2 + 2S

2Cl

2

(3) 3CH3COOH+PCl

3 ¾¾®3CH3COCl+H

3PO

3

(4) Sn + PCl5 ¾¾® SnCl

2 + PCl

3

48. fuEu esa ls dkSulk dFku lgh ugha gSa :

(1) B2H

6 dk ngu] m"ek{ksih vfHkfØ;k gSa

(2) B2H

6, CO ds lkFk fØ;k dj BH

3.CO nsrk gS

(3) K4[Fe(CN)

6], lkUæ H

2SO

4 ds lkFk fØ;k dj

CO nsrk gS

(4) CO + N2 , flu xSl gS


19/30Kota/01CT214064


49. The anode mud in the electrolytic refining ofsilver contains -(1) Zn, Cu, Ag, Au (2) Zn, Ag, Au(3) Au only (4) Cu, Ag, Au

50. Select the INCORRECT statement regardingfollowing reaction is :

XeF6

excess H O2'X' + HF

'Y' + HF2 mole H O2

(1) 'X' is an explosive(2) 'Y' is an oxyacid(3) Both are examples of non-redox reaction(4) X is XeO

3

51. + CH CH CH Cl+AlCl3 2 2 3 X (Major product)HNO + H SO3 2 4

(Major product)

Sn + HCl

Z

Y

SO3W

W is :

(1)

CH CH CH2 2 3

NH2

SO H3

(2)

CH

SO H3

NH2

CH3

CH3

(3)

CH CH CH2 2 3

SO H3

NH2

(4)

CH

NH2

SO H3

CH3

CH3

49. flYoj ds oS|qr vi?kVuh; 'kqf¼dj.k esa ,uksM eM+ esamifLFkr gS -(1) Zn, Cu, Ag, Au (2) Zn, Ag, Au(3) dsoy Au (4) Cu, Ag, Au

50. fuEu vfHkfØ;k ds lUnHkZ esa xyr dFku gS :

XeF6

H O 2 vfk/kD;

2 H O2eksy

'X' + HF

'Y' + HF

(1) 'X' ,d foLQksVd gS(2) 'Y' ,d vkWDlh&vEy gS(3) nksuksa ukWu&jsMkWDl vfHkfØ;k ds mnkgj.k gSa(4) X, XeO

3 gS

51. + CH CH CH Cl+AlCl3 2 2 3 X (eq[; mRikn)

HNO + H SO3 2 4

(eq[; mRikn)

Sn+ HCl

Z

Y

SO3W

W gS&

(1)

CH CH CH2 2 3

NH2

SO H3

(2)

CH

SO H3

NH2

CH3

CH3

(3)

CH CH CH2 2 3

SO H3

NH2

(4)

CH

NH2

SO H3

CH3

CH3

Kota/01CT21406420/30

Target : JEE (Main) 2015/19-12-2014


52. X Hg / H O2+

3+

2 butanone(Symmetrical alkyne)

Which compound reacts with Ag powderto give X(1) H3C – CCl3 (2) CHCl3

(3) CH3–CH2Cl (4) H3C–CºC–CH2Cl53. Benzylamine & aniline can be distinguished

by(1) Isocynide test (2) Musturd oil test

(3) Dye-azo test (4) Hinsberg test

54. NH C—

OMe C — Cl

+ AlCl3

3

P H O3 + Q

R

NaNO + HCl2

NCH3H3C

S(Azodye)

0º–5ºC

(Amine)

Correct option :

(1) S is N = NH N2

(2) P is NH—CO CMe3

(3) S is N = N NMe2

(4) R to S is carried out at pH = 9.5

52. X Hg / H O2+

3+

2 C;wVsukWu(lefer ,Ydkbu)

dkSulk ;kSfxd Ag pw.kZ ds lkFk fØ;k djds X nsrk gS&

(1) H3C – CCl3 (2) CHCl3

(3) CH3–CH2Cl (4) H3C–CºC–CH2Cl

53. csfUty,sehu rFkk ,fuyhu esa fdlds }kjk foHksnu fd;ktk ldrk gS&(1) vkblkslk;ukbM ijh{k.k (2) eLVMZ vkW;y ijh{k.k(3) Mkb&, stks ijh{k.k (4) fgUlcxZ ijh{k.k

54. NH C—

OMe C — Cl

+ AlCl3

3

P H O3 + Q

R

NaNO + HCl2

NCH3H3C

S(,stksMkb)

0º–5ºC

(,sehu)

lgh fodYi gS&

(1) S , N = NH N2 gSA

(2) P , NH—CO CMe3

gSA

(3) S , N = N NMe2 gSA

(4) R ls S rd pH = 9.5 ij gksrh gSA


21/30Kota/01CT214064


55. O

+O

OH–

P (dicarbonyl compound)(mf = C H O )10 14 2

P is obtained by reductive ozonolysis of

(1) (2)

(3) (4)

56. CH – C –OCH3 3

O

PQ

CH – C –H3

O

P & Q respectively is -

(1) DI–BALH followed by H2O &

Al(OCH2CH3)3

(2) LiAlH4 followed by H2O &

Al O – CHCH3

CH3 3

(3) DI–BALH followed by H2O & m–CPBA

followed by CH3OH/H+

(4) None of these

55.O+

OOH–

P

= C H O

(MkbZdkcksZfuy ;kSfxd)

(v.kq lw= )10 14 2

P fdlds vipk;h vkstksuh vi?kVu }kjk izkIr gksrk gS&

(1) (2)

(3) (4)

56. CH – C –OCH3 3

O

PQ

CH – C –H3

O

P rFkk Q Øe'k% gS&

(1) DI–BALH ds ckn H2O rFkk Al(OCH2CH3)3

(2) LiAlH4 ds ckn H2O rFkk Al O – CHCH3

CH3 3

(3) DI–BALH ds ckn H2O rFkk m–CPBA

ds ckn CH3OH/H+

(4) buesa ls dksbZ ugha

Kota/01CT21406422/30

Target : JEE (Main) 2015/19-12-2014


57. The correct order of basic strength.

N NH

NH

NNH

H N2 NH2

I II III IV

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

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

58. C7H6O (molecular formula of X which gives

–ve Fehling test but +ve tollen's test)

X Ac O2AcONa Y

1. SOCl22. Me NH2

Z

Correct option :

(1) X to Y is called claisen condensation

(2) Y is PhCH = CH–C

O

–OH

(3) Z is PhCH = CH–C

O

–NHMe

(4) X is

OH

57. {kkjh; lkeF;Z dk lgh Øe gS&

N NH

NH

NNH

H N2 NH2

I II III IV

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

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

58. C7H6O (X dk v.kq lw= tks –ve Qsgfyax ijh{k.k

ijUrq +ve VkWysUl ijh{k.k nsrk gSA)

X Ac O2AcONa Y

1. SOCl22. Me NH2

Z

lgh fodYi gS&

(1) X ls Y Dystu la?kuu dgykrk gSA

(2) Y , PhCH = CH–C

O

–OH gSA

(3) Z , PhCH = CH–C

O

–NHMe gSA

(4) X ,

OH gSA


23/30Kota/01CT214064


59. pH at isoeletric point of three amino acids are

given as follow.

pH at Isoelectric point

(Amino acid)I 6

(Amino acid)II 9.8

(Amino acid)III 3.9

(Amino acid)II is —(1) Glycine (2) Lysine

(3) Aspartic acid (4) Alanine

60.OHO

HOOH

OH

CH OH2

H+

MeOH

OHOHO

OHOMe

CH OH2

+ OHO

HOOH OMe

CH OH2

The intermediate formed in this reaction.

(1) Carbocation which does not undergo

resonance

(2) Carbocation which undergo

rearrangement

(3) Carbocation which undergoes resonance

(4) Free radical which does not undergo

resonance

59. rhu vehuksa vEyksa dh lefoHko fcUnq ij pH bl izdkjnh xbZ gS&

lefoHko fcUnq ij pH

(vehuksa vEy)I 6

(vehuksa vEy)II 9.8

(vehuksa vEy)III 3.9

(vehuksa vEy)II gS —¥

(1) Xykblhu (2) ykblhu

(3) ,sLikfVZd vEy (4) ,sykuhu

60.OHO

HOOH

OH

CH OH2

H+

MeOH

OHOHO

OHOMe

CH OH2

+ OHO

HOOH OMe

CH OH2

bl vfHkfØ;k esa cuus okyk e/;orhZ gS&

(1) dkcZ/kuk;u ftlesa vuqukn ugha gksrk gSA

(2) dkcZ/kuk;u tks iqufoZU;kflr gksrk gSA

(3) dkcZ/kuk;u ftlesa vuqukn gksrk gSA

(4) eqDr ewyd ftlesa vuqukn ugha gksrk gSA

Kota/01CT21406424/30

Target : JEE (Main) 2015/19-12-2014


61. Let z = i2i, then |z| is (where i = 1- )-

(1) 1 (2) ep (3) e–p (4) ep/2

62. Let a & b are roots of x2 + wx + w2 = 0, where w is

an imaginary cube root of unity and z = a9 + ib9,

then value of |z| is -

(1) 2 (2) 2 (3) 1 (4) 152

63. Let point P = a + ib, a, b > 0 undergoes thefollowing three transformations successively onargand plane

(I) Reflection about amp(z) = 4p

(II) Transformation through a distance 'b' unitalong the positive direction of real axis.

(III)Rotation through an angle 4p

about origin

in counter clockwise direction.

If final position of the point is given byQ 2 i 6= - + , then

(1) 1 32 2

a = - + (2) 3 1- = b

(3) 1 32 2

b = + (4) 3 1a = +

61. ekuk z = i2i gks] rks |z| gksxk (tgk¡ i = 1- )-

(1) 1 (2) ep (3) e–p (4) ep/2

62. ekuk a rFkk b lehdj.k x2 + wx + w2 = 0 ds ewy gSa]tgk¡ w bdkbZ dk dkYifud ?kuewy rFkk z = a9 + ib9 gks]rks |z| dk eku gksxk -

(1) 2 (2) 2 (3) 1 (4) 152

63. ekuk fcUnq P = a + ib, a, b > 0 vkxZ.M lery esafuEu rhu :ikarj.kksa ls xqtjrs gSa

(I) amp(z) = 4p

ds lkis{k ijkorZu

(II) okLrfod v{k dh /kukRed fn'kk ds vuqfn'k 'b'bdkbZ nwjh ls :ikarj.kA

(III)ewyfcUnq ds lkis{k okekorZ fn'kk esa 4p

dks.k ?kwerk

gSA

;fn fcUnq dh vfUre fLFkfr Q 2 i 6= - + }kjk nhtkrh gks] rks

(1) 1 32 2

a = - + (2) 3 1- = b

(3) 1 32 2

b = + (4) 3 1a = +

PART C - MATHEMATICS


25/30Kota/01CT214064


64. Let [.], {.} and sgn(.) denotes greatest integerfunction, fractional part function and signumfunction respectively, then value of determinant

1

[ ] amp(1 i 3) 11 0 2

sgn(cot x) 1 { }-

p +

p, is -

(1) 256

3 3p p

- + - (2) 25 5

3 3p p

- -

(3) 25 6

3 3p p

+ + (4) 255

3 3

3p p- + -

65. If 1 cot

P( )cot 1

qé ùq = ê ú- që û

and PQ = I, then

(cosec2q)Q is given by -(where I is an identity matrix of 2 × 2 order)

(1) P(q) (2) P(–q) (3) P(2q) (4) I

66. If system of equations kx + 2y – z = 2,(k – 1)x + ky + z = 1, x + (k – 1)y + kz = 3 hasonly one solution, then number of possible realvalue(s) of k is -

(1) 0 (2) 1 (3) 2 (4) infinite67. Let a, b, g, d are distinct imaginary roots of

z5 = 1, then value of

2 3 1

2 3 1

2 3 1

e e e ee e e ee e e e

a a a+ -d

b b b+ -d

g g g+ -d

---

is -

(1) 0 (2) e (3) 1 (4) e5

64. ekuk [.], {.} rFkk sgn(.) Øe'k% egÙke iw.kk±d iw.kk±dQyu] fHkUukRed Hkkx Qyu rFkk flXue Qyu dks n'kkZrkgks] rks lkjf.kd

1

[ ] amp(1 i 3) 11 0 2

sgn(cot x) 1 { }-

p +

pdk eku gksxk-

(1) 256

3 3p p

- + - (2) 25 5

3 3p p

- -

(3) 25 6

3 3p p

+ + (4) 255

3 3

3p p- + -

65. ;fn 1 cot

P( )cot 1

qé ùq = ê ú- që û

rFkk PQ = I gks] rks

(cosec2q)Q fuEu }kjk fn;k tk;sxk -(tgk¡ I, 2 × 2 dksfV dk rRled vkO;wg gS)

(1) P(q) (2) P(–q) (3) P(2q) (4) I

66. ;fn lehdj.k fudk; kx + 2y – z = 2,(k – 1)x + ky + z = 1, x + (k – 1)y + kz = 3 dkdsoy ,d gy gks] rks k ds lEHko okLrfod ekuksa dhla[;k gksxh -(1) 0 (2) 1 (3) 2 (4) vuUr

67. ekuk a, b, g, d; z5 = 1 ds fofHkUu dkYifud ewy gksa]

rks

2 3 1

2 3 1

2 3 1

e e e ee e e ee e e e

a a a+ -d

b b b+ -d

g g g+ -d

---

dk eku gksxk -

(1) 0 (2) e (3) 1 (4) e5

Kota/01CT21406426/30

Target : JEE (Main) 2015/19-12-2014


68. If a, b, g are roots of x3 – 2x2 + 3x – 2 = 0, then

the value of æ öab ag bg

+ +ç ÷a + b a + g b + gè ø is

(1) 134

(2) 2518 (3)

92 (4) None

69. If polynomial P(x) = x2 + ax + b has factors(x – a)(x – b), a, b Î R, then value of P(2) is -(1) 8 (2) 7 (3) 6 (4) 4

70. The number of arrangements of the letters ofthe word SATAYPAUL such that no two A aretogether and middle letter is consonant, is(1) (5!)2 (2) 5!6! (3) 5!4! (4) (60) × 5!

71. Mr. A has six children and atleast one child isa girl, then probability that Mr. A has 3 boysand 3 girls, is -

(1) 2063 (2)

16 (3)

511 (4)

132

72. If 33! is divisible by 2n, n Î N, then sum of allpossible values of n is-(1) 31 (2) 30 (3) 496 (4) 465

73. If m and s2 are the mean and variance ofrandom variable x, whose distribution isgiven by

X x 0 1 2 3 41 1 1P(X x) 0 03 2 6

=

=, then

(1) m = s2 = 2 (2) m = 1, s2 = 2(3) m = s2 = 1 (4) m = 2, s2 = 1

68. ;fn a, b, g lehdj.k x3 – 2x2 + 3x – 2 = 0 ds ewy

gksa] rks æ öab ag bg

+ +ç ÷a + b a + g b + gè ø dk eku gksxk

(1) 134

(2) 2518 (3)

92 (4) dksbZ ugha

69. ;fn cgqin P(x) = x2 + ax + b dk xq.ku[k.M(x – a)(x – b), a, b Î R gks] rks P(2) dk eku gksxk -(1) 8 (2) 7 (3) 6 (4) 4

70. 'kCn SATAYPAUL ds lHkh v{kjksa ds O;oLFkkvksa dhla[;k] rkfd dksbZ Hkh nks A lkFk&lkFk uk gks rFkk eè;v{kj O;atu gks] gksxh -(1) (5!)2 (2) 5!6! (3) 5!4! (4) (60) × 5!

71. fe- A ds N% cPps gSa rFkk de ls de ,d yM+dh gS] rcfe- A ds 3 yM+ds rFkk 3 yM+fd;k¡ gksus dh izkf;drkgksxh -

(1) 2063 (2)

16 (3)

511 (4)

132

72. ;fn 33!, 2n, n Î N }kjk foHkkftr gks] rks n ds lHkhlaHko ekuksa dk ;ksxQy gksxk&(1) 31 (2) 30 (3) 496 (4) 465

73. ;fn m rFkk s2 LoSPN vpj x, ftldk caVu fuEu gS] dsek/; rFkk izlj.k gS] rc

X x 0 1 2 3 41 1 1P(X x) 0 03 2 6

=

=

(1) m = s2 = 2 (2) m = 1, s2 = 2(3) m = s2 = 1 (4) m = 2, s2 = 1


27/30Kota/01CT214064


74. The interior angle of a 'n' sided convex polygonare in G.P.. The smallest angle is 1° and commonratio is 2° then number of possible valuesof 'n' is-(1) 0 (2) 1(3) 2 (4) none of these

75. If a + b + c = 50 and a, b, c are non negativeeven integers, then greatest value of ab2c is -(1) 97344 (2) 97656(3) 94864 (4) 94972

76. Let A(2, 3), B(4, 5) and let C = (x, y) be a pointsuch that (x – 2)(x – 4) + (y – 3)(y – 5) = 0. If

area of DABC = 2 sq. unit, then maximumnumber of positions of C in the xy plane is -(1) 1 (2) 2 (3) 3 (4) 4

77. Distance between two lines represented by thepair of straight lines 9x2 – 24xy + 16y2 + 3x–4y – 6 = 0 is -

(1) 1 (2) 15 (3) 2 (4)

25

78. If x2 + y2 + 2gx + 2ƒy + c = 0 is equation ofsmallest circle which is passing through (1, 2)and touches line x + y – 7 = 0, then value of(g + 2ƒ + 3c) is -(1) 25 (2) 17 (3) 30 (4) 23

79. The centres of a set of circles, each ofradius 2, lie on the circle x2 + y2 = 36. The locusof any point in the set is -

(1) 4 < x2 + y2 < 16 (2) 16 < x2 + y2 < 64

(3) 36 < x2 + y2 < 64 (4) 16 < x2 + y2 < 36

74. n Hkqth; mÙky cgqHkqt ds vUr%dks.k xq.kksÙkj Js.kh esa gSAlcls NksVk dks.k 1° rFkk lkoZvuqikr 2° gks] rks n dslEHko ekuksa dh la[;k gksxh -

(1) 0 (2) 1

(3) 2 (4) buesa ls dksbZ ugha

75. ;fn a + b + c = 50 rFkk a, b, c v½.kkRed leiw.kk±d gksa] rks ab2c dk egÙke eku gksxk -(1) 97344 (2) 97656(3) 94864 (4) 94972

76. ekuk A(2, 3), B(4, 5) rFkk C = (x, y) dksbZ fcUnq blizdkj gS fd (x – 2)(x – 4) + (y – 3)(y – 5) = 0 gSA;fn f=Hkqt ABC dk {ks=Qy = 2 oxZ bdkbZ gks] rksxy lery esa C dh fLFkfr;ksa dh vf/kdre la[;k gksxh -(1) 1 (2) 2 (3) 3 (4) 4

77. ljy js[kk ;qXe 9x2 – 24xy + 16y2 + 3x – 4y – 6 = 0

}kjk iznf'kZr nks js[kkvksa ds e/; nwjh gksxh -

(1) 1 (2) 15 (3) 2 (4)

25

78. ;fn fcUnq (1, 2) ls xqtjus okys rFkk js[kk x + y – 7 = 0dks Li'kZ djus okys lcls NksVs o`Ùk dk lehdj.kx2 + y2 + 2gx + 2ƒy + c = 0 gk s] rk s(g + 2ƒ + 3c) dk eku gksxk -(1) 25 (2) 17 (3) 30 (4) 23

79. 2 bdkbZ f=T;k ds o`Ùkksa dk ,d leqPp; ftuds dsUæ]o`Ùk x2 + y2 = 36 ij fLFkr gSA leqPp; ds fdlh fcUnqdk fcUnqiFk gksxk -(1) 4 < x2 + y2 < 16 (2) 16 < x2 + y2 < 64(3) 36 < x2 + y2 < 64 (4) 16 < x2 + y2 < 36

Kota/01CT21406428/30

Target : JEE (Main) 2015/19-12-2014


80. Total number of terms in the expansion of[(1 + x)100 + (1 + x2)100 + (1 + x3)100] is -

(1) 303 (2) 201 (3) 196 (4) 301

81. Coefficient of t12 in (1 + t2)6(1 + t6)(1 + t12) is -

(1) 24 (2) 21 (3) 22 (4) 23

82. The number of solution(s) of the equationln(lnx) = logxe is -

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

83. The tangents drawn at end points of latusrectumof parabola S = 0 intersect on x + y = 2 and(3, 2) is focus of parabola, then axis of parabolaS = 0 is -(1) x + y = 5

(2) 2x – y = 4

(3) x – y = 1

(4) cannot be determined

84. Normal at P, Q, R drawn to y2 = 4x whichintersect at (3, 0), then DPQR is -

(1) acute angled but not equilateral

(2) obtuse angled triangle

(3) equilateral triangle

(4) scalene triangle.

85. Length of common chord of the

ellipse 2 2(x 2) (y 2) 1

9 4- +

+ = and the circle

x2 + y2 – 4x + 2y + 4 = 0 is -

(1) 0 (2) 12

(3) 1 (4) 2

80. O;atd [(1 + x)100 + (1 + x2)100 + (1 + x3)100] dsizlkj esa inksa dh dqy la[;k gksxh -(1) 303 (2) 201 (3) 196 (4) 301

81. (1 + t2)6(1 + t6)(1 + t12) esa t12 dk xq.kkad gksxk -(1) 24 (2) 21 (3) 22 (4) 23

82. lehdj.k ln(lnx) = logxe ds gyks a dh la[;kgksxh -(1) 0 (2) 1 (3) 2 (4) vuUr

83. ijoy; S = 0 ds ukfHkyEc ds fljksa ij [khaph Li'kZjs[kk;sax + y = 2 ij izfrPNsn djrh gS rFkk (3, 2) ijoy; dhukfHk gks] rks ijoy; S = 0 dh v{k gksxh -

(1) x + y = 5

(2) 2x – y = 4

(3) x – y = 1

(4) Kkr ugha fd;k tk ldrkA

84. y2 = 4x ds fcUnq P, Q, R ij [khaps x;s vfHkyEc(3, 0) ij izfrPNsn djrs gks] rks f=Hkqt PQR gksxk -(1) U;wudks.k ijUrq leckgq f=Hkqt ugha(2) vf/kddks.k f=Hkqt(3) leckgq f=Hkqt(4) fo"keckgq f=Hkqt

85. nh?k Zo ` Ùk 2 2(x 2) (y 2) 1

9 4- +

+ = rFkk o `Ù k

x2 + y2 – 4x + 2y + 4 = 0 dh mHk;fu"B thok dh

yEckbZ gksxh -

(1) 0 (2) 12

(3) 1 (4) 2


29/30Kota/01CT214064


86. If a, b, g are positive number such that a + b = pand b + g = a, then tana is equal to -(where g ¹ np, n Î I)

(1) tan tan2

tanb + g

-g (2)

2 tan tantanb + g

g

(3) 2 tan tan

tanb + g

-g (4)

tan tantanb + g

g

87. If 2 2(3x 4y 1) (4x 3y 1) 1

100 225- - + -

- = , then

length of latusrectum of hyperbola is -

(1) 92 (2)

403 (3) 9 (4)

83

88. If line ax + by = 1 is normal to the hyperbola2 2

2 2

x y 1p q

- = , then 2 2

2 2

p qa b

- is equal to

(where a,b,p,q Î R+)-(1) 0 (2) 1

(3) (a2 + b2)2 (4) (p2 + q2)2

89. Number of solution(s) of the equation

1sin 2 cos 2 , 0,2 2

pé ùq + q = - qÎ ê úë û, is -

(1) 0 (2) 1 (3) 2 (4) 390. The curve represented by x = 5(cost + sint),

y = 3(cost – sint) is (where t is a parameter) -(1) pair of straight line (2) parabola

(3) ellipse (4) hyperbola

86. ;fn a, b, g /kukRed la[;k;sa bl izdkj gS fd a + b = prFkk b + g = a gks] rks tana dk eku gksxk -(tgk¡ g ¹ np, n Î I)

(1) tan tan2

tanb + g

-g (2)

2 tan tantanb + g

g

(3) 2 tan tan

tanb + g

-g (4)

tan tantanb + g

g

87. ;fn 2 2(3x 4y 1) (4x 3y 1) 1

100 225- - + -

- = gk s] rk s

vfrijoy; ds ukfHkyEc dh yEckbZ gksxh -

(1) 92 (2)

403 (3) 9 (4)

83

88. ;fn js[kk ax + by = 1, vfrijoy; 2 2

2 2

x y 1p q

- = ij

vfHkyEc gks] rks 2 2

2 2

p qa b

- cjkcj gksxk -

(tgk¡ a,b,p,q Î R+)(1) 0 (2) 1(3) (a2 + b2)2 (4) (p2 + q2)2

89. lehdj.k 1sin 2 cos 2 , 0,2 2

pé ùq + q = - qÎ ê úë û, ds

gyksa dh la[;k gksxh -

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

90. x = 5(cost + sint), y = 3(cost – sint) }kjk iznf'kZroØ gksxk (tgk¡ t izkpy gS) -(1) ljy js[kk ; qXe (2) ijoy;(3) nh?kZoÙk (4) vfrijoy;

Kota/01CT21406430/30

Target : JEE (Main) 2015/19-12-2014


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

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