allen t m paper code 01ct3140 68 · paper code 01ct3140 68 p nstructio ully. y ll 5 in specifical ....

CLASSROOM CONTACT PROGRAMME

FORM NUMBER

PAPER CODE 0 1 C T 3 1 4 0 6 8

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

DATE : 27 - 03 - 2015ALLEN JEE (Main) TEST

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

(ACADEMIC SESSION 2014-2015)

Leader & Enthusiast Course/Score/27-03-2015

H-1/31Kota/01CT314068

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

1. In a Fraunhofer's diffraction obtained by a

single slit aperture, the value of path difference

for nth order of minima is :–

(1) nl

(2) 2nl

(3) (2n 1)

2- l

(4) (2n–1)l

2. A positive charge q is placed in a spherical

cavity made in a positively charged sphere. The

centres of sphere and cavity are displaced by a

small distance lr

. Force on charge q is :

(1) in the direction parallel to vector lr

(2) in radial direction for all positions of q.

(3) in a direction which depends on the

magnitude of charge density in sphere

(4) direction can not be determined.

1. ,dy js[kk fNæ ls izkIr ÝkWugkWQj foorZu esa n osa Øe

ds fufEu"B ds fy, iFkkUrj dk eku gksrk gS :–

(1) nl

(2) 2nl

(3) (2n 1)

2- l

(4) (2n–1)l

2. ,d /kukos'k q dks /kukosf'kr xksys ds vUnj cuh xksykdkj

xqfgdk esa j[kk tkrk gSA xksys rFkk xqfgdk ds dsUæks dks vYi

nwjh lr

}kjk foLFkkfir fd;k tkrk gSA vkos'k q ij cy gS%&

(1) lfn'k lr

dh lekUrj fn'kk esa

(2) q dh lHkh fLFkfr;ksa ds fy;s f=T;h; fn'kk esa

(3) ml fn'kk esa tks xksys esa vkos'k ?kuRo ds ifjek.k ij

fuHkZj djrh gSA

(4) bl cy dh fn'kk Kkr ugh dh tk ldrhA

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

Kota/01CT314068H-2/31

Target : JEE (Main) 2015/27-03-2015


3. In a uniform electric field, a cube of side 1 cmis placed. The total energy stored in the cube is8.85µJ. The electric field is parallel to four ofthe faces of the cube. The electric flux throughany one of the remaining two faces is :-

(1) -1

V m5 2

(2) -100 2V m

(3) -5 2V m

(4) -10 2V m

4. A constant potential difference is applied to the

ends of a graphite rod, whose resistance

decreases with a rise of temperature. The rod

can be (1) covered with asbestos or (2) left open

to atmosphere. Answer for steady state.

(1) in both cases power lost is same but

temperature of rod is more in case-1

(2) in case-1 power lost as well as temperature

of rod is more

(3) in case-2 power lost as well as temperature

of rod is more

(4) in case-1 power lost is more but temperatureof rod is less

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

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

{ks= ?ku ds pkj Qydksa ds lekUrj gSA 'ks"k cps nksuksa

Qydksa esa ls fdlh ,d ls fuxZr fo|qr ¶yDl gksxk%&

(1) -1

V m5 2

(2) -100 2V m

(3) -5 2V m

(4) -10 2V m

4. ,d xzsQkbV NM+ ftldk izfrjks/k rkieku esa of¼ ds lkFk

?kVrk tkrk gS] ds fljksa ij ,d fu;r foHkokUrj yxk;k tkrk

gSA bl NM+ dks (1) ,LcsLVkWl ls <dk tkrk gS vFkok (2)

okrkoj.k esa [kqyk NksM+ fn;k tkrk gSA LFkk;h voLFkk esa :-

(1) nksuksa fLFkfr;kssa esa 'kfä ákl leku gS ijUrq fLFkfr-1

esa NM+ dk rkieku vf/kd gSA

(2) fLFkfr-1 esa 'kfä ákl rFkk NM + dk rkieku nksuksa

vf/kd gSA

(3) fLFkfr-2 esa 'kfä ákl rFkk NM + dk rkieku nksuksa

vf/kd gSA

(4) fLFkfr-1 esa 'kfä ákl vf/kd ijUrq NM+ dk rkieku

de gSA


H-3/31Kota/01CT314068


5. The logic circuit shown has the inputwaveforms 'A' and 'B' as shown. Pick out theCORRECT output waveform :-

A

B

Y

Input A

Input B

(1)

(2)

(3)

(4)

6. The amount of heat liberated in the circuit afterclosing the switch S.

(1) zero (2) 2

2CV (3) CV2 (4) none

5. uhps n'kkZ;s ykWftd ifjiFk ds fuos'k rjax :i 'A' ,oa

'B' fuEu gSaA lgh fuxZe rjax :i dk p;u djsa :-

A

B

Y

Input A

Input B

(1)

(2)

(3)

(4)

6. ifjiFk esa fLop S dks cUn djus ds ckn mRiUu Å"ek dh

ek=k gksxh

(1) 'kwU; (2) 2

2CV(3) CV2 (4) dksbZ ugha

Kota/01CT314068H-4/31

Target : JEE (Main) 2015/27-03-2015


7. The magnetic field existing in a region is given

by 0

x ˆB B 1 ké ù= +ê úë ûr

lA square loop of edge l and

carrying current I0, is placed with its edgesparallel to the x-y axis. Find the magnitude ofthe net magnetic force experienced by the loop :-

(1) 2I0B0l (2) 12 I0B0l

(3) I0B0l (4) Zero

8. What is the range of frequency of EM wave thatare reflected back by ionosphere ?

(1) 100 to 200 MHz

(2) 500 to 1000 MHz

(3) 2 to 10 MHz

(4) All range of frequency

9 In the circuit show below, the key K is closed att = 0. The current through the battery is :

K

R =51 WL

V = 10V

R =102 W

(1) 5 A at t = 0 and 7 A at t = ¥(2) 3 A at t = 0 and 1 A at t = ¥(3) 1 A at t = 0 and 3 A at t = ¥(4) 2 A at t = 0 and 6 A at t = ¥

7. fdlh Hkkx esa pqEcdh; {ks= 0

x ˆB B 1 ké ù= +ê úë ûr

l }kjk fn;k

tkrk gSA l Hkqtk rFkk I0 /kkjk okys oxkZdkj ywi dks bl

izdkj j[kk tkrk gS bldh Hkqtk,sa x-y v{k ds lekUrj jgsaAywi }kjk vuqHko fd;k x;k ifj.kkeh pqEcdh; cy dkifjek.k gksxk

(1) 2I0B0l (2) 12 I0B0l

(3) I0B0l (4) 'kwU;

8. EM rajxks dh vko`fr dh og ijkl tks vk;u e.My lsokil ijkofrZr gks tkrh gS] gksxh\(1) 100 ls 200 MHz(2) 500 ls 1000 MHz

(3) 2 ls 10 MHz

(4) lHkh vkofr dh ijkl9 iznf'kZr ifjiFk esa t = 0 ij dqath K dks cUn dj nsrs gSaA

cSVjh ls izokfgr /kkjk gksxh %&

K

R =51 WL

V = 10V

R =102 W

(1) t = 0 ij 5A rFkk t = ¥ ij 7A

(2) t = 0 ij 3A rFkk t = ¥ ij 1A(3) t = 0 ij 1A rFkk t = ¥ ij 3A

(4) t = 0 ij 2A rFkk t = ¥ ij 6A


H-5/31Kota/01CT314068


10. In the given figure, the current amplitude is :-

L R

~V = V sin t + (V /2)cos t0 0w w

(1) ( )20

2 2 2

5V

4 L Rw + (2) ( )0

2 2 2

V

4 L Rw +

(3) 0

2 2 2

5V

2 L Rw -(4)

0

2 2 2

5V

2 R L-w

11. A radioactive sample decays by b-emission.

In first two seconds ‘n’ b-particles are emitted

and in next 2 seconds, ‘0.25n’ b-particles are

emitted. The half life of radioactive nuclei is

(1) 2 sec (2) 4 sec

(3) 1 sec (4) None of these

12. For a substance the average life for a-emission

is 1620 years and for b emission is 405 years.

After how much time the 1/4 of the material

remains by simultaneous emission :-

(1) 648 years (2) 324 years

(3) 449 years (4) 810 years

10. iznf'kZr fp= esa /kkjk vk;ke gS %&

L R

~V = V sin t + (V /2)cos t0 0w w

(1) ( )20

2 2 2

5V

4 L Rw + (2) ( )0

2 2 2

V

4 L Rw +

(3) 0

2 2 2

5V

2 L Rw -(4)

0

2 2 2

5V

2 R L-w

11. ,d jsfM;kslfØ; izfrn'kZ b-mRltZu }kjk {kf;r gksrk gSA

izFke nks lsd.Mksa esa ‘n’ b-d.k rFkk vxys nks lsd.Mksa esa

‘0.25n’ b-d.k mRlftZr gksrs gSaA jsfM;kslfØ; ukfHkd dh

v/kZ vk;q gksxh%&

(1) 2 sec (2) 4 sec

(3) 1 sec (4) buesa ls dksbZ ugha

12. fdlh inkFk Z dh a-mRltZu ds fy, vkSlr vk;q

1620 o"kZ gS ,oa b-mRltZu ds fy, 405 o"kZ gS rks a rFkk

b nksuksa izdkj ds d.kksa ds ,dlkFk mRltZu ls fdrus le;

i'pkr~ ,d&pkSFkkbZ inkFkZ 'ks"k jgsxk :-

(1) 648 years (2) 324 years

(3) 449 years (4) 810 years

Kota/01CT314068H-6/31

Target : JEE (Main) 2015/27-03-2015


13. A concave mirror of radius h is placed at the

bottom of a tank containing a liquid of refractive

index µ upto a depth d. An object P is placed at

height h above the bottom of the mirror. Outside

the liquid, an observer O views the object and its

image in the mirror. The apparent distance

between these two will be :-(1) 0

(2) 2hm

O

h P

hh

d(3) 1

h 1æ ö-ç ÷mè ø

(4) 2h

1m -

14. The PV diagram shows four different possiblereversible processes performed on a monatomicideal gas. Process A is isobaric(constant pressure). Process B is isothermal(constant temperature). Process C is adiabatic.Process D is isochoric (constant volume). Forwhich process(es) does the temperature of thegas decrease ?(1) Process A only

(2) Process C only

(3) Only Processes C and D

(4) Only Processes B, C and D

13. f=T;k h okys vory niZ.k dks d xgjkbZ rd m viorZukad

okys nzo ls Hkjs Vsad ds iSans ij j[kk x;k gSA niZ.k ds iSans

ls h ÅapkbZ ij ,d fcEc P j[kk gqvk gSA nzo ds ckgj

fLFkr ,d izs{kd O fcEc rFkk niZ.k esa blds izfrfcEc

dk s n s[krk g SA bu nk suk s a d s e/; vkHk klh n w jh

gksxh(1) 0

(2) 2hm

O

h P

hh

d(3) 1

h 1æ ö-ç ÷mè ø

(4) 2h

1m-

14. fdlh ,dijek.oh; vkn'kZ xSl ij fd;s x;s pkj fofHkUu

laHkkfor mRØe.kh; izØeksa dks PV vkjs[k ij n'kkZ;k x;k

gSA izØe A lenkch; (fu;r nkc) gSA izØe B

lerkih; (fu;r rki) gSA izØe C :¼ks"e gSA izØe

D levk;rfud (fu;r vk;ru) gSA fdl izØe esa xSl

dk rki ?kVrk gSA

(1) dsoy A esa

(2) dsoy C esa

(3) dsoy C o D esa

(4) dsoy B, C o D esa


H-7/31Kota/01CT314068


15. The figure given below shows the variation inthe internal energy U with volume V of2.0 mole of an ideal gas in a cyclic processabcda. The temperatures of the gas during theprocesses ab and cd are 500K and 300Krespectively, the heat absorbed by the gasduring the complete process is :-(Take R = 8.3 J/mol–K and ln 2 = 0.69)

(1) 3200 J

(2) 0 a b

cd

V0 2Vu

U

V

(3) 2100 J

(4) 2291 J

16. A plank with a small block on top of it is undergoing vertical SHM. Its period is 2 sec.The minimum amplitude at which the blockwill separate from plank is :

(1) 210p

(2) 10

2p(3) 2

20p

(4) 10p

17. What is the period of small oscillations of theblock of mass m if the springs are ideal andpulleys are massless ?

(1) m

2 kp

(2)m

2 2kp

m

k

(3) 2m

2 kp

(4) m2k

p

15. bl fp= esa ,d vkn'kZ xSl dh 2.0 eksy ek=k ds fy;s

pØh; izØe abcda esa vk;ru V ds cnyus ij vkUrfjd

ÅtkZ U esa ifjorZu dks fn[kk;k x;k gSA izØe pj.kksa ab

vksj cd esa xSl dk rki Øekuqlkj 500K vkSj 300K jgrk

gSA iwjs izØe esa xSl }kjk vo'kk sf"kr Å"ek dk eku

gksxk %& (R = 8.3 J/mol–K rFkk ln 2 = 0.69)

(1) 3200 J

(2) 0 a b

cd

V0 2Vu

U

V(3) 2100 J

(4) 2291 J

16. ,d r[rs ij NksVk CykWd j[kk gqvk gS rFkk ;g fudk;2 lsd.M vkorZdky ds lkFk Å/okZ/kj ljy vkorZ xfrdj jgk gSA fdl U;wure vk;ke ij CykWd r[rs ls vyxgks tk;sxk\

(1) 210p

(2) 10

2p(3) 2

20p

(4) 10p

17. ;fn fLizxsa vkn'kZ rFkk f?kjfu;k¡ nzO;ekughu gS rks m

nzO;eku ds CykWd ds vYi nksyuksa dk vkorZdky D;k

gksxk \

(1) m

2 kp

(2)m

2 2kp

m

k

(3) 2m

2 kp

(4) m2k

p

Kota/01CT314068H-8/31

Target : JEE (Main) 2015/27-03-2015


18. A string fixed at one end is vibrating in itssecond overtone. The length of the string is10 cm and maximum amplitude of vibrationof particles of the string is 2mm. Then theamplitude of the particle at 9cm from theopen end is:

(1) Ö3 mm

(2) Ö2 mm

(3) 23

mm

(4) None of these

19. Consider the snapshot of a wave traveling inpositive x–direction :-

A

B

(1) The particle A is moving in –ve y–directionand particle B is moving in +y–direction

(2) The particle B is moving in –ve y–directionand particle A is moving in +y–direction

(3) Both are moving in the +ve y–direction

(4) Both are moving in the –ve y–direction

18. ,d fljs ls ca/kh gqbZ jLlh blds f}rh; vf/kLojd esa

dEiUu dj jgh gSA jLlh dh yEckbZ 10 cm gS rFkk jLlh

ds d.kksa ds dEiUu dk vf/kdre vk;ke 2mm gSA rc

[kqys fljs ls 9cm ij fLFkr d.k dk vk;ke gksxk %&

(1) Ö3 mm

(2) Ö2 mm

(3) 23

mm

(4) buesa ls dksbZ ugha

19. /kukRed x fn'kk esa xfr'khy rjax ds ,d izs{k.k fp= ij

fopkj dhft;sA

A

B

(1) d.k A ½.kkRed y fn'kk esa xfr'khy gS rFkk d.kB /kukRed y fn'kk esa xfr'khy gSA

(2) d.k B ½.kkRed y fn'kk esa xfr'khy gS rFkk d.kA /kukRed y fn'kk esa xfr'khy gSA

(3) nksukas /kukRed y fn'kk eas xfr'khy gSA

(4) nksuksa ½.kkRed y fn'kk esa xfr'khy gSA


H-9/31Kota/01CT314068


20. A train whistling at constant frequency ismoving towards a station at a constant speed V.The train goes past a stationary observer on thestation. The frequency n' of the sound as heardby the observer is plotted as a function of timet (Fig.) . Identify the expected curve.

(1)

n'

t

(2)

n'

t

(3)

t

n'

(4)

t

n'

21. Shown in the diagram is a system of two bodies,a block of mass m and a disc of mass 4m, heldin equilibrium. If the string 3 is burnt, find theacceleration of the disc. Neglect the masses ofthe pulleys P and Q. The co–efficient of frictionbetween the block and horizontal surface is 0.5and friction between disc and string is zero.

(1) 4 m/s2 (2) 5 m/s2 (3) 3 m/s2 (4) 2 m/s2

20. ,d Vªsu fu;r vko`fr ls lhVh ctkrs gq, fu;r pky V

ls ,d LVs'ku dh vksj xfr dj jgh gSA ;g Vªsu LVs'ku ij

[kM+s ,d fLFkj izs{kd ds ikl ls xqtjrh gSA bl izs{kd }kjk

lquh xbZ /ofu dh vkofr n' dks le; ds Qyu t ds :i

esa n'kkZ;k x;k gSA lgh oØ pqfu, %&

(1)

n'

t

(2)

n'

t

(3)

t

n'

(4)

t

n'

21. fp= esa nks oLrqvksa ds fudk; dks n'kkZ;k x;k gSA m nzO;eku

ds CykWd rFkk 4m nzO;eku dh pdrh dks lkE;koLFkk esa

j[kk x;k gSA ;fn jLlh 3 VwV tk;s rks pdrh dk Roj.k Kkr

dhft,A f?kjfu;ksa P rFkk Q ds nzO;ekuksa dks ux.; ekfu;sA

CykWd rFkk {kSfrt lrg ds e/; ?k"kZ.k xq.kkad 0.5 gS rFkk

pdrh o jLlh ds e/; ?k"kZ.k 'kwU; gSA

(1) 4 m/s2 (2) 5 m/s2 (3) 3 m/s2 (4) 2 m/s2

Kota/01CT314068H-10/31

Target : JEE (Main) 2015/27-03-2015


22. A small mass 'm' rests at the edge of a horizontaldisc of radius 'R'. The coefficient of staticfriction between mass and the disc is m. The discis rotated about its axis at an angular velocitysuch that the mass slides off the disc and landson the floor 'h' meters below. What was itshorizontal distance of travel from the point itleft the disc?

(1) hm (2) ( )2R hm +

(3) Rhm (4) 2 Rhm

23. A block of mass M placed on rough surface ofcoefficient of friction equal to 3. If F is the(4/5) of the minimum force required to justmove. Find out the force exerted by ground onthe block.

(1) 2.6 Mg

(2) MgM F

m=3(3) 4 Mg

(4) 3.4 Mg

24. The colour are characterised by which offollowing character of light-

(1) Frequency (2) Amplitude

(3) Wavelength (4) Velocity

22. ,d NksVs nzO;eku m dks f=T;k R okyh {kSfrt pdrh

ds fdukjs ij j[kk x;k gSA nzO;eku rFkk pdrh ds eè;

LFkSfrd ?k"kZ.k xq.kkad m gSA pdrh dks bldh v{k ds lkis{k

dks.kh; osx ls bl izdkj ?kw.kZu djk;k tkrk gS fd nzO;ekupdrh ls fQlydj h ehVj uhps Q'kZ ij fxjrk gSA pdrh

dks NksM+us okys fcUnq ls nzO;eku }kjk r; dh xbZ {kSfrt

nwjh D;k Fkh\

(1) hm (2) ( )2R hm +

(3) Rhm (4) 2 Rhm

23. M nzO;eku dk ,d CykWd] ?k"kZ.k xq.kkad 3 okys ,d [kqjnjsry ij fLFkr gSA ;fn F dk eku] CykWd dks xfr djkusds fy;s vko';d U;wure cy dk (4/5) gks rks lrg }kjkCykWd ij vkjksfir cy Kkr dhft;sA

(1) 2.6 Mg

(2) Mg M F

m=3(3) 4 Mg

(4) 3.4 Mg

24. izdk'k ds fdl xq.k ls jaxksa dk fu/kkZj.k djrs gS&

(1) vkofÙk ls (2) vk;ke ls

(3) rjaxnS/;Z ls (4) osx ls


H-11/31Kota/01CT314068


25. A beam of light strikes a glass plate at an angleof incident 60° and reflected light is completelypolarised than the refractive index of the plateis :-

(1) 1.5 (2) 3 (3) 2 (4) 3

2

26. When three electric dipoles are near each other,they each experience the electric field of theother two, and the three dipole system has acertain potential energy. Figure below showsthree arrangements (1), (2) and (3) in whichthree electric dipoles are side by side. All threedipoles have the same magnitude of electricdipole moment, and the spacings betweenadjacent dipoles are identical. If U

1, U

2 and U

3

are potential energies of the arrangements (1),(2) and (3) respectively then-

(a) (b) (c)

(1) U1 > U

2 > U

3

(2) U1 > U

3 > U

2

(3) U1 > U

2 = U

3

(4) U1 = U

2 = U

3

25. ,d izdk'k iqat fdlh dk¡p dh IysV ij 60° ds vkirudks.k ij vkifrr gksrk gS rFkk ijkofrZr izdk'k iw.kZr% èkzqforizkIr gksrk gS rks IysV dk viorZukad D;k gksxk \

(1) 1.5 (2) 3 (3) 2 (4) 3

2

26. tc rhu fo|qr f}/kzqoksa dks ,d nwljs ds utnhd ykrs gS

rks os izR;sd vU; nks ds dkj.k fo|qr {ks= vuqHko djrs

gS rFkk rhu f}/kzqo fudk; dh ,d fuf'fpr fLFkfrt ÅtkZ

gksrh gSA uhps fp= esa rhu O;oLFkk,¡ (1), (2) rFkk (3)

n'kkZ;h x;h gS ftlesa rhu fo|qr f}/kzqoksa dks ,d ds ckn

,d djds j[kk x;k gSA lHkh rhuksa f}èkzqoksa ds fo|qr f}èkzqo

vk?kw.kZ ds ifjek.k leku gS rFkk laxr f}/kz qoksa ds eè;

vUrjky ,dleku gSA ;fn O;oLFkkvksa (1), (2) rFkk (3)

dh fLFkfrt ÅtkZ,¡ Øe'k% U1, U

2 rFkk U

3 gS rks&

(a) (b) (c)

(1) U1 > U

2 > U

3

(2) U1 > U

3 > U

2

(3) U1 > U

2 = U

3

(4) U1 = U

2 = U

3

Kota/01CT314068H-12/31

Target : JEE (Main) 2015/27-03-2015


27. What is net force on the small dipole placed

inside the capacitor at steady state if the platesare separated by 1cm ?

2W2W16V

4µf–

+

(1) 0N (2) 4N

(3) 8N (4) 16 N

28. A rectangular loop has a sliding connector PQof length 2 m and resistance 10 W and it ismoving with a speed 5 m/s as shown.The set-up is placed in a uniform magnetic field3T going into the plane of the paper. The threecurrents I1, I2 and I are :-

(1) I1 = I2 = 3A, I = 1A P

10W

I I2

Q

10W 10W

I1

5m/s

× 3T(2) I1 = I2 = 5A, I = 2A

(3) I1 = I2 = 1A, I = 2A

(4) I1 = I2 = I = 2A

27. iznf'kZr ifjiFk esa LFkk;h voLFkk ij la/kkfj= ds vUnj fLFkr

NksVs f}/kzqo ij ifj.kkeh cy D;k gksxk ;fn bldh IysVksa

ds e/; nwjh 1cm gks ?

2W2W16V

4µf–

+

(1) 0N (2) 4N

(3) 8N (4) 16 N

28. ,d vk;rkdkj ywi esa folihZ la;kstd PQ yxk gqvk gSA

bldh yEckbZ 2 m rFkk izfrjks/k 10 W gS rFkk ;g

5 m/s dh pky ls xfr'khy gS] fp= ns[ksaA bl lEiw.kZ

O;oLFkk dks 3T okys le:i pqEcdh; {ks= esa j[k nsrs

gSa ftldh fn'kk dkxt ds ry eas vUnj dh vksj gSA èkkjk

I1, I2 ,oa I ds eku gksxsa %&

(1) I1 = I2 = 3A, I = 1A

P

10W

I I2

Q

10W 10W

I1

5m/s

× 3T(2) I1 = I2 = 5A, I = 2A

(3) I1 = I2 = 1A, I = 2A

(4) I1 = I2 = I = 2A


H-13/31Kota/01CT314068


29. Initially AB is the interface of slab of refractive

index µ and air. A ray of light strikes at AB at

the critical angle qC. Now a slab ABCD of

refractive index µ1 is placed so that total internal

reflection takes place at surface CD (not at

surface AB). What should be the value of µ1?

(1) Greater than µq C

P µDBA

Cµ1

(2) Lesser than µ

(3) Equal to µ

(4) Any value, greater, smaller or equal to µ

30. The equation of a wave is g iven by

(all quantity expressed in m.k.s units)

Y= 5 sin10p (t - 0.01x) along the x-axis. The

phase diff erence between the poin ts

separated by a distance of 10 m along

x- axis is

(1) p/2 (2) p

(3) 2p (4) p/4.

29. izkjEHk esa viorZukad µ okyh ifV~Vdk rFkk ok;q dk

vUrjki`"B AB gSA ,d izdk'k fdj.k AB ij ØkfUrd

dks.k qC ij Vdjkrh gSA vc viorZukad µ

1 okyh ,d

ifV~Vdk ABCD dks bl izdkj j[kk tkrk gS fd lrg CD

(lrg AB ij ugha) ij iw.kZ vkUrfjd ijkorZu gksrk gSA

µ1 dk eku gksxk

(1) µ ls vf/kd

(2) µ ls de q C

P µDBA

Cµ1

(3) µ ds cjkcj

(4) µ ls cM+k] NksVk ;k cjkcj dksbZ Hkh eku

30. x-v{k d s vu q fn' k ,d rj ax dh lehdj.k

Y= 5 sin10p (t - 0.01x) }kjk nh tkrh gS tgk¡ lHkh

jkf'k;k¡ m.k.s bdkbZ eas iznf'kZr dh x;h gSA x-v{k ds

vuqfn'k 10 m dh nwjh ij fLFkr fcUnqvks a ds eè;

dykUrj gksxk%&

(1) p/2 (2) p

(3) 2p (4) p/4

Kota/01CT314068H-14/31

Target : JEE (Main) 2015/27-03-2015


31. Activity of a radioactive substance can be

represented by various unit. Select correct

option

(1) 1dps = 106 Bq (2) 1Ci=3.7×1010 dps

(3) 1Ci = 1 Bq (4) 1 Bq =106 Rd

32. Select the incorrect statement for schottky

defect

(1) Vacant cation & anion sites in ratio 1 : 1

in an MX lattice


in an MX2 lattice


in an M2X lattice

(4) In each unit cell either cation or anion must

be missing.

33. A solution of 36g of glucose in 1000 gm of

water is cooled to –0.5ºC. How many gram of

ice would have separated from the solution

Kf = 1.86 deg/molal

(1) 744 g (2) 300 g

(3) 256 g (4) 180 g

31. jsfM;kslfØ; inkFkZ dh lfØ;rk dks dbZ bdkbZ;ks }kjk

iznf'kZr fd;k tk ldrk gS lgh fodYi dk p;u

dhft;sA

(1) 1dps = 106 Bq (2) 1Ci=3.7×1010 dps

(3) 1Ci = 1 Bq (4) 1 Bq =106 Rd

32. 'kkWV~dh =qfV ds fy, xyr dFku dk p;u dhft;sA

(1) MX tkyd esa /kuk;u rFkk ½.kk;u fjfDr;ksa dk

vuqikr 1 : 1 gksrk gSA

(2) MX2 tkyd esa /kuk;u rFkk ½.kk;u fjfDr;ksa dk

vuqikr 1 : 2 gksrk gSA

(3) M2X tkyd esa fjDr /kuk;u rFkk ½.kk;u fjfDr;ksa

dk vuqikr 2 : 1 gksrk gSA

(4) izR;sd bdkbZ lsy esa ;k rks /kuk;u ;k ½.kk;u yqIr

(missing) gksus pkfg,A

33. 1000 gm ty esa 36g Xywdkst dk ,d foy;u –0.5ºC

rd B.Mk gksrk gSA foy;u ls fdrus xzke cQZ iFkd fd;k

tk ldrk gSA Kf = 1.86 deg/molal

(1) 744 g (2) 300 g

(3) 256 g (4) 180 g

PART B - CHEMISTRY


H-15/31Kota/01CT314068


34. A salt MX has Ksp = 4 × 10–10. What value of

Ksp must another salt MX3 have if the molar

solubility of the two salts is to be identical -

(1) 3.2 × 10–10 (2) 1.024 × 10–19

(3) 1.78 × 10–5 (4) 4.32 × 10–18

35. Consider the cell :

Pt|H2(P1 atm) | H+(x1M) || H+ (x2M) | H2(P2 atm) |Pt.

The cell reaction will be spontaneous if -

(1) P1 = P2 and x1 > x2

(2) P1 = P2 and x1 = x2

(3) x1 = x2 and P1 > P2

(4) x1 = x2 and P1 < P2

36. Which of the following is false regarding

adsorption -

(1) Physical adsorption is usually multilayer

(2) Chemisorption is usually monolayer

(3) Physical adsorption requires high

activation energy

(4) Chemisorptions are usually irreversible

37. Potential energy of electron present in He+ is

(1) 2e

2pÎ0r(2)

23e4pÎ0r

(3) 2e

2-pÎ0 r (4)

2e4-pÎ 2

0r

34. ,d yo.k MX dk Ksp = 4 × 10–10 gSA vU; yo.k

MX3 ds Ksp dk eku D;k gksuk pkfg,] ;fn nksuksa yo.kksa

dh eksyj foys;rk leku gks-

(1) 3.2 × 10–10 (2) 1.024 × 10–19

(3) 1.78 × 10–5 (4) 4.32 × 10–18

35. lsy ij fopkj dhft;s :

Pt|H2(P1 atm) | H+(x1M) || H+ (x2M) | H2(P2 atm) |Pt.

lsy vfHkfØ;k Lor% gksxh] ;fn -

(1) P1 = P2 rFkk x1 > x2

(2) P1 = P2 rFkk x1 = x2

(3) x1 = x2 rFkk P1 > P2

(4) x1 = x2 rFkk P1 < P2

36. vf/k'kks"k.k ds lanHkZ esa fuEu esa ls dkSulk dFku vlR; gS

(1) HkkSfrd vf/k'kks"k.k lkekU;r% cgqijrh; gksrk gS

(2) jlk;fud vf/k'kks"k.k lkekU;r% ,dy ijrh; gksrk gS

(3) HkkSfrd vf/k'kks"k.k ds fy, mPp lfØ;.k ÅtkZ dh

vko';drk gksrh gSA

(4) jlk;fud vf/k'kks"k.k lkekU;r% vuqRØe.kh; gksrk gS

37. He+ esa mifLFkr bySDVªkWu dh fLFkfrt ÅtkZ gS-

(1) 2e

2pÎ0r(2)

23e4pÎ0r

(3) 2e

2-pÎ0 r (4)

2e4-pÎ 2

0r

Kota/01CT314068H-16/31

Target : JEE (Main) 2015/27-03-2015


38. Critical constant for a real gas are given as

TC = 180 K ; Vc = 0.123 L/mol

PC = 45 atm ; R = 0.082 Latm/mol K

The correct statement for the real gas is -

(1) The volume of single gas molecule is3

23

0.123 10 cc6.023 10

-æ ö´ç ÷´è ø

(2) b = 3 × 0.123L mol–1

(3) The Boyle’s temperature is less than 180K

(4) Gas cannot be liquified at 200K

39. 1mol of NH3 (g = 4/3) gas at 27ºC is expanded

under reversible adiabatic condition to make

volume 8 times. Calculate work done -

(1) –900 cal (2) –450 cal

(3) –1000 cal (4) –800 cal

40. For the the reaction

I– + ClO3– + H2SO4

® Cl– + HSO4– + I2

(no free H+ is present).

Select the correct statement in the balancedequation (Coefficients in smallest wholenumber ratio.)(1) Iodide ion is reduced(2) Sulphur is oxidised(3) Stoichiometric coefficient of HSO4

– is 6(4) Stoichiometric coefficient of H2O is 2

38. ,d okLrfod xSl ds fy, ØkfUrd fu;rkad fuEu izdkj

izdkj fn;s x;s gS

TC = 180 K ; Vc = 0.123 L/mol

PC = 45 atm ; R = 0.082 Latm/mol K

okLrfod xSl ds fy, lgh dFku gS

(1) xSl ds ,d v.kq dk vk;ru 3

23

0.123 10 cc6.023 10

-æ ö´ç ÷´è ø

gS

(2) b = 3 × 0.123 L mol–1

(3) ckW;y rki 180K ls de gS(4) 200K ij xSl nzohr ugha gks ldrh gS

39. 27ºC ij NH3 (g = 4/3) xSl ds 1 eksy dk mRØe.kh;

:¼ks"eh; ifjfLFkfr;ksa esas vk;ru 8 xquk rd izlkfjr fd;k

x;kA fd;s x;s dk;Z dh x.kuk dhft;s(1) –900 cal (2) –450 cal(3) –1000 cal (4) –800 cal

40. vfHkfØ;k

I– + ClO3– + H2SO4

® Cl– + HSO4– + I2

(dksbZ Hkh eqDr H+ mifLFkr ugha gS).

ds fy, lUrqfyr lehdj.k esa lgh dFku dk p;udhft;sA (xq.kkad U;wure iw.kk±d la[;k vuqikr esa gS)(1) vk;ksMkbZM vk;u vipf;r gksrk gS

(2) lYQj vkWDlhd`r gksrk gS

(3) HSO4– dk jllehdj.kehrh; xq.kkad 6 gS

(4) H2O dk jllehdj.kehrh; xq.kkad 2 gS


H-17/31Kota/01CT314068


41. Two homo-diatomic molecules 'X'2 and 'Y'

2

have same bond order then select the

CORRECT statement :

(1) Both must have same number of total

antibonding electrons

(2) Both must have same number of total

bonding electrons

(3) Difference of total number of antibonding

electron and bonding electron must be

same for both molecules

(4) Both (1) and (2)

42. Total number of molecules which have

cation-anion pair out of I2(s), dry ice, PCl

5(s),

N2O

5(s), PBr

5(g) is :

(1) 3 (2) 2

(3) 1 (4) 0

43. During the race of complex compounds, finish

line is fixed by the achieving E.A.N. value equal

to the atomic number of corresponding inert

gas. Then which of the following complex is

present at finish line :

(1) [Co(NO2)

6]4– (2) [Mn(CO)

5]

(3) K4[Fe(CN)

6] (4) Fe

3[Fe(CN)

6]

2

41. nks le&f}ijekf.od v.kqvksa 'X'2 rFkk 'Y'

2 ds ca/k Øe

leku gS rks lgh dFku pqfu, :

(1) nksuksa esa dqy izfrca/kh bysDVªkWuksa dh la[;k leku

gksuh pkfg,

(2) nksuksa esa dqy ca/kh bysDVªkWuksa dh la[;k leku gksuh

pkfg,

(3) nksuks a v.kqvksa ds fy,] izfrca/kh bysDVªkWuks a rFkk

ca/kh bysDVªkWuksa dh dqy la[;k dk vUrj leku

gksuk pkfg,

(4) (1) rFkk (2) nksuksa

42. I2(s), 'kq"d cQZ, PCl

5(s), N

2O

5(s), PBr

5(g) esa ls

,sls v.kqvksa dh dqy la[;k tks /kuk;u&½.kk;u ;qXe j[krs

gSa] gS :

(1) 3 (2) 2

(3) 1 (4) 0

43. ladqy ;k Sfxdks a dh nk SM + ds nk Sjku vfUre js[kk ]

lEcfU/kr vfØ; xSl ds ijek.kq Øekad ds cjkcj E.A.N.

eku izkIr djus ds fy, j[kh x;h gS rks fuEu esa ls dkSulk

ladqy vafre js[kk ij gS :

(1) [Co(NO2)

6]4– (2) [Mn(CO)

5]

(3) K4[Fe(CN)

6] (4) Fe

3[Fe(CN)

6]

2

Kota/01CT314068H-18/31

Target : JEE (Main) 2015/27-03-2015


44. Which of the following statement is

CORRECT :

(1) Ce+4 act as oxidising as well as reducing

agent

(2) Pure PH3 is produced by heating of PH

4I

with dil. CH3COOH

(3) Urea, sulphamic acid and thiourea are

used to destroy nitrate ion

(4) Titration of boric acid solution is not

satisfactory with NaOH solution

45. Which of the following change involve more

than two metallurgical process :

(1) ZnCO3 (Concentrated ore) ® Zn (Impure)

(2) CuFeS2 ® Cu (Pure)

(3) PbS (Roasted ore) ® Pb (Impure)

(4) All of these46. In s-block on moving down the group reactivity

of metals with water and air increases, whichof the following factor has no significance forthis trend of reactivity :(1) On moving down the group size of atom

increases(2) On moving down the group melting point

decreases(3) On moving down the group electron gain

enthalpy decreases(4) All of these

44. fuEu esa ls dkSulk dFku lgh gS :

(1) Ce+4, vkWDlhdkjd ds lkFk&lkFk vipk;d dk

dk;Z djrk gS

(2) PH4I dks ruq CH

3COOH ds lkFk xeZ djds

'kq¼ PH3 fufeZr fd;k tkrk gS

(3) ukbVª sV vk;u dk s u"V djus ds fy , ; w fj;k,

lYQsfed vEy rFkk Fkk;ks&;wfj;k dk iz;ksx fd;k

tkrk gS

(4) NaOH foy;u ds lkFk cksfjd vEy foy;u

dk vuqekiu larks"ktud ugha gksrk gS

45. fuEu esa ls dkSuls ifjorZu esa nks ls vf/kd /kkrqdehZ;izØe lfEefyr gS :(1) ZnCO

3 (lkfUær v;Ld) ® Zn (v'kq¼)

(2) CuFeS2 ® Cu ('kq¼)

(3) PbS (HkftZr v;Ld) ® Pb (v'kq¼)

(4) mijksDr lHkh46. s-CykWd esa] oxZ esa uhps dh vksj c<+us ij /kkrqvksa dh

ty rFkk ok;q ds lkFk fØ;k'khyrk c<+rh gSaA fØ;k'khyrkdh bl izo`fr ds fy,] fuEu esa ls dkSuls dkjd dk dksbZegRo ugha gS :(1) oxZ esa uhps dh vksj c<+us ij ijek.kq dk vkdkj

c<+rk gS(2) oxZ esa uhps dh vksj c<+us ij xyukad ?kVrk gS(3) oxZ esa uhps dh vksj c<+us ij bysDV ªkWu xzg.k ,UFkSYih

?kVrh gS(4) mijksDr lHkh


H-19/31Kota/01CT314068


47. Which of the following statement isINCORRECT :(1) Catenation tendency of phosphorus is

more than the catenation tendency ofnitrogen

(2) PCl3 is produced by the reaction of SOCl

2

with white phosphorus(3) According to bronsted theory water act as

amphoteric substance(4) By boiling the water permanent hardness

is removed

48. Salt(soluble)

'X' + dil. H SO gas(B) + C + 'Y'2 4 ®(neutraloxide)

BaCl solution2'Z'

(White ppt.)

aq. sol. of gas (B) 2H S(g)¾¾¾® 'Y' + (C)

neutral oxide

Select CORRECT statement :(1) Salt 'X' contains either SO

32– or CO

32–

anion(2) Salt 'X' is PbS

2O

3

(3) Salt 'X' contains an anion which haveaverage oxidation state of 'S' = +2

(4) 'C' is nitric oxide49. Which of the following can act as reducing as

well as oxidising agent :

(1) H2O

2(2) SO

2

(3) H2S (4) Both (1) and (2)

47. fuEu esa ls dkSulk dFku xyr gS :

(1) ukbV ªkstu dh rqyuk esa QkWLQksjl dh J`a[kyu izo`fr

vf/kd gS

(2) SOCl2 dh] 'osr QkWLQksjl ds lkFk vfHkfØ;k

ls PCl3 curk gS

(3) czkULVsM fl¼kUr ds vuqlkj ty mHk;/kehZ inkFkZ

ds :i esa dk;Z djrk gS

(4) ty dks mcky dj] LFkk;h dBksjrk nwj dh tkrh gS

48. yo.k( )foys;'khy

'X' + H SO (B) + C + 'Y'2 4 ® xSlruq(

)mnklhu

vkWDlkbM

BaCl 2 foy;u 'Z'( )'osr vo{ksi

xSl(B) dk tyh; foy;u 2H S(g)¾¾¾® 'Y' + (C)

mnklhu vkWDlkbM

lgh dFku pqfu, :

(1) yo.k 'X' esa ;k rks SO3

2– ;k CO

3

2– ½.kk;u mifLFkr gS

(2) yo.k 'X' , PbS2O

3 gS

(3) yo.k 'X' esa ,d ,slk ½.kk;u mifLFkr gS ftlesa

'S' dh vkSlr vkWDlhdj.k voLFkk] +2 gS

(4) 'C', ukbfVªd vkWDlkbM gS

49. fuEu esa ls dkSu vipk;d ds lkFk&lkFk vkWDlhdkjddk Hkh dk;Z dj ldrk gS :(1) H

2O

2(2) SO

2

(3) H2S (4) (1) rFkk (2) nksuksa

Kota/01CT314068H-20/31

Target : JEE (Main) 2015/27-03-2015


50. Select the correct order of the propertyindicated against them :

(1) Li+(g)<Be2+(g) < Mg2+(g) (Ionisation

energy)

(2) Li+ < Mg2+ > Al3+ (Polarisation power)

(3) [Ma3b

3] < [M(AA)

2cd] < [M(AB)

2cd]

(Number of stereo isomers)

(4) Mg < Fe < Al (Ascending position of

element in Ellingham graph, element

nearer to DGº = 0 line is kept at last)

51. Identify the major product ?

NH C—

: Conc.H SO2 4+ Conc.HNO3

O Product

(1) NH C—

NO2

O

(2) NH C—

NO2

O

(3) NH C— NO2

O

(4) NH C—O2NO

50. iznf'kZr fd;s x;s xq.kksa ds lUnHkZ esa lgh Øe pqfu, :

(1) Li+(g) < Be2+(g) < Mg2+(g) (vk;uu ÅtkZ)

(2) Li+ < Mg2+ > Al3+ (/kzqohdj.k {kerk)

(3) [Ma3b

3] < [M(AA)

2cd] < [M(AB)

2cd]

(f=foe leko;fo;ksa dh la[;k)

(4) Mg < Fe < Al (,fy?kae xzkQ esa rRoksa dh c<+rh

gqbZ fLFkfr] rRo tks fd DGº = 0 js[kk ds ikl

gSa mUgsa var esa j[kk x;k gS)

51. eq[; mRikn igpkfu,s&

NH C—

: lkUnz H SO2 4+ HNO3lkUnz

O

mRikn

(1) NH C—

NO2

O

(2) NH C—

NO2

O

(3) NH C— NO2

O

(4) NH C—O2NO


H-21/31Kota/01CT314068


52. Which of the following carbohydrate can notreduce tollen's reagent

(1) Fructose (2) Maltose

(3) Sucrose (4) Lactose

53. Identify the basic amino acid.

(1) Glycine (2) Alanine

(3) Arginine (4) Aspartic acid

54. Which of the following compound do not gives

isocyanide test.

(1) NH2 (2)

NH2

(3) CH3–NH–CH2–CH3 (4)

NH2

55. Identify the compound which rotate the plane

polarised light.

(1)

Cl

Cl

(2)

CH3

CH3

ClClH

H

(3)

ClF

Br(4) CH2ClF

52. fuEu esa ls dkSulk dkcksZgkbMªsV VkWysUl vfHkdeZd dks

vipf;r ugha dj ldrk gS&

(1) ÝDVksl (2) ekYVksl(3) lwØksl (4) ysDVksl

53. {kkjh; vehuksa vEy igpkfu,s&

(1) Xykbflu (2) ,sykfuu

(3) vkthZfuu (4) ,LikfVZd vEy

54. fuEu esa ls dkSulk ;kSfxd vkblkslk;ukbM ijh{k.k ughansrk gS&

(1) NH2 (2)

NH2

(3) CH3–NH–CH2–CH3 (4)

NH2

55. ,slk ;kSfxd igpkfu,s tks lery /kzqoh; izdk'k dks ?kqekrk

gS&

(1)

Cl

Cl

(2)

CH3

CH3

ClClH

H

(3)

ClF

Br(4) CH2ClF

Kota/01CT314068H-22/31

Target : JEE (Main) 2015/27-03-2015


56. Which of the following compound has least

pKa value.

(1)

OH

NO2

(2)

OHNO2

(3)

OH

(4)

OH

OCH3

57. C–NH2

OCH3

CH –CH2 3

H KOBr Product

Correct statement about product is :

(1) It is optically inactive

(2) It is optically active and having one less

carbon than reactant

(3) Product do not give isocyanide test

(4) Product is carboxylic acid

56. fuEu esa ls fdl ;kSfxd dk U;wure pKa eku gksrk gS&

(1)

OH

NO2

(2)

OHNO2

(3)

OH

(4)

OH

OCH3

57. C–NH2

OCH3

CH –CH2 3

H KOBr mRikn

mRikn ds lUnHkZ esa lgh dFku gS&

(1) ;g izdkf'kd vfØ; gS

(2) ;g izdkf'kd lfØ; gS rFkk blesa vfHkdkjd dh

rqyuk esa ,d dkcZu de gS

(3) mRikn] vkblkslk;ukbM ijh{k.k ugha nsrk gS

(4) mRikn] dkcksZfDlfyd vEy gS


H-23/31Kota/01CT314068


58. Which of the following compound do not

release CO2 on oxidative ozonolysis

(1) 1-butene (2) 2-butyne

(3) Propyne (4) Ethene

59. Which of the following do not give aliphatic

primary amine as product.

(1) CH –C–OH3

ON H3 , H SO2 4

(2) CH3–CH = N–OH H+

(3) CH –C–NH3 2

OBr , OH2

–

(4) CH3—NC H O3

+

60. In which of the following compound

hyperconjugation effect is not present :

(1)

CH2

(2)

(3) CH3–CH=CH2 (4)

58. fuEu esa ls dkSulk ;kSfxd vkWDlhdr vkstksuh vi?kVu

djkus ij CO2 eqDr ugha djrk gS&

(1) 1-C;qVhu (2) 2-C;qVkbu

(3) izksikbu (4) ,sfFku

59. fuEu es a ls dkSulh vfHkfØ;k es a mRikn ds :i esa

,sfyQsfVd izkFkfed ,sehu izkIr ugha gksrh gS&

(1) CH –C–OH3

ON H3 , H SO2 4

(2) CH3–CH = N–OH H+

(3) CH –C–NH3 2

OBr , OH2

–

(4) CH3—NC H O3

+

60. fuEu esa ls dkSuls ;kSfxd esa vfrla;qXeu izHkko mifLFkr

ugha gS&

(1)

CH2

(2)

(3) CH3–CH=CH2 (4)

Kota/01CT314068H-24/31

Target : JEE (Main) 2015/27-03-2015


61. Let r be a relation from R (Set of real number)to R defined by r = {(x, y) | x, y Î R and xy isan irrational number}, then relation r is -(1) reflexive and symmetric only(2) symmetric only(3) symmetric and transitive only(4) equivalence relation

62. A and B are two subsets of set S = {1,2,3,4}such that A È B = S, then number of orderedpair of (A, B) is -(1) 72 (2) 81 (3) 16 (4) 96

63. The negation of the statement "96 is divisibleby 2 and 3" is(1) 96 is not divisible by 2 and 3(2) 96 is not divisible by 3 or 96 is not divisible by 2(3) 96 is divisible by 2 or 96 is divisible by 3(4) none of these

64. A tower stands vertically inside an acute angledtriangular park PQR. If angle of elevation ofthe pole from each corner of the park is same,then in DPQR, the foot of the tower is at the(1) centroid (2) circumcentre(3) incentre (4) orthocentre

65. Variance of 10C0, 10C1,

10C2,.... 10C10

is -

(1) 20 10

1010. C 2100

-(2)

20 101011 C 211

-

(3) 20 20

1010. C 2100

-(4)

20 201011. C 2

121-

61. ekuk r, r = {(x, y) | x, y Î R rFkk xy ,d vifjes;la[;k gS} }kjk R ls R esa ifjHkkf"kr lEcU/k gks] rks rgksxk -(1) dsoy LorqY; rFkk lefer(2) dsoy lefer(3) dsoy lefer rFkk laØked(4) rqY;rk lEcU/k

62. A rFkk B leqPp; S = {1,2,3,4} ds nks mileqPp;bl izdkj gS fd A È B = S gks] rks (A, B) ds Øfer;qXeksa dh la[;k gksxh -(1) 72 (2) 81 (3) 16 (4) 96

63. "96, 2 rFkk 3 ls foHkkftr gS " dk fu"ks/k dFkugksxk(1) 96, 2 rFkk 3 ls foHkkftr ugha gksxkA(2) 96, 3 ls foHkkftr ugha ;k 96, 2 ls foHkkftr ugha gksxkA(3) 96, 3 ls foHkkftr ;k 96, 2 ls foHkkftr gksxkA(4) buesa ls dksbZ ugha

64. ,d ehukj] ,d U;wudks.k f=dks.kh; cxhps ds vUnj dhvksj mG/okZ/kj [kM+k gqvk gSA ;fn cxhps ds izR;sd dksusls [kEHks dk mUu;u dks.k leku gks] rks f=Hkqt PQR esa]ehukj dk ikn fuEu ij fLFkr gksxk(1) dsUæd (2) ifjdsUæ(3) vUrdsUæ (4) yEcdsUæ

65. 10C0, 10C1,

10C2,.... 10C10

dk izlj.k gksxk -

(1) 20 10

1010. C 2100

-(2)

20 101011 C 211

-

(3) 20 20

1010. C 2100

-(4)

20 201011. C 2

121-

PART C - MATHEMATICS


H-25/31Kota/01CT314068


66. If

k cos x x cos k x 0,2

ƒ(x)k sin x x sin k x ,

2

ì pé ù- Îï ê úï ë û= ípæ ùï + Î pç úï è ûî

is

differentiable in (0, p), then -

(1) k 2, 2é ùÎ -ë û (2) k ,2 2p pé ùÎ -ê úë û

(3) k = 0 (4) k Î f (Null set)

67. The value of x

x

e 9cos x 2sin x 7 dxe 7sin x 11cos x 14

+ - ++ + +ò

is -

(1) ( )x1 x n(e 7sin x 11cos x 14) C2

+ + + + +l

(2) ( )x1 x n(e 7sin x 11cos x 14) C2

- + + + +l

(3) x1x n(e 7sin x 11cos x 14) C

2+ + + + +l


2- + + + +l

(where C is constant of integration)

68. The value of 22

2

2

x dx1 tan x 1 tan x

p

-p + + +ò is -

(1) p3 (2) 3

12p

(3) 3

24p

(4) 3

48p

66. ;fn

k cos x x cos k x 0,2

ƒ(x)k sin x x sin k x ,

2

ì pé ù- Îï ê úï ë û= ípæ ùï + Î pç úï è ûî

vUrjky (0, p) esa vodyuh; gks] rks -

(1) k 2, 2é ùÎ -ë û (2) k ,2 2p pé ùÎ -ê úë û

(3) k = 0 (4) k Î f (fjDr leqPp;)

67.x

x

e 9cos x 2sin x 7 dxe 7sin x 11cos x 14

+ - ++ + +ò dk eku gksxk -

(1) ( )x1 x n(e 7sin x 11cos x 14) C2

+ + + + +l

(2) ( )x1 x n(e 7sin x 11cos x 14) C2

- + + + +l


2+ + + + +l


2- + + + +l

(tgk¡ C lekdyu vpj gS)

68.22

2

2

x dx1 tan x 1 tan x

p

-p + + +ò dk eku gksxk -

(1) p3 (2) 3

12p

(3) 3

24p

(4) 3

48p

Kota/01CT314068H-26/31

Target : JEE (Main) 2015/27-03-2015


69. Let x 8y ƒ(x) 8ƒ(y)ƒ9 9+ +æ ö =ç ÷

è ø for all real x &

y. If ƒ'(0) exist and equal to 2 and ƒ(0) = –5,

then ƒ(7) is equal to -

(1) 3 (2) 7 (3) 5 (4) 9

70. Number of solutions of 8cosx = x will be -

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

71. If 'a' is non real complex number forwhich system of equations ax – a2y + a3z = 0,–a2x + a3y + az = 0 and a3x + ay – a2z = 0 hasnon trivial solutions, then |a| is -

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

72. The number of real roots of the equation2 2P Q 1

x x 1+ =

-, where P and Q are non-zero real

numbers, is-

(1) 1 (2) 2 (3) 3 (4) 473. What is the sum of all two digit numbers which

give a remainder of 4 when divided by 6 ?(1) 777 (2) 776 (3) 780 (4) 784

74. In the given figure BC = AC, angle AFD = 40°and CE = CD. Then the value of angle BCE isequal to -

BE

C

F

A D

(1) 50° (2) 60° (3) 40° (4) 100°

69. ekuk lHkh okLrfod x o y ds fy,

x 8y ƒ(x) 8ƒ(y)ƒ9 9+ +æ ö =ç ÷

è ø gSA ;fn ƒ'(0) fo|eku

rFkk 2 ds cjkcj rFkk ƒ(0) = –5 gks] rks ƒ(7) cjkcj gksxk-

(1) 3 (2) 7 (3) 5 (4) 9

70. 8cosx = x ds gyksa dh la[;k gksxh -(1) 3 (2) 4 (3) 5 (4) 6

71. ;fn 'a' vokLrfod lfEeJ la[;k gS] ftlds fy,lehdj.k fudk; ax – a2y + a3z = 0,–a2x + a3y + az = 0 rFkk a3x + ay – a2z = 0 dslkFkZd gy gks] rks |a| gksxk -

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

72. lehdj.k 2 2P Q 1

x x 1+ =

- ds okLrfod ewyks a dh

la[;k gksxh(tgk¡ P rFkk Q v'kwU; okLrfod la[;k;s gSa)-

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

73. mu lHkh nks vadksa dh la[;kvksa dk ;ksxQy] ftudks 6 lsfoHkkftr djus ij 'ks"kQy 4 vkrk gS] gksxk ?(1) 777 (2) 776 (3) 780 (4) 784

74. fn, x, fp= esa BC = AC, dks.k AFD = 40° rFkk CE= CD gSA rc dks.k BCE dk eku gksxk -

BE

C

F

A D

(1) 50° (2) 60° (3) 40° (4) 100°


H-27/31Kota/01CT314068


75. If 2p is length of perpendicular from the origin

to line x y 1a b

+ = , then

(1) a2, 8p2, b2 are in A.P.(2) a2, 8p2, b2 are in G.P.

(3) a2, 8p2, b2 are in H.P.(4) none of these

76. Tangent to the circle x2 + y2 = 5 at thepoint (1, –2) also touches the circlex2 + y2 – 8x + 6y + 20 = 0. Then its point ofcontact is -(1) (–2, 1) (2) (–1, –1)(3) (–3, 0) (4) (3, –1)

77. If C is the centre of the ellipse 9x2 + 16y2 = 144and S is one focus. The ratio of CS to majoraxis, is -

(1) 7 : 16 (2) 7 : 4

(3) 5: 7 (4) 7 : 878. All face cards from pack of 52 playing cards

are removed. From remaining 40 cards two aredrawn randomly without replacement, thenprobability of drawing a pair(same denominations) is -

(1) 1

13 (2) 178 (3)

239 (4)

413

79. The length of chord of parabola x2 = 4aypassing through its vertex and having slopetana, is -(1) 4a coseca cota (2) 4a tana seca(3) 4a cosa cota (4) 4a sina tana

75. ;fn 2p ewyfcUnq ls js[kk x y 1a b

+ = ij [khaps x,] yEc

dh yEckbZ gks] rks -(1) a2, 8p2, b2 lekUrj Js.kh esa gksaxsA(2) a2, 8p2, b2 xq.kksÙkj Js.kh esa gksaxsA(3) a2, 8p2, b2 gjkRed Js.kh esa gksaxsA(4) buesa ls dksbZ ugha

76. o`Ùk x2 + y2 = 5 ds fcUnq (1, –2) ij Li'kZ js[kk o`Ùkx2 + y2 – 8x + 6y + 20 = 0 dks Li'kZ djrh gSA rcbldk Li'kZ fcUnq gksxk -(1) (–2, 1) (2) (–1, –1)

(3) (–3, 0) (4) (3, –1)

77. ;fn C nh?kZo`Ùk 9x2 + 16y2 = 144 dk dsUæ rFkkS bldh ,d ukfHk gSA CS rFkk nh?kZv{k dh yEckbZ dkvuqikr gksxk -

(1) 7 : 16 (2) 7 : 4

(3) 5: 7 (4) 7 : 878. 52 iÙkksa dh rk'k dh xM~Mh ls lHkh fpf=r iÙkksa dks

fudkyk tkrk gSA 'ks"k 40 iÙkksa esa ls nks iÙkksa dks fcukiquLFkkZiu ds ;kn`PN;k fudkyrs gS] rks iÙkksa dk ;qXe(leku Øe) ds fudyus dh izkf;drk gksxh -

(1) 1

13 (2) 178 (3)

239 (4)

413

79. ijoy; x2 = 4ay dh thok dh yEckbZ] tks blds'kh"kZ ls xqtjrh gS rFkk ftldh izo.krk tana gS]gksxh -(1) 4a coseca cota (2) 4a tana seca(3) 4a cosa cota (4) 4a sina tana

Kota/01CT314068H-28/31

Target : JEE (Main) 2015/27-03-2015


80. If A is skew symmetric matrix of order 3 andX be another matrix of same order, then|XA + AXT| is (where |P| denotes determinantof matrix P) -

(1) |X + XT| (2) |A + X|

(3) |A – X| (4) 0

81. In a parallelogram OACB, OA a, OB b= =uuur uuur rr

&

foot of perpendicular drawn from point B to

AC is M. If a.b 1=rr

& | a | | b | 2= =rr

,

then BMuuuur

is-

(1) 15 (2) 5

2(3) 5 (4)

152

82. If P,Q & R are foot of perpendiculars drawn

from point A(1,1,1) to planes P1 : x + 2y + 2z = 2,

P2 : 2x – 2y + z = –8 & to line of intersection of

P1 & P2 respectively, then area of DPQR is -

(1) 3 (2) 32

(3) 2 (4) 34

83. If a, b, c are non zero real numbers, thenminimum value of the expression

4 2 4 2 4 2

2 2 2

(a a 1)(b 7b 1)(c 11c 1)a b c

æ ö+ + + + + +ç ÷è ø

, is -

(1) 315 (2) 351

(3) 415 (4) 451

80. ;fn A, dksfV 3 dk fo"ke lefer vkO;wg rFkk Xleku dksfV dk vU; vkO;wg gks] rks |XA + AXT|gksxk (tgk¡ |P|, vkO;wg P ds lkjf.kd ds eku dks n'kkZrk gS)-

(1) |X + XT| (2) |A + X|

(3) |A – X| (4) 0

81. lekUrj prqHkqZt OACB esa] OA a, OB b= =uuur uuur rr

rFkkfcUnq B ls Hkqtk AC ij [kh aps x, yEc dk ikn

M gSA ;fn a.b 1=rr

rFkk | a | | b | 2= =rr

gks] rks BMuuuur

gksxk -

(1) 15 (2) 5

2(3) 5 (4)

152

82. ;fn P, Q rFkk R Øe'k% fcUnq A(1,1,1) ls leryP1 : x + 2y + 2z = 2, P2 : 2x – 2y + z = –8 rFkk P1

,oa P2 dh izfrPNsnh js[kk ij [khaps x, yEc ds ikn gks]

rks f=Hkqt PQR dk {ks=Qy gksxk -

(1) 3 (2) 32

(3) 2 (4) 34

83. ;fn a, b, c v'kwU; okLrfod la[;k;sa gksa] rks O;atd4 2 4 2 4 2

2 2 2

(a a 1)(b 7b 1)(c 11c 1)a b c

æ ö+ + + + + +ç ÷è ø

dk

U;wure eku gksxk -

(1) 315 (2) 351

(3) 415 (4) 451


H-29/31Kota/01CT314068


84. Let ƒ(x) = x3 + px + 1 and consider followingthree statements(i) for p > 0, ƒ(x) = 0 has one negative root

and ƒ(x) is monotonic(ii) for – 1 < p < 0, ƒ(x) = 0 has one negative

root and ƒ(x) is nonmonotonic(iii) for p < 0, ƒ(x) = 0 has three real and distinct

roots.Then -(1) Statements (i) & (ii) are false and (iii) is

true(2) Statements (i) & (ii) are true and (iii) is false(3) Statements (ii) & (iii) are true and (i) is false(4) Statement (i) & (iii) are true and (ii) is false

85. Area bounded by curve x(x2 + p) = y – 1 withy = 1 is -

(1) 2p

4(2)

p2 (3)

2p2

(4) p4

86. Number of solutions of equation |x2 – 2|x|| = 2x,is -(1) 1 (2) 2 (3) 3 (4) 4

87. z1 and z2 are two complex numbers such that

|z1 + z2| = 1 and 2 21 2z z 25+ = , then minimum

value of 3 31 2z z+ is-

(1) 24 (2) 42 (3) 37 (4) 33

88.n

2n

sin(n) en 1lim logn n e®¥

é + ùæ ö+ ç ÷ê ú+è øë û is equal to -

(1) 1ee

- (2) 1eee

-(3)

1 eee-

(4) 1

ee-

84. ekuk ƒ(x) = x3 + px + 1 rFkk fuEu rhu dFkuksa ijfopkj dhft,(i) p > 0 ds fy, ƒ(x) = 0 dk ,d ½.kkRed ewy

rFkk ƒ(x) ,dfn"V gksxkA(ii) – 1 < p < 0 ds fy, ƒ(x) = 0 dk ,d ½.kkRed

ewy rFkk ƒ(x) ,dfn"V ugha gksxkA(iii) p < 0 ds fy, ƒ(x) = 0 ds rhu okLrfod rFkk

fHkUu ewy gksaxsArc &(1) dFku (i) ,oa (ii) vlR; rFkk (iii) lR; gksxkA(2) dFku (i) ,oa (ii) lR; rFkk (iii) vlR; gksxkA(3) dFku (ii) ,oa (iii) lR; rFkk (i) vlR; gksxkA(4) dFku (i) ,oa (iii) lR; rFkk (ii) vlR; gksxkA

85. oØ x(x2 + p) = y – 1 rFkk y = 1 }kjk ifjc¼ {ks=Qygksxk -

(1) 2p

4(2)

p2 (3)

2p2

(4) p4

86. lehdj.k |x2 – 2|x|| = 2x ds gyksa dh la[;k gksxh -

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

87. z1 rFkk z2 nks lfEeJ la[;k;s a bl izdkj gS fd

|z1 + z2| = 1 rFkk 2 21 2z z 25+ = gks] rks 3 3

1 2z z+ dk

U;wure eku gksxk -

(1) 24 (2) 42 (3) 37 (4) 33

88.n

2n

sin(n) en 1lim logn n e®¥

é + ùæ ö+ ç ÷ê ú+è øë û cjkcj gksxk -

(1) 1ee

- (2) 1eee

-(3)

1 eee-

(4) 1

ee-

Kota/01CT314068H-30/31

Target : JEE (Main) 2015/27-03-2015


89. If ƒ(x) is invertible and twice differentiable

function satisfying

ƒ(x)1

0

ƒ '(x) ƒ (t)dt, x R-= " Îò and ƒ'(0) = 1

then ƒ'(1) can be -

(1) e (2) e2

(3) 1e (4) e

90. Consider a pair of circles (|x| – 1)2 + y2 = 1,

Ram is moving away from origin along one

circle in clockwise direction at the rate 2 m/s

and Shyam is moving away from origin along

other circle in anticlockwise direction at the rate

1 m/s. If Ram and Shyam start their journey

from origin, then rate of change of distance

between Ram and Shyam at the instant when

Ram crosses x-axis first time, is -

(1) 10 (2) 5

2

(3) 52

(4) 2 10p

89. ;fn ƒ(x) O;qRØe.kh; rFkk

ƒ(x)1

0

ƒ '(x) ƒ (t)dt, x R-= " Îò dks lUrq "V djus okyk

nks ckj vodyuh; Qyu rFk ƒ'(0) = 1 gks] rks ƒ'(1) gks

ldrk gS -

(1) e (2) e2

(3) 1e (4) e

90. ekuk o`Ùkksa dk ;q Xe (|x| – 1)2 + y2 = 1 gSA jke ewyfcUnqls nwj dh rjQ ,d o`Ùk ds vuqfn'k nf{k.kko`Ùk fn'kk esa

2 m/s dh nj ls xfr djrk gS rFkk ';ke ewyfcUnq ls nwj

dh rjQ nwljs o`Ùk ds vuqfn'k okekorZ fn'kk esa 1 m/s

dh nj ls xfr djrk gSA ;fn jke rFkk ';ke mudh ;k=k

ewyfcUnq ls izkjEHk djrs gS] rks jke rFkk ';ke ds e/; nwjh

esa ifjorZu dh nj tc jke izFke ckj x-v{k ikj djrk

gS] gksxh -

(1) 10 (2) 5

2

(3) 52

(4) 2 10p


H-31/31Kota/01CT314068



allen t m paper code 01ct3140 68 · paper code 01ct3140 68 p nstructio ully. y ll 5 in specifical ....

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