CAREER INSTITUTE Pat S KOTA (RAJASTHAN) CLASSROOM …€¦ · SCORE – I DATE : 18 - 01 - 2015...

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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

SCORE – I DATE : 18 - 01 - 2015

0 1 C E 2 1 4 0 2 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 # 09 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

Enthusiast Course/Score-I/18-01-2015

1/27Kota/01CE214024

SPACE FOR ROUGH WORK

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

1. A charge Q is distributed over a line of length L.Another point charge q is placed at a distance r fromthe centre of the line distribution. Then the forceexpericed by q is :-

+ + ++ + +r

qQ

L

(1) 2 20

qQ4 (r L )pÎ - (2) 2 2

0

qQ4 (r (L / 2) )pÎ -

(3) 20

qQ4 rpÎ (4) 3

0

qQL4 rpÎ

2. Three identical metal plates with large surface areasare kept parallel to each other as shown in figure.The left most plates is given a charge Q and theright most plate C is given a charge –2Q. Themiddle plate B is neutral. Then the charge appearingon the outer surface S of the plate C is :-

A B C

–2QQ

S

(1) –Q/2 (2) –Q (3) +Q (4) +2Q

1. ,d L yEckbZ dh iryh NM+ ij Q vkos'k ,d leku forfjrgSA ,d fcUnq vkos'k q bl NM+ dh yEckbZ ds vuqfn'k NM+ds dsUæ ls r nwjh ij j[kk gqvk gSA rc q }kjk vuqHko fd;kx;k cy gS :-

+ + ++ + +r

qQ

L

(1) 2 20

qQ4 (r L )pÎ - (2) 2 2

0

qQ4 (r (L / 2) )pÎ -

(3) 20

qQ4 rpÎ (4) 3

0

qQL4 rpÎ

2. rhu /kkrq dh ,d leku ifêdk;sa ftudh i"B {ks=Qy vf/kdgS] ,d nwljs ds lekukUrj fp=kuqlkj j[kh gqbZ gSaA lcls ck;havksj dh IysV A ij vkos'k Q fn;k x;k gS vkSj lcls nk;havksj dh IysV ij vkos'k –2Q fn;k x;k gSA e/; okyh IysVB mnklhu gSA rc IysV C dh cká lrg S ij vkos'kgksxk :-

A B C

–2QQ

S

(1) –Q/2 (2) –Q (3) +Q (4) +2Q

Kota/01CE2140242/27

Target : JEE (Main) 2015/18-01-2015


3. A proton is fired at an initial velocity of 150 m/sat an angle of 60º above the horizontal into auniform electric field of 2 × 10–4 N/C between twocharged parallel plates as shown in figure. Then thetotal time the particle is in motion is :-

+ + + + + + +

– – – – – – –q q

E

(1) 1.35 × 10–2 s (2) 6.76 × 10–3 s

(3) 2.70 × 10–2 s (4) None of the above

4. In the circuit diagram of figure, E = 5 volt,r = 1W, R2 = 4W, R1 = R3 = 1W and C = 3 µF. Thenthe magnitude of the charge on each capacitor plateis :-

+ – E,r

CC

CC

R3

R2

R1

(1) 6 µC (2) 12 µC

(3) 24 µC (4) 0 µC

3. nks {kSfrt vkosf'kr IysVksa ds e/; fo|eku ,dleku Å/okZ/kjfo|qr~ {ks= rhozrk 2 × 10–4 N/C esa ,d izksVkWu] {kSfrt ls60º ds dks.k ij 150 eh/ls ds osx ls izsf{kr fd;k tkrk gS(ns[ksa fp=)A rc izksVkWu ds xfr'khy jgus dk dqy le;gS :-

+ + + + + + +

– – – – – – –q q

E

(1) 1.35 × 10–2 s (2) 6.76 × 10–3 s

(3) 2.70 × 10–2 s (4) mijksDr esa ls ugha

4. fp= esa fn[kk;s x;s ifjiFk esa E = 5 oksYV, r = 1W,R2 = 4W, R1 = R3 = 1W vkSj C = 3 µF gSA rc izR;sdla/kkfj= dh IysV ij vkos'k dk ifj.kke gS :-

+ – E,r

CC

CC

R3

R2

R1

(1) 6 µC (2) 12 µC

(3) 24 µC (4) 0 µC


3/27Kota/01CE214024


5. Five indentical capacitor plates, each of area A, arearranged such that adjacent plates are at distanced apart. The plates are connected to a source of emfV as shown in fig. Then the charges 1 and 4 are,respectively :-

+–

V1 2 3 4 5

(1) Î0AV/d, 2Î0AV/d

(2) 2Î0AV/d, –2Î0AV/d

(3) Î0AV/d, –2Î0AV/d

(4) Î0AV/d, –Î0AV/d

6. The percentage error in the reading of the

12kW

+ –

4V4kW 4kW

V

voltmeter in the figure shown here is nearly :-

(1) 14% (2) 28% (3) 7% (4) 0.71%

5. ,d tSlh ik¡p IysVsa] izR;sd dk {ks=Qy A bl izdkj lekUrjj[kh xbZ gSa fd utnhdh IysVksa ds e/; leku nwjh d gSA buIysVksa dks fp=kuqlkj ,d V fo|qr okgd cy ds lzksr ls tksM+kx;k g SA rc Iy sV 1 vk Sj Iy sV 4 ij vkos'k gk sxk]Øe'k% :-

+–

V1 2 3 4 5

(1) Î0AV/d, 2Î0AV/d

(2) 2Î0AV/d, –2Î0AV/d

(3) Î0AV/d, –2Î0AV/d

(4) Î0AV/d, –Î0AV/d

6. ;gk¡ fp= dh O;oLFkk ds fy;s oksYVehVj dh ikB~;kad esaizfr'kr = qfV gS] yxHkx :-

12kW

+ –

4V4kW 4kW

V

(1) 14% (2) 28% (3) 7% (4) 0.71%

Kota/01CE2140244/27

Target : JEE (Main) 2015/18-01-2015


7. In a potentiometer experiment shown here, for theposition X of the jockey J, there occurs a nulldeflection in the galvanometer. Then the potentialdifference between points A and X is :-

8W

J1.5VA

G

B

2VRh

XK

(1) 1.0V (2) 1.5V (3) 2.0V (4) 1.75V8. A potentiometer consists of two wires AC and CB

of same material and of equal lengths but diameters

A B

Ep

Rh

C

K

in the ratio 3 : 1. Then the potential gradients onthe two wires will be in the ratio :-(1) 3 : 1 (2) 1 : 3 (3) 9 : 1 (4) 1 : 9

9. A plastic disc of radius R has a charge q uniformlydistributed over its surface. If the disc is rotated atan angular frequency w about it axis, the inductionat the center of the disc is :-

(1) 0µ q2 Rp

(2) 0µ q2 R

wp

(3) 0µ (2 q)Rp

w(4) 0µ q

2wp

7. fp= esa fn[kk;s x;s foHkokekih ds ,d iz;ksx esa tkWdh J dhfLFkfr X ij /kkjkekih G esa 'kwU; fo{ksi izkIr gksrk gSA rcfcUnqvksa A vkSj X ds e/; foHkokUrj gS :-

8W

J1.5VA

G

B

2VRh

XK

(1) 1.0V (2) 1.5V (3) 2.0V (4) 1.75V8. ,d foHkoekih esa nks rkj AC vkSj CB Js.khØe esa yxs gq;s

gSaA bu rkjksa dh yEckbZ;k¡ leku gSa ysfdu buds O;klksa dkvuqikr 3 : 1 gSA rc bu nks rkjksa ij foHko izo.krkvksa dkvuqikr gksxk :-

A B

Ep

Rh

C

K

(1) 3 : 1 (2) 1 : 3 (3) 9 : 1 (4) 1 : 99. R f=T;k dh ,d IykfLVd dh pdrh ij q vkos'k dh ek=k

,dleku :i ls i`"B ij forfjr dh gqbZ gSA pdrh viuhv{k ij w dks.kh; osx ls ?kwe jgh gSA rc pdrh ds dsUæij pqEcdh; izsj.k lfn'k dk ifjek.k gksxk :-

(1) 0µ q2 Rp

(2) 0µ q2 R

wp

(3) 0µ (2 q)Rp

w(4) 0µ q

2wp


5/27Kota/01CE214024


10. A particle of mass m = 1.67 × 10–27 kg and chargeq = 1.6 × 10–19 C enters a region of uniformmagnetic field of strength 1 tesla along the directionshown in the figure. The speed of the particle is

X X X X

B

D

C

q

mq 45º

X X X X

X X X X

X X X X

107 m/s. The magnetic field is directed along theinward normal to the plane of the paper. The particleenters the field at C and leaves at D. Then the angleq must be :-(1) 0º (2) 30º (3) 45º (4) 60º

11. In the above question, the radius of the circularportion of the path is :-(1) 0.104m (2) 0.148m(3) 0.074m (4) None of the above

12. In question 10, the particle leaves the magnetic fieldat point D, then the distance CD is :-(1) 0.148m (2) 0.104m(3) 0.208m (4) None of the above

13. In question 10, the time spent by the particle in themagnetic field is :-(1) 16.3 ns (2) 32.6 ns(3) 8.2 ns (4) 24.5 ns

10. æO;eku m = 1.67 × 10–27 fdxz k rFk k vkos'k1.6 × 10–19 dwykWe dk ,d d.k] 1 Vslyk rhozrk ds ,dlekuyEcor~ pqEcdh; {ks= esa] fp= esa fn[kkbZ xbZ fn'kk ds vuqlkjizos'k djrk gSA d.k dh pky 107 eh@ls gSA ry ds vfHkyEcuhps dh vksj gSA d.k] pqEcdh; {ks= esa fcUnq C ij izos'kdjrk gS rFkk fcUnq D ij ckgj fudyrk gSA rc dks.k q (ns[ksafp=) gksuk pkfg;s :-

X X X X

B

D

C

q

mq 45º

X X X X

X X X X

X X X X

(1) 0º (2) 30º (3) 45º (4) 60º11. mijksDr iz'u esa pqEcdh; {ks= ds vUnj d.k ds o`Ùkkdkj iFk

dh f=T;k gS :-(1) 0.104m (2) 0.148m(3) 0.074m (4) mijksDr esa ls dksbZ ugha

12. iz'u 10 ds fy;s nwjh CD gksuh pkfg;s :-(1) 0.148 eh (2) 0.104 eh

(3) 0.208 eh (4) mijksDr esa ls ugha

13. iz'u 10 esa d.k }kjk pqEcdh; {ks= esa O;rhr fd;k x;k le;gS :-(1) 16.3 ns (2) 32.6 ns(3) 8.2 ns (4) 24.5 ns

Kota/01CE2140246/27

Target : JEE (Main) 2015/18-01-2015


14. If the direction of the magnetic field in the question10, is along the outward normal to the plane of thepaper, then the time spent by the particle in theregion of the magnetic field after entering it at Cis nearly :-(1) 16 ns (2) 44 ns (3) 49 ns (4) 32 ns

15. A particle of charge q and mass m is moving alongthe x-axis with a velocity v, and enters a region ofelectric field E and magnetic field B as shown infigures below. For which figure the net force on thecharge may be zero :-

(1)

Y

vq

E

Z

B

X(2)

Y

vq

B

Z

E

X

(3)

Y

vq

Z

E

X

B

(4)

Y

vq

Z

EX

B

16. A magnetized steel wire of length L has a magneticmoment M. It is then bent into a semicircular arc.The new magnetic moment is :-(1) M (2) M/p (3) Mp (4) 2M/p

14. iz'u 10 esa ;fn pqEcdh; {ks= dh fn'kk myV dj] i`"B ryds yEcor~ Åij dh vksj dj nh tk;s rc d.k }kjk pqEcdh;

{ks= esa O;rhr fd;k x;k le; gksxk :-

(1) 16 ns (2) 44 ns(3) 49 ns (4) 32 ns

15. q vkos'k rFkk m æO;eku dk ,d d.k x-v{k ds vuqfn'kv osx ls xfr dj jgk gSA ;g ,d ,sls {ks= esa izos'k djrkgS tgk¡ fo|qr {ks= E vkSj pqEcdh; {ks= B nksuksa mifLFkr gSaAfdl fp= dh O;oLFkk ds fy;s vkos'k ij yxus okys ifj.kkehcy dk eku 'kwU; gks ldrk gS :-

(1)

Y

vq

E

Z

B

X(2)

Y

vq

B

Z

E

X

(3)

Y

vq

Z

E

X

B

(4)

Y

vq

Z

EX

B

16. ,d pqEcdh; LVhy rkj dh yEckbZ L rFkk pqEcdh; vk?kw.kZM gSA vc bls v/kZo`Ùkkdkj vkdkj esa eksM + fn;k x;k gSAbldk u;k pqEcdh; vk?kw.kZ gS :-(1) M (2) M/p (3) Mp (4) 2M/p


7/27Kota/01CE214024


17. Angle of dip at a place is 30º, and the horizontalcomponent of earth's field is 5 × 10–5 tesla, then thevertical component and the total magnetic field atthat place is :-

(1) 5 × 10–5 T, 10 × 10–5 T

(2) (5 × 3 ) × 10–5 T, (10/ 3 ) × 10–5 TT

(3) 5 3 × 10–5 T, 10 3 × 10–5 T

(4) None of the above

18. The magnetic suceptibility of a material is7.54 × 10–3. The nature and relative premeabilityof the substance is :-

(1) diamagnetic, 0.99246

(2) paramagnetic, 1.00754

(3) ferromagnetic, 7.5 × 103

(4) none of these

19. The work done in turning a magnet of moment Mby an angle q from initial position parallel tomagnetic meridian is :-

(1) MBH cosq (2) MBH (1 – cos q)

(3) MBH (1 + cos q) (4) MBH sin q

17. fdlh LFkku ij ufrdks.k 30º gS rFkk i`Foh ds pqEcdh; {ks=

dk {kSfrt ?kVd 5 × 10–5 Vslyk gSA ml LFkku ij i`Foh ds

pqEcdh; {ks= dk Å/okZ/kj ?kVd D;k gS :-

(1) 5 × 10–5 T, 10 × 10–5 T

(2) (5 × 3 ) × 10–5 T, (10/ 3 ) × 10–5 TT

(3) 5 3 × 10–5 T, 10 3 × 10–5 T

(4) None of the above

18. ,d inkFkZ dh pqEcdh; izo`fÙk 7.54 × 10–3 gSA bl inkFkZ

dh izd`fr vkisf{kr ikjxE;rk fuEu gS :-

(1) izfrpqEcdh;, 0.99246

(2) vuqpqEcdh;, 1.00754

(3) yksgpqEcdh;, 7.5 × 103

(4) mijksDr esa ls dksbZ ugha

19. M pqEcdh; vk?kw.kZ okyk ,d pqEcd pqEcdh; ;kE;ksÙkjds vuqfn'k j[kk gqvk gSA bls q dks.k ls ?kqekus ds fy;s fd;kx;k dk;Z gksxk :-

(1) MBH cosq (2) MBH (1 – cos q)

(3) MBH (1 + cos q) (4) MBH sin q

Kota/01CE2140248/27

Target : JEE (Main) 2015/18-01-2015


20. The magnetic flux passing perpendicular to theplane of the coil and directed into the paper isvarying according to the relation.

f = 3t2 + 2t + 3where f is in milliwebers and t is in seconds. Thenthe magnitude of emf induced in the loop whent = 2 second is :-

X X X X X

R

X X X X X

X X X X X

X X X X X

(1) 31 mV (2) 19 mV (3) 14 mV (4) 6 mV21. A conductor of length l and mass m can slide along

a pair of vertical metal guides connected by aresistance R, as shown in fig. Friction, resistanceof conductor and guide rails are negligible. Thereexist a horizontal uniform magnetic field of strengthB normal to the plane of the page and is directedoutward. The terminal speed of fall under theunfluence of gravity is, then :-(1) Zero

l

v

(2) mg/BlR

(3) 2 22mg

B l

(4) 2 2mgRB l

20. ,d dq.Myh ds ry ds yEcor rFkk uhps dh vksj xqtjusokyk pqEcdh; ¶yDl le; ds lkFk fuEu izdkj ifjofrZrgksrk gS,

f = 3t2 + 2t + 3tgk¡ f feyh oscj esa rFkk t lsd.M esa gSA rc t = 2 lsd.Mle; ij ywi esa izsfjr fo|qr okgd cy gS :-

X X X X X

R

X X X X X

X X X X X

X X X X X

(1) 31 mV (2) 19 mV(3) 14 mV (4) 6 mV

21. l yEckbZ m æO;eku dh ,d pkyd NM+, Å/okZ/kj j[kh /kkrqdks nks LFkkid NM+ksa ftuds Åijh fljs ,d izfrjks/k R ls tqM+s

gq;s gSa] ij fQly ldrh gSA ?k"k.kZ] pkyd NM+ vkSj LFkkidNM+ksa dk izfrjks/k ux.; gSA ml LFkku ij ,d ,dleku {kSfrtpqEcdh; {ks= B mifLFkr gSa (ns[ksa fp=) tks NM+ksa ls cuus

okys ry ds yEcor~ gSA ;fn NM+ xq:Roh; cy ds dkj.k

uhps fxjus yxs rc bldh vfUre pky gksxh :-(1) 'kwU;

l

v

(2) mg/BlR

(3) 2 22mg

B l

(4) 2 2mgRB l


9/27Kota/01CE214024


22. In an AC circuit shown in fig., which AC voltmeterwould show zero reading at the time of resonance:-

V1 V2 V3

V4

~E

L RC

(1) V1 and V2 (2) V3

(3) V4 (4) None of the above23. What is the reading in the AC ammeter for the ac

circuit shown in fig. :-

~240V X

= 4

8C

W

X =

24

LWA

(1) 2.0 A (2) 2.4 A(3) 5.0 A (4) 10.0 A

24. A long solenoid connected to a 12V dc sourcepasses a steady current of 2A. When the solenoidis connected to a source of 12V rms at 50Hz, thecurrent flowing is 1A rms. Then the inductance ofthe solenoid :-(1) 11 mH(2) 22 mH(3) 33 mH(4) None of the above

22. fp= esa ,d ,lh ifjiFk fn[kyk;k x;k gSA bl ifjiFk dhvuquknh voLFkk esa dkSu ls ,lh oksYVehVj dk ikB~;kad 'kwU;gksxk :-

V1 V2 V3

V4

~E

L RC

(1) V1 vkSj V2 (2) V3

(3) V4 (4) fdlh dk Hkh ugha23. fp= esa fn[kyk;s x;s ,lh ifjiFk ds fy;s ,lh vehVj dk

ikB~;kad D;k gS :-

~240V X

= 4

8C

W

X =

24

LWA

(1) 2.0 A (2) 2.4 A(3) 5.0 A (4) 10.0 A

24. ,d yEch ifjukfydk ds fljksa ij tc 12V fookc dk fn"V/kkjk lzksr yxk;k tkrk gS rc blesa 2A /kkjk cgrh gSA blh

ifjukfydk ds fljksa ij fn"V /kkjk lzksr ds LFkku ij tc 12VoxZ-ek/;-ewy eku izR;korhZ /kkjk lzksr (vkofÙk = 50 Hz)yxk;k tkrk gS rc 1A oxZ-ek/;-ewy eku dh /kkjk cgrh

gSA bl ifjukfydk dk izsjdRo D;k gS :-(1) 11 mH (2) 22 mH(3) 33 mH (4) mijksDr esa ls dksbZ ugha

Kota/01CE21402410/27

Target : JEE (Main) 2015/18-01-2015


25. The threshold wavelength of tungsten is 2300 Å.If ultraviolet light of wavelength 1800 Å is incidenton it, then the maximum kinetic energy ofphotoelectrons would be about,(1) 1.5 eV (2) 2.2 eV (3) 3.0 eV (4) 5.0 eV

26. The stopping potential for a source of wavelength4000 Å, when kept at a distance of 10 cm is1.5 V. If now the distance of the source is increasedto 20 cm, then the stopping potential will be :-(1) 0.75 V (2) 1.5 V (3) 3.0 V (4) 0.37 V

27. A rectangular glass slab ABCD, of refractive indexn1, is immersed in water of refractive indexn2 (n1 > n2). A ray of light in incident at the surfaceAB of the slab as shown. The maximum value ofthe angle of incidence amax, such that the ray comesout only from the other surface CD is given by :-

A D

C B

n1 n2 amax

(1)1 11 2

2 1

n nsin cos sinn n

- -é ùæ öê úç ÷

è øë û

(2) 1 1

12

1sin n cos sinn


è øë û

(3) 1 1

2

nsinn

- æ öç ÷è ø

(4) 1 2

1

nsinn

- æ öç ÷è ø

25. VaXLVsu /kkrq ds fy;s nsgyh rjaxnS/;Z 2300 Å gSA bl ij ;fn1800 Å rjaxnS/;Z dk ijkcSaxuh izdk'k vkifrr fd;k tk;src mRlftZr QksVks-bysDVªkWuksa ds fy, vf/kdre xfrt ÅtkZgksxh] yxHkx,(1) 1.5 eV (2) 2.2 eV (3) 3.0 eV (4) 5.0 eV

26. tc 4000 Å rjaxnS/;Z ds ,d izdk'k lzks= dks QksVks dSFkksMls 10 lseh nwjh ij j[kk tkrk gS rc fujks/kh foHko 1.5 Vik;k tkrk gSA tc ;fn lzksr dh nwjh c<+dj 20 lseh djnh tk;s rc fujks/kh foHko gks tk;sxk :-(1) 0.75 V (2) 1.5 V (3) 3.0 V (4) 0.37 V

27. n1 viorZukad okyh ,d vk;rkdkj dk¡p dh iV~Vh ABCD,dks n2 (n1 > n2) viorZukad okys ty esa Mqcks;k x;k gSA,d fdj.k iV~Vh ds i`"B AB ij fp=kuqlkj vkifrr gksrhgSA vkiru dks.k dk vf/kdre eku amax D;k gksxk rkfdfdj.k dsoy CD i`"B ls ckgj fudys :-

A D

C B

n1 n2 amax

(1)1 11 2

2 1

n nsin cos sinn n


è øë û

(2) 1 1

12

1sin n cos sinn


è øë û

(3) 1 1

2

nsinn

- æ öç ÷è ø

(4) 1 2

1

nsinn

- æ öç ÷è ø


11/27Kota/01CE214024


28. The mean life of a radioactive material for alphadecay and beta decay are, respectively, 1620 yearsand 520 years. What is the half life of the sample(in years) ?

(1) 1483

(2) 394

(3) 273

(4) 1123

29. The energy released per fission of uranium-235 isabout 200 MeV. A reactor using U-235 as fuel isproducing 1000 kilowatts power. The number ofU-235 nuclei undergoing fission per sec is,approximately :-

(1) 106

(2) 2 × 108

(3) 3 × 1016

(4) 931

30. In a common base circuit of a transistor, theamplification factor is 0.95. The base current whenthe emitter current is 2mA, is :-

(1) 0.1 mA

(2) 0.2 mA

(3) 0.19 mA

(4) 1.9 mA

28. jsfM;ks,sfDVo inkFkZ ds ,d uewus ds fy,] a-{k; ds fy,vkSlr-vk;q 1620 o"kZ vkSj b-{k; ds fy, vkSlr-vk;q520 o"kZ ik;h tkrh gSA bl uewus dh v/kZ-vk;q (o"kks ± esa) D;kgS ?

(1) 1483

(2) 394

(3) 273

(4) 1123

29. U-235 ds fy, izfr fo[k.Mu fueq ZDr ÅtkZ yxHkx200 MeV gksrh gSA ,d fj,sDVj] ftlesa U-235 dks b ±/kuds :i esa mi;ksx esa yk;k x;k gS] 1000 kW 'kfDr dk mRiknudj jgk gSA blesa izfr lsd.M fo[k.Mu gksus okys U-235ukfHkdksa dh la[;k] yxHkx gS :-

(1) 106

(2) 2 × 108

(3) 3 × 1016

(4) 931

30. ,d VªkaftLVj ds mHk;fu"B vk/kkj foU;kl esa /kkjk izo/kZuxq.kkad 0.95 gSA bl foU;kl esa ;fn mRltZd /kkjk 2mA gksrc vk/kkj /kkjk gksxh :-

(1) 0.1 mA

(2) 0.2 mA

(3) 0.19 mA

(4) 1.9 mA

12/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

PART B - CHEMISTRY

31. Based on the following information arrange fourmetals A, B, C and D in order of decreasing abilityto act as reducing agents :-[I] Only A, B and C react with 1MHCl to give

H2(g)[II] When C is added to solutions of the other metal

ions, metallic B and D are formed.[III]Metal C does not reduce An+

(1) C > A > B > D (2) C > A > D > B(3) A > C > D > B (4) A > C > B > D

32. The standard electrode potential for the followingreaction is +1.33V. What is the potential at pH = 2.0Cr2O7

2– (aq 1M) + 14H+(aq) + 6e– ® 2Cr+3 (aq,1M)+ 7H2O (l)(1) 1.820 V (2) + 1.990 V(3) + 1.608 V (4) + 1.0542 V

33. If the anion (A) from hexagonal closest packing andcation (C) occupy only 2/3 octahedral voids in it,then the general formula of the compound is :-(1) CA (2) CA2

(3) C2A3 (4) C3A2

34. Which of the following electrolyte havingmaximum coagulation power for AS2S3 sol.(1) NaCl(2) MgCl2

(3) Al2(SO4)3

(4) NH4Cl

31. fuEufyf[kr lwpukvksa ds vk/kkj A, B, C rFkk D dks mudsvip;u {kerk ds ?kVrs Øe esa O;ofLFkr dhft, :-[I] A, B rFkk C 1MHCl ls fØ;k dj H2(g) nsrs gS[II] tc C dks nwljh /kkrq vk;u ds foy;u esa feyk;k rks

B rFkk D curs gSaA[III] /kkrq C] An+ dks vipf;r ugha djrs gSaA(1) C > A > B > D(2) C > A > D > B(3) A > C > D > B(4) A > C > B > D

32. fuEufyf[kr vfHkfØ;k dk ekud bysDVªksM + foHko dk eku+1.33V gS rks pH = 2.0 ij foHko dk eku D;k gksxk

Cr2O72– (aq 1M) + 14H+(aq) + 6e– ® 2Cr+3 (aq,1M)

+ 7H2O (l)(1) 1.820 V (2) + 1.990 V(3) + 1.608 V (4) + 1.0542 V

33. ;fn ½.kk;u (A) h.c.p. O;oLFkk j[krk gS rFkk /kuk;u (C)blds 2/3 v"VQydh; fjDr gS rks bl ;kSfxd dk lkekU;lw= gksxk :-(1) CA (2) CA2

(3) C2A3 (4) C3A2

34. fuEu esa ls dkSulk fo|qrvi?kV; vf/kdre LdUnu {kerkAS2S3 lkWy ds fy, j[krk gSA

(1) NaCl(2) MgCl2

(3) Al2(SO4)3

(4) NH4Cl


13/27


Kota/01CE214024

35. If S1, S2, S3 and S4 are the solubilities of AgCl inwater, in 0.01M CaCl2, in 0.01M NaCl and in0.05 M AgNO3 respectively at a certaintemperature, the correct order of solubilities is(1) S1 > S2 > S3 > S4 (2) S1 > S3 > S2 > S4

(3) S1 > S2 = S3 > S4 (4) S1 > S3 > S4 > S2

36. The equilibrium constants 1PK and

2PK for the

reaction X �� 2Y and Z �� P + Q,

respectively are in the ratio of 1 : 4. If the degreeof dissociation of X is 2 times that of Z, then theratio of total pressure (P1 : P2) at these equilibriais : (Assume degree of dissociation for both reactionsare very very small)(1) 1 : 36 (2) 1 : 16(3) 1 : 64 (4) None of these

37. At 100°C, the gaseous reaction A ® 2B + C isfound to be first order. Starting with pure A, if atthe end of 10 minutes, the total pressure of thesystem is 160 mm and after a long time, whendissociation of A was complete, it was 300 mm. Thepartial pressure of A at the end of 10 min is:-(1) 70 mm (2) 160 mm (3) 60 mm (4) 80 mm

38. Two solutions of KNO3 and CH3COOH are preparedseparately. Molarity of both is 0.1 M and osmoticpressures are P1 and P2 respectively. The correctrelationship between the osmotic pressures is :-(1) P1 = P2 (2) P1 > P2

(3) P2 > P1 (4) 1 2

1 2 1 2

P PP P P P

=+ +

35. ;fn AgCl dh foys;rk ty es a] 0.01M CaCl2 es a]0.01M NaCl ,oa 0.05 M AgNO3 esa Øe'k% S1, S2, S3

,oa S4 gS rks bldh foys;rk ds fy;s lgh Øe gksxk %&

(1) S1 > S2 > S3 > S4 (2) S1 > S3 > S2 > S4

(3) S1 > S2 = S3 > S4 (4) S1 > S3 > S4 > S2

36. vfHkfØ;kvksa X �� 2Y rFkk Z �� P + Q ds fy;s

lkE; fu;rkad 1PK ,oa

2PK ds ekuksa dk vuqikr Øe'k%

1 : 4 gSA ;fn X ds fo;kstu dh ek=k] Z ds fo;kstu dh

ek=k dh 2 xquh gS rks lkE; ij bu vfHkfØ;kvksa ds dqy nkcksa

ds vuqikr (P1 : P2) dk eku gksxk] (nksuksa vfHkfØ;kvksa ds

fo;kstu dh ek=k cgqr vYi gS)

(1) 1 : 36 (2) 1 : 16

(3) 1 : 64 (4) buesa ls dksbZ ugha

37. 100°C ij] xSlh; vfHkfØ;k A ® 2B + C, izFke dksfVdh izkIr dh xbZA 'kqð A ls izkjEHk djrs gq;s] 10 feuV dsckn] fudk; dk dqy nkc 160 mm gS vkSj vf/kd le ;ds ckn tc A iw.k Zr% fo;ksftr gks tkrk gS] rc nkc300 mm ik;k x;kA 10 feuV ds ckn] A dk vkaf'kd nkcgksxk :-(1) 70 mm (2) 160 mm (3) 60 mm (4) 80 mm

38. KNO3 rFkk CH3COOH ds foy;u i`Fkd rS;kj fd;ktkrk gSA nksuksa foy;uksa fd lkUnzrk 0.1 M rFkk ijklj.k nkcØe'k% P1 rFk k P2 gSA ijklj.k nkc es a lghlEcU/k :-(1) P1 = P2 (2) P1 > P2

(3) P2 > P1 (4) 1 2

1 2 1 2

P PP P P P

=+ +

14/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

39. The bond energy of C = C and C – C at 298 K are590 and 331 kJ mol–1 respectively. The enthalpyof polymerisation per mole of ethylene is :-(1) –70 kJ (2) – 72 kJ(3) + 72 kJ (4) –68 kJ

40. One mole of an ideal diatomic gas (CV = 5 cal) wastransformed from initial 25°C and 1L to the statewhen temperature is 100°C and volume 10L. Theentropy change of the process can be expressed as(R = 2 calories /mol / k) :-

(1) 3ln 298373

+ 2ln 10 (2) 5ln 373298

+ 2ln 10

(3) 7ln 373298

+ 2ln 1

10(4) 5ln

373298

+ 2ln 1

1041. Select the correct statements of the following :-

(a) Effective nuclear charge for nitrogen is 3.90(b) IP of Ne is more than Na+

(c) Order of electron negativity sp > sp2 > sp3

(d) Order of acidic character NH3 < PH3 < AsH3

(1) a, b, d (2) b, c(3) a, c, d (4) a, b, c, d

42. The Incorrect order of property of the following is:-(a) Basic nature Na2O > MgO > Al2O3 > SiO2

(b) 2nd I.P. Na > S > P > Si(c) Electron affinity O > S > Se(d) Size I– < I < I+

(1) a, c, d (2) b, c, d(3) c, d (4) a, b, c

39. 298 K rki ij C = C rFkk C – C ca/k dh ca/k ÅtkZ dkeku Øe'k% 590 rFkk 331 kJ mol–1 gSA ,Fkhfyu dscgqyhdj.k dh ,UFkSYih dk eku izfr mol esa gksxh %&(1) –70 kJ (2) – 72 kJ(3) + 72 kJ (4) –68 kJ

40. ,d eksy vkn'kZ f}vkf.od xSl (CV = 5 cal) dks izkjfEHkd25°C rki rFkk 1L nkc dks 100°C rki rFkk vk;ru 10Lrd ifjofrZr djrs gSaA izØe ds ,.Vªksih ifjorZu dks fuEu

ds }kjk n'kkZrs gSa (R = 2 calories /mol / k) :-

(1) 3ln 298373

+ 2ln 10 (2) 5ln 373298

+ 2ln 10

(3) 7ln 373298

+ 2ln 1

10(4) 5ln

373298

+ 2ln 1

1041. fuEufyf[kr esa ls lR; dFku dk p;u dhft;s :-

(a) ukbVªkstu dk izHkkoh ukfHkdh; vkos'k 3.90 gksrk gSA

(b) Ne ds IP dk eku Na+ ls vf/kd gksrk gS(c) fo|qr½.krk dk Øe sp > sp2 > sp3

(d) vEyh; y{k.kksa dk Øe NH3 < PH3 < AsH3 gksrk gS(1) a, b, d (2) b, c(3) a, c, d (4) a, b, c, d

42. fuEufyf[kr esa muds xq.kksa dk xyr Øe gS :-(a) {kkjh; izÏfr Na2O > MgO > Al2O3 > SiO2

(b) 2nd I.P. Na > S > P > Si(c) bysDVkWu ca/kqrk O > S > Se(d) vkdkj I– < I < I+

(1) a, c, d (2) b, c, d(3) c, d (4) a, b, c


15/27


Kota/01CE214024

43. How many S–S bond, S–O–S bond, s bonds andp bonds are present in trimer of SO3.(1) 0, 3, 16, 2 (2) 0, 3, 12, 6(3) 0, 6, 12, 16 (4) 0, 4, 12, 6

44. The bond angles of NH3, 4NH+ and 2NH- are in

the order :-

(1) NH2 > NH3 > 4NHÅ

(2) NH3 > NH2 > 4NHÅ

(3) 4NHÅ

> NH3 > NH2 (4) NH3 > 4NHÅ

> NH2

45. One mole of the complex compound Co(NH3)5Cl3,gives 3 moles of ions on dissolution in water. Onemole of the same complex reacts with two molesof AgNO3 solution to yield two moles of AgCl(s).The structure of the complex is :-(1) [Co(NH3)5Cl]Cl2(2) [Co(NH3)3Cl3]2NH3(3) [Co(NH3)4Cl2]Cl.NH3(4) [Co(NH3)4Cl]Cl2.NH3

46. Match list-I with list-II and select List-I List-II

(A) Zeiglernatla (i) Fe4[Fe(CN)6]3(B) Brown ring complex (ii) Fe[(H2O)5NO]SO4(C) Prussian Blue (iii) Al(C2H5)3 + TiCl4(D) Turnbull Blue (iv) Fe3(Fe(CN)6]2Correct code is :-

A B C D(1) (iii) (ii) (iv) (i)(2) (iii) (i) (iv) (ii)(3) (iii) (ii) (i) (iv)(4) (i) (ii) (iii) (iv)

43. SO3 ds f=yd esa S–S ca/kksa] S–O–S ca/kksa] s ca/kksa rFkkp ca/kksa dh la[;k Øe'k% gksxh :-(1) 0, 3, 16, 2 (2) 0, 3, 12, 6(3) 0, 6, 12, 16 (4) 0, 4, 12, 6

44. NH3, 4NH+ o 2NH- es a c a /k dk s. k k s a dk lgh Øe

gksxk :-

(1) NH2 > NH3 > 4NHÅ

(2) NH3 > NH2 > 4NHÅ

(3) 4NHÅ

> NH3 > NH2 (4) NH3 > 4NHÅ

> NH2

45. Co(NH3)5Cl3 ladqy ;kSfxd dk ,d eksy] ty esa foys;gksus ij vk;uksa ds rhu eksy nsrk gSA blh ladqy dk ,d eksyAgNO3 foy;u ds nks eksy ls fØ;k djrk gS ,oa AgCl(s)ds nk s ek sy nsrk g SA rk s l adqy dh lajpukgksxh :-(1) [Co(NH3)5Cl]Cl2(2) [Co(NH3)3Cl3]2NH3(3) [Co(NH3)4Cl2]Cl.NH3(4) [Co(NH3)4Cl]Cl2.NH3

46. fyLV-I rFkk fyLV-II dk lgh feyku dhft,A lwpha-I lwpha-II

(A) ftXyj ukVk mRizsjd (i) Fe4[Fe(CN)6]3(B) Hkwjh oy; ladqy (ii) Fe[(H2O)5NO]SO4(C) iz;qfl;u Cyw (iii) Al(C2H5)3 + TiCl4(D) VuZcwy Cyw (iv) Fe3(Fe(CN)6]2lgh dksM + gS :-

A B C D(1) (iii) (ii) (iv) (i)(2) (iii) (i) (iv) (ii)(3) (iii) (ii) (i) (iv)(4) (i) (ii) (iii) (iv)

16/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

47. Poling process is used :-(1) for the removal of Cu2O from Cu(2) for the removal of Al2O3 from Al(3) for the removal of Fe2O3 from Fe(4) in all the above

48. µ 15= is true for the pair :-

(1) Co+2, Cr+3 (2) Fe+2, Cr+3

(3) Fe+3, Fe+2 (4) Mn+2, Fe+2

49. Colour of the bead in borax bead test is mainly dueto the formation of :-(1) metal oxides(2) boron oxide(3) metal metaborates(4) elemental boron

50. Gypsum CaSO4.2H2O on heating to about 120°Cforms a compound which has the chemicalcomposition represented by :-(1) CaSO4

(2) 2CaSO4.H2O(3) CaSO4.H2O(4) 2CaSO4.3H2O

51. The IUPAC name of the compound havingstructure C H – C – CH – CH2 5 3

CH CH2 3

is :-

(1) 3-methyl-2-ethyl butene-1(2) 2-ethyl-3-methyl butene-1(3) 3-ethyl-3-methyl butene-1(4) Ethyl isopropyl ethane

47. ikWfyax izØe dke ysrs gSa :-(1) Cu ls Cu2O gVkus ds fy,(2) Al ls Al2O3 gVkus ds fy,(3) Fe ls Fe2O3 gVkus ds fy,(4) mijksDr lHkh

48. µ 15= fuEu ;qXe ds fy, lR; gS :-

(1) Co+2, Cr+3 (2) Fe+2, Cr+3

(3) Fe+3, Fe+2 (4) Mn+2, Fe+2

49. lqgkxk eudk ijh{k.k esa eudk dk jax fuEu dh mifLFkrhds dkj.k gksrk gS %&

(1) /kkrq vkWDlkbM(2) cksjkWu vkWDlkbM

(3) /kkrq esVk cksjsV(4) ijek.oh; cksjkWu

50. ftIle CaSO4.2H2O dks 120°C rd xeZ djus ls ,d

;kSfxd curk gS ftldk jklk;fud laxBu gS :-

(1) CaSO4

(2) 2CaSO4.H2O

(3) CaSO4.H2O

(4) 2CaSO4.3H2O51. fuEu ;kSfxd dk IUPAC uke gS

C H – C – CH – CH2 5 3

CH CH2 3

(1) 3-esfFky-2-,fFky C;wVhu-1(2) 2-,fFky-3-esfFky C;wVhu-1(3) 3-,fFky-3-esfFky C;wVhu-1(4) ,fFky vkblksizksfiy ,Fkhu


17/27


Kota/01CE214024

52. What is the decreasing order of reactivity amongestthe following compounds towards aromaticelectrophilic substitution.(i) Chlorobenzene(ii) Benzene(iii) Anilinium chloride(iv) Toluene(1) i > ii > iii > iv(2) iv > ii > i > iii(3) ii > i > iii > iv(4) iii > i > ii > iv

53. C7H9N has how many isomeric forms that containa benzene ring(1) 4 (2) 5(3) 6 (4) 7

54. Action of hydrogen chloride on CH – C = CH3 2

CH3

and

on CH º CH will predominantly give thecompounds, respectively :-

(1) CH – CH – CH Cl3 2

CH3

and CH2Cl–CH2Cl

(2) CH – CCl – CH3 3

CH3

and CH3–CHCl2

(3) CH – CH – CH Cl3 2

CH3

and CH3–CHCl2

(4) CH – CH – CH3 3

CH3

and CH2Cl–CH2Cl

52. ,jkseSfVd bysDVªkWuLusgh izfrLFkkiu ds izfr fuEu ;kSfxdksa dschp fØ;k'khyrk dk ?kVrk Øe gSA

(i) DyksjkscsUthu

(ii) csUthu

(iii) ,uhfyfu;e DyksjkbM

(iv) VkWyqbZu(1) i > ii > iii > iv(2) iv > ii > i > iii(3) ii > i > iii > iv(4) iii > i > ii > iv

53. csUthu pØ;qDr C7H9N ds leko;fo;k s a dh la[;kgS(1) 4 (2) 5(3) 6 (4) 7

54. CH – C = CH3 2

CH3

vkSj CH º CH ij gkbMªkstu DyksjkbM

dh vfHkfØ;k ls izkIr gksus okys ;kSfxd izeq[kr % gksaxs]Øe'k%

(1) CH – CH – CH Cl3 2

CH3

and CH2Cl–CH2Cl

(2) CH – CCl – CH3 3

CH3

and CH3–CHCl2

(3) CH – CH – CH Cl3 2

CH3

and CH3–CHCl2

(4) CH – CH – CH3 3

CH3

and CH2Cl–CH2Cl

18/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

55. At low temperatures, the slow addition of molecularbromine to CH2 = CH–CH2–C º CH gives :-(1) CH2=CH–CH2–CBr = CHBr(2) BrCH2–CHBr–CH2–C º CH(3) CH2=CH–CH2–CH2–CBr3(4) CH3–CBr2–CH2–C º CH

56. The compound CH – C = CH – CH3 3

CH3

on reactionwith NaIO4 in the presence of KMnO4 gives :-(1) CH3CHO + CO2(2) CH3COCH3(3) CH3COCH3 + CH3COOH(4) CH3COCH3 + CH3CHO

57. Replacement of Cl of chlorobenzene to givephenol requires drastic conditions but chlorine of2, 4-dinitrochlorobenzene is readily replacedbecause.(1) NO2 make ring electron rich at ortho and para(2) NO2 withdraws e– from meta position(3) denotes e– at meta position(4) NO2 withdraws e– from ortho/para positions

58. Which will undergo a Friedel-Craft's alkylation reaction:CH3

NO2

(i) (ii) (iii) (iv)

CH CH2 3 COOH OH

(1) i, ii and iv (2) i and iii(3) ii and iv (4) i and ii

55. fuEu rkieku ij CH2 = CH–CH2–C º CH esa /khjs-/khjsczksehu feykus ij ;g nsrk gS :-(1) CH2=CH–CH2–CBr = CHBr(2) BrCH2–CHBr–CH2–C º CH(3) CH2=CH–CH2–CH2–CBr3(4) CH3–CBr2–CH2–C º CH

56. ;kSfxd CH – C = CH – CH3 3

CH3

, KMnO4 dh mifLFkfr esaNaIO4 ds lkFk vfHkfØ;k djus ij nsrk gS :-(1) CH3CHO + CO2(2) CH3COCH3(3) CH3COCH3 + CH3COOH(4) CH3COCH3 + CH3CHO

57. DyksjkscsUthu ls Cl dk foLFkkiu fo"ke ifjfLFkfr;ksa esa gksrkgS rFkk fQukWy curk gSA ijUrq 2, 4-MkbZukbVªksDyksjkscsUthudk Dyksjhu ljyrk ls foLFkkiuh; gS] D;ksafd(1) NO2 pØ esa vkWFkksZ rFkk iSjk dsUæksa ij bysDVªkWu ?kuRo

c<+krk gSA(2) NO2 esVk fLFkfr ls bysDVªkWu ys ysrk gSA(3) NO2 esVk fLFkfr ij bysDVªkWu iznku djrk gSA(4) NO2 vkWFkksZ rFkk iSjk fLFkfr ls bysDVªkWu ys ysrk gSA

58. dkSulk QhMy-ØkW¶V ,Ydkbyhdj.k esa Hkkx ysrk gS :-CH3

NO2

(i) (ii) (iii) (iv)

CH CH2 3 COOH OH

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


19/27


Kota/01CE214024

59. Which is not true about acetophenone :-

(1) Reacts to form 2,4-dinitorphenyl hydrazine

(2) Reacts with Tollen's reagent to form silvermirror

(3) Reacts with I2/NaOH to form iodoform

(4) On oxidation with alkaline KMnO4 followedby hydrolysis gives benzoic acid

60. The structural formula of an amino acid, isoleucineis :-

(1)CH – CH.COOH3

NH2

(2)

NH2

CH–CH.COOHCH3

CH3

(3)

NH2

CH–CH.COOHCH3

C H2 5

(4)

NH2

CH–CH.COOHC H2 5

C H2 5

59. ,lhVksfQukWu ds lEcU/k esa xyr dFku gS :-

(1) 2,4-MkbZukbVªks Qsfuy gkbMªkthu cukus ds fy;s fØ;k

djrk gSA

(2) VkWysu vfHkdeZd ls fØ;k djds jtr niZ.k cukrk gSA

(3) I2/NaOH ds lkFk fØ;k djds vk;ksMksQkWeZ cukrk gSA

(4) {kkjh; KMnO4 ds lkFk vkWDlhdj.k ls cus mRikn dk

ty vi?kVu djus ij csUtksbd vEy cukrk gSA

60. vkblksY;wflu ,ehuks vEy dk lajpukRed lw= gS :-

(1)CH – CH.COOH3

NH2

(2)

NH2

CH–CH.COOHCH3

CH3

(3)

NH2

CH–CH.COOHCH3

C H2 5

(4)

NH2

CH–CH.COOHC H2 5

C H2 5

20/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

PART C - MATHEMATICS

61. If A is a square matrix of 3 × 3 order, and |A| = 2then |(A–AT)6| + |(AT–A)7| is equal to (where AT

donotes the transpose of matrix A).(1) 0 (2) 26 + 27 (3) 26 – 27 (4) 2

62. A = {x1, x2, x3, x4}; B = {y, y2, y3, y4}. A function isdefined from set A to set B. Number of one-one functionsuch that f(xi) ¹ yi for i = 1, 2, 3, 4 is equal to:(1) 2 (2) 9 (3) 44 (4) 256

63. All the five digits numbers in which each successivedigit exceeds its predecessor are arranged in theincreasing order of their magnitude. The 97 th

number in the list does not contain the digit:(1) 4 (2) 5 (3) 7 (4) 8

64. There are 3 bags, each containing 5 white balls and3 black balls. Also there are 2 bags, each containing2 white balls and 4 black balls. A white ball is drawsat rondom. Find the probability that this white ballis from a bag of the first group.

(1) 1661

(2) 1561

(3) 4561

(4) None of these

65. If logtan30°

æ ö+ -ç ÷+è ø

22 | z | 2 | z | 3| z | 1 < –2 then

(1) | z | < 32

(2) | z | > 32

(3) | z | > 2 (4) | z | < 2

61. ;fn A ,d 3 × 3 dksfVh dk oxZ esfVªDl gks rFkk |A| = 2,rks |(A–AT)6| + |(AT–A)7| dk eku (tgk¡ AT esfVªDl A dkifjorZ esVªhDl iznf'kZr djrk gS) gS&(1) 0 (2) 26 + 27 (3) 26 – 27 (4) 2

62. A = {x1, x2, x3, x4}; B = {y, y2, y3, y4} ,d QyuleqPp; A ls leqPp; B esa ifjHkkf"kr gSA i = 1, 2, 3, 4ds fy,] ,dSdh Qyuksa dh la[;k tc f(xi) ¹ yi gS] gSa&(1) 2 (2) 9 (3) 44 (4) 256

63. ik¡p vadks dh cuus okyh oks lHkh la[;k, ftudk izR;sdmÙkjksrj vad vius iwoksZÙkj vad ls cM+k gks] dks o/kZeku (c<+rsgq,) Øe esa O;ofLFkr djus ij 97 osa LFkku ij vkus okyhla[;k esa dkSulk vad ugha gksxkA(1) 4 (2) 5 (3) 7 (4) 8

64. rhu csx gS ftuesa izR;sd esa 5 lQsn rFkk 3 dkyh xsans gS rFkknks vkSj csx gS ftuesa izR;sd esa 2 lQsn rFkk 4 dkyh xsans gSA;fn ;kn`P;k ,d lQsn xsan fudkyh tkrh gS] rks ;s fudkyhxbZ xsans ds igys lewg ds csx esa ls vkus dh izkf;drk gS&

(1) 1661

(2) 1561

(3) 4561

(4) buesa ls dksbZ ugha

65. ;fn logtan30°

æ ö+ -ç ÷+è ø

22 | z | 2 | z | 3| z | 1 < –2 rks

(1) | z | < 32

(2) | z | > 32

(3) | z | > 2 (4) | z | < 2


21/27


Kota/01CE214024

66. Let a and b be the two disinct roots of the equationx3 + 3x2 –1 = 0. The equation which has (ab) asits root is equal to(1) x3 – 3x – 1 =0(2) x3 – 3x2 + 1 = 0(3) x3 + x2 – 3x + 1 = 0(4) x3 + x2 + 3x – 1 = 0

67. The coefficient of x8 in the expansion of (x–1) (x–2) (x–3)...............(x–10) is :(1) 2640 (2) 1320(3) 1370 (4) 2740

68. The coefficient of x37 in the expansion of(1–x)30 (1 + x + x2)29 is :(1) 0 (2) 29C12 (3) – 29C12 (4) None

69. Let R be a relation defined on N × N by(a, b) R(c, d) Û a(b + c) = c(a + d). Then R is

(1) reflexive, symmetric

(2) symmetric, transitive

(3) transitive only

(4) equivalence

70. Domain of the definition of function

f(x) = 24 x

[x] 2-

+ is (where [.] ® G.I.F.)

(1) (–¥, –2) È [–1, 2](2) [0, 2](3) [–1, 2](4) (0, 2)

66. ;fn a, b lehdj.k x3 + 3x2 –1 = 0 ds nks fofHkUu ewygks rks og lehdj.k ftldk ewy(ab) gksxsa] gksxh&

(1) x3 – 3x – 1 =0

(2) x3 – 3x2 + 1 = 0

(3) x3 + x2 – 3x + 1 = 0

(4) x3 + x2 + 3x – 1 = 0

67. (x–1) (x–2) (x–3)...............(x–10) ds izlkj esa x8 dkxq.kkad gkssxk&

(1) 2640 (2) 1320(3) 1370 (4) 2740

68. (1–x)30 (1 + x + x2)29 ds izlkj esa x37 dk xq.kkad gksxk&(1) 0 (2) 29C12

(3) – 29C12 (4) dksbZ ugha69. ekuk N × N ij (a, b) R(c, d) Û

a(b + c) = c(a + d) }kjk ifjHkkf"kr gSA rc R gS :-(1) LorqY;] lefer(2) lefer] laØked(3) dsoy laØked(4) rqY;rk

70. Qyu f(x) = 24 x

[x] 2-

+ dh ifjHkk"kk dk izkar gS

(tgk¡ [.] ® G.I.F.)

(1) (–¥, –2) È [–1, 2]

(2) [0, 2]

(3) [–1, 2]

(4) (0, 2)

22/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

71. nlim

®¥

2 2 2 2 2 2 2 2

3[1 x 1 ] [2 x 2 ] [3 x 3 ] .... [n x n ]

n+ + + + + + + +

cjkcj gksxk :- (tgk¡ [.] egÙe iw.kk±d Qyu gS :-

(1) x3 (2) x +

13

(3) x3 +

13

(4) x3

– 13

72. oØ xmyn = am+n ds fdlh fcUnq ij v/k% Li'khZ lekuqikrhgS&(1) dksfV ds (2) Hkqt ds(3) (dksfV)n ds (4) (Hkqt)n ds

73. Qyu f(x)2

| x 1 |x-

= fuEu esa ls fdl varjky esa ,dfn"V

áleku gS&(1) (–¥,¥) (2) (0,1)(3) (2,¥) (4) (0,1) È (2,¥)

74. a O;kl ds xksys ls dkVs tkus okys vf/kdre vk;ru ds yEco`Ùkh; 'kadq dh Å¡pkbZ gS&(1) (2/3)a (2) (3/4)a (3) (1/3)a (4) (1/4)a

75. ;fn y = tan–1 12 2

1 1tanx x 1 x 3x 3

-æ ö æ ö+ç ÷ ç ÷è ø è ø+ + + +

12

1tanx 5x 7

- æ ö+ ç ÷è ø+ + + ..... n inksa rd] rc

dydx cjkcj

gksxk

(1) 2 21 1

1 (x n) (1 x )+

+ + + (2) 2 21 1

1 (x n) (1 x )-

+ + +

(3) 21

(1 x)-

+ (4) 0

71. nlim

®¥

2 2 2 2 2 2 2 2

3[1 x 1 ] [2 x 2 ] [3 x 3 ] .... [n x n ]

n+ + + + + + + +

is equal to :- (where [.] greatest integer function)

(1) x3 (2) x +

13

(3) x3 +

13

(4) x3

– 13

72. The length of the subtangent at any point of thecurve xmyn = am+n is proportional to-(1) Ordinate (2) Abscissa(3) (Ordinate)n (4) (Abscissa)n

73. Function f(x)2

| x 1 |x-

= is monotonic decreasing

in-(1) (–¥,¥) (2) (0,1)(3) (2,¥) (4) (0,1) È (2,¥)

74. The height of a right circular cone of maximumvolume inscribed in a sphere of diameter a is-(1) (2/3)a (2) (3/4)a (3) (1/3)a (4) (1/4)a

75. If y = tan–1 12 2

1 1tanx x 1 x 3x 3

-æ ö æ ö+ç ÷ ç ÷è ø è ø+ + + +

12

1tanx 5x 7

- æ ö+ ç ÷è ø+ + + ..... up to n terms, then

dydx

is equal to

(1) 2 21 1

1 (x n) (1 x )+

+ + + (2) 2 21 1

1 (x n) (1 x )-

+ + +

(3) 21

(1 x)-

+ (4) 0


23/27


Kota/01CE214024

76. A vertical pole consists of two parts, the lower partbeing one third of the whole. At a point in thehorizontal plane through the base of the pole anddistance 20 meters from it, the upper part of the pole

subtends an angle whose tangent is 12

. The possible

heights of the pole are

(1) 20 m and 20 3 m(2) 20 m and 60 m(3) 16 m and 48 m(4) None of these

77.p p p p

+ + + =4 4 4 43 5 7sin sin sin sin8 8 8 8

(1)12

(2) 14

(3) 32

(4) 34

78. The following data gives the distribution of heightof students

Height(in cm) 160 150 152 161 156 154 155

Numberof students 12 8 4 4 3 3 7

The median of the distribution is(1) 154 (2) 155(3) 160 (4) 161

76. ,d Å/okZ/kj NM+ ds nks Hkkx gSa] fupyk Hkkx lEiw.kZ Å¡pkbZdk ,d&frgkbZ gSA NM+ ds vk/kkj ls tkus okys {kSfrt lery

esa NM + ls 20 ehVj nwj fLFkr ,d fcUnq ij NM+ dk Åijh

Hkkx tks dks.k vUrfjr djrk gS mldh Li'kZT;k 12

gS] rks NM +

dh lEHkkfor yEckbZ gS

(1) 20 ehVj rFkk 20 3 ehVj

(2) 20 ehVj rFkk 60 ehVj

(3) 16 ehVj rFkk 48 ehVj


77.p p p p

+ + + =4 4 4 43 5 7sin sin sin sin8 8 8 8

(1)12

(2) 14

(3) 32

(4) 34

78. Nk=ksa dh Å¡pkbZ;ksa ds caVu ds vk¡dM+s fuEukuqlkj gS

Å¡pkbZ;k¡(lseh esa) 160 150 152 161 156 154 155

Nk=ksa dhla[;k 12 8 4 4 3 3 7

caVu dh ekf/;dk gSa(1) 154 (2) 155(3) 160 (4) 161

24/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

79. (p Ù ~ q) Ù (~ p Ú q) is :-(1) A contradiction(2) A tautology(3) Either (1) or (2)(4) Neither (1) nor (2)

80. Consider the family of lines x(a + b) + y = 1, wherea, b and c are the roots of the equationx3 – 3x2 + x + l= 0 such that c Î [1,2]. If the givenfamily of lines makes triangle of area 'A' withcoordinate axis, then maximum value of 'A' (in sq.units) will be -

(1) 14

(2) 1

(3) 18

(4) 12

81. If two tangents drawn from a point P to the parabolay2 = 4x be such that the slope of one tangent isdouble of the other, then P lies on the curve. :-(1) 9y = 2x2

(2) 9x = 2y2

(3) 2x = 9y2

(4) None of these82. The length of the shortest path that begins at the

point (2,5), touches the x-axis and then ends at apoint on the circlex2 + y2 + 12x – 20 y + 120 = 0

(1) 13 (2) 4 10

(3) 15 (4) 6 + 89

79. (p Ù ~ q) Ù (~ p Ú q) gS :-(1) iqufjfDr(2) O;k?kkr(3) ;k rks (1) ;k (2)(4) uk rks (1) uk gh (2)

80. ekuk js[kkvksa dk fudk; x(a + b) + y = 1 gS] tgk¡ a, brFkk c lehdj.k x3 – 3x2 + x + l = 0 ds ewy gS rFkk

c Î [1,2] gSA ;fn fn;k x;k js[kk fudk; funsZ'kh v{kksa ds

lkFk {ks=Qy 'A' dk f=Hkqt cukrk gS] rks 'A' dk vfèkdre

eku (oxZ bdkbZ esa) gksxk -

(1) 14

(2) 1

(3) 18

(4) 12

81. ;fn ,d fcUnq P ls ijoy; y2 = 4x ij [khaph xbZ Li'kZjs[kkvksa esa ,d dh izo.krk] vU; dh izo.krk dh nksxquh gS]rks P ftl oØ ij gS] og gS :-(1) 9y = 2x2

(2) 9x = 2y2

(3) 2x = 9y2


82. fcUnq (2,5) dh o`Ùk x2 + y2 + 12x – 20 y + 120 = 0ds fdlh fcUnq ls x-v{k dks Li'kZ djrh gq;h U;wure nwjh

Kkr djsa

(1) 13 (2) 4 10

(3) 15 (4) 6 + 89


25/27


Kota/01CE214024

83. In the figure shown, radius of circle C1 be r and

that of C2 be r2 , where

1r3

= PQ, then length of

AB is (where P and Q being centres of C1 & C2

respectively)

P

Q

A

B

C1

C2

(1) 2 3 r (2) 3 3 r

4

(3) 3 3 r (4) 3 3 r

284. The equation of the tangents to the hyperbola

4x2 – y2 = 12 are y = 4x+ c1 & y = 4x + c2, then|c1 – c2| is equal to -(1) 1 (2) 4 (3) 6 (4) 12

85. Let the points P, Q and R have position vectors

1r 3i 2 j k,= - -r

2r i 3j 4k= + +r and

3r 2i j – 2k= +r respectively relative to an origin O.Then the distance of P from the plane OQR is :-(1) 2 (2) 3 (3) 1 (4) 5

86. A plane passes through the point A(2, 1, –3). Ifdistance of this plane from origin is maximum, thenits equation is(1) 2x + y – 3z = 14 (2) 2x + y + 3z = 2(3) x + y – z = 1 (4) None

83. fn;s x;s fp= esa] o`Ùk C1 rFkk o`Ùk C2 dh f=T;k Øe'k% r

rFkk r2 gS] tgk¡

1r3

= PQ, rks AB dh yEckbZ gksxk (tgk¡

P rFkk Q Øe'k% C1 rFkk C2 ds dsUæ gS)

P

Q

A

B

C1

C2

(1) 2 3 r (2) 3 3 r

4

(3) 3 3 r (4) 3 3 r

284. vfrijoy; 4x2 – y2 = 12 dh Li'kZ js[kk dk lehdj.k

y = 4x+ c1 rFkk y = 4x + c2 gks] rks |c1 – c2| dk ekugksxk -(1) 1 (2) 4 (3) 6 (4) 12

85. ekuk rhu fcUnq P, Q rFkk R ftuds fLFkfr lfn'k

1r 3i 2 j k,= - -r

2r i 3j 4k= + +r rFkk

3r 2i j – 2k= +r ewy fcUnq ds lkis{k gks] rks P dh lery

OQR ls nwjh gksxh :-(1) 2 (2) 3 (3) 1 (4) 5

86. fcUnq A(2, 1, –3) ls gksdj tkus okys rFkk ewy fcanq lsegÙke nwjh ij fLFkr lery dh lehdj.k gS

(1) 2x + y – 3z = 14 (2) 2x + y + 3z = 2

(3) x + y – z = 1 (4) dksbZ ugha

26/27

Target : JEE (Main) 2015/18-01-2015


Kota/01CE214024

87. If m is a non-zero number and - -++ +ò

5m 1 4m 1

2m m 3

x 2x dx(x x 1)

= f(x) + c, then f(x) is:-

(1) ++ +

5m

2m m 2

x c2m(x x 1)

(2) ++ +

4m

2m m 2

x c2m(x x 1)

(3) +

++ +

5m 4m

2m m 2

2m(x x ) c(x x 1)

(4) -

++ +

5m 4m

2m m 2

x x c2m(x x 1)

88.2

21

[x] dx1 x-

é ùê ú+ë ûò , where [·] GIF is equal to:-

(1) –2 (2) –1 (3) 0 (4) None

89. ®¥

ì üæ öæ ö æ ö-ï ïæ ö+ + + +í ýç ÷ç ÷ ç ÷ç ÷è øï ïè øè ø è øî þ

12 2 2 n

2 2 2 2n

1 2 3 (n 1)lim 1 1 1 ........ 1n n n n

Equals to:-(1) e(4–p)/2 (2) e(p–4)/2 (3) 2e(p–4)/2 (4) None

90. -

= +1sin xy ke 3 is a solution of differeulial equation

(1) - = -2 dy1 x y 3dx

(2) 2 dy1 x y 3dx

+ = -

(3) 2 dy1 x y 3dx

+ = + (4) 2 dy1 x y 3dx

- = +

87. ;fn m ,d v'kwU; l[;k¡ gS rFkk - -++ +ò

5m 1 4m 1

2m m 3

x 2x dx(x x 1)

= f(x) + c, rc f(x) gS:-

(1) ++ +

5m

2m m 2

x c2m(x x 1)

(2) ++ +

4m

2m m 2

x c2m(x x 1)

(3) +

++ +

5m 4m

2m m 2

2m(x x ) c(x x 1)

(4) -

++ +

5m 4m

2m m 2

x x c2m(x x 1)

88.2

21

[x] dx1 x-

é ùê ú+ë ûò , tgk¡ [·] egÙkeiw.kkZd Qyu cjkcj gS:-

(1) –2 (2) –1 (3) 0 (4) None

89. ®¥

ì üæ öæ ö æ ö-ï ïæ ö+ + + +í ýç ÷ç ÷ ç ÷ç ÷è øï ïè øè ø è øî þ

12 2 2 n

2 2 2 2n

1 2 3 (n 1)lim 1 1 1 ........ 1n n n n

cjkcj gS:-(1) e(4–p)/2 (2) e(p–4)/2 (3) 2e(p–4)/2 (4) None

90. -

= +1sin xy ke 3 vody lehdj.k dk gy gS og gS:-

(1) - = -2 dy1 x y 3dx

(2) 2 dy1 x y 3dx

+ = -

(3) 2 dy1 x y 3dx

+ = + (4) 2 dy1 x y 3dx

- = +


27/27


Kota/01CE214024

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

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