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Your Target is to secure Good Rank in JEE (Main) 2015

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

SCORE – I DATE : 11 - 03 - 2015

0 1 C E 3 1 4 0 6 0

PHASE – ELC, ELD, ELP

Corporate OfficeALLEN CAREER INSTITUTE

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

+91-744-2436001 [email protected]

CLASSROOM CONTACT PROGRAMME

www.allen.ac.in

JEE (Main) : LEADER COURSE

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 ofChemistry, Mathematics and Physics having 30 questions in eachpart of equal weightage. Each question is allotted 4 (four) marks forcorrect response.

6. One Fourth mark will be deducted for indicated incorrect responseof each question. No deduction from the total score will be madeif no response is indicated for an item in the Answer Sheet.

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

8. No candidate is allowed to carry any textual material, printed or written,

bits of papers, pager, mobile phone any electronic device etc, except

the Identity Card inside the examination hall/room.

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

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

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

bl ijh{kk iqfLrdk dks rc rd u [kk sysa tc rd dgk u tk,A

1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys@dkys ckWy ikbaV isuls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA

2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkji= ij dgha vkSj u fy[ksaA

3. ijh{kk dh vof/k 3 ?k aVs gSA

4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 360 gSaA

5. bl ijh{kk iqfLrdk es a rhu Hkkx A, B, C gSa] ftlds izR;sd Hkkx esajlk;u foKku] xf.kr ,oa HkkSfrd foKku 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;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls½.kkRed vadu ugha gksxkA

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

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

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

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

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

MAJOR TEST Test Pattern : JEE (Main)IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

1/25Kota/01CE314060

SPACE FOR ROUGH WORK

Leader Course/Phase-ELC, ELD & ELP/Score-I/11-03-2015

1. m-Bromoanisole reacts with NaNH2 in liquidammonia to give mainly.(1) m-methoxyaniline(2) o-methoxyaniline(3) p-methoxyaniline(4) o, m and p-methoxyaniline in equal amounts

2. 4-Hydroxybenzenesulphonic acid is treated withbromine water. The product formed is -(1) 2, 4, 6-tribromophenol(2) 3, 5-dibromo-4-hydroxybenzenesulphonic acid(3) 3-bromo-4-hydroxybenzenesulphonic acid(4) 2, 6-dibromophenol

3. The final product (B) formed in the reaction sequence

CH3

4 3NBS/ CCl CH ONaheat A B¾¾¾¾® ¾¾¾¾® , product (B) is

(1) —CO CH2 3

(2) —OCH3

(3) —CH3CH O—3

(4) —CH OCH2 3

1. esVk&czkseks,fulksy dh fØ;k nzo veksfu;k dh mifLFkfr esaNaNH2 ls djkus ij eq[; mRikn gksxk(1) m-methoxyaniline(2) o-methoxyaniline(3) p-methoxyaniline(4) o, m and p-methoxyaniline in equal amounts

2. 4-gkbMªksDlh csUthu lYQksfud vEy dh fØ;k czksehu tyls djkus ij curk gS -(1) 2, 4, 6-tribromophenol(2) 3, 5-dibromo-4-hydroxybenxenesulphonic acid(3) 3-bromo-4-hydroxybenzenesulphonic acid(4) 2, 6-dibromophenol

3. fØ;k dk vfUre mRikn (B) gksxk &

CH3

4 3NBS/ CCl CH ONaheat A B¾¾¾¾® ¾¾¾¾® , mRikn (B) gS

(1) —CO CH2 3

(2) —OCH3

(3) —CH3CH O—3

(4) —CH OCH2 3

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

PART A - CHEMISTRY

Take it Easy and Make it Easy

Kota/01CE3140602/25


Target : JEE(Main) 2015/11-03-2015

4. Benxoylation of phenol with benzoyl chloride inthe presence of dilute NaOH gives phenyl benzoate.This reaction is an example of(1) Friedel – Crafts reaction(2) Reimer – Tiemann reaction(3) Claisen – Schmidt reaction(4) Schotten – Baumann reaction

5. Mutarotation involves -(1) Racemization(2) optical resolution(3) conformational inversion(4) Diastereoisomerization

6. Globular protein is present in -(1) Blood(2) Milk(3) Eggs(4) All of these

7. In a nucleotide the phosphate linkage is generallyattached to -(1) C-1 of the sugar(2) C-2 of the sugar(3) C-5 of the sugar(4) nitrogen atom of the base

8. Which of the following does not reduce fehling'ssolution ?(1) Glucose(2) Benzaldehyde(3) Sucrose(4) Both (2) and (3)

4. ruq NaOH dh mifLFkfr esa QhukWy dk csUtks;fydj.kcsUtks;y DyksjkbM + ls djkus ij Qsfuy csUtks,V curk gS ;g

fuEu esa ls fdl fØ;k dk mnkgj.k gSa(1) Friedel – Crafts reaction(2) Reimer – Tiemann reaction(3) Claisen – Schmidt reaction(4) Schotten – Baumann reaction

5. E;wVkjksVs'ku esa gksrk gS -(1) jsflfedj.k(2) izdkf'k; i`FkDdj.k(3) la:i.k izfriu(4) foofje leko;fodj.k

6. XyksC;wyj izksVhu fdlesa mifLFkr gS -(1) jDr(2) nw/k(3) v.Ms(4) mijksä lHkh

7. U;wfDy;ksVkbM esa QkLQsV ca/ku lkekU;r;k fdl dkcZu lstqM+k gksrk gS -(1) 'kdZjk ds C-1 ds lkFk(2) 'kdZjk ds C-2 ds lkFk(3) 'kdZjk ds C-5 ds lkFk(4) {kkj ds ukbVªkstu ds lkFk

8. fuEu esa ls dkSulk Qsgfyax foy;u dks vipf;r ugha djrkgS ?(1) Glucose(2) Benzaldehyde(3) Sucrose(4) Both (2) and (3)

fdlh iz'u ij nsj rd :dks ugha A

3/25Kota/01CE314060



9. fLFkj nkc ij vkn'kZ xSl ds fn;s x, æO;eku dk vk;rurkieku xzkQ uhps iznf'kZr fd;k x;k gSA

P2 P3

P1

V

O 273 T(K)nkc dk lgh Øe D;k gS(1) P1 > P3 > P2 (2) P1 > P2 > P3

(3) P2 > P3 > P1 (4) P2 > P1 > P3

10. ok.Mj okWy lehdj.k dk okLrfod xSlksa }kjk ikyu fd;ktkrk gS] okLrfod xSl ds n eksyksa ds fy;s fuEu esa ls dkSulklehdj.k lgh gS :-

(1) 2

P na VRT

n V n bæ öæ ö+ =ç ÷ç ÷

-è øè ø

(2) 2

aP (V b) nRT

Væ ö+ - =ç ÷è ø

(3) 2

naP (nV b) nRT


(4) 2

2

n aP (V nb) nRT

V

æ ö+ - =ç ÷

è ø

11. ;fn 298 K ,oa fLFkj nkc ij fuEu vfHkfØ;k ds fy, Å"ekesa ifjorZu +7.3 kcal gSA2B(s) ® 2A(s) + 1/2B2(g), DH = +7.3 kcalrks fLFkj vk;ru ij Å"ek esa ifjorZu gksxk :-(1) 7.3 kcal (2) 7.3 ls vf/kd(3) 'kwU; (4) buesa ls dksbZ ugha

9. The volume-temperature graph of a given mass ofan ideal gas at constant pressure are shown below.

P2 P3

P1

V

O 273 T(K)What is the correct order of pressures(1) P1 > P3 > P2 (2) P1 > P2 > P3

(3) P2 > P3 > P1 (4) P2 > P1 > P3

10. Vander Waal's equation of state is oberyed by realgases. For n moles of a real gas, the expression willbe :-

(1)2

P na VRT

n V n bæ öæ ö+ =ç ÷ç ÷

-è øè ø

(2) 2

aP (V b) nRT


(3) 2

naP (nV b) nRT


(4) 2

2

n aP (V nb) nRT

V

æ ö+ - =ç ÷

è ø

11. The heat change for the following reaction at 298Kand at constant pressure is +7.3 kcalA2B(s) ® 2A(s) + 1/2B2(g), DH = +7.3 kcalThe heat change at constant volume would be:-(1) 7.3 kcal (2) More than 7.3(3) Zero (4) None of these

Kota/01CE3140604/25


Target : JEE(Main) 2015/11-03-2015

12. fdlh vkn'kZ xSl ds 6 eksy dks 27ºC ij 1 yhVj vk;ruls 10 yhVj vk;ru rd lerkih; ,oa mRØe.kh; :i lsizlkfjr fd;k tkrk gS rk s fd;k x;k vf/kdre dk;Zgksxk :-(1) 47 kJ (2) 100 kJ(3) 0 (4) 34.465 kJ

13. 298 K ,oa 1 ok;qe.Myh; nkc ij H2(g), Cl2(g) ,oa HCl(g)

ds ,UVªkWih eku (JK–1 mol–1) Øe'k% 130.6, 223.0 ,oa186.7 gSA rc vfHkfØ;k H2(g) + Cl2(g) ® 2HCl(g) ds fy;s,UVªkWih ifjorZu gksxk :-(1) +540.3 (2) +727.3(3) –166.9 (4) +19.8

14. fn;k x;k gS2C(s) + 2O2(g) ® 2CO2(g) ; DH = –787 kJ

H2(g) + 12

O2(g) ® H2O(l) ; DH = –286 kJ

C2H2(g) + 212

O2(g) ® 2CO2(g) + H2O(l) ;

DH = –1301 kJ,lhfVyhu dh laHkou Å"ek gS :-(1) –1802 kJ (2) +1802 kJ

(3) –800 kJ (4) +228 kJ15. 0ºC vkSj 1 atm nkc ij ,d xSl dk vk;ru 100 cc

gksrk gS ;fn nkc dks 1½ (Ms<) xquk dj fn;k tk;s vkSj izkjfEHkd

rki dks ,d frgkbZ xquk ls c<+k;k tk;s rks xSl dk vfUre

vk;ru gksxk :-(1) 80 cc (2) 88.9 cc

(3) 66.7 cc (4) 100 cc

12. 6 moles of an ideal gas expand isothermally and

reversibly from a volume of 1 litre to a volume of

10 litres at 27ºC. What is the maximum work done:-

(1) 47 kJ (2) 100 kJ

(3) 0 (4) 34.465 kJ13. The entropy values (in JK–1 mol–1) of

H2(g) = 130.6, Cl2(g) = 223.0 and HCl(g) = 186.7 at298 K and 1 atm pressure. Then entropy change forthe reaction H2(g) + Cl2(g) ® 2HCl(g) is :-(1) +540.3 (2) +727.3(3) –166.9 (4) +19.8

14. Given that2C(s) + 2O2(g) ® 2CO2(g) ; DH = –787 kJ

H2(g) + 12

O2(g) ® H2O(l) ; DH = –286 kJ

C2H2(g) + 212

O2(g) ® 2CO2(g) + H2O(l) ;

DH = –1301 kJHeat of formation of acetylene is :-(1) –1802 kJ (2) +1802 kJ

(3) –800 kJ (4) +228 kJ15. At 0ºC and one atm pressure, a gas occupies

100 cc. If the pressure is increased to one and ahalf-time and temperature is increased by one-thirdof absolute temperature, then final volume of thegas will be :-(1) 80 cc (2) 88.9 cc(3) 66.7 cc (4) 100 cc

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

5/25Kota/01CE314060



16. 1 ok;qe.My nkc vkSj 100ºC ij 9.0 xzke H2O okf"ir gksrkgSA ;fn ty ds ok"iu dh xqIr Å"ek xJ / gm gS, rc DSfn;k tkosxk :-

(1) x

373(2)

18x100

(3) 18x373

(4) 1 18x2 373´

17. X2, Y2 rFkk XY3 ds ekud ,UVªk Wih Øe'k% 60, 40

rFkk 50 JK–1 mol–1 gSA vfHkfØ;k 12

X2 + 32

Y2 ® XY3,

DH = –30kJ, dks lkE;oLFkk esa gksus ds fy;s rki gksxk :-

(1) 500 K (2) 750 K

(3) 1000 K (4) 1250 K

18. ,d xSl ds ?kuRo vyx&vyx fLFkfr;ksa esa 1 : 2 ds vuqikresa gSa vkSj mlds rkiØe 2 : 1, ds vuqikr esa gSa] rc mldsnkcksa dk vuqikr gksxk:-

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

19. xSl ekWLd (ftlesa lfØ;rk pkjdksy gksrk gSa)] okrkoj.k lsfo"kSyh xSlksa dks nwj djrk gSa] fdl fl¼kUr ij vk/kkfjrgSa :-(1) vf/k'kks"k.k (2) vo'kks"k.k(3) 'kks"k.k (4) buesa ls dksbZ ugha

20. fuEu esa ls dkSulk dFku lR; ugha gSa :-

(1)ØkfUrd felSy lkUnzrk ij lkcqu dk foy;u laxqf.krdksykWbM cukrk gSa

(2)nzoLusgh dksykbM] vuqRØe.kh; dksykWbM Hkh dgykrs gSaA(3) viksgu izfØ;k ds }kjk jDr dk 'kqf¼dj.k fd;k tkrk gSa(4) Ca+2 ,oa K+ jDr dk LdUnu dj ldrs gSa

16. 9.0 gm of H2O is vaporised at 100ºC and 1 atmpressure. If the latent heat of vaporisation of wateris xJ / gm, then DS is given by :-

(1) x

373(2)

18x100

(3) 18x373

(4) 1 18x2 373´

17. Standard entropy of X2, Y2 and XY3 are 60, 40 and50 JK–1 mol–1, respectively. For the reaction,

12

X2 + 32

Y2 ® XY3, DH = –30kJ, to be at

equilibrium, the temperature will be :-(1) 500 K (2) 750 K(3) 1000 K (4) 1250 K

18. Densities of a gas at different conditions are in theratio 1 : 2 and their temperatures are in the ratio2 : 1, then the ratio of their respective pressureis :-(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1

19. Gas masks containing activated charcoal to removepoisonous gases from atmosphere acts on theprinciple of :-(1)Adsorption (2) Absorption(3) Sorption (4) None of these

20. Which of the following statements is/are not true:-(1)Soap solution form associated colloids at CMC(2) Lyophilic colloids are irreversible sols(3) Blood is purified by the process of dialysis(4) Ca+2 and K+ cause coagulation of blood if added

in excess

Kota/01CE3140606/25


Target : JEE(Main) 2015/11-03-2015

21. logxm

æ öç ÷è ø

,oa log P ds e/; xzkQ ,d ljy js[kk gSA ftldh

<+ky 45° dks.k cukrh gSaA ;fn Ýsadfyd izkWpy (K) dk eku

10 gks ,oa tc nkc 0.5 atm gS rks vf/k'kks"kd ds izfr xzkeij vf/k'kk sf"kr] foys; vf/k'kk s"; dh ek=k D;k gk sxh(log 5 = 0.699) :-

(1) 1 xzke (2) 6.99 xzke (3) 3 xzke (4) 5 xzke22. fuEufyf[kr esa ls dkSulk jklk;fud vf/k'kk s"k.k dk

vfHkyk{kf.kd xq.k ugha gSa :-(1) vf/k'kks"k.k vuqRØe.kh; gSa(2) vf/k'kks"k.k fof'k"V gSa(3) Å"ek dh ijkl 540 KJ ds dksfV dh gksrh gSa(4) i`"Bh; {ks=Qy c<+us ds lkFk gh] vf/k'kks"k.k esa Hkh of¼

gksrh gSa23. izR;sd lewg ls v.kq vFkok vk;u dk p;u dhft, ftlesa

ca/k dks.k U;wure gSa :(i) NH3,PH3 ;k AsH3 (ii) O3

+, O3(iii)NO2

– ;k O3

(iv)X—S—X angle in SOCl2 and SOF2(1) NH3, O3

+, O3 SOCl2(2) PH3, O3

+, NO2–, SOF2

(3) AsH3, O3, NO2–, SOF2

(4) AsH3, O3+, O3, SOF2

24. fuEufyf[kr v.kqvksa esa B—B ca/k yackb Z dh rqyuk dhft, :

BF F

F FBx

BCl Cl

Cl ClB

y

(solid)

(1) x > y (2) y > x(3) x = y (4) buesa ls dksbZ ugha

21. Plot of log xm

æ öç ÷è ø

against log P is a straight line

inclined at an angle of 45°. When the pressure is0.5 atm and freundlich parameter (K) is 10, thenthe amount of solute adsorbed pergram of adsorbentwill be (log 5 = 0.699) :-(1) 1gm (2) 6.99 gm(3) 3 gm (4) 5 gm

22. Which of the following is not a characteristic ofchemi-sorption :-(1)Adsorption is irreversible(2) Adsorption is specific(3) DH is of the order of 540 KJ(4) Adsorption increases with increase of surface

area23. Select from each set the molecule or ion having

the smallest bond angle :(i) NH3,PH3 or AsH3 (ii) O3

+, O3(iii)NO2

– or O3(iv)X—S—X angle in SOCl2 and SOF2(1) NH3, O3

+, O3 SOCl2(2) PH3, O3

+, NO2–, SOF2

(3) AsH3, O3, NO2–, SOF2

(4) AsH3, O3+, O3, SOF2

24. Compare B—B bond length in following molecules :

BF F

F FBx

BCl Cl

Cl ClB

y

(solid)

(1) x > y (2) y > x(3) x = y (4) None of these

7/25Kota/01CE314060



25. PCl5 v.kq esa ,d ry esa vf/kdre ijek.kqvksa dh la[;kfdruh gksrh gS :(1) 3 (2) 5(3) 4 (4) 2

26. f u Eu fy f[k r e s a l s fdl v. k q e s a v U rj k f . odH-bonding gk srh g S ?(1) Orth-nitrophenol(2) Ortho-boric acid(3) (1) o (2) nksuksa(4) buesa ls dksbZ ugha

27. fuEufyf[kr ;kSfxdksa dks rkih; LFkkf;Ro ds c<+rs gq, Øe esayxk;k x;k gSaA lgh Øe dks igpkfu, :

K2CO3(I), MgCO3(II), CaCO3(III), BeCO3(IV)

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

(3) IV < II < I < III (4) II < IV < III < I28. {kkjh; e`nk /kkrqvksa ds lYQsV yo.kksa dh foys;rk dk lgh

Øe gSa :(1) Be > Ca > Mg > Ba > Sr(2) Mg > Be > Ba > Ca > Sr(3) Be > Mg > Ca > Sr > Ba(4) Mg > Ca > Ba > Be > Sr

29. ckW'k izØe esa H2 xSl mRiknu ds fy, dkSulh xSl iz;qDrgksrh gS(1) mRiknd xSl (2) ty xSl(3) dksy xSl (4) izkd`frd xSl

30. ty esa vLFkk;h dBksjrk fuEufyf[kr esa ls fdldh mifLFkfrds dkj.k gksrh gSa(1) CaCl2, MgSO4

(2) Ca+2, Mg+2

(3) K+, CaCO3

(4) Ca(HCO3)2, Mg (HCO3)2

25. Find the maximum number of atoms that lie inthesame plane in PCl5 molecule :(1) 3 (2) 5(3) 4 (4) 2

26. Which of the following molecule hasintramolecular H-bonding ?(1) Orth-nitrophenol(2) Ortho-boric acid(3) Both (1) & (2)(4) None of these

27. The following compounds have been arranged inorder of their increasing thermal stabilities. Identifythe correct order :K2CO3(I), MgCO3(II), CaCO3(III), BeCO3(IV)(1) I < II < III < IV (2) IV < II < III < I(3) IV < II < I < III (4) II < IV < III < I

28. The right order of the solubility of sulphates ofalkaline earth metals is :(1) Be > Ca > Mg > Ba > Sr(2) Mg > Be > Ba > Ca > Sr(3) Be > Mg > Ca > Sr > Ba(4) Mg > Ca > Ba > Be > Sr

29. In Bosch process which of the following gas usesfor the production of H2 gas(1) Producer gas (2) Water gas(3) Coal gas (4) Natural gas

30. Temporary hardness of water is due to presenceof(1) CaCl2, MgSO4(2) Ca+2, Mg+2

(3) K+, CaCO3(4) Ca(HCO3)2, Mg (HCO3)2

Kota/01CE3140608/25


Target : JEE(Main) 2015/11-03-2015

PART B - MATHEMATICS31. If a,b,g are the roots of equation x3 + 2x – 5 = 0

and if equation x3 + bx2 + cx + d = 0 has roots

2a + 1, 2b + 1, 2g + 1 , then value of |b + c + d|

is (where b,c,d are coprime)-

(1) 41 (2) 39

(3) 40 (4) 43

32. Area of region enclosed by locus of z given by

Arg(z + i) – Arg(z – i) = 23p

and imaginary axis

is-

(1) 2 19 3

p- (2)

4 19 3

p-

(3) 2 29 3

p- (4)

4 29 3

p-

33. There are 3 bags A, B & C. Bag A contains1 Red & 2 Green balls, bag B contains 2 Red &1 Green balls and bag C contains only one greenball. One ball is drawn from bag A & put into bagB then one ball is drawn from B & put into bag C& finally one ball is drawn from bag C & put intobag A. When this operation is completed,probability that bag A contains 2 Red & 1 Greenballs, is -

(1) 14

(2) 12

(3) 13

(4) 16

31. ;fn lehdj.k x3 + 2x – 5 = 0 ds ewy a, b, g rFkk lehdj.kx3 + bx2 + cx + d = 0 ds ewy 2a + 1, 2b + 1, 2g + 1

gks] rks |b + c + d| dk eku gksxk (tgk¡ b,c,d lgvHkkT; la[;k;sa

gksxha)-(1) 41 (2) 39(3) 40 (4) 43

32. Arg(z + i) – Arg(z – i) = 23p

dks larq"V djus okys z

dk fcUnqiFk rFkk dkYifud v{k }kjk ifjc¼ {ks= dk {ks=Qy

gksxk&

(1) 2 19 3

p- (2)

4 19 3

p-

(3) 2 29 3

p- (4)

4 29 3

p-

33. 3 FkSys A, B rFkk C gSA FkSys A esa 1 yky ,oa 2 gjh xsnsa]FkSys B esa 2 yky ,oa 1 gjh xsan rFkk FkSys C esa dsoy 1 gjh

xsan gSA FkSys A ls 1 xsan fudky dj mldks FkSys B esa j[krs

gS] mlh le; FkSys B ls 1 xsan fudky dj FkSys C esa j[krs

gS rFkk vUr esa FkSys C ls 1 xsan fudky dj FkSys A esa j[krs

gSA ;g izfØ;k iw.k± gksus ij izkf;drk rkfd FkSys A esa 2 yky

rFkk 1 gjh xsan gks] gksxh -

(1) 14

(2) 12

(3) 13

(4) 16

J ges'kk eqLdjkrs jgsa A

9/25Kota/01CE314060



34. Number of different words that can be formed fromall letters of word APPLICATION such that twovowels never come together is -(1) (45)7! (2) 8!(3) 6!7! (4) (32)6!

35. If

2 4 2 2 2 2

2 2 2 4 2 2

2 2 2 2 2 4

1 a a 1 ab a b 1 ac a c

A 1 ab a b 1 b b 1 bc b c

1 ac a c 1 bc b c 1 c c

é ù+ + + + + +ê ú= + + + + + +ê úê ú+ + + + + +ë û

and det(A) = det(4I), where I is 3 × 3 identity matrix,then (a – b)3 + (b – c)3 + (c – a)3 can be equalto -(1) –24 (2) 6(3) –6 (4) 12

36. Given A and C are involutary matrices and B is anon-singular matrix, then (AB–1C)–1 is equal to -(1) A–1BC–1 (2) ABC(3) ABC–1 (4) CBA

37. If a,b are the roots of x2 – ax + b = 0 and ifan + bn = Vn, then -(1) Vn+1 = aVn + bVn–1

(2) Vn+1 = aVn + aVn–1

(3) Vn+1 = aVn – bVn–1

(4) Vn+1 = aVn–1 – bVn

38. If (1 +x + x2)n = a0 + a1x + a2x2 +......+a2nx

2n, thena0 + a3 + a6 +..... =(1) 3n (2) 3n–1

(3) 3n–2 (4) 3

34. 'kCn APPLICATION ds lHkh v{kjksa ls fufeZr fd;s tkldus okys fHkUu 'kCnksa dh la[;k rkfd nks Loj dHkh Hkh lkFkuk gks] gksxh&(1) (45)7! (2) 8!(3) 6!7! (4) (32)6!

35. ;fn

2 4 2 2 2 2

2 2 2 4 2 2

2 2 2 2 2 4

1 a a 1 ab a b 1 ac a c

A 1 ab a b 1 b b 1 bc b c

1 ac a c 1 bc b c 1 c c

é ù+ + + + + +ê ú= + + + + + +ê úê ú+ + + + + +ë û

rFkk det(A) = det(4I), tgk¡ I, 3 × 3 dk rRled vkO;wggS] rks (a – b)3 + (b – c)3 + (c – a)3 dk eku gks ldrkgS-(1) –24 (2) 6(3) –6 (4) 12

36. ;fn A rFkk C vUroZyuh; vkO;wg rFkk B O;qRØe.kh; vkO;wggks] rks (AB–1C)–1 dk eku gksxk -(1) A–1BC–1 (2) ABC(3) ABC–1 (4) CBA

37. ;fn a,b lehdj.k x2 – ax + b = 0 ds ewy rFkkan + bn = Vn gks] rks -(1) Vn+1 = aVn + bVn–1

(2) Vn+1 = aVn + aVn–1

(3) Vn+1 = aVn – bVn–1

(4) Vn+1 = aVn–1 – bVn

38. ;fn (1 +x + x2)n = a0 + a1x + a2x2 +......+a2nx

2n gks]rks a0 + a3 + a6 +..... =(1) 3n (2) 3n–1

(3) 3n–2 (4) 3

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

Kota/01CE31406010/25


Target : JEE(Main) 2015/11-03-2015

39. ;fn z1,z2,z3 vkx±M lery ij f=Hkqt ds 'kh"kZ bl izdkj

gS fd |z1 – z2| = |z1 – z3| gks] rks dks.kkad 1 2 3

3 2

2z z zz z

æ ö- -ç ÷-è ø

gksxk -

(1) 3p

± (2) 0 (3) 2p

± (4) 6p

±

40. AAAABBBC ds lHkh v{kjksa ds Øep;ksa dh la[;k] ftuesalHkh A, 4 v{kjksa ds lewg esa ,dlkFk ;k lHkh B, 3 v{kjksads lewg esa ,dlkFk gks] gksxh -(1) 44 (2) 50(3) 60 (4) 89

41. Js.kh 1.2015 + 2.2014 + 3.2013 +.....2015.1 dk ;ksxgksxk&(1) 336 × 2015 × 2016 (2) 336 × 2015 × 2017(3) 336 × 2016 × 2017 (4) dksbZ ugha

42. ;fn 7 ik¡lksa dks ,d lkFk QSadk tkrk gS] rks mGijh Qyd

ij lHkh vadksa ds izkIr gksus dh izkf;drk gksxh -

(1) 3

356 3´

(2) 125

6C

(3) 3

706 3´

(4) 126

6C

43. ;fn z1 rFkk z2 bdkbZ ekik¡d okyh nks lfEeJ la[;k;sa gS tksz1

2 + z22 = 5 dks larq"V djrh gS]

rks ( ) ( )2 2

1 1 2 2z z z z- + - dk eku gksxk -

(1) 6 (2) 5 (3) 9 (4) 1044. ;fn A dksfV 3 dk oxZ vkO;wg gS ftlesa |A| = 2 gS] rks

|(A – AT)5| + |(AT – A)3| dk eku gksxk&(1) 24 (2) 16 (3) 0 (4) 8

39. If z1,z2,z3 are vertices of a triangle in argand planesuch that |z1 – z2| = |z1 – z3|, then

1 2 3

3 2

2z z zarg

z z

æ ö- -ç ÷-è ø

is-

(1) 3p

± (2) 0 (3) 2p

± (4) 6p

±

40. The number of permutations of all the lettersAAAABBBC in which all the A's appear togetherin a block of 4 letters or all the B's appear togetherin the block of 3 letters, is-(1) 44 (2) 50 (3) 60 (4) 89

41. Sum of the series 1.2015 + 2.2014 + 3.2013+.....2015.1 is equal to :-(1) 336 × 2015 × 2016 (2) 336 × 2015 × 2017(3) 336 × 2016 × 2017 (4) None

42. If 7 dice are thrown simultaneously, thenprobability that all six digit appears on the upperface is equal to -

(1) 3

356 3´

(2) 125

6C

(3) 3

706 3´

(4) 126

6C

43. If z1 and z2 are two unimodular complex numbersthat satisfy z1

2 + z22 = 5,

then ( ) ( )2 2

1 1 2 2z z z z- + - is equal to -

(1) 6 (2) 5 (3) 9 (4) 1044. If A is a square matrix of order 3 with |A| = 2, then

the value of |(A – AT)5| + |(AT – A)3| is-(1) 24 (2) 16 (3) 0 (4) 8

11/25Kota/01CE314060



45. The minimum value of (8sec2q + 2cos2q) is equalto :-(1) 10 (2) 16 (3) 8 (4) None

46. The probability of hitting a target by three marks

men is 1 1

,2 3

and 14

respectively. If the probability

that exactly two of them will hit the target is l andthat at least two of them hit the target is mthen l + µ is equal to :-

(1) 1324

(2) 624

(3) 7

24(4) None

47. If 1z 2= , 2z 3= , 3z 4= and 1 2 32z 3z 4z 9+ + = ,

then value of 2 3 3 1 1 28z z 27z z 64z z+ + is equal to:-

(1) 216 (2) 18

(3) 64 (4) None

48. If sum of the coefficient of the first, second and third

terms of the expansion of æ ö+ç ÷è ø

m2 1

xx

is 46, then the

coefficient of the term that doesnot contain x is :-(1) 84 (2) 92(3) 98 (4) 106

49. The coefficient of t50 in(1 + t2)25 (1 + t25) (1 + t40) (1 + t45) (1 + t47) is(1) 1 + 25C5 (2) 1 + 25C5 + 25C7

(3) 1 + 25C7 (4) None of these

45. (8sec2q + 2cos2q) dk U;wure eku gksxk :-(1) 10 (2) 16(3) 8 (4) dksbZ ugha

46. rhu y{; Hksnh;ksa ds }kjk y{; dks Hksnus dh i zkf;drk,¡ Øe'k%

1 1,

2 3 ,

14

gSA ;fn muesa ls Bhd nks ds }kjk y{; Hksnus dh

izkf;drk l gS rFkk muesa ls de ls de nks ds }kjk y{;

Hksnus dh izkf;drk m gks rc l + µ cjkcj gksxk :-

(1) 1324

(2) 624

(3) 7

24(4) dksbZ ugha

47. ;fn 1z 2= , 2z 3= , 3z 4= rFkk 1 2 32z 3z 4z 9+ + =

gks] rks 2 3 3 1 1 28z z 27z z 64z z+ + cjkcj gS :-(1) 216 (2) 18(3) 64 (4) dksbZ ugha

48. æ ö+ç ÷è ø

m2 1

xx

ds izlkj esa ;fn igys] nwljs rFkk rhljs inksa ds

xq.kkadksa dk ;ksx 46 gS rks ml in dk xq.kkad ftlesa x ugha

gS] gS :-

(1) 84 (2) 92

(3) 98 (4) 106

49. (1 + t2)25 (1 + t25) (1 + t40) (1 + t45) (1 + t47) ds izlkjesa t50 dk xq.kkad gS(1) 1 + 25C5 (2) 1 + 25C5 + 25C7

(3) 1 + 25C7 (4) buesa ls dksbZ ugha

Kota/01CE31406012/25


Target : JEE(Main) 2015/11-03-2015

50. If z1, z2, z3, z4 are the roots of equation

z4+ z3+ z2 + z + 1 = 0 then=

+Õ4

ii 1

(z 2) is equal to:-

(1) 1 (2) 8(3) 11 (4) 17

51. There are 10 points in a row. In how many way can4 points be selected such that no two of them areconsecutive :-(1) 140 (2) 35(3) 104 (4) None

52. Number of positive integral solution of the equationxyz = 90 is equal to :-(1) 60 (2) 108(3) 54 (4) 120

53. A fair dice is thrown up to 20 times. The probabilitythat on the 10th throw, the fourth six apears is :-

(1) 6

10

84 56´

(2) 6

10

112 56

´

(3) 6

20

84 56´

(4) None

54. It is 5 : 2 against a husband who is 65 years old

living till he is 85 and 4 : 3 against his wife whois now 58, living till she is 78. If the probabilitythat atleast one of them will be alive for 20 years,

is 'k', then the value of '49k' -(1) 20 (2) 31 (3) 29 (4) 6

55. Three numbers selected from the set{31, 32, 33, .....320} then, the number of ways thatselected numbers form a increasing G.P. :-(1) 45 (2) 90 (3) 20 (4) 20C3

50. ;fn z1, z2, z3, z4 lehdj.k z4+ z3+ z2 + z + 1 = 0 ds

ewy gks rks =

+Õ4

ii 1

(z 2) dk eku gksxk :-

(1) 1 (2) 8(3) 11 (4) 17

51. ,d js[kk esa 10 fcUnq vafdr gS rks bu vafdr fcUnqvksa esa lspkj fcUnqvksa ds p;u ds rjhdksa dh la[;k Kkr djks ;fn dksbZHkh nks fcUnq Øekxr ugha gks :-(1) 140 (2) 35(3) 104 (4) dksbZ ugha

52. lehdj.k xyz = 90 ds ?kukRed iw.kk±dh; gyksa dh la[;kgksxh&(1) 60 (2) 108(3) 54 (4) 120

53. ,d Qs;j (fair) iklk 20 ckj mNkyk tkrk gS rks 10 osa mNkyesa pkSFkh ckj 6 vkus dh izkf;drk gksxh :-

(1) 6

10

84 56´

(2) 6

10

112 56

´

(3) 6

20

84 56´

(4) dksbZ ugha

54. ;fn ,d ifr ftldh orZeku vk;q 65 o"kZ gS] ds 85 o"kZvk;q rd thfor jgus ds foi{k esa la;ksxkuqikr 5 : 2 gS rFkkmldh iRuh] ftldh orZeku vk;q 58 o"kZ gS] ds 78 o"kZ vk;qrd thfor jgus ds foi{k esa la;ksxkuqikr 4 : 3 gSA ;fn muesals de ls de fdlh ,d ds vxys 20 o"kZ rd thfor jgusdh izkf;drk k gS] rks 49k dk eku gksxk -(1) 20 (2) 31 (3) 29 (4) 6

55. leqPp; {31, 32, 33, .......320} esa ls rhu la[;kvksa ds p;uds rjhds esa ;fn p;fur la[;k,sa o/kZeku xq- Js- cukrh gks]gksxsa :-(1) 45 (2) 90 (3) 20 (4) 20C3

13/25Kota/01CE314060



56. ;fn z1 = 6 + i vkSj z2 = 4 – 3i rFkk z ,d lfEeJ la[;k

gS tks fd dks.kkad -æ ö

ç ÷-è ø1

2

z zz z

= p2

dks lUrq"V djrh gS]

rks-

(1) |z – (5 – i)| = 5 (2) |z – (5 – i)| = 5

(3) |z – (5 + i)| = 5 (4) |z – (5 + i)| = 5

57. ;fn f=?kkr lehdj.k esa x2 dk xq.kkad 'kwU; gks rFkk 'ks"k xq.kkadokLrfod gks rFkk ,d ewy a = 3 + 4 i gks ,oa vU; ewy

b o g gksa rks abg gksxk :-(1) 150 (2) – 150

(3) 25 (4) buesa ls dksbZ ugha58. ;fn lehdj.k x2 + (a – 1)x + 2a = 0 dk Bhd

,d ewy vUrjky (0,3) esa fLFkr gks rks 'a' dk vUrjkygksxk :-(1) (–¥, 0) È (6, ¥) (2) (–¥, 0] È (6, ¥)(3) (–¥, 0] È [6, ¥) (4) (0, 6)

59. ;fn 3z

2 cos isin=

+ q + q gks rks z dk fcUnqiFk gksxk :-

(1) ljy js[kk(2) ,d o`Ùk ftldk dsUæ x-v{k ij fLFkr gks(3) ,d o`Ùk ftldk dsUæ y-v{k ij fLFkr gks(4) ijoy;

60. Js.kh1 1 1 1 1 .........9 18 30 45 63

+ + + + + ¥ ds vuUr inks a dk

;ksx gksxk :-

(1) 13

(2) 14

(3) 15

(4) 23

56. Let z1 = 6 + i and z2 = 4 – 3i. Let z be a complex

number such that arg-æ ö

ç ÷-è ø1

2

z zz z

= p2

, then

z satisfies -

(1) |z – (5 – i)| = 5 (2) |z – (5 – i)| = 5

(3) |z – (5 + i)| = 5 (4) |z – (5 + i)| = 5

57. In a cubic equation coefficient of x2 is zero andremaining coefficient are real has one roota = 3 + 4 i and remaining roots are b and g thenabg is :-(1) 150 (2) – 150(3) 25 (4) None of these

58. If exactly one root of the equationx2 + (a – 1)x + 2a = 0 lies in the interval (0,3), thenset of value of 'a' is given by :-(1) (–¥, 0) È (6, ¥) (2) (–¥, 0] È (6, ¥)(3) (–¥, 0] È [6, ¥) (4) (0, 6)

59. If 3z

2 cos isin=

+ q + q, then locus of z is :-

(1) a straight line(2) a circle having centre on x-axis(3) a circle having centre on y-axis(4) a parabola

60. The sum of the infinite series

1 1 1 1 1 .........9 18 30 45 63

+ + + + + ¥ is equal to :-

(1) 13

(2) 14

(3) 15

(4) 23

Kota/01CE31406014/25


Target : JEE(Main) 2015/11-03-2015

61. A light beam is traveling from Region I to RegionIV (Refer Figure). The refractive index in Regions

I, II, III and IV are n0, 0

2n

, 0

6n

and 0

8n

,

respectively. The angle of incidence q for whichthe beam just misses entering Region IV is figure

Region I Region II Region III Region IV

2n0

qn0 6n0

8n0

0 0.2 m 0.6 m

(1) sin–1 34

æ öç ÷è ø

(2) sin–1 18

æ öç ÷è ø

(3) sin–1 14

æ öç ÷è ø

(4) sin–1 13

æ öç ÷è ø

62. A ray of light travelling in water is incident on itssurface open to air. The angle of incidence is q,which is less than the critical angle. Then there willbe(1) only a reflected ray and no refracted ray(2) only a refracted ray and no reflected ray(3) a reflected ray and a refracted ray and the angle

between them would be less than 180° – 2q(4) a reflected ray and a refracted ray and the angle

between them would be greater than 180° – 2q

61. izdk'k dh ,d fdj.k iqat {ks= I ls IV dh vksj tk jgh gS (fp=ns[ks a)A {k s= I, II, III rFkk IV ds viorZuk ad Øe'k%

n0 , 0

2n

, 0

6n

rFkk 0

8n

gSA og vkiru dks.k q ftl ij izdk'k

iqat {ks= IV esa igqapus ls pwd tkrk gS] dk eku fuEu gS%&

Region I Region II Region III Region IV

2n0

qn0 6n0

8n0

0 0.2 m 0.6 m

(1) sin–1 34

æ öç ÷è ø

(2) sin–1 18

æ öç ÷è ø

(3) sin–1 14

æ öç ÷è ø

(4) sin–1 13

æ öç ÷è ø

62. ikuh esa ,d xfr'khy ,d izdk'k dh fdj.k mldh gok esa[kqyh gqbZ lrg ij vkifrr gksrh gSA vkiru dks.k q] ØkfUrddks.k ls NksVk gSA rc ogk¡ gksxh(1) dsoy ,d ijkofrZr fdj.k rFkk viofrZr fdj.k ugha(2) dsoy ,d viofrZr fdj.k rFkk ijkofrZr fdj.k ugha(3) ,d ijkofrZr fdj.k rFkk ,d viofrZr fdj.k rFkk muds

e/; dks.k 180°–2q ls de gksxk(4) ,d ijkofrZr fdj.k rFkk ,d viofrZr fdj.k rFkk muds

e/; dks.k 180°–2q ls vf/kd gksxk

PART C - PHYSICS

15/25Kota/01CE314060



63. A point object is placed at a distance of 20 cm froma thin plano-convex lens of focal length 15 cm, ifthe plane surface is silvered. The image will format :-

(1) 60 cm left of AB (2) 30 cm left of AB

(3) 12 cm left of AB (4) 60 cm right of AB

64. A point object is placed at the centre of a glasssphere of radius 6 cm and refractive index 1.5. Thedistance of the virtual image from the surface ofthe sphere is(1) 2 cm (2) 4 cm(3) 6 cm (4) 12 cm

65. A quarter cylinder of radius R and refractive index1.5 is placed on a table. A point object P is kept ata distance of mR from it. Find the value of m forwhich a ray from P will emerge parallel to the tableas shown in the figure.

mR R

P

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

63. 15 lseh Qksdl nwjh ds ,d irys leryksÙky ySal ls 20 lsehnwjh ij ,d fcEc j[kk x;k gSA vc ySal dh lery lrgjtfrr dj nh tkrh gSA izfrfcEc cusxk

(1) AB ds cka;h vksj 60 lseh ij(2) AB ds cka;h vksj 30 lseh ij(3) AB ds cka;h vksj 12 lseh ij(4) AB ds nka;h vksj 60 lseh ij

64. ,d fcUnq fcEc dks 6 lseh f=T;k rFkk 1-5 viorZukad ds dkap

ds xksys ds dsUæ ij j[kk x;k gSA xksys dh lrg ls vkHkklh

izfrfcEc dh nwjh gksxh&

(1) 2 cm (2) 4 cm

(3) 6 cm (4) 12 cm

65. 1-5 viorZukad rFkk R f=T;k ds ,d csyu ds ,d pkSFkkbZ Hkkxdks est ij j[kk x;k gSA blls mR nwjh ij ,d fcUnq fcEc Pj[kh xbZ gSA m dk og eku Kkr dhft;s ftlds fy, fcUnq Pls pyus okyh fdj.k est ds lekUrj fuxZr gks tSlk fd fp=esa n'kkZ;k x;k gSA

mR R

P

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

izR;sd iz'u dks vtqZu cudj djksA

Kota/01CE31406016/25


Target : JEE(Main) 2015/11-03-2015

66. A diverging lens of focal length 10 cm is placed10 cm in front of a plane mirror as shown in thefigure. Light from a very far away source falls onthe lens. The final image is at a distance :–

(1) 20 cm behind the mirror(2) 7.5 cm in front of the mirror(3) 7.5 cm behind the mirror(4) 2.5 cm in front of the mirror

67. The graph between angle of deviation (d) and angleof incidence (i) for a triangular prism is representedby :-

(1) (2)

(3) (4)

66. 10 lseh Qksdl nwjh dk vilkjh ySal fp=kuqlkj lery niZ.k

ds lkeus 10 lseh nwjh ij j[kk gSA cgqr nwj fLFkr lzksr ls

ySal ij izdk'k vkifrr gksrk gSA vafre izfrfcEc dh nwjh gksxh&

(1) niZ.k ds ihNs 20 lseh

(2) niZ.k ds lkeus 7-5 lseh

(3) niZ.k ds ihNs 7-5 lseh

(4) niZ.k ds lkeus 2-5 lseh

67. ,d f=Hkqtkdkj fizTe ds fy;s fopyu dks.k (d) vkSj vkiru

dks.k (i) ds chp xzkQ blls n'kkZ;k tkrk gS:-

(1) (2)

(3) (4)

17/25Kota/01CE314060



68. Two coherent narrow slits emitting light ofwavelength l in the same phase are placed parallelto each other at a small separation of 3l. The lightis collected on a screen S which is placed at adistance D (>>l) from the slits. The smallestdistance x such that the P is a maxima.

× ×

×

S1 S2

P

O

x

D

(1) 3D (2) 8D (3) 5D (4) D

52

69. Minimum thickness of a mica sheet having

m = 32

which should be placed in front of one of

the slits in YDSE is required to reduce the intensityat the centre of screen to half of maximum intensityis-(1) l/4 (2) l/8 (3) l/2 (4) l/3

70. Statement–1 : If white light is used in YDSE, thenthe central bright fringe will be whiteandStatement–2 : In case of white light used in YDSE,all the wavelengths produce their zero ordermaxima at the same position(1) Statement–1 is True, Statement–2 is True ;

Statement–2 is a correct explanation forStatement–1

(2) Statement–1 is True, Statement–2 is True ;Statement–2 is not a correct explanation forStatement–1

(3) Statement–1 is True, Statement–2 is False.(4) Statement–1 is False, Statement–2 is True.

68. nks ladjh dyk laca¼ fLyVsa 3l dh y?kq nwjh ij ,d nwljsls lekUrj :i ls fLFkr gSa rFkk leku dyk esa l rjaxnS/;Zdk izdk'k mRlftZr djrh gSaA ;g izdk'k fLyV ls D (>>l)nwjh ij fLFkr ,d insZ S ij ,df=r gksrh gSA vR;Ur y?kqnwjh x gksxh rkfd P mfPp"B gks&

× ×

×

S1 S2

P

O

x

D

(1) 3D (2) 8D (3) 5D (4) D

52

69. ;ax f}fLyV iz;ksx esa m = 32

okyh ekbdk 'khV dks ,d

fLyV ds lkeus j[kk tkrk gSA bl 'khV dh U;wure eksVkbZ

fdruh gksuh pkfg;s rkfd inZs ds dsUnz ij izkIr rhozrk bldh

vf/kdre eku dh vk/kh izkIr gksA(1) l/4 (2) l/8 (3) l/2 (4) l/3

70. dFku-1 : ;fn YDSE esa 'osr izdk'k dk iz;ksx fd;k tk;s]rks dsUnzh; nhIr fÝUt] 'osr gksxhAvkSjdFku-2 : YDSE esa 'osr izdk'k ds iz;ksx dh fLFkfr esa lHkhrjaxsa] leku fLFkfr ij 'kwU; Øe dk mfPp"B mRiUu djrhgSA(1) dFku - 1 rFkk dFku - 2 nks vyx&vyx lR; gSa rFkk

dFku - 2, dFku - 1 dk lgh Li"Vhdj.k gSA(2) dFku - 1 rFkk dFku - 2 nks vyx&vyx lR; gSa ijUrq

dFku - 2, dFku - 1 dk lgh Li"Vhdj.k ugha gSA(3) dFku - 1 lR; gS ijUrq dFku - 2 vlR; gSaA(4) dFku - 1 vlR; gS ijUrq dFku - 2 lR; gSaA

Kota/01CE31406018/25


Target : JEE(Main) 2015/11-03-2015

71. Spherical wavefronts shown in figure, strike a planemirror. Reflected wavefront will be as shown in

(1) (2)

(3) (4)

72. A point source of light is placed on the principleaxis between F and 2 F of a concave lens. On theother s ide very far, a screen is placedperpendicular to principal axis. As the screen isbrought close towards lens(1) the light intensity on screen continuously

decreases(2) the light intensity on screen continuously

increases(3) the light intensity on screen first increases,

then decreases(4) the light intensity on screen first decreases,

then increases

71. xksyh; rjaxkxz fp=kuqlkj lery niZ.k ls Vdjkrs gSa rks ijkofrZr

rjaxkxz iznf'kZr gksxk&

(1) (2)

(3) (4)

72. izdk'k dk ,d fcUnq lzksr vory ySal ds eq[; v{k ij F o

2F ds e/; j[kk x;k gSA nwljh rjQ cgqr nwj] eq[; v{k ds

yEcor ,d inkZ j[kk x;k gSA tc insZ dks ySal ds ikl yk;k

tkrk gS&

(1) insZ ij izdk'k dh rhozrk yxkrkj ?kVrh gS

(2) insZ ij izdk'k dh rhozrk yxkrkj c<+rh gS

(3) insZ ij izdk'k dh rhozrk igys c<+rh gS] fQj ?kVrh

gS

(4) insZ ij izdk'k dh rhozrk igys ?kVrh gS] fQj c<+rh

gS

19/25Kota/01CE314060



73. When an inert gas is filled in place of vacuum ina photo cell, then

(1) Photo-elecric current is decreased

(2) Photo-electric current is increased

(3) Photo-electric current remains the same

(4) Decrease or increase in photo-electric currentdoes not depend upon the gas filled

74. When a point source of monochromatic light is ata distance of 0.2 m from a photoelectric cell, thecut-off voltage and the saturation current are 0.6volt and 18 mA respectively. If the same source isplaced 0.6 m away from the photoelectric cell, then

(1) The stopping potential will be 0.2 V

(2) The stopping potential will be 0.6 V

(3) The saturation current will be 6 mA

(4) The saturation current will be 18 mA

75. The potenrial energy of a particle of mass m isgiven by

{ 0E ; 0 x 1U(x) 0; x 1£ £= >

l1 and l2 are the de-Broglie wavelengths of theparticle, when 0 £ x £ 1 and x > 1 respectively. If the

total energy of particle is 2E0, the ratio 1

2

ll

will be

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

2

73. ;fn QksVks lsy esa fuokZr~ ds LFkku ij dksbZ vfØ; xSl Hkj nhtkrh gS] rks

(1) izdk'k fo|qr /kkjk dk eku ?kV tkrk gS

(2) izdk'k fo|qr /kkjk dk eku c<+ tkrk gS

(3) izdk'k fo|qr /kkjk ogh jgrh gS

(4) izdk'k fo|qr /kkjk dk ?kVuk ;k c<+uk QksVks lsy esa HkjhxSl ij fuHkZj ugha djrk gS

74. tc ,d ,do. kh Z; i zd k' k dk fcUn q ò k s r ,dQksVk sbysfDVªd lsy ls 0.2 m dh nwjh ij gS rk s fujk s/khfoHko vkSj larÌr /kkjk Øe'k% 0.6 volt rFkk 18 mA gSA;fn ogh òksr QksVksbysfDVªd lsy ls 0.6 m nwj j[kk tkrk gSrk s

(1) fujks/kh foHko 0.2 V gksxk

(2) fujks/kh foHko 0.6 V gksxk

(3) larÌr /kkjk 6 mA gksxh

(4) larÌr /kkjk 18 mA gksxh

75. fdlh m nzO;eku ds d.k dh fLFkfrt mGtkZ fuEu izdkj nhtkrh gS

{ 0E ; 0 x 1U(x) 0; x 1£ £= >

0 £ x £ 1 ,oa x > 1 ds fy;s Mh&czksXyh rjaxnS/; Z Øe'k% l1

rFkk l2 gSaA ;fn d.k dh dqy mGtkZ 2E0 gS rks vuqikr 1

2

ll

gksxk

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

2

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

Kota/01CE31406020/25


Target : JEE(Main) 2015/11-03-2015

76. In the following arrangement y = 1.0 mm, d = 0.24 mmand D = 1.2 m. The work function of the material ofthe emitter is 2.2 eV. The stopping potential Vneeded to stop the photo current will be

D

dy

ABright

Dark

Dark

Bright

Bright

S

(1) 0.9 V (2) 0.5 V(3) 0.4 V (4) 0.1 V

77. Hydrogen (H), deuterium (D), singly ionizedhelium (He+) and doubly ionized lithium (Li++)all have one electron around the nucleus.Consider n = 2 to n = 1 trans ition. Thewavelengths of emitted radiations are l1, l2 l3

and l4 respectively.(1) l1 = l2 = 4l3 = 9l4

(2) 4l1 = 2l2 = 2l3 = l4

(3) l1 = 2l2 = 2 2 l3 = 3 2 l4

(4) l1 = l2 = 2l3 = 3 2 l4

78. The activity of a radiactive sample is measuredas N0 counts per minute at t = 0 and N0/e countsper minute at t = 5 minutes. The time (in minutes)at which the activity reduces to half its value is(1) 5 loge 2 (2) loge 2/5(3) 5/loge 2 (4) 5 log10 2

76. fuEu O;oLFkk esa y = 1.0 mm, d = 0.24 mm ,oa D = 1.2 m

mRltZd ds inkFkZ dk dk;ZQyu 2.2 eV gSA izdk'k /kkjk dks

jksdus ds fy;s vko';d fujks/kh foHko V gksxk

D

dy

ABright

Dark

Dark

Bright

Bright

S

(1) 0.9 V (2) 0.5 V(3) 0.4 V (4) 0.1 V

77. gkbMªkstu (H), M~;wVhfj;e (D), ,dy vk;uhd`r ghfy;e(He+) ,oa f}vk;uhd`r yhfFk;e (Li++) lHkh ds ukfHkd dspkjk s a vksj ,d bysDVªk Wu gSA ;fn n = 2 ls n = 1 rd ds

laØe.k dks ekuk tk;s ,oa mRlftZr fofdjdksa dh rjaxnS/; Z

l1, l2 l3 ,oa l4 gSa rks

(1) l1 = l2 = 4l3 = 9l4

(2) 4l1 = 2l2 = 2l3 = l4

(3) l1 = 2l2 = 2 2 l3 = 3 2 l4

(4) l1 = l2 = 2l3 = 3 2 l4

78. fdlh jsfM;ks,sfDVo uewus dh lfØ;rk t = 0 le; ij N0

dkmaV izfr feuV vkSj t = 5 feuV ij N0/e dkmaV izfr feuVekih xbZ gSA fdrus le; (feuVksa esa) ij lfØ;rk vius ekudh vk/kh gks tk;sxh(1) 5 loge 2 (2) loge 2/5(3) 5/loge 2 (4) 5 log10 2

21/25Kota/01CE314060



79. Half-life of radiactive substance is 20 mintes.

Difference between points of time when it is 33%

disintegrated and 67% disintegrated is approximately

(1) 10 min (2) 20 min

(3) 30 min (4) 40 min

80. A small quantity of solution containing Na24

radio nuclide of activity 1 microcurie is injected

into the blood of a person. A sample of the blood

of volume 1 cm3 taken after 5 hours shows an

activity of 298 disintegration per minute. What

will be the total volume of the blood in the body

of the person. Assume that the radioactive

solution mixes uniformly in the blood of the

person

(Take 1 curie = 3.7 × 1010 disintegration per second

and e–lt = 0.7927; where l = disintegration constant)

t = 5 hrs.

(1) 5.94 litre (2) 2 litre

(3) 317 litre (4) 1 litre

81. Which sample contains greater number of nuclei :

a 5.00-mCi sample of 240Pu (half-life 6560 y) or a

4.45-mCi sample of 243Am (half-life 7370 y)

(1) 240Pu (2) 243Am

(3) Equal in both (4) None of these

79. ,d jsfM;ks/kehZ inkFkZ dh v¼Zvk;q 20 feuV gS mu le;

fcUnqvks a dk vUrj D;k gS tc og Øe'k% 33% o 67%

fo?kfVr gS

(1) 10 min (2) 20 min

(3) 30 min (4) 40 min

80. ,d ?kksy dh FkksM+h lh ek=k ,d O;fDr ds jDr esa izos'k djk

nh tkrh gS] bl ?kksy es a mifLFkr jsfM;ks ukfHkd Na24 dh

lfØ;rk 1 ekbØksD;wjh gSA 5 ?k.Vs ckn O;fDr ds 'kjhj ls

1 cm3 jDr uewus ds rkSj ij fy;k tkrk gS] ftldh lfØ;rk

298 fo?kVu izfr feuV gSA O;fDr ds 'kjhj esa mifLFkr dqy

jDr dk vk;ru gS] ekuk fd jsfM;kslfØ; ?kksy jDr esa ,d

leku :i ls fefJr gS

(1 D;wjh = 3.7 × 1010 [email protected] ,os e–lt = 0.7927;

;gk¡ l = fo?kVu fu;rkad) t = 5 ?k.Vs

(1) 5.94 litre

(2) 2 litre

(3) 317 litre

(4) 1 litre

81. fuEu esa ls dkSu ls uewus esa ukfHkdksa dh la[;k vf/kd gS :240Pu (v¼Zvk;q 6560 o"kZ) dk 5.00-mCi ;k243Am (v¼Zvk;q 7370 o"kZ) dk 4.45-mCi

(1) 240Pu (2) 243Am

(3) nksuks a esa cjkcj (4) buesa ls dksbZ ugha

Use stop, look and go method in reading the question

Kota/01CE31406022/25


Target : JEE(Main) 2015/11-03-2015

82. When a sample of solid lithium is placed in aflask of hydrogen gas then following reactionhappened1 7 4 41 3 2 2H Li He He+ ® +

This statement is

H2Li

(1) True(2) False(3) May be true at a particular pressure(4) None of these

83. Nuclear reaction are given as(i) � (n, p)15p

32 (ii) � (p, a)8O16

(iii) 7N14 (� p) 6C

14

missing particle or nuclide (in box �) in thesereactions are respectively(1) S32, F19, 0n

1 (2) F19, S32, 0n1

(3) Be9, F19, 0n1 (4) None of these

84. Current in the circuit will be

20W 5V

30W

20W

i

(1)5

A40

(2) 5

A50

(3) 5

A10

(4) 5

A20

82. ;fn jsfM;kslfØ; uewus dk nzO;eku nksxquk dj fn;k tk;src uewus dh lfØ;rk ,oa bldk fo?kVu fLFkjkad Øe'k%1 7 4 41 3 2 2H Li He He+ ® +

H2Li

(1) lR; gS

(2) xyr gS

(3) fdlh fuf'pr nkc ij lR; gks ldrk gS

(4) mijksä esa ls dksb Z ugha83. ukfHkdh; vfHkfØ;k,sa uhps nh xbZ gSa

(i) � (n, p)15p32 (ii) � (p, a)8O

16

(iii) 7N14 (� p) 6C

14

bu vfHkfØ;kvksa esa NksM+ fn;s x;s ukfHkd (vFkkZr~ ckWDl esa �)Øe'k% gS(1) S32, F19, 0n

1 (2) F19, S32, 0n1

(3) Be9, F19, 0n1 (4) buesa ls dksbZ ugha

84. ifjiFk dh /kkjk gksxh

20W 5V

30W

20W

i

(1)5

A40

(2) 5

A50

(3) 5

A10

(4) 5

A20

23/25Kota/01CE314060



85. A sinusoidal voltage of peak value 200 volt is

connected to a diode and resistor R in the circuit

shown so that half wave rectification occurs. If

the forward resistance of the diode is negligible

compared to R the rms voltage (in volt) across R

is approximately

~E = 200 Volt0 R

(1) 200 (2) 100

(3)200

2(4) 280

86. Find VAB

30V

10W 10W

10W

VAB

(1) 10 V (2) 20 V

(3) 30 V (4) None of these

85. ,d v¼Zpkyd Mk;k sM v¼Zrj ax fn"Vdkjh ds :i es a

dk;Zjr gS ftlls ,d izfrjk s/k R tqM+k g S ,oa 200 volt

f'k[kj eku dk ,d izR;korhZ ok sYV st vkjk sfir gSA ;fn

Mk;ksM ds vxz izfrjks/k dk eku izfrjks/k R dh rqyuk esa de

gks rk s R ij mRiUu rms oksYVst (oksYVes a) dk eku yxHkx

gksxk

~E = 200 Volt0 R

(1) 200 (2) 100

(3)200

2(4) 280

86. VAB dk eku Kkr djsa

30V

10W 10W

10W

VAB

(1) 10 V (2) 20 V

(3) 30 V (4) buesa ls dksbZ ugha

Kota/01CE31406024/25


Target : JEE(Main) 2015/11-03-2015

87. Figure gives a system of logic gates. From thestudy of truth table i t can be found that toproduce a high output (1) at R, we must have

XY

O

PR

(1) X = 0, Y = 1 (2) X = 1, Y = 1(3) X = 1, Y = 0 (4) X = 0, Y = 0

88. Fermi level of energy of an intrinsic semiconductor lies(1) In the middle of forbidden gap(2) Below the middle of forbidden gap(3) Above the middle of forbiden gap(4) Outside the forbidden gap

89. A Ge specimen is doped with Al. The concentrationof acceptor atoms is ~1021 atoms/m3. Given that theintrinsic concentration of electron hole pairs is~1019 /m3, the concentration of electrons in thespecimen is(1) 1017 /m3 (2) 1015 /m3 (3) 104 /m3 (4) 102 /m3

90. Although carbon, silicon and germanium havesame lattice structure and four valence electronseach, their band structure leads to the energygaps as(1) Eg(Si) < Eg(Ge) < Eg(C)(2) Eg(Si) > Eg(Ge) < Eg(C)(3) Eg(Si) < Eg(Ge) > Eg(C)(4) Eg(Si) > Eg(Ge) > Eg(C)

87. fp= esa ykWftd xsVk s a ds ,d fudk; dks fn[kk;k x;k gSAlR;lkj.kh ds vk/kkj ij ckr;sa fd R fljs ij fuxZr dk mPpeku (1) izkIr djus ds fy;s fuos'kh gksuk pkfg;s

XY

O

PR

(1) X = 0, Y = 1 (2) X = 1, Y = 1(3) X = 1, Y = 0 (4) X = 0, Y = 0

88. uSt v¼Zpkyd esa QehZ mGtkZ Lrj gksrk gS(1) oftZr mGtkZ varjky ds e/;(2) oftZr mGtkZ varjky ds e/; ds uhps(3) oftZr mGtkZ varjky ds e/; ds mGij(4) oftZr mGtkZ varjky ds ckgj

89. ,d Ge tesZ fu;e fØLVy uewus es a Al v'kqf¼ ds :i es a

feyk;k x;k gSA xzkgh ijek.kqvksa dk ?kuRo ~1021 atoms/m3

gSA ;fn 'kq¼ voLFkk esa bysDVªk Wu&gksy dk ; qXe dk ?kuRo

~1019 /m3 gks rks uewus esa bysDV ªk Wuksa dk ?kuRo gksxk

(1) 1017 /m3 (2) 1015 /m3 (3) 104 /m3 (4) 102 /m3

90. ;|fi dkcZu] flfydk Wu rFkk tes Z fu;e leku tkydlajpuk ds gksrs gS ,oa izR;sd esa pkj la;ksth bysDVªkWu gksrs gSAbuds cSaM lajpuk esa mGtkZ vUrjky O;Dr gksrk gS(1) Eg(Si) < Eg(Ge) < Eg(C)

(2) Eg(Si) > Eg(Ge) < Eg(C)

(3) Eg(Si) < Eg(Ge) > Eg(C)

(4) Eg(Si) > Eg(Ge) > Eg(C)

Your moral dutyis to prove that ALLEN is ALLEN

25/25Kota/01CE314060



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

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CAREER INSTITUTE Pat S KOTA (RAJASTHAN ......N candidat i all t carr an textu materi, printe written, bit papers, pager, mobil hon electroni devic etc, except th Identit Car insid