CLASSROOM CONTACT PROGRAMME

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egRoiw.kZ funs Z'k

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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

SCORE – I DATE : 06 - 01 - 2015

0 1 C E 2 1 4 0 1 9

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 # 07 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/06-01-2015

1/28Kota/01CE214019

SPACE FOR ROUGH WORK

1. Two identical glass bulbs are interconnected by a

thin glass tube at 0ºC. A gas is filled in these bulb

is placed in ice and another bulb is placed in hot

bath, then the pressure of the gas becomes

1.5 times. The temperature of hot bath will be

Ice Hot water

(1) 100°C

(2) 182°C

(3) 256°C

(4) 546°C

2. A gaseous mixture consists of 16 g of helium and

16 g of oxygen. The ratio p

v

C

C of the mixture is

(1) 1.59 (2) 1.62

(3) 1.4 (4) 1.54

1. nks leku dk¡p ds cYc ,d iryh dk¡p dh uyh }kjk tqM+s

gSaA buesa lkekU; rki o nkc (N.T.P.) ij dksbZ xSl Hkjh gSA

;fn ,d cYc dks cQZ esa rFkk nwljs cYc dks xeZ ik= esa

j[kk tkrk gS] rks xSl dk nkc 1.5 xquk gks tkrk gSA xeZ ik=

dk rki gksxk-

Ice Hot water

(1) 100°C

(2) 182°C

(3) 256°C

(4) 546°C

2. fdlh xSlh; feJ.k esa 16 g ghfy;e rFkk 16 g vkWDlhtu

gSA bl feJ.k dk p

v

C

C vuqikr gS

(1) 1.59 (2) 1.62

(3) 1.4 (4) 1.54

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

Kota/01CE2140192/28

Target : JEE (Main) 2015/06-01-2015


3. Two rigid boxes containing different ideal gases areplaced on a table. Box A contains one mole ofnitrogen at temperature T0, while box B contains onemole of helium at temperature (7/3) T0. The boxesare then put into thermal contact with each other,and heat flows between them until the gases reacha common final temperature (Ignore the heatcapacity of boxes). Then, the final temperature ofgases, Tf, in terms of T0 is

(1) Tf = 37

T0 (2) Tf = 73

T0

(3) Tf = 32

T0 (4) Tf = 52

T0

4. The temperature of the two outer surfaces of acomposite slab, consisting of two materials havingcoefficients of thermal conductivity K and 2K andthickness x and 4x, respectively are T2 andT1(T2 > T1). The rate of heat transfer through the

slab, in a steady state is æ öç ÷è ø

2 1A(T -T )Kx

f, with f

equals to(1) 1(2) 1/2

4xx

2KKT2 T1(3) 2/3(4) 1/3

5. Two spheres of the same material have radii1 m and 4 m and temperatures 4000 K and2000 K respectively. The ratio of the energyradiated per second by the first sphere to that bythe second is(1) 1 : 1 (2) 16 : 1 (3) 4 : 1 (4) 1 : 9

3. fofHkék vkn'kZ xSlksa ls ;qDr nks n`<+ cDls ,d est ij j[ks gSaAcDls A esa T0 rkieku ij ukbVªkstu dk ,d eksy j[kk gS tcfdoDls B esa rkieku (7/3) T0 ij ghfy;e dk ,d eksy j[kkgSA oDlksa dks vc ,d nwljs ds rkih; Li'kZ esa yk;k tkrkgS vkSj Å"ek ,d ls nwljs esa rc rd izokfgr gksrh gS tcrd fd mudk ifj.kkeh rkieku ,dleku ugha gks tkrk gSA(cDlksa dh Å"ek /kkfjrk ux.; ekus) rc xSlksa dk ifj.kkehrkieku] Tf, T0 ds inksa esa gS-

(1) Tf = 37

T0 (2) Tf = 73

T0

(3) Tf = 32

T0 (4) Tf = 52

T0

4. nks inkFkks± ftuds Å"ek pkydrk xq.kkad K o 2K rFkk eksVkbZØe'k% x o 4x, gSa] dks tksM+dj ,d la;qDr ifV~Vdk cuk;h

x;h gS] ftlds nk s cká i`"Bks a ds rki Øe'k% T2 rFkk

T1(T2 > T1) gSaA LFkk;h voLFkk esa bl ifV~Vdk ls izokfgr

Å"ek dh nj æ öç ÷è ø

2 1A(T -T )Kx

f gS] ftlesa f dk eku gS

(1) 1

(2) 1/2

4xx

2KKT2 T1

(3) 2/3

(4) 1/3

5. ,d gh inkFkZ ds nks xksyks a dh f=T;k,¡ Øe'k % 1 eh o4 eh rFkk rki 4000 K o 2000 K gSA nk su k s axk syk s ls i zfr lsd.M mRlftZr Å"ekvks a dk vuqikrgksxk(1) 1 : 1 (2) 16 : 1(3) 4 : 1 (4) 1 : 9


3/28Kota/01CE214019


6. If the temperature of the sun were to increase fromT to 2T and its radius from R to 2R, then the ratioof the radiant energy received on earth to what itwas previously, will be-(1) 4 (2) 16 (3) 32 (4) 64

7. The figure shows a system of two concentricspheres of radii r1 and r2 and kept at temperaturesT1 and T2 , respectively. The radial rate of flow ofheat in a substance between the two concentricspheres, is proportional to

• T1

r1

r2 T2

(1) 2 1

1 2

(r r )(r r )

-(2) ln

2

1

rr

æ öç ÷è ø

(3) 1 2

2 1

r r(r r )- (4) (r2 – r1)

8. Assuming the sun to be a spherical body of radius

R at a temperature of T K, evaluate the total radiant

power, incident on earth, at a distance r from the

sun- (when radius of earth is r0)

(1) 2 204 r Rp sT4/r2

(2) 2 20r Rp sT4/r2

(3) 2 2 4 20r R T / 4 Rs p

(4) 2R sT4/r2

6. ;fn lw;Z dk rki T ls c<+dj 2T rFkk f=T;k R ls c<+dj 2Rgks tk, rks iFoh ij vc izkIr fofdfjr ÅtkZ vkSj iFoh ijigys izkIr gksus okyh ÅtkZ dk vuqikr gksxk-(1) 4 (2) 16(3) 32 (4) 64

7. vkjs[k esa r1 rFkk r2 f=T;kvksa ds nks ladsUnzh xksyksa dk ,dfudk; n'kkZ;k x;k gS ftUgsa Øe'k% T1 rFkk T2 rkiksa ij j[kkx;k gSA nksuksa ladsUnzh xksyksa ds chp ds inkFkZ esa Å"ek ds f=T;h;izokg dh nj lekuqikrh gS

• T1

r1

r2 T2

(1)2 1

1 2

(r r )(r r )

-(2) ln

2

1

rr

æ öç ÷è ø

(3) 1 2

2 1

r r(r r )- (4) (r2 – r1)

8. lw;Z dks T K rkieku ij R f=T;k dh ,d xksyh; oLrq ekudj]lw;Z ls r nwjh ij iFoh ij vkifrr lEiw.kZ fofdfjr 'kfDr dh

x.kuk dhft,-

(r0iFoh dh f=T;k gS)

(1) 2 204 r Rp sT4/r2

(2) 2 20r Rp sT4/r2

(3) 2 2 4 20r R T / 4 Rs p

(4) 2R sT4/r2

Kota/01CE2140194/28

Target : JEE (Main) 2015/06-01-2015


9. One end of a thermally insulated rod is kept at atemperature T1 and the other at T2. The rod iscomposed of two sections of lengths l1 and l2 andthermal conductivities K1 and K2 respectively. Thetemperature at the interface of the two sections is

l2l1 T2T1

K1 K2

(1) (K2l2T1 + K1l1T2)/(K1l1 + K2l2)(2) (K2l1T1 + K1l2T2)/(K2l1 + K1l2)(3) (K1l2T1 + K2l1T2)/(K1l2 + K2l1)(4) (K1l1T1 + K2l2T2)/(K1l1 + K2l2)

10. A long metallic bar is carrying heat from one of itsends to the other end under steady-state. The vari-ation of temperature q along the length x of the barfrom its hot end is best described by which of thefollowing figures ?

(1)

q

x

(2)

q

x

(3)

q

x

(4)

q

x

9. ,d Å"ekjks/kh NM+ dk ,d fljk T1 rki ij vkSj nwljk fljk

T2 rki ij gSA NM + Øe'k% l1o l2 yEckbZ vkSj K1 o K2

Å"ek pkydrkvksa dh NM+ksa dk la;ksx gSA nksuksa Hkkxksa dh lfU/k

dk rki gS-

l2l1 T2T1

K1 K2

(1) (K2l2T1 + K1l1T2)/(K1l1 + K2l2)(2) (K2l1T1 + K1l2T2)/(K2l1 + K1l2)(3) (K1l2T1 + K2l1T2)/(K1l2 + K2l1)(4) (K1l1T1 + K2l2T2)/(K1l1 + K2l2)

10. ,d yEch /kkrq dh NM + esa ,d fljs ls nwljs fljs rd fLFkjvoLFkk esa Å"ek izokfgr gks jgh gSA rkieku q dk NM+ ds xeZfljs ls yEckbZ x ls ifjorZu fuEufyf[kr fp=ksa esa ls fdlds}kjk loksZÙke iznf'kZr fd;k tkrk gS ?

(1)

q

x

(2)

q

x

(3)

q

x

(4)

q

x


5/28Kota/01CE214019


11. At what temperature is the rms velocity of ahydrogen molecule equal to that of an oxygenmolecules at 47° C ?(1) 80 K (2) –73 K(3) 3 K (4) 20 K

12. 1 mole of a gas with g = 7/5 is mixed with1 mole of a gas with g = 5/3, then the value of gfor the resulting mixture is-(1) 7/5 (2) 2/5 (3) 24/16 (4) 12/7

13. One kg of a diatomic gas is at a pressure of8 × 104 N/m2. The density of the gas is4 kg/m3. What is the energy of the gas due to itsthermal motion ?(1) 6 × 104 J (2) 7 × 104 J(3) 3 × 104 J (4) 5 × 104 J

14. 100 g of water is heated from 30°C to 50°CIgnoring the slight expansion of the water, thechange in its internal energy is (specific heat ofwater is 4184 J/kg/K) :-(1) 84 kJ (2) 2.1 kJ(3) 4.2 kJ (4) 8.4 kJ

15. P-V diagram of a diatomic gas is a straight linepassing through origin. The molar heat capacity ofthe gas in the process will be -(1) 4 R (2) 2.5 R(3) 3 R (4) 4R/3

16. Find the amount of work done to increase thetemperature of one mole of an ideal gas by 30°Cif it is expanding under the condition V µ T

2/3

[R = 1.99 cal/mol-K](1) 176 J (2) 167 J (3) 671 J (4) 761 J

11. fdl rki ij gkbMªkstu ds v.kqvksa dk oxZ ek/; ewy osx 47°Cij vkWDlhtu ds v.kqvksa ds oxZ ek/; ewy osx ds cjkcj gkstk,xk ?(1) 80 K (2) –73 K(3) 3 K (4) 20 K

12. g = 7/5 okyh ,d eksy xSl dks g = 5/3 okyh ,d eksyxSl ds lkFk fefJr fd;k tkrk gSA feJ.k ds fy;s g dkeku gksxk %&(1) 7/5 (2) 2/5 (3) 24/16 (4) 12/7

13. ,d f}ijek.kqd xSl dk ,d fdyksxzke 8 × 104 N/m2 dsnkc ij gSA xSl dk ?kuRo 4 kg/m3 gSA bldh Å"eh; xfr

ds dkj.k xSl dh ÅtkZ D;k gS ?

(1) 6 × 104 J (2) 7 × 104 J

(3) 3 × 104 J (4) 5 × 104 J

14. 100 g ikuh dks 30°C ls 50°C rd xeZ fd;k tkrk gSaA ikuh

ds lw{e izlkj dks ux.; eku dj] mldh vkUrfjd ÅtkZ

esa ifjorZu gS (ikuh dh fof'k"V Å"ek 4184 J/kg/K gS):-

(1) 84 kJ (2) 2.1 kJ

(3) 4.2 kJ (4) 8.4 kJ

15. ,d f}ijek.kqd xSl dk P-V oØ ,d ljy js[kk gS] tks ewyfcUnq ls xqtjrh gSA bl izfØ;k esa xSl dh eksyj Å"eh;/kkfjrk gksxh -(1) 4 R (2) 2.5 R(3) 3 R (4) 4R/3

16. ,d eksy vkn'kZ xSl dk rki 30°C ls c<+kus ds fy, ftldkizlj.k fuEu lehdj.k }kjk iznf'kZr V µ T

2/3

[R = 1.99 dSyksjh/eksy-K](1) 176 J (2) 167 J(3) 671 J (4) 761 J

Kota/01CE2140196/28

Target : JEE (Main) 2015/06-01-2015


17. A monoatomic ideal gas, initially at temperatureT1, is enclosed in a cylinder fitted with africtionless piston. The gas is allowed to expandadiabatically to a temperature T2 by releasing thepiston suddenly. If L1 and L2 are the lengths ofthe gas column before and after expansionrespectively, then T1/T2 is given by

(1)

2 / 3

1

2

LL

æ öç ÷è ø

(2) 1

2

LL

æ öç ÷è ø

(3) 2

1

LL

æ öç ÷è ø

(4)

2 / 3

2

1

LL

æ öç ÷è ø

18. An ideal gas is taken through the cycle A ® B ®C ® A, as shown in the figure. If the net heatsupplied to the gas in the cycle is 5J, the work doneby the gas in the process C ® A is

V(m )3

C B

A

2

110 P(N/m )2

(1) –5J (2) –10 J (3) –15 J (4) –20 J19. Which of the following graphs correctly represents

the variation of b = –(dV/dP)/V with P for an idealgas at constant temperature ?

(1)

P

b

(2)

P

b

(3)

P

b

(4)

P

b

17. ,d ,dijek.kqd vkn'kZ xSl izkjfEHkd rki T1 ij] ,d?k"kZ.kjfgr fiLVu ;qDr flys.Mj esa Hkjh gSA fiLVu dks vpkudLora= djds xSl dks :¼ks"e :i ls T2 rki rd izlkfjrgksus nsrs gSa ;fn flys.Mj esa xSl ds izlkj ls igys ,oa cknesa xSl LrEHkksa dh yEckbZ;ka Øe'k % L1 rFkk L2 gS] rc T1/T2 dk eku gksxk

(1)

2 / 3

1

2

LL

æ öç ÷è ø

(2) 1

2

LL

æ öç ÷è ø

(3) 2

1

LL

æ öç ÷è ø

(4)

2 / 3

2

1

LL

æ öç ÷è ø

18. ,d vkn'kZ xSl dks fp=kuqlkj pØ A ® B ® C ® Als ys tkrs gSaA pØ esa xSl dks nh xbZ dqy Å"ek 5 twy gSAizØe C ® A esa xSl }kjk fd;k x;k dk;Z gS &

V(m )3

C B

A

2

110 P(N/m )2

(1) –5J (2) –10 J (3) –15 J (4) –20 J19. fLFkj rki ij] vkn'kZ xSl ds fy;s b = –(dV/dP)/V dk

P ds lkFk ifjorZu dk lgh xzkQ gS&

(1)

P

b

(2)

P

b

(3)

P

b

(4)

P

b


7/28Kota/01CE214019


20. An ideal gas expands isothermally from a volumeV1 to V2 and then compressed to original volumeV1 adiabatically. Initial pressure is P1 andfinal pressure is P3. The total work done is W.Then

(1) P3 > P1, W > 0

(2) P3 < P1, W < 0

(3) P3 > P1, W < 0

(4) P3 = P1, W = 0

21. The plots of intensity versus wavelength for threeblack bodies at temperature T1, T2 and T3

respectively are as shown. Their temperatures aresuch that

T1

T3 T2

l

I

(1) T1 > T2 > T3

(2) T1 > T3 > T2

(3) T2 > T3 > T1

(4) T3 > T2 > T1

22. 2 kg ice at – 20°C is mixed with 5 kg water at 20°C.Then final amount of water in the mixture wouldbe; Given specific heat of ice = 0.5cal/g°C, specificheat of water = 1 cal/g°C, Latent heat of fusion ofice = 80 cal/g.(1) 6 kg(2) 5 kg(3) 4 kg(4) 2 kg

20. ,d vkn'kZ xSl lerkih; izØe esa V1ls V2 vk;ru rdizlkfjr gksrh gS rFkk ;g :¼ks"e :i ls ewy vk;ru V1 rdlEihfM+r gksrh gSA izkjfEHkd voLFkk esa nkc P1 rFkk vfUrevoLFkk esa nkc P3 gks tkrk gSA ;fn iwjs izØe esa xSl }kjkfd;k x;k dk;Z W gks rks–

(1) P3 > P1, W > 0

(2) P3 < P1, W < 0

(3) P3 > P1, W < 0

(4) P3 = P1, W = 0

21. rki T1, T2 rFkk T3 ij rhu d`f".kdk oLrqvksa ds fy, rhozrk

rFkk rjaxnS/;Z ds chp oØ [khaps x;s gSaA muds rki bl izdkj

gSa fd

T1

T3 T2

l

I

(1) T1 > T2 > T3

(2) T1 > T3 > T2

(3) T2 > T3 > T1

(4) T3 > T2 > T1

22. ux.; Å"eh; /kkfjrk okys ,d dqpkyd ik= esa –20°C okys2kg cQZ dks 20°C okys 5 kg ty ds lkFk fefJr fd;ktkrk gS rks ik= esa ikuh dh vfUre ek=k gksxh (fn;k gS cQZ dhfof'k"V Å"ek = 0.5 cal/g°C rFkk ikuh dh fof'k"V m"ek =1cal/g°C tcfd cQZ ds xyu dh xqIr m"ek 80 cal/g)(1) 6 kg(2) 5 kg(3) 4 kg(4) 2 kg

Kota/01CE2140198/28

Target : JEE (Main) 2015/06-01-2015


23. Liquid oxygen at 50 K is heated to 300 K at constantpressure of 1 atm. The rate of heating is constant.Which of the following graphs represents thevariation of temperature with time ?

(1)

Temp.

Time(2)

Temp.

Time

(3)

Temp.

Time(4)

Temp.

Time

24. Two identical conducting rods are first connectedindependently to two vessels, one containing waterat 100°C and the other containing ice at 0° C. Inthe second case, the rods are joined end to end andconnected to the same vessels. Let q1 and q2 g/sbe the rate of melting of ice in the two casesrespectively. The ratio q2/q1 is(1) 1/2 (2) 2/1(3) 4/1 (4) 1/4

25. 1 cc of water at its boiling point (100° C) isconverted into steam by supplying it 540 caloriesof heat the volume of steam is 1671 cc. Ifatmospheric pressure is 1.013 × 105 Nm-2 andJ = 4.18 joule/cal the approximate value of heat thatis required in overcoming the molecular attractionis

(1) 110 cal (2) 500 cal(3) 40 cal (4) zero cal

23. nzfor vkWDlhtu dks ,d ok;qe.Myh; fu;r nkc ij 50K

ls 300K rd xeZ djrs gSaA nh xbZ m"ek ;fn fu;r nj ls

gks] rks rki o le; d e/; oØ gksxk ?

(1)

Temp.

Time(2)

Temp.

Time

(3)

Temp.

Time(4)

Temp.

Time

24. nks le:i pkyd NM+ksa dks 100°C okys ikuh ds ik= rFkk00C okys cQZ ds ik= ls tksM+k tkrk gSA nwljh fLFkfr esa nksuksa

NM+ksa ds fljksa dks tksM+dj leku ik=ksa ls tksM+rs gSaA ekuk nksuksa

fLFkfr;ka esa cQZ ds xyu dh nj q1 rFkk q2 g/s gS rks vuqikr

1

2

qq gksxk

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

25. ty ds DoFkukad (1000C) ij 1 c.c. ty 540 dSyksjh Å"ekvo'kksf"kr djds ok"i esa ifjofrZr gksrk gS] ftldk vk;ru

1671 c.c. gSA ;fn ok;qe.M+yh; nkc = 1.013 × 105 U;wVu/

eh2 vkSj J = 4. 18 twy/dSyksjh gks rks vk.kfod vkd"kZ.k cyksa

dk mYya?ku djus esa yxus okyh Å"ek (dSyksjh esa) gksxh

yxHkx

(1) 110 dSyksjh (2) 500 dSyksjh

(3) 40 dSyksjh (4) 'kwU; dSyksjh


9/28Kota/01CE214019


26. 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). For which process(es) does thetemperature of the gas decrease?

P

P0

P0

A

BC

D

V0 2V0V

(1) Process A only

(2) Process C only

(3) Only Processes C and D

(4) Only Processes B, C and D

27. The specific heat of a gas at constant pressure ismore than that of the same gas at constant volumebecause

(1) work is done in the expansion of the gas atconstant pressure

(2) work is done in the expansion of the gas atconstant volume

(3) the molecular attraction incrcase under constantpressure

(4) the vibration of molecules increases underconstant pressure

26. fdlh ,dijek.oh; vkn'kZ xSl ij fd;s x;s pkj fofHkUumRØe.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

P

P0

P0

A

BC

D

V0 2V0V

(1) dsoy A esa

(2) dsoy C esa

(3) dsoy C o D esa

(4) dsoy B, C o D esa

27. fdlh xSl dh fof'k"V Å"ek fuf'pr nkc ij mlh xSl ds

fuf'pr vk;ru dh fof'k"V Å"ek dh vfèkd gksxh

D;ksfd

(1) fuf'pr nkc ij xSl ds izlkj esa dk;Z gksrk gS

(2) fuf'pr vk;ru ij xSl ds izlkj esa dk;Z gksrk gS

(3) fuf'pr nkc ij v.kqvksa ds chp vkd"kZ.k c<rk gS

(4) fuf'pr nkc ij v.kq dEiu dk c<+ko gksrk gS

Kota/01CE21401910/28

Target : JEE (Main) 2015/06-01-2015


28. A train whistling at constant frequency is movingtowards a station at a constant speed V. The traingoes past a stationary observer on the station. Thefrequency n' of the sound as heard by the observeris plotted as a function of time t (Fig.) . Identifythe expected curve.

(1)

n'

t

(2)

n'

t

(3)

t

n'

(4)

t

n'

28. ,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'


11/28Kota/01CE214019


29. Following statements are given for a stationarywave(a) Every particle has a fixed amplitude which is

different from the amplitude of its nearestparticle.

(b) All the particles cross their mean position at thesame time.

(c) All the particles are oscillating with sameamplitude.

(d) There is no net transfer of energy across anyplane.

(e) There are some particles which are always atrest.

Which of the following is correct(1) a, b, c, d, e (2) a, c, d, e(3) b, c, d, e (4) a, b, e

30. The pattern of standing waves formed on astretched string at two instants of time (extreme,mean) are shown in figure. The velocity of twowaves superimposing to form stationary waves is360 ms–1 and their frequencies are 256 Hz. Whichis not possible value of t (in sec) :-

(1) 9.8 × 10–4 (2) 10–3

(3) 2.94 × 10–3 (4) 4.9 × 10–3

29. fdlh vizxkeh rjax ds fy, fuEu dFku fn, x, gSa(a) izR; sd d.k dk vk;ke fLFkj gksxk ijUrq ;g fudVre d.k

ds vk;ke ls vyx gksxkA

(b) lHkh d.k mudh ek/; fLFkfr;ksa dks ,d gh le; ij ikj

djrs gSaA

(c) lHkh d.k leku vk;ke ls nksyu dj jgs gSaA

(d) fdlh Hkh ry ls ÅtkZ dk dqy LFkkukUrj.k ugha gks jgk

gSA

(e) dqN d.k ,sls gSa tks lnSo fojkekoLFkk esa jgrs gSaA

lgh fodYi pqfu,(1) a, b, c, d, e(2) a, c, d, e(3) b, c, d, e(4) a, b, e

30. le; ds nks {k.kksa (mPpre] ek/;) ij fdlh ruh gqbZ jLlhesa cuh vizxkeh rjaxksa ds izfr:i dks fp= esa n'kkZ;k x;k gSA

vizxkeh rjaxksa dks cukus ds fy, vè;kjksfir nksuksa rjaxksa dks

osx 360 ms–1 rFkk mudh vko`fÙk;k¡ 256 Hz gSA fuEu esa

ls t (sec esa) dk dkSulk eku laHko ugha gS

(1) 9.8 × 10–4 (2) 10–3

(3) 2.94 × 10–3 (4) 4.9 × 10–3

Kota/01CE21401912/28

Target : JEE (Main) 2015/06-01-2015


PART B - CHEMISTRY

31. Adsorption is multilayer in case of

(1) Physical adsorption

(2) Chemical adsorption

(3) Both (1) and (2)

(4) None of these

32. Plot of logxm

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 per gram ofadsorbent be (log 5 = 0.699)

(1) 1 gm

(2) 6.99 gm

(3) 3 gm

(4) 5 gm

33. A schematic plot of ln Keq verses inverse oftemperature for a reaction is shown below. Thereaction must be

6.0

2.01.5 × 10 1/T (K )–3 –1 2.0 × 10–3

ln Keq

(1) endothermic(2) exothermic(3) highly spontaneous at ordinary temperature(4) one with negligible enthalpy change

31. fuEu eas ls fdlesa vf/k'kks"k.k cgqvkf.od gksrk gS(1) HkkSfrd vf/k'kks"k.k(2) jklk;fud vf/k'kks"k.k(3) (1) o (2) nksuksa(4) dksbZ ugha

32. logxm

æ öç ÷è ø ,oa log P ds e/; xzkQ ,d ljy js[kk gSA ftldh

<+ky 45° dks.k cukrh gSA ;fn ÝsaMfyd izkWpy (K) dk eku10 gks ,oa tc nkc 0.5 atm gS rks vf/k'kks"kd ds izfr xzkeij vf/k'kksf"kr] foys; vf/k'kks"; dh ek=k D;k gksxh\ (log5 = 0.699)

(1) 1 xzke(2) 6.99 xzke(3) 3 xzke(4) 5 xzke

33. vfHkfØ;k ds fy, ln Keq rFkk rki ds O;qRØe dse/; fuEu xzkQ fn;k x;k gS rks vfHkfØ;k gksxh

6.0

2.01.5 × 10 1/T (K )–3 –1 2.0 × 10–3

ln Keq

(1) Å"ek'kks"kh

(2) Å"ek{ksih

(3) lkekU; rki ij mPp Lor%

(4) ,sUFksYih esa ifjorZu ux.;


13/28Kota/01CE214019


34. The half-life of a first order reaction X ® Y, is 100minutes. The concentration of X would be reducedto 10% of the initial concentration in(1) 100 minutes (2) 332 minutes(3) 900 minutes (4) 700 minutes

35. For a zero order reaction, the plot of conc.(a–x) Vs time is linear with(1) + ive slope and zero intercept(2) – ive slope and zero intercept(3) + ive slope and non zero intercept(4) – ive slope and non-zero intercept

36. A study of chemical kinetics of the reactionA + B ¾¾® Products, gave the following data at25°C.

Exp. No. [A] [B] Rate1. 1.0 0.15 4.2 × 10–6

2. 2.0 0.15 8.4 × 10–6

3. 1.0 0.20 5.6 × 10–6

Find out rate law(1) r = K[A](2) r = K[B](3) r = K[A][B](4) r = K[A][B]2

37. 8O18 has a half life of 4650 years. 200 mg of a

sample of 8O18 is reduced to 25 mg in

(1) 3050 years (2) 9300 years(3) 13950 years (4) 4650 years

38. 99% of first order was completed in 64 minutes.When 99.9% of the reaction will complete(1) 50 min (2) 48 min(3) 32 min (4) 96 min

34. izFke dk sfV vfHkfØ;k X ® Y dk v¼Z vk;qdky100 feuV gSA X dh lkUærk] fdrus le; i'pkr~ viuhizkjfEHkd lkUærk dh 10% 'ks"k jg tk;sxh(1) 100 feuV (2) 332 feuV(3) 900 feuV (4) 700 feuV

35. 'kwU; dksfV dh vfHkfØ;k ds fy, lkUærk (a–x) rFkk le;ds chp fuEu ds lkFk jsf[kd (linear) gksrk gS(1) + ive izo.krk ,oa 'kwU; vUr%[k.M(2) – ive izo.krk ,oa 'kwU; vUr%[k.M(3) + ive izo.krk ,oa v'kwU; vUr%[k.M(4) – ive izo.krk ,oa v'kwU; vUr%[k.M

36. A + B ¾¾® mRikn, vfHkfØ;k dh cy xfrdh dkv/;;u 25°C ij fuEu vk¡dM+s nsrk gS

Exp. No. [A] [B] Rate1. 1.0 0.15 4.2 × 10–6

2. 2.0 0.15 8.4 × 10–6

3. 1.0 0.20 5.6 × 10–6

osx fu;e Kkr djks :-(1) r = K[A]

(2) r = K[B]

(3) r = K[A][B](4) r = K[A][B]2

37. 8O18 dh v¼Zvk;q 4650 o"kZ gSA 8O

18 ds fdlh uewus ds200 mg ls ?kVdj 25 mg gksus esa le; yxsxk(1) 3050 o"kZ (2) 9300 o"kZ(3) 13950 o"kZ (4) 4650 o"kZ

38. izFke dksfV vfHkfØ;k 64 feuV esa 99% iw.kZ gksrh gSA blvfHkfØ;k ds 99.9% iw.kZ gksus esa fdruk le; yxsxk(1) 50 feuV (2) 48 feuV

(3) 32 feuV (4) 96 feuV

Kota/01CE21401914/28

Target : JEE (Main) 2015/06-01-2015


39. In an exothermic reaction A ® B, evolved heatis 280 kJ mol–1 and activation energy is200kJ mole–1. The activation energy of the reversereaction B ® A is(1) 80 kJ mole–1

(2) 480 kJ mole–1

(3) 200 kJ mole–1

(4) None of these40. For the reaction, 2SO2 + O2 ¾¾® 2SO3, the rate

of disappearance of O2 is 2.0 × 10–4 mol L–1s–1.The rate of appearance of SO3 is(1) 2.0 × 10–4 mol L–1s–1

(2) 4.0 × 10–4 mol L–1s–1

(3) 1.0 × 10–4 mol L–1s–1

(4) 6.0 × 10–4 mol L–1s–1

Paragraph : (Q. NO. 41)Carbon mono-oxide is neutral oxide of carbon butit act as very strong field ligand and associated withmetal to form complex compounds, these complexcompounds are very useful compounds and usesfor metallergical process.

41. Water gas is a mixture of(1) H2 + CO2

(2) CO + H2

(3) H2O + CO2

(4) CO + CO2

42. The correct increasing order of polarising power is(1) Ca2+ < Mg2+ < Be2+ < K+

(2) Mg2+ < Be2+ < K+ < Ca2+

(3) Be2+ < K+ < Ca2+ < Mg2+

(4) K+ < Ca2+ < Mg2+ < Be2+

39. Å"ek{k sih vfHkfØ;k A ® B es a fu"dkflr Å"ek280 kJ mol–1 gS ,oa lfØ;.k ÅtkZ 200 kJ mole–1 gS rks

i'p vfHkfØ;k B ® A ds fy, lfØ;.k ÅtkZ gksxh

(1) 80 kJ mole–1

(2) 480 kJ mole–1

(3) 200 kJ mole–1

(4) buesa ls dksbZ ugha40. ,d vfHkfØ;k 2SO2 + O2 ¾¾® 2SO3, ds fy,] O2 ds

foyksiu dh nj 2.0 × 10–4 mol L–1s–1 gS] rks SO3 dsfuekZ.k gksus dh nj gksxh(1) 2.0 × 10–4 mol L–1s–1

(2) 4.0 × 10–4 mol L–1s–1

(3) 1.0 × 10–4 mol L–1s–1

(4) 6.0 × 10–4 mol L–1s–1

x|ka'k : (iz- la- 41)dkcZu eksuks vkWDlkbM ,d mnklhu vkWDlkbM gS dkcZu dkysfdu ;g izcy {ks= ligand ds :i esa dk;Z djrk gS vkSj/kkrqvks ls caf/kr gksdj ladqy ;kSfxd cukrk gS] ;g ladqy;kSfxd vR;Ur mi;ksxh ;kSfxd gksrs gSa ,oa /kkrqdeZ esa iz;qDrgksrs gSA

41. ty xSl feJ.k gS(1) H2 + CO2

(2) CO + H2

(3) H2O + CO2

(4) CO + CO2

42. /kzqo.k {kerk dk lgh c<+rk gqvk Øe gS(1) Ca2+ < Mg2+ < Be2+ < K+

(2) Mg2+ < Be2+ < K+ < Ca2+

(3) Be2+ < K+ < Ca2+ < Mg2+

(4) K+ < Ca2+ < Mg2+ < Be2+


15/28Kota/01CE214019


43. Arrange Ce3+, La3+, Pm3+, and Yb+3 in increasingorder of size(1) Yb+3 < Pm3+ < Ce3+ < La3+

(2) Ce+3 < Yb3+ < Pm3+ < La3+

(3) Yb+3 < Pm3+ < La3+ < Ce3+

(4) Pm+3 < La3+ < Ce3+ < Yb3+

44. In the case of alkali metals, the covalent characterdecreases in the order(1) MI > MBr > MCl > MF(2) MCl > MI > MBr > MF(3) MF > MCl > MBr > MI(4) MF > MCl > MI > MBr

45. Find the species having the magnetic moment valueof 5.92 BM(1) Fe3+ (2) Mn2+

(3) Cr+1 (4) All of these46. Lattice energy (numerical value) of chloride of

alkali metals is in order(1) LiCl > NaCl > KCl > RbCl > CsCl(2) LiCl < NaCl < KCl < RbCl < CsCl(3) NaCl < KCl < LiCl < RbCl < CsCl(4) NaCl < KCl < RbCl < CsCl < LiCl

47. Which of the following metal carbonate isdecomposed on heating(1) Na2CO3

(2) Li2CO3

(3) Cs2CO3

(4) K2CO3

48. In Clark's method chemical used is(1) NaOH (2) Na2CO3

(3) CaCO3 (4) Ca(OH)2

43. Ce3+, La3+, Pm3+, vkSj Yb+3 dks vkdkj ds c<+rs gq, Øeesa O;ofLFkr djks\(1) Yb+3 < Pm3+ < Ce3+ < La3+

(2) Ce+3 < Yb3+ < Pm3+ < La3+

(3) Yb+3 < Pm3+ < La3+ < Ce3+

(4) Pm+3 < La3+ < Ce3+ < Yb3+

44. {kkjh; èkkrq ds gSykbZM esa lgla;ksth xq.k dk ?kVrk Øe dksulkgS(1) MI > MBr > MCl > MF(2) MCl > MI > MBr > MF(3) MF > MCl > MBr > MI(4) MF > MCl > MI > MBr

45. Lih'kht crkb;s ftldk pqEcdh; vk?kq.k Z 5.92 BMgSA(1) Fe3+ (2) Mn2+

(3) Cr+1 (4) mijksDr lHkh46. {k kj / k kr q ds g SykbM dh tkyd Åtk Z dk Øe

gksxk(1) LiCl > NaCl > KCl > RbCl > CsCl(2) LiCl < NaCl < KCl < RbCl < CsCl(3) NaCl < KCl < LiCl < RbCl < CsCl(4) NaCl < KCl < RbCl < CsCl < LiCl

47. fuEu esa ls dkSulk /kkrq dkcksZusV Å"ek nsus ij fo?kfVr gksrkgSA(1) Na2CO3

(2) Li2CO3

(3) Cs2CO3

(4) K2CO3

48. DykdZ fof/k esa iz;qDr jlk;u gS(1) NaOH (2) Na2CO3

(3) CaCO3 (4) Ca(OH)2

Kota/01CE21401916/28

Target : JEE (Main) 2015/06-01-2015


49. Which of the following gives H2O2 on hydrolysis(1) Na2O (2) Na2O2

(3) KO2 (4) 2 & 3 both50. Which element exist in free state mainly

(1) Hg (2) Fe(3) Au (4) Zn

51. Which of the following is maximum stable freeradical

(1) CH2 = CH – 2CH

·(2) CH2 = CH

·

(3) (4)

52. Among the following the weakest base is(1) C6H5CH2NH2

(2) C6H5CH2NHCH3

(3) NO2CH2NH2

(4) CH3NHCHO53. Which of the following is the most stable alkene

(1) R2C = CR2 (2) RCH = CHR(3) R2C = CHR (4) CH2 = CH2

54. In the following compounds

The order of basicity is :-(1) IV > I > III > II(2) III > I > IV > II(3) II > I >III > IV(4) I > III > II > IV

49. fuEu esa dkSu ty vi?kVu ij H2O2 nsxk(1) Na2O (2) Na2O2

(3) KO2 (4) 2 o 3 nksuksa50. dkSulk rRo eq[;r% eqDr voLFkk esa ik;k tkrk gS

(1) Hg (2) Fe(3) Au (4) Zn

51. fuEu esa ls dkSulk vf/kdre LFkk;h eqä ewyd gS

(1) CH2 = CH – 2CH·

(2) CH2 = CH·

(3) (4)

52. fuEu esa ls nqcZyre {kkj gksrk gS(1)C6H5CH2NH2

(2) C6H5CH2NHCH3

(3)NO2CH2NH2

(4) CH3NHCHO53. fuEu esa lokZf/kd LFkk;h ,Ydhu gS

(1) R2C = CR2 (2) RCH = CHR(3) R2C = CHR (4) CH2 = CH2

54. fuEu ;kSfxdksa dh {kkjdrk dk Øe gS

(1) IV > I > III > II

(2) III > I > IV > II

(3) II > I >III > IV

(4) I > III > II > IV


17/28Kota/01CE214019


55. Decreasing order of stability of

(a) CH3 – C HÅ

– CH3

(b) CH3 – C HÅ

–OCH3

(c) CH3 – C HÅ

– CO – CH3

(1) a > b > c

(2) b > c > a

(3) c > a > b

(4) b > a > c

56. Which of the following is better nucleophile

(1) H2O

(2) HO

(3) CH3O

(4) CH3COO

57. Which of the following is correct according to theIUPAC nomenclature

(1) 1, 2, 3-trihydroxy propane CH –OH2

CH –OH2

CH –OH2

(2) 2-Ethyl–3–methyl butane ,CH –CH–CH–CH3 3

Et Me

(3) 2–Carboxypropanoic acid,CH –CH–COOH3

COOH

(4) 6–Methylbicyclo [3.2.1]oct–2–ene Me

55. LFkkf;Ro dk ?kVrk Øe gS

(a) CH3 – C HÅ

– CH3

(b) CH3 – C HÅ

–OCH3

(c) CH3 – C HÅ

– CO – CH3

(1) a > b > c

(2) b > c > a

(3) c > a > b

(4) b > a > c

56. fuEu es ls dkSulk vPNk ukfHkd Lusgh gksrk gS(1) H2O

(2) HO

(3) CH3O

(4) CH3COO

57. IUPAC ukedj.k ds vuqlkj fuEufyf[kr esa dkSu lghgS

(1) 1, 2, 3-VªkbZ gkbMªkDlh izksisu CH –OH2

CH –OH2

CH –OH2

(2) 2-,fFky–3–esfFky C;wVsu ,CH –CH–CH–CH3 3

Et Me

(3)2–dkcksZDlh izksisuksbd vEy, CH –CH–COOH3

COOH

(4) 6–esfFky ckblkbDyks [3.2.1] vkDV–2–bZu Me

Kota/01CE21401918/28

Target : JEE (Main) 2015/06-01-2015


58. The correct IUPAC name of given compound

CH CH –O–C–CH –C–OH 3 2 2

O

O

(1) Ethyl carboxylate ethanoic acid(2) Ethoxy carbonyl ethanoic acid

(3) Ethyl–2–oxo propanoic acid

(4) 3–oxo–3–ethoxy propanoic acid

59. IUPAC name of the structure

CH3CH2 – NH

CHO

(1) N–formyl amino ethane

(2) N–ethyl formylamine

(3) Ethyl amino methanal

(4) N–ethyl methanamide

60. Correct name of

COOC H2 5

C=OH

is

(1) 2-oxocyclopentanecarboxylate

(2) 2-Formylcyclopentanecarboxylate

(3) Ethyl-2-formylcyclopentanecarboxylate

(4) Ethyl-2-oxocyclopentanecarboxylate

58. fn, x, ;kSfxd dk lgh IUPAC uke gS

CH CH –O–C–CH –C–OH 3 2 2

O

O

(1) ,fFky dkcksZfDlysV ,Fksuksbd vEy

(2) ,FkkDlh dkcksZfuy ,Fksuksbd vEy

(3) ,fFky–2–vkDlks izksisuksbZd vEy

(4) 3–vkDlks–3–,FkkDlh izksisuksbZd vEy

59. fuEu dk IUPAC uke gS %&

CH3CH2 – NH

CHO

(1) N–QkWfeZy ,ehuks , sFksu

(2) N–,fFky QkWfeZy ,ehu

(3) ,fFky ,ehuks esFksusy

(4) N–,fFky esFksu ,ekbM

60.

COOC H2 5

C=OH

dk lgh uke gS

(1) 2-vksDlkslkbDyksisUVsudkcksZfDlysV

(2) 2-QkfeZylkbDyksisUVsudkcksZfDlysV

(3) ,fFky-2-QkfeZylkbDyksisUVsudkcksZfDlysV

(4) ,fFky-2-vksDlkslkbDyksisUVsudkcksZfDlysV


19/28Kota/01CE214019


PART C - MATHEMATICS

61. If x 3x

x x

a eb cò dx =

x 3x

x x

1 a eP b c

æ öç ÷è ø

+ K ; then P = ??

(1) 3 loga + log (b/c)

(2) loga + 3 – log bc

(3) log (e3abc)

(4) logb + logc – loga – 3

62. (sinx cosx cos2x cos4x cos8x) dxò equal

(1) 1

cos16x C128-

+

(2) 1

cos16x C256

+

(3) 1

sin16x C256

- +

(4) 1

cos16x C256

- +

63. x 1 sin xe

1 cosx-æ ö

ç ÷è ø-ò dx is equal to

(1) ex tanx2

+ C (2) –ex tanx2

+ C

(3) –ex cotx2

+ C (4) ex cotx2

+ C

61. ;fn x 3x

x x

a eb cò dx =

x 3x

x x

1 a eP b c

æ öç ÷è ø

+ K ; rks P = ??

(1) 3 loga + log (b/c)

(2) loga + 3 – log bc

(3) log (e3abc)

(4) logb + logc – loga – 3

62. (sinx cosx cos2x cos4x cos8x) dxò cjkcj gS

(1) 1

cos16x C128-

+

(2) 1

cos16x C256

+

(3) 1

sin16x C256

- +

(4) 1

cos16x C256

- +

63. x 1 sin xe

1 cosx-æ ö

ç ÷è ø-ò dx cjkcj gS

(1) ex tanx2

+ C (2) –ex tanx2

+ C

(3) –ex cotx2

+ C (4) ex cotx2

+ C

Kota/01CE21401920/28

Target : JEE (Main) 2015/06-01-2025


64.2

2 2

x(x 1)(x 4)+ +ò dx is equal to

(1) –tan–1 x + 13

tan–1 x2

+ C

(2) –13

tan–1x + 23

tan–1 x2

+ C

(3) tan–1x + 23

tan–1 x2

+ C

(4) 13

tan–1x – 23

tan–1x2

+ C

65.2

2x2

1

1 1e

x 2xæ ö-ç ÷è øò dx is equal to

(1) 2e

4(e2 + 2) (2) e2(e2 – 2)

(3) 2 2e (e 2)

2-

(4) 2 2e (e 2)

4-

66.e 1

21

tan x nxx (1 x )

-æ ö+ç ÷è ø+òl dx is equal to

(1) 1e

tan–1 e (2) tan–1e

(3) e tan–11e

æ öç ÷è ø (4) tan–1(lne)

64.2

2 2

x(x 1)(x 4)+ +ò dx cjkcj gS

(1) –tan–1 x + 13

tan–1 x2

+ C

(2) –13

tan–1x + 23

tan–1 x2

+ C

(3) tan–1x + 23

tan–1 x2

+ C

(4) 13

tan–1x – 23

tan–1x2

+ C

65.2

2x2

1

1 1e

x 2xæ ö-ç ÷è øò dx cjkcj gS

(1) 2e

4(e2 + 2) (2) e2(e2 – 2)

(3) 2 2e (e 2)

2-

(4) 2 2e (e 2)

4-

66.e 1

21

tan x nxx (1 x )

-æ ö+ç ÷è ø+òl

dx cjkcj gS

(1) 1e

tan–1 e (2) tan–1e

(3) e tan–11e

æ öç ÷è ø (4) tan–1(lne)


21/28Kota/01CE214019


67.4

e

4

log (sin x cosx) dx

p

p-

+ò is equal to

(1) –4pln2 (2)

4pln2

(3) 8pln2 (4) –

8pln2

68. Area of one region included between the sine andcosine curves is given by

(1) 2 – 1 (2) 2

(3) 2 2 (4) 4 2

69. Area between the curves y = x3 and y = x is

(1) 53

(2) 54

(3) 5

12(4) None

70. Area common to the curves 5x2 – y = 0 and2x2 – y + 9 = 0 is equal to

(1) 12 3 (2) 6 3

(3) 36 (4) 1871. Solution of the differential equation

2 32 3dy (xy) dy (xy) dyx 1 xy ....

dx 2! dx 3! dxæ ö æ ö= + + + +ç ÷ ç ÷è ø è ø

is

(1) y = logex + C

(2) y = (logex)2 + C

(3) y = ± 2e(log x) 2C+

(4) xy = xy + K

67.4

e

4

log (sin x cosx) dx

p

p-

+ò cjkcj gS

(1) –4pln2 (2)

4pln2

(3) 8pln2 (4) –

8pln2

68. sine vkSj cosine oØksa ds chp ifjc¼ ,d {ks= (oneregion) dk {ks=Qy gksxk

(1) 2 – 1 (2) 2

(3) 2 2 (4) 4 2

69. oØksa y = x3 rFkk y = x ds e/; dk {ks=Qy gksxk

(1) 53

(2) 54

(3) 5

12 (4) dksbZ ugha

70. oØksa 5x2 – y = 0 rFkk 2x2 – y + 9 = 0 ds e/; mHk;fu"B{ks=Qy gksxk

(1) 12 3 (2) 6 3

(3) 36 (4) 1871. vody lehdj.k

2 32 3dy (xy) dy (xy) dyx 1 xy ....

dx 2! dx 3! dxæ ö æ ö= + + + +ç ÷ ç ÷è ø è ø

dk gy gS(1) y = logex + C(2) y = (logex)2 + C

(3) y = ± 2e(log x) 2C+

(4) xy = xy + K

Kota/01CE21401922/28

Target : JEE (Main) 2015/06-01-2025


72. Solve the differential equation

( ) ( )2 2 2 21 x x y dx x y 1 ydy 0+ + + + - =

(1) 2

2 2 3 / 2y 1x (x y ) C

2 3+ + + =

(2) 2

2 2 3 / 2y 1x (x y ) C

2 3- + + =

(3) 2

2 2 3/ 2y 1x (x y ) C

2 3- - + =

(4) None of these

73. The solution of dy ax bdx cy d

æ ö+= ç ÷+è ø

represents a

parabola if(1) a = 0, c = 0

(2) a = 1, c = 2

(3) a = 0, c ¹ 0

(4) a = 1, c = 1

74. A solution of y = 2x. dydx

æ öç ÷è ø

+ x2

4dydx

æ öç ÷è ø

is

(1) y = 2C1/2 x1/4 + C

(2) y = 2 C x2 + C2

(3) y = 2 C (x + 1)

(4) y = 2 Cx + C2

72. vody lehdj.k dk gy gksxk

( ) ( )2 2 2 21 x x y dx x y 1 ydy 0+ + + + - =

(1) 2

2 2 3 / 2y 1x (x y ) C

2 3+ + + =

(2) 2

2 2 3/ 2y 1x (x y ) C

2 3- + + =

(3) 2

2 2 3 / 2y 1x (x y ) C

2 3- - + =

(4) buesa ls dksbZ ugha

73. vody lehdj.k dy ax bdx cy d

æ ö+= ç ÷+è ø

dk gy ijoy;

fu:fir djsxk ;fn(1) a = 0, c = 0

(2) a = 1, c = 2

(3) a = 0, c ¹ 0

(4) a = 1, c = 1

74. vody lehdj.k y = 2x.dydx

æ öç ÷è ø

+ x2

4dydx

æ öç ÷è ø

dk gy gS

(1) y = 2C1/2 x1/4 + C

(2) y = 2 C x2 + C2

(3) y = 2 C (x + 1)

(4) y = 2 Cx + C2


23/28Kota/01CE214019


75.f - f

- f - fò 2

(3sin 2)cos5 cos 4sin

df is equal to

(1) 3log(2 – sinf) 4

(sin 2)f - + C

(2) 3log(sinf – 2) + 4

(2 sin )- f + C

(3) log(2 – sinf) + 4

(2 sin )- f + C

(4) 3log(2 – sinf) + 4

(2 sin )- f + C

76. é ù+ë ûò9

0

x 2 dx is equal to (where [·] is G.I.F.)

(1) 31 (2) 23

(3) 22 (4) 27

77.4

x x x x

4

(2 2 )(3 3 ) dx- -

-

+ -ò is equal to

(1) 10 (2) 16(ln2)(ln3)

(3) 16ln(2/3) (4) 0

75.f - f

- f - fò 2

(3sin 2)cos5 cos 4sin

df cjkcj gS

(1) 3log(2 – sinf) 4

(sin 2)f - + C

(2) 3log(sinf – 2) + 4

(2 sin )- f + C

(3) log(2 – sinf) + 4

(2 sin )- f + C

(4) 3log(2 – sinf) + 4

(2 sin )- f + C

76. é ù+ë ûò9

0

x 2 dx cjkcj gS (tgk¡ [×] ,d egÙke iwa.kkZd

Qyu gS)

(1) 31 (2) 23 (3) 22 (4) 27

77.4

x x x x

4

(2 2 )(3 3 ) dx- -

-

+ -ò cjkcj gS

(1) 10 (2) 16(ln2)(ln3)

(3) 16ln(2/3) (4) 0

Kota/01CE21401924/28

Target : JEE (Main) 2015/06-01-2025


78. Solution of differential equation

(x + 2y3)dydx

– y = 0 is

(1) y(1 – xy) = kx

(2) y3 – x = ky

(3) x(1 – xy) = ky

(4) x(1 + xy) = ky

79. The area of the region {(x, y) : x2 + y2 £ 1 £ x + y} is

(1)p2

2(2)

12

(3) 2

3p

(4) 1

4 2pæ ö-ç ÷

è ø

80. f(x) =

2

2

2cos 2x sin2x –sin x

sin2x 2sin x cosx

sinx –cosx 0, the value of

/ 2'

0

f (x)p

ò dx is equal to

(1) –2 (2) –1 (3) 2 (4) 0

81. If I1 = sin

21

x1 x

q

+ò dx and I2 = cosec

21

dxx(x 1)

q

+ò ; then

the value of 1 2

21 1 2

I I 22

2 21 2

I I I

e I 1

1 I I 1

+ -+ -

is

(1) 1 (2) –112

(3) 9 (4) 0

78. vody lehdj.k dk gy gksxk

(x + 2y3)dydx

– y = 0

(1) y(1 – xy) = kx

(2) y3 – x = ky

(3) x(1 – xy) = ky

(4) x(1 + xy) = ky

79. {ks= {(x, y) : x2 + y2 £ 1 £ x + y} dk {ks=Qy gSA

(1)p2

2(2)

12

(3) 2

3p

(4) 1

4 2pæ ö-ç ÷

è ø

80. f(x) =

2

2

2cos 2x sin2x –sin x

sin2x 2sin x cosx

sinx –cosx 0,rks

/ 2'

0

f (x)p

ò dx

cjkcj gS

(1) –2 (2) –1 (3) 2 (4) 0

81. ;fn I1 = sin

21

x1 x

q

+ò dx rFkk I2 = cosec

21

dxx(x 1)

q

+ò ; rks

1 2

21 1 2

I I 22

2 21 2

I I I

e I 1

1 I I 1

+ -+ -

dk eku gksxk

(1) 1 (2) –112

(3) 9 (4) 0


25/28Kota/01CE214019


82. Let I1 = 1 x

0

e(1 x)+ò dx and I2 = 3

1 2

x 30

xdx

e (2 x )-ò value

of 1

2

II is equal to

(1) 3e

(2) e3

(3) 3e (4) 13e

83.2

0

x x x .... dxé ù

+ + + ¥ê úë ûò is equal to

(where [·] is G.I.F.)

(1) 196

(2) 4

(3) 3 (4) 32

84. The solution of differential equation(x2 – xy)dy = (xy + y2)dx is(1) xy = ce–y/x (2) xy = ce–x/y

(3) yx2 = ce1/x (4) None of these85. The degree of the differential equation

2/ 33 2

3 2

d y d y dy4 3 5 0

dx dx dxæ ö

+ - + =ç ÷è ø is

(1) 1 (2) 2(3) 3 (4) none of these

82. I1 = 1 x

0

e(1 x)+ò dx rFkk I2 = 3

1 2

x 30

xdx

e (2 x )-ò

1

2

II dk eku cjkcj gS

(1) 3e

(2) e3

(3) 3e (4) 13e

83.2

0

x x x .... dxé ù

+ + + ¥ê úë ûò cjkcj gS

(tgk¡ [·] G.I.F. gS)

(1) 196

(2) 4

(3) 3 (4) 32

84. vody lehdj.k(x2 – xy)dy = (xy + y2)dx dk gy gS(1) xy = ce–y/x (2) xy = ce–x/y

(3) yx2 = ce1/x (4) buesa ls dksbZ ugha85. vody lehdj.k dh ?kkr gS

2/ 33 2

3 2

d y d y dy4 3 5 0

dx dx dxæ ö

+ - + =ç ÷è ø(1) 1 (2) 2(3) 3 (4) buesa ls dksbZ ugha

Kota/01CE21401926/28

Target : JEE (Main) 2015/06-01-2025


86. The differential equation whose solution is

(x – h)2 + (y – k)2 = a2 is (a is a constant)

(1)

32 22

2

dy d y1 a

dx dx

é ùæ ö+ =ê ç ÷ úè øë û

(2) 3 22 2

2

2

dy d y1 a

dx dx

é ù æ öæ ö+ =ê ç ÷ ú ç ÷è ø è øë û

(3)

3 222

2

dy d y1 a

dx dxé ù æ öæ ö+ =ç ÷ ç ÷ê úè ø è øë û

(4) None of these

87. The area of the smaller portion between curves x2

+ y2 = 8 and y2 = 2x is

(1) p + 23

(2) 2p + 23

(3) 2p + 43

(4) p + 43

88.x

3

(x 3)edx

(x 1)--ò is equal to

(1) xe

c(x 1)

+-

(2) x

2

ec

(x 1)+

-

(3) x

2

ec

(x 1)-

+- (4)

x

3

2ec

(x 1)+

-

86. og vody lehdj.k ftldk gy

(x – h)2 + (y – k)2 = a2 gS] gksxh (a is a constant)

(1)

32 22

2

dy d y1 a

dx dx

é ùæ ö+ =ê ç ÷ úè øë û

(2) 3 22 2

2

2

dy d y1 a

dx dx

é ù æ öæ ö+ =ê ç ÷ ú ç ÷è ø è øë û

(3)

3 222

2

dy d y1 a

dx dxé ù æ öæ ö+ =ç ÷ ç ÷ê úè ø è øë û

(4) buesa ls dksbZ ugha

87. oØksa x2 + y2 = 8 rFkk y2 = 2x ds chp NksVs Hkkx dk {ks=Qygksxk

(1) p + 23

(2) 2p + 23

(3) 2p + 43

(4) p + 43

88.x

3

(x 3)edx

(x 1)--ò cjkcj gS

(1) xe

c(x 1)

+-

(2) x

2

ec

(x 1)+

-

(3) x

2

ec

(x 1)-

+- (4)

x

3

2ec

(x 1)+

-


27/28Kota/01CE214019


89. Let f(x) be a non-negative continuous function suchthat the area bounded by the curve y= f(x), x-axis

and the ordinates x = 4p

and x = b > 4p

is

sin cos 24pæ öb b+ b+ bç ÷

è ø. Then f

2pæ ö

ç ÷è ø

is

(1) 2 14pæ ö+ -ç ÷

è ø(2) 2 1

4pæ ö- +ç ÷

è ø

(3) 1 24pæ ö- -ç ÷

è ø(4) 1 2

4pæ ö- +ç ÷

è ø90. Area of smaller region bounded by x2 + y2 = 4 and

line y = x above x-axis is

(1) 2p (2) p (3) 2p

(4) 4p

89. ekuk ,d v½.kkRed Qyu f(x) bl izdkj gS fd oØ

y= f(x), x-v{k rFkk dksfV;ksa x = 4p

rFkk x = b > 4p

}kjk

ifjc¼ {ks=Qy sin cos 24pæ öb b+ b+ bç ÷

è ø gS] rc f

2pæ ö

ç ÷è ø

dk eku gS

(1) 2 14pæ ö+ -ç ÷

è ø(2) 2 1

4pæ ö- +ç ÷

è ø

(3) 1 24pæ ö- -ç ÷

è ø(4) 1 2

4pæ ö- +ç ÷

è ø90. x2 + y2 = 4 o js[kk y = x ls f?kjs NksVs {ks= dk {ks=Qy

tks x-v{k ds Åij gS

(1) 2p (2) p (3) 2p

(4) 4p

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

Kota/01CE21401928/28

Target : JEE (Main) 2015/06-01-2025


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

CLASSROOM CONTACT PROGRAMME

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