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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 : 07 - 03 - 2015
0 1 C T 3 1 4 0 6 3
PHASE – ELS, 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 ofPhysics, Chemistry and Mathematics having 30 questions in eachpart of equal weightage. Each question is allotted 4 (four) marks forcorrect response.
6. One Fourth mark will be deducted for indicated incorrect responseof each question. No deduction from the total score will be madeif no response is indicated for an item in the Answer Sheet.
7. Use Blue/Black Ball Point Pen only for writting particulars/markingresponses on Side–1 and Side–2 of the Answer Sheet.Use of pencil is strictly prohibited.
8. No candidate is allowed to carry any textual material, printed or written,
bits of papers, pager, mobile phone any electronic device etc, 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;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
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
1/35Kota/01CT314063
SPACE FOR ROUGH WORK
PART A - PHYSICSBEWARE OF NEGATIVE MARKING
HAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS
1. The specific heat of alcohol is about half thatof water. Suppose you have identical massesof alcohol and water. The alcohol is initially attemperature TA. The water is initially at adifferent temperature TW. Now the two fluidsare mixed in the same container and allowedto come into thermal equilibrium, with no lossof heat to the surroundings. The finaltemperature of the mixture will be :-(1) Closer to TA than TW(2) Closer to TW than TA(3) Exactly halfway between TA and TW(4) Dependent on the volume of alcohol used
2. A pan balance has a container of water with anoverflow spout on the right-hand pan as shown.It is full of water right up to the overflow spout.A container on the left-hand pan is positionedto catch any water that overflows. The entireapparatus is adjusted so that it’s balanced.A brass weight on the end of a string is thenlowered into the water, but not allowed to reston the bottom of the container. What happensnext ?
1. ,Ydksgy dh fof'k"V Å"ek] ty dh rqyuk esa vk/khgksrh gSA vkids ikl ,Ydksgy rFkk ty ds cjkcj æO;ekugSA ,Ydksgy izkjEHk esa TA rkieku ij gS] tcfd tyizkjEHk esa TW rkieku ij gSA vc bu nksuksa æoksa dks ,dgh ik= esa feykdj rkih; lkE;koLFkk esa vkus fn;k tkrkgSA ;gk¡ ifjos'k esa dksbZ Å"ek âkl ugh gksrkA bl feJ.kdk vafre rkieku gksxk%&(1) TW dh rqyuk esa TA ds vf/kd lehi(2) TA dh rqyuk esa TW ds vf/kd lehi(3) TA rFkk TW ds Bhd e/; esa(4) ;g iz;qDr ,Ydksgy ds vk;ru ij fuHkZj djsxkA
2. ,d HkkSfrd rqyk ds nk¡;s iyM+s esa ty ls Hkjk ,d ik=
j[kk gSA ;g ty ls fp=kuqlkj VksaVh rd Hkjk gqvk gSA
ck¡;s iyM+s esa ,d vU; ik= j[kk gS tks ckgj fudyus
okys (over flow) ty dks xzg.k djrk gSA ;g iwjk
midj.k larqyu dh voLFkk esa j[kk gqvk gSAjLlh ls tqM+s
gq, ihry ds ,d Hkkj dks ty esa mrkjk tkrk gS ijUrq bls ik=
ds iSans ij ugha j[kk tkrk] rc %&
Kota/01CT3140632/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
(1) Water overflows and the right side of thebalance tips down.
(2) Water overflows and the left side of thebalance tips down.
(3) Water overflows but the balance remainsbalanced.
(4) Water overflows but which side of thebalance tips down depends on whether thebrass weight is partly or completelysubmerged.
3. In the given figure, two elastic rods A & B arerigidly joined to end supports. A small mass‘m’ is moving with velocity v between the rods.All collisions are assumed to be elastic & thesurface is given to be frictionless. The timeperiod of small mass ‘m’ will be :[A=area ofcross section, Y = Young’s modulus, L=lengthof each rod ; here, an elastic rod may be treated
as a spring of spring constant YAL ]
Y Y
(1) 2L mL2v AY
+ p (2) 2L 2mL2v AY
+ p
(3) 2L mLv AY
+ p (4) vL2
(1) ty ckgj fudy vk;sxk rFkk rqyk nka;h vksj >qdtk;sxh
(2) ty ckgj fudy vk;sxk rFkk rqyk cka;h vksj > qdtk;sxh
(3) ty ckgj fudy vk;sxk ijUrq rqyk larqyu dh voLFkkesa gh cuh jgsxhA
(4) ty ckgj fudy vk;sxk ijUrq rqyk fdl vksj > qdsxh;g bl ckr ij fuHkZj djrk gS fd ihry dk Hkkjvkaf'kd :i ls Mwck gS vFkok iw.kZr;kA
3. iznf'kZr fp= esa nks izR;kLFk NM+s A o B n`<+rk iwoZdnhokjks ls tqM+h gqbZ gSA ,d NksVk æO;eku ‘m’ bu NM+ks dse/; v osx ls xfr dj jgk gSA lHkh gksus okyh VDdjkas dksizR;kLFk o lrg dks fpduk ekuk x;k gSA bl NksVs æO;eku‘m’ dk vkorZdky gksxk [A = vuqizLFk dkV {ks=Qy,Y = ;ax izR;kLFkrk xq.kkad, L = izR;sd NM+ dh yEckbZ ;;gk¡ ,d izR;kLFk NM+ dks fLizax ds :i esa ekuk tk ldrk
gS] ftldk fLizax fu;rkad YAL gSA]
Y Y
(1) 2L mL2v AY
+ p (2) 2L 2mL2v AY
+ p
(3) 2L mLv AY
+ p (4) vL2
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
3/35Kota/01CT314063
SPACE FOR ROUGH WORK
4. Find the center of mass (x,y,z) of the followingstructure of four identical cubes if the lengthof each side of a cube is 1 unit.
(1) (1/2,1/2,1/2) (2)(1/3,1/3,1/3)(3) (3/4,3/4,3/4) (4)(1/2,3/4,1/2)
5. An electromagnetic wave of wavelengthl0 (in vacuum) passes from P towardsQ crossing three different media of refractiveindex m, 2m and 3m respectively as shown infigure. fP and fQ be the phase of the wave atpoints P and Q. Find the phase difference fQ –fP. [Take : m=1]
2.25 0l 3.5l0 3 0l
µ 2µ 3µ
P Q
(1) 0 (2) 4p
(3) 2p
(4) p
4. ;fn ?ku dh izR;sd Hkqtk dh yEckbZ 1 bdkbZ gks rks pkj,dtSls ?kuksa }kjk cuh fuEu lajpuk dk æO;eku dsUæ (x,y,z)gksxk
(1) (1/2,1/2,1/2) (2)(1/3,1/3,1/3)(3) (3/4,3/4,3/4) (4)(1/2,3/4,1/2)
5. fuokZr esa rjaxnS/;Z l0 okyh ,d fo|qr pqEcdh; rjax
fp=kuqlkj Øe'k% m, 2m rFkk 3m viorZukadksa okys rhu
fofHkUu ek/;eksa dks ikj djrs gq, fcUnq P ls fcUnq Q dh
rjQ xeu djrh gSA fcUnqvksa P rFkk Q ij rjax dh dyk
fP rFkk fQ gS] dykUrj fQ – fP Kkr dhft;sA [m=1 ysa]
2.25 0l 3.5l0 3 0l
µ 2µ 3µ
P Q
(1) 0 (2) 4p
(3) 2p
(4) p
Kota/01CT3140634/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
6. A disc is rotating with constant angular velocityabout an axis passing through centre C andperpendicular to the plane of disc. An insect ismoving over the disc along radial direction withconstant velocity with respect to the disc.Acceleration of the insect at the instant whenits distance from centre is r, will be :-
C I w
(1) rw2 towards the centre(2) rw2 away from the centre(3) Less than rw2 in magnitude(4) Greater than rw2 in magnitude
7. Two small bodies of mass of 2 kg each attachedto each other using a thread of length 10 cm,hang on a spring whose force constant is200 N/m, as shown in the figure. We burn thethread. What is the distance between the twobodies when the top body first arrives at itshighest position? (Take p2 = 10)(1) 60 cm(2) 70 cm (3) 80 cm(4) None of these
6. ,d pdrh fu;r dks.kh; osx ls blds ry ds yEcor~
rFkk dsUæ C ls xqtjus okyh v{k ds lkis{k ?kw.kZu dj jgh
gSA ,d dhM+k pdrh ds Åij pdrh ds lkis{k fu;r osx
ls f=T;h; fn'kk esa xfr dj jgk gSA tc ;g dsUæ ls r nwjh
ij gksxk rc bldk Roj.k gksxk :-
C I w
(1) rw2 ; dsUæ dh vksj
(2) rw2 ; dsUæ ls nwj
(3) ifjek.k esa rw2 ls de
(4) ifjek.k esa rw2 ls vf/kd
7. cy fu;rkad 200 N/m okyh fLizax ij izR;sd 2 kgnzO;eku ds nks NksVs fi.Mksa dks yVdk;k x;k gSA fi.M
10 cm yEch jLlh }kjk fp=kuqlkj ,d nwljs ls tqM+s gq,
gSaA ge jLlh dks tyk nsrs gSaA tc Åij okyk fi.M igyh
ckj bldh mPpre fLFkfr ij igqaprk gS rks nksuksa fi.Mksa ds
e/; nwjh fdruh gksxh\ (p2 = 10 ysa)(1) 60 cm(2) 70 cm (3) 80 cm
(4) buesa ls dksbZ ugha
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
5/35Kota/01CT314063
SPACE FOR ROUGH WORK
8. A steel wire is used to stretch a spring.An oscillating magnetic field drives the steelwire up and down. A standing wave with threeantinodes is created when the spring is stretchedby 4.0 cm. What stretch of the spring producesa standing wave with two antinodes with samefrequency?
steel wire Pull
Spring
(1) 4 cm (2) 12 cm (3) 9 cm (4) 6 cm9. The limbs of a U-tube glass are lowered into
vessels A and B, A containing water. Some airis pumped out through the top of the tube C.The liquids in the left hand limb A and the righthand limb B rise to heights of 10 cm and12 cm respectively. The density of liquid B is
12
BA
(water) (liquid)
C
(1) 0.75 g/cm3 (2) 0.83 g/cm3
(3) 1.2 g/cm3 (4) 0.25 g/cm3
8. ,d LVhy rkj dh lgk;rk ls fdlh fLizax dks [khapk tkrk
gSA bl rkj dks ,d nksyuh pqEcdh; {ks= }kjk Åij rFkk
uhps dh vksj xfr djkbZ tkrh gSA tc fLizax 4 cm foLrkfjr
gks tkrh gS rks rhu izLiUnksa okyh ,d vizxkeh rjax curh
gSA fLizax esa fdruk dqy foLrkj gks rkfd leku vkofr ij
nks izLiUnksa okyh vizxkeh rjax cus\
steel wire Pull
Spring
(1) 4 cm (2) 12 cm (3) 9 cm (4) 6 cm9. fdlh dkap ls cuh U- uyh dh Hkqtkvksa dks nks ik=ksa A o
B esa Mqck;k tkrk gSA A esa ikuh Hkjk gqvk gSA uyh C ds
Åijh Hkkx ls dqN ok;q ckgj fudky nh tkrh gSA cka;h
Hkqtk A o nka;h Hkqtk B esa nzoksa dk Lrj Øe'k% 10cm o12 cm gSA nzo B dk ?kuRo gksxk
12
BA
(water) (liquid)
C
(1) 0.75 g/cm3 (2) 0.83 g/cm3
(3) 1.2 g/cm3 (4) 0.25 g/cm3
Kota/01CT3140636/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
10. Two masses m1 and m2 are connected by astring of length l. They are held in a horizontalplane at a height H above two heavy plates Aand B made of different material placed on thefloor. Initially distance between two masses isa < l. When the masses are released undergravity they make collision with A and B withcoefficient of restitut ion 0.8 and 0.4respectively. The time after the collision whenthe string becomes tight is :-(Assume H>>l)
(1) 2 25 a
2 2gH-l
e = 0.8
A B
e = 0.4
m1 m2
a < l
H
(2) 2gH
(3) 2 23 a
2 2gH-l
(4) None of these
11. Two thin parallel slits are made in an opaquescreen. When a monochromatic beam of lightpasses through them at normal incidence, thefirst bright fringe in the transmitted light occursat ±45° with the original direction of the lightbeam on a distant screen when the apparatus isin air. When the apparatus is immersed in aliquid, the same bright fringe now occurs at± 30°. The index of refraction of the liquid is :-
(1) 2 (2) 3 (3) 43 (4)
32
10. nks æO;eku m1 o m2 ,d l yEckbZ dh jLlh }kjk tqM+sgq, gSA bUgsa Q'kZ ij j[kh vyx&vyx inkFkks± ls cuh nksHkkjh IysVksa A o B ds Åij H Å¡pkbZ ij {kSfrt ry esajksd dj j[kk x;k gSA izkjEHk esa nksuksa æO;ekuksa ds e/;nwjh a < l gSA tc bu æO;ekuksa dks xq:Ro ds v/khufojkekoLFkk ls NksM+ fn;k tkrk gS rks ;s A o B ds lkFkVdjkrs gSa ftuds fy;s izR;koLFkku xq.kkad Øe'k% 0.8 o0.4 gSA VDdj gksus ds fdrus le; i'pkr~ ;g jLlh rutkrh gS (ekuk H>>l gS)
(1) 2 25 a
2 2gH-l
e = 0.8
A B
e = 0.4
m1 m2
a < l
H
(2) 2gH
(3) 2 23 a
2 2gH-l
(4) buesa ls dksbZ ugha
11. ,d vikjn'khZ insZ esa nks iryh lekUrj fLyVsa cuk;htkrh gSA tc ,d ,do.khZ; izdk'k iqat bu fLyVksa lsyEcor~ vkiru dh fLFkfr ls xqtjrk gS rks ikjxfer izdk'kesa nwj fLFkr insZ ij izFke pedhyh fÝUt] izdk'k iqat dhewy fn'kk ls ±45° ij izkIr gksrh gS tc fd ;g midj.kok;q esa j[kk gSA tc bl midj.k dks ,d æo esa Mqck fn;ktkrk gS rks ;gh pedhyh fÝUt ± 30° ij izkIr gksrh gSAæo dk viorZukad gS %&
(1) 2 (2) 3 (3) 43 (4)
32
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
7/35Kota/01CT314063
SPACE FOR ROUGH WORK
12. Given that ax m axe dx a e C= +ò , then which
statement is incorrect (Dimension of x = L1) ?(1) m = –1(2) Dimension of C = L1
(3) Dimensions of a = L–1
(4) None of these13. In the P-V diagram shown, the gas does 5 J of
work in isothermal process ab and 4 J in adiabaticprocess bc. What will be the change in internalenergy of the gas in straight path c to a?
a b
cV
P
(1) 9J (2) 1 J (3) 4 J (4) 5 J14. There are three sources of sound of equal
intensities with frequencies 101, 103 and 106Hz. What is the beat frequency heard if all aresounded simultaneously?(1) 2 Hz (2) 4 Hz (3) 6 Hz (4) 8 Hz
15. It is found that an increase in pressure of100 kPa causes a certain volume of water todecrease by 5 × 10–3 percent of its originalvolume. Then the speed of sound in the wateris about (density of water 103 kg/m3)(1) 330 m/s (2) 1414 m/s(3) 1732 m/s (4) 2500 m/s
12. fn;k x;k gS ax m exe dx a e C= +ò , rks dkSulk dFku xyr
gS (x dh foek = L1) %&(1) m = – 1(2) C dh foek = L1
(3) a dh foek = L–1
(4) buesa ls dksbZ ugha
13. iznf'kZr P-V vkjs[k esa xSl lerkih; izØe ab esa 5 J rFkk
:¼ks"e izØe bc esa 4J dk;Z djrh gSA lh/ks iFk c ls ard xSl dh vkarfjd ÅtkZ esa gq, ifjorZu dk eku gksxk
a b
cV
P
(1) 9J (2) 1 J (3) 4 J (4) 5 J14. leku rhozrk ijUrq 101, 103 rFkk 106 Hz vko`fÙk okys
rhu /ofu lzksr gSaA ;fn lHkh dks ,d lkFk ctk;k tk;s rks
lqukbZ nsus okyh /ofu dh foLiUn vko`fÙk Kkr dhft,A(1) 2 Hz (2) 4 Hz (3) 6 Hz (4) 8 Hz
15. ;g ns[kk x;k fd nkc ds eku esa 100 kPa dh o`f¼ djus
ij ty dk ,d fo'ks"k vk;ru vius izkjfEHkd vk;ru
dk 5 × 10–3 izfr'kr de gks tkrk gSA ty esa /ofu dh
pky yxHkx gksxh (ty dk ?kuRo = 103 kg/m3)(1) 330 m/s (2) 1414 m/s(3) 1732 m/s (4) 2500 m/s
Kota/01CT3140638/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
16. At shallow depth h, the pressure in the oceanis simply given by P = P0 + rgh, in which r isthe density of water and P0 is the air pressure.As we go deeper, the high pressure causes thewater to compress and become denser. Whichof the following sketches illustrates the correctdependence of the pressure on the depth h ?
(1)
P
P0
h
(2)
P
P0
h
(3)
P
P0
h
(4)
P
P0
h
17. Find maximum amplitude for safe SHM (blockdoes not topple during SHM) of a cubical blockof side 'a' on a smooth horizontal floor as shownin figure (spring is massless)
2a/3
K
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
\\\\\\
\\\\\\
\\\\\\
\\\\\\
\\\\\\\\
(1) 3mgK
(2) 3mgK
(3) 23mgK
(4) None
16. leqæ e s a ikuh dh fNNyh xgjkb Z h ij nkc
P = P0 + rgh }kjk fn;k tkrk gS] tgk¡ r ty dk ?kuRo
rFkk P0 ok;qnkc gSA vfèkd xgjkbZ ij tkus ij mPp nkc
ds dkj.k ty laihfM+r gksrk gS] ftlds QyLo:i ;g
l?ku gksrk tkrk gSA fuEu esa ls dkSulk vkjs[k nkc dh
xgjkbZ h ij fuHkZjrk dks n'kkZrk gS ?
(1)
P
P0
h
(2)
P
P0
h
(3)
P
P0
h
(4)
P
P0
h
17. fp= esa n'kkZ;s vuqlkj ,d fpdus {kSfrt Q'kZ ij j[ks aHkqtk okys ?kukdkj CykWd dh lqjf{kr ljy vkorZ xfr
(ljy vkorZ xfr ds nkSjku CykWd iyVrk ugha gS) ds fy,
vf/kdre vk;ke Kkr dhft,A (fLizax æO;ekughu gS) :-
2a/3
K
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
\\\\\\
\\\\\\
\\\\\\
\\\\\\
\\\\\\\\
(1) 3mgK
(2) 3mgK
(3) 23mgK
(4) dksbZ ugha
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
9/35Kota/01CT314063
SPACE FOR ROUGH WORK
18. A rod of length 1000 mm and coefficient oflinear expansion a = 10–4 per degree is placedsymmetrically between fixed walls separatedby 1001 mm. The Young’s modulus of the rodis 1011 N/m2. If the temperature is increased by20°C, then the stress developed in the rod is :-
1000mm
1001mm
(1) 100 MPa (2) 50 MPa(3) 200 MPa (4) 400 MPa
19. Kinetic energy of a particle executing simpleharmonic motion in straight line is pv2 andpotential energy is qx2, where v is speed atdistance x from the mean position. It timeperiod is given by the expression
(1) 2 qp
p (2) 2 pq
p
(3) 2 qp q
p+ (4) 2 p
p qp
+
18. yEckbZ 1000 mm rFkk a = 10–4 izfr fMxzh js[kh; izlkj
xq.kkad okyh ,d NM+ dks 1001 mm nwj fLFkr nks fLFkj
nhokjksa ds e/; lefer :i ls j[kk tkrk gSA NM+ dk ; ax
izR;kLFkrk xq.kkad 1011 N/m2 gSA ;fn rkieku dks 20°C
c<+k fn;k tk, rks NM+ esa mRiUu izfrcy gksxk%&
1000mm
1001mm
(1) 100 MPa (2) 50 MPa(3) 200 MPa (4) 400 MPa
19. lh/kh js[kk esa ljy vkorZ xfr dj jgs d.k dh xfrt
ÅtkZ pv2 rFkk fLFkfrt ÅtkZ qx2 gS] tgka v d.k dh pky
rFkk x ekè; fLFkfr ls foLFkkiu gSA bl ljy vkorZ xfr
dk vkorZdky D;k gksxk\
(1) 2 qp
p (2) 2 pq
p
(3) 2 qp q
p+ (4) 2 p
p qp
+
Kota/01CT31406310/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
20. An engine is moving with uniform speed along
a circular track emitting a sound of frequency
400 Hz as shown in the figure. Speed of engine
is 30 m/sec and speed of sound is 330 m/sec.
An observer is standing inside the track.
Maximum frequency observed by the
observer is
Observer
Train
(1) Less than 440 Hz
(2) Greater than 440 Hz
(3) Equal to 440 Hz
(4) Equal to 400 Hz
20. ,d batu fp=kuqlkj o`Ùkkdkj iFk esa ,dleku pky ls
xfr djrk gqvk 400 Hz vkofÙk dh /ofu mRlftZr djrk
gSA baftu dh pky 30 m/sec o /ofu dh pky
330 m/sec gSA bl iFk ds vanj [kM+s fdlh izs{kd ds
lkis{k izsf{kr vf/kdre vko`fÙk gksxh %&
Observer
Train
(1) 440 Hz ls de
(2) 440 Hz ls vf/kd
(3) 440 Hz ds cjkcj
(4) 400 Hz ds cjkcj
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
11/35Kota/01CT314063
SPACE FOR ROUGH WORK
21. For the situation shown in the figure, waterflows on the surface of a fixed plate. Thevelocity of water as a function of distance 'y' is
given as : 2y yu 2
h hé ùæ ö= a -ê úç ÷
è øê úë û. Determine the
magnitude of the shear stress that the waterapplies at the base of the plate. Coefficient ofviscosity is h.
hy x u
Water
Free surface
(1) 3
hah
(2) 2
hah
(3) 4
hah
(4) hah
22. A ball is of mass m, strikes a smooth ground atangle a as shown in figure and is deflected atangle b. The coefficient of restitution will be
a b
(1) tana/tanb (2) cosa/cosb
(3) sina/sinb (4) tanb/tana
21. çnf'kZr fp= esa ,d fLFkj IysV dh lrg ij ty izokfgr
gks jgk gSA nwjh 'y' ds Qyu ds :i esa ty dk osx
2y yu 2h h
é ùæ ö= a -ê úç ÷è øê úë û
}kjk fn;k tkrk gSA ty }kjk IysV
ds isans ij yxk;s tkus okys vi:i.k izfrcy dk ifjek.k
D;k gksxk ;fn ';kurk xq.kkad h gks\
hy x u
Water
Free surface
(1) 3
hah
(2) 2
hah
(3) 4
hah
(4) hah
22. m nzO;eku dh xsan fp=kuqlkj a dks.k ij ,d fpduhlrg ls Vdjkrh gS rFkk la?kkr ds i'pkr~ b dks.k ij tkrhgS] rks izR;koLFkku xq.kkad gksxk&
a b
(1) tana/tanb (2) cosa/cosb
(3) sina/sinb (4) tanb/tana
Kota/01CT31406312/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
23. In the given figure light is incident at an angleq with the normal to a plane containing twoslits of separation d. Select the expression thatcorrectly describes the positions of theinterference maxima in terms of the incomingangle q and outgoing angle f.
d
fq
(1) 1sin sin m2 d
læ öf + q = +ç ÷è ø
(2) dsinq = ml
(3) ( )sin sin m 1dl
f - q = +
(4) sin sin mdl
f + q =
24. x-t graph for a uniformly accelerated particleis as shown in the figure. Then find the averagevelocity between points (i) and (ii)
x
t
q2=tan (3)-1(ii)
(i) 45°
(1) 3m/s (2) 2m/s (3) 4 m/s (4) 1.5 m/s
23. fp= esa izdk'k vfHkyEc ls q dks.k ij ,d ry ij
vkifrr gksrk gS ftlesa nks fLyVsa d nwjh ij j[kh gqbZ gSA
vkiru dks.k q rFkk fuxZeu dks.k f ds inksa esa O;frdj.k
mfPp"B dh fLFkfr dks n'kkZus okyk O;atd gksxk %&
d
fq
(1) 1sin sin m2 d
læ öf + q = +ç ÷è ø
(2) dsinq = ml
(3) ( )sin sin m 1dl
f - q = +
(4) sin sin mdl
f + q =
24. ,dleku Rofjr xfr dj jgs d.k dk foLFkkiu≤vkjs[k fp= esa iznf'kZr gSA fcUnq (i) rFkk (ii) ds e/;vkSlr osx Kkr dhft,A
x
t
q2=tan (3)-1(ii)
(i) 45°
(1) 3m/s (2) 2m/s (3) 4 m/s (4) 1.5 m/s
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
13/35Kota/01CT314063
SPACE FOR ROUGH WORK
25. 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 wavesis 360 ms–1 and their frequencies are 256 Hz.Which is not possible value of t (in sec) :-
(1) 9.8 × 10–4 (2) 10–3
(3) 2.9 × 10–3 (4) 4.9 × 10–3
26. A chain of length L and mass m is placed upona smooth surface. The length of BA is L–b.Calculate the velocity of the chain when its endreaches B.
q
A B
(1) 2 22gsin(L b )
Lq
- (2) 2 2gsin2 (L b )
Lq
-
(3) 2 2gsin(L b )
Lq
- (4) 2 2gsin(L b )
2Lq
-
25. le; ds nks {k.kksa (mPpre] ek/;) ij fdlh ruh gqbZ jLlh
esa 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 dk
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) 2.9 × 10–3 (4) 4.9 × 10–3
26. L yEckbZ rFkk m nzO;eku dh ,d tathj fpduh lrg ds
Åij j[kh gSA BA dh yEckbZ L–b gSA tc bldk fljk
B ij igq¡prk gS] rks tathj ds osx dh x.kuk dhft,A
q
A B
(1) 2 22gsin(L b )
Lq
- (2) 2 2gsin2 (L b )
Lq
-
(3) 2 2gsin(L b )
Lq
- (4) 2 2gsin(L b )
2Lq
-
Kota/01CT31406314/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
27. A ring of mass m is rolling without slippingwith linear velocity v as shown is figure. A rodof identical mass is fixed along one of itsdiameter. The total kinetic energy of thesystem is :-
(1) 75 mv2 (2)
25 mv2 (3)
53 mv2 (4)
54 mv2
28. A bob of mass m is attached to a string whoseother end is tied to a light vertical rod as shownin figure. The bob is swinging in horizontalplane with constant angular speed w. Thevertical rod is supported on a block of mass Mwhich is placed on a rough surface. What isminimum friction coefficient between groundand block for which block does not slip ?
(1)m cosm M
q+
M
q l
mw
(2) m tanm M
q+
(3) 3m tanm M
q+
(4) M tanm M
q+
27. ,d m æO;eku dh oy; fp=kuqlkj fcuk fQlys js[kh;osx v ds lkFk yq<+d jgh gSA blds leku æO;eku okyh,d NM + blds fdlh ,d O;kl ds vuqfn'k yxh gqbZ gSAfudk; dh dqy xfrt ÅtkZ gksxh %&
(1) 75 mv2 (2)
25 mv2 (3)
53 mv2 (4)
54 mv2
28. ,d m æO;eku dk xksyd fp=kuqlkj ,d jLlh ls ca/kk
gqvk gS rFkk jLlh dk nwljk fljk ,d gYdh Å/okZ/kj NM+
ls tqM+k gSA ;g xksyd {kSfrt ry esa fu;r dks.kh; pky
w ls xfr'khy gSA ;g Å/okZ/kj NM+ ,d M æO;eku ds
CykWd ij yxh gqbZ gS rFkk ;g CykWd [kqjnjh lrg ij j[kk
gSA CykWd rFkk lrg ds e/; U;wure ?k"kZ.k xq.kkad D;k
gksuk pkfg, rkfd CykWd fQlys ugha \
(1)m cosm M
q+
M
q l
mw
(2) m tanm M
q+
(3) 3m tanm M
q+
(4) M tanm M
q+
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
15/35Kota/01CT314063
SPACE FOR ROUGH WORK
29. Three copper blocks of masses M1, M2 and M3
kg respectively are brought into thermal contacttill they reach equilibrium. Before contact, theywere at T1, T2, T3(T1> T3). Assuming there isno heat loss to the surroundings, the equilibriumtemprature T is (s is specific heat of copper)
(1) 1 2 3
3T T T
T+ +
=
(2) 1 1 2 2 3 3
1 2 3
M T M T M TT
M M M
+ +=
+ +
(3) ( )1 1 2 2 3 3
1 2 33M T M T M T
TM M M
+ +=
+ +
(4) 1 1 2 2 3 3
1 2 3
M T s M T s M T sT
M M M
+ +=
+ +
30. In an industrial process 10 kg of water per houris to be heated from 20°C to 80°C . To do thissteam at 200°C is passed from a boiler into acopper coil immersed in water. The steamcondenses in the coil and is returned to theboiler as water at 90°C. How many kg of steamis required per hour.(Specific heat of steam = 0.5 cal/g°C, Latentheat of vaporisation = 540 cal/g)(1) 1g (2) 1 kg (3) 10 g (4) 10 kg
29. æO;eku M1, M2 o M3 kg okys rhu rkacs ds CykWdksa dks,d&nwljs ds lkFk rkih; laidZ esa ykdj lkE;koLFkk esayk;k tkrk gSA laidZ esa ykus ls iwoZ buds rkieku Øe'k%T1, T2, T3(T1> T3) FksA ekuk ifjos'k esa dksbZ Å"ek âklugha gksrk gSA ;fn rkacs dh fof'k"V Å"ek s gks rks lkE;koLFkkrkieku T gksxk %&
(1) 1 2 3
3T T T
T+ +
=
(2) 1 1 2 2 3 3
1 2 3
M T M T M TT
M M M
+ +=
+ +
(3) ( )1 1 2 2 3 3
1 2 33M T M T M T
TM M M
+ +=
+ +
(4) 1 1 2 2 3 3
1 2 3
M T s M T s M T sT
M M M
+ +=
+ +
30. ,d vkS|ksfxd izØe esa izfr ?k.Vs 10 kg ty dks 20°Cls 80°C rd xeZ fd;k tkrk gSA ,slk djus ds fy;s
200°C rki okyh Hkki dks ,d ckW;yj ls ikuh esa Mwch
gqbZ rkez dq.Mfy;kas esa izokfgr fd;k tkrk gSA Hkki
dq.Mfy;kas esa la?kfur gks tkrh gS ,oa ck;yj dks 90°Cij ty ds :i esa okil dj nh tkrh gSA izfr ?k.Vs fdrus
fdxzk Hkki dh vko';drk gksxh (Hkki dh fof'k"V Å"ek
0.5 cal/g × °C ,oa ok"iu dh x q Ir Å"ek= 540 cal/gm)(1) 1g (2) 1 kg (3) 10 g (4) 10 kg
Kota/01CT31406316/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
PART B - CHEMISTRY31. For the given series of reaction-
NH3(g) + O2(g) ® NO(g) + H2O(l)
NO(g) + O2(g) ® NO2(g)
To obtain maximum mass of NO2 from a given
mass of mixture NH3 and O2, the ratio of mass
of NH3 to O2 should be -
(1) 47 (2)
1716
(3) 1740 (4)
1756
32. A metal have work function 4 eV is exposedto photon of l = 1240Å. If a acceleratingpotential of 7.6 volt is applied, then the electronwith maximum kinetic energy will have speedequal to -
(1) 2.18 × 106 m/s
(2) 3 × 108 m/s
(3) 9.61 × 105 m/s
(4) 8.72 × 106 m/s
31. vfHkfØ;k dh nh xbZ Js.kh ds fy,-
NH3(g) + O2(g) ® NO(g) + H2O(l)
NO(g) + O2(g) ® NO2(g)
NH3 rFkk O2 feJ.k ds fn;s x;s nzO;eku ls NO2 dk
vf/kdre nzO;eku izkIr djus ds fy, NH3 o O2 ds
nzO;ekuksa dk vuqikr gksuk pkfg,-
(1) 47 (2)
1716
(3) 1740 (4)
1756
32. 4 eV dk;Z Qyu okyh ,d /kkrq ij l = 1240Å dk
QksVksu vkifrr gksrk gS] ;fn 7.6 oksYV dk Roj.k foHko
yxk;k tk,s] rc vf/kdre xfrt ÅtkZ ds bySDVªkWu dk
osx gksxk -
(1) 2.18 × 106 m/s
(2) 3 × 108 m/s
(3) 9.61 × 105 m/s
(4) 8.72 × 106 m/s
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
17/35Kota/01CT314063
SPACE FOR ROUGH WORK
33. When 1.685 gram of an alkali metal chlorideis dissolved in 200gram water, the boiling pointof the solution is measured to be 100.051ºC.If the ionic solid has a crystal lattice with cationand anion radius 1.70Å and 1.80Å respectively.Find the edge length of solid assuming nodefect in the cyrstal -Given : Kb(H2O) = 0.51 Kkg mol–1
NA = 6 × 1023
[Li = 7, Na = 23, K = 39, Rb = 85.5,Cs = 133, Cl = 35.5]
(1) 7 Å (2)7 Å3
(3) 14 Å
3 (4) 3.5 Å
34. Following are the critical temperature of thesome gases :
Gases H2 He O2
TC(K) 33.2 5.3 154.3
From the above data what would be order ofliquefication of these gases.Start writing the order from the gas liquifyingfirst(1) H2, He, O2 (2) He, O2, H2
(3) O2, He, H2 (4) O2, H2, He
33. tc 1.685 xzke {kkfj; /kkrq DyksjkbM dks 200 xzke ty
esa foys; fd;k tkrk gS] rc foy;u dk ekik x;k DoFkukad
100.051ºC gS ;fn vk;uhd Bksl esa fØLVy tkyd ds
/kuk;u rFkk ½.kk;u fd f=T;k,sa Øe'k% 1.70Å rFkk
1.80Å gSA eku fyft;s fØLVy esa dksbZ =qfV ugha gS rks
Bksl ds fdukjs dh yEckbZ Kkr dhft,
fn;k gS : Kb(H2O) = 0.51 Kkg mol–1
NA = 6 × 1023
[Li = 7, Na = 23, K = 39, Rb = 85.5,
Cs = 133, Cl = 35.5]
(1) 7 Å (2)7 Å3
(3) 14 Å
3 (4) 3.5 Å
34. dqN xSlkas ds Økafrd rki fuEu izdkj gS :
xSlsa H2 He O2
TC(K) 33.2 5.3 154.3
mijksDr vk¡dM+ks ls bu xSlksa ds nzohdj.k dk Øe D;k
gksxkA
Øe dks igys nzohr gksus okyh xSl ls izkjEHk dhft;sa
(1) H2, He, O2 (2) He, O2, H2
(3) O2, He, H2 (4) O2, H2, He
Kota/01CT31406318/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
35. An alloy of metal 'A' 'B' and 'C' is found to have'A' constituting ccp lattice. If 'B' atom occupythe edge centres and 'C' is present at the bodycentre, the formula of the alloy is :-(1) A
4B
2C (2) A
4B
4C
(3) A4B
3C (4) ABC
36. A(g) ® 2B(g)initially 2 moles of A are taken in 5 litre vessel.
After 20 min, [A]t = t[B]2 . Find half life time
of A in the first order reaction(1) 20 min (2) 10 min(3) 40 min (4) 5 min
37. Calculate Ecell
Pt(s)| 21atm
H (g) |HA-7
a(K =10 )1M || HB
–5a(K =10 )1M |
21atm
H (g) |Pt(s)
(1) 0.06 V (2) 0.03 V(3) 0.04 V (4) 0.05 V
38. At certain temperature (T) if conductivity ofpure water is 5.5 × 10–7Scm–1, then calculatepH of water at given temp.
Given : 2 1350Scm eq+H
¥ -l =
2 1200Scm eq-OH
¥ -l =
(1) 5 (2) 6 (3) 7 (4) 8
35. 'A' 'B' rFkk 'C' /kkrqvksa dh ,d feJ /kkrq esa 'A' ccptkyd dk fuekZ.k djrk gS] ;fn 'B' ijek.kq fdukjs dsdsUnz dks ?ksjrk gS vkSj 'C' dk; dsUnz ij mifLFkr gks] rksfeJ /kkrq dk lq= gS :-(1) A
4B
2C (2) A
4B
4C
(3) A4B
3C (4) ABC
36. A(g) ® 2B(g)izkjEHk esa A ds 2 eksy 5 yhVj ds ik= esa fy;s x;s gSA
20 ehuV ds i'pkr~ [A]t = t[B]2 . izFke dksfV vfHkfØ;k
esa A dh v/kZ vk;q crkb;sA(1) 20 min (2) 10 min
(3) 40 min (4) 5 min
37. Ecell dh x.kuk dhft;s
Pt(s)| 21atm
H (g) |HA-7
a(K =10 )1M || HB
–5a(K =10 )1M |
21atm
H (g) |Pt(s)
(1) 0.06 V (2) 0.03 V(3) 0.04 V (4) 0.05 V
38. fuf'pr rki (T) ij ;fn 'k q¼ ty dh pkydrk5.5 × 10–7Scm–1 gS rk s fn;s x;s rki ij ty dhpH Kkr dhft;s
fn;k gS : 2 1350Scm eq+H
¥ -l =
2 1200Scm eq-OH
¥ -l =
(1) 5 (2) 6 (3) 7 (4) 8
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
19/35Kota/01CT314063
SPACE FOR ROUGH WORK
39. 10 gram mixture of two gases A2(mol.wt. = 20)and B2 (mol. wt. = 30), which decompose byfirst order kinetics, was taken in a vessel. Thehalf life of decomposition of A2 and B2 are 2and 3 hours respectively. After 6 hours weightof mixture of A2 and B2 is found to be 2 gram.Find the weight of A2 in the initial mixture-(1) 4 gm (2) 6 gm (3) 8 gm (4) 2 gm
40. Which of the following have highest osmaticpressure at 300 K(1) 0.1 M CH3COOH (a = 0.7)(2) 0.1 M KCl (a = 1)(3) 0.1 M Na2SO4 (a = 1)(4) 0.1 M K2Zn[Fe(CN)6] (a = 1)
41. Consider the following statements :(a) When the lead-silver alloy rich in silver,
lead is removed by the cupellation process(b) Froth flotation can be applied for non
sulphide ore also using suitableactivator
(c) PbSO4 + H2SO4, electrolyte is used for therefining of Pb by electrolysis
(d) Any metal will not reduce the oxide of othermetals which lie above it in the Ellinghamdiagram.
Using 'T' for true and 'F' for false statementsin the given sequence, select the correct setof code(1) T T F T (2) T T F F(3) F T F T (4) T T T F
39. nks xSlksa A2 (vkf.od Hkkj = 20 ) rFkk B2 (vkf.odHkkj = 30) dk 10 xzke feJ.k] tks izFke dksfV xfrdhls fo;ksftr gksrk gS] dks ,d ik= esa fy;k x;kA A2 rFkkB2 ds fo;kstu dh v/kZvk;q Øe'k% 2 rFkk 3 ?k.Vs gS6 ?k.Vs ds i'pkr~ A2 rFkk B2 ds feJ.k dk Hkkj2 xzke ik;k x;kA izkjfEHkd feJ.k esa A2 dk Hkkj Kkrdhft,-(1) 4 gm (2) 6 gm (3) 8 gm (4) 2 gm
40. fuEu esa ls fdldk ijklj.k nkc 300K ij vf/kdregksxkA(1) 0.1 M CH3COOH (a = 0.7)(2) 0.1 M KCl (a = 1)(3) 0.1 M Na2SO4 (a = 1)(4) 0.1 M K2Zn[Fe(CN)6] (a = 1)
41. fuEu dFkuksa ij fopkj dhft, :(a) tc ySM&flYoj ,sykW; esa flYoj dh vf/kdrk gks
rks ySM dks D;wisyhdj.k izØe }kjk gVk;k tkrk gS(b) mi;qDr lfØ;d dk iz;ksx djds >kx Iykou
fof/k dk iz;ksx uku&lYQkbM v;Ld ds fy, Hkhfd;k tk ldrk gS
(c) PbSO4 + H2SO4, oS|qrvi?kV~; dk iz;k sxoS|qrvi?kVu }kjk Pb ds 'kqf¼dj.k ds fy, fd;ktkrk gS
(d) dksbZ Hkh /kkrq vU; ml /kkrq ds vkWDlkbM dks vipf;rugha djsxh tks ,fya?ke fp= esa blls Åij fLFkr gks
lR; ds fy, 'T' dk rFkk vlR; dFku ds fy, 'F' dkiz;ksx djrs gq;s ] dwV dk lgh leqPp; pqfu,(1) T T F T (2) T T F F(3) F T F T (4) T T T F
Kota/01CT31406320/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
42. Blue colouration is not produced when :
(1) Cr2O72– + H2O2
H+
¾¾®
(2) NO2 + NO Cooling to23ºC-¾¾¾¾®
(3) Fe+3 + SCN¯ ¾¾®
(4) CuSO (s)4(anhydrous) + H2O(g) ¾¾®
43. Which of the following compounds arepartially soluble or insoluble in NH4OHsolution ?(I) Fe(OH)3 (II) Ag2CrO4
(III) Al(OH)3 (IV) Ag2CO3
(1) II & III (2) I & III(3) I, III & IV (4) II, III & IV
44. Identify the INCORRECT statement(s)among the following :(I) Moist ammonia gas can be dried by using
anhydrous CaCl2(II) Phosphine is a weaker base than ammonia(III) R3SiCl produces very complex cross
linked polymeric silicone on hydrolysis(IV) When metallic copper react with 6M
HNO3 NO2 is the only product(1) (III) & (IV) (2) (II), (III) & (IV)(3) (I), (III) & (IV) (4) (IV) Only
42. fuEu esa ls dkSulh vfHkfØ;k esa uhyk jax fufeZr ugha gksrkgS :
(1) Cr2O72– + H2O2
H+
¾¾®
(2) NO2 + NO -¾¾¾¾¾®23ºCB.Mk djus ij
rd
(3) Fe+3 + SCN¯ ¾®
(4) CuSO (s)4(futZyh;)
+ H2O(g) ¾¾®
43. fuEu esa ls dkSulk ;kSfxd NH4OH foy;u esa vkaf'kd
foys; ;k vfoys; gS ?
(I) Fe(OH)3 (II) Ag2CrO4
(III) Al(OH)3 (IV) Ag2CO3
(1) II rFkk III (2) I rFkk III
(3) I, III rFkk IV (4) II, III rFkk IV
44. fuEu esa ls xyr dFku igpkfu,sa :
(I) ue veksfu;k xSl dks futZyh; CaCl2 dk iz;ksx
dj 'kq"d fd;k tk ldrk gS
(II) veksfu;k dh rqyuk esa QkWLQhu nqcZy {kkj gS
(III) R3SiCl, tyvi?kVu ij cgqr tfVy ØkWl&caf/kr
cgqydh; flfydksu cukrk gS
(IV) tc /kkfRod dkWij 6M HNO3, ds lkFk fØ;k
djrk gS rks mRikn ] dsoy NO2 gS
(1) (III) rFkk (IV) (2) (II), (III) rFkk (IV)
(3) (I), (III) rFkk (IV) (4) dsoy (IV)
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
21/35Kota/01CT314063
SPACE FOR ROUGH WORK
45. Which of the following is not halide ore ?(1) Cryolite (2) Fluorspar(3) Horn silver (4) Limonite
46. In which of the ƒ-block element number ofelectron in (n – 2)ƒ subshell is zero :(1) Ce (2) U(3) Th (4) None of these
47. Select the correct matching :(1) Pyro metallurgy : Extraction of Fe(2) Electro metallurgy : Extraction of Al(3) Hydro metallurgy : Extraction of Au(4) All above are correct
48. Compound A on borax bead test in reducingflame gives green colour bead. Compound Aon treatment with H2O2 followed by treatmentwith Pb(OAc)2 gives yellow ppt. The metal ionin the compound A is :
(1) Fe+3 (2) Mn2+ (3) Cr+3 (4) Ba2+
49. Thermal stability of BaCO3, CaCO3, SrCO3 andMgCO3 decreases in the order of :
(1) BaCO3 > SrCO3 > MgCO3 > CaCO3
(2) CaCO3 > SrCO3 > MgCO3 > BaCO3
(3) MgCO3 > CaCO3 > SrCO3 > BaCO3
(4) BaCO3 > SrCO3 > CaCO3 > MgCO3
45. fuEu esa ls dkSulk gSykbM v;Ld ugha gSa ?(1) Øk;ksykbV (2) ¶yksjLikj
(3) gkWuZ flYoj (4) fyeksukbV
46. fuEu esa ls dkSuls] ƒ-CykWd rRo ds (n – 2)ƒ midks'kesa bysDVªkWuksa dh la[;k 'kwU; gS :(1) Ce (2) U(3) Th (4) buesa ls dksbZ ugha
47. lgh feyku pqfu, :(1) rki /kkrqdeZ : Fe dk fu"d"kZ.k
(2) oS|qr /kkrqdeZ : Al dk fu"d"kZ.k
(3) ty /kkrqdeZ : Au dk fu"d"kZ.k
(4) mijksDr lHkh lgh gS
48. ;kSfxd A, vipk;d Tokyk esa cksjsDl eudk ijh{k.k fd;s
tkus ij gjs jax dh eudk cukrk gSA ;kSfxd A, H2O2
ds lkFk rFkk mlds ckn Pb(OAc)2 ds lkFk mipkfjr
fd;s tkus ij ihyk vo{ksi nsrk gSA ;kSfxd A esa mifLFkr
/kkrq vk;u gS :(1) Fe+3 (2) Mn2+ (3) Cr+3 (4) Ba2+
49. BaCO3, CaCO3, SrCO3 rFkk MgCO3 ds rkih;
LFkkf;Ro dk ?kVrk gqvk Øe gSa :(1) BaCO3 > SrCO3 > MgCO3 > CaCO3
(2) CaCO3 > SrCO3 > MgCO3 > BaCO3
(3) MgCO3 > CaCO3 > SrCO3 > BaCO3
(4) BaCO3 > SrCO3 > CaCO3 > MgCO3
Kota/01CT31406322/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
50. X NaOHD Y HCl White fumes
Colourlesssalt
ZWhite ppt.
KMnO4 Colourless solution
CaCl2
Salt X is :
(1) BaC2O4 (2) (NH4)2C2O4
(3) K2C2O4 (4) CaC2O4
51. Consider the following :
CH –CH –NH3 2 2(I)
Ph—NH2(III)
Ph–CH –NH2 2(II)
Correct order of their basic strength is :
(1) III > II > I (2) I > II > III
(3) I > III > II (4) II > I > III
52. The decreasing order of nucleophilicity amongthe following nucleophiles is :
(I) CH3COO1 (II) CH3O1
(III) CH3CH21 (IV) CH3SO3
1
(1) IV > I > II > III (2) II > I > IV > III
(3) III > I > II > IV (4) III > II > I > IV
50. X NaOHD Y HCl 'osr /kqez
jaxghuyo.k
Z'osr vo{ksi
KMnO4 jaxghu foy;u
CaCl2
yo.k X gS :
(1) BaC2O4 (2) (NH4)2C2O4
(3) K2C2O4 (4) CaC2O4
51. fuEu ij fopkj dhft;sA
CH –CH –NH3 2 2(I)
Ph—NH2(III)
Ph–CH –NH2 2(II)
budh {kkjh; lkeF;Z dk lgh Øe gS &
(1) III > II > I (2) I > II > III
(3) I > III > II (4) II > I > III
52. fuEu ukfHkdLusgh;ksa esa ukfHkdLusghrk dk ?kVrk g qvk ØegS&
(I) CH3COO1 (II) CH3O1
(III) CH3CH21 (IV) CH3SO3
1
(1) IV > I > II > III (2) II > I > IV > III
(3) III > I > II > IV (4) III > II > I > IV
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SPACE FOR ROUGH WORK
53. Which of the following compound can showtautomerism :
(1)O
O
(2) O
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Ph (4) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Me
54. Relation between given pair is :
Cl
HC H2 5
CH OH2
andHOCH CH2 2
HCH3
Cl
(1) Enantiomers (2) Diastereomers
(3) Homomers (4)Structural isomers
55. The major product of the reaction is :
OH
EtConc.H PO3 4 Product
(1)Et
(2) OH
Et
O
(3)Et
(4) EtO
53. fuEu esa ls dkSulk ;kSfxd pyko;ork iznf'kZr djldrk gS&
(1)O
O
(2) O
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Ph (4) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Me
54. fn;s x;s ;qXe esa lEcU/k gS&Cl
HC H2 5
CH OH2
rFkkHOCH CH2 2
HCH3
Cl
(1) izfrfcEc:i leko;oh (2) foofje leko;oh
(3) leyd (leku) (4) lajpukRed leko;oh
55. fuEu vfHkfØ;k dk eq[; mRikn gS&
OHEt
lkUnz H PO3 4 mRikn
(1)Et
(2) OH
Et
O
(3)Et
(4) EtO
Kota/01CT31406324/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
56. Which of the following compounds will giveracemic mixture by SN1 reaction ?
(1)
CH3
Br
H
Me (2) Ph–CH2–Br
(3)
CD3
Br
H
Me (4) CH –CH–CH3 3
CH –Br2
57. In the given reaction sequence
Ph–CH–OH
CH3
PCC Ai) NH OH2 B + Cii) H /+
B and C are :
(1) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–CH3 , Ph–CH2–CHO
(2) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–NH–CH3 , CH3–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–NH–Ph
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–CH3 , Ph–C=N
CH3
OH
(4) C = NPh
H C3
OH, C = N
Ph
H C3 OH
56. fuEu ;kSfxdksa esa ls dkSu] SN1 vfHkfØ;k }kjk jslsfedfeJ.k nsxk&
(1)
CH3
Br
H
Me (2) Ph–CH2–Br
(3)
CD3
Br
H
Me (4) CH –CH–CH3 3
CH –Br2
57. fn;s x;s vfHkfØ;k Øe esa
Ph–CH–OH
CH3
PCC Ai) NH OH2 B + Cii) H /+
B rFkk C gS&
(1) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–CH3 , Ph–CH2–CHO
(2) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–NH–CH3 , CH3–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–NH–Ph
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–CH3 , Ph–C=N
CH3
OH
(4) C = NPh
H C3
OH, C = N
Ph
H C3 OH
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SPACE FOR ROUGH WORK
58.
C–NH2
Et
O
Br + KOH2 Product ;
Major product is :
(1)Et
Br(2)
Et
NH2
(3)Et
NH2
(4) Et
NH2
59. Which of the following compound decolouriseBr2 / H2O and also give positive test with neutralFeCl3 :
(1) (2) OH
(3) (4) Ph–OH
60. Compound 'A' give positive test with 2,4-DNPand with I2/NaOH. Compound 'A' may be :
(1) Ph–CHO (2) O
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Ph (4) O
58.
C–NH2
Et
O
Br + KOH2 mRikn ;
eq[; mRikn gS&
(1)Et
Br(2)
Et
NH2
(3)Et
NH2
(4) Et
NH2
59. fuEu ;kSfxdksa esa ls dkSu Br2 / H2O dks jaxghu djrk
gS ,oa mnklhu FeCl3 ds lkFk /kukRed ijh{k.k Hkh nsrk gS&
(1) (2) OH
(3) (4) Ph–OH
60. ;k Sfxd 'A' , 2,4-DNP rFkk I2/NaOH ds lkFk
/kukRed ijh{k.k nsrk gS ;kSfxd 'A' gks ldrk gS&
(1) Ph–CHO (2) O
(3) Ph–COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
–Ph (4) O
Kota/01CT31406326/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
PART C - MATHEMATICS61. Length of latus rectum of hyperbola
2 2
2 2
x y 4cos sin
- =a a
is n , n I2pæ öa ¹ Îç ÷
è ø
(1) 1 cos 22
cos- a
a(2)
1 cos 2sin
+ aa
(3) 1 cos 22
sin+ a
a(4)
1 cos 2cos
- aa
62. The differential equation representing the
family of curves 2y c(x 2c)= + , where c is apositive parameter, is of
(1) order = 1, degree = 2
(2) order = 2, degree = 2
(3) order = 1, degree = 3
(4) order = 2, degree = 4
63. The origin and the points where the line L1
intersect the x-axis and y-axis are vertices of
right angled triangle T whose area is 8.
Also the line L1 is perpendicular to line
L2 : 4x – y = 3, then perimeter of triangle T is -
(1) 10 68+ (2) 8 32+
(3) 17 257+ (4) 4 2 4+
61. vfrijoy;2 2
2 2
x y 4cos sin
- =a a
, n , n I2pæ öa ¹ Îç ÷
è øds ukfHkyEc dh yEckbZ gksxh
(1) 1 cos 22
cos- a
a(2)
1 cos 2sin
+ aa
(3) 1 cos 22
sin+ a
a(4)
1 cos 2cos
- aa
62. oØ fudk; 2y c(x 2c)= + (tgk¡ c ,d /kukRedizkpy gS) dks iznf'kZr djus okyh vody lehdj.k dh
(1) dksfV = 1, ?kkr = 2 gksxhA
(2) dksfV = 2, ?kkr = 2 gksxhA
(3) dksfV = 1, ?kkr = 3 gksxhA
(4) dksfV = 2, ?kkr = 4 gksxhA
63. ewyfcUnq rFkk fcUnq] tgk¡ js[kk L1, x-v{k rFkk y-v{k dks
dkVrh gS] ,d ledks.k f=Hkqt T ds 'kh"kZ gS ftldk
{ks=Qy 8 oxZ bdkbZ gS rFkk js[kk L1, js[kk L2 : 4x – y = 3
ds yEcor~ gS] rks f=Hkqt T dk ifjeki gksxk -
(1) 10 68+ (2) 8 32+
(3) 17 257+ (4) 4 2 4+
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SPACE FOR ROUGH WORK
64. If ˆ ˆ ˆ ˆ ˆ ˆa 2i j k, b i j 2k= - + = + -rr and
2ˆ ˆ ˆc i 3j ( 3 )k= + - l + lr
(where l is a constant)
and ar is perpendicular to c b- lrr , then sum of
different values of l is
(1) –1 (2) 1 (3) 4 (4) –4
65. If 2
2 4
sec8 1 a b tan 2sec 4 1 1 c tan 2 d tan 2
q - + q=
q - + q + q
(where n , n I16
pq ¹ Î ), then value of
(a – b + c – d) is -
(1) 0 (2) 1 (3) 7 (4) 13
66. If tan(psinq) = cot(pcosq), then cot4pæ öq -ç ÷
è øis -
(1) 17
(2) 7 (3) 27
(4) 2 7
67. If z = z(x) and dz(2 cos x) (sin x)z sin xdx
+ + = ,
z(x) > 0 & z 32pæ ö =ç ÷
è ø, then z
3pæ ö
ç ÷è ø
equals -
(1) 72 (2)
32 (3)
52 (4)
12
64. ;fn ˆ ˆ ˆ ˆ ˆ ˆa 2i j k, b i j 2k= - + = + -rr rFkk
2ˆ ˆ ˆc i 3j ( 3 )k= + - l + lr
(tgk¡ l vpj gS) rFkk ar ,
c b- lrr ds yEcor~ gks] rks l ds fofHkUu ekuksa dk ;ksxQy
gksxk
(1) –1 (2) 1 (3) 4 (4) –4
65. ;fn 2
2 4
sec8 1 a b tan 2sec 4 1 1 c tan 2 d tan 2
q - + q=
q - + q + q
(tgk¡ n , n I16
pq ¹ Î ) gks] rks (a – b + c – d) dk eku
gksxk -(1) 0 (2) 1 (3) 7 (4) 13
66. ;fn tan(psinq) = cot(pcosq) gks] rks cot4pæ öq -ç ÷
è øgksxk -
(1) 17
(2) 7 (3) 27
(4) 2 7
67. ;fn z = z(x) ,oa dz(2 cos x) (sin x)z sin xdx
+ + = ,
z(x) > 0 rFkk z 32pæ ö =ç ÷
è ø gks] rks z
3pæ ö
ç ÷è ø
cjkcj gksxk -
(1) 72 (2)
32 (3)
52 (4)
12
Kota/01CT31406328/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
68. If equation in variable q, 3tan(q – a) = tan(q + a),(where a is constant) has no real solution, thena can be (wherever tan(q - a) & tan(q + a)both are defined)
(1) 15p
(2) 518
p(3)
512
p(4)
1718
p
69. In DABC, 8D = (b + c)(bc + 1), thencircumradius of DABC is (D denotes area oftriangle and b, c are length of sides AC and ABrespectively) -
(1) D (2) 12D
(3) 2D (4) 1D
70. If y
2y
dtx1 9t-
=+
ò and 2
2
d y kydx
= , then k equals
(1) 9 (2) 94 (3)
92 (4) 18
71. Consider ellipse E, hyperbola H and parabolaP such that each curve has focus (2, 3) andcorresponding directrix is x + y – 10 = 0. If(a, a1), (b, b1), (g, g1) are nearest vertices ofellipse, hyperbola & parabola to the givendirectrix, then
(1) a > b > g (2) b > g > a
(3) a > g > b (4) a < b < g
68. ;fn pj q esa lehdj.k, 3tan(q – a) = tan(q + a),(tgk¡ a vpj gS) dk dksbZ okLrfod gy uk gks] rks a gksldrk gS (tan(q - a) rFkk tan(q + a) nksuksa tgk¡ HkhifjHkkf"kr gS)
(1) 15p
(2) 518
p(3)
512
p(4)
1718
p
69. f=Hkqt ABC esa, 8D = (b + c)(bc + 1) gks] rks f=HkqtABC dh ifjf=T;k gksxh (D, f=Hkqt ds {ks=Qy dksn'kkZrk gS rFkk b, c Øe'k% Hkqtkvksa AC rFkk AB dhyEckbZ;k¡ gSa) -
(1) D (2) 12D
(3) 2D (4) 1D
70. ;fn y
2y
dtx1 9t-
=+
ò rFkk 2
2
d y kydx
= gks] rks k gksxk -
(1) 9 (2) 94 (3)
92 (4) 18
71. ekuk nh?kZoÙk E, vfrijoy; H rFkk ijoy; P blizdkj gS fd izR;sd oØ dh ukfHk (2, 3) rFkk blds laxr
fu;rk x + y – 10 = 0 gSA ;fn (a, a1), (b, b1), (g. g1)nh xbZ fu;rk ds fy, nh?kZoÙk] vfrijoy; rFkk ijoy;
ds lehiorhZ 'kh"kZ gks] rks
(1) a > b > g (2) b > g > a(3) a > g > b (4) a < b < g
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SPACE FOR ROUGH WORK
72. If sinx + cosx = a, a [ 2, 2] { 1,1}Î - - - , then
n n
n 1(sin x cos x)
¥
=
+å is equal to -
(1) 2
2
2(1 a a )(a 1)+ -
+ (2) 2
2
2(a a 1)(a 1)
- +-
(3) 2
2
2(a a 1)(a 1)
- ++
(4) 2
2
2(1 a a )(a 1)+ -
-
73. Let 7 1/ 7
xƒ(x)(1 x )
=+
and g(x) = (ƒoƒoƒoƒoƒoƒoƒ)(x),
then 5x g(x)dxò equals
(1) 7 6 / 71 (1 6x ) C42
+ +
(2) 7 5 / 71 (1 7x ) C35
+ +
(3) 7 5/ 71 (1 5x ) C35
+ +
(4) 7 6 / 71 (1 7x ) C42
+ +
(where C is constant of integration)
72. ;fn sinx + cosx = a, a [ 2, 2] { 1,1}Î - - - gS]
rks n n
n 1
(sin x cos x)¥
=
+å dk eku gksxk -
(1) 2
2
2(1 a a )(a 1)+ -
+ (2) 2
2
2(a a 1)(a 1)
- +-
(3) 2
2
2(a a 1)(a 1)
- ++
(4) 2
2
2(1 a a )(a 1)+ -
-
73. ekuk 7 1/ 7
xƒ(x)(1 x )
=+
rFkk g(x) = (ƒoƒoƒoƒoƒoƒoƒ)(x) gks]
rks 5x g(x)dxò dk eku gksxk
(1) 7 6 / 71 (1 6x ) C42
+ +
(2) 7 5 / 71 (1 7x ) C35
+ +
(3) 7 5/ 71 (1 5x ) C35
+ +
(4) 7 6 / 71 (1 7x ) C42
+ +
(tgk¡ C lekdyu vpj gS)
Kota/01CT31406330/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
74. Each set Xr contains 5 elements and each set
Yr contains 4 elements and 24 n
r rr 1 r 1
X S Y= =
= =U U .
If each element of set S belong to exactly 10 ofthe Xr's and to exactly 6 of Yr's, then n is (where
24
rr 1
X=U denotes X1 È X2 È X3 È ....... È X24)
(1) 18 (2) 15 (3) 20 (4) 24
75.1 3
21
x | x | 3 dxx 4 | x | 3-
+ ++ +ò is equal to -
(1) / 2
0
4 log(sin )dp
a ap ò
(2) / 2
0
4 log(sin )dp
- q ap ò
(3) / 2
0
2 log(sin 2 )dp
- a ap ò
(4) / 2
0
2 (log(sin ) log(cos ))dp
- a + a ap ò
76. Maximum number of equivalence relations onset A = {1, 2, 3, 4} is N, then -
(1) 14 < N < 20 (2) 21 < N < 28
(3) 29 < N < 36 (4) N > 37
74. izR;sd leqPp; Xr esa 5 vo;o rFkk izR;sd leqPp; Yr
esa 4 vo;o rFkk 24 n
r rr 1 r 1
X S Y= =
= =U U gSA ;fn leqPp;
S dk izR;sd vo;o] Xr ds Bhd 10 rFkk Yr dsBhd 6 leqPp;k s a l s lEcfU/kr g S] rk s n gk sx k
(tgk¡ 24
rr 1
X=U , X1 È X2 È X3 È ....... È X24 dks
iznf'kZr djrk gS)
(1) 18 (2) 15 (3) 20 (4) 24
75.1 3
21
x | x | 3 dxx 4 | x | 3-
+ ++ +ò cjkcj gksxk -
(1) / 2
0
4 log(sin )dp
a ap ò
(2) / 2
0
4 log(sin )dp
- q ap ò
(3) / 2
0
2 log(sin 2 )dp
- a ap ò
(4) / 2
0
2 (log(sin ) log(cos ))dp
- a + a ap ò
76. leqPp; A = {1, 2, 3, 4} esa ifjHkkf"kr rqY;rk lEcUèkksadh vf/kdre la[;k N gks] rks -
(1) 14 < N < 20 (2) 21 < N < 28(3) 29 < N < 36 (4) N > 37
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SPACE FOR ROUGH WORK
77. Let P(n) : 3n < 1 × 2 × 3 × ..... × n, n Î N isalways true for n > l, then smallest value of lis
(1) 7 (2) 9
(3) 13 (4) can't determine
78. a, b, crr r are non coplanar vectors such that
P a b c,Q 4a 3b 4c= + + = + +r rrr r r r r
and R a b c= + a + brr r r are linearly dependent
vectors, then number of possible values of a is
(1) 0 (2) 1 (3) 2 (4) infinite
79. The converse of the statement
"If p < q, then p – x < q – x" is -
(1) If p < q, then p – x > q – x
(2) If p > q, then p – x > q – x
(3) If p – x > q – x, then p > q
(4) If p – x < q – x, then p < q
80. If line 2x 8 y sin z 1 , Rsin 1 cos
- - a -= = bÎ
b a,
sin 1b ¹ lies in the plane 2x – (sinb)y + (cosb)z= k " a Î R, then
(1) k = 8 – sina (2) k = 8 + sina
(3) k = 8 – cosb (4) None of these
77. ekuk n > l ds fy, P(n) : 3n < 1 × 2 × 3 × ..... × n,n Î N lnSo lR; gks] rks l dk U;wure eku gksxk
(1) 7 (2) 9
(3) 13 (4) Kkr ugha fd;k tk ldrk
78. a, b, crr r vleryh; lfn'k bl izdkj gS fd
P a b c,Q 4a 3b 4c= + + = + +r rrr r r r r
rFkk R a b c= + a + brr r r js[kh; vkfJr lfn'k gks] rks a
ds lEHko ekuksa dh la[;k gksxh
(1) 0 (2) 1 (3) 2 (4) vuUr
79. fuEu dFku dk O;qRØe (converse) gksxk
";fn p < q gks] rks p – x < q – x gS" -
(1) ;fn p < q gks] rks p – x > q – x
(2) ;fn p > q gks] rks p – x > q – x
(3) ;fn p – x > q – x gks] rks p > q
(4) ;fn p – x < q – x gks] rks p < q
80. ;fn js[k k 2x 8 y sin z 1 , Rsin 1 cos
- - a -= = bÎ
b a ,
sin 1b ¹ lery 2x – (sinb)y + (cosb)z = k" a Î R esa fLFkr gks] rks
(1) k = 8 – sina (2) k = 8 + sina
(3) k = 8 – cosb (4) buesa ls dksbZ ugha
Kota/01CT31406332/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
81. The line x y 1a b
+ = moves in such a way that
2 2 2
1 1 1a b 2c
+ = , where a, b, c Î R0 and c is
constant, then locus of the foot of theperpendicular from the origin on the givenline is -(1) x2 + y2 = c2 (2) x2 + y2 = 2c2
(3) 2
2 2 cx y2
+ = (4) x2 + y2 = 4c2
82. In three dimensional space, ƒ(x, y, z) = xy + xz,then locus of all points which satisfies theequation ƒ(x, y, z) = 0 is -(1) pair of perpendicular lines(2) pair of a line and a plane which are parallel to each other(3) pair of perpendicular planes(4) pair of a line and a plane which are perpendicular to each other
83. Planet M orbits around its sun, S, in an ellipticalorbit with the sun at one of the foci. When M isclosest to S, it is 2 unit away. When M is farthestfrom S, it is 18 unit away, then the equation ofmotion of planet M around its sun S, assumingS at the centre of the coordinate plane and theother focus lie on negative y-axis, is -
(1) 2 2x (y 8) 1
36 100-
+ = (2) 2 2x (y 8) 1
36 100+
+ =
(3) 2 2x (y 8) 1
64 100-
+ = (4) 2 2x (y 8) 1
64 100+
+ =
81. j s[kk x y 1a b
+ = bl i zdkj xfr djrh g S fd
2 2 2
1 1 1a b 2c
+ = gS] tgk¡ a, b, c Î R0 rFkk c vpj gks]
rks nh xbZ js[kk ij ewy fcUnq ls [khaps x, yEc ds ikn dk
fcUnqiFk gksxk -
(1) x2 + y2 = c2 (2) x2 + y2 = 2c2
(3) 2
2 2 cx y2
+ = (4) x2 + y2 = 4c2
82. f=foeh; lef"V esa] ƒ(x, y, z) = xy + xz gks] rks mulHkh fcUnqvksa dk fcUnqiFk tks lehdj.k ƒ(x, y, z) = 0dks lUrq"V djrs gSa] gksxk -(1) yEcor~ js[kkvksa dk ; qXe(2) js[kk rFkk lery dk ; qXe] tks ,d nwljs ds lekUrj gSA(3) yEcor~ leryksa dk ;qXe(4) js[kk rFkk lery dk ; qXe] tks ,d nwljs ds yEcor~ gSA
83. xzg M dk blds lw;Z S ds pkjksa vksj ifjØek iFk nh?kZo`Ùkh;gS ftldh ,d ukfHk ij bldk lw;Z fLFkr gSA tc M, Sds lehi gS] rks ;g nks bdkbZ nwjh ij gS] tc M, S lsvfèkdre nwjh ij gS] rks ;g 18 bdkbZ nwjh ij gS] rks xzgM ds }kjk lw;Z ds pkjksa vksj xfr dk lehdj.k] ;g ekursgq, fd S funsZ'kh lery ds dsUæ ij fLFkr gS rFkk nwljhukfHk ½.kkRed y v{k ij fLFkr gS] gksxh
(1) 2 2x (y 8) 1
36 100-
+ = (2) 2 2x (y 8) 1
36 100+
+ =
(3) 2 2x (y 8) 1
64 100-
+ = (4) 2 2x (y 8) 1
64 100+
+ =
Leader Course/Phase-ELS, ELC, ELD & ELP/Score-I/07-03-2015
33/35Kota/01CT314063
SPACE FOR ROUGH WORK
84. Maximum number of points on parabolay2 = 16x which are equidistant from a variablepoint P (which lie inside the parabola), is -(1) 2 (2) 3(3) 4 (4) more than 4
85. Let in equilateral DABC,A(–1 + acosq, 2 + asinq),B(–1 + acosa, 2 – asina),C(–1 + asinb, 2 + acosb)
and length of median through vertex A is 2b,then equation of circumcircle of triangle ABCis (where a is consant) -
(1) x2 + y2 + 18x – 36y + 5 – b2 = 0(2) 9x2 + 9y2 + 18x – 36y + 45 – 16b2 = 0
(3) 9x2 + 9y2 + 18x – 36y + 45 – 4b2 = 0
(4) 9x2 + 9y2 – 18x + 36y + 45 – 4b2 = 086. If x + by + c = 0 is normal to parabola y2 = 12x,
then complete set of all values of c is -(1) (–¥, –6) (2) (9, ¥)(3) ( , 6) (9, )-¥ - È ¥ (4) ( , ) {0}-¥ ¥ -
87. Let PQ and RS be the tangent at the extremitiesof the diameter PR of a circle of radius r. If PSand RQ intersect at a point X on thecircumference of the circle, then (PQ.RS) isequal to(1) (PX).(RX) (2) (QX).(SX)
(3) (PX)2 + (RX)2 (4) (QX)2 + (SX)2
84. ijoy; y2 = 16x ij fLFkr vf/kdre fcUnqvksa dh la[;k]tks pj fcUnq P (tks ijoy; ds vUnj dh vksj fLFkr gS) lsleku nwjh ij fLFkr gS] gksxh -(1) 2 (2) 3(3) 4 (4) 4 ls vf/kd
85. ekuk leckgq f=Hkqt ABC esa]A(–1 + acosq, 2 + asinq),B(–1 + acosa, 2 – asina),C(–1 + asinb, 2 + acosb)
rFkk 'kh"kZ A ls xqtjus okyh ekf/;dk dh yEckbZ 2bgks] rks f=Hkqt ABC ds ifjo`Ùk dk lehdj.k gksxk(tgk¡ a vpj gS) -(1) x2 + y2 + 18x – 36y + 5 – b2 = 0(2) 9x2 + 9y2 + 18x – 36y + 45 – 16b2 = 0(3) 9x2 + 9y2 + 18x – 36y + 45 – 4b2 = 0(4) 9x2 + 9y2 – 18x + 36y + 45 – 4b2 = 0
86. ;fn x + by + c = 0, ijoy; y2 = 12x dk vfHkyEcgks] rks c ds lHkh ekuksa dk iw.kZ leqPp; gksxk -(1) (–¥, –6) (2) (9, ¥)(3) ( , 6) (9, )-¥ - È ¥ (4) ( , ) {0}-¥ ¥ -
87. ekuk PQ rFkk RS, r f=T;k ds o`Ùk ds O;kl PR ds fljksaaij [khaph xbZ Li'kZ js[kk;sa gSA ;fn PS rFkk RQ, o`Ùk dh
ifjf/k ij fLFkr fcUnq X ij izfrPNsn gksrh gS] rks (PQ.RS)cjkcj gksxk -
(1) (PX).(RX) (2) (QX).(SX)
(3) (PX)2 + (RX)2 (4) (QX)2 + (SX)2
Kota/01CT31406334/35
Target : JEE(Main) 2015/07-03-2015
SPACE FOR ROUGH WORK
88. The angle of elevation of the top of a mobiletower from three points P, Q and R (on a straightline through the foot of the tower) are a, b andg respectively. If all three points are lie on sameside of foot of tower and a : b : g = 1 : 2 : 3 andPQ = l, then height of tower is -(1) ltana (2) lsinb(3) lsing (4) ltan(a + b + g)
89. Let ƒ(x) = max{sin–1x, cos–1x}, then areabounded by x = –1, x = 1, y = ƒ(x) and y = 0is -
(1) 3 22p
- (2) 22 2
p+
(3) 22 2
p p+ (4) None of these
90. Let h(x) = x
0
g(t)dtò , where g(x) is a
differentiable and odd function " x Î R andg(x) is periodic with period 3.Statement 1 : h(x) + h(–x) = 0 " xÎR
Statement 2 : h(x) + h(–x) = x
0
2 g(t)dtò " xÎR
Statement 3 : h(3n) = 0 " nÎ Ithen which of the following statement(s) is/aretrue ?(1) Statement 1 & Statement 3(2) Statement 2 & Statement 3(3) Only Statement 1(4) Only Statement 2
88. rhu fcUnqvksa P, Q rFkk R (ehukj ds ikn ls xqtjus okyhljy js[kk ij) dk eksckbZy ehukj ds 'kh"kZ ls mUu;u
dks.k Øe'k% a, b rFkk g gSA ;fn lHkh rhu fcUnq ehukj
ds ikn dh leku fn'kk esa gks rFkk a : b : g = 1 : 2 : 3,oa PQ = l gks] rks ehukj dh m¡GpkbZ gksxh -
(1) ltana (2) lsinb
(3) lsing (4) ltan(a + b + g)89. ekuk ƒ(x) = vf/kdre{sin–1x, cos–1x} gks] rks x = –1,
x = 1, y = ƒ(x) rFkk y = 0 }kjk ifjc¼ {ks=Qy gksxk -
(1) 3 22p
- (2) 22 2
p+
(3) 22 2
p p+ (4) buesa ls dksbZ ugha
90. ekuk h(x) =x
0
g(t)dtò , tgk¡ g(x) lHkh x Î R ds
fy, vodyuh; rFkk fo"ke Qyu gS ,oa g(x) dk
vkorZdky 3 gSAdFku 1 : h(x) + h(–x) = 0 " xÎR
dFku 2 : h(x) + h(–x) = x
0
2 g(t)dtò " xÎR
dFku 3 : h(3n) = 0 " nÎ Irc fuEu esa ls dkSulk@dkSuls dFku lgh gksxk@gksaxs\(1) dFku 1 rFkk dFku 3(2) dFku 2 rFkk dFku 3(3) dsoy dFku 1(4) dsoy dFku 2