CAREER INSTITUTE Pat S KOTA (RAJASTHAN) CLASSROOM …€¦ · N candidat i all t carr an textu...

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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

PAPER CODE

SCORE – II DATE : 19 - 02 - 2015

0 1 C T 2 1 4 0 8 7

TARGET : JEE (Main) 2015

Corporate OfficeALLEN CAREER INSTITUTE

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

+91-744-2436001 [email protected]

dlp.allen.ac.in, dsat.allen.ac.in

Full Syllabus 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 Booklet

with 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,A1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys@dkys ckWy ikbaV isu

ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkj

i= ij dgha vkSj u fy[ksaA3. ijh{kk dh vof/k 3 ?k aVs gSA4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 360 gSaA

5. bl ijh{kk iqfLrdk es a rhu Hkkx A, B, C gSa] ftlds izR;sd Hkkx esa

HkkSfrd foKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj lHkh iz'uksa ds vad

leku gSaA izR;sd iz'u ds lgh mÙkj ds fy, 4 (pkj)vad fuèkkZfjr fd;s x;s gSaA6. izR;sd xyr mÙkj ds fy, ml iz'u ds dqy vad dk ,d pkSFkkbZ vad dkVk

tk;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa 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 fcYdqy 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

Note : In case of any correction in the test paper, please mail [email protected] within 2 days along with YourForm No. & Complete Test Details.

uk sV % ;fn bl iz'u i= esa dksbZ Correction gks rks Ïi;k vkidsForm No. ,oa iw.k Z Test Detai ls ds lkFk 2 fnu ds [email protected] ij mail djsaA

IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

(ACADEMIC SESSION 2014-2015)CLASSROOM CONTACT PROGRAMME

JEE (Main) : ENTHUSIAST COURSE

Enthusiast Course/Score-II/19-02-2015

1/35Kota/01CT214087

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

1. A rod P of length 1m is hinged at one end A

and there is a ring attached to the other end.

Another long rod Q is hinged at B and it passes

through the ring. The rod P is rotated about an

axis which is perpendicular to plane in which

both the rods are present and the variation

between the angles q and f are plotted as

shown. The distance between the hinges A and

B is :-

P

qB A

Q

f

60°q

f

(1) 3m (2) 2m

(3) 1m (4) 0.5m

1. yEckbZ 1m okyh NM+ P dks ,d fljs A ls dhydhr

fd;k tkrk gS rFkk blds nwljs fljs ls ,d oy; tqM+h gqbZ

gSA ,d vU; yEch NM+ Q dks fcUnq B ls dhydhr

fd;k tkrk gS rFkk ;g oy; ls gksdj xqtjrh gSA NM+ P

dks ,d v{k ds lkis{k ?kqek;k tkrk gS tks ml ry ds

yEcor~ gS ftlesa nksuksa NM+s fo|eku gSA dks.k q o f ds

e/; ifjorZu fp= esa n'kkZ;k x;k gSA dhyd A o B ds

e/; nwjh gksxh :-

P

qB A

Q

f

60°q

f

(1) 3m (2) 2m

(3) 1m (4) 0.5m

PART A - PHYSICSBEWARE OF NEGATIVE MARKING

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

Kota/01CT2140872/35

Target : JEE (Main) 2015/19-02-2015


2. Two concentric conducting thin spherical shells

of radii a and b (b > a) are given charges Q and

–2Q respectively. The electric field along a line

passing through centre as a function of distance

(r) from centre is given by :-

(1) (2) a b r

E

(3) a b r

E

(4) a b r

E

3. The maximum possible acceleration of a train

moving on a straight track is 10 m/s2 and

maximum possible retardation is 5 m/s2. If

maximum achievable speed of train is 10m/s

then minimum time in which train can complete

a journey of 135m starting from rest and ending

at rest, is :-

(1) 5s (2) 10s (3) 15s (4) 20s

2. f=T;k a rFkk b (b > a) okys nks ladsUæh; pkyd irys

xksykdkj dks'kksa dks Øe'k% Q o –2Q vkos'k fn;s tkrs

gSaA dsUæ ls nwjh r rFkk dsUæ ls xqtjus okyh js[kk ds

vuqfn'k fo|qr {ks= ds e/; vkjs[k gksxk %&

(1) (2) a b r

E

(3) a b r

E

(4) a b r

E

3. ,d lh/ks Vªsd ij xfr'khy Vªsu dk vf/kdre laHkkfor

Roj.k 10 m/s2 rFkk vf/kdre laHkkfor eanu 5 m/s2

gSA ;fn Vªsu dh vf/kdre izkIr dh tk ldus okyh pky

10m/s gks rks ;g fojkekoLFkk ls izkjEHk gksdj 135m dh

nwjh ij tkdj :dus e s a U; wure fdruk le;

ysxh %&(1) 5s (2) 10s (3) 15s (4) 20s


3/35Kota/01CT214087


4. A metal wire PQ slides on parallel metallic railshaving separation 0.25 m, each havingnegligible resistance. There is a 2W resistor and10V battery as shown in figure. There is auniform magnetic field directed into the planeof the paper of magnitude 0.5 T. A force of0.5N to the left is required to keep the wire PQmoving with constant speed to the right. Withwhat speed is the wire PQ moving ? (Neglectself inductance of the loop)

0.25 m

P

Q

10V

2W

(1) 8 m/s (2) 16 m/s(3) 24 m/s (4) 32 m/s

5. A rod of length L is held vertically on a smoothhorizontal surface. The top end of the rod isgiven a gentle push. At a certain instant of time,when the rod makes an angle q with horizontalthe velocity of COM of the rod is v0. Thevelocity of the end of the rod in contact withthe surface at that instant is :(1) v0 cot q (2) v0 cos q

(3) v0 sin q (4) v0 tan q

4. ,d /kkfRod rkj PQ ux.; izfrjks/k okyh rFkk 0.25 m

nwjh ij fLFkr nks lekUrj /kkfRod iVfj;ksa ij xfr djrk

gSA ;gk¡ 2W izfrjks/k o 10V dh cSVjh fp=kuqlkj yxs

gq, gSaA ;gk¡ 0.5 T ifjek.k dk ,d le:i pqEcdh;

{ks= dkxt ds ry esa vanj dh vksj dk;Zjr gSA rkj PQ

dks nka;h vksj fu;r pky ls xfr'khy cuk;s j[kus 0.5N

dk cy cka;h vksj yxkuk iM+rk gSA rkj PQ fdl pky ls

xfr'khy gS\ (ywi dk LoizsjdRo ux.; ekfu;s)

0.25 m

P

Q

10V

2W

(1) 8 m/s (2) 16 m/s

(3) 24 m/s (4) 32 m/s

5. yEckbZ L okyh ,d NM+ dks fpduh {kSfrt lrg ij

ÅèokZ/kj jksddj j[kk x;k gSA NM + ds Åijh fljs ij ,d

gYdk /kDdk fn;k tkrk gSA fdlh {k.k tc NM+ {kSfrt ds

lkFk q dks.k cukrh gS] ml {k.k NM + ds æO;eku dsUæ dk

osx v0 izkIr gksrk gSA bl {k.k lrg ds lkFk lEidZ esa

fLFkr NM + ds fljs dk osx gksxk %&(1) v0 cot q (2) v0 cos q

(3) v0 sin q (4) v0 tan q

Kota/01CT2140874/35

Target : JEE (Main) 2015/19-02-2015


6. In a YDSE setup the intensity due to twocoherent beams differ from each other by 1%.If one of the beam has intensity I, then intensityof minima is :-

(1) I(10–2) (2) ( )2I10

2-

(3) ( )2I10

3-

(4) ( )4I10

4-

7. Correct output for given logic circuit and inputsis :-

CBA

T0 2T0

t

B

2T0

A

(1) t

output

(2)

output

(3) T0 2T0

output

(4) t

output

6. ,d YDSE O;oLFkk esa nks dyklaxr iqatksa ds dkj.k

mRiUu rhozrkvksa esa 1% dk varj gSA ;fn ,d iqat dh

rhozrk I gks rks fufEu"B dh rhozrk gksxh %&

(1) I(10–2) (2) ( )2I10

2-

(3) ( )2I10

3-

(4) ( )4I10

4-

7. fn, x, rkfdZd ifjiFk rFkk fuos'kh ds fy, lgh fuxZr

gksxk %&

CBA

T0 2T0

t

B

2T0

A

(1) t

output

(2)

output

(3) T0 2T0

output

(4) t

output


5/35Kota/01CT214087


8. A plane electromagnetic wave is travelling in

the positive X-axis. At the instant shown

electric field at the extremely narrow dashed

rectangle is in the -ve z direction and its

magnitude is increasing. Which diagram

correctly shows the direction and relative

magnitudes of magnetic field at the edges of

rectangle :-

y

small

x

z

E

(1)

®Ex (2)

®Ex

(3)

®Ex (4)

®Ex

8. ,d lery fo|qr pqEcdh; rjax /kukRed X-v{k esa

xfr'khy gSA iznf'kZr {k.k ij lhekUr ladjs fcUnqdhr

vk;r ij fo|qr {ks= ½.kkRed z fn'kk esa gS rFkk bldk

ifjek.k c<+ jgk gSA fuEu esa ls dkSulk vkjs[k vk;r dh

Hkqtkvksa ij pqEcdh; {ks= dh fn'kk rFkk lkisf{kd ifjek.kksa

dks n'kkZrk gS %&

y

small

x

z

E

(1)

®Ex (2)

®Ex

(3)

®Ex (4)

®Ex

Kota/01CT2140876/35

Target : JEE (Main) 2015/19-02-2015


9. A person lives in a high rise building at a height30m on the bank of a river. Across the river is awell lit tower of height 60m. When person looksthrough a polarizer at an appropriate angle atlight from top of tower reflecting from the riversurface, he finds intensity of light is least. If

refractive index of water is 43

, width of river is :-

60m30m

(1) 120m (2) 80m (3) 60m (4) 40m10. The drawing shows a top view of a frictionless

horizontal surface, where there are twoindentical springs with particles of mass m

1

and m2 attached to them. Each spring has a

spring constant of 1200 N/m. The particles arepulled to the right and then released from thepositions shown in the drawing. How muchtime passes before the particles are again sideby side for the first time if m

1 = 3.0 kg and

m2 = 27 kg ?

Natural length

10cmm1

m2

(1) 40p

sec (2) 20p

sec (3) 403p

sec (4) 10p

sec

9. ,d O;fä unh ds fdukjs ij fLFkr 30m Å¡ph bZekjr esajgrk gSA unh ds nwljh vksj ,d 60m Å¡pkbZ dk izdk'kLrEHk gSA tc ;g O;fä ,d /kzqod dh lgk;rk ls unhdh lrg ls ijkofrZr gksdj bl izdk'k LrEHk ds 'kh"kZ lsvkus okys izdk'k dks ,d mfpr dks.k ij ns[krk gS rks mlsizdk'k dh rhozrk U;wure izkIr gksrh gSA ;fn ty dk

viorZukad 43

gks rks unh dh pkSM+kbZ gksxh %&

60m30m

(1) 120m (2) 80m (3) 60m (4) 40m

10. fp= esa ,d ?k"kZ.kjfgr {kSfrt lrg dk Åij ls ns[kk x;kn`'; n'kkZ;k x;k gS] tgk¡ nks ,d tSlh fLizaxsa m

1 o m

2

æO;eku okys d.kksa ls tqM+h gqbZ gSA izR;sd fLizax dk fLiazxfu;rkad 1200 N/m gSA d.kksa dks nk¡;h vksj f[kapdjfp= esa n'kkZ;h x;h fLFkfr;ksa ls fojkekoLFkk ls NksM+ktkrk gSA ;fn m

1 = 3.0 kg rFkk m

2 = 27 kg gks rks ; s

d.k izFke ckj iqu% ,d gh lh/k esa fdrus le; i'pkr~vk;sxsa ?

Natural length

10cmm1

m2

(1) 40p

sec (2) 20p

sec (3) 403p

sec (4) 10p

sec


7/35Kota/01CT214087


11. A smooth rightangled wedge of mass M is kept

on a smooth horizontal surface as shown. A

mass m is released from top of wedge when m

reaches ground its speed is V. Work done by

normal contact force on m, while it comes down

to ground is :-

(1) 21

mv2

-

mµ = 0

M

µ = 0(2) 21

Mv2

-

(3) 2 21 m v

2 M- (4)

2 21 M v2 m

-

12. A transverse wave is passing through a

stretched string with a speed of 20 m/s. The

tension in the string is 20 N. At a certain point

P on the string, it is observed that energy is

being transferred at a rate of 40 mW at a given

instant. Find the speed of point P.

(1) 40 cm/s (2) 20 cm/s

(3) 2 mm/s (4) 20 mm/s

11. æO;eku M okyk ,d fpduk vk;rkdkj ost fp=kuqlkj

fpduh {kSfrt lrg ij j[kk gqvk gSA ,d æO;eku m dks

ost ds Åij ls fojkekoLFkk ls NksM+k tkrk gS rFkk tc ;g

/kjkry ij igq¡prk gS rks bldh pky V gksrh gSA èkjkry

ij vkus ij m ij vfHkyEc lEidZ cy }kjk fd;k x;k

dk;Z gksxk %&

(1) 21

mv2

-

mµ = 0

M

µ = 0(2) 21

Mv2

-

(3) 2 21 m v

2 M- (4)

2 21 M v2 m

-

12. ,d vuqizLFk rjax fdlh ruh gqbZ jLlh ls 20 m/s dh

pky ls gksdj xqtjrh gSA jLlh esa ruko dk eku 20 N

gSA jLlh ij ,d fcUnq P ij ;g ik;k x;k fd bl {k.k

ij ÅtkZ 40 mW dh nj ls LFkkukUrfjr gksrh gSA fcUnq P

dh pky gksxh %&

(1) 40 cm/s (2) 20 cm/s

(3) 2 mm/s (4) 20 mm/s

Kota/01CT2140878/35

Target : JEE (Main) 2015/19-02-2015


13. In the circuit shown, power factor of circuit is

1 and power factor of box is 35

. Find reading

of ammeter :-(1) 5A

(2) 6A

C = 1mFBox

A W90V, 25 rad/s

(3) 4A

(4) 3A14. A thin plate separates two liquids of coefficients

of viscosity h and 4 h kept between two fixedplates as shown. If plate has to be pulled by

applying minimum force then 2

1

dd is :-

F4h

h d1

d2

(1) 1 (2) 2

(3) 3 (4) 415. Two particles of equal masses are moving with

equal speeds at an angle 60°. de-Brogliewavelength of these particles is l . Findde-Brooglie wavelength of the particles in theframe of centre of mass of particles :-(1) l (2) 2 l(3) 3 l (4) 4 l

13. iznf'kZr ifjiFk esa ifjiFk dk 'kfä xq.kkad 1 rFkk ckWDl

dk 'kfä xq.kkad 35

gSA vehVj dk ikB~;kad gksxk %&

(1) 5A

(2) 6A

C = 1mFBox

A W90V, 25 rad/s

(3) 4A

(4) 3A

14. nks fLFkj IysVksa ds e/; Hkjs gq, h rFkk 4 h ';kurk xq.kkadksaokys nks æoksa dks ,d iryh IysV fp=kuqlkj i`Fkd&i`Fkddjrh gSA ;fn bl IysV dks U;wure cy yxkdj [khapk

tk, rks 2

1

dd dk eku gksxk :-

F4h

h d1

d2

(1) 1 (2) 2

(3) 3 (4) 4

15. leku æO;eku okys nks d.k leku pky ls 60° ds dks.k

ij xfr'khy gSaA bu d.kksa dh Mh&czksXyh rjaxnS/; Z l gSAd.kksa ds æO;eku dsUæ ra= esa bu d.kksa dh Mh&czksXyhrjaxnS/;Z gksxh %&

(1) l (2) 2 l

(3) 3 l (4) 4 l


9/35Kota/01CT214087


16. A container of mass m is pulled by a constantforce in which a second block of same mass mis placed connected to the wall by a mass-lessspring of constant k. Initially the spring is inits natural length. Velocity of the container atthe instant compression in spring is maximumfor the first time :-

m

m

F

(1)1

F2km

p (2) F 12 2km

p

(3) 1

Fkm

p (4) F 12 km

p

17. When the gap is closed without placing anyobject in the screw gauge whose least count is0.005 mm, the 5th division on its circular scalewith the reference line on main scale, and whena small sphere is placed reading on main scaleadvances by 4 divisions, whereas circular scalereading advances by five times to thecorresponding reading when no object wasplaced. There are 200 divisions on the circularscale. The radius of the sphere is :-

(1) 4.10 mm (2) 4.05 mm

(3) 2.10 mm (4) 2.05 mm

16. æO;eku m okys ,d ik= dks fu;r cy ls [khapk tkrkgSA bl CykWd ds vUnj leku æO;eku m okyk ,d vU;CykWd k fu;rkad okyh æO;ekughu fLizax }kjk nhokj lstqM+k gqvk gSA izkjEHk esa fLizax viuh izkdfrd yEckbZ esagSA fLiazx esa izFke ckj laihM+u vf/kdre gksus ij ik= dkosx gksxk %&

m

m

F

(1)1

F2km

p (2) F 12 2km

p

(3) 1

Fkm

p (4) F 12 km

p

17. tc 0.005 mm vYirekad okys fdlh LØwxst ds tcM+ksa

ds e/; dksbZ oLrq ugha j[kh tkrh rks blds o`Ùkkdkj iSekus

dk ik¡pok Hkkx eq[; iSekus ij lanHkZ js[kk ds lEikrh

gksrk gS rFkk tc ,d NksVs xksys dks tcM+ksa ds eè; j[krs gSa

rks eq[; iSekus dk ikB~;kad pkj Hkkx vkxs c<+ tkrk gS

tcfd o`Ùkkdkj iSekus dk ikB~;kad ik¡p Hkkx vkxs c<+

tkrk gSA ;fn o`Ùkkdkj iSekus ij 200 Hkkx cus gksa rks xksys

dh f=T;k gksxh %&

(1) 4.10 mm (2) 4.05 mm

(3) 2.10 mm (4) 2.05 mm

Kota/01CT21408710/35

Target : JEE (Main) 2015/19-02-2015


18. In the figure shown, amount of heat suppliedto one mole of an ideal gas is plotted on thehorizontal axis and amount of work done bygas is drawn on vertical axis. Assuming processbe isobaric i.e. gas can be :-

Q(J)80604020

10203032

W(J)

(1) He (2) CO (3) CO2 (4) NH3

19. Point charges are fixed at the hour marks of awall clock as shown. If net dipole moment ofthe system is along the direction of hour handthen time shown by clock is :-

(1) 12'O' Clock

(2) 3'O' Clock

(3) 6'O' Clock

(4) 9'O' Clock

20. A household refrigerator with a coefficientof performance 1.2 removes heat from therefrigerated space at the rate of 60kJ/min.What would be cost of running this fridgefor one month (30 days) (assuming each dayit is used for 4 hours and cost of one electricalunit is 6Rs.) :-(1) 180 Rs. (2) 300 Rs.(3) 480Rs. (4) 600 Rs.

18. iznf'kZr vkjs[k esa ,d eksy vkn'kZ xSl dks nh xbZ Å"ekdh ek=k dks {kSfrt v{k ij rFkk xSl }kjk fd;s x;s dk;Zdh ek=k dks Å/okZ/kj v{k ij fy;k x;k gSA izØe dkslenkch; ekuus ij ;g xSl gks ldrh gS %&

Q(J)80604020

10203032

W(J)

(1) He (2) CO (3) CO2 (4) NH3

19. fdlh nhokj ?kM+h ij vafdr ?k.Vksa dks n'kkZus okys fpUg~ksa

ij fcUnq vkos'k fp=kuqlkj j[ks x;s gSaA ;fn bl fudk;

dk dqy f}/kzqo vk?kw.kZ ?k.Vs okyh lqbZ dh fn'kk ds vuqfn'k

gks rks ?kM+h }kjk n'kkZ;k x;k le; gksxk %&(1) 12'O' Clock

(2) 3'O' Clock

(3) 6'O' Clock

(4) 9'O' Clock

20. ,d ?kjsyw jsfÝtjsVj dk n{krk xq.kkad 1.2 gS rFkk ;g60kJ/min dh nj ls Å"ek gVkrk gSA ;fn bls izR;sdfnu pkj ?k.Vs pyk;k tk, rFkk ,d fo|qr bdkbZ dkewY; 6 :i;s gks rks bl fÝt dks 30 fnuksa rd pykus dsfy, fdruk [kpZ ogu djuk gksxk %&

(1) 180 Rs. (2) 300 Rs.

(3) 480Rs. (4) 600 Rs.


11/35Kota/01CT214087


21. A meter bridge set up as shown to determine

end correction at A and B. When a resistance

of 15W is used in left gap and of 20W in right

gap, then null point comes at a distance 42cm

from A. When these resistances are

interchanged null point comes at a distance

57cm from A. Values of end corrections are :-

G

A B

(1) 1 cm, 2 cm (2) 2 cm, 3 cm

(3) 3 cm, 4 cm (4) 3 cm, 2 cm

22. The total intensity of earth's magnetic field at

the magnetic equator is 5 units. Its value at a

magnetic latitude of 37° is equal to :-

(1) 73 units (2) 52 units


21. fp= esa A rFkk B ij vUR; la'kks/ku Kkr djus ds fy,

ehVj lsrq O;oLFkk n'kkZ;h xbZ gSA tc ck¡;s fjä&LFkku esa

15W izfrjks/k rFkk nk¡;s fjä&LFkku esa 20W izfrjks/k iz;qä

fd;k tkrk gS rks larqyu fcUnq A ls 42cm dh nwjh ij

izkIr gksrk gSA bu izfrjks/kksa dks vkil esa varZifjofrZr

djus ij lUrqyu fcUnq A ls 57cm dh nwjh ij izkIr gksrk

gSA vUR; la'kks/ku ds eku gS %&

G

A B

(1) 1 cm, 2 cm (2) 2 cm, 3 cm

(3) 3 cm, 4 cm (4) 3 cm, 2 cm

22. pqEcdh; fo"kqor~ ij i`Foh ds pqEcdh; {ks= dh dqy

rhozrk 5 bdkbZ gSA bldk eku 37° pqEcdh; v{kka'k ij

gksxk %&



Kota/01CT21408712/35

Target : JEE (Main) 2015/19-02-2015


23. A hollow sphere of radius R is filled completelywith an ideal liquid of density r. sphere ismoving horizontally with an acceleration 2g,where g is acceleration due to gravity in thespace. If minimum pressure of liquid is P0, thenpressure at the centre of sphere is :-

2g

(1) P0 + rgR (2) P0 + rg R 2

(3) P0 + rg R 5 (4) 0gR

P5

r+

24. In AM transmission, band width of

transmission is 10 kHZ and carrier frequency

is 100 kHZ, then side band frequencies are

given by :-

(1) 90 kHZ & 110 kHZ

(2) 95 kHZ, 105 kHZ

(3) 100 kHZ, 110 kHZ

(4) 90 kHZ, 100 kHZ

23. f=T;k R okys ,d [kks[kys xksys dks r ?kuRo ds vkn'kZ

æo ls iw.kZr;k Hkj fn;k tkrk gSA xksyk 2g Roj.k ds lkFk

{kSfrt :i ls xfr'khy gS] tgk¡ g xq:Roh; Roj.k gSA ;fn

æo dk U;wure nkc P0 gks rks xksys ds dsUæ ij nkc

gksxk %&

2g

(1) P0 + rgR (2) P0 + rg R 2

(3) P0 + rg R 5 (4) 0gR

P5

r+

24. fdlh AM lEiz s"k.k esa lEiz s"k.k dh cS.M pkSM+kbZ

10 kHZ rFkk okgd vko`fÙk 100 kHZ gSA rc ik'oZ

cS.M vkofÙk;k¡ gksxh :-

(1) 90 kHZ & 110 kHZ

(2) 95 kHZ, 105 kHZ

(3) 100 kHZ, 110 kHZ

(4) 90 kHZ, 100 kHZ


13/35Kota/01CT214087


25. A sample of paramagnetic salt contains

20 × 1024 atomic dipoles each of dipole moment

1.5 × 10–23J T –1. The sample is placed under a

homogeneous magnetic field of 0.84 T, and

cooled to a temperature of 4.2 K. The degree

of magnetic saturation achieved is equal to

15%. What is the total dipole moment

(approximate) of the sample for a magnetic

field of 0.98 T and a temperature of 2.8 K?

(Assume Curie's law).

(1) 7.9 J T –1 (2) 52.5 J T –1

(3) 30 J T –1 (4) 4.6 J T –1

26. Figure shows the variation of the gravitatioalacceleration ag of four planets with the radialdistance r from the centre of the planet forr ³ radius of the planet. Plots 1 and 2 coincidefor r ³ R2 and plots 3 and 4 coincide for r ³ R4.The sequence of the planets in the descendingorder of their densities is :

43

21

R1 R2 R4

R3

ag

rO

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

25. fdlh vuqpqEcdh; yo.k ds izfrn'kZ esa 20 × 1024

ijekf.od f}/kzqo gSa] ftuesa ls izR;sd dk f}/kzqo vk?kw.kZ

1.5 × 10–23J T –1 gSA bl izfrn'kZ dks 0.84 T ds le:i

pqEcdh; {ks= esa j[kdj 4.2 K rkieku rd B.Mk fd;k

tkrk gSA izkIr dh xbZ pqEcdh; larIrk dh dksfV 15%

gSA pqEcdh; {k s= 0.98 T rFk k r kieku

2.8 K ds fy, bl izfrn'kZ dk dqy f}/kz qo vk?kw.kZ

(yxHkx) D;k gksxk ? (D;wjh dk fu;e yxk;s).

(1) 7.9 J T –1 (2) 52.5 J T –1

(3) 30 J T –1 (4) 4.6 J T –1

26. fp= esa pkj xzgksa ds xq:Roh; Roj.k ag esa xzg ds dsUæ ls

f=T;h; nwjh r ds lkFk ifjorZu dks n'kkZ;k x;k gS] tgk¡

r ³ xzg dh f=T;k gSA vkjs[k 1 rFkk 2, r ³ R2 ds fy,

rFkk 3 o 4, r ³ R4 ds fy, lEikrh gSaA bu xzgksa dks buds

?kuRo ds ?kVrs gq, Øe esa O;ofLFkr dhft, %&

43

21

R1 R2 R4

R3

ag

rO

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

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

Kota/01CT21408714/35

Target : JEE (Main) 2015/19-02-2015


27. In a city sound intensity level increases

uniformly by about 1dB annually. How many

years will it take for intensity of sound to

become double :-

(1) 1 year

(2) 2 year

(3) 3 year

(4) 4 year

28. A radioactive sample decays by two modes by

a decay and by b-decay. 66.6% of times it

decays by a-decay and 33.3% of times, it

decays by b-decay. If half life of sample is 60

years then what will be half life of sample, if it

decays only by a - decay.

(1) 30 years

(2) 90 years

(3) 120 years

(4) 180 years

27. fdlh 'kgj esa /ofu rhozrk dk Lrj izfr o"kZ ,dleku

:i ls yxHkx 1dB c<+ tkrk gSA /ofu dh rhozrk dks

nqxuk gksus esa fdruk le; yxsxk %&

(1) 1 o"kZ

(2) 2 o"kZ

(3) 3 o"kZ

(4) 4 o"kZ

28. ,d jsfM;kslfØ; izfrn'kZ nks fo/kkvksa a fo?kVu rFkk

b-fo?kVu }kjk {kf;r gksrk gSA bldk 66.6% Hkkx

a-fo?kVu }kjk rFkk 33.3% Hkkx b-fo?kVu }kjk fo?kfVr

gksrk gSA ;fn izfrn'kZ dh v/kZvk;q 60 o"kZ gks rks izfrn'kZ

dh v/kZ&vk;q D;k gksxh ;fn ;g dsoy a -fo?kVu }kjk

gh fo?kfVr gksrk gks %&

(1) 30 o"kZ

(2) 90 o"kZ

(3) 120 o"kZ

(4) 180 o"kZ


15/35Kota/01CT214087


29. One mole of an ideal gas undergoes a cyclicprocess, consisting of two isochores and twoisobars. Temperature at 1 and 3 equal to T

1 and

T3 respectively. The work done by the gas over

the cycle, if the point 2 and 4 lie on the sameisotherm :-

2

1 4

3

V

P

(1) 1 3R(T T )2+

(2) 23 1R( T T )-

(3) 21 3

R( T T )

2+ (4) 1 2R T T

30. Two steel wires of same length but radii r and

2r are connected together end to end and tied

to a wall as shown. The force stretches the

combination by 10 mm. How far does the

midpoint A move :-

Radius= r Radius= 2 r

LL

A

(1) 2 mm (2) 4 mm

(3) 6 mm (4) 8 mm

29. ,d eksy vkn'kZ xSl ,sls pØh; izØe ls gksdj xqtjrh gS

tks nks levk;rfud rFkk nks lenkch; js[kkvksa ls feydj

cuk gSA 1 rFkk 3 ij rkieku Øe'k% T1 o T

3 gSA ;fn

fcUnq 2 rFkk 4 ,d gh lerkih; js[kk ij gks rks lEiw.kZ

pØ esa xSl }kjk fd;k x;k dk;Z gksxk %&

2

1 4

3

V

P

(1) 1 3R(T T )2+

(2) 23 1R( T T )-

(3) 21 3

R( T T )

2+ (4) 1 2R T T

30. leku yEckbZ rFkk r o 2 r f=T;kvksa okys LVhy ds nks

rkjksa dks fljs ls fljk tksM+dj fp=kuqlkj ,d nhokj ls

cka/k fn;k tkrk gSA bl la;kstu dks cy }kjk 10 mm

rd f[kapk tkrk gSA e/; fcUnq A fdruh nwjh rd

foLFkkfir gksxk %&

Radius= r Radius= 2 r

LL

A

(1) 2 mm (2) 4 mm

(3) 6 mm (4) 8 mm

Kota/01CT21408716/35


Target : JEE (Main) 2015/19-02-2015

31. Observe the following graph for the de-Broglie

wavelength of a hypothetical charged particle

(q = 1.6 × 10–19C). Find the mass of the particle

(m)

V –1/2

(volt )–1/2

q= tan (0.5)–1

(h = 6.0 × 10–34 J-s)

(1) 45 × 10–49 kg (2) 45 × 10–45 kg

(3) 4.5 × 10–45 kg (4) 4.5 × 10–45 gm

32. 4.3 gm of an alkane is burnt in sufficient

oxygen. The CO2 formed reacts completely

with 300 ml, 2N NaOH solution producing

Na2CO3. The alkane should be

(1) C3H8 (2) C12H26

(3) C6H14 (4) C2H6

31. ,d dkYifud vkosf'kr d.k (q = 1.6 × 10–19C) dhMh&czksXyh rjaxnS/; Z ds fy, fn;s x;s fuEu xzkQ ij fopkj

dhft,A d.k dk nzO;eku Kkr dhft,A

(m)

V –1/2

(volt )–1/2

q= tan (0.5)–1

(h = 6.0 × 10–34 J-s)

(1) 45 × 10–49 kg (2) 45 × 10–45 kg

(3) 4.5 × 10–45 kg (4) 4.5 × 10–45 gm

32. 4.3 gm ,sYdsu dks i;kZIr vkWDlhtu esa tyk;k x;kAfufeZr CO2 ] 300 ml, 2N NaOH foy;u ds lkFkiw.kZ vfHkfØ;k djds Na2CO3 cukrh gSA , sYdsu gksuhpkgh,

(1) C3H8 (2) C12H26

(3) C6H14 (4) C2H6

PART B - CHEMISTRY


17/35Kota/01CT214087


33. Z vs P graph of four gases is given below at

same temperature

DC

AB

z = 1

Z

PWhich of the gases will have the highest valueof critical temperature.(1) D (2) C(3) B (4) A

34. For the reaction :

NH3(g) 12 N2(g) +

32 H2(g) ; KP

The degree of dissociation (a) of NH3 is relatedto total equilibrium pressure (Pº) as :

(1)

12

P

3 3 Pº14K

-æ ö

a = +ç ÷ç ÷è ø

(2)

12

P

3 3 Pº14K

æ öa = +ç ÷ç ÷

è ø

(3)

12

0

p

3P14K

æ öa = +ç ÷ç ÷

è ø(4)

12

0

p

3P14K

-æ ö

a = +ç ÷ç ÷è ø

35. What is the pH of solution obtained by mixing

5.076 gm of methyl ammonium nitrate

(CH3NH3NO3) to 120 ml, 0.225 M

methylamine (CH3NH2 ; Kb = 4 × 10–4).

(1) 3.7 (2) 4.3

(3) 10.3 (4) 11

33. pkj xSlksa ds leku rki ij Z vs P oØ fups fn;s x;sgS

DC

AB

z = 1

Z

PfuEu esa ls fdl xSl ds ØkfUrd rki dk eku vf/kdregksxk(1) D (2) C(3) B (4) A

34. vfHkfØ;k :

NH3(g) 12 N2(g) +

32 H2(g) ; KP

ds fy, NH3 ds fo;kstu dh ek=k (a) dqy lkE; nkc(Pº) ls fuEu izdkj lEcfU/kr gS :

(1)

12

P

3 3 Pº14K

-æ ö

a = +ç ÷ç ÷è ø

(2)

12

P

3 3 Pº14K

æ öa = +ç ÷ç ÷

è ø

(3)

12

0

p

3P14K

æ öa = +ç ÷ç ÷

è ø(4)

12

0

p

3P14K

-æ ö

a = +ç ÷ç ÷è ø

35. 5.076 gm esfFky veksfu;e ukbVªsV (CH3NH3NO3)

dk s 120 ml ] 0.225 M e sfFky , s ehu

(CH3NH2 ; Kb = 4 × 10–4) esa fefJr dj cuk;s x;s

foy;u dh pH D;k gksxh

(1) 3.7 (2) 4.3

(3) 10.3 (4) 11

Kota/01CT21408718/35


Target : JEE (Main) 2015/19-02-2015

36. For the 2nd order reaction ; R ¾® P, the

following graph holds true.

1[R]

(L mol )–1

2

0 Time (min)

= tan (0.5)–1

Identify the incorrect statement.

(1) The half life (t1/2) of reaction is 4 min

(2) Initial concentration of R is 2 M

(3) Initial concentration of R is 0.5 M

(4) Time required for 75% completion

of reaction is 0.2 hours

37. PbCl2 is dissolved in water to make its saturatedsolution. What will be the freezing point of thissolution.

Given : Kf (H2O) = 2 K kg mole–1,

Ksp (PbCl2) = 4 × 10–6

(Assume molarity to be equal to molality)

(1) –0.04ºC (2) –0.06ºC

(3) –0.02ºC (4) –0.6ºC

36. 2nd dksfV vfHkfØ;k ; R ¾® P, ds fy, fuEu oØlR; ik;k x;k

1[R]

(L mol )–1

2

0 Time (min)

= tan (0.5)–1

xyr dFku dk p;u dhft,

(1) vfHkfØ;k dh v/kZ vk;q (t1/2) = 4 min gS

(2) R dh izkjEHkhd lkUnzrk 2 M gS

(3) R dh izkjEHkhd lkUnzrk 0.5 M gS

(4) vfHkfØ;k dks 75% iw.kZ gksus ds fy, vko';dle; 0.2 ?k.Vs gS

37. PbCl2 dks ty esa ?kksydj bldk larIr foy;u cuk;kx;k A bl foy;u dk fgekad fcUnq D;k gksxk

fn;k gS : Kf (H2O) = 2 K kg mole–1,

Ksp (PbCl2) = 4 × 10–6

(eku fyft;s eksyjrk] eksyyrk ds leku gS)

(1) –0.04ºC (2) –0.06ºC

(3) –0.02ºC (4) –0.6ºC


19/35Kota/01CT214087


38. A current of 0.01 amp. is passed through asolution of a metal bromide (M Brx). The onlyreaction at the cathode is the deposition of metal(M). After 3 hours, 0.072 gm of metal (M) isdeposited. The value of 'x' is approximately(Given : Atomic wt. of M = 192)(1) 1 (2) 2

(3) 3 (4) 4

39. Which of the following does not changes the

density of SiC solid.

(1) By substitution of some Si atoms by

some carbon atoms

(2) Vacancy defect

(3) By substitution of "Si" or "C" by "Ge"

(4) By interchanging position of Si and

C-atoms

40. Which of the following statement is incorrect

when a system consisting of ideal gas

undergoes change of state.

(1) In reversible change, DSuniv. = 0

(2) In an irreversible change, DSuniv. ¹ 0

(3) In an adiabatic process always DSsyst.= 0

(4) In an isochoric process,

DSsyst. = nCv ln2

1

TT

æ öç ÷è ø

38. 0.01 amp. dh /kkjk dks /kkrq czksekbM (M Brx) ds ,dfoy;u ls izokghr fd;k x;kA vfHkfØ;k esa dsoy dSFkksMij /kkrq (M) ,df=r gksrh gSA 3 ?k.Vk s ds i'pkr~0.072 gm /kkrq (M) ,df=r gksrh gSA 'x' dk ekuyxHkx gS (fn;k gS : M dk ijek.kq Hkkj = 192)

(1) 1 (2) 2

(3) 3 (4) 4

39. fuEu esa ls dkSulk dkjd SiC Bksl dk ?kuRo ifjofrZrugha djrk gS(1) dqN Si ijek.kqvks a dks dqN C ijek.kqvksa }kjk

izfrLFkkiu ls(2) fjfDrdk =qfV(3) "Ge" ds }kjk "Si" ;k "C" ds i zfrLFkkiu

ls(4) Si rFkk C-ijek.kqvks a dh fLFkfr;ks a esa vkilh

ifjorZu ls

40. fuEu esa ls dkSulk dFku xyr gS tc ,d ra= esa mifLFkr

vkn'kZ xSl dk voLFkk ifjorZu gksrk gS

(1) mRØe.kh; ifjorZu eas, DSuniv. = 0

(2) vuqRØe.kh; ifjorZu eas, DSuniv. ¹ 0

(3) ,d :¼ks"eh; izØe eas lnSo , DSsyst. = 0

(4) ,d levk;rfud izØe esa

DSsyst. = nCv ln2

1

TT

æ öç ÷è ø

Kota/01CT21408720/35


Target : JEE (Main) 2015/19-02-2015

41. The reagent used for toning in photographyis :

(1) AgNO3

(2) CuSO4

(3) HgCl2

(4) AuCl3

42. Which of the following pair of complex isco-ordination isomers :

(1) [Co(NH3)

5Br]SO

4 & [Co(NH

3)

5SO

4]Br

(2) [Pt(NH3)

4] [PtCl

4] & [Pt(NH

3)

2Cl

2]

[Pt(NH3)

2Cl

2]

(3) [Co(H2O)

6]Cl

3 & [Co(H

2O)

5Cl]Cl.2H

2O

(4) None of these

43. Which of the following is INCORRECTstatement :

(1) Above 373 K, Na2CO

3.10H

2O converted

into anhydrous form

(2) Hemihydrate of CaSO4 is known as

Gypsum

(3) NaHCO3 act as weak antiseptic for

skin infections

(4) All of these

44. Which of the following molecule has lowestbond angle :

(1) PH3

(2) H2S (3) H

2O (4) SbH

3

45. Which of the following mineral does notcontain iron :

(1) Copper pyrite (2) Iron pyrite

(3) Illemenite (4) Tincal

41. QksVksxzkWQh esa Vksfuax ds fy, iz;ksx fd;k tkus okykvfHkdeZd gS :

(1) AgNO3

(2) CuSO4

(3) HgCl2

(4) AuCl3

42. fuEu fdl ; qXe ds ladqy ;kSfxd milgla;kstu leko;ohgS :(1) [Co(NH

3)

5Br]SO

4 rFkk [Co(NH

3)

5SO

4]Br

(2) [Pt(NH3)

4] [PtCl

4] rFkk [Pt(NH

3)

2Cl

2]

[Pt(NH3)

2Cl

2]

(3) [Co(H2O)

6]Cl

3 rFkk [Co(H

2O)

5Cl]Cl.2H

2O

(4) buesa ls dksbZ ugha43. fuEu esa ls dkSulk dFku xyr gS :

(1) 373 K ls vf/kd ij Na2CO

3.10H

2O, futZyh;

:i esa ifjofrZr gks tkrk gS

(2) CaSO4 ds gsehgkbMªsV dks "ftIle" ds uke ls tkuk

tkrk gS

(3) peZ laØe.k ds fy, NaHCO3 dk iz;ksx nqcZy

,UVhlsfIVd ds :i esa djrs gS

(4) mijksDr lHkh

44. fuEu esa ls dkSuls v.kq esa U;wure ca/k dks.k mifLFkrgS :(1) PH

3(2) H

2S (3) H

2O (4) SbH

3

45. fuEu esa ls dkSuls [kfut es a vk;ju mifLFkr ughagS :(1) dkWij ik;jkbV (2) vk;ju ik;jkbV(3) bysfeukbV (4) fVady


21/35Kota/01CT214087


46. 'X'Simple

Salt

Flame test

BaCl solution2'Y'White ppt.(insoluble in dil. HCl)

Golden yellow flame

When Y is treated with KMnO4 then final ppt

is of -

(1) BaSO4

(2) MnSO4

(3) K2SO

4(4) None of these

47. Which of the following process does notproduce ammonia gas

(1) NH4Cl KOH¾¾¾®

(2) NH4Cl D¾¾®

(3) NH4NO

2 D¾¾®

(4) NH4NO

2 Zn / NaOH¾¾¾¾®

48. Which of the following order is incorrect :

(1) H2 < D

2 < T

2(Number of protons)

(2) H2 < D

2 < T

2(Bond energy)

(3) H2 < D

2 < T

2(Boiling point)

(4) H2 < D

2 < T

2(No. of neutrons)

49. Which of the following planar complex isdiamagnetic and show stereo isomerism :

(1) [Ni(gly)2] (2) [Pt(NH

3)

3Cl]+

(3) [Co(H2O)

3Cl

3] (4) [Co(en)

3]3+

46. 'X'lkekU;yo.k

Tokyk ijh{k.k

BaCl 2 foy;u 'Y''osr vo{ksi( HCl )ruq esa vfoys; gS

lqugjh ihyh Tokyk

tc Y dks KMnO4 ds lkFk mipkfjr fd;k tkrk gS rks

vfUre vo{ksi gksxk :(1) BaSO

4 dk (2) MnSO

4 dk

(3) K2SO

4 dk (4) buesa ls dksbZ ugha

47. fuEu esa ls dk Suls izØe es a vek sfu;k ugh a curhgSa

(1) NH4Cl KOH¾¾¾®

(2) NH4Cl D¾¾®

(3) NH4NO

2 D¾¾®

(4) NH4NO

2 Zn / NaOH¾¾¾¾®

48. fuEu esa ls dkSulk dFku xyr gS :

(1) H2 < D

2 < T

2(izksVksuksa dh la[;k)

(2) H2 < D

2 < T

2(ca/k ÅtkZ)

(3) H2 < D

2 < T

2(DoFkukad)

(4) H2 < D

2 < T

2(U;wVªkWuksa dh la[;k)

49. fuEu esa ls dkSulk leryh; ladqy izfrpqEcdh; gS rFkkf=foe leko;ork iznf'kZr djrk gS :

(1) [Ni(gly)2] (2) [Pt(NH

3)

3Cl]+

(3) [Co(H2O)

3Cl

3] (4) [Co(en)

3]3+

Kota/01CT21408722/35


Target : JEE (Main) 2015/19-02-2015

50. Which of the following statement is incorrect:

(1) MnO2 act as catalyst for decomposition of

KClO3 to O

2

(2) Conversion of SO2 to SO

3 is exothermic

reaction in contact process

(3) White phosphorous having garlic odour and

poisonous in nature

(4) g-SO3 having chain structure

51. Which of the following is correct order of acidic

strength for the given compounds

OH

NO2

I

COOH

II

COOH

IIICH3

OH

IVCH3

(1) I > II > III > IV

(2) II > I > III > IV

(3) II > III > I > IV

(4) I > III > II > IV

50. fuEu esa ls dkSulk dFku xyr gS :

(1) KClO3 ds fo?kVu ls O

2 izkIr djus esa] MnO

2 ,d

mRizsjd dh rjg dk;Z djrk gS

(2) lEidZ izØe esa SO2 dk SO

3 esa :ikUrj.k ,d

Å"ek{ksih vfHkfØ;k gS

(3) 'osr QkWLQksjl yglqu tSlh xa/k j[krk gS rFkk fo"kSyh

izd`fr dk gS

(4) g-SO3 , Ja`[kyk lajpuk j[krk gS

51. fn;s x;s ;kSfxd ds fy;s vEyh; lkeF;Z dk fuEu esa ls

dkSulk lgh Øe gS&

OH

NO2

I

COOH

II

COOH

IIICH3

OH

IVCH3

(1) I > II > III > IV

(2) II > I > III > IV

(3) II > III > I > IV

(4) I > III > II > IV


23/35Kota/01CT214087


52. What is the relation between given compounds.

OH

CHOCH3

H

HHO

CHOHH C3

HOH

OH

(1) Enantiomers (2) Diastereomers

(3) Conformers (4) Identical

53. Which of the following statement is incorrect

about a-D-glucopyranose & b-D-glucopyranose

(1) Both are anomers of each other

(2) Both are reducing sugar

(3) Both shows mutarotation

(4) Both can raduce schiff reagent under

normal conditions

54. Which of the following is a copolymer

(1) Natural rubber (2) Polystyrene

(3) PAN (4) Bakelite

52. fn;s x;s ;kSfxdksa ds e/; D;k lEcU/k gS\

OH

CHOCH3

H

HHO

CHOHH C3

HOH

OH

(1) izfrfcEc:i leko;oh (2) foofje leko;oh

(3) la:i.k leko;oh (4) leku

53. a-D-Xywdksik;jsuksl rFkk b-D-Xywdksik;jsuksl ds lUnHkZ

esa fuEu esa ls dkSulk dFku xyr gS&

(1) nksuksa ,d nwljs ds ,suksej gS

(2) nksuksa vipk;h 'kdZjk gS

(3) nksuksa ifjorhZ ?kw.kZu (mutarotation) n'kkZrs gSa

(4) nksuksa lkekU; ifjfLFkfr;ksa esa f'kQ vfHkdeZd dks

vipf;r djrs gS

54. fuEu esa ls dkSulk lgcgqyd gS&

(1) izkd`frd jcj (2) ikWfyLVk;fju

(3) PAN (4) csdsykbV

Kota/01CT21408724/35


Target : JEE (Main) 2015/19-02-2015

55. Out of five carbonyl aromatic isomers of

molecular formula containing C8H8O. One of

the isomer X can give iodoform test while Y

can give Benedict's test. What are the structure

of X & Y respectively

(1)

CHO

CH3

CHO

CH3

(2)

COCH3 CH –CHO2

(3)

COCH3 CHO

CH3

(4)

CHO CHO

CH3

CH3

55. C8H8O v.kqlw= j[kus okys ik¡p dkcksZfuy ,sjkseSfVd

leko;oh;ksa esa ls ,d leko;oh X vk;ksMksQkeZ ijh{k.k

ns ldrk gS tcdh Y csusfMDV ijh{k.k ns ldrk gSA X

rFkk Y dh lajpuk Øe'k% gS&

(1)

CHO

CH3

CHO

CH3

(2)

COCH3 CH –CHO2

(3)

COCH3 CHO

CH3

(4)

CHO CHO

CH3

CH3


25/35Kota/01CT214087


56.Me

Cl

D

H NaIAcetone X aq.KOH Y

(Major) (Major)S 2N

X & Y should be :

(1)Me

ID

HMe

OHD

H

(2) Me

ID

HMe

HD

HO

(3)Me

HD

IMe

OHD

H

(4) Me

H

D

IMe

HD

HO

57. Which of following does not represent correct

major product.

(1) Me3C –Cl NaDry-ether CH =C—CH2 3

CH3

+ CH3–CH–CH3

CH3

(2) MeCOONa(aq.) Electrolysis CH3–CH3

(3) Me3COONa(aq.) Electrolysis

CH = C—CH + CH2 3 4

CH3

(4) Me—Cl NaDry-ether CH3 – CH3

56.

Me

Cl

D

H NaI,slhVksu

X tyh; KOH Y(eq[;) (eq[;)S 2N

X rFkk Y gksus pkfg,s&

(1)Me

ID

HMe

OHD

H

(2)Me

ID

HMe

HD

HO

(3)Me

HD

IMe

OHD

H

(4)Me

H

D

IMe

HD

HO

57. fuEu es ls dkSu lgh eq[; mRikn dks ugha n'kkZrk gS&

(1) Me3C –Cl Na'kq"d bZFkj

CH =C—CH2 3

CH3

+ CH3–CH–CH3

CH3

(2) MeCOONa(tyh;) oS|qr vi?kVu

CH3–CH3

(3) Me3COONa(tyh;) oS|qr vi?kVu

CH = C—CH + CH2 3 4

CH3

(4) Me—Cl Na'kq"d bZFkj

CH3 – CH3

Kota/01CT21408726/35


Target : JEE (Main) 2015/19-02-2015

58. Which of the following can be used to

differentiate pentan-2-one & pentan-3-one

(1) Iodoform test (2) NaHSO3

(3) Both 1 & 2 (4) Tollen's reagent

59. How many aldol products are possible

excluding stereoisomers

CH3 –C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O– H + CH3–CH2–CHO OH/H O2

–

(1) 2 (2) 4

(3) 6 (4) 8

60. Which of the following reaction will not give

aldehyde as major product.

(1) CH3 – CºN H , Pd2BaSO4

H O3+

(2) CH – C – Cl3

O

H , Pd2BaSO4

(3) CH3MgBr + CH3CºNH O3

+

(4) CH3–C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O–Cl

(i) DIBAL-H(ii) H O3

+

58. fuEu es a l s fdldk mi;k sx i sUV su -2-vk Wu rFk k

isUVsu-3-vkWu dks foHksfnr djus ds fy;s fd;k tk ldrk

gS&

(1) vk;ksMksQkeZ ijh{k.k (2) NaHSO3

(3) 1 rFkk 2 nksuksa (4) VkWysUl vfHkdeZd

59. f=foe~ leko;fo;ksa dks NksMdj fdrus ,sYMksy mRiknlEHko gS&

CH3–C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O–H + CH3–CH2–CHO OH/H O2

–

(1) 2 (2) 4

(3) 6 (4) 8

60. fuEu esa ls dkSulh vfHkfØ;k ,sfYMgkbM ds :i esa eq[;mRikn ugha nsxh&

(1) CH3 – CºN H , Pd2BaSO4

H O3+

(2) CH – C – Cl3

O

H , Pd2BaSO4

(3) CH3MgBr + CH3CºNH O3

+

(4) CH3–C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O–Cl

(i) DIBAL-H(ii) H O3

+


27/35Kota/01CT214087


61. The geometric and harmonic means of two

numbers x1 and x2 are 18 and 816

13 respectively.

The value of |x1 – x2| is

(1) 5 (2) 10 (3) 15 (4) 20

62. The coefficient of x9 in the polynomial given

by 11

r 1

(x r)(x r 1)(x r 2)...(x r 9)=

+ + + + + + +å is -

(1) 5511 (2) 5151 (3) 1515 (4) 115563. ar and br (ar < br) are the roots of

x2 – r2(r + 1)x + r5 = 0. The value ofn

r rr 1

(3 2 )=

a + bå is -

(1) 21 n(n 1)(n 3n 1)

2+ + +

(2) 21 n(n 1)(3n n 1)

2+ + +

(3)23 n(n 1)(n n 1)

2+ + +

(4) 21 n(n 1)(n n 3)

2+ + +

64. The solutions of equation in z,2| z | (z z) i(z z) 2 0- + + - + = are (i 1)= -

(1) 2 + i, 1 – i (2) 1 + i, 1 – i

(3)1 + 2i, –1 – i (4) 1 + i, 1 + i

61. nks la[;kvksa x1 rFkk x2 dk xq.kksÙkj ek/; rFkk gjkRed

ek/; Øe'k% 18 rFkk 81613

gSA rc |x1 – x2| dk eku

gksxk -(1) 5 (2) 10 (3) 15 (4) 20

62.11

r 1

(x r)(x r 1)(x r 2)......(x r 9)=

+ + + + + + +å}kjk fn, x, cgqin esa x9 dk xq.kkad gksxk -

(1) 5511 (2) 5151 (3) 1515 (4) 1155

63. ar rFkk br (ar < br) lehdj.k x2 – r2(r + 1)x + r5 = 0

ds ewy gSA rc n

r rr 1

(3 2 )=

a + bå dk eku gksxk -

(1) 21 n(n 1)(n 3n 1)

2+ + +

(2) 21 n(n 1)(3n n 1)

2+ + +

(3)23 n(n 1)(n n 1)

2+ + +

(4) 21 n(n 1)(n n 3)

2+ + +

64. z esa lehdj.k 2| z | (z z) i(z z) 2 0- + + - + =

(i 1)= - dk gy gksxk(1) 2 + i, 1 – i (2) 1 + i, 1 – i(3)1 + 2i, –1 – i (4) 1 + i, 1 + i

PART C - MATHEMATICS

Kota/01CT21408728/35


Target : JEE (Main) 2015/19-02-2015

65. If 1 x ysin2 y x

æ öq = +ç ÷ç ÷

è ø, (x,yÎ R – {0}).

Then(1) x = y

(2) x < y

(3) x > y

(4) x + y = 1" x,yÎR

66. The general value of q that satisfies both the

equations cot3q +3 3 0= & 5cosec 32 0q + =is (n Î I)

(1) 2n6p

p - (2) n6p

p -

(3) nn ( 1)6p

p - - (4) n3p

p +

67. S1 and S2 are two concentric circles of radii1 and 2 respectively. Two parallel tangents to S1

cut off an arc from S2. The length of the arc is -

(1) 2p

(2) 23p

(3) 34p

(4) 4p

68. 1 1 13 1tan cos 2 tan sin 2cot4 2

- - -é ùæ ö æ ö+ç ÷ ç ÷ê úè ø è øë û is -

(1) not real (2) equal to 4p

(3) greater than 4p

(4) less than 4p

65. ;fn 1 x ysin2 y x

æ öq = +ç ÷ç ÷

è ø,(x,yÎ R – {0}) gSA

rc(1) x = y

(2) x < y

(3) x > y

(4) x + y = 1" x,yÎR

66. lehdj.k cot3q +3 3 0= rFkk 5cosec 32 0q + =(n Î I) dks lUrq"V djus okys q dk O;kid ekugksxk -

(1) 2n6p

p - (2) n6p

p -

(3) nn ( 1)6p

p - - (4) n3p

p +

67. nks ledsUæh; o`Ùk S1 rFkk S2 ftudh f=T;k;sa Øe'k% 1rFkk 2 gSA S1 ds fy, [khaph xbZ nks lekUrj Li'kZjs[kk;sa]S2 ls ,d pki cukrk gSA rc pki dh yEckbZ gksxh -

(1) 2p

(2) 23p

(3) 34p

(4) 4p

68. 1 1 13 1tan cos 2 tan sin 2cot4 2

- - -é ùæ ö æ ö+ç ÷ ç ÷ê úè ø è øë û gksxk -

(1) okLrfod ugha (2) 4p

ds cjkcj

(3) 4p

ls cM+k (4) 4p ls NksVk


29/35Kota/01CT214087


69. Consider two statements S1 and S2.S1 : If ƒ(x) is a differentiable function withƒ'(x) = 0 in (a, b) and ƒ(x) is increasing in (a,

b), then ƒ(x)ƒ '(x)

is also increasing in (a, b).

S2 : Both sinx and tanx are increasing function

in 0,2pæ ö

ç ÷è ø

.

Which of the following is true -

(1) both S1 and S2 are wrong

(2) S1 is correct and implies S2.

(3) S1 is wrong and S2 is right.

(4) both S1 and S2 are right.

70. An integer x is chosen at random from 1 to 50.

The probability that 336x 50

x+ £ is -

(1) 7

10 (2) 3

10 (3) 1725 (4)

825

71. If x

2 2

0

ƒ(x) (2cos 3t 3sin 3t)dt= +ò ,

then ƒ(x + p) is equal to -

(1) ƒ(x) + ƒ(p) (2) ƒ(x) 2ƒ2pæ ö+ ç ÷

è ø

(3) 2ƒ(x) 3ƒ3pæ ö+ ç ÷

è ø(4) ƒ(x) 4ƒ

4pæ ö+ ç ÷

è ø

69. ekuk nks dFku S1 rFkk S2 gSAS1 : ;fn ƒ(x), ƒ'(x) = 0 vUrjky (a, b) esa vodyuh;Qyu rFkk ƒ(x) vUrjky (a, b) esa o/kZeku Qyu gks] rksƒ(x)ƒ '(x)

Hkh vUrjky (a, b) esa o/kZeku Qyu gksxkA

S2 : vUrjky 0,2pæ ö

ç ÷è ø

esa sinx rFkk tanx nksuksa o/kZeku

Qyu gksaxsAfuEu esa ls dkSulk lR; gS-

(1) S1 rFkk S2 nksuksa xyr gSa

(2) S1 lgh rFkk S2 Hkh lgh gS

(3) S1 xyr rFkk S2 lgh gS

(4) S1 rFkk S2 nksuksa lgh gSa

70. 1 ls 50 esa ls ;kn`PN;k ,d iw.kk±d x dk p;u djrs gS]

rc izkf;drk rkfd 336x 50x

+ £ gks] gksxh -

(1) 7

10 (2) 3

10 (3) 1725 (4)

825

71. ;fn x

2 2

0

ƒ(x) (2cos 3t 3sin 3t)dt= +ò gS]

rks ƒ(x + p) cjkcj gksxk

(1) ƒ(x) + ƒ(p) (2) ƒ(x) 2ƒ2pæ ö+ ç ÷

è ø

(3) 2ƒ(x) 3ƒ3pæ ö+ ç ÷

è ø(4) ƒ(x) 4ƒ

4pæ ö+ ç ÷

è ø

Kota/01CT21408730/35


Target : JEE (Main) 2015/19-02-2015

72. The derivative of 1 1 xtan1 x

- -+

with respect to

sin–1x is -

(1) 1 (2) 12

-

(3) 12

(4) –1

73. Let ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆp 2i 3j ak,q bi 5j k, r i j 3k.= + + = + - = + +r r r

If p, q, rr r r are coplanar and p.q 20=

r r, then the

ordered pair (a, b) is -

(1) (1, 3) or (13, 9) (2) (9, 7)

(3) (5, 5) or (7, 3) (4) (7, 3)

74. The tangent(s) from the point of intersection

of the lines 2x – 3y + 1 = 0 and 3x – 2y – 1 = 0

to circle x2 + y2 + 2x – 4y = 0 will be -

(1) x + 2y = 0, x – 2y + 1 = 0

(2) 2x – y – 1 = 0

(3) y = x, y = 3x – 2

(4) 2x + y + 1 = 0

75. If q1, q2, q3 are roots of the equation

x3 + 64 = 0, then the value of 1 2 3

2 3 1

3 1 2

q q qq q qq q q

is -

(1) 1 (2) 4 (3) 16 (4) 0

72. 1 1 xtan1 x

- -+

dk sin–1x ds lkis{k vodyt

gksxk -

(1) 1 (2) 12

-

(3) 12

(4) –1

73. ekuk ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆp 2i 3j ak,q bi 5j k, r i j 3k= + + = + - = + +r r r

gSA ;fn p, q, rr r r leryh; rFkk p.q 20=

r r gks] rks Øfer

;qXe (a, b) gksxk -

(1) (1, 3) or (13, 9) (2) (9, 7)

(3) (5, 5) or (7, 3) (4) (7, 3)

74. js[kk 2x – 3y + 1 = 0 rFkk 3x – 2y – 1 = 0 ds

izfrPNsn fcUnq ls o`Ùk x2 + y2 + 2x – 4y = 0 ij [khaph

xbZ Li'kZjs[kk gksxh -

(1) x + 2y = 0, x – 2y + 1 = 0

(2) 2x – y – 1 = 0

(3) y = x, y = 3x – 2

(4) 2x + y + 1 = 0

75. ;fn q1, q2, q3 lehdj.k x3 + 64 = 0 ds ewy gks] rks

1 2 3

2 3 1

3 1 2

q q qq q qq q q

dk eku gksxk -

(1) 1 (2) 4 (3) 16 (4) 0


31/35Kota/01CT214087


76. xr occurs in the expansion of n

34

1xx

æ ö+ç ÷è ø

provided -(1) 2n – r is divisible by 5

(2) 3n – r is divisible by 5



77. Let [x] stand for greatest integer function and

ƒ(x) =

2

2

14x + [2x]x, if x , 021ax bx, if x 0,2

ì é öÎ - ÷ï êï ë øí

é öï - Î ÷êï ë øî

. Then

(1) ƒ(x) is continuous in 1 1,2 2

æ ö-ç ÷è ø

, iff a = 4

and b = 0.

(2) ƒ(x) is continuous and differentiable in

1 1,2 2

æ ö-ç ÷è ø

iff a = 4, b = 1.

(3) ƒ(x) is continuous and differentiable in

1 1, a R & b 12 2

æ ö- " Î =ç ÷è ø

(4) ƒ(x) is not differentiable in 1 1,2 2

æ ö-ç ÷è ø

for

any value of a and b.

76.n

34

1xx

æ ö+ç ÷è ø

ds izlkj esa izkIr in xr gksxk] ;fn -

(1) 2n – r, 5 ls foHkkftr gks




77. ekuk [x] egÙke iw.kk±d Qyu rFkk

ƒ(x) =

2

2

14x + [2x]x, x , 021ax bx, x 0,2

ì é öÎ - ÷ï êï ë øí

é öï - Î ÷êï ë øî

;fn

;fn gSA rc

(1) ƒ(x), 1 1,2 2

æ ö-ç ÷è ø

esa larr~ gksxk] dsoy vkSj dsoy

;fn a = 4 rFkk b = 0 gSA

(2) ƒ(x), 1 1,2 2

æ ö-ç ÷è ø

esa larr~ gksxk] dsoy vkSj dsoy

;fn a = 4 rFkk b = 1 gSA

(3) ƒ(x), 1 1, a R2 2

æ ö- " Îç ÷è ø

e sa l arr~ rFk k

vodyuh; rFkk b = 1 gSA

(4) ƒ(x), vUrjky 1 1,2 2

æ ö-ç ÷è ø

esa a rFkk b ds fdlh Hkh

eku ds fy, vodyuh; ugha gksxkA

Kota/01CT21408732/35


Target : JEE (Main) 2015/19-02-2015

78. The number of integers greater than amillion that can be formed using the digits2, 3, 0, 3, 4, 2, 3 is -(1) 240 (2) 320 (3) 360 (4) 300

79. If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â i 2 j 3k, b 2i 3j k, c 3i j 2k= + + = + + = + +rr r

and ˆ â b c 3(i k)a + b + g = - -rr r

. Then the triplet(a, b, g) is -(1) (2, –1, –1) (2) (–2, 1, 1)(3) (–2, –1, 1) (4) (2, 1, –1)

80. Let 0 < x < p and y(x) be given by(1 + sinx)y3 – (cosx)y2 + 2(1 + sinx)y – 2cosx=0.

The derivative of y w.r.t. xtan2

at x2p

= is -

(1) 12

(2) 12

- (3) 2 (4) –2

81.cos x sin x 0

A ƒ(x) sin x cos x 00 0 1

é ùê ú= = -ê úê úë û

.Then A–1 is

equal to -

(1) ƒ(–x) (2) ƒ(x)

(3) –ƒ(x) (4) –ƒ(–x)

82. Product of length of the perpendiculars drawn

from foci on any tangent to hyperbola 2

2 yx 14

- =

is -

(1) 2 (2) 4 (3) 1 (4) 12

78. vadk s a 2, 3, 0, 3, 4, 2, 3 ds iz;k sx ls fufeZriw.kk±d vadksa dh la[;k tks fd nl yk[k ls cM+s gSa]gksxh -(1) 240 (2) 320 (3) 360 (4) 300

79. ;fn ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â i 2j 3k, b 2i 3j k, c 3i j 2k= + + = + + = + +rr r

rFkk ˆ â b c 3(i k)a + b + g = - -rr r

gSA rc f=d (a,b,g)gksxk -(1) (2, –1, –1) (2) (–2, 1, 1)(3) (–2, –1, 1) (4) (2, 1, –1)

80. ekuk 0 < x < p rFkk y(x), (1 + sinx)y3 – (cosx)y2

+ 2(1 + sinx)y – 2cosx = 0 }kjk fn;k x;k gS] rks y dk

xtan2

ds lkis{k x2p

= ij vodyt gksxk -

(1) 12

(2) 12

- (3) 2 (4) –2

81.

cos x sin x 0A ƒ(x) sin x cos x 0

0 0 1

é ùê ú= = -ê úê úë û

gSA rc A–1

cjkcj gksxk -(1) ƒ(–x) (2) ƒ(x)

(3) –ƒ(x) (4) –ƒ(–x)

82. vfrijoy; 2

2 yx 14

- = dh fdlh Hkh Li'kZ js[kk ij

ukHkh ls [khaps x;s yEcksa dh yEckbZ dk xq.kuQy gksxk -

(1) 2 (2) 4 (3) 1 (4) 12


33/35Kota/01CT214087


83. The area of triangle ABC is 84 sq. units. IfAB = 13 and AC = 15, then BC can be -

(1) 14 units (2) 4 14 units

(3) 37 units (4) 2 37 units

84. The shadow of the tower standing on a levelplane is found to be 100 m longer when thesun's is angle of elevation is 30°, than when itis 45°. The height of the tower is -

(1) 50( 3 1)- (2) 50( 3 1)+

(3) 100( 3 1)- (4) 100( 3 1)+

85. The digit in unit place in the number843843 + 492295 is -

(1) 0 (2) 1 (3) 3 (4) 5

86. If ƒ(x) is a quadratic in x, then 1

0

ƒ(x)dxò is -

(1) 1 1(ƒ(0) 4ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

(2) 1 1(4ƒ(0) ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

(3) 1 1(ƒ(0) ƒ 4ƒ(1))6 2

æ ö+ +ç ÷è ø

(4) 1 1(ƒ(0) ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

83. f=Hkqt ABC dk {ks=Qy 84 oxZ bdkbZ gSA ;fn AB = 13rFkk AC = 15 gks] rks BC dk eku gks ldrk gS -

(1) 14 bdkbZ (2) 4 14 bdkbZ

(3) 37 bdkbZ (4) 2 37 bdkbZ

84. le lery ij fLFkr ,d ehukj dh ijNkbZ dh yEckbZ100 ehVj gS]tc lw;Z dk mUu;u dks.k 30° gS] rks ehukjdh m¡GpkbZ] tc lw;Z dk mUu;u dks.k 45° gS] gksxh -

(1) 50( 3 1)- (2) 50( 3 1)+

(3) 100( 3 1)- (4) 100( 3 1)+

85. la[;k 843843 + 492295 esa bdkbZ LFkku ij vadgksxk -

(1) 0 (2) 1 (3) 3 (4) 5

86. ;fn ƒ(x), x esa ,d f}?kkr gks] rks 1

0

ƒ(x)dxò gksxk -

(1) 1 1(ƒ(0) 4ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

(2) 1 1(4ƒ(0) ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

(3) 1 1(ƒ(0) ƒ 4ƒ(1))6 2

æ ö+ +ç ÷è ø

(4) 1 1(ƒ(0) ƒ ƒ(1))6 2

æ ö+ +ç ÷è ø

Kota/01CT21408734/35


Target : JEE (Main) 2015/19-02-2015

87. The differential equation of family of ellipse2 2

2 2

x y ca b

+ = is 2

2

dy d yy ' , y ''dx dx

æ ö= =ç ÷

è ø

(1) y '' y ' 1 0y ' y x

+ - = (2) y '' y ' 1 0y ' y x

+ + =

(3) y '' y ' 1 0y ' y x

- - = (4) y '' y ' 0y ' y

- =

88. If n(U) = 600, n(A) = 100, n(B) = 200 and

n(A Ç B) = 50, then n(A B)Ç is -

(U is universal set and A and B are subsets of U)

(1) 300 (2) 350 (3) 250 (4) 200

89. If mean deviations about median of

x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x is 30, then|x| equals -

(1) 12 (2) 11 (3) 10 (4) 9

90. Which of the following statement is true -

(1) ~(p « ~q) is tautology

(2) ~(p « ~q) is equivalent to p « q

(3) (p Ù ~q) is a fallacy

(4) (p Ù ~q) Ù (~p Ù q) is a tautology

87. nh?kZoÙk ds fudk;

2 2

2 2

x y ca b

+ =2

2

dy d yy ' , y ''dx dx

æ ö= =ç ÷

è ø

dk vody lehdj.k gksxk

(1) y '' y ' 1 0y ' y x

+ - = (2) y '' y ' 1 0y ' y x

+ + =

(3) y '' y ' 1 0y ' y x

- - = (4) y '' y ' 0y ' y

- =

88. ;fn n(U) = 600, n(A) = 100, n(B) = 200 rFkk

n(A Ç B) = 50 gks] rks n(A B)Ç dk eku gksxk -

(tgk¡ U O;kid leqPp; rFkk A o B, U ds mileqPp; gS)

(1) 300 (2) 350 (3) 250 (4) 200

89. ;fn x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x dhekfè;dkvksa ds lkis{k ek/; fopyu 30 gks] rks |x| cjkcjgksxk -

(1) 12 (2) 11 (3) 10 (4) 9

90. fuEu esa ls dkSulk dFku lgh gksxk -

(1) ~(p « ~q) iqu:fDr gksxhA

(2) ~(p « ~q), p « q ds leku gksxkA

(3) (p Ù ~q) ,d O;k?kkr gksxhA

(4) (p Ù ~q) Ù (~p Ù q) iqu:fDr gksxhA


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