e-paper-25012015 (1)

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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 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 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]eksckby Qksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU;izdkj dh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA

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

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

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

IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

ALL INDIA OPEN TEST # 01 DATE : 25 - 01 - 2015TARGET : JEE (Main) 2015

ENTHUSIAST & LEADER COURSE

Test Type : Major Test Pattern : JEE (Main)

FORM NUMBER

PAPER CODE 0 0 C E 3 1 4 0 0 1Hindi

(ACADEMIC SESSION 2014-2015)CLASSROOM CONTACT PROGRAMME

H-1/38

SPACE FOR ROUGH WORK

ALL INDIA OPEN TEST/JEE (Main)/25-01-2015

00CE314001

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 that ofwater. Suppose you have identical masses ofalcohol 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 allowed tocome into thermal equilibrium, with no loss ofheat to the surroundings. The final temperatureof 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. Consider a uniformly charged hemisphericalshell of radius R and charge Q. If field at point

A (0, 0, –z0) is Er

then field at point (0, 0, z0) is[z0 < R] :-

(1) E-r

(2) 0

KQ ˆE kz

- +r

O

A

B

z

x

y

(3) + Er

(4) None of these

1. ,Ydksgy dh fof'k"V Å"ek ikuh dh fof'k"V Å"ek dh

yxHkx vk/kh gksrh gSA ekukfd vkids ikl ,Ydksgy

rFkk ikuh ds ,dleku æO;eku gSaA ,Ydksgy izkjEHk esa

rki TA ij gSA ikuh izkjEHk esa fHkUu rki TW ij gSA vc

nksuksa rjyksa dks ,d gh ik= esa fefJr fd;k tkrk gS rFkk

ifjos'k esa Å"ek dh gkfu ds fcuk mUgsa ijLij rkih;

lkE;koLFkk esa vkus fn;k tkrk gS feJ.k dk vfUre rki

gksxk %&

(1) TW dh rqyuk esa TA ds utnhd

(2) TA dh rqyuk esa TW ds utnhd

(3) TA rFkk TW ds Bhd e/; esa

(4) iz;qä ,Ydksgy ds vk;ru ij fuHkZj djrk gSA

2. f=T;k R rFkk vkos'k Q okys le:i vkosf'kr v/kZxksyh;

dks'k ij fopkj dhft,A ;fn fcUnq A (0, 0, –z0) ij

{ks= Er

gks rks fcUnq (0, 0, z0) ij {ks= gksxk [z0 < R] :-

(1) E-r

(2) 0

KQ ˆE kz

- +r

O

A

B

z

x

y

(3) + Er

(4) None of these

H-2/38



00CE314001

3. A particle move on an circular overbridge withconstant speed. The friction coefficient variesso that the speed remains constant. Which ofthe following graph shows the magnitude offriction.

q

(1) q

f

(2) q

f

(3) q

f

(4) q

f

4. A beaker of height H is made up of a materialwhose coefficient of linear thermal expansionis 3a. It is filled up to the brim by a liquid whosecoefficient of thermal expansion is a. If nowthe beaker along with its contents is uniformlyheated through a small temperature T the levelof liquid will reduce by (given a << 1) :-(1) 5aTH (2) 3aTH

(3) 9aTH (4) 8aTH

3. ,d d.k o`Ùkh; iFk ij fu;r pky ls fp=kuqlkj xfr

dj jgk gSA ?k"kZ.k xq.kkad bl izdkj ifjofrZr gksrk gS fd

bldh pky fu;r jgrh gSA ?k"kZ.k ds ifjek.k dks fuEu esa

ls dkSulk xzkQ iznf'kZr djrk gS\

q

(1) q

f

(2) q

f

(3) q

f

(4) q

f

4. Å¡pkbZ H okys ,d chdj dks ml inkFkZ ls cuk;k x;k gS

ftldk js[kh; rkih; izlkj xq.kkad 3a gSA bls iwjk eq¡g

rd ml æo }kjk Hkjk tkrk gS ftldk rkih; izlkj

xq.kkad a gSA ;fn vc chdj rFkk mlesa Hkjs æo dks ,d

vYi rki T ls ,dleku :i ls xje djrs gSa rks æo dk

Lrj fdruk ?kV tk,xk (fn;k gS a << 1) :-(1) 5aTH (2) 3aTH

(3) 9aTH (4) 8aTH

H-3/38



00CE314001

5. Consider telecommunication through opticalfibres. Which of the following statements isNOT true ?

(1) Optical fibres can be of graded refractive index

(2) Optical fibres are subjected to electromagneticinterference from outside

(3) Optical fibres have extremely lowtransmission loss

(4) Optical fibres may have homogeneous corewith a suitable cladding

6. A circular conducting loop of radius R carries acurrent I. Another straight infinite conductorcarrying current I passes through the diameterof this loop as shown in the figure. Themagnitude of force exerted by the straightconductor on the loop is :-

RO

I

(1) pµ0I2 (2) µ0I

2

(3) 2

0I2

mp

(4) 2

0Imp

5. izdk'kh; rarqvkas }kjk gksus okys nwjlapkj ij fopkj dhft,A

blds lanHkZ esa xyr dFku pqfu;s%&

(1) izdk'kh; rarq Js.khdr viorZukad okys gks ldrs gSA

(2) izdk'kh; rarq] ckgj ls gksus okys fo|qrpqEcdh;

O;frdj.k ls lEikfnr gksrs gSA

(3) izdk'kh; rarqvksa }kjk cgqr de izs"k.k âkl gksrk gSA

(4) izdk'kh ; rarqvksa esa mi;qDr vkoj.k okys le:i

ØksM gks ldrs gSA

6. f=T;k R okys ,d o`Ùkh; pkyd ywi esa I /kkjk izokfgr

gksrh gSA ,d vU; lh/ks vifjfer pkyd esa /kkjk I

izokfgr gks jgh gS rFkk ;g fp=kuqlkj bl ywi ds O;kl ls

gksdj xqtjrk gSA ywi ij lh/ks pkyd }kjk vkjksfir cy

dk ifjek.k gksxk%&

RO

I

(1) pµ0I2 (2) µ0I

2

(3) 2

0I2

mp

(4) 2

0Imp

H-4/38



00CE314001

7. The figure shows the cross section of a long

cylindrical conductor through which an axial

hole of radius r is drilled with its centre at point

A. O is the centre of the conductor. If an

identical hole were to be drilled centred at point

B while maintaining the same current density

the magnitude of magnetic field at O :-

A

120°

30°

B

O

r

(1) will increase

(2) will decrease

(3) will remains the same

(4) May increase or decrease depending on the

value of r.

8. There are 10 turns in coil M and 15 turns in coilN. If a current of 2A is passed through coil Mthen the flux linked with coil N is1.8 × 10–3 Wb. If a current of 3A is passedthrough coil N then flux linked with coil Mis:-(1) 1.2 × 10–3 Wb (2) 2.7 × 10–3 Wb(3) 1.8 × 10–3 Wb (4) 4.05 × 10–3 Wb

7. fp= esa ,d yEcs csyukdkj pkyd ds vuqizLFk dkV

dks fn[kk;k x;k gS ftlesa fcUnq A ij dsfUær f=T;k r

okys ,d v{kh; fNæ dks cuk;k x;k gSA pkyd dk

dsUæ O gSA ;fn blh rjg ds ,d fNæ dks fcUnq B ij

dsUæ ysrs gq;s cuk;k tk, tcfd /kkjk ?kuRo leku jgs rks

O ij pqEcdh; {ks= dk ifjek.k :-

A

120°

30°

B

O

r

(1) c<+sxkA

(2) ?kVsxkA

(3) leku jgsxkA

(4) ?kV Hkh ldrk gS vFkok c<+ Hkh ldrk gS ;g r ds

eku ij fuHkZj djrk gSA

8. dq.Myh M esa 10 ?ksjs gSa rFkk dq.Myh N esa 15 ?ksjs gSaA

;fn dq.Myh M esa 2A /kkjk izokfgr djrs gSa rks dq.Myh

N ls lEc¼ ¶yDl 1.8 × 10–3 Wb gSA ;fn dq.Myh

N esa 3A dh /kkjk izokfgr djsa rks dq.Myh M ls lEc¼

¶yDl gksxk %&

(1) 1.2 × 10–3 Wb (2) 2.7 × 10–3 Wb

(3) 1.8 × 10–3 Wb (4) 4.05 × 10–3 Wb

H-5/38



00CE314001

9. A radio station has two channels. One is AM at

1020 kHz and the other is FM at 89.5 MHz.

For good results you will use

(1) longer antenna for the AM channel and

shorter for the FM

(2) shorter antenna for the AM channel and

longer for the FM

(3) Same length antenna with work for both

(4) Information given is not enough to say which

one to use for which 10. An electron is projected

wall

xa b

v0

B

normally from the surfaceof a sphere with speed v0

in a uniform magneticfield perpendicular to theplane of the paper such thatits strikes symmetrically opposite on the spherewith respect to the x-axis. Radius of the sphereis 'a' and the distance of its centre from the wallis 'b'. What should be magnetic field such thatthe charge particle just escapes the wall:-

(1) ( )

02 2

2bmvB

b a e=

-(2)

( )=

+0

2 2

2bmvB

a b e

(3)

( )=

-

0

2 2

mvB

b a e(4)

( )=

-

0

2 2

2mvB

b a e

9. ,d jsfM;ks LVs'ku ds nks pSuy gSA buesa ls ,d 1020kHz

okyk AM pSuy gS rFkk nwljk 89.5 MHz ij pyus

okyk FM pSuy gSA vPNs ifj.kkeksa ds fy;s vki D;kmi;ksx djsxsa\

(1) AM pSuy ds fy;s yEck ,sfUVuk rFkk FM pSuy ds

fy;s NksVk ,sfUVuk

(2) AM pSuy ds fy;s NksVk ,sfUVuk rFkk FM pSuy ds

fy;s yEck , sfUVuk

(3) nksuksa ds fy, leku yEckbZ dk ,sfUVuk(4) ;g Kkr djus ds fy, vkdM+s vi;kZIr gSA

10. ,d bysDVªkWu dks dkxt ds ry wall

xa b

v0

B

ds yEcor~ le:i pqEcdh;

{ks= esa xksys dh lrg ls v0

pky ls yEcor~ :i ls bl

izdkj iz{ksfir djrs gSa fd ;g

x-v{k ds lkis{k xksys ds nwljh vksj lefer :i ls

Vdjkrk gSA xksys dh f=T;k 'a' gS rFkk nhokj ls blds

dsUæ dh nwjh 'b' gSA pqEcdh; {ks= dk eku D;k gksuk

pkfg;s rkfd vkosf'kr d.k nhokj ls cp tk;s%&

(1) ( )

02 2

2bmvB

b a e=

-(2)

( )=

+0

2 2

2bmvB

a b e

(3)

( )=

-

0

2 2

mvB

b a e(4)

( )=

-0

2 2

2mvB

b a e

H-6/38



00CE314001

11. In the situation as shown in figure time periodof vertical oscillation of block for smalldisplacements will be :-

kk m

qq

(1) m

2 cos2k

p q (2) m

2 sec2k

p q

(3) m

2 sin2k

p q (4) m

2 cosec2k

p q

12. A charged particle with charge q and mass mstarts with an initial kinetic energy K at thecentre of a uniformly charged spherical regionof total charge Q and radius R. Charges q andQ have opposite signs. The spherically chargedregion is not free to move and kinetic energy Kis just sufficient for the charge particle to reachboundary of the spherical charge. How muchtime does it take the particle to reach theboundary of the region?

(1) 3

o4 mRqQ

pep (2)

3o4 mR

2 qQpep

(3) 3

o4 mR4 qQ

pep(4) None of these

11. fp= esa iznf'kZr fLFkfr esa vYi foLFkkiuksa ds fy, CykWdds Å/okZ/kj nksyu dk vkorZdky D;k gksxk%&

kk m

qq

(1) m

2 cos2k

p q (2) m

2 sec2k

p q

(3) m

2 sin2k

p q (4) m

2 cosec2k

p q

12. vkos'k q rFkk nzO;eku m dk ,d vkosf'kr d.k iz kjfEHkdxfrt ÅtkZ K ls ,d R f=T;k rFkk dqy vkos'k Q dsle:i vkosf'kr xksyh; {ks= ds dsUnz ls xfr izkjEHkdjrk gSA vkos'k q rFkk Q foijhr fpUg ds gSaA xksyh;vkosf'kr {ks= xfr ds fy;s Lora= ugha gS rFkk xfrt ÅtkZK dk eku bruk gh gS fd vkosf'kr d.k] xksyh; vkos'kdh ifjlhek rd igqap tk;sA d.k dks {ks= dh ifjlhekrd igqapus ds fy;s fdruk le; yxsxk\

(1) 3

o4 mRqQ

pep (2)

3o4 mR

2 qQpep

(3) 3

o4 mR4 qQ

pep(4) None of these

H-7/38



00CE314001

13. A silicon specimen is made into a p-typesemiconductor by doping, on an average, oneindium atom per 5 × 107 silicon atoms. If thenumber density of atoms in the silicon specimenis 5 × 1028 atom m–3, then the number of acceptoratoms in silicon per cubic centimeter will be(1) 2.5 × 1030 atom cm–3

(2) 2.5 × 1035 atom cm–1

(3) 1 × 1013 atom cm–3

(4) 1 × 1015 atom cm–3

14. Two different coils have inductances L1 = 4mHand L2 = 2mH. At a certain instant the currentin the two coils is increasing at the same constantrate and power supplied to the first coil is fourtime that of in second coil. If e, I and U indicatethe potential difference, current and energystored of the inductance respectively than whichof the following is correct.

(1) 3

1 1 1

2 2 2

U I e4

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(2) 3

1 1 1

2 2 2

U I e4

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(3) 3

1 1 1

2 2 2

U I e2

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(4) 1/3

1 1 1

2 2 2

U I eU I e

æ ö= =ç ÷

è ø

13. ,d flfydkWu izfrn'kZ dks vifeJ.k }kjk ,d p- izdkjds v¼Zpkyd esa ifjofrZr fd;k tkrk gSA blds fy;s5 × 107 flfydkWu ijek.kqvksa esa ,d bf.M;e ijek.kqfeyk;k tkrk gSA ;fn flfydkWu izfrn'kZ esa ijek.kqvksadk la[;k ?kuRo 5 × 1028 atom m–3 gks rks flfydkWu esaizfr?ku ls.VhehVj xzkgh ijek.kqvksa dh la[;k gksxh%&(1) 2.5 × 1030 atom cm–3

(2) 2.5 × 1035 atom cm–1

(3) 1 × 1013 atom cm–3

(4) 1 × 1015 atom cm–3

14. nks fHkUu dq.Mfy;ksa ds izsjdRo L1 = 4mH rFkkL2 = 2mH gSA fdlh {k.k nksuksa dq.Mfy;ksa esa /kkjkleku fu;r nj ls c<+ jgh gS rFkk igyh dq.Myh dks nhxbZ 'kfä nwljh dq.Myh dh rqyuk esa pkj xquk vf/kdgSA ;fn e, I o U Øe'k% foHkokUrj] /kkjk rFkk izsjddq.Myh dh laxzfgr ÅtkZ dks O;ä djrs gSa rks fuEu esals dkSulk lEcU/k lR; gSa %&

(1) 3

1 1 1

2 2 2

U I e4

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(2) 3

1 1 1

2 2 2

U I e4

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(3) 3

1 1 1

2 2 2

U I e2

U I e

æ ö æ ö= =ç ÷ ç ÷

è ø è ø

(4) 1/3

1 1 1

2 2 2

U I eU I e

æ ö= =ç ÷

è ø

H-8/38



00CE314001

15. In figure a point charge +Q1 is at the centre ofan imaginary spherical surface and another pointcharge +Q2 is outside it. Point P is on the surfaceof the sphere. Let FS be the net electric flux

through the sphere and PE

r be the electric field

at point P on the sphere. Which of the followingstatements is TRUE?

Å Q1 Å Q2

P

(1) Both charges +Q1 and +Q2 make nonzerocontributions to FS but only the charge +Q1

makes a nonzero contribution to PE

r.

(2) Both charges +Q1 and +Q2 make nonzerocontributions to FS but only the charge +Q2

makes a nonzero contribution to PE

r.

(3) Only the charge +Q1 makes a nonzerocontribution to FS but both charges +Q1 and

+Q2 make nonzero contributions to PE

r.

(4) Only the charge +Q2 makes a nonzerocontribution to FS but both charges +Q1 and

+Q2 make nonzero contributions to PE

r.

15. fp= esa ,d fcUnq vkos'k +Q1 dkYifud xksyh; lrg

ds dsUæ ij gS rFkk nwljk fcUnq vkos'k +Q2 blds ckgj

gSA fcUnq P xksys dh lrg ij gSA ekuk FS xksys ls fuxZr

ifj.kkeh fo|qr ¶yDl rFkk PE

r xksys ds fcUnq P ij

fo|qr {ks= gks rks fuEu esa ls dkSulk dFku lR; gS\

Å Q1 Å Q2

P

(1) nksuksa vkos'k +Q1 rFkk +Q2 , FS esa v'kwU; lg;ksx

nsrs gSa ijUrq dsoy vkos'k +Q1, PEr

esa v'kwU; lg;ksx

nsrk gSA

(2) nksuksa vkos'k +Q1 rFkk +Q2] FS esa v'kwU; lg;ksx

nsrs gSa ijUrq dsoy vkos'k +Q2, PEr

esa v'kwU; lg;ksx

nsrk gSA

(3) dsoy vkos'k +Q1] FS esa v'kwU; lg;ksx nsrk gS

ijUrq nksuksa vkos'k +Q1 rFkk +Q2, PEr

esa v'kwU;

lg;ksx nsrs gSA

(4) dsoy vkos'k +Q2] FS esa v'kwU; lg;ksx nsrk gS

ijUrq nksuksa vkos'k +Q1 rFkk +Q2, PEr

esa v'kwU;

lg;ksx nsrs gSA

H-9/38



00CE314001

16. A composite heavy rope of two materials issuspended vertically from a high ceiling. Theratios of different quantities for upper to lowerrope are :

length uL 1L 2

=l

, cross sectional area uA 2A 1

=l

,

density ud 2d 3

=l

. What is the ratio of maximum

stress in the two ropes.

Lu

Ll

(1) 23

(2) 34

(3) 56

(4) 45

17. Two polaroids are placed in the path ofunpolarized beam of intensity I0 such that nolight is emitted from the second polaroid. If athird polaroid whose polarization axis makes anangle q with the polarization axis of firstpolaroid, is placed between these polaroids, thenthe intensity of light emerging from the lastpolaroid will be :-

(1) 20I

sin 28

æ ö qç ÷è ø

(2) 20I

sin 24

æ ö qç ÷è ø

(3) 20I

cos2

æ ö qç ÷è ø

(4) I0 cos4 q

16. nks inkFkks± ls cuh ,d la;qä Hkkjh jLlh dks ,d Å¡phNr ls Å/okZ/kj yVdk;k x;k gSA Åijh rFkk fupyhjLlh ds fy, fHkUu jkf'k;ksa dk vuqikr fuEu gS %

yEckbZ uL 1L 2

=l

, vuqizLFk dkV {ks=Qy uA 2A 1

=l

,

?kuRo ud 2d 3

=l

. nksuksa jfLl;ksa esa vf/kdre izfrcy dk

vuqikr D;k gksxk %&

Lu

Ll

(1) 23

(2) 34

(3) 56

(4) 45

17. nks /kzqodksa dks I0 rhozrk okys v/kzqfor iqat ds iFk esa blizdkj ls j[kk tkrk gS fd f}rh; /kzqod ls dksbZ çdk'kckgj uk fudysA ;fn bu nksuksa /kzqodksa ds e /; ,drhljs èkqzod] ftldh /kzqo.k v{k izFke /kzqod dh /kzqo.kv{k ls q dks.k cukrh gS] dks j[kk tkrk gS rks vafreèkzqod ls fudyus okys izdk'k dh rhozrk gksxh%&

(1) 20I

sin 28

æ ö qç ÷è ø

(2) 20I

sin 24

æ ö qç ÷è ø

(3) 20I

cos2

æ ö qç ÷è ø

(4) I0 cos4 q

H-10/38



00CE314001

18. A point particle of mass 0.5 kg is moving along

the x-axis under a force described by the

potential energy V shown below. It is projected

towards the right from the origin with a speed

v. What is the minimum value of v for which

the particle will escape infinitely far away from

the origin ?

–4 –3 –2 –1 1 2 3 4 5

1

2

3

4

V(in J)

x(in m)

(1) 2 2 ms–1

(2) 2 ms–1

(3) 4 ms–1

(4) The particle will never escape

18. æO;eku 0.5 kg dk ,d fcUnqd.k fp= esa iznf'kZr fLFkfrt

ÅtkZ V }kjk O;ä fd;s x;s cy ds v/khu x-v{k ds

vuqfn'k xfr dj jgk gSA bls ewyfcUnq ls nk¡;h vksj v

pky ls iz{ksfir fd;k tkrk gSA v dk U;wure eku fdruk

gksuk pkfg, rkfd d.k ewyfcUnq ls vuUr nwjh rd iyk;u

dj lds%&

–4 –3 –2 –1 1 2 3 4 5

1

2

3

4

V(in J)

x(in m)

(1) 2 2 ms–1

(2) 2 ms–1

(3) 4 ms–1

(4) d.k dHkh Hkh iyk;u ugha djsxkA

H-11/38



00CE314001

19. A wheel of radius R with an axle of radius R/2is shown in the figure and is free to rotate abouta frictionless axis through its centre andperpendicular to the page. Three forces(F, F, 2F) are exerted tangentially to therespective rims as shown in the figure. Themagnitude of the net torque acting on the systemis nearly :-

FR/2

R2F

45°F

(1) 3.5 FR (2) 3.2 FR(3) 2.5 FR (4) 1.5 FR

20. In the figure shown a mass 1 kg is connected toa string of mass per unit length 1.2 gm/m. Lengthof string is 1 m and its other end is connected tothe top of a ceiling which is accelerating up withan acceleration 2 m/s2. A transverse pulse isproduced at the lowest point of string. Timetaken by pulse to reach the top of string is :

1kg1m

a=2m/s2

(1) 0.1 s (2) 0.01 s (3) 0.05 s (4) 0.5 s

19. f=T;k R okys ,d ifg;s dks fp= esa n'kkZ;k x;k gS rFkkbldh /kqjh R/2 f=T;k okyh gSA ;g blds dsUæ lsgksdj xqtj jgh rFkk dkxt ds ry ds yEcor~ ?k"kZ.kjfgrv{k ds ifjr% eqä :i ls ?kw.kZu djrk gSA rhu cyksa(F, F, 2F) dks fp=kuqlkj bldh ifjf/k;ksa ij Li'kZjs[kh;:i ls vkjksfir fd;k tkrk gSA fudk; ij dk;Zjrifj.kkeh cyk?kw.kZ dk ifjek.k yxHkx gksxk %&

FR/2

R2F

45°F

(1) 3.5 FR (2) 3.2 FR(3) 2.5 FR (4) 1.5 FR

20. fp= esa 1 kg æO;eku dks izfr ,dkad yEckbZ æO;eku1.2 gm/m okyh jLlh ls tksM+k x;k gSA jLlh dh yEckbZ1 m gS rFkk bldk nwljk fljk Nr ds 'kh"kZ ls tqM+k gS tks2 m/s2 Roj.k ls Åij dh vksj Rofjr gks jgh gSA jLlhds fuEure fcUnq ij ,d vuqizLFk Lian mRiUu fd;ktkrk gSA LiUn }kjk jLlh ds 'kh"kZ rd igq¡pus esa fy;kx;k le; gksxk %&

1kg1m

a=2m/s2

(1) 0.1 s (2) 0.01 s (3) 0.05 s (4) 0.5 s

H-12/38



00CE314001

21. A parallel beam of fast moving electrons is

incident normally on a narrow slit. A screen is

placed at a large distance from the slit. If the

speed of the electrons is increased, which of the

following statement is CORRECT ?

(1) Diffraction pattern is not observed on the

screen in the case of electrons

(2) The angular width of the central maximum

of the diffraction pattern will increase


of the diffraction pattern will decrease


of the diffraction pattern will remain the

same

22. A closed organ pipe of length l is sounded

together with another closed organ pipe of length

l + x (x << l) both in fundamental mode. If

v = speed of sound, the beat frequency heard is:

(1) 2

vx2l

(2) 2

vx4l

(3) 2vx

4l(4) 2

vxl

21. rhozxkeh bysDVªkWuksa dk ,d lekUrj iqat fdlh ladjh

fLyV ij yEcor~ vkifrr gksrk gSA ,d insZ dks fLyV

ls vf/kd nwjh ij j[kk tkrk gSA ;fn bysDVªkWuksa dh pky

c<+k nh tk, rks lgh dFku pqfu, %&

(1) bysDVªkWuksa ds izdj.k esa insZ ij foorZu izfr:i izkIr

ugha gksxkA

(2) foorZu iz fr:i ds dsUæh; mfPp"B dh dks.kh; pkSM+kbZ

c<+ tk,xhA


?kV tk,xhA


ogh jgsxhA

22. ,d l yEckbZ ds can vkWxZu ikbi dks yEckbZ l + x (x << l)

okys ,d vU; can vkWxZu ikbi ds lkFk ewyHkwr fo/kk esa

èofur fd;k tkrk gSA ;fn v = /ofu dh pky gks rks

lqukbZ nsus okyh foLian vkofr gksxh%&

(1) 2

vx2l

(2) 2

vx4l

(3) 2vx

4l(4) 2

vxl

H-13/38



00CE314001

23. A space consists of two uniform magnetic fields,

as shown. A conducting circular loop is placed

symmetrically at the boundary :-

× × ×

× × ×

× × ×

× × ×

Boundary

(1) If loop is moved towards right current is

initially anticlockwise

(2) If loop is moved towards left current is

initially anticlockwise

(3) If loop is moved towards right current is zero

initially only

(4) Current will always be zero irrespective of

direction of motion and time.

24. In Young’s double slit experiment, the fringes

are displaced by a distance x when a glass plate

of refractive index 1.5 is introduced in the path

of one of the beams. When this plate is replaced

by another plate of the same thickness, the shift

of fringes is (3/2)x. The refractive index of the

second plate is

(1) 1.75 (2) 1.50 (3) 1.25 (4) 1.00

23. ,d lef"V esa fp=kuqlkj nks le:i pqEcdh; {ks=fo|eku gSA ,d pkyd o`Ùkkdkj ywi dks bldh ifjlhek

ij lefer :i ls j[kk x;k gS %&

× × ×

× × ×

× × ×

× × ×

Boundary

(1) ;fn ywi dks nk¡;h vksj xfr djkrs gSa rks /kkjk iz kjEHk esa

okekorZ fn'kk esa gksxhA

(2) ;fn ywi dks ck¡;h vksj xfr djkrs gSa rks /kkjk izkjEHk

esa okekorZ fn'kk esa gksxhA

(3) ;fn ywi dks nk¡;h vksj xfr djkrs gSa rks /kkjk dsoy

izkjEHk esa 'kwU; gksxhA

(4) xfr dh fn'kk rFkk le; ds vuisf{kr /kkjk lnSo

'kwU; gksxhA

24. ;ax f}&fLyV iz;skx esa tc 1.5 viorZukad okys dkap

dh IysV dks fdlh ,d iqat ds iFk esa j[kk tkrk gS rks

fÝUtsa x nwjh rd foLFkkfir gks tkrh gSA tc bl IysV ds

LFkku ij leku eksVkbZ okyh vU; IysV dks j[kk tkrk gS

rks fÝUtsa (3/2)x nwjh rd foLFkkfir gks tkrh gSA f}rh;

IysV dk viorZukad gksxk %&

(1) 1.75 (2) 1.50 (3) 1.25 (4) 1.00

H-14/38



00CE314001

25. If the B-H curves of two samples of P and Q ofiron are as shown below, then which one of thefollowing statements is CORRECT ?

H H

B B

Sample P Sample Q

(1) Both P and Q are suitable for makingpermanent magnet

(2) P is suitable for making permanent magnetand Q for making electromagnet

(3) P is suitable for making electromagnet andQ is suitable for permanent magnet

(4) Both P and Q are suitable for makingelectromagnets

26. For a certain hypothetical one electron atom thewavelength (in Å) for the spectral lines fortransition from n = p to n = 1 are given by

2

2

1500pp 1

l =-

(where p > 1), then the ionization

potential of this element must be :- (Take hc =12420 eV-Å)(1) 0.95 V (2) 2.05 V(3) 8.25 V (4) None of these

25. ;fn yksgs ds nks izfrn'kks ± P o Q ds fy, B-H oØfp=kuqlkj izkIr gksrs gks rks lgh dFku pqfu, %&

H H

B B

Sample P Sample Q

(1) P o Q nksuksa gh LFkk;h pqEcd ds fuekZ.k ds fy,

mi;qä gSaA(2) LFkk;h pqEcd ds fuekZ.k ds fy, P rFkk fo|qr

pqEcd ds fuekZ.k ds fy, Q mi;qä gSA

(3) fo|qr pqEcd ds fuekZ.k ds fy, P rFkk LFkk;hpqEcd ds fuekZ.k ds fy, Q mi;qä gSA

(4) P o Q nksuksa gh fo|qr pqEcd ds fuekZ.k ds fy,mi;qä gSaA

26. fdlh dkYifud ,dy bysDVªkWu ijek.kq ds n = p ls

n = 1 laØe.k ds fy;s LisDVªeh js[kkvksa dh rjaxnS/; Z

(Å esa) 2

2

1500pp 1

l =-

}kjk nh tkrh gS tgka p > 1 gS rks

bl rRo dk vk;uu foHko gksxk%& (hc = 12420 eV-Å)

(1) 0.95 V (2) 2.05 V

(3) 8.28 V (4) buesa ls dksbZ ugha

H-15/38



00CE314001

27. What amount of heat will be generated in a coil

of resistance R due to a charge q passing through

it if the current in the coil decreases to zero

uniformly during a time interval Dt

(1) tRq

34 2

D(2) ln

t2Rq2

D(3)

t3Rq2 2

D (4) ln

( )Rqt2

2D

28. The switch in circuit shifts from 1 to 2 whenV

C > 2V/3 and goes back to 1 from 2 when

VC < V/3. The voltmeter reads voltage as plotted.

What is the period T of the wave form in terms of Rand C?

VC

21

CV

R

T

2V/3

V/3

t2 t1

Voltage

time

(1) RC ln3 (2) 2RC ln 2

(3) RC2 ln3 (4)

RC3 ln3

27. izfrjks/k R okyh fdlh dq.Myh esa vkos'k q izokfgr gksus

ds dkj.k fdruh Å"ek mRiUu gksxh ;fn dq.Myh esa

èkkjk le;kUrjky Dt ds nkSjku ,dleku :i ls 'kwU;

rd ?kVrh gS %&

(1) tRq

34 2

D(2) ln

t2Rq2

D(3)

t3Rq2 2

D (4) ln

( )Rqt2

2D

28. ifjiFk esa tc VC > 2V/3 gS rks fLop 1 ls 2 ij foLFkkfir

gks tkrk gS rFkk tc VC < V/3 gS rks fLop okil 2 ls 1

ij foLFkkfir gks tkrk gSA oksYVehVj dk ikB~;kad vkjsf[kr

fd;k x;k gSA R rFkk C ds inksa esa rjax dk vkorZdky T

D;k gksxk\

VC

21

CV

R

T

2V/3

V/3

t2 t1

Voltage

time

(1) RC ln3 (2) 2RC ln 2

(3) RC2 ln3 (4)

RC3 ln3

H-16/38



00CE314001

29. Statement-1 : A photodiodes operates in reversebias.Statement-2 : The fractional change due to thephoto-effects on the minority carrier dominatedreverse bias current is more easily measurable thanthe fractional change in the forward bias current.(1) Statement-1 is true, Statement-2 is true,

Statement-2 is the correct explanation ofStatement-1.

(2) Statement-1 is true, Statement-2 is true,Statement-2 is not the correct explanationof Statement-1.

(3) Statement-1 is true, Statement-2 is false.(4) Statement-1 is false, Statement-2 is true.

30. Starting from rest on her swing at initial heighth

0 above the ground, Saina swings forward. At

the lowest point of her motion, she grabs herbag that lies on the ground. Saina continuesswinging forward to reach maximum height h

1.

She then swings backward and when reachingthe lowest point of motion again, she simple letsgo off the bag, which falls freely. Saina'sbackward swing then reaches maximum heighth

2. Neglecting air resistance, how are the three

heights related?

h0

(1) h0 > h

1 > h

2(2) h

0 = h

1= h

2

(3) h0 > h

1 = h

2(4) h

0 = h

2 > h

1

29. çdFku-1 : ,d izdk'k Mk;ksM dk izpkyu i'p ck;lesa fd;k tkrk gSAçdFku-2 : izdk'k izHkkoksa ds dkj.k vYika'k okgdks }kjki'pfnf'kd ck;l /kkjk esa fHkUukRed varj vxzfnf'kdck;l /kkjk esa fHkUukRed varj dh vis{kk vf/kd vklkuhls ukik tk ldrk gSA(1) çdFku-1 lR; gS] izdFku -2 lR; gS] izdFku -2

çdFku-1 dh lgh O;k[;k djrk gSA(2) izdFku-1 lR; gS] izdFku -2 lR; gS] izdFku -2

izdFku-1 dh lgh O;k[;k ugha djrk gSA(3) çdFku-1 lR; gS] izdFku-2 xyr gSA(4) çdFku-1 xyr gS] izdFku-2 lR; gSA

30. lk;uk ,d djrc ds nkSjku ,d jLlh dks idM+s gq,èkjkry ls h

0 ÅapkbZ ij fojkekoLFkk esa gSA vc og vkxs

dh vksj > wyrh gqbZ viuh xfr ds fuEure fcUnq ijèkjkry ij j[ks ,d cSx dks mBkrh gSA og mlh fn'kk esa>wyrs gq, vf/kdre ÅapkbZ h

1 rd igqap tkrh gSA vc

og iqu% ihNs dh fn'kk esa ykSVrh gS rFkk xfr ds fuEurefcUnq ij igqapus ij og cSx dks NksM+ nsrh gS tks fd eqDr:i ls fxjrk gSA vc og ihNs dh fn'kk esa vf/kdreh

2 ÅapkbZ rd igqap tkrh gSA ok;q izfrjks/k dks ux .;

ekurs gq, bu rhuksa ÅapkbZ;ksa esa D;k laca/k gS\

h0

(1) h0 > h

1 > h

2(2) h

0 = h

1= h

2

(3) h0 > h

1 = h

2(4) h

0 = h

2 > h

1

H-17/38


00CE314001


31. Which of the following statement is FALSE

(1) Number of spherical node for 4d orbitalis 3

(2) For py orbital xz plane is angular node

(3) For ‘s’ orbital , Y2 is maximum at nucleus.

(4) Angular wave function of Pz orbital isproportional to cos q (Given : q is anglefrom z-axis)

32. A heat engine is operating between 500K to300K. If the engine absorbs 100J heat, thenwhich of the following is impossible amountof heat rejected by the engine.

(1) 80 J (2) 75 J

(3) 70 J (4) 20 J

33. In which of the following case percentageincrease in rate constant will be maximum.

Ea(Kcal/mol) Temp. Change (K)

(I) 40 200 – 210

(II) 80 200 – 210

(III) 40 300 – 310

(IV) 80 300 – 310

(1) I (2) II

(3) III (4) IV

31. fuEu esa ls dkSulk dFku xyr gS -

(1) 4d d{kd ds fy, xksyh; uksMks dh dqy la[;k 3

gksrh gS

(2) py d{kd ds fy, xz ry dks.kh; uksM gksrk gS

(3) 's' d{kd ds fy,, ukfHkd ij Y 2 vf/kdre gksrk gS

(4) Pz d{kd dk dks.kh; rjax Qyu cos q ds lekuqikrhgksrk gS (fn;k gS: q, z-v{k ls dks.k gS)

32. ,d Å"eh; bZtau dks 500K ls 300K ds e/; lapkfyrfd;k tkrk gSA ;fn bZtau 100 J Å"ek vo'kksf"kr djrkgS] rc bZtau }kjk R;kx dh xbZ Å"ek dh vlEHko ek=kdkSulh gksxh

(1) 80 J (2) 75 J

(3) 70 J (4) 20 J

33. fuEu esa ls fdl fLFkfr esa nj fu;rkad esa izfr'kr o`f¼vf/kdre gksxhA

Ea(Kcal/eksy) rki ifjorZu (K)

(I) 40 200 – 210

(II) 80 200 – 210

(III) 40 300 – 310

(IV) 80 300 – 310

(1) I (2) II

(3) III (4) IV

PART B - CHEMISTRY

H-18/38


00CE314001


34. In the decay,A

ZX ¾® 12C + b+

(1) A = 13, Z = 7 (2) A = 13, Z = 6(3) A = 12, Z = 7 (4) A = 12, Z = 5

35. Which of the following solutions has themaximum pH value ?

(1) 0.2 M HNO3

(2) 0.2 M HCl

(3) 0.2 M CH3COOH

(4) 0.2 M CH3COONa

36. During electrolysis of aq.NaCl, if 3mole of H2Oare electrolysed then how much charge isrequired if current efficiency is 75%-

(1) 1 F (2) 2 F

(3) 4 F (4) 8 F37. Select the incorrect statement -

(1) F-centres are a factor for imparting thecolour to the crystal

(2) Stoichiometry of crystal remains uneffecteddue to schottky defect

(3) In orthorhombic crystal each interfacialangle is 90º

(4) There are four unit cells in tetragonal crystalsystem.

34. fuEu {k; esa]AZX ¾® 12C + b+

(1) A = 13, Z = 7 (2) A = 13, Z = 6(3) A = 12, Z = 7 (4) A = 12, Z = 5

35. fuEu esa ls dkSuls foy;u eas pH dk eku vf/kdre gS?

(1) 0.2 M HNO3

(2) 0.2 M HCl

(3) 0.2 M CH3COOH(4) 0.2 M CH3COONa

36 tyh; NaCl ds oS|qr vi?kVu ds nkSjku] ;fn 3 eksyH2O dks oS|qr vi?kfVr fd;k x;k gks rks fdrus vkos'kdh vko';drk gksxh ] ;fn /kkjk n{krk 75% gks-(1) 1 F (2) 2 F

(3) 4 F (4) 8 F37. xyr dFku dk p;u dhft, -

(1) fØLVy esa jax ykus ds fy, F-lsUVj Hkh ,d dkjd

gksrk gS

(2) 'kkWVdh = qfV ds dkj.k fØLVy dh LVªkbfd;ksesVªh

vizHkkfor jgrh gS

(3) vkFkksZjksfEcd fØLVy esa izR;sd vUrjki`"Bh; dks.k

90º dk gksrk gS

(4) prq"dks.kh; fØLVyh; ra= esa pkj bdkbZ lsy gksrs gS

H-19/38


00CE314001


38. Select the incorrect statement using phasediagram of a substance

C

O

BAP2

P1

P

TT6 T1T4 T2 T5 T3

(1) At pressure P2, T4 is melting point(2) Substance will be in liquid state if

temperature T is 'T4 < T < T5' at P2 pressure(3) At T3 temperature vapour pressur of liquid

is P1

(4) Sublimation temperature is always less thantriple point

39. According to the kinetic theory of gases, in anideal gas, between two successive collisions agas molecule travels -(1) In a straight line path(2) with an accelerated velocity(3) In a circular path(4) In a wavy path

40. To obtain maximum mass of NO2 from a givenmass of a mixture of NH3 and O2, the ratio ofmass of NH3 to O2 should be

NH3 + O2 ¾¾® NO2 + H2O

(1) 1740

(2) 1764

(3) 1756

(4) 1732

38. ,d inkFkZ ds voLFkk oØ dk mi;ksx djrs gq, xyrdFku dk p;u dhft,

C

O

BAP2

P1

P

TT6 T1T4 T2 T5 T3

(1) P2 nkc ij] T4 xyukad fcUnq gS

(2) inkFkZ nzo vOkLFkk esa gksxk ;fn P2 nkc ij rki T,

'T4 < T < T5' gks

(3) T3 rki ij nzo dk ok"i nkc P1 gksrk gS

(4) Å/oZikru rki lnSo f=d fcUnq ls de gksrk gS39. xSlksa ds xfrd fl¼kUr ds vuqlkj ,d vkn'kZ xSl esa nks

Øekxr VDdjksa ds e/; xSl dk ,d v.kq xfr djrk gS-(1) ,d lh/kh js[kk iFk esa(2) ,d Rofjr osx ds lkFk(3) ,d o`Ùkh; iFk esa

(4) ,d rjax iFk esa

40. NH3 rFkk O2 ds ,d feJ.k ds ,d fn;s x;s nzO;ekuls NO2 dk vf/kdre nzO;eku izkIr djus ds fy, NH3

ls O2 ds nzO;eku dk vuqikr gksuk pkfg,NH3 + O2 ¾¾® NO2 + H2O

(1) 1740

(2) 1764

(3) 1756

(4) 1732

H-20/38


00CE314001


41. Na H O2 (A) (B) (C) (D)

(E) + F(g)

CO2 SO2 Evaporation

D

In excess SO2Excess

D

(G) + HCompound D is :

(1) Na2SO3

(2) Na2S2O5.xH2O

(3) Na2S

(4) Na2SO4

42. In Which of the following process nitrogenundergoes in oxidation process :

(1) N2 ® HN3 (2) N2O4 ® 2NO2

(3) NO¯3 ® N2O5 (4) N2O ® NO

43. Case hardening is a process of heating steel inatmosphere of :

(1) Carbon dioxide

(2) Ammonia

(3) Charcoal

(4) Oxygen

44. For an octahedral complex, which of thefollowing d-electronic configuration will givemaximum magnitude of crystal fieldstabilisation energy, in terms of DO :(1) Low spin d5 (2) Low spin d4

(3) High spin d7 (4) High spin d6

41. Na H O2 (A) (B) (C) (D)

(E) + F(g)

CO2 SO2

D

SO2 ds vkf/kD;esas ok"iuvkf/kD;

D

(G) + H

;kSfxd D gS :

(1) Na2SO3

(2) Na2S2O5.xH2O

(3) Na2S

(4) Na2SO4

42. fuEu esa ls dkSuls izØe esa ukbVªkstu dk vkWDlhdj.k gksjgk gS :

(1) N2 ® HN3 (2) N2O4 ® 2NO2

(3) NO3 ® N2O5 (4) N2O ® NO

43. fuEu esa ls fdlds ok;qeaMy esa LVhy dks xeZ djus dsizØe dks dsl dBksjhdj.k dgk tkrk gS :

(1) dkcZu MkbvkWDlkbM

(2) veksfu;k

(3) pkjdksy

(4) vkWDlhtu

44. fuEu esa ls dkSulk d-bysDVªkWfu; foU;kl ,d v"VQydh;ladqy ds fy,] DO ds inksa esa fØLVy {ks= fLFkjhdj.kÅtkZ dk lokZf/kd ifjek.k nsxk :(1) fuEu pØ.k d5 (2) fuEu pØ.k d4

(3) mPp pØ.k d7 (4) mPp pØ.k d6

H-21/38


00CE314001


45. Dichloronitronium ion [ONCl2]+ and thionyl

chloride (OSCl2), which has higher

Cl – µX – Cl bond angle.(where X = central atom N and S)

(1) µCl N Cl- - > $Cl S Cl- -

(2) µCl N Cl- - < $Cl S Cl- -

(3) µCl N Cl- - = $Cl S Cl- -(4) None of these

46. Which of the following molecule has zerodipole moment :

(1)

OH

OH(2)

Cl

Cl

(3)

OH

NO2

(4) All of these

47. Which of the following species show(s)synergic bonding :

(1) [Mo(CO)6]

(2) [Mn(CO)6]¯

(3) [Ni(CN)4]–4

(4) All above species have synergic bonding

45. MkbDyksjksukbVªksfu;e vk;u [ONCl2]+ rFkk Fkk;ksfuy

DyksjkbM (OSCl2), esa ls fdlesa Cl – µX – Clca/k dks.k vf/kd gS(tgk¡ X = dsUæh; ijek.kq N rFkk S)

(1) µCl N Cl- - > $Cl S Cl- -

(2) µCl N Cl- - < $Cl S Cl- -

(3) µCl N Cl- - = $Cl S Cl- -(4) buesa ls dksbZ ugha

46. fuEu esa ls dkSuls v.kq dk f}/kzqo vk?kw.kZ 'kwU; gS :

(1)

OH

OH(2)

Cl

Cl

(3)

OH

NO2

(4) mijksDr lHkh

47. fuEu esa ls dkSulh Lih'kht esa fluftZd ca/ku mifLFkrgS :

(1) [Mo(CO)6]

(2) [Mn(CO)6]¯

(3) [Ni(CN)4]–4

(4) mijksDr lHkh Lih'kht esa fluftZd ca/ku mifLFkr gS

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48. Purple solution of KMnO42

2 4

BaClNa SO¾¾¾® Pink ppt.

Select the correct statement about pink ppt.(1) Pink ppt. is of MnS(2) Pink ppt. is of BaSO4

(3) Pink ppt. is of Mn(OH)2

(4) Pink ppt. does not obtained49. Select the correct statement :

(1) Photography is based upon photosensitivity

of silver halide mainly AgBr

(2) Sodium thiosulphate is used to remove the

unexposed AgBr from photographic film

(3) AuCl3 is used in toning

(4) All are correct

50. Predict the physical state of metal obtained inthe following reduction reaction shown by thisdiagram :

C CO®

MM

O®

DGº

TMO(s) + C(s) ® M( ? ) + CO(g)(1) Solid(2) Gas(3) May be liquid or gas(4) May be liquid or solid

48. KMnO4 dk cSaxuh foy;u 2

2 4

BaClNa SO¾¾¾® xqykch vo{ksi

xqykch vo{ksi ds lUnHkZ esa lgh dFku pqfu, :(1) xqykch vo{ksi] MnS dk gS(2) xqykch vo{ksi] BaSO4 dk gS(3) xqykch vo{ksi] Mn(OH)2 dk gS

(4) xqykch vo{ksi izkIr ugha gksrk gS49. lgh dFku pqfu;sa :

(1) QksVksxzkQh] flYoj gSykbM eq[;r% AgBr dh izdk'kh;lfØ;rk ij vk/kkfjr gSA

(2) QksVksxzkfQd fQYe ls] 'ks"k jgs (unexposed) AgBrdks gVkus ds fy, lksfM;e Fkk;kslYQsV dk iz;ksx fd;ktkrk gS

(3) Vksfuax (toning) ds fy, AuCl3 dk iz;ksx fd;ktkrk gS

(4) lHkh lgh gS50. fuEu fp= esa iznf'kZr vip;u vfHkfØ;k esa izkIr /kkrq

dh HkkSfrd voLFkk crkbZ;sa :

C CO®

MM

O®

DGº

TMO(s) + C(s) ® M( ? ) + CO(g)(1) Bksl(2) xSl(3) æo ;k xSl gks ldrh gS

(4) æo ;k Bksl gks ldrh gS

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51.

OHH SO2 4 A

(Major)Major product A is :

(1) (2)

(3) (4)

52. Which is the most acidic hydrogen on propylacetoacetate ?CH – CO – CH – CO – CH – CH – CH3 2 2 2 2 3V W X Y

(1) V (2) W(3) X (4) Y

53. (a)

OH

C–Cl

OO = C

H

x PhMgBr(excess) (A)

(b)

O

OCO2H

(B) + (C)gas

Number of moles of PhMgBr consumed (x) in(a) + molecular mass of (C)(1) 42 (2) 44(3) 46 (4) 48

51.

OHH SO2 4 A

(eq[;)eq[; mRikn A gS&

(1) (2)

(3) (4)

52. izksfiy ,slhVks,slhVsV ij lokZf/kd vEyh; gkbMªkstudkSulk gS&CH – CO – CH – CO – CH – CH – CH3 2 2 2 2 3V W X Y

(1) V (2) W(3) X (4) Y

53. (a)

OH

C–Cl

OO = C

H

x PhMgBr(vkf/kD;) (A)

(b)

O

OCO2H

(B) + (C)xSl

(a) esa PhMgBr ds [kpZ gq,s eksyksa dh la[;k (x) + (C)dk vkf.od nzO;eku gS&

(1) 42 (2) 44

(3) 46 (4) 48

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54.O3Zn

(1 mole)

(A) conc.KOH (B)

End product (B) of above reaction is :

(1)CHO

CH OH2

(2) CO K2

CH OH2

– +

(3)CH OH2

CH OH2(4)

– +CO K2

– +CO K2

55. Give the product from the following reactionsequence :

O

OHEthyleneglycol

H+

PCC

EtMgBrH+

H O2

(1) OH

HO

(2) H

O

O

(3) (4)

OH

54.O3Zn

(1 mole)

(A) lkUnz KOH (B)

mijksDr vfHkfØ;k dk vfUre mRikn (B) gS&

(1)CHO

CH OH2

(2) CO K2

CH OH2

– +

(3) CH OH2

CH OH2(4)

– +CO K2

– +CO K2

55. fuEu vfHkfØ;k Øe ls mRikn nhft,s&

,sfFkyhu XykbdkWy

O

OH H+

PCC

EtMgBrH+

H O2

(1) OH

HO

(2) H

O

O

(3) (4)

OH

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56. Product obtained in the given reactions areshown below :

CO H2

CO H2 CO +x,2

CO H2

CO H2 Br CO +y2

Br Br Br

The number of possible product for x and y is

(1) 1, 1 (2) 1, 2

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

57.

NMe3+

HO–

A(Major)

Major product A is :-

(1) (2)

(3) (4)

56. nh x;h vfHkfØ;kvksa esa izkIr mRikn uhps fn[kk,s x;s gSaA

CO H2

CO H2 CO +x,2

CO H2

CO H2 Br CO +y2

Br Br Br

x rFkk y ds fy, mRikn dh lEHkkfor la[;k gS&

(1) 1, 1 (2) 1, 2

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

57.

NMe3+

HO–

A(eq[;)

eq[; mRikn A gS :-

(1) (2)

(3) (4)

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58. Give the major product of the followingreaction

O

H

N CH COCl3AlCl3

Major product

(1)O

H

N

COCH3

(2)

O

H

NCOCH3

(3)O

H

N

H COC3

(4)O

H

N

COCl

58. fuEu vfHkfØ;k dk eq[; mRIkkn nhft,s&

O

H

N CH COCl3AlCl3

eq[; mRikn

(1)O

H

N

COCH3

(2)O

H

NCOCH3

(3)O

H

N

H COC3

(4)O

H

N

COCl

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59.1. BH / THF3

2. H O / OH2 2–

Product of reaction is -

(1)

OH

HH

CH3

(2) CH3

CH3

(3) CH3OH (4)

H

HOH

CH3

60. Product of reaction is :O

O

N–H (i) NaOBr

(ii) H+

(1)COOH

NH2

(2)

CH –COOH2

NH2

(3)

COOH

CH –NH2 2

(4)

O

NH

59.1. BH / THF3

2. H O / OH2 2–

vfHkfØ;k dk mRikn gS&

(1)

OH

HH

CH3

(2) CH3

CH3

(3) CH3OH (4)

H

HOH

CH3

60. vfHkfØ;k dk mRikn gS&

O

O

N–H (i) NaOBr

(ii) H+

(1)COOH

NH2

(2)

CH –COOH2

NH2

(3)

COOH

CH –NH2 2

(4)

O

NH

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61. A variable straight line AB divides thecircumference of the circle x2 + y2 = 25 in theratio 1 : 2. If a tangent CD is drawn to thesmaller arc parallel to AB, such that ABCD isa rectangle, then locus of C & D is (as shownin the figure) :-

D C

A BO

(1) 2 2 175

x y4

+ = (2) x2 + y2 = 36

(3) x2 + y2 = 40 (4) x2 + y2 = 2062. The number of shortest

A

M

N

Dpaths from point A to D(as shown in figure)(1) 276(2) 186(3) 150(4) 126

63. If n is a factor of 72, such that xy = n,then number of ordered pairs (x, y) are :-(where x, y Î N)(1) 40 (2) 50(3) 60 (4) 70

61. ,d pj ljy js[kk AB, o`Ùk x2 + y2 = 25 dh ifjfèk dks1 : 2 vuqikr esa foHkkftr djrh gSA ;fn y?kq pkiij AB ds lekUrj ,d Li'kZjs[kk CD bl izdkj [khaphtkrh gS fd ABCD ,d vk;r gS] rks C rFkk D dkfcUnqiFk gksxk (fp=kuqlkj)

D C

A BO

(1) 2 2 175

x y4

+ = (2) x2 + y2 = 36

(3) x2 + y2 = 40 (4) x2 + y2 = 2062. fp=kuqlkj fcUnq A ls D rd

A

M

N

Dds lcls NksVs iFkksa dh la[;kgksxh(1) 276(2) 186(3) 150(4) 126

63. ;fn 72 dk ,d xq.ku[k.M n bl izdkj gS fdxy = n gS] rks Øfer ; qXeksa (x, y) dh la[;k gksxh %&(tgk¡ x, y Î N)(1) 40 (2) 50(3) 60 (4) 70

PART C - MATHEMATICS

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64. If variable point (x, y) satisfies the equation

x2 + y2 – 8x – 6y + 9 = 0, then range of yx

is

(1) 7 7

,24 24

é ù-ê úë û(2)

7,

24é ö- ¥ ÷êë ø

(3) 7

,24

é ö¥÷êë ø(4) (–¥,¥)

65. Coefficient of xn–6 in the expansion

n n n n0 1 1 2n n

0 1

C C x C Cn xC 2 C

é ù é ùæ ö æ ö+ +- -ê ú ê úç ÷ ç ÷

è øè ø ë ûë û

n n n n2 3 n 1 nn n

2 n 1

C Cx x C C.....3 C n C

-

-


è ø è øë û ë û

is equal to

(where n = n . (n – 1) . (n – 2).... 3.2.1)(1) nC6(n + 1)6 (2) nC6.n

6

(3) nC6(n + 2)6 (4) nC5(n + 1)5

66. If 2

n 2

2n n 1x2n 3n 2

+ +=

- +,

then rn r

ir 1 i 1i 1

x 2 (2i 1)= ==

é ùæ ö- -ê úç ÷

è øë ûå åÕ is equal to -

(1) n(n 1)

2+

(2) n(n 3)

2+

(3) n(n 1)

2-

(4) 2n(n + 1)

64. ;fn pj fcUnq (x,y) lehdj.k x2 + y

2 – 8x – 6y + 9 = 0 dks

lUrq"V djrk gks] rks yx

dk ifjlj gksxk -

(1) 7 7

,24 24

é ù-ê úë û(2)

7,

24é ö- ¥ ÷êë ø

(3) 7

,24

é ö¥÷êë ø(4) (–¥,¥)

65.

n n n n0 1 1 2n n

0 1

C C x C Cn xC 2 C


è øè ø ë ûë û

n n n n2 3 n 1 nn n

2 n 1

C Cx x C C.....3 C n C

-

-


è ø è øë û ë û

ds izlkj esa xn–6 dk xq.kkad gksxk

(tgk¡ n = n . (n – 1) . (n – 2).... 3.2.1)

(1) nC6(n + 1)6 (2) nC6.n6

(3) nC6(n + 2)6 (4) nC5(n + 1)5

66. ;fn 2

n 2

2n n 1x2n 3n 2

+ +=

- + gks]

rks rn r

ir 1 i 1i 1

x 2 (2i 1)= ==

é ùæ ö- -ê úç ÷

è øë ûå åÕ dk eku gksxk -

(1) n(n 1)

2+

(2) n(n 3)

2+

(3) n(n 1)

2-

(4) 2n(n + 1)

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67. If z is a complex number satisfying |z – 3| < 5,

then range of |z + 3i| is (where i 1= - ).

(1) 5 3 2, 5 3 2é ù- +ë û

(2) 3 2 5,3 2 5é ù- +ë û

(3) 0, 5 3 2é ù+ë û

(4) 0, 5 3 2é ù-ë û68. The range of values of the function

1ƒ(x)2 3sin x

=-

is -

(1) 11,5

é ù-ê úë û(2) [–1, 5]

(3) 1( , 1] ,5

é ö-¥ - È ¥÷êë ø(4) [ )1, 1,

5æ ù-¥ È ¥ç úè û

69. If P lies in the first quadrant on the ellipse2 2

2 2

x y 1a b

+ = (where a > b), and tangent &

normal drawn at P meets major axis at thepoints T & N respectively, then the value of

2 1 2 1

2 1 2 1

(| F N | | F N |)(| F T | | FT |)(| F N | | F N |)(| F T | | FT |)

+ -- +

is equal to

(where F1 & F2 are the foci (ae, 0) & (–ae, 0)respectively)

(1) 1 (2) 2a (3) 2b (4) ae

67. ;fn lfEeJ la[;k z, |z – 3| < 5 dks lUrq"V djrh gS] rks

|z + 3i| dk ifjlj gksxk (tgk¡ i 1= - )

(1) 5 3 2, 5 3 2é ù- +ë û

(2) 3 2 5,3 2 5é ù- +ë û

(3) 0, 5 3 2é ù+ë û

(4) 0, 5 3 2é ù-ë û68. Qyu

1ƒ(x)2 3sin x

=-

ds ekuksa dk ifjlj gksxk -

(1) 11,5

é ù-ê úë û(2) [–1, 5]

(3) 1( , 1] ,5

é ö-¥ - È ¥÷êë ø(4) [ )1, 1,

5æ ù-¥ È ¥ç úè û

69. ;fn P, nh?kZoÙk 2 2

2 2

x y 1a b

+ = (tgk¡ a > b) ij izFke

prqFkk±'k esa fLFkr gS rFkk fcUnq P ij [khaph xbZ Li'kZ js[kko vfHkyEc nh?kZ v{k dks Øe'k% T rFkk N ij feyrs gSa]

rks 2 1 2 1

2 1 2 1

(| F N | | F N |)(| F T | | FT |)(| F N | | F N |)(| F T | | FT |)

+ -- +

dk eku gksxk

(tgk¡ F1 rFkk F2 Øe'k% ukfHk;k¡ (ae,0) rFkk (–ae,0) gS)

(1) 1 (2) 2a (3) 2b (4) ae

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70. Let a1, a2, ........ a101 is a group of real numberssuch that ai > ai+1 for all values of i and meansquare deviation of the group is minimum aboutthe number 'a51', then mode of the group willbe -(1) 2a51 (2) a51

(3) a50 (4) a52

71. Which of the following statement is atautology?

(1) ( )( ){ } ( ) ( ){ }p q t p q r p té ùÙ Ú Ù ® Ú Ù Úë û

( )~ q r ~ p« Ú ®é ùë û

(2) ( )( ){ } [ ]p q t p (q r) pÙ Ú Ù « Ú ®

(3) ( )( ){ } [ ]p q t p q r pÙ Ú Ù « Ù Ù

(4) ( )( ){ }p q t p tÙ Ú Ù « (where t denotes tautology)

72.{ }( ){ }

{ }

{ }

1x

x

x 1

e1 x

elim1 cos x+®

+ -

-

(where {.} denotes fractional part function)

(1) 0 (2) 2e3

(3) 3e2

(4) does not exist

70. ekuk a1, a2, ........ a101 okLrfod la[;kvksa dk ,d lewgbl izdkj gS fd i ds lHkh ekuksa ds fy, ai > ai+1 gS rFkkla[;k a51 ds lkis{k lewg dk oxZ ek/; fopyu U;wuregS] rks lewg dk cgqyd gksxk -

(1) 2a51 (2) a51

(3) a50 (4) a52

71. fuEu esa ls dkSulk dFku iqu#fDr gS ?

(1) ( )( ){ } ( ) ( ){ }p q t p q r p té ùÙ Ú Ù ® Ú Ù Úë û

( )~ q r ~ p« Ú ®é ùë û

(2) ( )( ){ } [ ]p q t p (q r) pÙ Ú Ù « Ú ®

(3) ( )( ){ } [ ]p q t p q r pÙ Ú Ù « Ù Ù

(4) ( )( ){ }p q t p tÙ Ú Ù «

(tgk¡ t, iqu#fDr dks n'kkZrk gS)

72.{ }( ){ }

{ }

{ }

1x

x

x 1

e1 x

elim1 cos x+®

+ -

-

(tgk¡ {.} fHkUukRed Hkkx Qyu dks n'kkZrk gS)

(1) 0 (2) 2e3

(3) 3e2

(4) fo|eku ugha gSA

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73. Number of points of discontinuity of thefunction ƒ(x) = sin({ 2x + [2x] + [3–x]}) forx Î [0,4] is(where [.] and {.} denotes greatest integer andfractional part function respectively) -(1) 5 (2) 4 (3) 15 (4) 16

74.( )

( ) ( )

2

2

d sin

1 sin cos 3

q

- q qò is equal to-

(1)

p

q

pæ öq +ç ÷+ç ÷pç ÷q -ç ÷

è ø

l

tan6

seccos cos2 6n e C

3 cos cos6

(2)

p

q


è ø

l

tan6

coscos cos2 6n e C

3 cos cos6

(3)

p

p-q


è ø

l

tan6

sec( )cos cos2 6n e C

3 cos cos6

(4)

p

p-q


è ø

l

tan6

cos( )cos cos2 6n e C

3 cos cos6

(where C is constant of integration)

73. mu fcUnqvksa dh la[;k] tgk¡ Qyu

ƒ(x) = sin({2x + [2x] + [3–x]}), x Î [0,4] esa vlarr~

gS (tgk¡ [.] rFkk {.} Øe'k% egÙke iw.kk±d Qyu rFkk

fHkUukRed Hkkx Qyu dks n'kkZrs gS) -

(1) 5 (2) 4 (3) 15 (4) 16

74.( )

( ) ( )

2

2

d sin

1 sin cos 3

q

- q qò dk eku gksxk -

(1)

p

q


è ø

l

tan6

seccos cos2 6n e C

3 cos cos6

(2)

p

q


è ø

l

tan6

coscos cos2 6n e C

3 cos cos6

(3)

p

p-q


è ø

l

tan6

sec( )cos cos2 6n e C

3 cos cos6

(4)

p

p-q


è ø

l

tan6

cos( )cos cos2 6n e C

3 cos cos6

(tgk¡ C lekdyu dk vpj gS)

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75. Solution of differential equation(1 + e2y) etan–1x dx – (1 + x2) (ey + (ey–1)2)dy = 0is-

(1) ( ) ( )-

= - +l1tan xn y tan y e C

(2) ( )( )-

= - +l1tan xy n tan y e C

(3) ( ) ( )-

= - +l1tan xn y tan e y C

(4) ( )( )-

= - +l1tan xy n tan e y C

(where C is constant of integration)76. Area bounded on the left by y-axis, below by

x-axis, right by x2p

= above left by y = cosx

and above right by y = sinx is -

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

77. If ( ) ( ) ( )+= Înx 1ƒ x e ; n N , then value of 'n' for

which ( ) ( )=nn 2ƒ" 1 67 2 e is-

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

78. If ƒ is a differentiable function such thatƒ(2x + 1) = ƒ(1 – 2x) " xÎR then minimumnumber of roots of the equation ƒ'(x) = 0 inx Î (–5,10), given that ƒ(2) = ƒ(5) = ƒ(10), is-(1) 2 (2) 3 (3) 4 (4) 5

75. vody lehdj.k(1 + e2y) etan–1x dx – (1 + x2) (ey + (ey–1)2)dy = 0dk gy gksxk -

(1) ( ) ( )-

= - +l1tan xn y tan y e C

(2) ( )( )-

= - +l1tan xy n tan y e C

(3) ( ) ( )-

= - +l1tan xn y tan e y C

(4) ( )( )-

= - +l1tan xy n tan e y C

(tgk¡ C lekdyu dk vpj gS)

76. ck;sa ls y-v{k }kjk] uhps ls x-v{k }kjk] nk;sa ls x2p

=

}kjk] mGij ck;s a ls y = cosx rFkk mGij nk¡;s ls

y = sinx }kjk ifjc¼ {ks=Qy gksxk -

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

77. ;fn ( ) ( ) ( )+= Înx 1ƒ x e ; n N gks] rks n dk eku ftlds

fy, ( ) ( )=nn 2ƒ" 1 67 2 e gks] gksxk -

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

78. ;fn ƒ ,d vodyuh; Qyu bl izdkj gS fdƒ(2x + 1) = ƒ(1 – 2x) " x Î R gS] rks x Î (–5,10)esa lehdj.k ƒ'(x) = 0 ds ewyksa dh U;wure la[;k gksxh]fn;k gS ƒ(2) = ƒ(5) = ƒ(10) gS -(1) 2 (2) 3 (3) 4 (4) 5

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79. Let a function

n(3x [3x]) ; 3x n;n Nƒ(x)

n(sgn(3x)) ; 3x n;n N- - ¹ Îì

= í = Îî

l

l,

(where [.] and sgn(x) denotes greatest integerfunction and signum function respectively)then number of point(s), where ƒ(x) isminimum in x Î (0, 5), is -(1) 0 (2) 4(3) 5 (4) 14

80. Area bounded by the tangents of the curvegiven by y = sinq cos2q ; x = sin2q cosq, whichare parallel to co-ordinate axes (other thanaxes), is-

(1) 4

27 (2) 274 (3)

1627 (4)

2716

81. If n

2n

n

I tan {x}dx-

= ò , then (where {.} denotes

fractional part function and n Î N) -(1) I1I2 = 8(sec21 – 2 – I1)

(2) I1I2 = 8(sec21 – 2 + I1)

(3) I1I2 = 8(sec21 + 2 – I1)

(4) I1I2 = 8(sec21 + 2 + I1)82. If focus divides a focal chord of the parabola

y2 = 16x into 2 parts having lengths a and c,such that a,b,c are in H.P., then value of b isequal to -(1) 2 (2) 4(3) 6 (4) 8

79. ekuk Qyu

n(3x [3x]) ; 3x n;n Nƒ(x)

n(sgn(3x)) ; 3x n;n N- - ¹ Îì

= í = Îî

l

l gks

(tgk¡ [.] rFkk sgn(x) Øe'k% egÙke iw.kk±d Qyu rFkkflXue Qyu dks n'kkZrk gS)rks fcUnqvksa dh la[;k] tgk¡ ƒ(x) dk eku x Î (0, 5) esaU;wure gks] gksxk -(1) 0 (2) 4(3) 5 (4) 14

80. ;fn y = sinq cos2q ; x = sin2q cosq, }kjk fn, x,oØ ij funsZ'kh v{kksa ds lekUrj (funsZ'kh v{kksa dks NksM+dj)[khaph xbZ Li'kZ js[kkvksa }kjk ifjc¼ {ks=Qy gksxk -

(1) 4

27 (2) 274 (3)

1627 (4)

2716

81. ;fn n

2n

n

I tan {x}dx-

= ò gks] rks (tgk¡ {.} fHkUukRed

Hkkx Qyu dks n'kkZrk gS rFkk n Î N) -(1) I1I2 = 8(sec21 – 2 – I1)

(2) I1I2 = 8(sec21 – 2 + I1)

(3) I1I2 = 8(sec21 + 2 – I1)

(4) I1I2 = 8(sec21 + 2 + I1)82. ;fn ijoy; y2 = 16x dh ukfHk fdlh ukHkh; thok dks

a rFkk c yEckbZ ds nks Hkkxksa esa bl izdkj gS foHkkftrdjrh gS] fd a,b,c gjkRed Js.kh esa gks] rks b dk ekugksxk -(1) 2 (2) 4(3) 6 (4) 8

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83. Let F1 & F2 be the foci of an ellipse 2 2x y

14 9

+ =

such that a ray from F1 strikes the ellipticalmirror at the point P and get reflected. Thenequation of angle bisector of the angle betweenincident ray and reflected ray can be -

(1) 5y x13

= + (2) 5y 2x13

= -

(3) x + y – 5 = 0 (4) 3x – 4y – 5 = 0

84. Number of integral points interior to the circlex2 + y2 = 10 from which exactly one real tangentcan be drawn to the curve

( ) ( )2 22 2x 5 2 y x 5 2 y 10+ + - - + =

are (where integral point (x, y) means x, y Î I)

(1) 12 (2) 14

(3) 16 (4) 18

85. If angles A,B,C of a DABC are 75º, 45º and60º respectively. Then ratio of the areas ofDOBC, DCOA and DAOB respectively is[where O is the circumcentre of the triangle]-

(1) 3 1: 2 : 6+ (2) 1: 2 : 3

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

83. ekuk F1 rFkk F2 nh?kZoÙk 2 2x y

14 9

+ = dh ukfHk;k¡ bl

izdkj gS fd F1 ls ,d fdj.k] nh?kZoÙkkdkj niZ.k ls fcUnq

P ij ijkofrZr gksrh gSA rks vkifrr fdj.k rFkk ijkofrZr

fdj.k ds dks.kk¼Zd dk lehdj.k fuEu gks ldrk gS -

(1) 5y x13

= + (2) 5y 2x13

= -

(3) x + y – 5 = 0 (4) 3x – 4y – 5 = 0

84. o`Ùk x2 + y2 = 10 ds vUnj fLFkr iw.kk±d fcUnqvksa dhla[;k tgk¡ ls oØ

( ) ( )2 22 2x 5 2 y x 5 2 y 10+ + - - + = ij

Bhd ,d okLrfod Li'kZ js[kk [khaph tk ldrh gks] gksxh(tgk¡ iw.kk±d fcUnqvksa (x, y) dk vFkZ x, y Î I gS)

(1) 12 (2) 14

(3) 16 (4) 18

85. ;fn f=Hkqt ABC ds dks.k A,B,C Øe'k% 75º, 45ºrFkk 60º gS] rks Øe'k% f=Hkqt OBC, COA rFkk AOBds {ks=Qy dk vuqikr gksxk [tgk¡ O f=Hkqt dk ifjdsUægS] -

(1) 3 1: 2 : 6+ (2) 1: 2 : 3

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

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86. If = q - + = + q -rr ˆ ˆ ˆ ˆˆ ˆa 2 sin i j 2k, b 2i 2 sin j k and

= - + qr 2ˆ ˆ ˆc 4i j 4cos k are coplanar. Then q canbe equal to

(1) ( )nn 1

6p

p + - , n Î I

(2) ( )nn 1

4p

p + - , n Î I

(3) ( )nn 1

3p

p + - , n Î I

(4) ( )2n 12p

+ , n Î I

87. Let z1,z2,z3,w,z0,z'0 are the fixed points on

complex plane such that no 3 are collinear,

satisfying the condition

31 2

2 3 3 1 1 2

zz zArg Arg Arg

z z z z z z 2

æ ö æ ö æ öw-w- w - p= = =ç ÷ ç ÷ ç ÷- - -è øè ø è ø

and z1,z2,z3 satisfies the equation |z – z0| = R1,

z2,w,z3 satisfies the equation |z – z0'| = R2 then

1

2

RR

is equal to-

(1) 1 (2) 2

(3) 3 (4) 4

86. ;fn = q - + = + q -rr ˆ ˆ ˆ ˆˆ ˆa 2 sin i j 2k, b 2i 2sin j k rFkk

= - + qr 2ˆ ˆ ˆc 4i j 4cos k leryh; gks] rks q dk ekufuEu gks ldrk gS -

(1) ( )nn 1

6p

p + - , n Î I

(2) ( )nn 1

4p

p + - , n Î I

(3) ( )nn 1

3p

p + - , n Î I

(4) ( )2n 12p

+ , n Î I

87. ekuk z1,z2,z3,w,z0,z'0 lfEeJ ry ij fLFkj fcUnq bl

izdkj gS fd dksbZ Hkh rhu fcUnq lejs[kh; ugha gS] izfrcUèk

31 2

2 3 3 1 1 2

zz zArg Arg Arg

z z z z z z 2

æ ö æ ö æ öw-w- w - p= = =ç ÷ ç ÷ ç ÷- - -è øè ø è ø

dks lUrq"V djrs gS rFkk z1,z2,z3 lehdj.k |z – z0| = R1

dks rFkk z2,w,z3 lehdj.k |z – z0'| = R2 dks lUrq"V

djrs gks] rks 1

2

RR

dk eku gksxk -

(1) 1 (2) 2

(3) 3 (4) 4

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88. Let = + +r ˆ ˆ ˆu ai bj ck

= + +r ˆ ˆ ˆv bi cj ak

& ˆ ˆ ˆw ci aj bk x y= + + = l + mr r r

where [u v w] 0=r r r

& (a + b + c), l, m ¹ 0 ,

then the vectors x, y, u,v,wr r r r r

are-

(1) collinear(2) coplanar(3) non-coplanar(4) nothing can be said

89. 3 numbers are chosen from first 15 naturalnumbers, then probability that the numbers arein arithmetic progression-

(1) 25

(2) 6

85

(3) 15

215

3

CC

(4) 7

65

90. If P & Q are two non-singular matrices of thesame order such that Qr = I, for some integerr > 1, then P–1Qr–1P – P–1Q–1P is equal to -(1) O (2) 2I(3) I (4) –I(where I is identity matrix and O is null matrix)

88. ekuk = + +r ˆ ˆ ˆu ai bj ck

= + +r ˆ ˆ ˆv bi cj ak

rFkk ˆ ˆ ˆw ci aj bk x y= + + = l + mr r r

tgk¡ [u v w] 0=r r r

rFkk (a + b + c), l, m ¹ 0 gks]

rks lfn'k x, y, u,v,wr r r r r

gksaxs -

(1) lejs[kh;(2) leryh;(3) vleryh;(4) dqN ugha dgk tk ldrk

89. izFke 15 izkd`r la[;kvksa esa ls 3 la[;kvksa dk p;ufd;k tkrk gS] rks p;fur la[;kvksa ds lekUrj Js.kh esagksus dh izkf;drk gksxh -

(1) 25

(2) 6

85

(3) 15

215

3

CC

(4) 7

65

90. ;fn P rFkk Q nks leku dksfV ds O;qRØe.kh; vkO;wg blizdkj gS fd dqN iw.kk±dksa r > 1 ds fy, Qr = I gks]rks P–1Qr–1P – P–1Q–1P dk eku gksxk -(1) O (2) 2I(3) I (4) –I(tgk¡ I rRled vkO;wg rFkk O 'kwU; vkO;wg gS)

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