hindi classroom contact programme jee(main)

Name of the Candidate (in Capitals)ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :Form Number : in figuresQkWeZ uEcj : vadksa esa

: in words: 'kCnksa esa

Centre of Examination (in Capitals) :ijh{kk dsUæ (cM+s v{kjksa esa) :Candidate’s Signature : Invigilator’s Signature :ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :

Read carefully the Instructions this Test Booklet.bl ijh{kk iq fLrdk ij fn, funsZ'kk s a dks /;ku ls i<+ s aA

Do not open this Test Booklet until you are asked to do so.bl ijh{kk iqfLrdk dks tc rd uk [kk sysa tc rd dgk u tk,A

)0000CJA102120026)(0000CJA102120026) Test Pattern

egRoiw.kZ funs Z'k :1. 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 gSaA

2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk

iqfLrdk@mÙkj i= ij dgha vkSj u fy[ksaA

3. ijh{kk dh vof/k 3 ?kaVs gSA4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 300 gSaA

5. bl ijh{kk iq fLrdk esa rhu Hkkx 1, 2, 3 gSa] ftlds izR;sd Hkkx esa HkkSfrdfoKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj izR;sd fo"k;esa 2 [k.M gSA(i) [k.M-I esa 20 cgqfodYih; iz'u gSA ftuds dsoy ,d fodYi

lgh gSaAvad ;k stuk : +4 lgh mÙkj ds fy,] 0 iz;kl ugha djus ijrFkk –1 vU; lHkh voLFkkvksa esaA

(ii) [k.M-II esa 10 la[;kRed eku izdkj ds iz'u gSA fdUgh 5 iz'uksadk mÙkj nhft,A fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksadks gh vad fn;s tk;saxsaAvad ;kstuk : +4 lgh mÙkj ds fy, rFkk 0 vU; lHkh voLFkkvksa esaA

6. mÙkj i= ds i`"B–1 ,oa i`"B–2 ij okafNr fooj.k ,oa mÙkj vafdr djusgsrq dsoy uhys@dkys ckWy ikbaV isu dk gh iz;ksx djsaA isfUly dk iz;ksxloZFkk oftZr gSA

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

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

9. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{kfujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iq fLrdkdks ys tk ldrs gSaA

10. mÙkj i= dks u eksM+ s a ,oa u gh ml ij vU; fu'kku yxk,s aA

Important Instructions :1. Immediately fill in the form number on this page of the

Test Booklet with Blue/Black Ball Point Pen. Use of pencilis strictly prohibited.

2. The candidates should not write their Form Numberanywhere else (except in the specified space) on the TestBooklet/Answer Sheet.

3. The test is of 3 hours duration.4. The Test Booklet consists of 90 questions. The maximum

marks are 300.5. There are three parts in the question paper 1,2,3

consisting of Physics, Chemistry and Mathematicshaving 30 questions in each subject and each subjecthaving Two sections.(i) Section-I contains 20 multiple choice questions

with only one correct option.Marking scheme : +4 for correct answer, 0 if notattempted and –1 in all other cases.

(ii) Section-II contains 10 Numerical Value Typequestions. Attempt any 5 questions.First 5 attempted questions will be considered formarking.Marking scheme : +4 for correct answer and 0 inall other cases.

6. Use Blue/Black Ball Point Pen only for writtingparticulars/marking responses on Side–1 and Side–2 of theAnswer Sheet. Use of pencil is strictly prohibited.

7. No candidate is allowed to carry any textual material,printed or written, bits of papers, mobile phone anyelectronic device etc, except the Identity Card inside theexamination hall/room.

8. Rough work is to be done on the space provided for thispurpose in the Test Booklet only.

9. On completion of the test, the candidate must hand overthe Answer Sheet to the invigilator on duty in the Room/Hall. However, the candidate are allowed to take awaythis Test Booklet with them.

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

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

JEE(Main+Advanced) : ENTHUSIAST & LEADER COURSE

Your Target is to secure Good Rank in JEE(Main) 2021

JEE(Main)AIOT

07-03-2021

Paper : Physics, Chemistry & Mathematicsç'u iq fLrdk : HkkSfrd foKku] jlk;u foKku rFkk xf.kr

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA-324005

+91-744-2757575 [email protected] www.allen.ac.in

Hin

di

ALLEN

/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026

SECTION–I : (Maximum Marks : 80)

� This section contains TWENTY questions.

� Each question has FOUR options (A), (B), (C)

and (D). ONLY ONE of these four options iscorrect.

� For each question, darken the bubble

corresponding to the correct option in the ORS.

� For each question, marks will be awarded in

one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. Two blocks A and B of masses m and 2mrespectively are connected by a spring offorce constant k. The masses are moving tothe right with uniform velocity v = 10m/seach, the heavier mass leading the lighterone. The spring in between them is of naturallength during the motion. Block B collideswith a third block C of mass m, at rest. Thecollision being completely inelastic. Findminimum speed (in m/s) of block A in thesubsequent motion.(A) 5 (B) 6 (C) 8 (D) 10

[k.M–I : (vf/kdre vad : 80)

� bl [k.M esa chl iz'u gSa

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj

(D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi

ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa

ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys

dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. æO;eku m rFkk 2m okys nks CykWd Øe'k% A rFkk B dks

cy fu;rkad k okyh ,d fLizax ls tksM+k x;k gSA izR;sd

æO;eku nk¡;h vksj ,dleku osx v = 10m/s ls xfr'khy

gS rFkk Hkkjh æO;eku] gYds æO;eku ls vkxs gSA xfr ds

nkSjku muds e/; yxh fLizax ewy yEckbZ esa gksrh gSA CykWd

B fojkekoLFkk esa fLFkr æO;eku m okys ,d rhljs CykWd

C ls Vdjkrk gSA VDdj iw.kZr;k vizR;kLFk gksrh gSA rnksijkUr

xfr esa CykWd A dh U;wure pky (m/s esa) gS%&

(A) 5 (B) 6 (C) 8 (D) 10

PART 1 - PHYSICS

ALLEN

07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg /37All India Open Test/Enthusiast & Leader Course 0000CJA102120026

2. In the given circuit diagram, switch wasconnected to position 1 for long time. Att = 0, switch is shifted from position 1 toposition 2. Find the final charge on capacitor2C.

C 2C

C

V 2

1

(A) 6CV

(B) 3CV

(C) 2

3CV

(D) 4

3CV

3. The region between two concentric spheresof radii R1 and R2(R1 < R2) has volume charge

density r =br , where b is constant and r is

the radial distance. A point charge q = 16 µCis placed at the origin, r = 0. Find the valueof b (in SI units) for which the electric fieldin the region between spheres is constant.

(Take : R2 = 2R1 = 4p

mm ).

R1R2 q

(A) 1 (B) 2 (C) 4 (D) 8

2. iznf'kZr ifjiFk esa fLop fLFkfr 1 ij yEcs le; ds fy,tqM+k gqvk FkkA t = 0 ij fLop dks fLFkfr 1 ls fLFkfr 2 ijfoLFkkfir fd;k tkrk gSA la/kkfj= 2C ij vfUre vkos'kKkr dhft;sA

C 2C

C

V 2

1

(A) 6CV

(B) 3CV

(C) 2

3CV

(D) 4

3CV

3. f=T;k R1 rFkk R2(R1 < R2) okys nks ladsUæh; xksyksa ds

eè; izHkkx dk vk;ru vkos'k ?kuRo r =br gS] tgk¡ b

,d vpj rFkk r f=T;h; nwjh gSA ,d fcUnq vkos'kq = 16 µC dks ewy fcUnq (r = 0 ij) j[kk tkrk gS A bdk og eku (SI bdkbZ esa ) Kkr dhft;s] ftlds fy;sxksyksa ds e/; izHkkx esa fo|qr {ks= fu;r gSA

(R2 = 2R1 = 4p

mm ysa)

R1R2 q

(A) 1 (B) 2 (C) 4 (D) 8

ALLEN


4. In a football game,a player wants to hit afootball from the ground to one of histeammates, who is running on the field. Takehitter position as origin & receiver's initial

position as ˆ ˆ2i 3j+ , where i & j are in theplane of field. Football's initial velocity

vector is ˆ ˆ ˆ2i 5 j 25k+ + & in the subsequent

run receiver displacement is ˆ5i & ˆ8j , then

ˆ ˆ2 i 4 j+ & then ˆ6 j . How far is the receiverfrom the football when football lands on

ground ? (assume ˆg 10k= -r )

(A) 10 (B) 17 (C) 26 (D) 135. A hollow hemi-sphere of mass m and radius

r is released from rest in the position shown.Knowing that the hemi-sphere rolls withoutsliding. Determine the reaction at thehorizontal surface at the instant when it hasrolled through 90°.

(A) 7mg

4 (B) 5mg

2

(C) 8mg

3 (D) 5mg

3

4. QqVckWy ds [ksy esa ,d f[kykM+h QqVckWy dks /kjkry lsmlds fdlh ,d lkFkh f[kykM+h dh vksj fdd ekjukpkgrk gS] tks fd eSnku ij nkSM+ jgk gSA xsan ij izgkj dhfLFkfr ewyfcUnq ij rFkk xsan idM+us okys f[kykM+h dh

izkjfEHkd fLFkfr ˆ ˆ2i 3j+ ij ekfu;s] tgk¡ i rFkk j eSnkuds ry es a g SA QqVck Wy dk i z kjfEHkd osx lfn'k

ˆ ˆ ˆ2i 5 j 25k+ + gS rFkk rnksijkUr nkSM+us esa xsan idM+us

okys f[kykM+h dk foLFkkiu ˆ5i rFkk ˆ8j , fQj ˆ ˆ2i 4 j+ ;

rFkk fQj ˆ6 j gSA tc QqVckWy /kjkry ij fxjrh gS rksxsan idM+us okys f[kykM+h dh QqVckWy ls nwjh gS%&

(ekuk ˆg 10k= -r )

(A) 10 (B) 17 (C) 26 (D) 135. æO;eku m rFkk f=T;k r okys ,d [kks[kys v¼Z xksys dks

iznf'kZr fLFkfr ls fojkekoLFkk ls NksM+k tkrk gSA ekuk fd

v¼Z xksyk fcuk fQlys yq<+drk gSA tc ;g 90° rd

yq<+d pqdk gksrk gS rks ml {k.k ij {kSfrt lrg ij izfrfØ;k

Kkr dhft;sA

(A) 7mg

4 (B) 5mg

2

(C) 8mg

3 (D) 5mg

3

ALLEN


6. Observer standing at the sea coast observes30 waves reaching the coast per minute. Thewavelength of each wave is 10m. Whenobserver drives a boat in sea against waves,with velocity 5 m/sec then n waves strikesthe boat per minute. Velocity of wave andvalue of n is respectively (in SI units)(A) 5, 30 (B) 10, 60(C) 10, 30 (D) 5, 60

7. In a meter Bridge experiment the resistanceof resistance box is 16 W , which is insertedin right gap. The null point is obtained at25cm from the left end. The leastcount formeter scale is 1mm. The percentage errorin calculating unknown resistance isapproximately:(A) 0.23 (B) 0.44 (C) 0.35 (D) 0.53

8. A source contains two phosphorus

radionuclides ( )3215 1/2 1P T T= & ( )33

15 1/2 2P T T= .

Initially 10% of decays comes from 3315P . How

long one must wait until 90% to do so?

(A)

1 2

4 n3t1 1n2T T

=æ ö

-ç ÷è ø

l

l

(B)

1 2

4 n3t1 1n2T T

=æ ö

+ç ÷è ø

l

l

(C)

1 2

2 n3t1 1n2T T

=æ ö

-ç ÷è ø

l

l

(D) None

6. leqæh rV ij [kM+k ,d izs{kd izfr feuV rV ij igq¡pus

okyh 30 rjaxksa dks izsf{kr djrk gSA izR;sd rjax dh rjaxnSè;Z

10m gSA tc izs{kd ,d uko dks leqæ esa rjaxksa ds fo:¼

5m/sec osx ls pykrk gS rks uko ls izfr feuV n rjaxs

Vdjkrh gSA rjax dk osx rFkk n dk eku (SI bdkbZ esa)Øe'k% gS%&(A) 5, 30 (B) 10, 60(C) 10, 30 (D) 5, 60

7. ,d ehVj lsrq iz;ksx esa izfrjks/k ckWDl dk izfrjks/k 16 W gS

tks fd nkfgus fjDr LFkku esa izfo"B djk;k x;k gSA 'kwU;

fcUnq cka;s fljs ls 25 cm nwjh ij izkIr gksrk gSA ehVj

iSekus dk vYirekad 1mm gSA vKkr izfrjks/k dh x.kuk

djus esa izfr'kr = qfV yxHkx gS%&

(A) 0.23 (B) 0.44 (C) 0.35 (D) 0.53

8. ,d L=ksr esa nks QkWLQksjl jsfM;ksU;wDykbM ( )3215 1/2 1P T T=

rFkk ( )3315 1/2 2P T T= gSA izkjEHk esa fo?kVu dk 10%,

3315P ls izkIr gksrk gSA 90% fo?kfVr gksus esa fdruk le;

yxksxk\

(A)

1 2

4 n3t1 1n2T T

=æ ö

-ç ÷è ø

l

l

(B)

1 2

4 n3t1 1n2T T

=æ ö

+ç ÷è ø

l

l

(C)

1 2

2 n3t1 1n2T T

=æ ö

-ç ÷è ø

l

l

(D) dksbZ ugha

ALLEN


9. Two blocks A and B of same mass M areconnected with each other with an idealstring of length 2l passing over an idealpulley. The block A is connected to a lightpan C with an ideal string as shown in figure.

A particle of mass 2M

is dropped on pan from

height 2l

as shown. If collision between

particle and pan is perfectly inelastic,acceleration of B just after the collision, is :-

MMA B

l

C

M2

l

2

(A) g (B) 9g

(C) 2g (D) g/18

9. leku æO;eku M okys nks CykWdksa A rFkk B dks ,d vkn'kZ

f?kjuh ij ls xqtjus okyh 2l yEckbZ dh fdlh vkn'kZ jLlh

ls fp=kuqlkj ,d&nwljs ls tksM+k tkrk gSA CykWd A dks ,d

gYds ik= C ls ,d vkn'kZ jLlh ls fp=kuqlkj tksM+k tkrk

gSA æO;eku 2M

okys ,d d.k dks 2l

Å¡pkbZ ls bl ik= ij

fxjk;k tkrk gSA ;fn d.k rFkk ik= ds e/; VDdj iw.kZr;k

vizR;kLFk gS rks VDdj ds Bhd i'pkr~ B dk Roj.k gS%&

MMA B

l

C

M2

l

2

(A) g (B) 9g

(C) 2g (D) g/18

ALLEN


10. Figure shows a large closed cylindrical tankcontaining water. Initially, the air trappedabove the water surface has a height h0 andpressure 2P0 where P0 is the atmosphericpressure. There is a hole in the wall of thetank at a depth h1 below the top from whichwater comes out. A long vertical tube isconnected as shown. Find the distance ofwater surface in the long tube from the topof the tank, when the water stops comingout of the hole.

2P0h0 h1

h2

(A) 2h0

(B) h0

(C) h2

(D) h1

10. fp=kuqlkj ikuh ls Hkjk ,d cM+k can csyukdkj Vsad n'kkZ;k

x;k gSA izkjEHk esa ty lrg ds Åij h0 Å¡pkbZ rd ok;q

fo|eku gS rFkk nkc 2P0 gS] tgk¡ P0 ok;qe.Myh; nkc gSA

Vsad dh nhokj esa 'kh"kZ ls uhps h1 xgjkbZ ij ,d fNæ gS

ftlls ty ckgj fudyrk gSA ,d yEch ÅèokZèkj uyh

fp=kuqlkj tqM+h gq;h gSA tc ty fNæ ls ckgj fudyuk

can gks tkrk gS rks yEch uyh esa ty lrg dh Vsad ds 'kh"kZ

ls nwjh Kkr dhft;sA

2P0h0 h1

h2

(A) 2h0

(B) h0

(C) h2

(D) h1

ALLEN


11. The K(alpha) X-ray of molybdenum haswavelength 71 pm. If the energy of amolybdenum atom with a K electronknocked out is 23.5 keV, what will be theapprox energy (in keV) of an anothermolybdenum atom when an L electron isknocked out?(A) 6 (B) 4(C) 2 (D) 8

12. A disc of radius 20 cm & mass 1kg is rollingwith slipping on a flat horizontal surface. Ata certain instant, the velocity of its center is4 m/s and its angular velocity is 10 rad/s. Thelowest contact point is O. Find angularmomentum about any point on the groundat shown instant.

O

4m/sP

10rad/sec

(A) 0.60 (B) 0.70(C) 1.00 (D) 1.20

11. eksfyCMsue dh K (,sYQk) X-fdj.k dh rjaxnS/;Z 71pm

gSA ;fn ,d K bysDVªkWu ckgj fudyus ds ckn eksfyCMsue

ijek.kq dh ÅtkZ 23.5 keV gS rks tc ,d L bysDVªkWu

fdlh vU; eksfyCMsue ijek.kq ls ckgj fudyrk gS rks bl

ijek.kq dh yxHkx ÅtkZ (keV esa) Kkr dhft;sA(A) 6 (B) 4(C) 2 (D) 8

12. f=T;k 20 cm rFkk æO;eku 1kg okyh ,d pdrh ,d

lery {kSfrt lrg ij fQlyrs gq, yq<+drh gSA fdlh

fo'ks"k {k.k ij blds dsUæ dk osx 4 m/s gS RkFkk bldk

dks.kh; osx 10 rad/s gSA fuEure laidZ fcUnq O gSA

iznf'kZr {k.k ij èkjkry ij fdlh fcUnq ds lkis{k dks.kh;

laosx Kkr dhft;sA

O

4m/sP

10rad/sec

(A) 0.60 (B) 0.70(C) 1.00 (D) 1.20

ALLEN


13. An airplane is in supersonic flight at analtitude h. At what smallest distance a (alongthe horizontal) from the observer on theground is there a point from which the soundemitted by the airplane motors reaches tothe observer earlier than sound emitted byplane when it is at point A that is directlyabove the observer. (vp = velocity of airplane,vs = velocity of sound.)

h

O

AS

a

(A) h

vv

vv

a

s

p

s

p

1

2

2

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

<

(B) h

vv

vv

a

s

p

s

p

12

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

<

(C) not possible for any real value of a

(D) h

vv

vv

a

s

p

s

p

1

2

2

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

>

13. ,d ok;q;ku h Å¡pkbZ ij ijk/ofud mM+ku (supersonic)esa mM+ jgk gSA /kjkry ij [kM+s ,d izs{kd ls {kSfrt dsvuqfn'k fdl U;wure nwjh a ij ,d fcUnq fo|eku gksxk]tgk¡ ls ok;q;ku dh eksVj }kjk mRlftZr /ofu] izs{kd ds BhdÅij fLFkr fcUnq A ij ok;q;ku ds gksus ij blls mRlftZr/ofu dh rqyuk esa izs{kd rd igys igq ¡psxh\ (vp = ok;q;kudk osx] vs = /ofu dk osx)

h

O

AS

a

(A) h

vv

vv

a

s

p

s

p

1

2

2

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

<

(B) h

vv

vv

a

s

p

s

p

12

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

<

(C) a ds fdlh okLrfod eku ds fy, laHko ugh gSA

(D) h

vv

vv

a

s

p

s

p

1

2

2

-÷÷ø

öççè

æ

÷÷ø

öççè

æ

>

ALLEN


14. A potential difference is applied between a

conducting sphere and a conducting plate

(“plus” on the sphere and “minus” on the

plate). The dimension of the plate are much

larger than the distance between sphere and

plate. A point positive charge is moved from

point 1 to point 2 parallel to the plate. Using

the above information choose the correct

statement.

1 2

+

(A) Work done by external agent in the

process is zero

(B) Net positive work will be done in moving

charge from 1 to 2 by external agent

(C) Net negative work will be done in

moving charge from 1 to 2 by external

agent

(D) Information is insufficient to give any

assertion regarding work.

14. ,d pkyd xksys rFkk ,d pkyd IysV ds e/; ,d foHkokUrj

vkjksfir fd;k tkrk gSA (xksys ij + rFkk IysV ij –) IysV

dh foek xksys rFkk IysV ds e/; nwjh dh rqyuk esa cgqr

vf/kd gSA ,d fcUnq /kukRed vkos'k dks IysV ds lekUrj

fcUnq 1 ls fcUnq 2 rd xfr djk;h tkrh gSA mijksDr tkudkjh

dk mi;ksx djrs gq;s lgh dFku pqfu;sA

1 2

+

(A) bl izfØ;k esa ckg~; dkjd }kjk fd;k x;k dk;Z 'kwU;

gSA

(B) vkos'k dks 1 ls 2 rd xfr djkus esa ckg~; dkjd }kjk

dqy /kukRed dk;Z fd;k tkrk gSA

(C) vkos'k dks 1 ls 2 rd xfr djkus esa ckg~; dkjd }kjk

dqy ½.kkRed dk;Z fd;k tkrk gSA

(D) dk;Z ds laca/k esa dksbZ Hkh tkudkjh nsus ds fy;s

tkudkjh vi;kZIr gSA

ALLEN


15. A long solenoid of cross-sectional radius Rhas a thin insulated copper wire ring tightlyput on its winding. One half of the ring hasthe resistance 10 times that of the other half.The magnetic induction produced by thesolenoid varies with time as B = bt, where bis a constant. Find the magnitude of theelectric field strength in the ring.

(A) Rb119

(B) Rb229

(C) 9 Rb(D) Rb

16. A particle of mass 1kg is moving wherepotential energy varies with displacementU = 10 (1 – cos 2x), then the time period ofsmall oscillation about origin will be

(A) 1220

p (B) 12

10p

(C) 140

p (D) 1

10p

15. vuqizLFk dkV f=T;k R okyh ,d yEch ifjukfydk esa

,d iryh dqpkyd rk¡cs ds rkj dh oy; gS ftls bldh

okbafMax ij dldj yisVk x;k gSA oy; ds vkèks Hkkx dk

izfrjksèk nwljs vk/ks Hkkx ds izfrjks/k ls 10 xquk gSA ifjukfydk

}kjk mRiUu pqEcdh; izsj.k] le; ds lkFk B = bt ds

vuqlkj ifjofrZr gksrk gS] tgk¡ b ,d fu;rkad gSA oy; esa

fo|qr {ks= lkeF;Z dk ifjek.k Kkr dhft;sA

(A) Rb119

(B) Rb229

(C) 9 Rb(D) Rb

16. æO;eku 1kg okyk ,d d. k , s l s L Fk k u ij

xfr'khy gS] tgk¡ fLFkfrt ÅtkZ] foLFkkiu ds lkFk

U = 10 (1 – cos 2x) ds vuqlkj ifjofrZr gksrh gS rks

ewyfcUnq ds lkis{k vYi nksyu dk vkorZdky gS%&

(A) 1220

p (B) 12

10p

(C) 140

p (D) 1

10p

ALLEN


17. Two blocks are connected by a string passingover a pulley as shown. Wedge is fixed. M issliding down with speed 12 m/s at an instantand suddenly the string breaks. Thecoefficient of friction between the blocks and

wedge is 12

. Find the velocity (in m/s) of

block M with respect to m after t = 2 secondsfrom the moment the string breaks.

[Take g = 10 ms–2 and 3 1.22

= . Assume that

the length of inclined is sufficiently long.]

30° 30°

Mm

(A) 10 (B) 15 (C) 2.5 (D) 2018. In an experiment, a boy draws graph

between 2v (y-axis) and 2a (x-axis) (wherev = velocity and a = tangential acceleration)for a simple pendulum. The graph is foundto be a straight line of negative slope makingan angle of 30° (with x-axis) whenexperiment was done on the ground and 60°(with x-axis) when experiment was done atheight h above the ground. Then h must be(R = radius of earth)(A) 0.5R (B) 0.24 R

(C) 0.73 R (D) R

17. nks CykWdksa dks ,d f?kjuh ij ls xqtjus okyh ,d jLlh }kjk

fp=kuqlkj tksM+k tkrk gSA ost fLFkj gSA M fdlh {k.k ij

12 m/s pky ls uhps dh vksj xfr'khy gS rFkk vpkud

jLlh VwV tkrh gSA CykWdksa rFkk ost ds e/; ?k"kZ.k xq.kkad

12 gSA jLlh ds VwVus ds {k.k ls t = 2 sec i'pkr~ CykWd

M dk m ds lkis{k osx (m/s esa) Kkr dhft;sA ekuk vkur

ry dh yEckbZ i;kZIr :i ls yEch gSA g = 10 ms–2

rFkk 3 1.22

= ysaA

30° 30°

Mm

(A) 10 (B) 15 (C) 2.5 (D) 2018. fdlh iz;ksx esa ,d yM+dk ,d ljy yksyd ds fy;s

2v (y-v{k) rFkk 2a (x-v{k) ds e/; vkjs[k cukrk gS]

tgk¡ v = osx rFkk a = Li'kZ js[kh; Roj.k gSA tc iz;ksx

èkjkry ij rFkk /kjkry ls h Å¡pkbZ ij fd;k tkrk gS rks

;g vkjs[k x v{k ls Øe'k% 30° o 60° dks.k cukus okyh

½.kkRed <ky dh lh/kh js[kk ds :i esa izkIr gksrk gSA hdk eku gksuk pkfg;s%& (R = i`Foh dh f=T;k)

(A) 0.5R (B) 0.24 R

(C) 0.73 R (D) R

ALLEN


19. In the given circuit, initial charge on thecapacitor is 1.5 C with polarity as shown infigure and current in the inductor is zero.Now at t = 0 if the key k is closed, then theminimum current through the key is

0.25F3WD

3VAk

2WC

6.8V 5H 3.8W 2W 2V

B

+ –

(A) 33/25 A (B) 34/25 A(C) 36/25 A (D) none

20. A stationary sound source ‘S’ of frequency334 Hz and a stationary observer ‘O’ areplaced near a reflecting surface moving awayfrom the source with velocity 2 m/sec shownin the figure. If the velocity of the soundwaves in air is V = 332 m/sec, the apparentfrequency of the echo is

SO

2m/s

(A) 332 Hz (B) 326 Hz(C) 334 Hz (D) 330 Hz

19. iznf'kZr ifjiFk esa la/kkfj= ij izkjfEHkd vkos'k dk eku 1.5

C gS rFkk /kzqork fp= esa n'kkZ;h x;h gS] tcfd izsjd dq.Myh

eas /kkjk dk eku 'kwU; gSA vc t = 0 ij dqath k dks can dj

fn;k tkrk gS rks dqath ls izokfgr U;wure /kkjk gksxh%&

0.25F3WD

3VAk

2WC

6.8V 5H 3.8W 2W 2V

B

+ –

(A) 33/25 A (B) 34/25 A(C) 36/25 A (D) dksbZ ugha

20. vkofr 334 Hz okyk ,d fLFkj /ofu L=ksr ‘S’ rFkk ,d

fLFkj izs{kd ‘O’ fp=kuqlkj L=ksr ls nwj 2 m/sec osx ls

xfr'khy ijkorZd lrg ds utnhd j[ks gq;s gSA ;fn ok;q

esa /ofu rjaxks dk osx V = 332 m/sec gS rks izfrèofu dh

vkHkklh vkofr gS%&

SO

2m/s

(A) 332 Hz (B) 326 Hz(C) 334 Hz (D) 330 Hz

ALLEN


SECTION-II : (Maximum Marks: 20)� This section contains TEN questions.

Attempt any 5 questions. First 5 attemptedquestions will be considered for marking.

� The answer to each question is aNUMERICAL VALUE.

� For each question, enter the correctnumerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimal places;e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ifanswer is 11.36777..... then both 11.36 and11.37 will be correct) by darken thecorresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

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•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases

[kaM-II : (vf/kdre vad : 20)� bl [kaM esa nl iz'u gSaA fdUgh 5 iz'uksa dk mÙkj nhft,A

fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksa dks gh vadfn;s tk;saxsaA

� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA

� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk %&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

ALLEN


1. A biconvex thin lens is prepared from glass

of refractive index µ2 = 32 . The two

converging surface have equal radii of 20cmeach. One of the surface is silvered fromoutside to make it reflecting. It is placed in

a medium of refractive index µ1 = 53 . This

system will behave as concave mirror of focal

length f, find value of f in cm.

2. Find the maximum kinetic energy (in eV) ofthe photoelectron liberated from the surface

of lithium (work function 2.15eVf = ) by

electromagnetic radiation whose electriccomponent varies with time as

E = a (1 + cos wt)cosw0t,where a is a constant, w = 12 × 1014 rads–1 andw0 = 3.6 × 1015 rads–1 (h = 6.6 × 10–34 inSI units)

3. A photosensitive surface is irradiated withlight of wavelength l, the stopping potentialis V. When the same surface is irradiatedwith the light of wavelength 2l, stopping

potential is 3V . Then the ratio of threshold

wavelength (lmax) and the l is

1. ,d f}mÙky irys ysal dks viorZukad µ2 = 32 okys dk¡p

ls cuk;k x;k gSA nksuksa vfHklkjh lrgksa esa izR;sd dh leku

f=T;k 20cm gSA buesa ls fdlh ,d lrg dks ijkorZd

cukus ds fy, ckgj dh vksj ls jtfrr fd;k tkrk gSA bls

viorZukad µ1 = 53 okys ek/;e esa j[kk tkrk gSA ;g

fudk; Qksdl nwjh f okys ,d vory niZ.k dh Hkk¡fr

O;ogkj djrk gSA f dk eku cm esa Kkr dhft;sA

2. fyFkh;e (dk;Z Qyu 2.15eVf = ) dh lrg ls ,sls fo|qr

pqEcdh ; fofdj.k }kjk mRlftZr izdk'k bysDVªkWu dh

vf/kdre xfrt ÅtkZ (eV esa) Kkr dhft;s ftldk fo|qr

?kVd] le; ds lkFk E = a (1 + cos wt)cosw0t,

ds vuqlkj ifjofr Zr gk srk gS _ tgk¡ a fu;rkad gS]

w = 12 × 1014 rads–1 rFkk w0 = 3.6 × 1015 rads–1

o SI bdkbZ esa h = 6.6 × 10–34 gSA

3. ,d izdk'klaosnh lrg ij rjaxnS/; Z l okyk izdk'k vkifrr

fd;k tkrk gS rFkk fujks/kh foHko V gSA ;fn blh lrg ij

rjaxnS/; Z 2l ds izdk'k dks vkifrr fd;k tkrk gS rks

fujksèkh foHko dk eku 3V gSA nSgyh rjaxnSè;Z ( )maxl

rFkk l dk vuqikr Kkr dhft;sA

ALLEN


4. One of the circuits for the measurement ofresistance by potentiometer is shown. Thegalvanometer is connected at point A andzero deflection is observed at lengthPJ = 30 cm. In second case the secondarycell is changed. Take ES = 10 V and r = 1W in1st reading and ES = 5V and r = 2W in2nd reading. In second case, the zerodeflection is observed at length PJ = 10 cm.What is the resistance R (in ohm) ?

EP( )

P Q

GJ

AR

rES

5. A uniform triangular plate of triangular area1m2, base length 60 cm and thickness 10 mm(prism like shape) is lying vertically on asmooth ground as shown in figure. Findmaximum value of cotq for which it doesnottopple.

qg

60cm

6. On a particular day, the maximumfrequency reflected from the ionosphere is8 MHz. On next day it was found to increaseto 9MHz. If ratio of maximum electrondensities of first day to maximum electrondensities of next day the ionosphere is n,find value of 81 × n.

4. fp=kuqlkj foHkoekih }kjk izfrjks/k ds ekiu ds fy, ,difjiFk n'kkZ;k x;k gSA xsYosuksehVj dks fcUnq A ij tksM+ktkrk gS rFkk yEckbZ PJ = 30 cm ij 'kwU; fo{ksi izsf{krfd;k tkrk gSA nwljs izdj.k esa f}rh;d lsy dks cnyfn;k tkrk gSA izFke ikB~;kad esa ES = 10 V ,oa r = 1WrFkk f}rh; ikB~;kad eas ES = 5V ,oa r = 2W yhft;sAnwljs izdj.k esa yEckbZ PJ = 10 cm ij 'kwU; fo{ksi izkIrgksrk gSA izfrjks/k R dk eku (vkse esa) Kkr dhft;sA

EP( )

P Q

GJ

AR

rES

5. f=Hkqtkdkj {ks=Qy 1m2, vk/kkj yEckbZ 60 cm rFkk eksVkbZ10 mm (fizTe ln`'; vkd`fr) okyh ,d le:if=Hkqtkdkj IysV ,d fpdus /kjkry ij ÅèokZèkj :i lsfp=kuqlkj j[kh gq;h gSA ;g IysV iyVs ugh blds fy,cotq dk vf/kdre eku Kkr dhft;sA

qg

60cm

6. fdlh fo'ks"k fnu vk;ue.My ls ijkofrZr vf/kdrevkofÙk dk eku 8 MHz gSA vxys fnu ;g c<+dj 9MHzgks tkrh gSA ;fn igys fnu rFkk blds vxys fnuvk;ue.My ds vfèkdre bysDVªkWu ?kuRoksa dk vuqikr ngks rks 81 × n dk eku Kkr dhft;sA

ALLEN


7. Find value of base resistance RB (in kW) inthe circuit as shown in figure, if bd,c = 90,VBE = 0.7V; VCE = 4V ?

9VRC 2kW

RB

3V

8. An electric field of 300 V/m is confined to acircular area 10 cm in diameter. If the fieldis increasing at the rate of 20 V/m-s and ifthe magnitude of magnetic field at a point15 cm from the centre of the circle isB × 10–18 T, find value of B.

9. In a given common Emitter transistor,Emitter current is changed by 2.1 mA. Thisresults in a change of 2 mA in the collectorcurrent and a change of 0.05 V in theEmitter-base voltage. The input resistance(in ohm) is :-

10. A rod of ferromagnetic material withdimensions 10cm × 0.5 cm × 0.2 cm isplaced in a magnetic field of strength0.5 × 104 amp/m as a result of which amagnetic moment of 5 amp-m2 is producedin the rod. Find the value of magneticinduction (in SI unit).

7. iznf'kZr ifjiFk esa vk/kkj izfrjks/k RB (kW esa) dk eku Kkr

dhft;s tcfd bd,c = 90, VBE = 0.7V; VCE = 4V gSA

9VRC 2kW

RB

3V

8. ,d 300 V/m dk fo|qr {ks= 10 cm O;kl ds o`Ùkkdkj

{ks=Qy esa ifjc¼ gSA ;fn {ks= dk eku 20 V/m-s dh

nj ls c<+ jgk gS rFkk ;fn o`Ùk ds dsUæ ls 15 cm nwj fLFkr

fcUnq ij pqEcdh; {ks= dk ifjek.k B × 10–18 T gks rks Bdk eku Kkr dhft;sA

9. ,d fn;s x;s mHk;fu"B mRltZd Vªk¡ftLVj esa mRltZd

èkkjk esa 2.1 mA dk ifjorZu fd;k tkrk gSA blds

ifj.kkeLo:i laxzkgd /kkjk esa 2 mA dk ifjorZu rFkk

mRltZd vk/kkj oksYVrk esa 0.05 V dk ifjorZu izsf{kr

gksrk gSA fuos'kh izfrjks/k (vkse esa) dk eku gS%&10. yk SgpqEcdh ; inkFk Z ls cuh ,d NM + dk vkdkj

10cm × 0.5 cm × 0.2 cm g S rFk k bl s0.5 × 104 amp/m lkeF;Z okys pqEcdh; {ks= eas j[kktkrk gS ftlds ifj.kkeLo:i NM+ esa 5 amp-m2 dkpqEcdh; vk?kw.kZ mRiUu gks tkrk gSA pqEcdh; izsj.k dkeku (SI bdkbZ esa) Kkr dhft;sA

ALLEN


SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B),

(C) and (D). ONLY ONE of these fouroptions is correct.

� For each question, darken the bubblecorresponding to the correct option in theORS.

� For each question, marks will be awardedin one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. Ratio of solubilities of gases N2 & O2 in water(as mole of gas in large volume of water) fromair at 25º & 1 atm will be if air is 20 % byvolume of O2 & 80% by volume of N2

Given : KH (N2) = 2 × 104 atmKH (O2) = 104 atm

(A) 8 : 1 (B) 1 : 8(C) 2 : 1 (D) 1 : 2

2. The resistance of 0.05 M solution of oxalic acidis 200 ohm and cell constant is 2.0 cm–1, theequivalent conductance (in S cm2 eq–1) of 0.05 Moxalic acid is :-(A) 100 (B) 0.2(C) 200 (D) 400

PART 2 - CHEMISTRY[k.M–I : (vf/kdre vad : 80)


� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj

(D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi ds

vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls

fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys

dks dkyk fd;k gSA



1. ;fn ok;q esa vk;ru }kjk 20% O2 rFkk vk;ru }kjk 80%N2 gS] rk s 25º rFkk 1 atm ij ok; q ls ty es a(ty ds vf/kd vk;ru esa xSl ds eksy ds :i esa)N2

rFkk O2 xSlksa dh foys;rkvksa dk vuqikr gksxk &

fn;k gS : KH (N2) = 2 × 104 atmKH (O2) = 104 atm

(A) 8 : 1 (B) 1 : 8(C) 2 : 1 (D) 1 : 2

2. vkWDlsfyd vEy ds 0.05 M foy;u dk izfrjk s/k200 vkse rFkk lsy fu;rkad 2.0 cm–1 gS rks 0.05 MvkWDlsfyd vEy dh rqY;kad pkydrk (S cm2 eq–1 esa)gS :-(A) 100 (B) 0.2(C) 200 (D) 400

ALLEN


3. Lyophilic sols are more stable than lyophobicsols because :(A) The colloidal particles have positive

charge(B) The colloidal particles have no charge(C) The colloidal particles are solvated(D) There are strong electrostatic

repulsions between the negativelycharged colloidal particles

4. The decomposition of a gaseous substance(A) to yield gaseous products (B), (C) followsfirst order kinetics. If initially only (A) ispresent and l0 minutes after the start of thereaction, the pressure of (A) is 200 mm Hgand the total pressure is 300mm Hg, thenthe rate constant for reaction A ® B + Cis(A) (1/600) ln 1.25 sec–1

(B) (2.303/10) log 1.5 min–1

(C) (1/10) ln 1.25 sec–1

(D) 2.303log 1.5

600sec–1

5. Which one is non-aromatic?

(A)

1

(B)

(C) (D)

Å

3. nzofojks/kh lkWy dh rqyuk esa nzoLusgh lkWy vf/kd LFkkbZ gksrs

gSa D;ksafd&

(A) dksykbMh d.kksa ij /ku vkos'k gksrk gSA

(B) dksykbMh d.kks a ij dkbZ vkos'k ugha gksrk gSA

(C) dksykbMh d.k foyk;dhÏr gksrs gSaA

(D) ½.kkosf'kr dksykbMh d.kks a ds e/; izcy fo|qr

fLFkSfrd izfrd"kZ.k gksrs gSaA

4. ,d xSlh; inkFkZ (A) ds fo?kVu ls izkIr xSlh; mRikn

(B) rFkk (C) dh vfHkfØ;k izFke dksfV cyxfrdh dk

vuqlj.k djrh gS ;fn izkjEHk esa dsoy (A) mifLFkr gS

rFkk vfHkfØ;k izkjaHk gksus ds l0 feuV ds ckn (A) dk

nkc 200mm Hg rFkk dqy nkc 300mm Hg gS rks

vfHkfØ;k] A ® B + C ds fy, nj fu;arkd gS&

(A) (1/600) ln 1.25 sec–1

(B) (2.303/10) log 1.5 min–1

(C) (1/10) ln 1.25 sec–1

(D) 2.303log 1.5

600sec–1

5. fuEu es ls dkSulk uksu ,sjksesfVd gS\

(A)

1

(B)

(C) (D)

Å

ALLEN


6.

OLDA CH =CH–CH –Cl2 2

HOH/H+A (1) LiAlH4(2) conc.H SO2 4

CB

The compound C is -

(A) CH–CH=CH2

(B) CH –CH=CH2 2

(C) CH–CH –CH2 3

(D) CH–CH –CH2 3

7. Which of the following pair(s) is/areEpimers?(A) D-glucose & D-mannose(B) D-mannose & D-galactose(C) D-glucose & D-fructose(D) D-fructose & L-galactose

8. Following conversion can be carried out byusing :-

CH–CH3

Br COOH

(A) Ag2O (B) (i) NaOBr (ii) H+

(C) Aq. KOH (D) Alc. KOH

6.O

LDA CH =CH–CH –Cl2 2HOH/H+A (1) LiAlH4

(2) conc.H SO2 4CB

;kSfxd C gS -

(A) CH–CH=CH2

(B) CH –CH=CH2 2

(C) CH–CH –CH2 3

(D) CH–CH –CH2 3

7. fuEu fdl ; qXe ds ;kSfxd ^,ihej* gS:-

(A) D-XyqdkWl rFkk D-esuksl

(B) D-esuksl rFkk D-xsysDVksl

(C) D-XyqdkWl rFkk D-ÝDVksl

(D) D-ÝDVksl rFkk L-xsysDVksl

8. fuEu :ikarj.k dks] fdlds iz;ksx }kjk djk;k tk ldrk

gS :-

CH–CH3

Br COOH

(A) Ag2O (B) (i) NaOBr (ii) H+

(C) Aq. KOH (D) Alc. KOH

ALLEN


9. Select the CORRECT order of size for Ln(III)ions :-

(A) Ce3+

Pr3+

Nd3+

Pm3+

r

(pm)

®

z ®

(size)

(B)

r ®

z ®

Ce3+ Pr3+ Nd3+ Pm3+

(size)(pm)

(C)

r ®

z ®

Ce3+

Pr3+

Nd3+

Pm3+(size)(pm)

(D)

r(size)(pm)

®

z ®

Pm3+

Nd3+

Pr3+

Ce3+

9. Ln(III) vk;uksa ds fy, vkdkj ds lgh Øe dks pqfu, :-

(A) Ce3+

Pr3+

Nd3+

Pm3+

r

(pm)

®

z ®

(size)

(B)

r ®

z ®

Ce3+ Pr3+ Nd3+ Pm3+

(size)(pm)

(C)

r ®

z ®

Ce3+

Pr3+

Nd3+

Pm3+(size)(pm)

(D)

r(size)(pm)

®

z ®

Pm3+

Nd3+

Pr3+

Ce3+

ALLEN


10. Which of the following reaction isINCORRECT?

(A) Na + (x+y)NH3(l) ¾® [Na(NH3)x]+ + [e(NH3)y]–

(B) 2NH3 + H2O + CO2 ¾® (NH4)2CO3

(C) Na2CO3.10H2O D¾¾® Na2O + CO2 + 10H2O

(D) Be(OH)2 + 2OH– ¾® [Be(OH)4]2–

11. When H2S gas is passed in acidic mediumwhich set of cations get precipitated?(A) Mn2+, Co2+, Zn2+

(B) Pb2+, Cu2+, Bi3+

(C) Ba2+, Sr2+, Ca2+

(D) Fe3+, Cr3+, Al3+

12. On heating compound X a gas Y is obtained,which is also manufactured by reaction ofBa(NO3)2 + Zn + KOH(aq.).(A) Compound X is (NH4)2Cr2O7

(B) Gas Y is used in Ostwald process(C) Gas Y is colourless and odourless(D) Gas Y produce blue solution in AgNO3(aq.)

13. Calculate % volume occupied by atoms inCsCl type structure assuming ideal crystalhaving anion-anion contact.

[Given +

-

rr = 0.7; p = 3 ; 3 = 1.7]

(A) 67.15 (B) 90.09(C) 78.5 (D) 84

10. fuEu eas ls dkSulh vfHkfØ;k xyr gS ?

(A) Na + (x+y)NH3(l) ¾® [Na(NH3)x]+ + [e(NH3)y]–

(B) 2NH3 + H2O + CO2 ¾® (NH4)2CO3

(C) Na2CO3.10H2O D¾¾® Na2O + CO2 + 10H2O

(D) Be(OH)2 + 2OH– ¾® [Be(OH)4]2–

11. fuEu eas ls dkSuls leqPp; ds /kuk;u] vEyh; ek/;eesa H2S xSl izokfgr fd;s tkus ij vo{ksfir gks tkrs gSa

(A) Mn2+, Co2+, Zn2+

(B) Pb2+, Cu2+, Bi3+

(C) Ba2+, Sr2+, Ca2+

(D) Fe3+, Cr3+, Al3+

12. ;kSfxd X dks xeZ fd;s tkus ij ,d xSl Y izkIr gksrhgSA ftls Ba(NO3)2 + Zn + KOH(aq.) dh vfHkfØ;k

}kjk Hkh cuk;k tk ldrk gS

(A) ;kSfxd X, (NH4)2Cr2O7 gS

(B) xSl Y, vksLVokYV izØe eas iz;ksx esa yh tkrh gS

(C) xSl Y, jaxghu rFkk xa/kghu gS

(D) xSl Y, AgNO3(aq.) esa uhyk foy;u cukrh gS13. ½.kk;u ½.kk;u lEidZ j[kus okys vkn'kZ fØLVy ekurs

gq, CsCl izdkj dh lajpuk esa ijek.kqvksa }kjk ?ksjs x;s

vk;ru dh izfr'krrk dh x.kuk dhft;s -

[Given +

-

rr = 0.7; p = 3 ; 3 = 1.7]

(A) 67.15 (B) 90.09

(C) 78.5 (D) 84

ALLEN


14. The DH° for the mutarotation of glucose inaqueous solution,a – D – glucose (aq) ¾® b – D – glucose (aq)has been measured in a microcalorimeterand found to be –1.16 kJ.mol–1. Theenthalpies of solution of the two forms ofglucose have been determined to bea – D – glucose (s) ¾® a – D – glucose (aq)

DH° = 10.72 kJ mol–1

b – D – glucose (s) ¾® b – D – glucose (aq)DH° = 4.68 kJ.mol–1

Calculate DH° (in kJ/mol) for themutarotation of solid a – D – glucose to solidb - D – glucose.

(A) +4.88 kJ/mol (B) –4.88 kJ/mol(C) –2.44 kJ/mol (D) +2.44 kJ/mol

15.

O(i) Cl + anh. AlCl(ii) Na(Hg) + HCl(iii) HNO + H SO

2 3

3 2 4Z (Major),

Z is :

(A)

O

Cl

O2N (B)

NO2

Cl

(C)

Cl

O2N (D)

O

ClON

14. tyh; foy;u esa Xywdkst ds ifjorhZ ?kw.kZu ds fy,]

a – D – Xywdkst (aq) ¾® b – D – Xywdkst (aq),d ekbØks dsykSjhehVj esa DH° ekik x;k gS rFkk

–1.16 kJ.mol–1 ik;k x;k gSA Xywdkst ds nks :iksa ds

foy;u dh Kkr dh xbZ ,UFkSYih fuEu gSa &

a – D – Xywdkst (s) ¾® a – D – Xywdkst (aq)DH° = 10.72 kJ mol–1

b – D – Xywdkst (s) ¾® b – D – Xywdkst (aq)DH° = 4.68 kJ.mol–1

Bksl a – D – Xywdkst ls Bksl b - D – Xywdkst ds

ifjorhZ ?kw.kZu ds fy, DH° (kJ/mol esa) dh x.kuk dhft,

(A) +4.88 kJ/mol (B) –4.88 kJ/mol(C) –2.44 kJ/mol (D) +2.44 kJ/mol

15.

O(i) Cl + anh. AlCl(ii) Na(Hg) + HCl(iii) HNO + H SO

2 3

3 2 4Z (eq[;),

Z gS :

(A)

O

Cl

O2N (B)

NO2

Cl

(C)

Cl

O2N (D)

O

ClON

ALLEN


16. Racemic mixture of 3-hydroxy-3-methylcyclohexanone is obtained as a product inreaction -

(A) (i) CH MgCl(1eq.)3

(ii) NH Cl4

O

(B)

O

O

(i) CH MgCl(1eq.)3

(ii) NH Cl4

(C) O O

dil.OH–

low temperature

(D)

O

O (i) CH MgCl(excess)3

(ii) NH Cl4

17. How many of the following will give whiteppt of silver salt when treated withammonical AgNO3 solution?(a) CH3–CH2–CºCH (b) CH3–CºC–CH3

(c) CH2=CH–CH2–Cl (d) CH =C–CH2 3

Cl

(e) Cl (f) Ph–CH2–Cl

(g) I

(A) 3 (B) 4 (C) 5 (D) 6

16. fuEu es ls dkSulh vfHkfØ;k esa 3-gkbMªksDlh-3-esfFkylkbDyksgsDlsuksu dk jslsfed feJ.k mRikn ds :i es izkIrgksrk gS -

(A) (i) CH MgCl(1eq.)3

(ii) NH Cl4

O

(B)

O

O

(i) CH MgCl(1eq.)3

(ii) NH Cl4

(C) O O

dil.OH–

low temperature

(D)

O

O (i) CH MgCl(excess)3

(ii) NH Cl4

17. fuEu ;kSfxdks dks tc veksfudy AgNO3 foy;u ds

lkFk mipkfjr fd;k tkrk gS rks fdrus ;kSfxd flYoj yo.kdk 'osr vo{ksi nsaxs\(a) CH3–CH2–CºCH (b) CH3–CºC–CH3

(c) CH2=CH–CH2–Cl (d) CH =C–CH2 3

Cl

(e) Cl (f) Ph–CH2–Cl

(g) I

(A) 3 (B) 4 (C) 5 (D) 6

ALLEN


18. Select INCORRECT statement regardingKMnO4 & K2Cr2O7.(A) Both act as oxidising agent in acidic

medium(B) Both are paramagnetic and coloured(C) O2 is used in the preparation of both(D) The central atom of the anion in both has

d3s hybridisation19. Select CORRECT option regarding

formation of molecular orbital from thecombination of 1s-1s orbitals.

E

®

0(energy)

(1)(2)

y(u)

y(g)

(r) Distance between atoms®

(A) Curve-(1) represents formation of BMO(B) Curve-(2) represents formation of ABMO(C) y(g) = yA + yB

(D) y(u) = yA + yB

20. For a d6 metal ion in an octahedral field, thecorrect electronic configuration is :-

(A) 6 02g gt e when (D0 < P)

(B) 4 22g gt e when (D0 > P)

(C) 6 02g gt e when (D0 > P)

(D) 3 32g gt e when (D0 < P)

18. KMnO4 rFkk K2Cr2O7 ds lUnHkZ esa xyr dFku dk p;udhft,

(A) nksuksa vEyh; ek/;e esa vkWDlhdkjd ds :i esa dk;Z

djrs gSa

(B) nksuksa vuqpqEcdh; rFkk jaxhu gS

(C) O2 dk iz;ksx nksuksa ds fuekZ.k esa fd;k tkrk gS

(D) nksuksa esa] /kuk;u ds dsUæh; ijek.kq dk laØe.k d3s gS

19. 1s-1s d{kdksa ds la;kstu ls vkf.od d{kd fuekZ.k ds

lUnHkZ esa lgh fodYi pqfu,

E

®

0(energy)

(1)(2)

y(u)

y(g)

(r) Distance between atoms®

(A) oØ-(1), BMO ds fuekZ.k dks iznf'kZr djrk gS

(B) oØ-(2), ABMO ds fuekZ.k dks iznf'kZr djrk gS

(C) y(g) = yA + yB

(D) y(u) = yA + yB

20. v"VQydh; {k s= esa d6 /kkrq vk;u ds fy, lghbysDVªkWfu; foU;kl gS :-

(A) 6 02g gt e tc (D0 < P)

(B) 4 22g gt e tc (D0 > P)

(C) 6 02g gt e tc (D0 > P)

(D) 3 32g gt e tc (D0 < P)

ALLEN






0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases.



� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA

� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

ALLEN


1. 1 mole of monoatomic ideal gas subjected toirreversible adiabatic expansion againstconstant external pressure of 1 atm startingfrom initial pressure of 5 atm and initialtemperature of 300 K till the final pressureis 2 atm. What is the final temperature(in K) in the process? (take R = 2 cal/mol K).

2. The solubility of Pb(OH)2 in water is 5 × 10-6 M.Calculate the pH of buffer solution in whichsolubility of Pb(OH)2 is 0.5 × 10–3 mol litre–1.

3. Total number of lone pair(s) present insulfanilamide

4. For the complex [Pt(H2O)(NH3)(Cl)(Br)].If oxidation number of Pt = xCoordination number = yand total possible geometrical isomers = Zthen find sum of x + y + z = ?

5. By a sample of ground state atomic hydrogen,

UV light of energy 13.6 48 eV49 quanta

´ is

absorbed. How many different wavelengthswill be observed in Balmer region ofhydrogen spectrum ?

6. Root mean square speed of an unknown gasat 727ºC is 105 cm/second. Calculate molarmass of unknown gas (in gram/mole)

[Take R = 253

J/mole-K].

1. 5 atm ds izkjfEHkd nkc rFkk 300 K ds izkjfEHkd rki ls

izkjEHk dj vafre nkc 2 atm rd] 1 eksy ,dy ijek.kq

vkn'kZ xSl dk 1 atm ds fu;r ckg~; nkc ds fo:¼

vuqmRØe.kh; :¼ks"eh; izlkj fd;k x;kA izØe esa vafre

rki (K esa) D;k gSA (fn;k gS R = 2 cal/mol K)

2. ty esa Pb(OH)2 dh ?kqyu'khyrk 5 × 10-6 M gSA cQj

foy;u] ftles a Pb(OH)2 dh ?k qyu'khyrk

0.5 × 10–3 mol litre–1 gS] dh pH Kkr dhft;sA

3. lYQsfuysekbM es mifLFkr ,dkadh bysDVªkWu ; qXeks dh

dqy la[;k crkbZ;s\4. ladqy [Pt(H2O)(NH3)(Cl)(Br)] ds fy,

;fn Pt dh vkWDlhdj.k la[;k = xmilgla;kstu la[;k = yrFkk lEHkkfor dqy T;kfefr; leko;oh = Zrks x + y + z dk ;ksx Kkr dhft,

5. vk| voLFkk ds ijekf.o; gkbMªkstu ds uewus }kjk]

13.6 48 eV49 quanta

´ ÅtkZ dk UV izdk'k vo'kksf"kr fd;k

tkrk gS rks gkbMªkstu LiSDVªe ds ckej {ks= es fdruh

fHkUu&fHkUu rjaxnS/; Z izsf{kr gksxhA

6. 727ºC ij ,d vKkr xSl dk oxZ ek/; ewy osx (rms)105 cm/second gSA vKkr xSl ds eksyj nzO;eku dh

x.kuk (gram/mole esa) dhft,A

[fn;k gS R = 253

J/mole-K].

ALLEN


7. The number of condensation polymersamong below mentioned polymers is:-(i) Nylon-6 (ii) Glyptal(iii) Buna-N (iv) PVC(v) Poly vinyl acetate (vi) Terylene(vii) Bakelite (viii) Nylon-6,6(ix) orlone

8. Total number of mono-chloro product(s)obtained (including stereo isomers), whenisopentane react with chlorine in thepresence of sun light?

9. Sum of % p-character used in hybridisationof central atom of given molecules.

XeO3F2 , SO3 & 4NF +

10. XeF6 + 2H2O ¾® X + 4HFNi + 4CO ¾® [Y] (Mond's process)Sum of maximum number of atoms presentin one plane of X & Y.

7. fuEufyf[kr esa ls la?kuu cgqydksa dh la[;k D;k gS \

(i) uk;yksu-6 (ii) fXyIVy

(iii) C;quk-N (iv) PVC

(v) iksyh fouk;y , slhVsV (vi) Vsjhyhu

(vii) csdsykbV (viii) uk;yksu-6,6

(ix) vkjyksu

8. tc vkblksisUVsu lw;Z ds izdk'k dh mifLFkfr esa Dyksfju

ds lkFk fØ;k djrk gS rks izkIr gksus okys eksuksDyksjks mRiknks

(f=foe leko;ofo;ksa dks lfEefyr djrs gq;s) dh dqy

la[;k crkbZ;s\

9. fn;s x;s v.kqvks ds dsUæh; ijek.kq ds ladj.k esa iz;ksx

eas fy;s x;s % p-y{k.k dk ;ksx crkbZ;s

XeO3F2 , SO3 rFkk 4NF +

10. XeF6 + 2H2O ¾® X + 4HFNi + 4CO ¾® [Y] (ekWM izde)X rFkk Y ds ,d ry esa mifLFkr ijek.kqvksa dh vf/kdre

la[;k dk ;ksx gS

ALLEN


SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B),

(C) and (D). ONLY ONE of these fouroptions is correct.

� For each question, darken the bubblecorresponding to the correct option in theORS.

� For each question, marks will be awardedin one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. Let A(1,3), B(a,b), C(4,7) are collinear points

and ( ) ( ) ( ) ( )- + - + - + - =2 2 2 2a 1 b 3 4 a 7 b k ,

then k CANNOT be-(A) 4 (B) 5 (C) 6 (D) 9

2. If a circle passes through point (1, 1) and cutsthe circle x2 + y2 = 8 orthogonally, then thelocus of its centre is -(A) x + y + 5 = 0 (B) x + y = 5(C) x + y = 10 (D) x + y + 10 = 0

3. If the two parabolas y2 = 4x and y2 = (x – k)have a common normal other than x-axis,then value of K cannot be -(A) 1 (B) 2 (C) 3 (D) 4

[k.M–I : (vf/kdre vad : 80)


� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj(D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYids vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esals fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqysdks dkyk fd;k gSA



1. ekuk A(1,3), B(a,b), C(4,7) lejs[kh; fcUnq gSa rFkk

( ) ( ) ( ) ( )- + - + - + - =2 2 2 2a 1 b 3 4 a 7 b k

gS] rks k ugha gks ldrk gS-(A) 4 (B) 5 (C) 6 (D) 9

2. ;fn ,d o`Ùk fcUnq (1, 1) ls xqtjrk gS rFkk o`Ùkx2 + y2 = 8 dks yEcdks.kh; dkVrk gS] rks blds dsUæ dkfcUnqiFk gksxk-(A) x + y + 5 = 0 (B) x + y = 5(C) x + y = 10 (D) x + y + 10 = 0

3. ;fn nks ijoy; y2 = 4x rFkk y2 = (x – k) dk x-v{k dsvykok ,d vU; mHk;fu"B vfHkyEc gks] rks K dk ekuugha gks ldrk gS-(A) 1 (B) 2 (C) 3 (D) 4

PART 3 - MATHEMATICS

ALLEN


4. Let S = 0 be an ellipse whose vertices arethe extremities of minor axis of the ellipse

E : 2 2

2 2x y 1, a ba b

+ = > . If S = 0 passes through

the foci of E, then its eccentricity is(considering the eccentricity of E as e)

(A) 2

21 2e1 e--

(B) 2

11 e+

(C) 2

21 2e1 e--

(D) 2

2e

1 e+

5.2x 3yx 1

+=

+ is a rectangular hyperbola.

Length of its latus rectum is -

(A) 1 (B) 2 (C) 2 (D) 2 26. Value of

cos20 cos50 cos110sin50 sin110 sin110 .sin20 sin20 sin50

° ° °+ +

° ° ° ° ° °

is -(A) 2 (B) 4 (C) 1 (D) 8

7. The smallest positive root of the equation

sin(1 x) cosx- = is equal to -

(A) 1 32 4

p+ (B)

1 72 4

p+

(C) 1 112 4

p+ (D)

12

4. ekuk nh?k Zo `Ùk S = 0, ftlds 'kh"k Z nh?k Zo `Ùk

E : 2 2

2 2x y 1, a ba b

+ = > ds y?kq v{k ds vfUre fljs

gSA ;fn S = 0, E dh ukfHk ls xqtjrh gks] rks bldhmRdsUærk gksxh (E dh mRdsUærk dks e ekurs gq,)

(A) 2

21 2e1 e--

(B) 2

11 e+

(C) 2

21 2e1 e--

(D) 2

2e

1 e+

5. ,d ledks.kh; vfrijoy; 2x 3yx 1

+=

+ gSA blds

ukfHkyEc dh yEckbZ gksxh -

(A) 1 (B) 2 (C) 2 (D) 2 2

6. cos20 cos50 cos110sin50 sin110 sin110 .sin20 sin20 sin50

° ° °+ +

° ° ° ° ° °

dk eku gksxk &(A) 2 (B) 4(C) 1 (D) 8

7. lehdj.k sin(1 x) cosx- = dk U; wure

èkukRed ewy gksxk-

(A) 1 32 4

p+ (B)

1 72 4

p+

(C) 1 112 4

p+ (D)

12

ALLEN


8. Let Ai, i = 1,2,3,......10 be vertices of regular

decagon P. If area of polygon P is 5sin5pæ ö

ç ÷è ø

square units then the value of(A1A2)(A1A3)(A1A4)......(A1A10) is-(A) 10 (B) 9

(C) 45sin5pæ ö

ç ÷è ø

(D) 45cos5pæ ö

ç ÷è ø

9. If A = {1, 3, 5, 7, 9, 11}, B = {2, 4, 6, 8, 10, 12},U is universal set, then A' È ((A È B) Ç B') is-(A) U (B) A'(C) A Ç B (D) A

10. Let I be the set of positve integers. R is arelation on the set I given by

R = {(a,b) Î I × I | 2alogb

æ öç ÷è ø

is a non-negative

integer}, then R is(A) neither symmetric nor transitive but

reflexive.(B) reflexive, transitive but not symmetric(C) neither reflexive nor transitive but

symmetric(D) equivalence relation.

11. If = - +r ˆˆ ˆa i j k and = + +

r ˆˆ ˆb 2i 4j 3k are one of

the sides and medians respectively, of atriangle through the same vertex, then areaof triangle is-

(A) 86 (B) 1 852 (C)

1 842 (D) 80

8. ekuk Ai, i = 1,2,3,......10 lenlHkqt P ds 'kh"kZ gSA ;fn

cgqHk qt P dk {k s=Qy 5sin5pæ ö

ç ÷è ø

oxZ bdkb Z g S]

rk s (A1A2)(A1A3)(A1A4)......(A1A10) dk ekugksxk -(A) 10 (B) 9

(C) 45sin5pæ ö

ç ÷è ø

(D) 45cos5pæ ö

ç ÷è ø

9. ;fn A = {1, 3, 5, 7, 9, 11}, B = {2, 4, 6, 8, 10, 12},, U lkoZf=d leqPp; gS]rks A' È ((A È B) Ç B') gksxk&(A) U (B) A'(C) A Ç B (D) A

10. ekuk I /kukRed iw.kk±dksa dk leqPp; gSA R, leqPp; I ij

,d lEcU/k gS] tks R = {(a,b) Î I × I | 2alogb

æ öç ÷è ø

,d

v½.kkRed iw.kk±d gS} }kjk fn;k x;k gS] rc R gksxk

(A) uk lefer uk gh laØked ijUrq LorqY; gksxkA

(B) LorqY;] laØked ijUrq lefer ugha gksxkA

(C) uk LorqY; uk gh laØked ijUrq lefer gksxkA

(D) rqY;rk lEcU/k gksxkA

11. ;fn = - +r ˆˆ ˆa i j k rFkk = + +

r ˆˆ ˆb 2i 4j 3k leku 'kh"kZ ls

xqtjus okys ,d f=Hkqt dh Øe'k% dksbZ ,d Hkqtk rFkk

ekf/;dk gks] rks f=Hkqt dk {ks=Qy gksxk -

(A) 86 (B) 1 852 (C)

1 842 (D) 80

ALLEN


12. Let A(a)r and B(b)r

be points on two skew

lines = + ar a pr r r and = + br b qrr r and the

shortest distance between them is 3 . Ifangle between AB and the line of shortestdistance is 30°, then AB =

(A) 12

(B) 2 (C) 3 (D) 4

13. The range of values of ‘p’ so that both the rootsof the equation (p – 5)x2 – 2px + (p – 4) = 0are positive, one is less than 2 and otheris lying between 2 & 3, is

(A) ÷ø

öçè

æ 24,4

49(B) (5, ¥)

(C) (–¥, 4) U ÷ø

öçè

æ ¥,4

49(D) None of these

14. If square root of 1 1 1 1a 2a 4a 8aa .(2a) .(4a) .(8a) ........ ¥

is 827

, then the value of 'a' is -

(A) 12

(B) 13

(C) 14

(D) 15

15. The negation of ( ) ( )( )® Úp ~ p ~ q is-

(A) Ùp q (B) ®p q

(C) ( )® Ùp p q (D) ( )Ù Úp p q

12. ekuk fcUnq A(a)r rFkk B(b)r

nks fo"ke js[kkvksa = + ar a pr r r

rFkk = + br b qrr r ij fLFkr gS rFkk fo"ke js[kkvksa ds e/;

U;wure nwjh 3 A ;fn AB rFkk y?kqÙke nwjh dh js[kk ds

e/; dks.k 30° gks] rks AB gksxk&

(A) 12

(B) 2 (C) 3 (D) 4

13. ‘p’ ds ekuksa dk ifjlj] rkfd lehdj.k(p – 5)x2 – 2px + (p – 4) = 0 ds nksuksa ewy èkukRedgS] ftlesa ls ,d ewy 2 ls de rFkk vU; ewy 2 rFkk 3ds e/; fLFkr gS] gksxk&

(A) ÷ø

öçè

æ 24,4

49(B) (5, ¥)

(C) (–¥, 4) U ÷ø

öçè

æ ¥,4

49(D) buesa ls dksbZ ugha

14. ;fn 1 1 1 1a 2a 4a 8aa .(2a) .(4a) .(8a) ........ ¥ dk oxZewy

827

gks] rks a dk eku gksxk -

(A) 12

(B) 13

(C) 14

(D) 15

15. ( ) ( )( )® Úp ~ p ~ q dk fu"ks/k gksxk -

(A) Ùp q (B) ®p q

(C) ( )® Ùp p q (D) ( )Ù Úp p q

ALLEN


16. An urn contains 3 red, 4 green and certainnumber of white balls. Two balls are drawnsimultaneously and found to be of differentcolour. If the chance that none of then was awhite ball is 2/9, then number of white ballsis equal to -(A) 3 (B) 4 (C) 6 (D) 8

17. The numerically greatest term in theexpansion of (2x + 5y)34, when x = 3 &y = 2, is(A) T21 (B) T22 (C) T23 (D) T24

18. Let ƒ : (3, 6) ® (4, 7) be a function defined by

ƒ(x) = 4x x3 3

ì ü- í ýî þ

(where {.} denotes the

fractional part function), then ƒ–1(x) is equalto -(A) x + 1 (B) x – 1

(C) x

x3

ì ü+ í ýî þ

(D) 3x x4 3

ì ü+ í ýî þ

19. ( )2

x

x2 x 1x

1 1 1lim x x 1 x x ...... x

2 2 2-

-®¥

ì üæ öæ ö æ ö+ + + +í ýç ÷ç ÷ ç ÷è øè ø è øî þ

is

equal to, (x Î N) -(A) 2 (B) ln2 (C) e2 (D) 1

20. If 2

xx 4x 3ƒ(x) tan

5sin x e 7æ öé ù- +

= pç ÷ê úç ÷+ +ë ûè ø,

then number of points, where ƒ(x) isdiscontinuous is ([.] is greatest integerfunction)(A) 0 (B) 1 (C) 2 (D) infinite

16. ,d ik= esa 3 yky, 4 gjh rFkk dqN la[;k esa lQsn

xsans gaSA nks xsankas dks ,d lkFk fudkyk tkrk gS rFkk og

fHkUu jaxks dh ikbZ tkrh gSA muds lQsn xsan ds u gksus

dh izkf;drk 2/9 gS rks lQsn xsanks dh la[;k gksxh&

(A) 3 (B) 4 (C) 6 (D) 817. (2x + 5y)34 ds çlkj esa la[;kRed egÙke in gksxk tc

x = 3 rFkk y = 2 gks(A) T21 (B) T22 (C) T23 (D) T24

18. ekuk ƒ : (3, 6) ® (4, 7) ,d Qyu ƒ(x) = 4x x3 3

ì ü- í ýî þ

ls ifjHkkf"kr gksrk gS (tgk¡ {.} fHkUukRed Hkkx Qyu dks

n'kkZrk gS), rks ƒ–1(x) cjkcj gksxk -(A) x + 1 (B) x – 1

(C) x

x3

ì ü+ í ýî þ

(D) 3x x4 3

ì ü+ í ýî þ

19. ( )2

x

x2 x 1x

1 1 1lim x x 1 x x ...... x

2 2 2-

-®¥

ì üæ öæ ö æ ö+ + + +í ýç ÷ç ÷ ç ÷è øè ø è øî þ

dk

eku gksxk (x Î N) -(A) 2 (B) ln2 (C) e2 (D) 1

20. ;fn 2

xx 4x 3ƒ(x) tan

5sin x e 7æ öé ù- +

= pç ÷ê úç ÷+ +ë ûè ø gks] rks fcUnqvksa

dh la[;k] tgk¡ ƒ(x) vlarr~ gks] gksxh

(tgk¡ [.] egÙke iw.kk±d Qyu gSA)

(A) 0 (B) 1 (C) 2 (D) vuUr

ALLEN






0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

••••••••••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

••••••••••

� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases



� izR; sd i z'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA

� izR; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rdVª ad sV@jkm aM vk WQ (truncate/round-off) djsa_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

••••••••••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

••••••••••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

ALLEN


1. If function ƒ(x) = x3 + ax2 + bx + c ismonotonically increasing " x Î R, where a& b are prime numbers less than 10, thennumber of possible ordered pairs (a,b) is

2. Suppose that ƒ is differentiable for all x suchthat ƒ'(x) £ 2 for all x. If ƒ(1) = 2 and ƒ(4) = 8then ƒ(2) has the value equal to -

3. Consider 2

31

0

I x 1 dx= +ò & 33 2

21

I x 1 dx= -ò ,

then I1 + I2 is equal to

4. If m =2

1

x sin(x(1 x))dx-

-ò and n =2

1

sin(x(1 x))dx-

-ò ,

then mn

is equal to :-

5. Let y = g(x) be the inverse of a bijectivemapping f : R ® R, f(x) = 3x3 + 2x. The areabounded by graph of g(x), the x-axis and theordinate at x = 5 is -

6. A curve satisfying the differentialequation (x2y2 – 1) dy + 2xy3dx = 0, ispassing through the points (1, 1) and(k, 2) then the value of (8k2) is

7. The number of 3 × 3 skew symmetricmatrices A whose entries are chosen from

{–1, 0, 1} and for which the system x 0

A y 0

z 0

é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë û

has infinite solution is

1. ;fn lHkh x Î R ds fy, Qyu ƒ(x) = x3 + ax2 + bx + c

,dfn"V o/kZeku gS] tgk¡ a rFkk b, 10 ls NksVh vHkkT;

la[;k,¡ gS] rks laHko Øfer ;qXeks (a,b) dh la[;k gksxh

2. ekuk ƒ lHkh x ds fy, vodyuh; gS rFkk ƒ'(x) £ 2

gSA ;fn ƒ(1) = 2 rFkk ƒ(4) = 8 gks] rks ƒ(2) dk eku

gksxk &

3. ekuk 2

31

0

I x 1 dx= +ò rFkk 33 2

21

I x 1 dx= -ò gks] rks

I1 + I2 dk eku gksxk

4. ;fn m =2

1

x sin(x(1 x))dx-

-ò rFkk n =2

1

sin(x(1 x))dx-

-ò ,

rks mn

dk eku gksxk :-

5. ekuk y = g(x), ,dSdh vkPNknd izfrfp=.k f : R ® R,

f(x) = 3x3 + 2x dk izfrykse gSA g(x) ds vkjs[k]

x-v{k rFkk x = 5 ij dksfV }kjk ifjc¼ {ks=Qy gS -

6. ,d oØ] vody lehdj.k (x2y2 – 1) dy + 2xy3dx = 0

dks larq"V djrk gS tks fcUnq (1, 1) rFkk (k, 2) ls xqtjrk

gS] rks (8k2) dk eku gksxk

7. 3 × 3 fo"ke lefer vkO;wg A dh la[;k] ftlds çfof"V;k

{–1, 0, 1} esa ls pquh tkrh gS rFkk ftlds fy;s fudk;

x 0

A y 0

z 0

é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë û

ds vuUr gy gks] gksxh

ALLEN


8. If [.] represents greatest integer function, thesystem of equations 2x – 3y = 4;7x – 2y = 2; 9x – 5y = [4a]

(a is some constant) is consistent, then rangeof values of 'a' is [a, b), where a + b is

9. A jar contains 7 white marbles and 3 bluemarbles. Given that the 4-marbles are chosenfrom the jar at the same time, then thestandard deviation of the number of the blue

marbles choosen is ab

where a and b are

co-prime numbers and a is square free thena + b is

10. If number of arrangements of letters of theword "DHARAMSHALA" taken all at a timeso that no two alike letters appear togetheris (4a.5b.6c.7d), (where a, b, c, d Î N), thena + b + c + d is equal to

8. lehdj.k fudk; 2x – 3y = 4; 7x – 2y = 2;9x – 5y = [4a], tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrkgS

(a dksbZ vpj gS )] laxr gS] rks 'a' ds ekuksa dk ifjlj[a, b) gS] tgk¡ a + b dk eku gksxk

9. ,d dy'k esa 7 lQsn ekcZy 3 uhys ekcZy gSA fn;k x;k

gS fd ,d gh ckj esa dy'k ls 4 ekcZy dk p;u fd;k

tkrk gS] rks uhys ekcZy ds pquus dh la [;k dk ekud

fopyu a

b gS] tgk¡ a rFkk b lg vHkkT; la[;k;sa rFkk a

iw.kZ oxZ ugha gks] rks a + b gksxk

10. 'kCn "DHARAMSHALA" ds lHkh v{kjksa dks ,d lkFk

ysdj O;ofLFkr djus dh la[;k rkfd dksbZ Hkh nks le:i

;fn v{kj lkFk&lkFk u vk; s (4a.5b.6c.7d) gSA

(tgk¡ a, b, c, d Î N), rks a + b + c + d dk eku gksxk

ALLEN


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

07032021 Page 37/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026

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