)1001cja102119076) - 1 file download

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Your Target is to secure Good Rank in JEE (Main) 2020

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1. Immediately fill in the form number on this page of the TestBooklet with Blue/Black Ball Point Pen. Use of pencil isstrictly prohibited.

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4. The Test Booklet consists of 75 questions. The maximummarks are 300.

5. There are three parts in the question paper 1, 2, 3 consistingof Physics, Chemistry and Mathematics having 25questions in each subject and each subject having Twosections.(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 5 Numerical Value TypequestionsMarking scheme : +4 for correct answer and 0 inall other cases.

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

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egRoiw.kZ funsZ'k

bl ijh{kk iqfLrdk dks rc rd u [kksys a tc rd dgk u tk,A

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 gSA

4. bl ijh{kk iqfLrdk esa 75 iz'u gaSA vf/kdre vad 300 gSaA

5. bl ijh{kk iqfLrdk esa rhu Hkkx 1, 2, 3 gSa] ftlds izR;sd Hkkx esa HkkSfrdfoKku] jlk;u foKku ,oa xf.kr ds 25 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 fodYilgh gSaA

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

(ii) [k.M-II esa 5 la[;kRed eku izdkj ds iz'u gSA

vad ;k stuk : +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 djus gsrqdsoy uhys@dkys ckWy ikbaV isu dk gh iz;ksx djsaA isfUly dk iz;ksx loZFkkoftZr gSA

7. ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkh izdkj dhikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa] eksckby Qksu ;kfdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU; izdkj dh lkexzhdks 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{k fujh{kd

dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iqfLrdk dks ys tk ldrs

g SaA

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

TEST DATE : 31 - 05 - 2020

Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main+Advanced) : ENTHUSIAST & LEADER COURSE [SCORE-II]

Hin

di

JEE(Main)FULL SYLLABUS

)1001CJA102119076)(1001CJA102119076)

PAPER – 2

SPACE FOR ROUGH WORK / dPps dk;Z ds fy, LFkku

H-1/34

Enthusiast & Leader Course/Score-II/31-05-2020/Paper-2ALLEN

1001CJA102119076

PART 1 - PHYSICSBEWARE OF NEGATIVE MARKING

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

[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 esaA1. iznf'kZr fp= esa OA v{k ds lkis{k tM+Ro vk?kw.kZ Kkr

djks&

m m

3l2ll

O

nl

mm

(A) l2m

6 (B) + +l

2m n(n 1)(2n 1)6

(C) 'kwU; (D) +l

2 n(n 1)m2

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. Find moment of inertia about axis OA

m m

3l2ll

O

nl

mm

(A) l2m

6 (B) + +l

2m n(n 1)(2n 1)6

(C) zero (D) +l

2 n(n 1)m2


H-2/34

Target : JEE (Main + Advanced) 2020/31-05-2020/Paper-2ALLEN

1001CJA102119076

2. Resistance of rod is calculated by measuringits length with help of meter scale of leastcount 1mm. Its radius is measured with helpof screw gauge having 50 division oncircular scale and pitch is of 1mm.Resistivity of material is exact. Length ofthe wire is found to be 20 cm and diameter ofwire is 4 mm. Find the percentage error incalculation of resistance:(A) 1.5% (B) 2 % (C) 2.5 % (D) 3.5%

3. Two straight infinitely long current carryingwires are kept along z-axis at the coordinates(0, a, 0) and (0, –a, 0) respectively as shownin the figure. The current in each of the wireis equal and along negative z-axis (into theplane of the paper). The variation of magneticfield on the x-axis will be approximately:-

a

ax

(A) x

B

(B) x

B

(C) x

B

(D) x

B

2. fdlh NM+ dk çfrjks/k 1 mm vYirekad okys ehVj iSekusdh lgk;rk ls bldh yEckbZ dh x.kuk dj Kkr fd;k tkrkgS rFkk bldh f=T;k dk ekiu 1mm pwM+hvUrjky okysLØwxst dh lgk;rk ls fd;k tkrk gS ftlds o`Ùkkdkj iSekusij Hkkxksa dh la[;k 50 gSA blds inkFkZ dh izfrjksèkdrk;FkkFkZ gSA rkj dh yEckbZ 20 cm rFkk O;kl 4 mm çkIrgksrk gSA çfrjks/k ds ekiu esa çfr'kr =qfV gksxh :-(A) 1.5% (B) 2 % (C) 2.5 % (D) 3.5%

3. nks lh/ks vuUr yEcs èkkjkokgh rkjksa dks z–v{k ds vuqfn'kfp=kuqlkj Øe'k% funsZ'kkadksa (0, a, 0) o (0, –a, 0) ij j[kktkrk gSA izR;sd rkj esa /kkjk dk eku leku gSa rFkk ½.kkRedz-v{k ds vuqfn'k (dkxt ds ry esa vUnj dh vksj) gSAx-v{k ij pqEcdh; {ks= dk ifjorZu yxHkx gksxk %&

a

ax

(A) x

B

(B) x

B

(C) x

B

(D) x

B


H-3/34


1001CJA102119076

4. A flexible wire loop in the shape of a circlehas a radius that grows linearly with time.There is a magnetic field perpendicular tothe plane of the loop that has a magnitudeinversely proportional to the distance from

the centre of loop i.e. ( ) µ1B rr

. How does the

emf E vary with time?(A) 2E tµ (B) E tµ

(C) E tµ (D) E is constant5. There are two particles P and Q separated

by distance 10 km and moving withvelocities 10 kmph as shown on a smoothhorizontal surface. The time elapsed beforethey come at minimum separation is :

10 kmph

10 km

10 kmphP

Q

(A) 1 hr (B) 1 hr2

(C) 1 hr4 (D) 2 hr

4. ,d yphys rkj ywi ls o`Ùkkdkj vkÏfr cukbZ tkrh gSftldh f=T;k] le; ds lkFk jSf[kd :i ls c<+rh tkrhgSA bl ywi ds ry ds yEcor~ ,d pqEcdh; {ks= fo|ekugS ftldk ifjek.k ywi ds dsUnz ls nwjh ds O;qRØekuqikrh gS

vFkkZr~ ( ) µ1B rr

gSA fo|qr okgd cy E, le; t ds

lkFk fdl izdkj ifjofrZr gksxk\(A) 2E tµ (B) E tµ

(C) E tµ (D) E fu;r gS

5. fp=kuqlkj nks d.k P rFkk Q ds chp dh nwjh 10 km gS

vkSj os ,d fpduh {kSfrt lrg ij 10 kmph osx ls py

jgs gSaA vkjaHk ls fdrus le; i'pkr~ muds e/; dh nwjh

U;wure gksxh&

10 kmph

10 km

10 kmphP

Q

(A) 1 hr (B) 1 hr2

(C) 1 hr4 (D) 2 hr


H-4/34


1001CJA102119076

6. Energy levels A, B and C of a certain atom

correspond to increasing values of energy, i.e.

EA < EB < EC. If l1, l2 and l3 are the

wavelengths of radiations corresponding to

transitions C to B, B to A and C to A

respectively, which of the following relations

is correct ?

(A) l3 = l1 + l2

(B) l3 = 21

21

l+lll

(C) l1 + l2 + l3 = 0

(D) l32 = l1

2 + l22

7. Two polaroids A and B are placed

perpendicularly in the path of a beam of

unpolarized light. In between these two a

third polaroid C is placed at an angle of 30°

with that of A. The percentage of intensity

of incident unpolarized light that emerges

from B :-

(A) 2.8 % (B) 9.4%

(C) 15.3 % (D)10.2 %

6. ÅtkZ ds c<+rs gq,s eku ds laxr fdlh ijek.kq ds ÅtkZ Lrj

A, B vkSj C gSa vFkkZr~ EA < EB < EC gSA ;fn l1, l2

rFkk l3 Øe'k% C ls B, B ls A rFkk C ls A laØe.k ds

laxr fofdj.kksa dh rjaxnS/; Z gS rks fuEu esa dkSulk laca/k

lgh gS\

(A) l3 = l1 + l2

(B) l3 = 21

21

l+lll

(C) l1 + l2 + l3 = 0

(D) l32 = l1

2 + l22

7. nks /kzqod A o B fdlh v/kzqfor çdk'k iaqt ds iFk esa

yEcor~ :i ls j[ks gq, gSA bu nksuksa ds e/; esa ,d rhljk

/kzqod C, A ls 30° dks.k ij j[kk gqvk gSA B ls fuxZr

vkifrr v/k q z for çdk'k dh rho zrk dk çfr'kr

gksxk :-

(A) 2.8 % (B) 9.4%

(C) 15.3 % (D) 10.2 %


H-5/34


1001CJA102119076

8. The capacitor ‘C’ is initially uncharged.Switch S1 is closed for a long time while S2

remains open. Now at t = 0, S2 is closed whileS1 is opened. All the batteries are ideal andconnecting wires are resistanceless. FindINCORRECT statement :

S1

A3R

2R

C

R0

E

S2

(A) At time t = 0 (just after S2 is closed),

reading of ammeter is E

5R

(B) At time t = 0 (just after S2 is closed),reading of ammeter is zero

(C) Heat developed till time t = 5 RC ln 2 in

resistance 3R is 29 CE40

(D) After time t > 0 charge on the capacitorfollows the equation CE e–t/5RC

8. fp= esa iznf'kZr la/kkfj= ‘C’ izkjEHk esa vukosf'kr gSA fLop

S1 dks ,d yEcs le; rd can j[kk tkrk gS tcfd S2

[kqyk jgrk gSA vc t = 0 ij S2 dks can dj nsrs gSa] tcfd

S1 [kksy fn;k tkrk gSA lHkh cSVfj;k¡ vkn'kZ gS rFkk la;ksth

rkj izfrjks/kghu gSA xyr dFku pqfu, %&

S1

A3R

2R

C

R0

E

S2

(A) le; t = 0 (S2 dks can djus ds rqjUr i'pkr~), ij

vehVj dk ikB~;kad E

5R gSA

(B) le; t = 0 (S2 dks can djus ds rqjUr i'pkr~), ij

vehVj dk ikB~;kad 'kwU; gSA

(C) le; t = 5 RC ln 2 rd izfrjks/k 3R esa mRiUu

Å"ek 29 CE40 gSA

(D) le; t > 0 ds i'pkr~ la/kkfj= ij vkos'k lehdj.k

CE e–t/5RC ds vuqlkj gksrk gSA


H-6/34


1001CJA102119076

9. In the figure shown find out the distance ofcentre of mass of a system of a uniformcircular plate of radius 3 R from O in whicha hole of radius R is cut whose centre is at2R distance from centre of large circularplate.

(A) R/4 (B) R/5(C) R/2 (D) None of these

10. A ring of mass m = 1 kg and radiusR = 1.25m is kept on a rough horizontalground. A small body of same mass m = 1kgis struck to the top of the ring. When it wasgiven a slight push forward, the ring startedrolling purely on the ground. What is themaximum speed of the centre of the ring (inm/s)?

(A) 1 (B) 5 (C) 2 (D) 10

9. fp= esa fn[kk;s vuqlkj fudk; ds nzO;eku dsUnz dh O ls

nwjh Kkr djks] ftlesa 3 R f=T;k dh ,dleku oÙkkdkj

IysV ls R f=T;k dk fNnz dkVk x;k gS rFkk ftlds dsUnz dh

nwjh cM+h o`Ùkkdkj IysV ds dsUnz ls 2R gks &

(A) R/4 (B) R/5(C) R/2 (D) buesa ls dksbZ ugh

10. æO;eku m = 1 kg o f=T;k R = 1.25 m okyh ,d

oy; [kqjnjs {kSfrt /kjkry ij j[kh gSA leku æO;eku

m = 1kg okyk ,d NksVk fi.M oy; ds 'kh"kZ ij Vdjkrk

gSA tc bls vkxs dh vksj gYdk /kDdk fn;k tkrk gS rks

oy; /kjkry ij 'kq¼ ykSVuh xfr djuk çkjEHk dj nsrh

gSA oy; ds dsUæ dh vf/kdre pky (m/s esa) gksxh :-

(A) 1 (B) 5 (C) 2 (D) 10


H-7/34


1001CJA102119076

11. The following four wires of length L andradius r are made of the same material.Which of these will have the largestextension, when the same tension is applied?(A) L = 100 cm, r = 0.2 mm(B) L = 200 cm, r = 0.4 mm(C) L = 300 cm, r = 0.6 mm(D)L = 400 cm, r = 0.8 mm

12. A 110 cm long wire is to be divided into 3segments by placing two bridges on asonometer so that their fundamentalfrequency are in the ratio 1 : 2 : 3. Thepositions of the two bridges should be :-(A) 40 and 50 cm from one end(B) 60 and 90 cm from one end(C) 30 and 50 cm from one end(D) 35 and 50 cm from one end.

13. An ideal gas is taken through the cycleA ® B ® C ® A, as shown in the figure. Ifthe net heat supplied to the gas in the cycleis 5 J, the work done by the gas in the processC ® A is

C

A

B

10

2

1

V(m3)

P(N/m2)

(A) –5 J (B) –10 J(C) –15 J (D) –20 J

11. yEckbZ L rFkk f=T;k r okys fuEu pkj rkj leku inkFkZ lscus gSA bu ij leku ruko yxk;s tkus ij fuEu esa lsfdlesa vf/kdre foLrkj mRiUu gksxk\(A) L = 100 cm, r = 0.2 mm(B) L = 200 cm, r = 0.4 mm(C) L = 300 cm, r = 0.6 mm(D)L = 400 cm, r = 0.8 mm

12. fdlh lksuksehVj ij ,d 110 cm yEcs rkj dks nks lsrqvksadks j[kdj 3 [k.Mksa esa bl izdkj foHkkftr fd;k tkuk gSrkfd budh ewyHkwr vkofÙk vuqikr 1 : 2 : 3 esa gksA nksuksalsrqvksa dh fLFkfr;k¡ gksuh pkfg;s%&(A) ,d fljs ls 40 rFkk 50 cm ij(B) ,d fljs ls 60 rFkk 90 cm ij(C) ,d fljs ls 30 rFkk 50 cm ij(D) ,d fljs ls 35 rFkk 50 cm ij

13. ,d vkn'k Z x Sl dk s fp= es fn[kk; s x;s pØ

A ® B ® C ® A, ls gksdj ys tk;k tkrk gSA ;fn pØ

ds nkSjku xSl dks nh xbZ dqy Å"ek 5 J gS rks xSl }kjk

AC ® çØe esa fd;k x;k dk;Z gS

C

A

B

10

2

1

V(m3)

P(N/m2)

(A) –5 J (B) –10 J(C) –15 J (D) –20 J


H-8/34


1001CJA102119076

14. The masses and radii of the earth and themoon are M1, R1 and M2, R2 respectively.Their centres are distance d apart. Theminimum speed with which particle of massm should be projected from a point midwaybetween the two centres so as to escape toinfinity is :

(A) ( )+= 1 24g M Mv

d

(B) ( )+= 1 24G M Mv

d

(C) ( )= 1 2v 4G M M

(D) ( )= +1 2v 4Gd M M

15. A parallel plate capacitor is charged to apotential difference of 100 V anddisconnected from the source of emf. A slabof dielectric is then inserted between theplates. Which of the following threequantities change?(i) The potential difference(ii) The capacitance(iii) The charge on the plates(A) only (i) and (ii)(B) only (i) and (iii)(C) only (ii) and (iii)(D) All (i), (ii) and (iii)

14. i`Foh rFkk pUæek ds æO;eku o f=T;k Øe'k% M1, R1

rFkk M2, R2 gSA buds dsUækas ds e/; nwjh d gSA nksuksa dsUæksa

ds e/; fLFkr fdlh fcUnq ls m æO;eku ds d.k dks fdl

U;wure pky ls iz{ksfir fd;k tk;s rkfd ;g vuUr eas

iyk;u dj tk;s\

(A) ( )+= 1 24g M Mv

d

(B) ( )+= 1 24G M Mv

d

(C) ( )= 1 2v 4G M M

(D) ( )= +1 2v 4Gd M M

15. ,d lekUrj iê la/kkfj= dks 100 V foHkokUrj rdvkosf'kr dj fo|qr okgd cy L=ksr ls vyx dj fn;ktkrk gSA vc bldh IysVksa ds e/; ,d ijkoS|qr ifêdkizfo"B djk;h tkrh gSA fuEu esa ls dkSulh j kf'k;k¡ ifjofrZrgksxha\(i) foHkokUrj(ii) /kkfjrk(iii) IysVksa ij vkos'k(A) dsoy (i) rFkk (ii)(B) dsoy (i) rFkk (iii)(C) dsoy (ii) rFkk (iii)(D) (i), (ii) rFkk (iii) lHkh


H-9/34


1001CJA102119076

16. A conducting bar PQ of length l carryingcurrent I is suspended from a rigid supportas shown in figure. A uniform magnetic fieldB perpendicular to PQ and directed awayfrom the reader (inside the plane) is applied.If the mass of the bar is M the tension ineach string is

T T

P l Q

(A) Mg/2(B) (Mg + IBL)/2(C) (Mg – IBL)/2(D)Mg – IBL/2

17. A ray of light is incident at the glass-waterinterface at an angle i, it emerges finallyparallel to the surface of water, then valueof mg will be:

rr

i

mw= 4/3Air

WaterGlass

(A) (4/3) sin i (B) 1/sin i(C) 4/3 (D) 1.0

16. fdlh n`<+ vk/kkj ls] ,d pkyd NM+ PQ ftldh yEckbZ

l rFkk ftlesa /kkjk I cg jgh gS] yVdkbZ xbZ gS tSlk fd fp=

esa fn[kk;k x;k gSA ,d ,dleku pqEcdh; {ks= B, NM+

PQ ds yEcor~ vkSj izs{kd ls nwj fn'kk esa (i`"B ds vUnj

dh vksj) yxk;k x;k gSA ;fn NM+ dk nzO;eku M gS rc

izR;sd Mksjh esa ruko gksxk&

T T

P l Q

(A) Mg/2(B) (Mg + IBL)/2(C) (Mg – IBL)/2(D)Mg – IBL/2

17. ,d izdk'k dh fdj.k dk¡p vkSj ikuh ds vUrjki`"B ij idks.k ij vkifrr gksrh gSA ;g vUrr% ikuh dh lrg ds

lekUrj fuxZr gksrh gS rks mg dk eku gksxk&

rr

i

mw= 4/3Air

WaterGlass

(A) (4/3) sin i (B) 1/sin i(C) 4/3 (D) 1.0


H-10/34


1001CJA102119076

18. In Young's double-slit experiment, the

separation between the slits is halved and

the distance between the slits and the screen

is doubled. The fringe width is:

(A) unchanged (B) halved

(C) doubled (D) quadrupled

19. The graph represents the decay of a newly

prepared sample of radioactive nuclide X to

a stable nuclide Y. The half-life of X is t. The

growth curve for Y intersects the decay curve

for X after time T. What is the

time T?

number ofatoms x Y

0 T time

(A) t2 (B) tæ ö

ç ÷è øIn

2

(C) t (D) 2t

18. ;ax f}&fNnz iz;ksx esa fNnzksa dh chp dh nwjh vk/kh rFkk

fNæ rFkk insZ ds chp dh nwjh nqxuh dj nh tkrh gSA fÝt

pkSM+kbZ gksxh&

(A) vifjofrZr (B) vk/kh

(C) nqxquh (D) pkj xq.kk

19. jsfM;kslfØ; ukfHkd X ds ,d rktk izfrn'kZ ls LFkk;h

ukfHkd Y esa fo?kVu dks vkjs[k esa n'kkZ;k x;k gSA X dh

v/kZvk;q t gSA Y dk o`f¼ oØ T le; i'pkr~ X ds

fo?kVu oØ dks izfrPNsfnr djrk gSA le; T dk eku

gksxk%&

number ofatoms x Y

0 T time

(A) t2 (B) tæ ö

ç ÷è øIn

2

(C) t (D) 2t


H-11/34


1001CJA102119076

20. A building has two overhead water tanks,each fitted with a sensor (S1, S2) which goesto 0 when the water level in the tank fallsbelow a set value and remains 1 otherwise.A common pump is used to raise water froman underground storage tank to theseoverhead tanks. Of the following circuits,which one will turn on (P =1) the pump whenat least one of the tanks have water levelbelow the set value?

(A)P

S1

S2

(B)P

S1

S2

(C)P

S1 S2

(D)P

S1 S2

20. fdlh Hkou esa nks ikuh dh Vafd;k¡ j[kh gq;h gSA izR;sdVadh esa ,d lsalj (S1, S2) yxk gqvk gSA tc Vadh esa ikuhdk Lrj ,d fuf'pr eku ls uhps vk tkrk gS rks laslj 'kwU;ij pyk tkrk gSA vU; fLFkfr;ksa esa ;g 1 ij cuk jgrk gSA,d Hkwfexr lap; Vasd ls bu Vafd;ksa rd ikuh dks p<+kusds fy;s ,d gh iEi dk mi;ksx fd;k tkrk gSA tc fdlhHkh ,d Vadh esa ty Lrj bl fuf'pr eku ls uhps vk tkrkgS rks fuEu esa ls dkSulk ifjiFk bl iEi dks pkyw (P=1)dj nsxk\

(A)P

S1

S2

(B)P

S1

S2

(C)P

S1 S2

(D)P

S1 S2


H-12/34


1001CJA102119076

SECTION-II : (Maximum Marks: 20)� This section contains FIVE questions.� The answer to each question is a

NUMERICAL VALUE.� For each question, enter the correct

numerical 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, if answer is 11.36777..... then both11.36 and 11.37 will be correct) by darkenthe corresponding 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

•

•

•

•

•

•

•

•

••

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 ik¡p iz'u gSaA� 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 LFkku gS] rksla[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaMvk WQ (truncate/round-off) dj sa _ 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:icqy cqy s d ks d ky kd j saAmnkgj.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 ;kstuk dsvuqlkj 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


H-13/34


1001CJA102119076

1. In the system shown in figure pulleys andstrings are ideal. Find the magnitude ofacceleration of m1 w.r.t. m2 (in m/s2) is(m1 = 2kg; m2 = 2kg)

m1

m2

2. A closed vessel have a small hole in one faceof vessel near the bottom as shown. Thevelocity (in m/s) of water coming out fromthe given hole at given instant is :

hole

h=9/4 m

air P=0.9 P0

v

P =10 N/m05 2

1. n'kkZ, x, fudk; esa f?kjfu;k¡ rFkk jfLl;k¡ vkn'kZ gSA

m1 dk m2 ds lki s{k Roj.k (m/s2 es a ) g S&

(m1 = 2kg; m2 = 2kg)

m1

m2

2. ,d can ik= dh ,d nhokj esa blds iSans ds fudV ,d NksVk

fNæ cuk gqvk gS] fp= ns[ksaA iznf'kZr {k.k ij fNæ ls ckgj

fudyus okys ty dk osx (m/s esa) gksxk%&

hole

h=9/4 m

air P=0.9 P0

v

P =10 N/m05 2


H-14/34


1001CJA102119076

3. A nucleus at rest undergoes a-decayaccording to equation 226

92X ® Y + a. Att = 0 the emitted a-particle enters in a regionof space where a uniform magnetic fieldB = B0 i and electric field =

r

0ˆE E i exists.

The a-particle enters in a region with

velocity v = v0 j at x = 0. At t = a

a

æ ö´ç ÷

è ø7

0

m3 10q E

time later the particle was found to havetwice the initial speed. If initial speed of thea-particle is 10k m/s then find the value of k.

4. A particle is placed at the lowest point of asmooth wire frame in the shape of a parabola,lying in the vertical xy-plane havingequation x2 = 5y (x, y are in meter). Afterslight displacement, the particle is set free.Find angular frequency of oscillation (in rad/sec) (take g = 10 m/s2) :

5. The potential difference across 8 ohmresistance is 48 volt as shown in the figure.The value of potential difference (in volt)across X and Y points will be:-

X

Y 1W

24W

20W 30W 60W

8W

3W

48V

3. fojkekoLFkk es a fLFkr ,d ukfHkd dk lehdj.k226

92X ® Y + a ds vuqlkj a fo?kVu gksrk gSA t = 0 ijmRlftZr a-d.k ,d ,sls LFkku esa izos'k djrk gS] tgk¡

le:i pqEcdh; {ks= B = B0 i rFkk fo|qr {ks=

0ˆE E i=

r fo|eku gSA ;g a-d.k x = 0 ij v = v0 j

osx ls bl Hkkx esa izos'k djrk gSA 7

0

m3 10

q Ea

a

æ ö´ç ÷

è øle; i'pkr~ d.k dh pky izkjfEHkd pky dh nqxuh gkstkrh gSA ;fn a-d.k dh izkjfEHkd pky 10k m/s gks rks kdk eku Kkr dhft;sA

4. ,d d.k Å/okZ/kj xy-ry esa fLFkr ijoy; x2 = 5y(x, y ehVj esa gS) dh vkd`fr okys ,d fpdus rkj Ýse ds

fuEure fcUnq ij fLFkr gSA d.k dks vYi foLFkkfir dj

Nk sM + fn;k tkrk g SA nk syu dh dk s.kh; vko`fÙk

(rad/sec esa) Kkr dhft, (g = 10 m/s2 ysa) :-

5. fp= esa 8 W izfrjks/k ds fljks ij 48 oksYV dk foHkokUrjgSA X vkSj Y fcUnqvksa ds fljksa ij foHkokUrj dk eku(oksYV esa) gksxk

X

Y 1W

24W

20W 30W 60W

8W

3W

48V


H-15/34


1001CJA102119076

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 fodYids 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 ughafd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA1. Xywdksl] Br2(aq) ds lkFk vkWDlhdj.k djkus ij nsrk gSA

(A) Xywdksfud vEy (B) lSdfjd vEy(C) VkVZfjd vEy (D) vkWDtsfyd vEy

2. vk;u Be2+ rFkk Al3+ dk vkdkj fcYdqy leku ughagS ysfdu os fod.kZ laca/k n'kkZrsa gS D;ksafd%

(A) fHkUu izHkkoh ukfHkdh; vkos'k

(B) fHkUu ijek.kq la[;k

(C) izfr bdkbZ {ks=Qy esa leku vkos'k

(D) ;g fod.kZ laca/k iznf'kZr ugha djrs gaS





1. Glucose, on oxidation with Br2(aq) gives(A) Gluconic acid (B) Saccharic acid(C) Tartric acid (D) Oxalic acid

2. The size of Be2+ and Al3+ is not so close butthey show diagonal relationship due to

(A) Different effective nuclear charge

(B) Different atomic number

(C) Similar charge per unit area

(D) they do not show diagonal relationship


H-16/34


1001CJA102119076

3. tc esfFky ,ehu] lksfM;e /kkrq ls vfHkfØ;k djrh gS rksmRlftZr xSl gS &

(A) NH3 (B) N2 (C) H2 (D) C2H6

4. lYQj fdlesa lokZf/kd foys; gS&

(A) CS2 (B) Et2O (C) ,sFksukWy (D) H2O

5. fuEu ,Ydksgyksa dh SN1 vfHkfØ;k ds futZyhdj.k dhnj gksxhA

(a)

OH

(b) OH

(c) OH

(d) OH

(A) a > b > c > d (B) c > d > b > a(C) d > c > b > a (D) a > c > d > b

6. fuEufyf[kr esa ls ca/k dks.k dk lgh Øe dkSulk gS\

(A) NH3 > PH3 > NF3

(B) NF3 > NH3 > PH3

(C) NH3 > NF3 > PH3

(D) PH3 > NH3 > NF3

3. The gas evolved when methyl aminereacts with sodium metal is

(A) NH3 (B) N2 (C) H2 (D) C2H6

4. Sulphur is most soluble in

(A) CS2 (B) Et2O (C) Ethanol (D) H2O

5. Write the dehydration rate of SN1 reactionof following Alcohols.

(a)

OH

(b) OH

(c) OH

(d) OH

(A) a > b > c > d (B) c > d > b > a

(C) d > c > b > a (D) a > c > d > b

6. Which of the following is correct order ofbond angle ?

(A) NH3 > PH3 > NF3

(B) NF3 > NH3 > PH3

(C) NH3 > NF3 > PH3

(D) PH3 > NH3 > NF3


H-17/34


1001CJA102119076

7. dkSulh vipk;d 'kdZjk gS\

(A) O

HO

HO H

OH

OHOH

(B)

OHO

HOO

OHOH

OOH OH

HOO

CH3

(C) OHO

OH OH

OH

O

(D)

OHO

HOO

OH

OHO

OH

HON

HO

8. AgF, AgCl, NaCl, NaBr, NaI jaxghu gS ysfduAgBr rFkk AgI jaxhu gksrs gSa D;ksfd(A) Ag+, Br– rFkk I– dks /kzqfor dj nsrk gS vkSj Cl–

,oa F– dks /kzqfor djus esa leFkZ ugha gSA(B) AgBr v;qfXer bysDVªkWu j[krk gSA(C) AgBr =qfV j[krk gSA(D) mijksDr lHkh

7. Which is reducing sugar ?

(A) O

HO

HO H

OH

OHOH

(B)

OHO

HOO

OHOH

OOH OH

HOO

CH3

(C) OHO

OH OH

OH

O

(D)

OHO

HOO

OH

OHO

OH

HON

HO

8. AgF, AgCl, NaCl, NaBr, NaI are colourlessbut AgBr and AgI are coloured because(A) Ag+ polarises Br– and I– and not able

to polarise Cl– and F–.(B) AgBr has unpaired electron(C) AgBr has defects(D) All of these


H-18/34


1001CJA102119076

9. From the following which are colloids -(i) NaCl(aq.) (ii) Fog(iii) Paint (iv) Aerosols(v) Mud (vi) Blood(A) All (B) (ii), (iii), (iv), (vi)(C) (ii), (iii), (vi) only (D) (ii), (iv), (vi) only

10. What is final product of this reaction ?NO2

HC O

+

CH3

CHO

¾¾¾®OHs ?

(A)

NO2

OH

CH3

OH

+ (B)

NO2

COO

CH3

OH

+

(C)

NO2

COO

CH3

COO

+ (D)

NO2 CH3

COO

+

OH11. Hydrogen is produced by the reaction :

(A) Na2O2 + 2HCl(B) Mg + hot H2O(C) BaO2 + HCl(D) H2S2O8 + H2O

9. fuEu esa ls dkSuls dksyksbM gSa &(i) NaCl(aq.) (ii) dksgjk(iii) isUV (iv) ,sjkslkWy(v) dhpM + (eM) (vi) jDr(A) lHkh (B) (ii), (iii), (iv), (vi)(C) dsoy (ii), (iii), (vi) (D) dsoy (ii), (iv), (vi)

10. fuEu vfHkfØ;k dk vafre mRikn gS\NO2

HC O

+

CH3

CHO

¾¾¾®OHs ?

(A)

NO2

OH

CH3

OH

+ (B)

NO2

COO

CH3

OH

+

(C)

NO2

COO

CH3

COO

+ (D)

NO2 CH3

COO

+

OH11. fuEu esa ls fdl vfHkfØ;k }kjk gkbMªkstu mRiUu gksxh&

(A) Na2O2 + 2HCl(B) Mg + xeZ H2O(C) BaO2 + HCl(D) H2S2O8 + H2O


H-19/34


1001CJA102119076

12. Excluded volume (v) per molecule is relatedwith vander waal constant 'b' by followingway.

(A) v = b (B) =A

bvN

(C) =A

bv4N (D) v = bNA

13.

O

+ -¾¾¾¾¾¾¾® ¾¾¾¾®2 3Br CH COOH Alc.KOHA B (Product with six membered ring)

Å- +D¾¾¾¾¾¾® ¾¾¾¾® ¾¾¾¾¾®32

(1) CH MgBr H HBr(1mole)(2) H O C D E

What is ‘E’ ?

(A)

Br

(B) Br

(C) Br

(D) Br

12. iz frv.k q ] vioftZr (excluded) vk;ru (v)ok.Mjoky fu;rkad 'b' ds lkFk fdl izdkj lslacaf/kr gS &

(A) v = b (B) =A

bvN

(C) =A

bv4N (D) v = bNA

13.

O

+ -¾¾¾¾¾¾¾® ¾¾¾¾®2 3Br CH COOH Alc.KOHA B (N% lnL;h; oy;

ds lkFk mRikn)Å- +D¾¾¾¾¾¾® ¾¾¾¾® ¾¾¾¾¾®3

2(1) CH MgBr H HBr(1mole)

(2) H O C D E

mRikn ‘E’ gS ?

(A)

Br

(B) Br

(C) Br

(D) Br


H-20/34


1001CJA102119076

14. In following two plots, Y2 is plotted againstthe distance ‘r’ from nucleus.

r

Y2(a)

r

Y2(b)

node

select the correct statement :

(A) ‘a’ is for 1s and ‘b’ for 2s

(B) ‘a’ is for 2s and ‘b’ for 1s

(C) ‘a’ is for 2s and ‘b’ for 2P

(D) ‘a’ is for 2p and ‘b’ for 2s

15. Write the IUPAC name of the followingstructure (E).

HO O

O O

(E)

(A) 2–Ethyl–3–Ethoxy carbonyl propanoicacid.

(B) 2–Ethyl–2–Ethoxy carbonyl ethanoicacid.

(C) 2–Methoxy carbonyl butanoic acid.(D) 2–Ethoxy carbonyl butanoic acid.

14. fuEUkfyf[kr nks oØksa esa Y2 dks ukfHkd ls nwjh ‘r’ dsfo:¼ vkjsf[kr fd;k x;k gS &

r

Y2(a)

r

Y2(b)

node

lgh dFku dk p;u dhft,&

(A) ‘a’] 1s ds fy, gS rFkk ‘b’, 2s ds fy,A

(B) ‘a’, 2s ds fy, gS rFkk ‘b’, 1s ds fy,A

(C) ‘a’, 2s ds fy, gS rFkk ‘b’ 2P ds fy,A

(D) ‘a’, 2p ds fy, gS rFkk ‘b’ 2s ds fy,A

15. fuEufyf[kr lajpuk (E) dk IUPAC ukefyf[k,A

HO O

O O

(E)(A) 2–,sfFky–3–,sFkk WDlh dkck s Z fuy iz k si suk sbd

vEy(B) 2–,sfFky–2–,sFkk WDlh dkck s Z fuy , sF k suk sbd

vEy(C) 2–esFkkWDlh dkcksZfuy C;wVsuksbd vEy(D) 2–,ssFkkWDlh dkcksZfuy C;wVsuksbd vEy


H-21/34


1001CJA102119076

16. In square planar complex of Co2+ ionunpaired electrons are present in

(A) -2 2x yd (B) 2z

d

(C) dxy (D) dzx

17. At temperature T a compound AB2(g)dissociates according to the reaction

+��2 22AB (g) 2AB(g) B (g)

with a degree of dissociation ‘x’ which isvery small compared to unity. The value ofx is

(A) p2KP

(B) p32K

P

(C) p3KP

(D) 3 pK

18. Which is optically active ?

(A) CH3C

HC C C

H

CH3

(B) H

H3C

CH3

CH3

(C)

(D)

COOHH OH

HHO COOH

16. Co2+ vk;u ds oxZ leryh; ladqy esa v;qfXer bysDVªkWumifLFkr gksrs gS&

(A) -2 2x yd esa (B) 2z

d esa

(C) dxy esa (D) dzx esa

17. ,d ;kSfxd AB2(g), T rki ij fuEu vfHkfØ;k vuqlkjfo;ksftr gksrk gS&

+��2 22AB (g) 2AB(g) B (g)

fo;kstu dh ek=k ‘x’, tks fd ,d dh rqyuk esa cgqr

de gSA ‘x’ dk eku Kkr dhft,A

(A) p2KP

(B) p32K

P

(C) p3KP

(D) 3 pK

18. dkSulk izdkf'kd lfØ; gS\

(A) CH3C

HC C C

H

CH3

(B) H

H3C

CH3

CH3

(C)

(D)

COOHH OH

HHO COOH


H-22/34


1001CJA102119076

19.

OH

COOH

+¾¾¾¾¾®2 2Br H Ovkf/kD; mRikn

(A)

OH

Br

BrBr

(B)

COOH

OH

Br

(C)

OH

COOH

BrBr

(D)

OH

COOHBr

20. jsMkWDl vfHkfØ;k esa :

2 3

4 3 2 2

x MnO y PbO z HNOa HMnO b Pb(NO ) c H O

+ + ¾¾®

+ +

(A) x = 2, y = 5, z = 10(B) x = 2, y = 7, z = 8(C) x = 2, y = 5, z = 8(D) x = 2, y = 5, z = 5

19.

OH

COOH

+¾¾¾¾¾®2 2Br H Oexcess Product

(A)

OH

Br

BrBr

(B)

COOH

OH

Br

(C)

OH

COOH

BrBr

(D)

OH

COOHBr

20. In the redox reaction :

+ + ¾¾®

+ +2 3

4 3 2 2

x MnO y PbO z HNOa HMnO b Pb(NO ) c H O

(A) x = 2, y = 5, z = 10(B) x = 2, y = 7, z = 8(C) x = 2, y = 5, z = 8(D) x = 2, y = 5, z = 5


H-23/34


1001CJA102119076



numerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimalplaces; e.g. 6.25, 7.00, –0.33, –.30, 30.27,–127.30, if answer is 11.36777..... then both11.36 and 11.37 will be correct) by darkenthe corresponding 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

•

•

•

•

•

•

•

•

••

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 markingscheme: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 ik¡p iz'u gSaA

� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku


(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkku gS]rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vkWQ (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 ;kstuk ds

vuqlkj gksxk%&

iw.k Z vad : +4 ;fn flQZ lgh la[;kRed eku

(Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA

'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA


H-24/34


1001CJA102119076

1. How many amperes must be passed througha Down’s cell to produce sodium metal at arate of 30.0 g/h ?

2. Number of species among the followingthat can be called as polyamide fibre.Nylon 6, Terylene, PHBV, Buna-N, Neo-prene, Nylon - 6,6, Novolac, PVC, Bakelite,Silk.

3. A metal having atomic mass 60.23 gm/molecrystallises in ABCABC close packing.Calculate the density of each metal if edgelength is 10 Å. (in g/ml)[Given : NA = 6.023 × 1023]

4. Calculate the molar solubility of AgCl at25ºC in 3.0 M NH3

(Ksp of AgCl = 2.0 × 10–10, Kf of [Ag(NH3)2]+

= 1.25 × 107)

5. Number of spin free complex among thefollowing -

[Fe(H2O)6]Cl2, [Co(ox)3]3–, K4[Fe(CN)6],

[Ni(CO)4], [Ni(CN)4]–2, [CoF3(H2O)3],

Na2[PtCl4]

1. Mkmu lSy ls 30.0 g/h dh nj ls lksfM;e /kkrq cukus dsfy, fdruh ,sEih;j /kkjk izokfgr djuh gksxhA

2. fuEu esa ls , slh Lih'kht dh la[;k crkbZ;s ftUgsaikWyh,sekbM js'ks ds uke ls Hkh tkuk tk ldrk gS

uk;ykWu 6, Vsjhyhu, PHBV, C;wuk-N, fuvksfizu,uk;ykWu - 6,6, uksoksysd, PVC, csdsykbV, flYd

3. ,d /kkrq ftldk ijek.kq æO;eku 60.23 gm/mole gS]ABCABC can ladqyu eas fØLVyhÏr gksrh gS rksizR;sd /kkrq ds ?kuRo dh x.kuk djsa ;fn fdukjk yEckbZ10 Å gS\ (g/ml esa)[fn;k gS : NA = 6.023 × 1023]

4. 25ºC ij AgCl dh eksyj foys;rk dh x.kuk 3.0 MNH3 esa dhft,A

(AgCl dh Ksp = 2.0 × 10–10, [Ag(NH3)2]+ dkKf = 1.25 × 107)

5. fuEu esa ls pØ.k eqDr ladqy dh la[;k crkbZ; -

[Fe(H2O)6]Cl2, [Co(ox)3]3–, K4[Fe(CN)6],

[Ni(CO)4], [Ni(CN)4]–2, [CoF3(H2O)3],

Na2[PtCl4]


H-25/34


1001CJA102119076

PART 3 - MATHEMATICS[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 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 ughafd;k gSA

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

1. 9 fHkUu&fHkUu izs{k.kksa dh ekf/;dk dk eku 20.5 gSA ;fn

lcls cM+s 4 izs{k.kksa dk eku 2 ls c<+k fn;k tkos] rc

bl izdkj izkIr u;s leqPp; dh ekf/;dk dk eku

(A) 2 ls c<+ tk;sxk

(B) 2 ls ?kV tk;sxk

(C) nqxquk gks tk;sxk

(D) vifjofrZr jgsxk





1. The median of a set of 9 distinct observations

is 20.5. If each of the largest 4 observations of

the set is increased by 2, then the median of the

new set

(A) is increased by 2

(B) is decreased by 2

(C) is two times the original median

(D) remains the same as that of the original set


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1001CJA102119076

2. dFku ^;fn eSa ,d v/;kid curk gWw] rks eSa ,d fo|ky;

[kksywaxk** dk fu"ks/k dFku gksxk&

(A) eSa v/;kid cuwaxk rFkk eSa fo|ky; ugha [kksywaxk

(B) ;k rks eSa v/;kid ugha cuwaxk ;k u eSa fo|ky;

[kksywaxk

(C) u eSa v/;kid cuwaxk vkSj u gh eSa fo|ky; [kksywaxk

(D) eSa v/;kid ugha cuwaxk ;k eSa fo|ky; [kksywaxk

3. ;fn A r`rh; ikn esa fLFkr gS rFkk 3 tan A – 4 = 0,

rc 5 sin 2A + 3sinA + 4 cosA dk eku gksxk&

(A) 0

(B) 1

(C) 2

(D) buesa ls dksbZ ugha

4. 'm' ds ekuksa dk lEiw.kZ leqPp; ftlds fy, Qyu

ƒ(x) = esinx + 2msinx + 1pæ ö" Î ç ÷

è øx 0,

2 o/kZeku gS]

gksxk-

(A) æ ö-¥ -ç ÷è ø

1,2 (B)

æ ö- ¥ç ÷è ø

1 ,2

(C) (–¥, –e) (D) (–e, ¥)

2. The negation of the statement"If I become a teacher, then I will open a school",is :(A) I will become a teacher and I will not open

a school(B) Either I will not become a teacher or I will

not open a school(C) Neither I will become a teacher nor I will

open a school.(D) I will not become a teacher or I will open

a school.

3. If A l ies in the third quadrant and3tanA – 4 = 0, then find the value of5 sin 2A + 3sinA + 4 cosA

(A) 0

(B) 1

(C) 2

(D) none of these4. Complete set of values of 'm' for which function

ƒ(x) = esinx + 2msinx + 1 is increasing

pæ ö" Î ç ÷è ø

x 0,2 , is-

(A) æ ö-¥ -ç ÷è ø

1,2 (B)

æ ö- ¥ç ÷è ø

1 ,2

(C) (–¥, –e) (D) (–e, ¥)


H-27/34


1001CJA102119076

5. ekuk ƒ(x) = sinx + 2sin2x + 3sin3x + 4sin4x +

... .¥ gk s] rk s pì üÎ -p p - ±í ý

î þx [ , ]

2 es a lehdj.k

ƒ(x) = 2 ds gyksa dh la[;k gksxh -

(A) 0 (B) 2 (C) 4 (D) 8

6. ;fn L : ax + by + c = 0 ,d pj ljy js[kk gS] tgk¡ a,

b rFkk c Øe'k% ,d lekUrj Js.kh dk nwljk] pkSFkk rFkk

lkrok¡ in gks] rks L, fuEu fu;r fcUnq ls xqtjsxh -

(A) æ ö-ç ÷è ø

3 5,2 2 (B)

æ ö-ç ÷è ø

3 5,2 2

(C) æ öç ÷è ø

3 5,2 2 (D)

æ ö- -ç ÷è ø

3 5,2 2

7. ;fn P(2, 8) o`Ùk x2 + y2 – 2x + 4y – p = 0 tks uk rks

v{kksa dks Li'kZ djrk gS] uk gh dkVrk gS] dk vUr% fcUnq

gS] rks p ds fy, leqPp; gS&

(A) p < –1 (B) p < – 4

(C) p > 96 (D) f

8. ;fn ,d lfEeJ la[;k z bl izdkj gS fd |z| = 4 rFkk

arg(z) = p5

6 gks] rks z cjkcj gksxk&

(A) - +2 3 2i (B) +2 3 i

(C) -2 3 2i (D) - +3 i

5. Let ƒ(x) = sinx + 2sin2x + 3sin3x + 4sin4x + ....¥,

then number of solution(s) of equation ƒ(x) = 2

in pì üÎ -p p - ±í ý

î þx [ , ]

2 is -

(A) 0 (B) 2 (C) 4 (D) 8

6. If L : ax + by + c = 0 is a variable straight line,where a, b and c are second, fourth and seventhterm of an AP respectively, then L passesthrough the fixed point -

(A) æ ö-ç ÷è ø

3 5,2 2 (B)

æ ö-ç ÷è ø

3 5,2 2

(C) æ öç ÷è ø

3 5,2 2 (D)

æ ö- -ç ÷è ø

3 5,2 2

7. If P(2, 8) is an interior point of a circlex2 + y2 – 2x + 4y – p = 0 which neither touchesnor intersects the axes, then set for p is

(A) p < –1 (B) p < – 4

(C) p > 96 (D) f

8. If z is a complex number such that |z|= 4 and

arg(z) = p5

6 , then z is equal to :

(A) - +2 3 2i (B) +2 3 i

(C) -2 3 2i (D) - +3 i


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1001CJA102119076

9. The value of -

+ + + +31 2 n

0 1 2 n 1

CC C C2. 3. .... nC C C C is

equal to :

(A) -n(n 1)

2

(B) - +(n 1)(n 1)

2

(C) +n(n 1)

2

(D) +2n n4

10. 110 triangles can be formed by joning 10 points

as vertices in which n points are collinear. Then

the value of n is

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

11. Solve -+ =

+2 2xdy ydxxdx ydy

x y

(A) -+ = +2 2 11 (x y ) tan (y / x) c2

(B) -+ + + =2 2 11 (x y ) tan (y / x) c 02

(C) -- = +2 2 11 (x y ) tan (y / x) c2

(D) -+ = +2 2 1(x y ) tan (y / x) c

9.-

+ + + +31 2 n

0 1 2 n 1

CC C C2. 3. .... nC C C C dk eku gS&:

(A) -n(n 1)

2

(B) - +(n 1)(n 1)

2

(C) +n(n 1)

2

(D) +2n n4

10. 10 fcUnqvksa] ftlesa ls n fcUnq ljs[kh; gS] dks tksM+ djdqy 110 f=Hkqt cuk;s tk ldrs gS] rc n dk ekugksxk&(A) 5 (B) 6 (C) 3 (D) 4

11. vody lehdj.k -

+ =+2 2

xdy ydxxdx ydyx y dk gy

gS

(A) -+ = +2 2 11 (x y ) tan (y / x) c2

(B) -+ + + =2 2 11 (x y ) tan (y / x) c 02

(C) -- = +2 2 11 (x y ) tan (y / x) c2

(D) -+ = +2 2 1(x y ) tan (y / x) c


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1001CJA102119076

12. If the variable line y = kx + 2h is tangent to anellipse 2x2 + 3y2 = 6, then locus of P(h, k) is aconic C whose eccentricity equals

(A) 5

2(B)

73

(C) 7

2(D)

73

13. If angle between asymptotes of hyperbola

- =2 2

2x y 4

3a is p3 , then its conjugate hyperbola

is

(A) - =2 2y x 1

12 9

(B) - =2 2y x 1

12 25

(C) - =2 2y x 1

12 36

(D) - =2 2y x 1

12 414. The sum of all the solution(s) of the equation

sin–1 2x = cos–1 x is–

(A) 0 (B) 25 (C)

15 (D)

-15

12. ;fn pj js[kk y = kx + 2h, nh?kZoÙk 2x2 + 3y2 = 6 dh

Li'kZ js[kk gS rc fcUnq P(h, k) dk fcUnq iFk ,d 'kkado

C gS ftldh mRdsUnzrk gS &

(A) 5

2(B)

73

(C) 7

2(D)

73

13. ;fn vfrijoy;

- =2 2

2x y 4

3a dh vuUrLi'khZ ds eè; dks.k p3 gks] rks

vfrijoy; dk la;qXeh vfrijoy; gksxk-

(A) - =2 2y x 1

12 9

(B) - =2 2y x 1

12 25

(C) - =2 2y x 1

12 36

(D) - =2 2y x 1

12 4

14. lehdj.k sin–1 2x = cos–1 x ds lHkh gyksa dk ;ksxQygksxk&

(A) 0 (B) 25 (C)

15 (D)

-15


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1001CJA102119076

15. The function given by y = ||x| – 1| isdifferentiable for all real numbers except thepoints(A) {0, 1, –1} (B) ±1(C) 1 (D) –1

16. If y = logsinx(tanx), then p

æ öç ÷è ø /4

dydx is equal to

(A) 4

ln 2

(B) –4 ln 2

(C) -4ln 2

(D) 4 ln 2

17. -+ò

7

71 x dx

x(1 x ) equals :

(A) ln x + 27 ln (1 + x7) + c

(B) ln x -27 ln (1-x7) + c

(C) ln x - 27 ln (1+x7) + c

(D) ln x + 27 ln (1-x7) + c

15. Qyu y = ||x| – 1| fn;k x;k gS tks lHkh okLrfodla[;kvksa ij fuEUkfyf[kr ds vfrfjDr vodyuh ; gS]og gS&(A) {0, 1, –1} (B) ±1

(C) 1 (D) –1

16. ;fn y = logsinx(tanx), rc p

æ öç ÷è ø /4

dydx ds cjkcj gS&

(A) 4

ln 2

(B) –4 ln 2

(C) -4ln 2

(D) 4 ln 2

17.-+ò

7

71 x dx

x(1 x ) dk eku gS&

(A) ln x + 27 ln (1 + x7) + c

(B) ln x -27 ln (1-x7) + c

(C) ln x - 27 ln (1+x7) + c

(D) ln x + 27 ln (1-x7) + c


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1001CJA102119076

18. If line = - - + l + +r ˆ ˆˆ ˆ ˆ ˆr (i 2j k) (2i j 2k) is parallel

to the plane - - =r ˆˆ ˆr.(3i 2j mk) 14 , then the

value of m is

(A) 2 (B) –2

(C) 0 (D) 1

19. If é ù

= ê úë û

1 1A

0 1 and

é ùê úê ú=ê ú-ê ú

ë û

3 12 2B1 32 2

, then

( )5TBB A is equal to

(A) é ù+ê ú

- -ê úë û

2 3 11 2 3 (B)

é ùê úë û

1 510 12

(C) é ùê úë û

1 50 1 (D)

é ùê úë û

5 10 1

20. Equation of tangent at p

=x2 to curve

y = xsinx is

(A) x + y = 0 (B) x – y = 0

(C) x + y = p (D) x – y = p

18. ;fn js[kk = - - + l + +r ˆ ˆˆ ˆ ˆ ˆr (i 2j k) (2i j 2k) lery

- - =r ˆˆ ˆr.(3i 2j mk) 14 ds lekukUrj gS rc m dk

eku gS&

(A) 2 (B) –2

(C) 0 (D) 1

19. ;fn é ù= ê ú

ë û

1 1A

0 1 rFkk

é ùê úê ú=ê ú-ê ú

ë û

3 12 2B1 32 2

, rc

( )5TBB A dk eku gksxk&

(A)é ù+ê ú

- -ê úë û

2 3 11 2 3 (B)

é ùê úë û

1 510 12

(C) é ùê úë û

1 50 1 (D)

é ùê úë û

5 10 1

20.p

=x2 ij] oØ y = xsinx dh Li'kZ js[kk dk lehdj.k

gksxk&(A) x + y = 0 (B) x – y = 0(C) x + y = p (D) x – y = p


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1001CJA102119076



numerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimalplaces; e.g. 6.25, 7.00, –0.33, –.30, 30.27,–127.30, if answer is 11.36777..... then both11.36 and 11.37 will be correct) by darkenthe corresponding 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

•

•

•

•

•

•

•

•

••

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 markingscheme: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 ik¡p iz'u gSaA

� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku


(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkku gS]rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vkWQ (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 ;kstuk ds

vuqlkj gksxk%&

iw.k Z vad : +4 ;fn flQZ lgh la[;kRed eku

(Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA

'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA


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1001CJA102119076

1. Suppose that x and y are number satisfying

=1xy9 ; ( )+ =

7x y 19 ; ( )+ =

5y x 118 .

If n = (x +1) (y +1), then the value of 9n is

2. If the equation

x2 + px + 2q = 0 and

x2 + qx + 2p = 0 (p ¹ q) have a common root

then the absolute value of (p + q) is

3. If n be the number of distinct solutions of

2e|x|tan–1|x| = 1, then the value of 1n

is

4. The value of ®

- -x

2x 0

e x 1limx is

5. If three students A, B, C independently solve a

problem with probabilitities 1 1,3 4 and

15

respectively, then the probability that theproblem will be solved is

1. ekuk x rFkk y la[;k;sa gS tks

=1xy9 ; ( )+ =

7x y 19 ; ( )+ =

5y x 118 dks larq"V

djrh gSA ;fn n = (x +1) (y +1) gks] rks 9n dk eku

gksxk

2. ;fn lehdj.k

x2 + px + 2q = 0 ,oa

x2 + qx + 2p = 0 (p ¹ q) dk ,d ewy mHk;fu"B gS

rc (p + q) dk fujis{k eku gksxk

3. ;fn 2e|x|tan–1|x|=1 ds fofHkUu gyksa dh la[;k n gks]

rks 1n

dk eku gksxk

4.®

- -x

2x 0

e x 1limx dk eku gS

5. ;fn rhu Nk=ks A, B, C }kjk LorU= :i ls ,d iz'u dks

gy djus dh izkf;drk,sa Øe'k% 1 1,3 4 ,oa

15 gS] rc

bl ckr dh izkf;drk D;k gksxh fd iz'u gy gks tkrk

gS


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1001CJA102119076


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