DISTANCE LEARNING PROGRAMME - SelfStudys

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

egRoiw.kZ funsZ'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 75 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 25 iz'u gSa vkSj zR;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 5 la[;kRed eku izdkj ds iz'u gSAvad ;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 75 questions. The maximum

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

consisting of Physics, Chemistry and Mathematicshaving 25 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 5 Numerical Value TypequestionsMarking 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.

JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE

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

Paper : Physics, Chemistry & Mathematics

ç'u iq fLrdk : HkkSfrd foKku] jlk;u foKku rFkk xf.kr

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Test Pattern

DISTANCE LEARNING PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main)AIOT

29-12-2019

12th Undergoing/Pass Students

MEDIIT

ALLEN

/35 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 29122019All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119026

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. In the given figure, a diode D is connectedto an external resistance R =100 W and ane.m.f of 3.5 V. If the barrier potentialdeveloped across the diode is 0.5 V, thecurrent in the circuit will be :

100WD

R

3.5V

(A) 35 mA (B) 30 mA(C) 40 mA (D) 20 mA

[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. fn;s x;s fp= esa ,d Mk;ksM D dks ,d cká izfrjks/kR = 100W rFkk 3.5V fo|qr okgd cy dh ,d cSVjh ls

tksM+k x;k gSA ;fn Mk;ksM ij mRiUu jksf/kdk foHko

0.5 V gS rks ifjiFk esa çokfgr /kkjk gksxh :

100WD

R

3.5V

(A) 35 mA (B) 30 mA(C) 40 mA (D) 20 mA

PART 1 - PHYSICS

MEDIIT

ALLEN

29122019 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg /35All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119026

2. A reflecting telescope is to be made by using

spherical mirror of radius of curvature 1.2m

and an eye piece with a focal length of

1.25 cm. For maximum strain position of eye

what will be the angular magnification?

(For normal eye least distance of vision is

25 cm) :

(A) 22.7

(B) 50.4

(C) 48.5

(D) 118.2

3. The electric potential in a certain region of

space depends only on x coordinate as

V = –ax3 + b, here a and b are constants. The

volume charge distribution of space charge

is given as :

(A) r(x) = 3ae0x2

(B) r(x) = –3ae0x2

(C) r(x) = 6ae0x2

(D) r(x) = 6ae0x

2. ,d ijkorhZ nwjn'khZ oØrk f=T;k 1.2 m okys xksyh;

niZ.k rFkk Qksdl nwjh 1.25 cm okyh uSf=dk dk mi;ksx

djds cukuk gSA vk¡[k dh vf/kdre ruko okyh fLFkfr

ds fy, dks.kh; vko/kZu dk eku D;k gksxk\ (lkekU;

vk¡[k ds fy, Li"V n`f"V dh U;wure nwjh 25 cm gk srh

gSA)

(A) 22.7

(B) 50.4

(C) 48.5

(D) 118.2

3. lef"V ds fdlh çHkkx esa fo|qr foHko dk eku dsoy x

funsZ'kkad ij V = –ax3 + b ds vuqlkj fuHkZj djrk gS]

tgk¡ a rFkk b vpj gSA lef"V&vkos'k dk vk;ru vkos'k

forj.k gS :

(A) r(x) = 3ae0x2

(B) r(x) = –3ae0x2

(C) r(x) = 6ae0x2

(D) r(x) = 6ae0x

MEDIIT

ALLEN


4. fdlh [kqjnjh lrg ij j[ks 10 kg æO;eku okys ,d

fi.M dks {kSfrt ls 30° dks.k cuk jgs cy F }kjk /kdsyk

tkrk gSA ;fn ?k"kZ.k dks.k Hkh 30° gS rks fi.M dks xfr

djkus ds fy, vko';d cy F dk ifjek.k gS :-(g = 10 m/s2)

f

30°

FN

(A) 100 N

(B) 50 2 N

(C) 100 2 N(D) 50 N

5. fp= esa n'kkZ;s x;s ifjiFk esa vehVj rFkk oksYVehVj vkn'kZ

gSaA ;fn E = 4V, R = 9W rFkk r = 1W gS] rc vehVj

rFkk oksYVehVj ds ikB~;kad gksxsa&

RE ,r1

V

R R

A

(A) 1A, 3V(B) 2A, 3V(C) 3A, 4V(D) 4A, 4V

4. A body of mass 10 kg placed on rough surfaceis pushed by force F making an angle of 30°to the horizontal. If the angle of friction isalso 30° then the magnitude of force Frequired to move the body is equal to(g = 10 m/s2)

f

30°

FN

(A) 100 N

(B) 50 2 N

(C) 100 2 N(D) 50 N

5. In the circuit shown in figure, ammeter andvoltmeter are ideal. If E = 4V, R = 9W andr = 1W, then readings of ammeter andvoltmeter are

RE ,r1

V

R R

A

(A) 1A, 3V(B) 2A, 3V(C) 3A, 4V(D) 4A, 4V

MEDIIT

ALLEN


6. A ray of light ab passing through air, entersa liquid of refractive index µ1, at theboundary XY. In the liquid, the ray is shownas bc. The angle between ab and bc is d (angleof deviation). The ray then passes through arectangular slab ABCD of refractive indexµ2 (µ2 > µ1), and emerges from the slab as rayde. The angle between XY and AB is q. Theangle between ab and de is :-

X Y

A

B

C

D

c

d

a

bair

e

µ1

µ2

q

(A) d(B) d + q

(C) 1 1

2

µsinµ

- æ öd + ç ÷

è ø

(D) 1 1

2

µsinµ

- æ öd + q - ç ÷

è ø

6. ,d izdk'k fdj.k ab ok;q esa ls gksrs gq, fp=kuqlkj XYifjlhek ls µ1 viorZukad okys nzo esa izos'k djrh gSA

æo esa ;g fdj.k bc ds :i esa fn[kkbZ xbZ gSA ab o bcds e/; fopyu dk s. k d gSA vc ;g fdj.k

µ2 (µ2 > µ1) viorZukad okyh ,d vk;rkdkj ifV~Vdk

ABCD ls xqtjrs gq, bl ifV~Vdk ls fdj.k de ds :i

esa ckgj fudy tkrh gSA XY o AB ds e/; dks.k q gSA

ab o de ds e/; dks.k gS :-

X Y

A

B

C

D

c

d

a

bair

e

µ1

µ2

q

(A) d(B) d + q

(C) 1 1

2

µsinµ

- æ öd + ç ÷

è ø

(D) 1 1

2

µsinµ

- æ öd + q - ç ÷

è ø

MEDIIT

ALLEN


7. A magnetic flux through stationary loop withresistance R varies during the time intervalt = 0 to t = n as f = at (n – t). The amount ofheat generated in the loop during this timeis :-

(A) n2a2R

(B) n2 3a3R

(C) n2 32a3R

(D) na

3R8. In gravity free space Iron Man of mass ‘M’

standing at height ‘h’ above the floor throwsa Precious Stone of mass ‘m’ straight downwith speed u. When stone reaches the floor,what is the distance of Iron Man above thefloor?

(A) mh 1M

æ ö+ç ÷è ø

(B) æ ö-ç ÷è ø

mh 1M

(C) mhM

æ öç ÷è ø

(D) h

7. çfrjks/k R okys ,d fLFkj ywi ls fuxZr pqEcdh; ¶yDl

t = 0 ls t = n le;kUrjky ds nkSjku f = at (n – t) ds

vuqlkj ifjofrZr gksrk gSA bl le; ds nkSjku ywi esa

mRiUu Å"ek dh ek=k gksxh :-

(A) n2a2R

(B) n2 3a3R

(C) n2 32a3R

(D) na

3R

8. xq:Ro eqDr lef"V esa æO;eku ‘M’ okyk ,d ykSg iq:"k

Q'kZ ds Åij ‘h’ Å¡pkbZ ij [kM+k gksdj æO;eku ‘m’ okyk

,d dherh iRFkj pky u ls lh/kk uhps Qsadrk gSA tc

iRFkj Q'kZ ij igq¡prk gS rks ykSg iq:"k dh Q'kZ ds Åij

nwjh D;k gksxh ?

(A) mh 1M

æ ö+ç ÷è ø

(B) æ ö-ç ÷è ø

mh 1M

(C) mhM

æ öç ÷è ø

(D) h

MEDIIT

ALLEN


9. A charged particle of specific charge a isreleased from origin at time t = 0 withvelocity = +

uur

o oˆ ˆV V i V j in magnetic field

=ur

oˆB B i . The coordinates of the particle at

time o

tB

p=

a are (specific charge a = q/m)

(A) o o o

o o o

V 2V V, ,2B B B

æ ö-ç ÷ç ÷a a aè ø

(B) o

o

V ,0,02B

æ ö-ç ÷aè ø

(C) o o

o o

2V V0, ,B 2B

æ öpç ÷a aè ø

(D) o o

o o

V 2V,0, ,B B

æ öp-ç ÷a aè ø

10. A black body, initially at temperature T,cools to temperature (T/2) in time Dt insurrounding which is near absolute zero. Itwill cool further to a temperature (T/4) inadditional time(A) 8Dt(B) 7Dt(C) 9Dt(D) None

9. fof'k"V vkos'k a okys ,d vkosf'kr d.k dks le;

t = 0 ij ewyfcUnq ls pqEcdh ; {ks= =ur

oˆB B i esa osx

= +uur

o oˆ ˆV V i V j ls fojkekoLFkk ls NksM+k tkrk gSA le;

o

tB

p=

a ij d.k ds funsZ'kkad gS

(fof'k"V vkos'k a = q/m)

(A) o o o

o o o

V 2V V, ,2B B B

æ ö-ç ÷ç ÷a a aè ø

(B) o

o

V ,0,02B

æ ö-ç ÷aè ø

(C) o o

o o

2V V0, ,B 2B

æ öpç ÷a aè ø

(D) o o

o o

V 2V,0, ,B B

æ öp-ç ÷a aè ø

10. ,d d`f".kdk izkjEHk esa T rkieku ij gS rFkk Dt le; esa

,sls okrkoj.k esa (T/2) rkieku rd B.Mh gksrh gS tks

yxHkx ije 'kwU; gSA ;g fuEu esa ls fdrus vfrfjDr

le; esa (T/4) rkieku rd B.Mh gksxh \(A) 8Dt(B) 7Dt(C) 9Dt(D) dksbZ ugha

MEDIIT

ALLEN


11. A ball rests upon a flat piece of paper on atable top. The paper is pulled quickly andhorizontally to the right as shown. Just afterthe paper is pulled, the ball

(A)moves to the left & rotates anticlockwise

(B)moves to the right & rotates anticlockwise

(C)moves to the left & rotates clockwise

(D)moves to the right & rotates clockwise12. Light incident normally on the short face of

a right angle prism as shown. A layer ofliquid is placed on the hypotenuse of prism.Refractive index of prism is µ = 1.5. Findpossible value of refractive index that liquidmay have if the light is totally reflected.

Liquid

5390

37°

(A) 1.4 (B) 1.3(C) 1.1 (D) All of these

11. ,d est ds Åij j[ks lery dkxt ds VqdM+s ij ,d xsanfojkekoLFkk esa gSA fp=kuqlkj dkxt dks nka;h vksj {kSfrt:i ls rsth ls [khapk tkrk gSA dkxt dks [khapus ds rqjUri'pkr~ xsan&

(A) cka;h vksj xfr djrh gS rFkk okekorZ fn'kk esa ?kw.kZudjrh gSA

(B) nka;h vksj xfr djrh gS rFkk okekorZ fn'kk esa ?kw.kZudjrh gSA

(C) cka;h vksj xfr djrh gS rFkk nf{k.kkorZ fn'kk esa ?kw.kZudjrh gSA

(D) nka;h vksj xfr djrh gS rFkk nf{k.kkorZ fn'kk esa ?kw.kZudjrh gSA

12. fp=kuqlkj ,d ledksf.kd fçTe dh fdlh NksVh Qyd

ij çdk'k yEcor~ :i ls vkifrr gksrk gSA fçTe ds d.kZ

Qyd ij æo dh ,d ijr j[kh tkrh gSA fçTe dk

viorZukad µ = 1.5 gSA ;fn çdk'k iw.kZ :i ls ijkofrZr

gkss tkrk gS rks bl æo ds viorZukad dk laHkkfor eku D;k

gks ldrk gS\

Liquid

5390

37°

(A) 1.4 (B) 1.3(C) 1.1 (D) mijksDr lHkh

MEDIIT

ALLEN


13. Ratio of charges at capacitor C1 after longtime when switch S is open and closedrespectively, in the given circuit, is

6W4W

20v

C1C2

S

(A) 1 : 1(B) 1 : 8(C) 5 : 2(D) data insufficient

14. Consider an excited hydrogen atom in nth

state moving with velocity v (v << c). It emitsa photon in the direction of its motion andchange its state to lower energy state ‘m’.Now if the frequency of photon emitted is fin this case, while if atom is stationary andfor same transition of electron from excitedstate ‘n’ to ‘m’, the frequency emitted is 0fthen which relation is correct after doingreasonable approximations?

(A) 0vf f 1c

æ ö= -ç ÷è ø

(B) ovf f 1c

æ ö= -ç ÷è ø

(C)2

o 2vf f 1c

æ ö= +ç ÷

è ø(D)

2

o 2vf f 1c

æ ö= -ç ÷

è ø

13. iznf'kZr ifjiFk esa fLop S dks [kksyus rFkk cUn djus ij

yEcs le; i'pkr~ la/kkfj= C1 ij Øe'k% vkos'kksa dk

vuqikr gksxk %&

6W4W

20v

C1C2

S

(A) 1 : 1(B) 1 : 8(C) 5 : 2(D) vkadM+s vi;kZIr gSaA

14. noha voLFkk esa osx v (v << c) ls xfr'khy ,d mÙksftrgkbMªkstu ijek.kq ij fopkj dhft;sA ;g bldh xfr dhfn'kk esa ,d QksVksu mRlftZr djrk gS rFkk viuh voLFkkdks ifjofrZr dj fuEu ÅtkZ voLFkk m esa pyk tkrk gSAvc ;fn bl fLFkfr esa mRlftZr QksVksu dh vkofr f gS]tcfd ;fn ijek.kq fLFkj gS rFkk mÙksftr voLFkk n ls mrd bysDVªkWu ds leku laØe.k ds fy, mRlftZr vkofr

0f gS rc mfpr lfUudVuksa ds mi;ksx }kjk fuEu esa ls

lgh laca/k dkSulk gS\

(A) 0vf f 1c

æ ö= -ç ÷è ø

(B) ovf f 1c

æ ö= -ç ÷è ø

(C)2

o 2vf f 1c

æ ö= +ç ÷

è ø(D)

2

o 2vf f 1c

æ ö= -ç ÷

è ø

MEDIIT

ALLEN


15. Through an ideal inductor coil of L = 0.2 H,an A.C current of amplitude 2A is passed,first with frequency ‘f1’ then with frequency‘f2’. Then ratio of maximum value of inducedemf in two cases e1 and e2 respectively. Find

1

2

ee

æ öç ÷è ø

in the coil, in the two cases is :

(A) 1

2

ff (B) 2

1

ff (C)

æ öç ÷è ø

21

2

ff (D) 1 : 1

16. A particle is moving along vertical circle ofradius R = 20 m with constant speedv = 31.4 m/s as shown. Straight line ABC ishorizontal and passes through the centre ofcircle. A shell is fired from point A at theinstant when particle is at ‘C’. If distance

between AB is 20 3 and shell collides withthe particle ‘B’ then what is the smallestpossible value of projection angle ‘q’ ?

q

u

AB CR=20m

(A) rad4p

(B) rad6p

(C) rad8p

(D) rad3p

15. çsjdRo L = 0.2 H okyh ,d vkn'kZ çsjd dq.Myh ls

vk;ke 2A okyh çR;korhZ /kkjk çokfgr dh tkrh gS_ igys

vkofÙk ‘f1’ rFkk fQj vkofÙk f2 okyhA nksuksa çdj.kksa esa

çsfjr fo|qr okgd cy ds vf/kdre eku dk vuqikr

Øe'k% e1 rFkk e2 gSA nksuksa çdj.kksa esa dq.Myh esa 1

2

ee

æ öç ÷è ø

Kkr dhft;sA

(A) 1

2

ff (B) 2

1

ff (C)

æ öç ÷è ø

21

2

ff (D) 1 : 1

16. ,d d.k f=T;k R = 20 m okys Å/okZ/kj o`Ùk ds vuqfn'k

fu;r pky v = 31.4 m/s ls fp=kuqlkj xfr'khy gSA

lhèkh js[kk ABC {kSfrt gS rFkk o`Ùk ds dsUæ ls xqtjrh gSA

ftl {k.k ij d.k fcUnq ‘C’ ij gksrk gS] fcUnq A ls ,d

dks'k nkxk tkrk gSA ;fn AB ds e/; nwjh 20 3 gks rFkk

dks'k d.k B ls Vdjkrk gS rks ç{ksi.k dks.k q dk U;wure

laHkkfor eku D;k gS\

q

u

AB CR=20m

(A) rad4p

(B) rad6p

(C) rad8p

(D) rad3p

MEDIIT

ALLEN


17. A satellite is in a circular orbit very close tothe surface of a planet. At some point it isgiven an impulse along its direction ofmotion, causing its velocity to increase htimes. It now goes into an elliptical orbit.The maximum possible value of h for this tooccur should be slightly less than :

(A) 2 (B) 2

(C) 2 1+ (D) 1

2 1-

18. An ideal liquid in streamline motion flowsthrough a frictionless duct with varyingcross section as shown in figure. Pressure Pat points along x-axis of duct is most likelyto be represented by.

x axis

(A)

P

x

(B)

P

x

(C)

P

x

(D)

P

x

17. ,d mixzg fdlh xzg dh lrg ds cgqr utnhd ,d

o`Ùkkdkj d{kk esa gSA fdlh fcUnq ij bls bldh xfr dh

fn'kk ds vuqfn'k ,d vkosx fn;k tkrk gS] ftlds dkj.k

bldk osx h xquk c<+ tkrk gSA vc ;g nh?kZoÙkkdkj d{kk

eas pyk tkrk gSA ,slk gksus ds fy;s h dk vf/kdre

lEHkkfor eku fuEu esa ls fdlls FkksM+k de gksuk pkfg;s\

(A) 2 (B) 2

(C) 2 1+ (D) 1

2 1-

18. ,d vkn'kZ æo /kkjk js[kh; izokg esa fp=kuqlkj ifjorhZ

vuqizLFk dkV okys ?k"kZ.kjfgr MDV ls gksdj xqtjrk gSA

MDV dh x-v{k ds vuq fn'k fcUnqvksa ij nkc P dks lokZfèkd

lgh :i ls n'kkZus okyk vkjs[k gksxk %&

x axis

(A)

P

x

(B)

P

x

(C)

P

x

(D)

P

x

MEDIIT

ALLEN


19. A siren (point source) creates a sound levelof 60 dB at location 500m from the speaker.The siren is powered by a battery that delivera total energy of 1 kJ. Assuming thatefficiency of siren is 30% then the total timefor which siren will produce sound is

(A) 180 s

(B) 95.5 s

(C) 1405 s

(D) 60 s

20. The equation of state of a gas is given byPV = nRT + Pb, here b = constant,n = number of moles. If 2 moles of a gas isisothermally expanded from volume V to 2Vthen work done during the process is

(A) -æ öç ÷-è ø

2V b2RT lnV b

(B) æ öç ÷è ø

2V2RTlnV

(C) æ öç ÷-è ø

2VRT ln2V b

(D) 2V bRT lnV b

-æ öç ÷-è ø

19. ,d Hkksaiw (fcUnq L=ksr) Lihdj ls 500m nwjh ij 60dB

Lrj dh /ofu mRiUu djrk gSA lkbju dks ,slh cSVjh }kjk

'kfDr nh tkrh gS tks dqy 1 kJ ÅtkZ iznku djrh gSA Hkksaiw

dh n{krk 30% ekusaA Hkksaiw dqy fdrus le; rd /ofu

mRiUu djsxk\

(A) 180 s(B) 95.5 s(C) 1405 s(D) 60 s

20. ,d xSl dh voLFkk lehdj.k PV = nRT + Pb }kjk

nh tkrh gS] tgk¡ b = vpj rFkk n = eksyksa dh la[;k gSA

;fn xSl ds 2 eksy vk;ru V ls 2V rd lerkih; :i

ls çlkfjr fd;s tkrs gS rks çØe ds nkSjku fd;k x;k dk;Z

gS :

(A) -æ ö

ç ÷-è ø

2V b2RTlnV b

(B) æ öç ÷è ø

2V2RTlnV

(C) æ öç ÷-è ø

2VRTln2V b

(D) 2V bRT lnV b

-æ öç ÷-è ø

MEDIIT

ALLEN


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, 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

•

•

•

•

•

•

•

•

••

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 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 nksuksa

lgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i

cqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksa

dks 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.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

MEDIIT

ALLEN


1. Two identical springs are attached to asmall block P. The outer ends of thesprings are fixed at A and B such that ABis vertical. When P is in equilibrium theextension of top spring is 20 cm andextension of bottom spring is 10 cm. If theperiod of small vertical oscillations of P

about its equilibrium position is secN 2

p ,

then fill the value of N.

2. Two long straight wires A and B are placed50 cm apart and carry current 20 A and 15Arespectively in same direction. A point Pis 40 cm from wire A and 30 cm from wireB. What is the magnitude of resultantmagnetic field at ‘P’ in (µT) unit ?

( )2 1.414=

1. nks ,dtSlh fLizaxs ,d NksVs CykWd P ls tqMh gq;h gSA

fLiazxksa ds ckgjh fljs A ,oa B ij bl izdkj tqM+s gq;s gS fd

AB m/okZ/kj gSA tc P lkE;koLFkk esa gS rks Åij okyh

fLizax dk izlkj 20 cm ,oa fupyh fLizax dk izlkj 10cm

gksrk gSA ;fn P ds bldh lkE;koLFkk ds lkis{k vYi

ÅèokZ/kj nksyuksa ds fy;s vkorZdky secN 2

p gks rks N

dk eku Kkr dhft;sA

2. nks yEcs lh/ks rkj A rFkk B ,d nwljs ls 50 cm nwjh ij

j[ks gq;s gS rFkk buesa leku fn'kk esa Øe'k% 20A rFkk 15A

/kkjk çokfgr gks jgh gSA ,d fcUnq P rkj A ls 40 cm ij

rFkk rkj B ls 30 cm dh nwjh ij fLFkr gSA fcUnq ‘P’ ij

ifj.kkeh pqEcdh; {ks= dk ifjek.k (µT) bdkbZ esa Kkr

dhft;sA ( )2 1.414=

MEDIIT

ALLEN


3. If 27.3 gm ice is at –136.5°C then findchange in its entropy in Cal/K if all icemelt and final temperature becomes 0°C.

Given that =icecals 0.5

gm.K, =water

cals 1.0gm.K ,

Lf = cal80

gm.K , ln 2 = 0.6

4. To one end of thread thrown over the pulleywe suspended load of mass M = 10kg, madefrom a material specific gravity 2.5. Themass M is suspended in a container withwater. By the other end of the thread wesuspended a mass m (see figure). For whatminimum value of m (in kg) mass M can floatin liquid in equilibrium? Friction negligibleeverywhere.

m

M

3. ;fn 27.3 gm cQZ –136.5°C ij gS rFkk ;fn lEiw.kZ

cQZ fi?ky tkrh gS ,oa vfUre rkieku 0°C gks tkrk gS

rks bldh ,UVªksih esa ifjorZu Cal/K esa Kkr dhft;sA

fn;k gS =icecals 0.5

gm.K, =water

cals 1.0gm.K rFkk

Lf = cal80

gm.K , ln 2 = 0.6

4. ,d f?kjuh ij ls gksdj xqtjus okyh jLlh ds ,d fljs ij

M = 10kg æO;eku dk ,d Hkkj yVdk;k tkrk gSA CykWd

ds inkFkZ dk fof'k"V xq:Ro 2.5 gSA æO;eku M dks ikuh

ls Hkjs ,d ik= esa yVdk;k x;k gSA jLlh ds nwljs fljs ij

,d æO;eku m yVdk;k tkrk gS] fp= ns[ksaA m (kg esa)

ds fdl U;wure eku ds fy, æO;eku M æo esa lkE;koLFkk

esa rSj ldrk gS\ ?k"kZ.k loZ= ux.; gSA

m

M

MEDIIT

ALLEN


5. In the figure S1 and S2 are two identical

coherent sources of light. D >> l

(wavelength) and screen is very large. If a

detector starts from A and moves along the

screen along line AP upto an infinite

distance then find total number of minima

detected by the detector.

37°

P

S1

S2

5lD

O

Alarge screen

5. fn;s x, fp= esa S1 rFkk S2 nks ,dtSls dyk&lEc¼ L=ksr

gSaA inkZ vR;f/kd cM+k gS rFkk D >> l (rjaxnSè;Z) gSA

;fn ,d izdk'k lalwpd fcUnq A ls xfr izkjEHk djrk gS

rFkk insZ ds vuqfn'k vuUr nwjh rd js[kk AP ds vuqfn'k

xfr djrk gS rks lalwpd }kjk lalwfpr fufEu"Bksa dh dqy

la[;k Kkr dhft;sA

37°

P

S1

S2

5lD

O

Alarge screen

MEDIIT

ALLEN


PART 2 - CHEMISTRYSECTION–I : (Maximum Marks : 80)

� This section contains TWENTY questions.� Each question has FOUR options (A), (B),



� 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. A 30 litre sample of moist air at 50oC and1 atm has 75% relative humidity. Then whatwill be the final volume if it is compressedisothermally untill 100% relativehumidity.(given: Vapour pressure of waterat 50oC is 0.2 atm)

(A) 25 litre

(B) 27 litre

(C) 22.5 litre

(D) 30 litre










dks dkyk fd;k gSA



1. 50oC rFkk 1 atm ij] ue ok;q ds 30 yhVj uewus dh

vkisf{kd vknzZrk] 75% gSA ;fn bls 100% vkisf{kd

vknzZrk gksus rd lEihfMr fd;k x;k gS rks vafre vk;ru

D;k gk sxk (fn;k gS : 50oC ij ty dk ok"i nkc

0.2 atm gS)

(A) 25 yhVj

(B) 27 yhVj

(C) 22.5 yhVj

(D) 30 yhVj

MEDIIT

ALLEN


2. Arrange the following compounds in the

order of decreasing acidic strength.

(i) CH3CHClCOOH

(ii) CH3CH2OH

(iii) CH3CH2COOH

(iv) HCOOH

(A) i > iv > iii > ii

(B) ii > iii > i > iv

(C) iii > i > ii > iv

(D) i > iii > iv > ii

3. Which of the following is incorrectly

matched ?

Order Corresponding

property

(A)N3–<Cl–<S2– Ionic radii

(B)Fe2+ < Co2+ < Ni2+ Ionisation energy

(C)Al < Ga < Zn Ionisation energy

(D)Sc > Y > La Atomic radii

2. fuEu ;kSfxdksa dks vEyh;rk ds ?kVrs Øe esa O;ofLFkr

dhft,

(i) CH3CHClCOOH

(ii) CH3CH2OH

(iii) CH3CH2COOH

(iv) HCOOH

(A) i > iv > iii > ii

(B) ii > iii > i > iv

(C) iii > i > ii > iv

(D) i > iii > iv > ii

3. fuEu esa ls dkSu] xyr :i ls lqesfyr gS ?

Øe lEcfU/kr xq.k

(A)N3–<Cl–<S2– vk;fud f=T;k

(B)Fe2+ < Co2+ < Ni2+ vk;uu ÅtkZ

(C)Al < Ga < Zn vk;uu ÅtkZ

(D)Sc > Y > La ijekf.o; f=T;k

MEDIIT

ALLEN


4. The normalised wave function of 1s orbital

is 0

ZraN e

æ ö-ç ÷è øY = × and radial distribution

function is 2 24 r ,p Y where N is normalisation

constant 3

30

ZNa

æ ö=ç ÷pè ø

. Which of the following

graph is correct for the radial distribution

( 2 24 rp Y ) of 1s electron with respect to ‘r’

for H-like specie of atomic number Z ?(Where r is radial distance from nucleus)

(A)

2 24 rp Y

r

(B)

2 24 rp Y

r

(C)

2 24 rp Y

r

(D)

2 24 rp Y

r

4. 1s d{kd dk lkekU;hÏr rjax Qyu 0

ZraN e

æ ö-ç ÷è øY = ×

rFkk f=T;h; forj.k Qyu 2 24 r ,p Y g S] tgk ¡ N

lkekU;hdj.k fu;rkad 3

30

ZNa

æ ö=ç ÷pè ø

gSA fuEu ea s ls

dkSu lk vkjs[k] Z ijek.kq Øekad dh H-leku Lih'kht

ds 1s bysDVªk Wu ds] f=T;h; forj.k ( 2 24 rp Y ) ds

fo:¼ r ds fy, lgh gS? (tgk¡ r ukfHkd ls f=T;h; nwjh gS)

(A)

2 24 rp Y

r

(B)

2 24 rp Y

r

(C)

2 24 rp Y

r

(D)

2 24 rp Y

r

MEDIIT

ALLEN


5. Which of the following statement is incorrectfor 3,4-dimethylcyclopentan-1-one

(A) It posses 2 chiral carbons

(B) It shows geometrical isomerism

(C) It posses 3 optical isomers

(D) I.U.P.A.C. name of its optically activestereoisomer is

(3R,4S)-3,4-dimethylcyclopentan-1-one

6. Following reactions occur in solvay process

(a) 3 2 22NH H O CO (A)+ + ¾¾®

(b) 4 2 3 2 2(NH ) CO H O CO (B)+ + ¾¾®

(c) 4 3NH HCO NaCl (C) (D)+ ¾¾® +

(d) D D¾¾® 3 number of products :

Which of the following statements isincorrect regarding compound A, B, C, D?

(A) Phenolphthalein indicator does not givepink colour in aqueous solution of D

(B) compound A is more basic than C

(C) Massive hydrogen bonding is present inD(s)

(D) Only co-ordinate and covalent bonds arepresent in ‘C’.

5. 3,4-MkbesfFkylkbDyksisUVsu-1-vkWu ds lUnHkZ esa] fuEu esa

ls dkSu lk dFku xyr gS

(A) blesa 2 fdjsy dkcZu mifLFkr gS

(B) ;g T;kfefr; lkeo;ork iznf'kZr djrk gS

(C) blds 3 izdkf'kd leko;oh gksrs gS

(D) blds izdkf'kd lfØ; f=foe leko;oh dk I.U.P.A.C.

uke (3R,4S)-3,4-MkbesfFkylkbDyksisUVsu-1-vkWu gS

6. lksYos izØe esa fuEu vfHkfØ;k,sa gksrh gSa &

(a) 3 2 22NH H O CO (A)+ + ¾¾®

(b) 4 2 3 2 2(NH ) CO H O CO (B)+ + ¾¾®

(c) 4 3NH HCO NaCl (C) (D)+ ¾¾® +

(d) D D¾¾® 3 mRiknksa dh la[;k :

;kSfxd A,B,C,D ds lUnHkZ eas] fuEu esa ls dkSu lk dFku

xyr gS?

(A) D ds tyh; foy;u esa fQuks¶Fksyhu] xqykch jax ugha

nsrk gS

(B) C dh rqyuk esa ;kSfxd A vf/kd {kkjh; gS

(C) D(s) esa o`gn (Massive) gkbMªkstu ca/ku mifLFkr gSa

(D) ‘C’ esa dsoy milgla;kstd rFkk lgla;kstd ca/k

mifLFkr gSa

MEDIIT

ALLEN


7. A particular water sample is saturated inCaF2 and has Ca+2 content of 120 ppm. Thenthe F– content of water in ppm would be:[Given Ksp(CaF2) = 6.0 × 10–9M3](A) 26.8 pm

(B) 268 ppm

(C) 2.68 ppm

(D) 0.268 ppm

8. BrO ONa

+

Major product obtained will be ?

(A)

O

O

(B) O O

(C)

(D) O O

7. ty dk ,d fo'ks"k uewuk CaF2 ls larIr gS rFkk blesaCa+2, 120 ppm ek=k esa mifLFkr gS rks ty esa F– dhek=k] ppm esa gksxh :

[fn;k gS Ksp(CaF2) = 6.0 × 10–9M3](A) 26.8 pm(B) 268 ppm(C) 2.68 ppm(D) 0.268 ppm

8. BrO ONa

+

izkIr eq[; mRikn gksxk ?

(A)

O

O

(B) O O

(C)

(D) O O

MEDIIT

ALLEN


9. Which of the following statements is/arecorrect ?

(i) Orange solution of sodium dichromate,can be crystallised as Na2Cr2O7.2H2O.

(ii) Cu+ ion disproportionate in aqueoussolution into Cu and Cu3+

(iii) Cr+2 can liberate hydrogen gas from adilute acid

(iv) Cr2O3 and MnO2 are amphoteric oxides

(v) Sc and Zn are transition metals

(A) i, ii, iv, v (B) i, iii, v

(C) i, iii, iv (D) ii, iii, iv

10. Following plots are obtained betweenworkdone(W) and ratio of volume V2(final) andV1((initial) for ideal gas in isothermal process.

T1

T2

T3

(V2 /V1)

WnR

æ ö-ç ÷è ø

The correct order of temperature is(A) T1 = T2 = T3

(B) T1 < T2 < T3

(C) T3 < T2 < T1

(D) Can not be predicted

9. fuEu esa ls dkSulk dFku lgh gS@gSa?

(i) lk s fM;e MkbØk sesV ds ukj axh foy;u dk s

Na2Cr2O7.2H2O ds :i esa fØLVyhÏr fd;k tk

ldrk gS

(ii) Cu+ vk;u tyh; foy;u esa Cu rFkk Cu3+ esa

fo"kekuqikfrr gks tkrs gSa

(iii) Cr+2, ,d ruq vEy ls gkbMªkstu xSl mRlftZr dj

ldrk gS

(iv) Cr2O3 rFkk MnO2 mHk;/kehZ vkWDlkbM gSa

(v) Sc rFkk Zn laØe.k /kkrq,sa gSa

(A) i, ii, iv, v (B) i, iii, v

(C) i, iii, iv (D) ii, iii, iv

10. fuEu vkjs[k] lerkih; izØe esa ,d vkn'kZ xSl ds fy,fd;s x;s dk;Z (W) rFkk vk;ru vuqikr] ds e/; izkIrfd;s x;s gSa tgk¡ V2(vafre) vkSj V1(izkjfEHkd) vk;ru gSa

T1

T2

T3

(V2 /V1)

WnR

æ ö-ç ÷è ø

rki dk lgh Øe gS(A) T1 = T2 = T3

(B) T1 < T2 < T3

(C) T3 < T2 < T1

(D) vuqeku ugha yxk ldrs

MEDIIT

ALLEN


11. Identify the product (P) of given reaction.

D+ ¾¾¾¾¾®dil NaOH6 5 3 2C H CHO CH CH CHO P

(A) C6H5CH=CHCH2CHO(B) C6H5CH2OH(C) C6H5COONa

(D) =6 5

3

C H CH CCHO|CH

12. Which of the following ion has greater bondorder than its parent molecule ?(A) C2

+ (B) O2+ (C) N2

+ (D) B2+

13. M+ form M3C60 where 360C- fulleride

(a superconductor) form octahedral holes. Ifradius of 3

60C- is 500 pm then the minimumpossible radius for M+ is(A) 307 pm (B) 367 pm(C) 207 pm (D) 500 pm

14. Which of the following statement isINCORRECT ?(A) Aniline and N,N-dimethylaniline can be

differentiated by hinsberg reagent .(B) Sucrose and glucose can be

differentiated by Tollen’s reagent.(C) Diethylamine and aniline can be

differentiated by carbylamine test.(D) Phenetole and anisole can be

differentiated by neutral ferric chloridetest

11. nh x;h vfHkfØ;k dk mRikn (P) igpkfu,

D+ ¾¾¾¾¾®dil NaOH6 5 3 2C H CHO CH CH CHO P

(A) C6H5CH=CHCH2CHO

(B) C6H5CH2OH

(C) C6H5COONa

(D) =6 5

3

C H CH CCHO|CH

12. fuEu esa ls dkSu ls vk;u dk ca/k Øe] blds ewy ;kSfxd

ls vf/kd gS?(A) C2

+ (B) O2+ (C) N2

+ (D) B2+

13. M+, M3C60 cukrk gS tgk¡ 360C- QqyjkbM(,d vfrpkyd)

v"VQydh; fNæ fufeZr djrs gSaA ;fn 360C- dh f=T;k

500 pm gS rks M+ dh lEHkkfor U;qure f=T;k gS

(A) 307 pm (B) 367 pm

(C) 207 pm (D) 500 pm

14. fuEu esa ls dkSu lk dFku xyr gS?

(A) ,fuyhu rFkk N,N-MkbZesfFky,sfuyhu dks fgalcxZ

vfHkdeZd }kjk foHksfnr fd;k tk ldrk gS

(B) lqØksl rFkk Xyqdksl dks VkWysu vfHkdeZd }kjk foHksfnr

fd;k tk ldrk gS

(C) MkbZ,sfFky,sehu rFkk ,uhyhu dks dkfcZy,ehu ijh{k.k

}kjk foHksfnr fd;k tk ldrk gS

(D) fQuksy rFkk ,sfulksy dks mnklhu Qsfjd DyksjkbM

ijh{k.k }kjk foHksfnr fd;k tk ldrk gS

MEDIIT

ALLEN


15. Correct representation at 1200 K accordingto the given plot is :

M MO®

C CO®–400

–600

1000K T(K)1200K

DG°(KJ/mol)

(A) + ® +(s) (g) (s) (s)M CO MO C ;oGD = –200kJ/mol

(B) + ® +(l ) (g) (s) (S)M CO MO C ;oGD = –200kJ/mol

(C) + ® +(s) (s) (l) (g)MO C M CO ;oGD = –200kJ/mol

(D) + ® +(s) (s) (s) (g )MO C M CO ;

oGD = –200kJ/mol

16. Which one of the following statement isINCORRECT ?

(A) The net increase in entropy of a systemis zero in any reversible Cyclic process.

(B) At constant temperature & pressureavailable energy present in system iscalled free energy.

(C) The change in Gibb’s free energy withpressure for one mole ideal gas at

constant temperature is 2

1

PG RT nP

D = l

(D) For a spontaneous change (DG)T,P > 0

15. fn;s x;s vkjs[k ds vuqlkj] 1200 K ij lgh izn'kZu gS

M MO®

C CO®–400

–600

1000K T(K)1200K

DG°(KJ/mol)

(A) + ® +(s) (g) (s) (s)M CO MO C ;oGD = –200kJ/mol

(B) + ® +(l ) (g) (s) (S)M CO MO C ;oGD = –200kJ/mol

(C) + ® +(s) (s) (l) (g)MO C M CO ;oGD = –200kJ/mol

(D) + ® +(s) (s) (s) (g )MO C M CO ;

oGD = –200kJ/mol

16. fuEu esa ls dkSu lk dFku xyr gS ?

(A) fdlh Hkh mRØe.kh; pØh; izØe esa ra= dh ,sUVªksih

esa dqy of¼ 'kwU; gksrh gS

(B) fu;r rki ,oa nkc ij] ra= esa mifLFkr miyC/k ÅtkZ]

eqDr ÅtkZ dgykrh gS

(C) fu;r rki ij ,d eksy vkn'kZ xSl ds fy, nkc ds

lkFk fxCl eqDr ÅtkZ es a gk su s okyk ifjorZu]

2

1

PG RT nP

D = l gksrk gS

(D) ,d Lor% ifjorZu ds fy, (DG)T,P > 0

MEDIIT

ALLEN


17. Which of the following is Antacid drug ?

(A) Ranitidine

(B) Penicillin

(C) Tetracycline

(D) NaOH

18. Which of the following statement(s) is/are correctregarding complex ion [Co(en)2(NH3)2]3+

(I) Central metal is sp3d2 hybridised

(II) Complex ion shows geometrical as wellas optical isomerism.

(III) It has two N–Co–N bond angles.

(A) I & II

(B) II & III

(C) Only II

(D) Only III

19. For a first order reaction. Rate constant is

represented by logk(s–1) = 15 – ´ 39.19 10 (K)T

,

then the temperature at which its half lifeis 28 min (ln2 = 0.7, log3 = 0.48, log2 = 0.30)

(A) 500K (B) 919 K

(C) 1000K (D) 638 K

17. fuEu esa ls dkSu] izfrvEy vkS"kf/k gS?

(A) jsfUVMhu

(B) isfuflyhu

(C) VsVªklkbfDyu

(D) NaOH

18. ladqy vk;u [Co(en)2(NH3)2]3+ ds fy,] fuEu esa ls

dkSu lk dFku lgh gS

(I) dsUæh; ijek.kq sp3d2 ladfjr gS

(II) ladqy vk;u T;kfefr; ds lkFk&lkFk izdkf'kd

leko;ork Hkh iznf'kZr djrk gS

(III) blesa nks N–Co–N ca/k dks.k mifLFkr gSa

(A) I rFkk II

(B) II rFkk III

(C) dsoy II

(D) dsoy III

19. ,d izFke dks fV vfHkfØ;k ds fy, nj fu;rk ad]

logk(s–1) = 15 – ´ 39.19 10 (K)T

, }kjk iznf'kZr fd;k tkrk

gS rks og rki ftl ij bldh v¼Z vk;q 28 min gksxh

(ln2 = 0.7, log3 = 0.48, log2 = 0.30)

(A) 500K (B) 919 K

(C) 1000K (D) 638 K

MEDIIT

ALLEN


20. How many of the following are aromatic ?

(a) N

N

(b) O

O

(c)

OOH

(d)

(e) B

HH

(f) BH

(A) 1 (B) 2

(C) 3 (D) 4

20. fuEu esa ls fdrus ;kSfxd ,sjksesfVd gSa ?

(a) N

N

(b) O

O

(c)

OOH

(d)

(e) B

HH

(f) BH

(A) 1 (B) 2

(C) 3 (D) 4

MEDIIT

ALLEN



NUMERICAL 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

•

•

•

•

•

•

•

•

••

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 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


MEDIIT

ALLEN


1. Number of geometrical isomers of a complexion [Pt(NH3)2(py)2F2]2+ are x, if 1 mole ofethylenediamine replaces two weakestligands, then the new complex formed has‘y’ number of geometrical isomers. Find the

value of xy .

2. The Ksp of AgCl is 10–10 M2 .Find E (volt) forAg(aq)

+/Ag(s) if Ag electrode is immersed in 1MKCl at 25°C. [Given:E°Ag(aq)

+/Ag(s) = 0.80 V,2.303RT 0.059

F= ]

3. How many of the following reactions canproduce ethane ?

(i) CH COOH2

CH COOH2

D¾¾®

(ii) D¾¾¾¾¾¾®NaOH & CaO3CH COONa

(iii) D¾¾®3 2(CH COO) Ca

(iv) D¾¾¾¾¾¾®NaOH & CaO3 2CH CH COONa

(v) CH CH3

COOH

COOH

D

4. Total number of lanthanoids (4f-elements)having 5d electrons in their electronicconfiguration are:

5. A solution of isopropyl alcohol and propylalcohol has a vapour pressure 200 mm of Hg ifit has 25% mole of isopropyl alcohol. Anothersolution of same components containing25% mole propyl alcohol has vapour pressure300 mm of Hg.Then vapor pressure ofisopropyl alcohol in mm of Hg is :

1. ,d ladqy vk;u [Pt(NH3)2(py)2F2]2+ ds T;kfefr;

leko;fo;ksa dh la[;k x gS, ;fn 1 eksy ,sfFkyhuMkb,sehu

nks nqcZyre fyxs.Mksa dks izfrLFkkfir dj nsrk gS rks fufeZr

u;s ladqy ds T;kfefr; leko;fo;ksa dh la[;k ‘y’ gks

tkrh gSA xy dk eku crkbZ;s

2. AgCl dk Ksp, 10–10 M2 gSA Ag(aq)+/Ag(s) ds fy,

E (oksYV) crkb;s ;fn Ag bysDVªk sM dks 25°C ij1M KCl esa Mqck;k x;k gSA [fn;k gS:

E°Ag(aq)+/Ag(s) = 0.80 V,

2.303RT 0.059F

= ]

3. fuEu esa ls fdruh vfHkfØ;kvksa esa ,sFksu mRikfnr gks ldrh gS?

(i) CH COOH2

CH COOH2

D¾¾®

(ii) D¾¾¾¾¾¾®NaOH & CaO3CH COONa

(iii) D¾¾®3 2(CH COO) Ca

(iv) D¾¾¾¾¾¾®NaOH & CaO3 2CH CH COONa

(v) CH CH3

COOH

COOH

D

4. ,sls ysUFksuk;Mksa (4f-rRo) dh dqy la[;k crkbZ;s ftuds

bysDVªkfu; foU;kl esa 5d bysDVªkWu mifLFkr gksrs gSa :

5. vkblksizksfiy , sYdksgkWy rFkk izksfiy , sYdksgkWy ds ,dfoy;u dk ok"i nkc 200 mm Hg gS] ;fn blesa 25%vkblksizksfiy mifLFkr gSA leku ?kVdksa ds vU; foy;uftlesa 25% izksfiy vEy mifLFkr gS] dk ok"i nkc300 mm Hg gS rks vkblksizksfiy ,sYdksgkWy dk] mm Hgesa ok"inkc D;k gS\

MEDIIT

ALLEN


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



� 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. The number of complex number(s) satisfying

the curves |z – 5|£ 5 and p

- =3arg(z 5i)4 is

(A) 0 (B) 1

(C) 2 (D) More than 22. Area of quadrilateral formed by angle

bisectors of lines 3x – 4y + 1 = 0,3x + 4y – 7 = 0 and co-ordiante axes is(in sq. units)

(A) 1 (B) 2 (C) 4 (D) 52

3. Sum of series (25C13 + 25C14 + .... + 25C25) isequal to(A) 225 (B) 224

(C) 224.24C11 (D) 25C12










dks dkyk fd;k gSA



1. oØksa |z – 5|£ 5 rFkk p

- =3arg(z 5i)4 dks larq"V

djus okyh lfEeJ la[;kvksa dh la[;k gS(A) 0 (B) 1(C) 2 (D) 2 ls vf/kd

2. js[kkvksa 3x – 4y + 1 = 0, 3x + 4y – 7 = 0 ds dks.kv¼Zdksa rFkk funsZ'kh v{kksa }kjk fufeZr prqHkqZt dk {ks=Qy(oxZ bdkbZ esa) gksxk

(A) 1 (B) 2 (C) 4 (D) 52

3. Js<+h (25C13 + 25C14 + .... + 25C25) dk ;k sxQygksxk

(A) 225 (B) 224

(C) 224.24C11 (D) 25C12

PART 3 - MATHEMATICS

MEDIIT

ALLEN


4. If æ ö

- +ç ÷ç ÷è øìï + ¹= íï =î

1 1x x(x 1)2 , x 0f(x)

0 , x 0, then

(A) f(x) is continuous at x = 0(B) f(x) is differentiable at x = 0(C) +®x 0

lim f (x) does not exist

(D) + -® ®¹

x 0 x 0lim f(x) lim f(x)

5. We roll three identical dice and we find thatthe sum of upper face number on the threedice is 10, then the probability that one dieshows ‘6’, is

(A) 27 (B)

33216

(C) 13 (D) None of these

6. The range of ‘a’ for which all the points oflocal extrema of the functionf(x) = x3 – 3ax2 + 3(a2 – 1)x + 1 lie in theinterval (–2, 4), is(A) (–1, 3) (B) (3, 4)(C) (–4, –2) (D) (–2, –1)

7. If the truth value of the statement(P Q) R® « is ‘F’then truth values of P, Q,R can be respectively(A) TFF (B) TFT (C) FFT (D) FTT

8. If A and B are non zero matrices of order 3such that 3A + 2B = AT, then det(A + B) is(A) 1 (B) –1 (C) 2 (D) 0

9. In DABC, if AB = x, BC = x + 1 andp

Ð =C3

, then least integral value of x is(A) 6 (B) 7 (C) 8 (D) 9

4. ;fn æ ö

- +ç ÷ç ÷è øìï + ¹= íï =î

1 1x x(x 1)2 , x 0f(x)

0 , x 0 gks] rks

(A) x = 0 ij f(x) larr gS(B) x = 0 ij f(x) vodyuh; gS(C)

+®x 0lim f(x) fo|eku ugha gS

(D) + -® ®¹


5. ge rhu le:i iklksa dks Qsdrsa gS rFkk ge ikrs gS fd rhuiklksa ds Åijh Qydksa dk ;ksxQy 10 gS] rks ,d ikls ij‘6’ izdV gksus dh izkf;drk gS

(A) 27 (B)

33216

(C) 13 (D) bueas ls dksbZ ugha

6. ‘a’ dk ifjlj] ftlds fy;s Qyu

f(x) = x3 – 3ax2 + 3(a2 – 1)x + 1 ds lHkh LFkuh;

pje fcUnq vUrjky (–2, 4) esa fLFkr gksa] gS(A) (–1, 3) (B) (3, 4)(C) (–4, –2) (D) (–2, –1)

7. ;fn dFku (P Q) R® « dh lR;rk eku ‘F’ gS] rks

P, Q, R ds lR;rk eku Øe'k% gks ldrs gS

(A) TFF (B) TFT (C) FFT (D) FTT

8. ;fn A rFkk B, dksVh 3 ds v'kwU; vkO;wg bl izdkj gS fd3A + 2B = AT gks] rks det(A + B) dk eku gksxk(A) 1 (B) –1 (C) 2 (D) 0

9. ;fn f=Hkqt ABC esa AB = x, BC = x + 1 rFkkp

Ð =C3 gks] rks x dk U;wure iw.kk±d eku gS

(A) 6 (B) 7 (C) 8 (D) 9

MEDIIT

ALLEN


10. An isosceles triangle is inscribed in theparabola y2 = 4ax with its base as the linesegment joining the vertex and positive endof the latus rectum of the parabola. If(at2, 2at) is the vertex of this triangle, then(A) 2t2 – 8t + 5 = 0(B) 2t2 + 8t – 5 = 0(C) 2t2 + 8t + 5 = 0(D) 2t2 – 8t – 5 = 0

11. Let x1,x2,....xn are 'n' observations such thatn

ii 1

x 10=

=å and n

2i

i 1x 260

=

=å and standard

deviation is 5, then n is equal to(A) 13 (B) 12 (C) 10 (D) 8

12. If 2 4sec x cosec x dx f(x)=ò and

pæ ö = -ç ÷è ø

8f6 3

, then pæ öç ÷è ø

f4 is

(A) 1 (B) -13 (C) -

43 (D) 0

13. The value of ( )( )

12

r 0

2r 1tan1 r r 1

¥-

=

æ ö+ç ÷

ç ÷ç ÷+ +è øå is

(A) 3p

(B) 2p

(C) 4p

(D) 6p

14. If ar and br are any two unit vectors and

range of 3 a b

2 a b2+

+ -

rrrr is [k1, k2], then

k1 + k2 is

(A) 5 (B) 3 (C) 8 (D) 9

10. ijoy; y2 = 4ax ds vUrxZr ,d leckgq f=HkqtgS ftldk vk/kkj] ijoy; ds ukfHkyEc ds /kukRedfljs rFkk 'kh"kZ dks feykus okyh js[kk[k.M gSA ;fnbl f=Hkqt dk 'kh"kZ (at2, 2at) gks] rks(A) 2t2 – 8t + 5 = 0(B) 2t2 + 8t – 5 = 0(C) 2t2 + 8t + 5 = 0(D) 2t2 – 8t – 5 = 0

11. ekuk x1,x2,....xn, 'n' izs{k.k bl izdkj gS fd n

ii 1

x 10=

=å

rFkk n

2i

i 1x 260

=

=å ,oa ekud fopyu 5 gks] rks n dk

eku gksxk(A) 13 (B) 12 (C) 10 (D) 8

12. ;fn 2 4sec x cosec x dx f(x)=ò rFkk

pæ ö = -ç ÷è ø

8f6 3

gks] rks pæ ö

ç ÷è ø

f4 dk eku gS

(A) 1 (B) -13 (C) -

43 (D) 0

13.( )( )

12

r 0

2r 1tan1 r r 1

¥-

=

æ ö+ç ÷

ç ÷ç ÷+ +è øå dk eku gS

(A) 3p

(B) 2p

(C) 4p

(D) 6p

14. ;fn ar rFkk br

nk s bdkb Z lfn'k gk s a rFk k

3 a b2 a b

2+

+ -

rrrr dk ifjlj [k1, k2] gk s] rk s

k1 + k2 dk eku gksxk

(A) 5 (B) 3 (C) 8 (D) 9

MEDIIT

ALLEN


15. There are three distinct boxes and sixdifferent balls. Any box can receive anynumber of balls. The number of ways inwhich these balls can be put into boxes sothat no box remains empty, is(A) 729 (B) 537 (C) 480 (D) 540

16. Number of solutions of equation

secx + tanx 13

= in [0,4p], is

(A) 0 (B) 3 (C) 2 (D) 417. General solution of differential equation

x xdye e y 1dx

+ = , is

(A) y = ex + ce–x (B) y = xex + cex

(C) y = e–x + cex (D) x xy xe ce- -= +

(where c is constant of integration)18. For the circle x2 + y2 = r2, the value of ‘r’ for

which area enclosed by the pair of tangentsdrawn from the point (6, 8) to the circle andthe chord of contact, is maximum, is(A) 5 (B) 10(C) 5 (D) None of these

19. Roots of quadratic equationp(x – 2) (x – 3) + e(x – 3) (x – 4) = 1, are(A) Imaginary (B) Real and equal(C) Real and distinct (D) Real and negative

20. The value of 2

0

sin x dxp

é ùë ûò , is

(where [.] is greatest integer function)

(A) p (B) -p (C) 0 (D) 2p

-

15. rhu fHkUu cDls rFkk N% fHkUu xsan nh xbZ gSaA fdlh HkhcDls esa fdruh Hkh xsan j[k ldrs gSaA mu rjhdksa dhla[;k ftlesa cDlksa esa xsansa j[kh tk ldrh gSa] rkfd dksbZHkh cDlk [kkyh u jgs] gS

(A) 729 (B) 537 (C) 480 (D) 540

16. vUrjky [0,4p] esa lehdj.k secx + tanx 13

= ds

gyksa dh la[;k gksxh(A) 0 (B) 3 (C) 2 (D) 4

17. vody lehdj.k x xdye e y 1dx

+ = dk O;kid gy gksxk

(A) y = ex + ce–x (B) y = xex + cex

(C) y = e–x + cex (D) x xy xe ce- -= +

(tgk¡ c lekdyu vpj gS)

18. ;fn o`Ùk x2 + y2 = r2 ij fcUnq (6, 8) ls [khaph Li'kZ js[kk

;qXe o budh Li'kZ thok ds e/; ifjc¼ {k s=Qy

vfèkdre gks] rc r dk eku gS

(A) 5 (B) 10

(C) 5 (D) buesa ls dksbZ ugha19. f}?kkr lehdj.k

p(x – 2) (x – 3) + e(x – 3) (x – 4) = 1 ds ewy gksaxs(A) dkYifud (B) okLrfod rFkk leku(C) okLrfod rFkk fHkUu (D) okLrfod rFkk ½.kkRed

20.2

0

sin x dxp

é ùë ûò dk eku gS

(tgk¡ [.], egÙke iw.kk±d Qyu dks n'kkZrk gS)

(A) p (B) -p (C) 0 (D) 2p

-

MEDIIT

ALLEN



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, 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

•

•

•

•

•

•

•

•

••

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 LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rd

Vª 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 nksuksa

lgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i

cqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksa

dks 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.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA


MEDIIT

ALLEN


1. The integer ‘n’ for which the value of

3x

nx 0

x(1 cosx)(e cosx)2lim

x®

- - - is a finite non

zero number, is

2. If -= 1f (x) sin x and ( )g x [cos(sin x)]= ,where [.] denotes greatest integer function.Then number of element(s) in the range off(g(x)) is

3. If normal at point (4, 1) of the curve xy = 4intersects the curve again at the point (a,b),then a + b is equal to

4. The value of

( )( )( )( )+ ° + ° ° + ° ° + °

+ ° + °2 2

1 3 tan1 1 3 tan2 tan1 tan59 tan2 tan58(1 tan 1 )(1 tan 2 )

is -

5. If =

+ +=

+ + +å19

5 4 3r 1

3(r 1)r 1 K(r 1)(r 2r r )

then 1000 K is

1. iw.kk±d ‘n’, ftlds fy;s

3x

nx 0

x(1 cosx)(e cosx)2lim

x®

- - - dk eku ifjfer

v'kwU; la[;k gks] gS

2. ;fn -= 1f (x) sin x rFkk ( )g x [cos(sin x)]= ,

tgk¡ [.], egÙke iw.kk±d Qyu dks n'kkZrk gSA rc f(g(x))

ds ifjlj esa vo;oksa dh la[;k gS3. ;fn oØ xy = 4 ds fcUnq (4, 1) ij [khapk x;k vfHkyEc]

oØ dks iqu% fcUnq (a,b) ij dkVrk gS] rks a + b dk ekugS

4. ( )( )( )( )+ ° + ° ° + ° ° + °

+ ° + °2 2

1 3 tan1 1 3 tan2 tan1 tan59 tan2 tan58(1 tan 1 )(1 tan 2 )

dk eku gS

5. ;fn =

+ +=

+ + +å19

5 4 3r 1

3(r 1)r 1 K(r 1)(r 2r r )

gks] rks 1000 K

dk eku gS

MEDIIT

ALLEN


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

29122019 Page 35/35All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119026

MEDIIT

Dear Student,

We request you to provide feedback for the test series till you have appeared. Kindly answer the questionsprovided on the reverse of paper with honesty and sincerely.

Although our test series questions are extremely well designed and are able to improve speed, accuracy &developing examination temperament, yet we are always open to improvements.

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MEDIIT

HS-1/8Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005

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SECTION-I1. Ans. (B)2. Ans. (B)Sol. For max. strain position of eye

o e

e

f fM (1 )f D

= +

= = =0R 120f 60 cm2 2

ef 1.25cm=

M = 50.43. Ans. (D)

Sol.Area = S

x x= x x + dx=

= - = a 2dvE 3 xdx

f = - a 2in (3 x )s

f = a + = a2out {3 (x dx) }s 3 s(2xdx)

f = anet (6 x)(sdx)f = = a

enet0

q (6 x)(sdx)

r = = aea 0

q 6 xsd

4. Ans. (A)

Sol.

f

30°

FN

tan q = µ

m =13

F sin 30 + mg = NF cos 30 = µ (N)F cos 30 = µ (F sin 30 + mg)

æ ö= +ç ÷è ø

3 1 FF 1002 23

F = 100 N

PART-1 : PHYSICS SOLUTION

(0000CJA102119026) Test Pattern

DISTANCE LEARNING PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE

JEE(Main)AIOT

29-12-2019

ANSWER KEYPART-1 : PHYSICS

PART-3 : MATHEMATICS

PART-2 : CHEMISTRY

Q. 1 2 3 4 5 6 7 8 9 10A. A A B D C A B D B BQ. 11 12 13 14 15 16 17 18 19 20A. C C B C D C D C C BQ. 1 2 3 4A. 4.00 2.00 –16.25 3.00

SECTION-I

SECTION-II 999.87 or 999.885

Q. 1 2 3 4 5 6 7 8 9 10A. C A D A D D A C C CQ. 11 12 13 14 15 16 17 18 19 20A. D B C D C D A C A BQ. 1 2 3 4 5

A. 1 .66 OR1.67 0 .21 1 .00 3 .00 350 .00

SECTION-I

SECTION-II

Q. 1 2 3 4 5 6 7 8 9 10A. B B D A A A B A D AQ. 11 12 13 14 15 16 17 18 19 20A. B C C A A B B A B AQ. 1 2 3 4 5A. 5.00 14.14 16.19 6.00 8.00

SECTION-I

SECTION-II

MEDIIT

ALLEN

LTS/HS-2/8 29122019All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119026

5. Ans. (A)

Sol. From nodal analysis : v 4 v3 01 9- æ ö+ =ç ÷

è øÞ V=3 Volt (voltmeter reading)

(Let) V

V

A0 V (let)

Ammeter reading = v 4

1-

= 1 Amp

6. Ans. (A)Sol. bc is parallal to de7. Ans. (B)

Sol.¶f

= = n -¶ 0

EMF a 2att

= = n -ò ò2

2(EMF) dt 1Heat (a 2at) dtR R

ì üï ï= n + - ní ýï ïî þò 2 2 2 21 (a ) 4a t 4a t dt

Ræ ö

= n n + n -ç ÷è ø

2 32 2 2 31 4 4a va a

R 3 2n æ ö= + -ç ÷

è ø

2 3a 41 2R 3

n=

2 3a3R

8. Ans. (A)

Sol. time take by stone to reach floor =htuTo find the speed of Iron man we can use

momentum conservation1M mun =

n =1muM

distance of Iron man moved above platform

is mu h mhyM u M

æ ö æ ö= =ç ÷ ç ÷è ø è ø

Total height mh 1M

æ ö= +ç ÷è ø

9. Ans. (D)10. Ans. (A)11. Ans. (B)

Sol. ff F

Þ The ball moves to the right and rotatesanti clockwise.

12. Ans. (C)Sol. Angle of incidence on lens-liquid interface.

i = 53°Now critical angle by snell’s law forcondition

1.5 sin i = µ sin 90m

=sin i1.5

i = 53° m

Þ =45 1.5

µ = 1.2Now for all value of µ £ 1.2it will show “TIR”

13. Ans. (C)Sol. In 1st case q¥ = C1 × 20

In 2nd case q¥ = C1 × 20Þ Ratio = 1 : 1

14. Ans. (A)

Sol. mv = mv ‘ + hl

h hfv ' v vM Mc

= - = -l

By energy cons.2 2

m1 1En mv E mv ' hf2 2

+ = + +

2 2n m

1 1hf mv (E E ) Mv '2 2

= + + - -

22

01 1 hmv m v hf2 2 mc

+æ ö= - - +ç ÷è ø

0vf f 1c

æ ö= -ç ÷è ø

15. Ans. (A)Sol. f = Li

f= - = -

d LdiEMFdt dt

If = p0 1i i sin(2 f t)

=

=

æ ö- ç ÷è ø

=æ ö- ç ÷è ø

1

2

f f1

2

f f

diLdtedie Ldt

=1 1

2 2

e fe f

16. Ans. (B)

Sol.( )

( )R 20t 2s

10p p

= = =n p

MEDIIT

ALLEN

29122019 LTS/HS-3/8All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119026

So time of flight 2V sin 2

gq

=

Vy = 10Now VxT = R

Vx (2) = 20 3

tanq = x

Vy 10 1V 10 3 3

= =

q = 30 °17. Ans. (B)18. Ans. (A)Sol. AV = constant

21P v2

+ r = constant

If velocity increases then pressuredecreases.

19. Ans. (B)Sol. L = 60 dB = 10 log I/I0

I = 10–6 w/m2

Power = (I) Area = (10-6) (4p × 5002)= 3.14 watt

( ) ( )3w 10 0.30t 95.5sp 3.14

= = =

20. Ans. (A)

Sol. = =-ò ò

nRTW PdV dV(V b)

= -2vvnRT ln(V b)

-æ ö= ç ÷-è ø

2V bW nRTlnV b

SECTION-II1. Ans. 5.002. Ans. 14.14

Sol.

A B50 cm

B2

40 cm

B1

30 cm

m=

p0

11

iB2 r

--´

= =7

5(2 10 )(20) 10 T(0.4)

--m ´

= = =p

750

22

i (2 10 )(15)B 10 T2 r 0.30

-= + = ´ =2 2 5net 1 2B B B 2 10 14.14

3. Ans. 16.19

Sol.dQsT

D = ò

273ice ice

136.5

dQM s dTT T

æ öç ÷+è øòò phase change

= MiceSice dT mLT T

+ò

= (27.3) (0.5) ln ( ) ( )

( )273 27.3 80

136.5 273æ ö +ç ÷è ø

DS = 16.19 cal/gm4. Ans. 6.00Sol. Mg = mg + B

Mg = mg + p2 × 1

Mp

g

M 2

1

p1p

æ ö-ç ÷

è ø = m = 6 kg

5. Ans. 8.00

37°S1

S2

5l O

A

37°

D ®x 3l

D ®x 5l

D =x 0

At A, x 5D = lAt O, x OD =At ¥ distance

x 5 sin37D = l °= 3l\ No. of minima between A and O = 5No. of minima beyond point O = 3\ Total no. of minima = 8

MEDIIT

ALLEN


PART–2 : CHEMISTRY SOLUTIONSECTION - I

1. Ans.(C)Initial pressure of water vapour P1 = 0.2x.75 = 0.15 atmP1V1 = P2V2

0.15 x 30 = 0.2xVV = 22.5 litre

2. Ans.(A)CH3CHCl – (–I), CH3 – CH2 – (+I),

Acidic strength ( I)µ - and 1

I+(i) CH3CHClCOOH>(iv) HCOOH>(iii)CH3CH2COOH>(ii) CH3CH2OH

3. Ans.(D) FactualSc < Y < La Atomic radii

4. Ans.(A)

p y2 24 r

-pp

03

( 2 z r / a )23

0

Z4 r .ea

above function has a maxima at (a0/z) andzero at infinite,so it follows 1st graph.

5. Ans.(D)

O

CH3CH3

O

CH3CH3

O

CH3CH3

S R S S R R

Optically inactive3 Stereo isomer2 chiral C

(3R,4S)-3,4-dimethylcyclopentan-1-one Optically inactive

6. Ans.(D)A ® (NH4)2CO3

B ® NH4HCO3

C ® NH4ClD ® NaHCO3

7. Ans.(A)2

2CaF (s) Ca (aq) 2F (aq)+ -+�

Ksp = 6.0 × 10–9

Ksp = [Ca+2] [ F- ]2

Given+ -= =2 3

31 2 0 1[C a ] x 3x1 0 m ol / L

4 01 0 L-

--

´=

92

36 .0 1 0[ F ]

3 x1 0[ F- ] = 1.414 × 10–3 mol/L[ F- ] = 1.414 × 10–3 × 19 × 103

[ F- ] = 26.84 ppm8. Ans.(C)

E2 reaction9. Ans.(C)

(ii) 2(aq) (aq) (s)Cu Cu Cu+ +¾¾® +

(iii) ( g)

2 3(aq) (aq) 22Cr 2H 2Cr H+ + ++ ¾¾® +

(v) Zn, Cd, Hg are not consider as transitionelements.

10. Ans.(C)Based on law of thermodynamics

2

1

VW nRTlnV

= - 2

1

W VTlnnR V

æ öæ ö- = ç ÷ç ÷è ø è ø

11. Ans.(D)It is aldol reaction and form product fromalpha hydrogen and produce

=6 5

3

C H CH CCHO|CH

.

12. Ans.(B)due to removal of e– from A.B.M.O (antibonding molecular orbital)

13. Ans.(C)

For octahedral structure = R 0.414R

+

- =

R+ = 0.414 × R–

= 0.414 × 500R+ = 207 pm

14. Ans.(D) Factual15. Ans.(C)

At 1200K C will reduce MO into M(l)

16. Ans.(D)T,P( G) 0D < for a spontaneous process

MEDIIT

ALLEN


17. Ans.(A)Ranitidine is only antacid drug whileNaOH is not drug.

18. Ans.(C)

CoIII

NH3

NN

NNH3

N

trans optically inactive

CoIII NH3

N

NH3

N

N

NCis (Optically active)

Co+3 ® [Ar], 3d6

Hybridisation d2sp3 in presence of strongfield ligand.Total 15 N–Co–N bond angles

19. Ans.(A)For 1st order Rxn.

= = = =´ ´1/ 2

ln 2 0.7 1 1kt 28 60 40 60 2400

31 9.19 10log 152400 T

´æ ö = -ç ÷è ø

´- - = -

39.19 102 1.38 15T

39.19 10 18.38T´

=

310T 500K2

= =

20. Ans.(B)a,c are aromatic

SECTION-II1. Ans.(1.66 or 1.67)

Ma2b2c2 : 5 GIM(AA)b2c2 : 3 GI

2. Ans.(0.21)

AgCl(s) Ag (aq) Cl (aq).+ -+�

ksp = [Ag+] [Cl–]10

10ksp 10[Ag ] 101[Cl ]

-+ -

-= = =

Ag e Ag(s)+ -+ ¾¾®

Cell 100.059 1E E log

1 10°

-= -

E = 0.80 – 0.59E = 0.21 V

3. Ans.(1.00) D¾¾¾¾¾®NaOH & CaO

3 2CH CH COONa CH3CH3+Na2CO3

4. Ans.(3.00)Atomic number: 58, 64, 71

5. Ans.(350.00)

Sol. i PP P 32004 4

´æ ö= +ç ÷è ø

....(1)

i P3P P300 34 4

æ ö= + ´ç ÷è ø

....(2)

ii i

9 P700 P 2P4 4

= - =

i700P 350

2= = mm of Hg.

PART-3 : MATHEMATICS SOLUTIONSECTION-I

1. Ans. (A)

Sol.

arg(z–5i)= 34p

|z–5| 5<

So number of common solution of both theequations is 0.

2. Ans. (A)

Sol. Angle bisectors are

( )3x 4y 1 3x 4y 75 5

- + + -= ±

Þ y = 1 or x = 1

3. Ans. (B)

Sol. S = 25C13 + 25C14 + .... + 25C25

Þ S = 25C12 + 25C11 + .... + 25C0 (Q nCr = nCn – r)

Þ 2S = 25C0 + 25C1 + .... + 25C25 = 225

Þ S = 224

MEDIIT

ALLEN


4. Ans. (D)

Sol.

2

x(x 1)2 , x 0

f(x) 0 , x 0

x 1 , x 0

-ì+ >ï

ï= =íï + <ïî

RHL =2

x

x 0lim (x 1)2

+

-

®+ = 0

LHL =x 0lim(x 1) 1

-®+ =


+ -® ®Þ ¹

5. Ans. (C)

Sol. A : Sum on the dice equal to 10

B : One die shows ‘6’

Sum of 10 is possible

When 6, 1, 3

6, 2, 2

5, 1, 4

5, 2, 3

4, 2, 4

4, 3, 3

so n(A) = 6

Also, n(A Ç B) = 2

Hence, B n(A B) 2 1

PA n(A) 6 3

Çæ ö = = =ç ÷è ø

6. Ans. (A)

Sol. ( )= - a + a - = - a + a -2 2 2 2f '(x) 3x 3 2x 3( 1) 3x 6 x 3( 1)

since, extrema lie in (–2, 4)so roots of f ¢(x) lie in (–2, 4)

4f ' (x) = 3(x – (a – 1)) (x –(a + 1))

Þ –2 < a – 1 and a + 1 < 4Þ –1 < a < 3

Second methodf '( 2) 0- >

Þ 12 + 12a + 3a2 – 3 > 0Þ 3a2 + 12a + 9 > 0Þ a2 + 4a + 3 > 0,Þ (a + 1) (a + 3) > 0Þ a Î (–¥, –3) È (–1, ¥) ..... (1)Also, f ¢(4) > 0Þ 48 – 24a + 3a2 – 3 > 0Þ 3a2 – 24a + 45 > 0Þ a2 – 8a + 15 > 0Þ (a – 5) (a – 3) > 0 Þ a Î ( ,3) (5, )-¥ È ¥

...... (2)And, D ³ 0Þ 36a2 – 4 × 3 × 3 (a2 – 1) ³ 0Þ a2 – a2 + 1 ³ 0 (this is always true)

Also 62 4

2 3

a- < <

× Þ ( 2, 4)a Î - ....(3)

Taking intersetion of (1), (2), (3) a Î (–1, 3)7. Ans. (B)Sol. P Q ®P Q R ® Û(P Q) R

T F F F TT F F T FF F T T TF T T T T

8. Ans. (D)Sol. 3A + 2B = AT .... (1)

2A + 2B = AT – AQ AT – A is skew symmetric matrix\ |2A + 2B| = 0Þ |A + B| = 0

9. Ans. (B)

Sol.

A

B

x60° C

x + 1Let AC = y

+ - +° =

+

2 2 2(x 1) x ycos602(x 1)y

Þ (x + 1)y = 2x + 1 + y2

Þ y2 – (x + 1)y + (2x + 1) = 0Now,D ³ 0Þ (x + 1)2 – 4(2x + 1) ³ 0 Þ ³ +x 3 12

MEDIIT

ALLEN


10. Ans. (B)

Sol.

(0, 0)O

A (a, 2a)

B(at , 2at)2

For isosceles triangleOB2 = AB2

Þ (at2)2 + (2at)2 = (at2 – a)2 + (2at – 2a)2

Þ t4 + 4t2 = t4 + 1 – 2t2 + 4t2 + 4 – 8tÞ 2t2 + 8t – 5 = 0

11. Ans. (C)

Sol. s2 =

2n n2i i

i 1 i 1x x

n n= =

æ öç ÷ç ÷- ç ÷è ø

å å

Þ 25 = 2260 100n n

-

Þ 25n2 – 260n + 100 = 0 Þ n = 25 , 10

12. Ans. (C)Sol. put tan x = t

2 2 2

4 2(1 t )(1 t ) dtI

t 1 t+ +

=+ò

I 2 2 4 2

4 4 4 2

(1 t ) 1 t 2t 1 2dt dt 1 dt

t t t t

+ + + æ ö= = = + +ç ÷è øò ò ò

3

1 2t c

t3t= + + +

--

3

1 2I tan x c

tan x3tan x= + - +

-

So f(x) = 3

1 2tan x

tan x3tan x+ -

-

Þ 1 4

f 1 24 3 3

pæ ö = - + - = -ç ÷è ø

13. Ans. (B)

Sol. Since, Tr = 2 2

12

(r 1) rtan1 ((r 1)r)

- æ ö+ -ç ÷

+ +è ø

Hence, 1 2 1nS tan (n ) tan (0) S

2- -

¥p

= - Þ =

14. Ans. (C)

Sol. + = + + = + + q2 2 2a b a b 2a b 1 1 2cos

2 2a b 2 2cos2q

Þ + = ×

a b 2cos2q

Þ + =

Also, q

- =a b 2sin2

So, q qæ ö æ ö= +ç ÷ ç ÷è ø è ø

3S 2cos 2 2sin2 2 2

3cos 4sin 3,52 2q q

= + Îé ùë û as 0,2 2q pé ùÎ ê úë û

15. Ans. (D)Sol. Number of onto function from A to B

= 36 – 3C126 + 3C21 = 729 – 192 + 316. Ans. (C)

Sol. secx + tanx = 13 .... (1)

secx – tanx = 3 .... (2)(1) – (2) :-

2tanx = 83

- Þ tanx = 43

-

Þ x lies in IVth quadrant(in IInd quadrant secx, tanx < 0)

17. Ans. (D)Sol. ex(y + y') = 1

Integrating :-yex = x + c Þ y = (x + c)e–x

18. Ans. (C)

Length of tangent 21s 100 r= = -

1

1Area s sin2

2= q

Now 1

rtan

sq =

1 1

2 21

1

r2

s 2r ssin2

r s r1

s

q = =+

+

MEDIIT

ALLEN


22

2 2

1 2r 100 rA (100 r )

2 100 r r

-= - ×

- + ( )

32 2

1r 100 r

100

é ù= -ê ú

ë û

1 32 22 2

dA 1 3r (100 r ) ( 2r) (100 r ) 1

dr 100 2

é ù= × - - + - ×ê ú

ë ûFor critical pointdA

0dr

= Þ 3

2 2 2 23r 100 r (100 r )- = -

Þ 2 23r 100 r= -

Þ 24r 100=Þ r2 = 25Þ r = 5

at this ‘r’ the area is maximum19. Ans. (C)Sol. Let f(x) = p(x – 2) (x – 3) + e(x – 3)(x – 4) – 1

Now f(2) = 2e – 1 > 0 f(3) = –1 < 0 f(4) = 2p – 1 > 0

So, f(x) has two distinct real roots one in(2, 3) and other in (3,4).

20. Ans. (B)2

0

I sin x dxp

= é ùë ûò2

0

I sin x dxp

Þ = -é ùë ûò

( )2

0

2I 1 dxp

Þ = -ò IÞ = -p

SECTION-II1. Ans. 4.00

The value of n is the least degree ofnumerator

Nr.3

x x(cosx 1)(cosx e )

2= - - -

(expanding this)

= 2 4 2 4 2 3 3x x x x x x x

1 ... 1 1 .... 1 x ...2 24 2 24 2 6 2

æ öæ ö æ ö æ ö- + - - + - + + + -ç ÷ç ÷ ç ÷ ç ÷ç ÷è ø è ø è øè ø

2 4 3 5 3

2x x x x x.... x x ...

2 24 6 120 2

æ öæ ö= - + - - - - -ç ÷ç ÷

è øè ø

3 5 4 6 3x x x x x

... ...2 24 2 24 2

= - + - -

So, minimum degree of numerator is 4So, the value of n is 4.

2. Ans. 2.00

p/2 p

cos11

0

y=cos(sinx)

3. Ans. –16.25

Normal at æ öç ÷è ø

cct,t intersect the curve again

æ öç ÷è ø

cct ',t ' then = -3t ' t 1 .

1(x ', y ') , 164

æ öÞ = - -ç ÷è ø

4. Ans. 3.00

q + - q =(60 ) 3

Þq + - q

=- q - q

tan tan(60 ) 31 tan tan(60 )

Þ q + - q = - q - qtan tan 60 3 3 tan tan(60 )

Þ q + - q + q - q =tan tan(60 ) 3 tan tan(60 ) 3Þ q + q + - q + q = q2 2tan 3 tan tan(60 )(1 3 tan ) 3 sec

Þ + q q + - q = q2(1 3 tan )(tan tan(60 )) 3 sec

Þ+ q q + - q

=+ q2

(1 3 tan )(tan tan(60 ) 3(1 tan )

Now, put q = 1° and q = 2°+ ° + ° ° + ° ° + °

= ×+ ° + °2 2

(1 3 tan1 )(1 3 tan2 )(tan10 tan59 )(tan2 tan54 ) 3 3(1 tan 1 )(1 tan 2 )

= 35. Ans. 999.87 or 999.88

= =

+ += -

+ + + +å å19 19

5 4 3 3 3r 1 r 1

3(r 1)r 1 1 1(r 1)(r 2r r ) r (r 1)

= -+r 3 3

1 1Tr (r 1)

-= - = =3

1 8000 1 7999S 18000 800020

= =79991000K 999.875

8

MEDIIT

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