TARGET : JEE (Main + Advanced) 2019 · Do not write any of these details anywhere else on the ORS. Darke th pr bble unde numb. viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa

READ THE INSTRUCTIONS CAREFULLY / Ïi;k bu funsZ'kks a dks /;ku ls i<+ s aGENERAL / lkekU; %1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.

;g eksgjcU/k iqfLrdk vkidk iz'ui= gSA bldh eqgj rc rd u rksM+s tc rd bldk funsZ'k u fn;k tk;sA2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

iz'uksa dk mÙkj nsus ds fy, vyx ls nh x;h vkWfIVdy fjLikal 'khV (vks- vkj- ,l-) (ORS) dk mi;ksx djsaA3. Blank spaces are provided within this booklet for rough work.

dPps dk;Z ds fy, bl iqfLrdk esa [kkyh LFkku fn;s x;s gSaA4. Write your name, form number and sign in the space provided on the back cover of this booklet.

bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke o QkWeZ uEcj fyf[k, ,oa gLrk{kj cukb;sA5. After breaking the seal of the booklet, verify that the booklet contains 38 pages and that all the

18 questions in each subject and along with the options are legible. If not, contact the invigilator forreplacement of the booklet.bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ysa fd blesa 38 i`"B gSa vkSj izR;sd fo"k; ds lHkh 18 iz'u vkSj muds mÙkj fodYi Bhdls i<+ s tk ldrs gSaA ;fn ugha] rks iz'ui= dks cnyus ds fy, fujh{kd ls lEidZ djsaA

6. You are allowed to take away the Question Paper at the end of the examination.ijh{kkFkhZ iz'ui= dks ijh{kk dh lekIrh ij ys tk ldrs gSaA

OPTICAL RESPONSE SHEET / vkWfIVdy fjLikal 'khV (vks-vkj-,l-) %7. The ORS will be collected by the invigilator at the end of the examination.

vks- vkj- ,l- dks ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk,xkA8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

vks- vkj- ,l- esa gsj&Qsj@foÏfr u djsaA vks-vkj-,l- dk dPps dke ds fy, iz;ksx u djs aA9. Write your name, form number and sign with pen in the space provided for this purpose on the ORS.

Do not write any of these details anywhere else on the ORS. Darken the appropriate bubbleunder each digit of your form number.viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa esa dye ls fy[ksa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh fooj.kvks-vkj-,l- es a dgha vkSj u fy[ks aA QkWeZ uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

DARKENING THE BUBBLES ON THE ORS / vks-vkj-,l- ij cqycqyks a dks dkyk djus dh fof/k %10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

vks-vkj-,l- ds cqycqyksa dks dkys ckWy ikWbUV dye ls dkyk djsaA11. Darken the bubble COMPLETELY. / cqycqys dks iw.k ± :i ls dkyk djsaA12. The correct way of darkening a bubble is as : / cqycqys dks dkyk djus dk mi;qDr rjhdk gS : 13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.

vks-vkj-,l- e'khu tk¡P; gSA lqfuf'pr djsa dh cqycqys lgh fof/k ls dkys fd, x;sa gSaA14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or

"un-darken" a darkened bubble.cqycqys dks rHkh dkyk djsa tc vki mÙkj ds ckjs esa fuf'pr gksA dkys fd, gq, cqycqys dks feVkus vFkok lkQ djus dk dksbZ rjhdk ugha gSA

15. Take g = 10 m/s2 unless otherwise stated. / g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksAQUESTION PAPER FORMAT AND MARKING SCHEME / iz'ui= dk izk:i vkSj vadu ;k stuk %16. The question paper has three parts : Physics, Chemistry and Mathematics.

bl iz'ui= esa rhu Hkkx gSa % HkkSfrd foKku] jlk;u foKku ,oa xf.krA17. Each part has two sections as detailed in table page 38.

izR;sd Hkkx esa nks [k.M gSa ftudk fooj.k i`"B 38 ij rkfydk esa fn;k x;k gSA

LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : JEE (Main + Advanced) 2019

DISTANCE LEARNING PROGRAMME(Academic Session : 2018 - 2019)

Please see the page 38 of this booklet for rest of the instructions/Ïi;k 'ks"k funs Z'kk sa ds fy, bl iqfLrdk ds i`"B 38 dks i<+ s aA

DO

NOT

BRE

AK

THE

SEA

LS W

ITHO

UT

BEIN

G IN

STRU

CTED

TO

DO

SO B

Y TH

E IN

VIGI

LATO

R / f

ujh{

kd d

s vuqn

s'kksa d

s fcuk

eqgj

sa u r

ksM+s

)0000CJA103118017)Paper Code

(0000CJA103118017)

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-AdvancedTEST # 13 TEST DATE : 15 - 05 - 2019Time : 3 Hours Maximum Marks : 180PAPER – 1

LTS-2/38

All India Open Test/Leader Test Series/Joint Package Course/JEE (Advanced)/15-05-2019/PAPER-1

0000CJA103118017

Note : In case of any correction in the test paper, please mail to [email protected] within 2 days along with Paper Code& Your Form No.(uksV % ;fn bl iz'u i= esa dksbZ Correction gks rks Ïi;k Paper Code ,oa vkids Form No. ,oa iw.kZ Test Details ds lkFk 2 fnu ds [email protected] ij mail djsaA)

Space for Rough Work / dPps dk;Z ds fy, LFkku

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58,Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1cm

· Planck constant h = 6.63 × 10–34 J–s

0000CJA103118017


LTS-3/38

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART-1 : PHYSICS Hkkx-1 : HkkSfrd foKku

SECTION–I(i) : (Maximum Marks : 24) [k.M–I(i) : (vf/kdre vad : 24)

� This section contains SIX questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

both of which are correct options.Partial Marks : +1 If two or more options are correct but ONLY one option is chosen

and it is a correct option.Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).Negative Marks : –2 In all other cases.

� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –2 marks.

� bl [kaM esa N% iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd

fodYi lgh gS(gSa)A� izR;sd iz'u ds fy,] iz'u dk (ds) mÙkj nsus gsrq lgh fodYi (fodYiksa) dks pqusA� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk %

iw.kZ vad : +4 ;fn dsoy (lkjs) lgh fodYi (fodYiksa) dks pquk x;k gSAvkaf'kd vad : +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSAvkaf'kd vad : +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gSa ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj

pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSaA'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ iz'u vuqÙkfjr gS)A½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feysaxs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi ) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi ), –2 vad feysaxs] pkgs lgh fodYi (fodYiks a) dks pquk x;k gks ;k u pquk x;k gksA

0000CJA103118017


LTS-4/38


1. A thin vertical tube of length l is inserted into mercury to a depth of l/2. Then the top end issealed. Find the height h of the mercury column left in the tube after the tube is pulled outof the mercury. The temperature is constant and the atmospheric pressure is equal to that ofa mercury column of height H.

yEckbZ l okyh ,d iryh Å/okZ/kj uyh dks l/2 xgjkbZ rd ikjs eas çfo"B djk;k tkrk gSA vc blds Åijh fljs dks

lhyc¼ dj fn;k tkrk gSA uyh dks ikjs ls ckgj fudkyus ds i'pkr~ uyh esa cps gq;s ikjn~ LrEHk dh Å¡pkbZ h Kkr dhft;sA

rkieku fu;r gS rFkk ok;qe.Myh; nkc] H Å¡pkbZ ds ikjn~ LrEHk ds cjkcj gSA

(A) 2 2H 1h H2 2+

= - +l

l (B) 2 2H 1h H2 2+

= + +l

l

(C) 2 2H H+ - +l l (D) 2 2H H+ + +l l

2. A charged particle of mass m = 1kg & charge Q = 1C, is projected from (0, 10) with velocity vr .

Uniform electric field ( )ˆ ˆ10i 5j- + N/C exist in the region. Gravity act along negative y direction.

(take : g = 10 m/s2) For ˆv 10i=r

(A) Particle passes through origin at some time.(B) Time spent by charged particle in I quadrant is 2 seconds.(C) While crossing origin, the trajectory makes equal angle with both negative x and negative

y axes.(D) K.E. of particle at (0, 10) will be half the K.E. at origin.

x(m)

g

y(m)

v(0, 10)

æO;eku m = 1kg rFkk vkos'k Q = 1C okys ,d vkosf'kr d.k dks fcUnq (0, 10) ls osx vr ls iz{ksfir fd;k tkrk gSA ;gk¡

le:i fo|qr {ks= ( )ˆ ˆ10i 5j- + N/C fo|eku gSA xq:Ro ½.kkRed y fn'kk ds vuqfn'k dk;Zjr gSA ˆv 10i=r ds fy;s

(g = 10 m/s2)(A) d.k fdlh {k.k ewy fcUnq ls xqtjrk gSA(B) vkosf'kr d.k }kjk izFke prqFkk±'k esa O;frr le; 2 lsd.M gSA(C) ewy fcUnq ls xqtjus ds nkSjku iz{ksI; iFk ½.kkRed x rFkk ½.kkRed y nksuksa v{kksa ls leku dks.k cukrk gSA(D) fcUnq (0, 10) ij d.k dh xfrt ÅtkZ ewy fcUnq ij xfrt ÅtkZ dh vk/kh gSA

0000CJA103118017


LTS-5/38


3. A hollow spherical conducting shell of radius a has thin long wires connected at the top(r = a, q = 0) and bottom (r = a, q = p). A direct current I flows down the upper wire, down thespherical surface, and out the lower wire. The sphere is centered at the origin and the wiresare along the x axis.(A) At the centre of the sphere, magnetic field is zero(B) At point (0.5r, 0.8r, 0) the magnetic field is m0I/1.6pr(C) At point (0.3r, 0.4r, 0) the magnetic field is m0I/1.2pr(D) At point (0.5r, 0.8r, 0) the magnetic field is zerof=T;k a okys ,d [kks[kys xksyh; pkyd dks'k ds 'kh"kZ (r = a, q = 0) o vk/kkj (r = a, q = p) ij yEcs irys rkj tqM+sgq, gSA Åij okys rkj esa ,d fn"V /kkjk I] xksyh; lrg ls uhps dh vksj rFkk fupys rkj ls ckgj dh vksj çokfgr gksrh gSAxksyk ewyfcUnq ij dsfUær gS rFkk rkj x v{k ds vuqfn'k fLFkr gSA(A) xksys ds dsUæ ij pqEcdh; {ks= 'kwU; gSA(B) fcUnq (0.5r, 0.8r, 0) ij pqEcdh; {ks= dk eku m0I/1.6pr gSA(C) fcUnq (0.3r, 0.4r, 0) ij pqEcdh; {ks= dk eku m0I/1.2pr gSA(D) fcUnq (0.5r, 0.8r, 0) ij pqEcdh; {ks= dk eku 'kwU; gSA

4. Two plane air capacitor with the same plate area are charged to the same charge. The distancebetween the plates of the first capacitor is half that of the second. Now the first capacitor isinserted between the plates of the second as shown in the figure.(A) the potential difference between the plates of the first capacitor will increase after the

insertion(B) the potential difference between the plates of the second capacitor will increase after

the insertion(C) the magnitude of charge on one of the facing surface of the plates of the first capacitor

will increase after the insertion(D) the magnitude of charge on one of the facing surface of the plates of the second capacitor

will increase after the insertion

A

B

C

D

Q

–Q

Q

–Q

C1C2

A

B

C

D

+

+

––

leku IysV {ks=Qy okys nks lery ok;q la/kkfj=ksa dks leku vkos'k rd vkosf'kr fd;k tkrk gSA çFke la/kkfj= dh IysVksads e/; nwjh] f}Ùkh; la/kkfj= dh rqyuk esa vk/kh gSA vc çFke la/kkfj= dks fp=kuqlkj f}rh; la/kkfj= dh IysVksa ds e/;çfo"B djk;k tkrk gS rks(A) çfo"B djkus ds ckn çFke la/kkfj= dh IysVksa ds e/; foHkokUrj c<+ tk;sxkA(B) çfo"B djkus ds ckn f}rh; la/kkfj= dh IysVksa ds e/; foHkokUrj c<+ tk;sxkA(C) çfo"B djkus ds ckn çFke la/kkfj= dh IysVksa dh fdlh ,d lEeq[k lrg ij vkos'k dk ifjek.k c<+ tk;sxkA(D) çfo"B djkus ds ckn f}rh; la/kkfj= dh IysVksa dh fdlh ,d lEeq[k lrg ij vkos'k dk ifjek.k c<+ tk;sxkA

0000CJA103118017


LTS-6/38

5. Assembled circuit shown in Figure. EMF of battery e1 is 1.5 V. EMF of battery e2 is variable.Both batteries are ideal. Initially e2 is such that current through it is zero.(A) If e2 = 2V keeping same polarity it is getting charged.(B) e2 = 0.9V(C) If e2 = 0.5V keeping same polarity it is getting charged.(D) If e2 is increased keeping the same polarity, e1 may be charged.

e1

e2

3W

4.5W

fp= esa çnf'kZr ifjiFk ij fopkj dhft;sA cSVjh e1 dk fo|qrokgd cy 1.5 V gSA cSVjh e2 dk fo|qr okgd cy ifjorhZ

gSA nksuksa cSVfj;k¡ vkn'kZ gSA çkjEHk esa e2 bl çdkj gS fd blls çokfgr /kkjk dk eku 'kwU; gSA

(A) ;fn leku /kzqork j[krs gq, e2 = 2V gks rks ;g vkosf'kr gks tkrh gSA(B) e2 = 0.9V(C) ;fn leku /kzqork j[krs gq, e2 = 0.5V gks rks ;g vkosf'kr gks tkrh gSA

(D) ;fn leku /kzqork j[krs gq, e2 dks c<+k;k tk;s rks e1 vkosf'kr gks ldrh gSA


0000CJA103118017


LTS-7/38

6. Find the forces in the hinged rods BC and AC, if AB = 60 cm, AC = 1.2 m, BC = 1.6 m(see figure). Hanging mass has a mass of 50 kg, the mass of the rods can be neglected. C ishinge.(A) AC is in tension with force of 1000 N.

(B) AC is in tension with force of 4000 N

3 .

(C) BC is in compression with force of 4000 N

3 .

(D) BC is in compression with force of 400 N.

A

B

C

çnf'kZr fp= eas dhydhr NM+ksa BC o AC esa cy Kkr dhft;s] ;fn AB = 60 cm, AC = 1.2 m, BC = 1.6 m gksA

yVd gq,s æO;eku dk eku 50 kg gS rFkk NM+ksa dk æO;eku ux.; ekuk tk ldrk gSA C dhyd gSA

(A) AC, 1000 N cy ds lkFk ruko esa gSA

(B) AC, 4000 N

3 cy ds lkFk ruko esa gSA

(C) BC, 4000 N

3 cy ds lkFk laihM+u dh voLFkk esa gSA

(D) BC, 400 N cy ds lkFk laihM+u dh voLFkk esa gSA


0000CJA103118017


LTS-8/38

SECTION–I(ii) : (Maximum Marks : 12)[k.M–I(ii) : (vf/kdre vad : 12)

� This section contains TWO paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

� bl [k.M esa nks vuqPNsn gSa� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d 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 : +3 ;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 gSA½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

Paragraph for Questions 7 and 8iz'u 7 ,oa 8 ds fy;s vuqPNsn

A cylindrical block of density 2 gm/cc is suspended by a thin wire 3m long. When the wire isvibrated, it produces fundamental frequency of 280 Hz. The mass of wire is so small thattension in the wire can be assumed to be constant. Now the block is dipped in a liquid. It isfound that the first overtone now is 280 Hz.?kuRo 2 gm/cc okyk ,d csyukdkj CykWd 3m yEcs irys rkj }kjk yVdk gqvk gSA rkj dks dafir djkus ij ;g 280 Hzdh ewy vko`fr mRiUu djrk gSA rkj dk æO;eku bruk vYi gS fd rkj esa ruko dk eku fu;r ekuk tk ldrk gSA vcCykWd dks ,d æo esa Mqcks;k tkrk gSA ;g ik;k tkrk gS fd çFke vf/kLojd dk eku vc 280 Hz gSA

7. What is the density of the liquid?æo ds ?kuRo dk eku D;k gS\(A) 1 gm/cc (B) 1.5 gm/cc (C) 0.75 gm/cc (D) 1.25 gm/cc

8. If the block is dipped to 1/3 of it's height in the liquid. What is the new fundamental frequency?;fn CykWd dks æo esa bldh Å¡pkbZ dk 1/3 xquk rd Mqcks;k tk;s rks u;h ewy vkofr D;k gksxh\(A) 140 Hz (B) 210 Hz (C) 140 3 Hz (D) 140 2 Hz


0000CJA103118017


LTS-9/38


De-broglie hypothesis states that every material particle is associated with a wave known

as matter wave. The wavelength of matter wave is h

mvl = , where h is Planck's constant.

When we use this in Bohr's theory, it shows that the angular momentum of electron in ahydrogen atom is quantized.The same principle can be applied to a particle of mass m moving in a circular path on asmooth inverted circular cone as shown. The speed of the particle at a height H from vertexis v. Reference level for potential energy is at the ground.Mh&czksXyh ifjdYiuk ds vuqlkj çR;sd inkFkZ d.k ,d rjax ls lacaf/kr gksrk gS ftls æO; rjax dgk tkrk gSA æO; rjax dh

rjaxnS/;Z h

mvl = gksrh gS] tgk¡ h Iykad fu;rkad gSA tc bls cksgj fl¼kUr esa ç;qDr fd;k tkrk gS ;g n'kkZrk gS fd

gkbMªkstu ijek.kq esa bysDVªkWu dk dks.kh; laosx DokUVhd`r gksrk gSA

;g fl¼kUr fp=kuqlkj ,d fpdus mYVs oÙkkdkj 'kadq ij oÙkkdkj iFk esa xfr'khy m æO;eku ds d.k ij yxk;k tk ldrk

gSA 'kh"kZ ls H Å¡pkbZ ij d.k dh pky v gSA fLFkfrt ÅtkZ ds fy;s lUnHkZ Lrj /kjkry ij fy;k x;k gSA

q H

v

9. The minimum height H of the particle is :d.k dh U;wure Å¡pkbZ H gS :-

(A) 2/3

1/3h 1

2 m tan gæ ö ´ç ÷p qè ø

(B) 2/3

1/3h 1tan

2 m gæ öq ´ç ÷pè ø

(C) 2/3

1/3h 1cos


(D) 2/3

1/3h 1sin


10. The total energy of particle is (n is an integer).d.k dh dqy ÅtkZ gS (n ,d iw.kk±d gS)

(A) 2/3

1/3 1/3 2 /32 hm g n tan3 2

æ ö´ qç ÷pè ø(B)

2/31/3 2 /3 2 /33 hm g n

2 2 tanæ öç ÷p qè ø

(C) 2/3

1/3 2 /3 2 /31 hm g n tan2 2

æ öqç ÷pè ø(D) None of these


0000CJA103118017


LTS-10/38

SECTION-II : (Maximum Marks: 24) [kaM-II : (vf/kdre vad : 24)

� This section contains EIGHT questions.� The answer to each question is a NUMERICAL VALUE.� For each question, enter the correct numerical value (in decimal notation, truncated/rounded-

off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubblesin the ORS.For Example : If answer is –77.25, 5.2 then fill 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 evaluated according to the following marking scheme:Full Marks : +3 If ONLY the correct numerical value is entered as answer.Zero Marks : 0 In all other cases.

� bl [kaM esa vkB iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_

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 lghgksxsa) 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 : +3 ;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


0000CJA103118017


LTS-11/38

1. In figure, a rod of mass m = 1kg is held vertically by an unknown mass M hanging over apulley as shown. Assuming h = 50 cm and H = 100 cm, find the minimum value of M (in kg)that makes the equilibrium stable for small displacement.çnf'kZr fp= esa m = 1kg æO;eku dh NM+ dks f?kjuh ij yVds gq, vKkr æO;eku M }kjk Å/okZ/kj j[kk x;k gSA ekukh = 50 cm o H = 100 cm gSA M dk U;wure eku (kg esa) D;k gksxk rkfd vYi foLFkkiu ds fy, lkE;koLFkkLFkk;h gks\

H

hm

M

2. An inductor L and resistor R are connected in a single loop. The current reduces to zero fromits maximum value. At time t = t0, the energy converted to heat equal the energy stored in the

inductor at that time. Find the value of t0. If 0xLtR

= , find x.

,d izsjdRo L rFkk ,d izfrjks/kd R fdlh ,dy ywi esa tqM+s gq;s gSA /kkjk dk eku blds vf/kdre eku ls 'kwU; rd

?kVrk gSA le; t = t0 ij Å"ek esa :ikUrfjr ÅtkZ dk eku ml le; ij izsjdRo esa lafpr ÅtkZ ds rqY; gSA t0 dk eku

Kkr dhft;sA ;fn 0xLtR

= gks rks x dk eku Kkr dhft;sA


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LTS-12/38

3. Capillary tube with very thin walls is suspended vertically in a cup containing water. Beforethe capillary is suspended, the cup was balanced against weights on a beam balance. Whencapillary is dipped to a small distance in the water, water rises in the capillary. Watercompletely wets the capillary walls. To restore the balance, we have to decrease the load onthe other scales by 0.14 g. If the surface tension of water is 0.07 N/m, determine the radius r(in mm) of the capillary.,d cgqr iryh nhokjksa okyh ds'kuyh dks ty ls Hkjs di esa Å/okZ/kj yVdk;k tkrk gSA ds'kuyh dks yVdkus ls iwoZ ;g

di ,d che larqyu ij Hkkjksa }kjk larqfyr FkkA tc ds'kuyh dks ty eas ,d vYi nwjh rd Mqcks;k tkrk gS rks ds'kuyh esa

ty p<+rk gSA ty ds'kuyh dh nhokjksa dks iw.kZr;k fHkxks nsrk gSA larqyu dks iqu% çkIr djus ds fy;s nwljs iSekuksa ij Hkkj dks

0.14 g ?kVkuk gksxkA ;fn ty dk i"B ruko 0.07 N/m gks rks ds'kuyh dh f=T;k r (mm) Kkr dhft;sA

4. The working substance of a heat engine is monoatomic ideal gas. Determine the efficiency(in percentage) of heat engine; the cycle is shown in the figure.,d Å"ek batu dk dk;Zdkjh inkFkZ ,dijekf.od vkn'kZ xSl gSA çnf'kr pØ ds fy;s Å"ek batu dh n{krk (çfr'kr esa)

Kkr dhft;sA

v

p

0 2V0 4V0

2p0

4p0

1

2

3p0

V0


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LTS-13/38

5. Find the equivalent resistance (in ohms) across the source (see figure), if emf E = 10 V.Resistance of all resistors is the same : R = 68 ohms.çnf'kZr ifjiFk esa L=ksr ij rqY; çfrjks/k (vkse esa) Kkr dhft;s] ;fn fo|qr okgd cy E = 10 V gSA lHkh çfrjks/kdks dkçfrjks/k leku gSA R = 68 vkse gSA

E

6. A boat runs between points A and B located on opposite banks of the river. The boat isconstantly on the line AB (see figure). Points A and B are at a distance of s = 1200 m fromeach other. The rate of flow of the river is u = 2 m/sec. Line AB makes an angle a = 30° withthe direction of flow of the river. What is the speed v (in m/s) of boat with respect to water ifboat has to go from A to B and get back in 20 minutes ?,d uko fdlh unh ds foijhr fdukjksa ij fLFkr fcUnq A rFkk B ds e/; xfr djrh gSA uko yxkrkj js[kk AB ij jgrh gS]fp= ns[ksaA fcUnq A o B ,d nwljs ls s = 1200 m nwjh ij fLFkr gSA unh ds çokg dh nj u = 2 m/sec gSA js[kk AB unhds çokg dh fn'kk ds lkFk a = 30° dks.k cukrh gSA ;fn uko dks A ls B rd tkdj iqu% 20 fefuV esa gh okil ykSVuk gksrks ty ds lkis{k uko dh pky v (m/s esa) D;k gksxh\

a

A

BC

u

v


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LTS-14/38

7. A homemade illuminator is a cylindrical tube of radius R = 10 mm with a black inner surface.A point source of light is located on the tube axis at a distance d = 20 cm from the lens. Thescreen is placed in front of illuminator at such a distance that a sharp image of the pointsource of light is produced on it. Focal length of the lens is F = 15 cm. What will be the innerradius of circle of light (in cm) directly from the source on the screen, if you remove the tubeand just keep the remaining arrangement as it is ?,d ?kjsyw çdk'kd R = 10 mm f=T;k dh ,d csyukdkj uyh ds :i esa gS] ftldh vkarfjd lrg dkyh gSA ,d çdk'kfcUnq L=ksr ySal ls d = 20 cm nwjh ij uyh dh v{k ij fLFkr gSA ,d inkZ çdk'kd ds lkeus bruh nwjh ij j[kk gS fd blij çdk'k fcUnq L=ksr dk ,d Li"V çfrfcEc cukrk gSA ySal dh Qksdl nwjh F = 15 cm gSA ;fn uyh dks gVk fn;k tk;srFkk 'ks"k midj.k dks iw.kZor~ gh j[kk tk;s rks insZ ij L=ksr ls lh/ks vkus okys çdk'k ds o`Ùk dh vkarfjd f=T;k (cm esa)D;k gksxh\

2R

d

||||||||

||||

||||||||||||||||||

8. A bi-metallic strip is made of two carefully polished flat plates: a silver and lithium. Theplate is placed in a vacuum, and the surface of silver is illuminated normally bymonochromatic beam of violet light with a wavelength l = 0.40 micron. Plate is now turnedby 180° so that the lithium surface faces the light. What is the ratio of saturation currentfrom silver to saturation current from lithium? Assume that the photoelectric efficiency inboth the cases is 0.01 and that all photoelectrons are emitted normal to the surface as fastas possible. Work function of lithium = 2.5 eV. Work function of silver = 4.5 eV. Mass ofelectron = 9 × 10–31 kg.

,d f}&/kkfRod iV~Vh nks lko/kkuhiwoZd ikWfy'k dh x;h lery IysVksa_ ,d pk¡nh dh rFkk nwljh yhfFk;e ls cuh gSA ;g

IysV fuokZr~ eas j[kh gS rFkk pk¡nh dh lrg dks rjaxnS/;Z l = 0.40 ekbØksu okys cSaxuh çdk'k ds ,do.khZ; iqat ls yEcor~

:i ls çdkf'kr fd;k tkrk gSA vc IysV dks 180° ij ?kqek;k tkrk gS rkfd çdk'k yhfFk;e lrg ij fxjsA pk¡nh rFkk

yhfFk;e ls çkIr larIr /kkjk dk vuqikr Kkr dhft;sA ekuk nksuksa çdj.kksa esa çdk'kfo|qr n{krk 0.01 gS rFkk lHkh

çdk'kbysDVªkWu lrg ds yEcor~ laHkkfor rsth ds lkFk mRlftZr gksrs gSA yhfFk;e dk dk;ZQyu = 2.5 eV, pk¡nh dk

dk;ZQyu = 4.5 eV o bysDVªkWu dk æO;eku = 9 × 10–31 kg gksrk gSA


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PART-2 : CHEMISTRY Hkkx-2 : jlk;u foKku

SECTION–I(i) : (Maximum Marks : 24) [k.M–I(i) : (vf/kdre vad : 24)

� This section contains SIX questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of









pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj


� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feysaxs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –2 vad feysaxs] pkgs lgh fodYi (fodYiks a) dks pquk x;k gks ;k u pquk x;k gksA

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1. Edison cell is represented as :Fe(s) | FeO(s) , 20% KOH(aq.) , Ni2O3(s), NiO(s) | Ni(s)The correct information(s) related with this cell is/are -(A) Anode reaction is Fe(s) + 2OH– (aq.) ® FeO(s) + H2O(l) + 2e–

(B) Cathode reaction is Ni2O3(s) + H2O(l) + 2e– ® 2NiO(s) + 2OH–(aq.)(C) Ecell will not change on changing the concentration of KOH solution(D) DG0 for the cell reaction is –96500 × E0

cell per mole of Fe(s),sfMlu lSy dks bl izdkj iznf'kZr fd;k x;k gS :Fe(s) | FeO(s) , 20% KOH(aq.) , Ni2O3(s), NiO(s) | Ni(s)bl lSy ls lEcfU/kr lgh lwpuk,sa gS@gSa-(A) ,suksM vfHkfØ;k Fe(s) + 2OH– (aq.) ® FeO(s) + H2O(l) + 2e– gS

(B) dSFkksM + vfHkfØ;k Ni2O3(s) + H2O(l) + 2e– ® 2NiO(s) + 2OH–(aq.) gS

(C) KOH foy;u dh lkUnzrk esa ifjorZu fd;s tkus ij Ecell ifjofrZr ugha gksxk

(D) lSy vfHkfØ;k ds fy;s DG0, –96500 × E0cell izfr eksy Fe(s) gS

2. The phase diagram of an one component system is shown below :Choose the correct option(s) from the following statements for triple point of the system(5.2 atm, 217 K)(A) All the three phases are in equilibrium(B) Molar Gibbs energy for the three phases is the same (C) Molar volume of the three phases is identical(D) Molar entropy of the three phases is the same

72.8

5.2

1.0

–195 217 T =304C

T(/K)

liquid solid

Gas

P(/atm)

S G�

S L�

L G�

,d ?kVd ra= dk izkoLFkk fp=.k uhpsa fn;k x;k gS :ra= ds f=d fcUnq (5.2 atm, 217 K) ds fy; lgh fodYiksa dk p;u dhft;s

(A) lHkh rhuksa izkoLFkk,sa lkE; esa gS (B) rhuksa izkoLFkk dh eksyj fxCl ÅtkZ leku gksrh gS

(C) rhuksa izkoLFkk dk eksyj vk;ru leku gksrk gS (D) rhuksa izkoLFkkvksa dh eksyj ,.VªkWih leku gksrh gS

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3. Which of the cation gives coloured ppt with K4[Fe(CN)6]?fuEu esa ls dkSu lk /kuk;u] K4[Fe(CN)6] ds lkFk jaxhu vo{ksi nsrk gS ?(A) Fe2+ (B) Cu2+ (C) Fe3+ (D) Ca2+

4. Which of the following compounds are associated with sp3-s-sp3 type of overlap?fuEu eas ls dkSuls ;kSfxd] sp3-s-sp3 izdkj ds vfrO;kiu ls lEcfU/kr gSa?(A) Al2(CH3)6 (B) Be2H4 (C) (BeH2)n (D) Al2H6

5. Which of the following order(s) is/are correct ?

fuEu esa ls dkSulk Øe lgh gS@gSa&

(A)CH3

NH2

>

NH2

(Basic strength / {kkjh; lkeF;Z)

(B) < (Water solubility / ty esa foys;rk)

(C) < (Heat of combustion / ngu dh Å"ek)

(D)>

(Resonance energy / vuqukn ÅtkZ)

6. Which of the following give pair of diastereomers on reaction with CH3MgBr followedby H2O :

fuEu esa ls dkSulk@dkSuls ;kSfxd dh CH3MgBr ds lkFk fØ;k ds i'pkr H2O ls vfHkfØ;k djkus ij foofjeleko;fo;ksa dk ; qXe izkIr gksrk gS&

(A)

O

(B) O

(C)

O

(D)

O

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LTS-18/38









Ligands has a role to make the compound overall symmetrical and unsymmetrical dependingupon the structure of it.fyxs.M] ;kSfxd dks iw.kZ :i ls lefer ;k vlefer cukus esa Hkwfedk vnk djrs gSa tks ;kSfxd dh lajpuk ij fuHkZjdjrk gS

7. Which of the following ligand itself is asymmetrical bidentate ligand?(A) tn (B) dmg– (C) acac– (D) None of thesefuEu eas ls dkSu lk fyxs.M] Lo;a vlefer f}nUrqd fyxs.M gS ?

(A) tn (B) dmg– (C) acac– (D) bueas ls dksbZ ugha

8. Which of the following bidentate ligand makes the given compound optically active?The compound is [MIII(AA)3]3–

(A) en (B) 22 2 2C S O - (C) tn (D) All of these

fuEu eas ls dkSu lk f}nUrqd fyxs.M fn;s x;s ;kSfxd dks izdkf'kd lfØ; cukrk gS ?;kSfxd [MIII(AA)3]3– gS

(A) en (B) 22 2 2C S O - (C) tn (D) mijksDr LkHkh

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LTS-19/38



Primary alkyl halide C4H9Br (P) reacted with alcoholic KOH to give compound (Q). Compound(Q) is reacted with HBr to give (R) which is an isomer of (P). (R) on reaction with waterproduces alcohol (S) with chiral centre.izkFkfed ,sfYdy gsykbM C4H9Br (P), ,sYdksgkWfy; KOH ds lkFk fØ;k djds ;kSfxd (Q) nsrk gS ;kSfxd (Q)HBr ds lkFk fØ;k djds (R) nsrk gS tks (P) dk ,d leko;oh gS (R), ty ds lkFk vfHkfØ;k djkus ij fdjsy

dsUæ ; qDr , sYdksgkWy (S) cukrk gSA

9. Compound (P) in above sequence is :

mijksDr Øe esa ;kSfxd (P) gS&

(A) Br (B) Br

(C)

Br

(D) Br

10. Compound (Q) on reaction with NBS produces (X) monobromo derivative then (X) is :;kSfxd (Q), NBS ds lkFk fØ;k djkus ij (X) eksuksczkseks O;qRiUu cukrk gS rks (X) gS&(A) 2 (B) 3 (C) 4 (D) 1

0000CJA103118017


LTS-20/38





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

••••••••••



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 lghgksxsa) 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 : +3 ;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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LTS-21/38


1. For the reaction : N2O4(g) �� 2NO2(g), DG0 = –1.223 kJ/mol at 40ºC. If the density of gaseous

mixture is 5.73 g/L at the same temperature, the average molar mass of the gaseous mixture

( in g/mol) is : (Given : 1223 = 8.314 × 313 × ln 85

æ öç ÷è ø

, 5.73 × 0.0821 × 313 = 1.6 × 92, 17 = 4.12)

vfHkfØ;k : N2O4(g) �� 2NO2(g) ds fy, 40ºC ij DG0 = –1.223 kJ/mol gSA leku rkiØe ij xSlh; feJ.k

dk ?kuRo 5.73 g/L gS rks xSlh; feJ.k dk vkSlr eksyj nzO;eku (g/mol esa) gS:

(fn;k gS : 1223 = 8.314 × 313 × ln 85

æ öç ÷è ø

, 5.73 × 0.0821 × 313 = 1.6 × 92, 17 = 4.12)

2. 0.05 moles of N2 gas dissolves completely in 900 gm water at 27ºC, as shown in figure.The available space between piston and liquid solution is 16.42 L. Moles of N2 in gaseous phase,above solution is (KH for N2 gas = 7.2 × 104 atm.) (R = 0.0821 atm-L/mole-K)

N (g)2

N + water2

0.05 eksy N2 xSl dks 27ºC ij 900gm ty esa iw.kZ :i ls foys; fd;k x;k gS

fiLVu rFkk nzOk foy;u ds e/; miyC/k LFkku 16.42 L gSA foy;u ds Åij xSlh; izkoLFkk esa N2 ds eksy gS&

(N2 xSl ds fy;s KH = 7.2 × 104 atm.) (R = 0.0821 atm-L/mole-K)

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3. For a reversible reaction, the rate of forward reaction is 20 times the rate of backward reactionat 27ºC. DrG at this stage of reaction ( in kcal /mol) is (ln10 = 2.3 , ln2 = 0.7),d mRØe.kh; vfHkfØ;k ds fy;s vxz vfHkfØ;k dh nj 27ºC ij i'p~ vfHkfØ;k dh nj dh 20 xquk gSA vfHkfØ;k

dh bl voLFkk ij DrG ( kcal /mol esa) gSA (ln10 = 2.3 , ln2 = 0.7)4. An imaginary AAAA..........type crystal is formed in 3D using ABAB.......type 2D packing

(hexagonal) of identical spheres of radius R. If 'r' is the radius of largest sphere which may

be fitted in the voids of 3D crystal formed, the value of rR

æ öç ÷è ø

is [Given : 21 = 4.59]

R f=T;k ds leku xksyksa ls 2D ladqyu] ABAB ..........izdkj dk ("kVQydh;) iz;ksx djrs gq, 3D esa AAAA.......izdkj

dk ,d dkYifud fØLVy cuk;k x;k gSA ;fn 'r' lcls cM+s xksys dh f=T;k gS tks fufeZr 3D fØLVy dh fjfDr;ksa

esa fQV gks ldsA rR

æ öç ÷è ø

dk eku gS [fn;k gS : 21 = 4.59]

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5. How many of the following processes are involved in the extraction of pure lead from PbShaving high impurity content.Calcination, Froth flotation, Roasting, Self reduction, Carbon reduction, Cupellation,Electrolytic refiningfuEu esa ls fdrus izØe] v'kqf¼ dh vf/kd ek=k j[kus okys PbS ls 'kq¼ ySM ds fu"d"kZ.k esa lfEefyr gksrs gSa\

fuLrkiu, >kx Iyou, HktZu, Lo vip;u, dkcZu vip;u, D;qisyhdj.k, oS|qr vi?kVuh; ifj'kks/ku

6. In H2S5O6 molecule, find the number of atoms having only bent shape geometry around it.H2S5O6 v.kq esa] , sls ijek.kqvksa dh la[;k crkbZ;s tks vius pkjksa vksj dsoy ca sV (bent) vkÏfr dh T;kfefr

j[krs gSa

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LTS-24/38

7. 2-Butanol H+

ProductsIf all products formed in the above reaction undergo reaction with Br2(CCl4) individuallythen number of products formed will be :

2-C;wVsukWy H+

mRikn

;fn mijksDr vfHkfØ;k esa fufeZr LkHkh mRiknksa dh i`Fkd&i`Fkd Br2(CCl4) ds lkFk vfHkfØ;k djk;h tkrh gS rks fufeZrmRiknksa dh la[;k gksxh&

8.

CH3

OH

H

Et Kmetal X CH CH –Br3 2

Y

X'TsClKCN

(DMF)

Y'

Number of steps in the reaction sequence giving retention of configurationCH3

OH

H

Et Kmetal X CH CH –Br3 2

Y

X'TsClKCN

(DMF)

Y'

mijksDr vfHkfØ;k Øe esa , sls inksa dh la[;k ftuesa] foU;kl dk vf/k/kkj.k gksrk gS


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PART-3 : MATHEMATICS Hkkx-3 : xf.krSECTION–I(i) : (Maximum Marks : 24)

[k.M–I(i) : (vf/kdre vad : 24)� This section contains SIX questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of









pqus gq , nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj


� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –2 vad feysaxs] pkgs lgh fodYi (fodYiks a) dks pquk x;k gks ;k u pquk x;k gksA

0000CJA103118017


LTS-26/38


1. Let Z and R be set of complex numbers and real numbers respectively,A = {x : x5 – 1 = 0, x Î Z} B = {x : 0 < Re(x) < 1 and x Î A}, P(x) = x + x4, x Î Z,Q(x) = x2 + x3, x Î Z, S(x)= x21 + x14 + x9 + x + 1, x Î Z and Re(x) denotes real part of x, x Î Z.Then, which of the following option(s) is/are correct ?(A) P(a) is real number, a Î B(B) {P(a) + Q(a)} Ì A, a Î B(C) |P(a)||Q(a)|Î A, a Î B(D) Locus of point x = S(a) + ib, x Î Z, a Î B, b Î R is a straight line.ekuk lfEeJ la[;kvksa rFkk okLrfod la[;kvksa ds leqPp; Øe'k% Z rFkk R gS,A = {x : x5 – 1 = 0, x Î Z} B = {x : 0 < Re(x) < 1 rFkk x Î A}, P(x) = x + x4, x Î Z,Q(x) = x2 + x3, x Î Z, S(x)= x21 + x14 + x9 + x + 1, x Î Z rFkk Re(x), x (x Î Z) ds okLrfod Hkkx dks n'kkZrkgSA rc fuEu esa ls dkSulk@dkSuls fodYi lgh gksxk@gksaxs\

(A) P(a) okLrfod la[;k gS, a Î B(B) {P(a) + Q(a)} Ì A, a Î B(C) |P(a)||Q(a)|Î A, a Î B(D) fcUnq x = S(a) + ib, x Î Z, a Î B, b Î R dk fcUnqiFk ,d ljy js[kk gS

2. In acute DABC, length of side BC is 5, inradius and circumradius of DABC are

( ) ( )+ -5 1 2 3 22 and 5 respectively. Then, which of the following option(s) is/are correct ?

(A) Length of internal angle bisector through A is ( )+5 2 32

(B) Area of DABC is ( )+25 2 32

(C) Length of altitude through vertex A is ( )+5 2 3(D) Perimeter of DABC greater than 20U;wudks.k f=Hkqt ABC esa, Hkqtk BC dh yEckbZ 5 bdkbZ gS rFkk f=Hkqt ABC dh vUr% f=T;k rFkk ifjf=T;k Øe'k%

( ) ( )+ -5 1 2 3 22 rFkk 5 gSA rc fuEu esa ls dkSulk@dkSuls fodYi lgh gksxk@gksaxs\

(A) fcUnq A ls xqtjus okys vUr%dks.k v¼Zd dh yEckbZ ( )5 2 32

+ gksxh

(B) f=Hkqt ABC dk {ks=Qy ( )+25 2 32 gksxk

(C) 'kh"kZ A ls xqtjus okys 'kh"kZ yEc dh yEckbZ ( )+5 2 3 gksxh

(D) f=Hkqt ABC dk ifjeki 20 ls vf/kd gksxk

0000CJA103118017


LTS-27/38


3. Let P1 : 2x + 2y – z = 3 and P2 : x + 2y + 2z = 2 be two planes. Then, which of the followingstatement(s) is(are) true ?(A) The line of intersection of P1 and P2 has direction ratios 6, –5, 2

(B) The acute angle between plane P1 and P2 is - æ öç ÷è ø

1 4cos9

(C) The acute angle between plane P1 and P2 is - æ öç ÷ç ÷è ø

1 5sin3

(D) Equation of plane passing through (1,0,0) and containing the line of intersection of planeP1 and P2 is x + y – 3z – 1 = 0

ekuk nks lery P1 : 2x + 2y – z = 3 rFkk P2 : x + 2y + 2z = 2 gSA rc fuEu esa ls dkSulk@dkSuls dFku lR;gksxk@gksaxs\

(A) lery P1 rFkk P2 dh izfrPNsfnr js[kk ds fnd~ vuqikr 6, –5, 2 gksaxs

(B) lery P1 rFkk P2 ds e/; U;wudks.k - æ öç ÷è ø

1 4cos9

gksxk

(C) lery P1 rFkk P2 ds e/; U;wudks.k - æ öç ÷ç ÷è ø

1 5sin3

gksxk

(D) ml lery dk lehdj.k] tks fcUnq (1,0,0) ls xqtjrk gS rFkk lery P1 rFkk P2 dh izfrPNsfnr js[kk ij fLFkr gS]x + y – 3z – 1 = 0 gksxk

0000CJA103118017


LTS-28/38


4. If ƒ : R – {0} ® [–1,1] is given by ( ) p=ƒ x cos

x. Then, which of the following statement(s) is(are)

correct ?

(A) ƒ(x) increases in the interval æ öç ÷+è ø

1 1,2k 1 2k for all non zero integral values of k.

(B) ƒ(x) increases in the interval æ öç ÷+ +è ø2 2

1 1,2k 3 2k 2 for all integral values of k.

(C) ƒ(x) decreases in the interval æ öç ÷+ +è ø2 2

1 1,2k 4 2k 3 for all integral values of k

(D) ƒ(x) decreases in the interval æ öç ÷+è ø

1 1,2k 2 2k for all non zero integral values of k

;fn ƒ : R – {0} ® [–1,1] gS] ( ) p=ƒ x cos

x }kjk fn;k gSA rc fuEu esa ls dkSulk@dkSuls dFku lgh gksxk@gksaxs\

(A) k ds lHkh v'kwU; iw.kk±d ekuksa ds fy, vUrjky æ öç ÷+è ø

1 1,2k 1 2k esa ƒ(x) o/kZeku gksxkA

(B) k ds lHkh iw.kk±d ekuksa ds fy, vUrjky æ öç ÷+ +è ø2 2

1 1,2k 3 2k 2 esa ƒ(x) o/kZeku gksxkA

(C) k ds lHkh iw.kk±d ekuksa ds fy, vUrjky æ öç ÷+ +è ø2 2

1 1,2k 4 2k 3 esa ƒ(x) Îkleku gksxkA

(D) k ds lHkh v'kwU; iw.kk±d ekuksa ds fy, vUrjky æ öç ÷+è ø

1 1,2k 2 2k esa ƒ(x) Îkleku gksxkA

0000CJA103118017


LTS-29/38


5. The slope of normal at any point(x,y) of a curve y = ƒ(x), is given by -+ +2 22xy

x y 1 and curvepasses through (1,0). Then, which of the following point(s) can lie on the curve y = ƒ(x) ?

oØ y = ƒ(x) ds fdlh fcUnq (x,y) ij vfHkyEc dh izo.krk, -+ +2 22xy

x y 1 }kjk nh xbZ gS rFkk oØ fcUnq (1,0) ls xqtjrk

gSA rc fuEu esa ls dkSuls fcUnq oØ y = ƒ(x) ij fLFkr gks ldrs gS\

(A) ( )3,2 2 (B) ( )5,3 2

(C) ( )3 2,5 (D) ( )4,3 2

6. Let ƒ : [1,¥) ® (–¥,–1] be a continuous and differentiable function such that

( ) ( )= - + ò

x2

1

ƒ tƒ x x 2x dt

x for all x Î [1,¥). Then, which of the following statement(s) is(are)

True ?(A) The curve y = ƒ(x) passes through the point (3,–24)

(B) Function ƒ(x) is an onto function

(C) Function ƒ(x) is an into function

(D) The area of the region {(x,y) Î [1,2] × R : ƒ(x) < y < – x} is 72

ekuk ƒ : [1,¥) ® (–¥,–1] ,d larr~ rFkk vodyuh; Qyu bl izdkj gS fd ( ) ( )= - + ò

x2

1

ƒ tƒ x x 2x dt

x " x Î

[1,¥) gSA rc fuEu esa ls dkSulk@dkSuls dFku lR; gksxk@gksaxs\

(A) oØ y = ƒ(x) fcUnq (3,–24) ls xqtjrk gS

(B) Qyu ƒ(x) vkPNknd Qyu gS

(C) Qyu ƒ(x) vUr%{ksih Qyu gS

(D) {ks= {(x,y) Î [1,2] × R : ƒ(x) < y < – x} dk {ks=Qy 72 gksxk

0000CJA103118017


LTS-30/38









Let S be the circle in the x-y plane defined by the equation x2 + y2 = 25.Consider line L : 3x + 4y – 5 = 0 intersects the circle S at points A and B.ekuk lehdj.k x2 + y2 = 25 }kjk ifjHkkf"kr x-y lery esa o`Ùk S gSA

ekuk js[kk L : 3x + 4y – 5 = 0, o`Ùk S dks fcUnq A rFkk B ij izfrPNsfnr djrh gSA7. The point of intersection of tangents to the circle S at point A and B is-

o`Ùk S ds fcUnq A rFkk B [khaph xbZ Li'kZ js[kkvksa dk izfrPNsn fcUnq gksxk –(A) (6,8) (B) (12,16) (C) (18,24) (D) (15,20)

8. Let P be the parabola in the x-y plane which passes through points A,B and (0,0) is vertex ofparabola P and line L is double ordinate (Line perpendicular to axis of parabola) of parabolaP, then length of latus rectum of parabola P is-ekuk x-y lery esa ijoy; P gS tks fcUnq A, B ls xqtjrk gS rFkk ijoy; P dk 'kh"kZ (0,0) gS rFkk js[kk L, ijoy; P dhf}dksfV gS (ijoy; ds v{k ds yEcor js[kk gS), rks ijoy; P ds ukfHkyEc dh yEckbZ gksxh –(A) 4 (B) 6 (C) 20 (D) 24

0000CJA103118017


LTS-31/38



An urn contains 4 red and 3 green balls. A ball is drawn at random from the urn. If the drawnball is green, then a red ball is added to the urn and if the drawn ball is red, then a greenball is added to the urn, the original ball is not returned to the urn and this process repeatsthree times.,d ik= esa 4 yky rFkk 3 gjh xsansa gSA ik= esa ls ;kn`PN;k ,d xsan fudkyh tkrh gSA ;fn fudkyh xbZ xsan gjh gS]

rks ik= esa ,d yky xsan feyk nh tkrh gS rFkk ;fn fudkyh xbZ xsan yky gS] rks ik= esa gjh xsan feyk nh tkrh gS]

okLrfod xsan dks ik= esa ugha feyk;k tkrk gS rFkk ;g izfØ;k rhu ckj nksgjk;h tkrh gSA9. The probability that urn contains 5 red and 2 green balls in the end, is-

vUr esa] ik= esa 5 yky rFkk 2 gjh xsansa gksus dh izkf;drk gksxh –

(A) 144343 (B)

108343 (C)

135343 (D)

129343

10. The probability that urn contains 5 red and 2 green balls in end given that first drawn ball isgreen, is-

vUr esa] ik= esa 5 yky rFkk 2 gjh xsanas gksus dh izkf;drk] fn;k x;k gS fd fudkyh xbZ xsan gjh gks] gksxh –

(A) 2749 (B)

8149 (C)

129143 (D) None of these

0000CJA103118017


LTS-32/38





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

••••••••••



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 lghgksxsa) 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 : +3 ;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

0000CJA103118017


LTS-33/38


1. If 2

xlog x

log 24 1 14 1 .... 54 18 6 2 ...3 3 9

é ùæ ö- - + = + + + +é ùê úç ÷ ë ûè øë û

, then sum of all possible value(s) of x is

;fn 2

xlog x

log 24 1 14 1 .... 54 18 6 2 ...3 3 9

é ùæ ö- - + = + + + +é ùê úç ÷ ë ûè øë û

gks] rks x ds lEHko ekuksa dk ;ksxQy gksxk

2. The number of 5 digit numbers which are divisible by 4 and sum of digits odd, with digits fromthe set {1,2,3,4,5,6} and repetition of digits is not allowed, is

5 vadksa okyh la[;kvksa dh la[;k] tks 4 ls foHkkftr gks rFkk ftuds vadksa dk ;ksxQy fo"ke gks] tks leqPp; {1,2,3,4,5,6}

esa ls gS] rFkk vadksa dh iqujko`fÙk u gks] gksxh

0000CJA103118017


LTS-34/38

3. Consider two sequence <an> such that a1 = 2 and ar+1 = 2ar"rÎN and <bn> such that

( ) ( )=

+ += " Îå

n

rr 1

n n 1 2n 1b n N

6, then number of possible value(s) of n for which an = bn, is-

vuqØe] <an> bl izdkj gS fd a1 = 2 rFkk a r+1 = 2ar"rÎN rFkk <bn> bl izdkj g S fd

( ) ( )=

+ += " Îå

n

rr 1

n n 1 2n 1b n N

6 gks] rks n ds lEHko ekuksa dh la[;k] ftlds fy, an = bn gks] gksxh –

4. If ( )( )( )- - += " Î

+

21 1

2x asin cot sin cot x , x Rx b

, then value of ab is

;fn ( )( )( )- - += " Î

+

21 1

2x asin cot sin cot x , x Rx b

gks] rks ab dk eku gksxk


0000CJA103118017


LTS-35/38

5. Let ƒ : (0,¥) ® [1, ¥), F : (0,¥) ® [0, ¥) and ( ) ( )= òx

0

F x ƒ t dt . If F(x2) = x(1 + x), then ƒ(4) is

ekuk ƒ : (0,¥) ® [1, ¥), F : (0,¥) ® [0, ¥) rFkk ( ) ( )= òx

0

F x ƒ t dt gSA ;fn F(x2) = x(1 + x) gks] rks ƒ(4)

dk eku gksxk

6. The vector = a + b + gr ˆˆ ˆP i j k , a ¹ 0, lies in the plane of the vector = +

r ˆ ˆQ i j and = +r ˆˆR i k and bisects

the angle between r

Q and r

R . Then the value of b + g

a3 4

2 is

lfn'k = a + b + gr ˆˆ ˆP i j k , a ¹ 0, lfn'k = +

r ˆ ˆQ i j rFkk = +r ˆˆR i k ds lery esa fLFkr gS rFkk

r

Q ,oa r

R ds eè;

dks.k lef}Hkkft gSA rc b + g

a3 4

2 dk eku gksxk


0000CJA103118017


LTS-36/38

7. Let a,b,g be three non zero real numbers such that the equationp pé ùa + b = g Î -ê úë û

3 cos x 2 sin x ,x ,2 2 has two distinct real roots a and b with

p+ =a b

3 . If range

of values of g

a + b2

3 2 is [q,r], then value of q + r is

ekuk a,b,g rhu v'kwU; okLrfod la[;k;sa bl izdkj gS fd lehdj.k

p pé ùa + b = g Î -ê úë û3 cos x 2 sin x ,x ,

2 2 ds nks fHkUu okLrfod ewy a rFkk b gS] ftlesa p

+ =a b3 gSA ;fn

ga + b2

3 2ds ekuksa dk ifjlj [q,r] gks] rks q + r dk eku gksxk

8. The number of 2 × 2 matrices A, whose entries are either 0 or 1 and for which the systemé ù é ù

=ê ú ê úë û ë û

x 1A

y 0 has exactly one solution, is

2 × 2 ds vkO;wgksa A dh la[;k ftudh izfof"V;ka 0 ;k 1 gS rFkk ftlds fy, fudk; é ù é ù

=ê ú ê úë û ë û

x 1Ay 0

ds Bhd ,d gy

gks] gksxk



LTS-37/380000CJA103118017


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I have verified the identity, name and Formnumber of the candidate, and that questionpaper and ORS codes are the same.eSaus ijh{kkFkhZ dk ifjp;] uke vkSj QkWeZ uEcj dks iwjh rjgtk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks a leku gSaA

____________________________Signature of the Invigilator / fujh{kd ds gLrk{kj

NAME OF THE CANDIDATE / ijh{kkFkhZ dk uke .............................................................................FORM NO. / QkWeZ uEcj .............................................

Your Target is to secure Good Rank in JEE 2019 0000CJA103118017LTS-38/38


Que. No. Category-wise Marks for Each Question / oxkZuqlkj izR;sd iz'u ds vad MaximumSection Type of Full Marks Partial Marks Zero Marks Negative Marks of the

Que. iw.kZ vad vkaf'kd vad 'kwU; vad Marks section[k.M iz'u dk iz'uksa ½.k vad [k.M esa

izdkj dh vf/kdÙke vadla[;k

+4 +1 0 –2One or more If only the bubble(s) For darkening a bubble If none In all

correct corresponding corresponding to each of the otheroption(s) to all the correct correct option, provide dbubbles is cases

I(i) ,dy ;k ,d ls 6 option(s) is(are) NO incorrect option darkened vU; lHkh 24vf/kd lgh darkened darkened ;fn fdlh Hkh ifjfLFkfr;ksa

fodYi ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i cqycqys dks esa(fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] dkyk ughacqycqys (cqycqyksa) dks ;fn dksbZ xyr fodYi dkyk fd;k gS

dkyk fd;k x;k gS ugha fd;k gSParagraph +3 0 –1

Based If only the bubble If none In all(Single corresponding to of the othercorrect the correct option bubbles is cases

I(ii) option) 4 is darkened — darkened vU; lHkh 12vuqPNsn ij ;fn flQZ lgh fodYi ds ;fn fdlh Hkh ifjfLFkfr;ksavk/kkfjr vuq:Ik cqycqys dks cqycqys dks dkyk esa

(,dy lgh dkyk fd;k gS ugha fd;k gSfodYi)

+3 0Numerical If only the bubble In allValue Type corresponding other

(Up to second to correct answer casesdecimal place) is darkened vU; lHkh

II la[;kRed eku 8 ;fn flQZ lgh mÙkj ds — ifjfLFkfr;ksa — 24izdkj vuq:Ik cqycqys dks esa

(n'keyo ds nks dkyk fd;k gSLFkku rd)

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TARGET : JEE (Main + Advanced) 2019 · Do not write any of these details anywhere else on the ORS. Darke th pr bble unde numb. viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa

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