(0999dja110319008) )) test pattern distance learning ... · vks- vkj- ,l- esa gsj&qsj@foÏfr u...

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 46 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 46 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 djsaA9. 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- esa 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 djsaA

12. 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 ;kstuk %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 three sections as detailed in table page 45.

izR;sd Hkkx esa rhu [k.M gSa ftudk fooj.k i`"B 45 ij rkfydk esa fn;k x;k gSA

Please see the page 45 of this booklet for rest of the instructions / Ïi;k 'ks"k funs Z'kk sa ds fy, bl iqfLrdk ds i`"B 45 dks i<+ s aA

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Time : 3 Hours Maximum Marks : 222

)0999DJA110319008)(0999DJA110319008) Test Pattern

DISTANCE LEARNING PROGRAMME(Academic Session : 2019 - 2020)

JEE(Advanced)MINOR TEST # 07

29-09-2019

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

Test Type : Unit Test # 0612th Undergoing/Pass Students

LTS-2/46

Target : JEE (Main + Advanced) 2020/29-09-2019

0999DJA110319008

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 = 58Atomic 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

1

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

SYLLABUS :PHYSICS : Rotational Motion, Alternating current (AC)

CHEMISTRY : Ionic Equilibrium, Acid Base theory, Nuclear chemistry, Reaction intermediate.

MATHEMATICS : Sequences and Series, Application of Derivatives (Maxima & Minima, Monotonicity,Tangent & Normal)

ALLEN

0999DJA110319008

Leader Test Series/Joint Package Course/29-09-2019

LTS-3/46

ALLEN

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 : 12)[k.M–I(i) : (vf/kdre vad : 12)

� This section contains FOUR 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 option is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

� bl [k.M esa pkj 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 : +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 gSA

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

1. In L-R series circuit, source voltage is 10V and voltage across R is 8V. Phase difference (f)between applied voltage and current is:,d L-R Js.kh ifjiFk esa L=ksr oksYVrk 10V gS rFkk R ds fljksa ij oksYVrk 8V gSA vkjksfir oksYVrk rFkk /kkjk ds e/;

dykUrj (f) gksxk :

(A) 1 3tan4

- æ öf = ç ÷è ø

(B) 1 1tan4


(C) 1 4tan1


(D) 1 4tan3



0999DJA110319008


LTS-4/46

ALLEN2. The circuit shown in the figure is in a steady state with the switch open. When the switch

is closed, which of the following will change immediately? (All of them eventually change,but three of them stay the same for an instant.)(A) The potential difference across the capacitor C(B) The current through the inductor L(C) The potential difference around the resistor R1(D) The current through the resistor R2

fp= esa fn;k x;k ifjiFk daqth dh [kqyh voLFkk esa LFkk;h voLFkk (steady state) esa gSA tc daqth dks cUn fd;k tkrk

gS rks fuEu esa ls dkSu rqjUr ifjofrZr gksxk \ (buesa ls lHkh ifjofrZr gks ldrs gS] ijUrq muesa rhu ,d {k.k ds fy;s leku

jg ldrs gS)(A) la?kkfj= C ds fljksa ij vkjksfir foHkokUrj (B) izsjd dq.Myh L esa izokfgr /kkjk

(C) izfrjks/kd R1 ij foHkokUrj (D) izfrjks/k R2 esa izokfgr /kkjk


0999DJA110319008


LTS-5/46

ALLEN3. Two rods of mass M and length L each are rigidly joined at point O to form a L-Shaped

composite rod. Assembly can turn about a frictionless hinge at O. An impulse acts at lowerend of rod. What is angular velocity immediately after impulse acts. Entire assembly lies inhorizontal plane.O fcUnq ij n`<+ :i ls tqM+h M nzO;eku ,oa L yEckbZ dh nks NM+ksa }kjk L-vkdkj dh la;qä NM+ cuk;h tkrh gSA

fudk; O ij ?k"kZ.kjfgr dhyd (hinge) ds lkis{k ?kwe ldrk gSA NM+ ds fupys fljs ij vkosx vkjksfir fd;k tkrk gSA

vkosx vkjksfir djus ds rqjUr ckn dks.kh; osx D;k gksxk\ lEiw.kZ fudk; {kSfrt ry esa fLFkr gSA

Hinge

J

O

(A) ML3J2

(B) ML2J3

(C) ML4J3

(D) MLJ3

4. A hollow thin walled pipe is projected on a rough horizontal surface with speed v. As soon asthe pipe begins to roll, what will be its speed?,d iryh nhokj ds [kks[kys ikbi dks [kqjnjh {kSfrt lrg ij v pky ls Qsadk tkrk gSA tSls gh ikbi yq<+duk izkjEHk djrk

gS bldh pky D;k gksxh \(A) (1/4) × v (B) (1/3) × v (C) (1/2) × v (D) v


0999DJA110319008


LTS-6/46

ALLENSECTION–I(ii) : (Maximum Marks : 16)

[k.M–I(ii) : (vf/kdre vad : 16)� This section contains TWO paragraphs.� Based on each paragraph, there will be TWO 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 [k.M esa nks vuqPNsn gSa

� izR;sd vuqPNsn ij nks iz'u fn, x;sa 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 feysax s 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 (fodYiksa ) dks pquk x;k gks ;k u pquk x;k gksA

0999DJA110319008


LTS-7/46

ALLENParagraph for Questions 5 and 6

iz'u 5 ,oa 6 ds fy;s vuqPNsnTwo uncharged identical capacitors A and B, each of capacitance C, and an inductor ofinductance L are arranged as shown in the adjacent figure. At t = 0, the switch S1 is closed

while switch S2 remains open. At time t = t0 = LC2p

, switch S2 is closed while switch S1 isopenednks vuvkosf'kr leku la/kkfj= A ,oa B, izR;sd dh /kkfjrk C gS ,oa ,d iszjd dq.Myh ftldk iszjdRo L gS fp=kuqlkj

O;ofLFkr gS t = 0 ij fLop S1 cUn gS tcfd S2 [kqyk jgrk gS le; t = t0 = LC2p

ij fLop dks S2 cUn fd;k tkrk

gS tcfd fLop S1 [kqyk gSA

A B

LE

S2S1

5. The charge on capacitor A at that time is :la/kkfj= A ij ml le; vkos'k gS :

(A) 2CE (B) 2CE

(C) CE (D) 4CE

6. The current flowing through the inductor at that time is :izsjd dq.Myh ls ml le; izokfgr /kkjk gS :

(A) LC1CE2 (B) 0 (C) LC

12

CE´ (D)

LC1CE ´


0999DJA110319008


LTS-8/46


iz'u 7 ,oa 8 ds fy;s vuqPNsnAs shown below, a uniform disc is mounted to an axle and is free to rotate without friction. Athin uniform rod is rigidly attached to the disc so that it will rotate with the disc. A block isattached to the end of the rod. Properties of the disc, rod, and block are as follows.fp=kuqlkj ,d le:i pdrh fdlh ,Dlsy ij yxh gqbZ gS rFkk ;g fcuk ?k"kZ.k ?kw.kZu ds fy, Lora= gSA ,d iryh le:iNM+ dks pdrh ds lkFk n`<+rk iwoZd bl izdkj tksM+ fn;k tkrk gS rkfd ;g pdrh ds lkFk ?kw.kZu dj ldsA NM+ ds fljs ls,d CykWd tksM+ nsrs gSA pdrh] NM+ rFkk CykWd ds xq.k fuEu izdkj gSA

String

Block

Rod

Disc

q0

• Disc : mass = 3m, radius = R, moment of inertia about center 2D

3I mR

2=

pdrh % nzO;eku = 3m, f=T;k = R, dsUnz ds lkis{k tM+Ro vk?kw.kZ 2D

3I mR

2=

• Rod : mass = m, length = 2R, moment of inertia about one end 2R

4I mR

3=

NM + % nzO;eku = m, yEckbZ = 2R, ,d fljs ds lkis{k tM+Ro vk?kw.kZ 2R

4I mR

3=

• Block : mass = 2mCykWd : nzO;eku = 2m

The system is held in equilibrium with the rod at an angle q0 to the vertical, as shown above,by a horizontal string of negligible mass with one end attached to the disc and the other to awall. The arrow represents the direction of vector quantities.fp=kuqlkj NM+ Å/okZ/kj ds lkFk q0 dks.k ij gS rFkk ux.; nzO;eku dh {kSfrt jLlh dk ,d fljk pdrh ls tqM+k gS ,oa nwljkfljk nhokj ls tqM+k gqvk gSA bl izdkj bl fudk; dks lkE;koLFkk esa j[kk x;k gS rFkk ;gk¡ rhj }kjk lfn'k jkf'k;ksa dh fn'kkdks iznf'kZr fd;k x;k gSA

7. Choose the CORRECT statement(s) :(A) The tension in the string is 5mgsinq0(B) Moment of inertia of system about an axis passing through centre of disc and perpendicular

to it is 265mR

6(C) Horizontal component of reaction from support is 5mgsinq0(D) Vertical component of reaction from support is 6 mg.lgh dFku@dFkuksa dks pqfu;s :(A) jLlh esa ruko dk eku 5mgsinq0 gSA

(B) pdrh ds dsaæ ls xqtjus okyh rFkk blds yEcor~ v{k ds lkis{k fudk; dk tM+Ro vk?kw.kZ 265mR

6 gSA

(C) liksVZ }kjk yxk;h x;h çfrfØ;k dk {kSfrt ?kVd 5mgsinq0 gSA(D) liksVZ }kjk yxkbZ x;h çfrfØ;k dk Å/okZ/kj ?kVd 6 mg gSA

0999DJA110319008


LTS-9/46

ALLEN8. The string is now cut, and the disc-rod-block system is free to rotate. As the disc rotates, the

rod passes the horizontal position shown below. Mark the correct statement(s) :vc bl jLlh dks dkV fn;k tkrk gS ,oa pdrh&NM+&CykWd fudk; ?kw.kZu ds fy, Lora= gSA pdrh ds ?kw.kZu djus ij ;g

NM+ fp= esa iznf'kZr {kSfrt fLFkfr ls xqtjrh gSA lgh dFku@dFkuksa dks pqfu;s %&

(A) Angular acceleration of the mass at the end of the rod, for the instant the rod is in the

horizontal position 6g

13R(B) Angular acceleration of the mass at the end of the rod, for the instant the rod is in the

horizontal position 8g

13R(C) The linear speed of the mass at the end of the rod, for the instant the rod is in the

horizontal position is 0

48gR cos

13q ¯

(D) The linear speed of the mass at the end of the rod, for the instant the rod is in the

horizontal position is 0

48gR cos

13q ®

(A) ftl {k.k NM + {kSfrt fLFkfr esa gksrh gS] ml {k.k ds fy, NM + ds fljs ij tqM+s æO;eku dk dks.kh; Roj.k 6g

13R gksrk gSA

(B) ftl {k.k NM + {kSfrt fLFkfr esa gksrh gS] ml {k.k ds fy, NM + ds fljs ij tqM+s æO;eku dk dks.kh; Roj.k 8g

13R gksrk gSA

(C) ftl {k.k NM + {kSfrt fLFkfr esa gksrh gS] ml {k.k ds fy, NM+ ds fljs ij tqM+s nzO;eku dh jsf[kd pky 0

48gR cos

13q ¯

gksxhA

(D) ftl {k.k NM+ {kSfrt fLFkfr esa gksrh gS] ml {k.k ds fy, NM+ ds fljs ij tqM+s nzO;eku dh jsf[kd pky 0

48gR cos

13q ®

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

0999DJA110319008


LTS-10/46

ALLENSECTION–I(iii) : (Maximum Marks : 06)

[k.M–I(iii) : (vf/kdre vad : 06)� This section contains TWO questions.� Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D)

out of which ONLY ONE is correct� For each question, marks will be awarded in one of the following categories :


� bl [k.M esa nks iz'u gSa

� izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA






0999DJA110319008


LTS-11/46

ALLEN9. Figure shows particles each connected rigidly by massless thin rods. Mass of each particle

is m. In each case calculate moment of inertia of assembly about axis shown in figure.List-I List-II

(P) edge length of cube = L (1) 8 mL2

Axis passing through centre of upper and lower faces

(Q) edge length of cube = L (2) 4 mL2

Axis passing through edge of cube

(R) radius of cylinder = L (3) 3 mL2

Axis passing through symmetry axis of cylinder

(S) edge of tetrahedron = L (4) mL2

Axis passing through the centroid of baseCodes :(A) P ® 2; Q ® 4; R ® 1; S ® 3 (B) P ® 2; Q ® 3; R ® 1; S ® 4(C) P ® 2; Q ® 1; R ® 1; S ® 4 (D) P ® 2; Q ® 1; R ® 1; S ® 3


0999DJA110319008


LTS-12/46

ALLEN

fp= esa æO;ekughu iryh NM+ksa ls n`<+rkiwoZd tqM+s d.kksa dks n'kkZ;k x;k gSA izR;sd d.k dk æO;eku m gSA izR;sd izdj.k

esa fudk; dk iznf'kZr v{k ds lkis{k tM+Rovk?kw.kZ Kkr dhft,A

lwph-I lwph-II

(P) ?ku dh Hkqtk yEckbZ = L (1) 8 mL2

Åijh rFkk fupys Qydksa ds dsUæ ls xqtjus okyh v{k ds lkis{k

(Q) ?ku dh Hkqtk yEckbZ = L (2) 4 mL2

?ku dh Hkqtk ls xqtjus okyh v{k ds lkis{k

(R) csyu dh f=T;k = L (3) 3 mL2

csyu dh lefer v{k ls xqtjus okyh v{k ds lkis{k

(S) prq"Qyd dh Hkqtk = L (4) mL2

vk/kkj ds dsUæd ls xqtjus okyh v{k ds lkis{k

dwV :(A) P ® 2; Q ® 4; R ® 1; S ® 3 (B) P ® 2; Q ® 3; R ® 1; S ® 4(C) P ® 2; Q ® 1; R ® 1; S ® 4 (D) P ® 2; Q ® 1; R ® 1; S ® 3


0999DJA110319008


LTS-13/46

ALLEN10. List-I contains some circuits and List-II contains some properties. In List-II current signifies

instantaneous current and potential difference refers instantaneous potential difference.List-I List-IICharge System Properties

(P)~

a bSR

e w= sin( t)0e

x

(1) In branch x, current varies

A closely separated conducting wire iswrapped over an insulating cylinder.Wire is connected to a resistance Rand an alternating source via aswitch S, which is closed at t = 0

(Q)r a

R

b

S

x

A charged metallic sphere of radius (2) Between points a and b magnituder is earthed with a resistance R and of potential difference increasesa switch S, which is closed at t = 0. always.

(R)

x R

r

L

e

S

a b

An inductor L and resistor R is (3) In branch x, current is constantconnected to a cell of emf e andinternal resistance r with a switch(s),which is closed at t = 0

(S)

GX

s

R X

a

e

b

Q

J

A circuit (used to measure an (4) Between points a and b ofunknown resistance) is in balanced potential difference zero always.conditions (Galvanometer reads zero)

Code :(A) P ® 1; Q ® 1; R ® 1; S ® 3 (B) P ® 1; Q ® 2; R ® 3; S ® 4(C) P ® 2; Q ® 1; R ® 3; S ® 4 (D) P ® 2; Q ® 3; R ® 1; S ® 4

0999DJA110319008


LTS-14/46

ALLENlwph-I esa dqN ifjiFk rFkk lwph-II esa buls lEcfU/kr dqN fo'ks"krk;sa n'kkZ;h xbZ gSA lwph-II esa of.kZr /kkjk rFkk foHkokUrjØe'k% rkR{kf.kd /kkjk o rkR{kf.kd foHkokUrj dks n'kkZrs gSA

lwph-I lwph-IIvkos'k fudk; fo'ks"krk

(P)~

a bSR

e w= sin( t)0e

x

(1) 'kk[kk x esa /kkjk ifjofrZr gksrh gSA

,d pkyd rkj dks dqpkyd csyu ij ikl&iklyisVk tkrk gSA rkj dks izfrjks/k R rFkk ,dizR;korhZ L=ksr ls fLop S }kjk tksM+k tkrk gSftls t = 0 ij can dj fn;k tkrk gSA

(Q)r a

R

b

S

x

f=T;k r okys ,d vkosf'kr /kkfRod xksys dks (2) fcUnq a o b ds e/; foHkokUrj dkizfrjks/k R rFkk fLop S dh lgk;rk ls HkwlaifdZr ifjek.k lnSo c<+rk gSAfd;k tkrk gSA fLop dks t = 0 ij can dj nsrs gSA

(R)

x R

r

L

e

S

a b

izsjd dq.Myh L rFkk izfrjks/kd R dks fLop S dh (3) 'kk[kk x esa /kkjk fu;r gSAlgk;rk ls vkarfjd izfrjks/k r rFkk e fo|qr okgdcy okyh cSVjh ds lkFk tksM+ nsrs gSA fLop dkst = 0 ij can dj fn;k tkrk gSA

(S)

GX

s

R X

a

e

b

Q

J

vKkr izfrjks/k ds ekiu esa iz;qDr ifjiFk (4) fcUnq a o b ds e/; foHkokUrj lnSolarqfyr fLFkfr esa gSA (xsYosuksehVj 'kwU; 'kwU; gksrk gSAikB~;kad i<+rk gS)

dwV :(A) P ® 1; Q ® 1; R ® 1; S ® 3 (B) P ® 1; Q ® 2; R ® 3; S ® 4(C) P ® 2; Q ® 1; R ® 3; S ® 4 (D) P ® 2; Q ® 3; R ® 1; S ® 4

0999DJA110319008


LTS-15/46

ALLENSECTION-II : Numerical Value Type (Up to second decimal place)

No question will be asked in section IISECTION–III : (Maximum Marks : 24)

[k.M–III : (vf/kdre vad : 24)� This section contains SIX questions.� The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

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

Full Marks : +4 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 In all other cases.

� bl [k.M esa N% iz'u gSa� izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh; iw.kk ±d gSA� izR;sd iz'u esa] vks-vkj-,l- ij lgh iw.kk±d 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 mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. A bank of two series connected capacitors of capacitance C = 500 µF each is charged to avoltage 20 V and is connected to a coil of inductance L = 1000 H so that an oscillatory circuit

is formed at the initial moment . After a time 4p

sec, a breakdown occurs in one of the

capacitors, and the resistance between its plates becomes zero. The amplitude of chargeoscillations on the undamaged capacitor is a mC. Fill a in OMR sheet.Js.khØe esa tqM+s nks la/kkfj=ksa ftuesa izR;sd dh /kkfjrk C = 500 µF gS rFkk izR;sd dks 20 V oksYVrk rd vkosf'kr dj ,dL = 1000 H izsjdRo okyh dq.Myh ds lkFk tksM+ fn;k tkrk gS ftlds QyLo:i izkjfEHkd {k.k ij ,d nksyuh ifjiFk

dk fuekZ.k gksrk gSA 4p

le; i'pkr~ buesa ls fdlh ,d la/kkfj= dk Hkatu (breakdown) gks tkrk gS rFkk bldh IysVksa ds

eè; izfrjks/k 'kwU; gks tkrk gSA nwljs la/kkfj= (undamaged) ij vkos'k nksyuksa dk vk;ke ;fn a mC gks rks a dk ekuviuh mÙkjiqfLrdk esa Hkfj,sA

LC

C


0999DJA110319008


LTS-16/46

ALLEN2. An AC source of angular frequency w is fed across a resistor R and a capacitor C in series.

The current registered is I. If now the frequency of source is changed to w/3 (but maintainingthe same voltage), the current in the circuit is found to be halved. The ratio of reactance to

resistance at the original frequency w is K5

. Find the value of K.

dks.kh; vkofÙk w okys ,d AC L=ksr dks izfrjks/kd R o la/kkfj= C ds Js.khØe esa tksM+k tkrk gSA ntZ dh xbZ /kkjk I gSA;fn L=ksr dh vkofÙk dks w/3 dj fn;k tk, ijUrq oksYVrk ogh j[kh tk, rks ifjiFk esa /kkjk dk eku vk/kk gks tkrk gSA ewy

vkofÙk w ij izfr?kkr o izfrjks/k dk vuqikr K5

gks rks K dk eku Kkr djsaA

3. In the circuit shown in figure XCX R,L 2

= = the peak value current i0 through source is 0xV2R

.

Find the value of x.

fn, x, ifjiFk esa] XCX R,L 2

= = rFkk L=ksr ls fuxZr f'k[kj /kkjk i0 dk eku 0xV2R gS rks x dk eku Kkr dhft,A

R

C

L

V = V sin t0 w


0999DJA110319008


LTS-17/46

ALLEN4. A uniform thin rod of mass M and length L is supported horizontally by two supports, one at

each end. The acceleration of gravity g is constant and in the downward direction. Attime t = 0, the left support is removed. Normal force on the right support immediately after

the left support is removed is Mgx . Fill x in OMR sheet.

æO;eku M rFkk yEckbZ L okyh ,d le:i iryh NM+ ds izR;sd fljs ij ,d liksVZ yxkdj bls {kSfrt :i ls lgkjkiznku fd;k tkrk gSA ;gk¡ xq:Roh; Roj.k g fu;r rFkk uhps dh fn'kk esa dk;Zjr gSA le; t = 0 ij cka;s liksVZ dks gVk fn;k

tkrk gSA cka;s liksVZ dks gVkus ds Bhd i'pkr~ nka;s liksVZ ij vfHkyEc cy ;fn Mgx gks rks x dk eku Kkr dhft;sA

Mg5. A mass m is moving at speed v perpendicular to a rod of length d and mass M = 6m which

pivots around a frictionless axle running through its centre. It strikes and sticks to the end ofthe rod. The moment of inertia of the rod about its centre is Md2/12.Then the angular speed ofthe system just after the collision is found to be k times v/3d. Find the value of k.yEckbZ d rFkk æO;eku M = 6m okyh ,d NM + blds dsUæ ls gksdj xqtj jgs ,d ?k"kZ.kjfgr ,Dlsy ds lkis{k dhydhr

gSA ,d m æO;eku bl NM+ ds yEcor~ v pky ls xfr djrs gq;s NM+ ds fljs ls Vdjkdj mlls fpid tkrk gSA bl NM+

dk blds dsUnz ds lkis{k tM+Ro vk?kw.kZ Md2/12 gSA ;fn VDdj ds rqjUr i'pkr~ fudk; dh dks.kh; pky v/3d dh k xquk

gks rks k dk eku Kkr dhft,A

v

d

M


0999DJA110319008


LTS-18/46

ALLEN6. Mass of the pulley is m & radius is R. Assume pulley to be disc. Block of mass m is released

from the position shown. String is massless & inextensible. There is no slipping betweenrope and pulley. The impulse exerted by the string on the pulley at the moment string

becomes taut is J. 2m gRJ

is equal to.

fdlh f?kjuh dk æO;eku m rFkk f=T;k R gSA ekuk fd f?kjuh ,d pdrh gSA çnf'kZr fLFkfr ls æO;eku m dk CykWdiznf'kZr fLFkfr ls fojkekoLFkk ls NksM+k tkrk gSA jLlh æO;ekughu rFkk vforkU; gSA jLlh rFkk f?kjuh ds e/; fQlyu ugha

gSA jLlh ds rfur gksus ds {k.k ij jLlh }kjk f?kjuh ij yxk;k x;k vkosx J gS rks 2m gRJ

dk eku Kkr dhft,A

m

R

Rm


0999DJA110319008


LTS-19/46

ALLENSECTION–IV : (Maximum Marks : 16)

[k.M–IV : (vf/kdre vad : 16)� This section contains TWO questions.� Each question contains two columns, Column-I and Column-II.� Column-I has four entries (A), (B), (C) and (D)� Column-II has five entries (P), (Q), (R), (S) and (T)� Match the entries in Column-I with the entries in column-II.� One or more entries in Column-I may match with one or more entries in Column-II.� The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

� For each entry in column-I, darken the bubbles of all the matching entries. For example,if entry (A) in Column-I matches with entries (Q), (R) and (T), then darken these threebubbles in the ORS. Similarly, for entries (B), (C) and (D).

� For each question, marks will be awarded in one of the following categories :For each entry in Column-IFull Marks : +2 If only the bubble(s) corresponding to all the correct match(es) is (are)

darkenedZero Marks : 0 In all other cases

� bl [k.M esa nks iz'u gSa� izR;sd iz'u esa nks dkWye gSa] dkWye-I rFkk dkWye-II� dkWye-I esa pkj izfo"V;k¡ (A), (B), (C) rFkk (D) gSa� dkWye-II esa ik¡p izfof"V;k¡ (P), (Q), (R), (S) rFkk (T) gSa� dkWye-I dh izfof"V;ksa dk dkWye-II dh izfof"V;ksa ls lqesfyr djsa� dkWye-I dh ,d ;k ,d ls vf/kd izfof"V;k¡] dkWye-II dh ,d ;k ,d ls vf/kd izfof"V;ksa ls lqesfyr gks ldrh gSa� vks-vkj-,l- esa s uhps n'kkZ;h x;h tSlh 4 × 5 vkO;wg nh x;h gS :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

� dkWye-I dh izR;sd izfof"V ds fy, lHkh lqesfyr izfof"V;ks a ds cqycqyksa dks dkyk djsaA mnkgj.k Lo:i] ;fndkWye-I dh izfof"V (A) izfof"V;ksa (Q), (R) rFkk (T) ls lqesfyr gks] rks bu rhuksa cqycqyksa dks vks-vkj-,l- esa dkykdjsaA blh izdkj izfof"V;ksa (B), (C) rFkk (D) ds fy;s Hkh djsa

� vadu ;kstuk :dkWye-I dh izR;sd izfof"V ds fy,

iw.kZ vad : +2 ;fn flQZ lHkh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k tk;'kwU; vad : 0 vU; lHkh voLFkkvksa esa

0999DJA110319008


LTS-20/46

ALLEN1. Each of the four wheels in column I has an outer ring having radius R and mass m. Other than

the outer ring the wheels comprise of some uniform rods (each of mass m and length R) or

some lamina (having the same mass m). Column II gives radius of gyration about axis passingthrough centre of ring perpendicular to its plane.dkWye–I esa izR;sd pkj ifg;ksa dh cká oy; dh f=T;k R rFkk nzO;eku m gSA cká oy; ds vfrfjDr ifg;ksa eas dqN

,dleku NM+s (izR;sd dk nzO;eku m rFkk yEckbZ R) ;k dqN iryh ijr (leku nzO;eku m dh) lfEefyr gSA

dkWye–II esa blds ry ds yEcor~ oy; ds dsUnz ls gksdj xqtjus okyh v{k ds lkis{k ?kw.kZu f=T;k nh xbZ gSAColumn–I : Wheels Column–II : Radius of gyrationdkWye&I (ifg;s) dkWye&II (?kw.kZu f=T;k)

(A) (P) 3R

4

(B) (Q) 11R

18

(C) (R)R

2

One disc + ring / ,d pdrh + oy;

(D) (S) R 715

Outer ring + half disc + one rodcká oy;+v¼Zpdrh+ ,d NM+

(T)R

3


0999DJA110319008


LTS-21/46

ALLEN2. A stick with mass m and length 2R is pivoted at one end on a vertical wall. It is held horizontal,

and a disc with mass m and radius R is placed beneath it, in contact with both it and wall asshown in figure. The coefficient of friction between disc and wall is mw, between disc and stickis ms. System is released and it does not fall. Now match column I and II.æO;eku m rFkk yEckbZ 2R okyh ,d NM + dks Å/okZ/kj nhokj ij ,d fljs ls dhydhr fd;k x;k gSA bls {kSfrt fLFkfr

esa j[kk x;k gS rFkk æO;eku m o f=T;k R okyh ,d pdrh blds uhps bl izdkj j[kh x;h gS fd ;g NM + rFkk nhokj nksuksa

ds lkFk laidZ esa jgrh gS] fp= ns[ksaA pdrh o nhokj ds e/; ?k"kZ.k xq.kkad mw rFkk pdrh o NM+ ds e/; ?k"kZ.k xq.kkad ms

gSaA fudk; dks fojkekoLFkk ls NksM+k tkrk gS rFkk ;g fxjrk ugha gSA dkWye-I dk dkWye-II ls feyku dhft;sAms m

mwR

m

Column–I/dkWye–I Column–II/dkWye–II(A) (mS) min (P) 1

(mS) U;wure(B) (mw) min (Q) 2

(mw) U;wure(C) Vertical component of contact (R) 2 mg

force between stick and discNM+ o pdrh ds e/; laidZ cy dk

Å/okZ/kj ?kVd(D) Vertical component of contact (S) mg

force between wall and discnhokj o pdrh ds e/; laidZ cy dk (T) 5 mg

Å/okZ/kj ?kVd


0999DJA110319008


LTS-22/46

ALLEN

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

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

� This section contains FOUR questions.� Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options









½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA1. For the nuclear reaction ;

AZ

X A – 8 – 4Z

Y ; t1/2= 1600 yrs.

If initial activity was 107 dps, how many a-particles will be emitted per second after4800 years ?(A) 1.25 × 106 s–1 (B) 2.5 × 106 s–1 (C) 1.25 × 107 s–1 (D) NonefuEu ukfHkdh; vfHkfØ;k ;

AZ

X A – 8 – 4Z

Y ; t1/2= 1600 yrs.

ds fy, ;fn izkjafHkd lfØ;rk 107 dps Fkh rks 4800 o"kksZa ds ckn] fdrus a-d.k izfr lSd.M mRlftZr gks jgs gksaxs ?

(A) 1.25 × 106 s–1 (B) 2.5 × 106 s–1 (C) 1.25 × 107 s–1 (D) buesa ls dksbZ ugha2. Concentration of OH– ion in 0.1 M RNH3

+Cl– solution (Given : Kb(RNH2) = 10–9)0.1 M RNH3

+Cl– foy;u esa OH– vk;uksa dh lkUnzrk gS (fn;k gS : Kb(RNH2) = 10–9)(A) 10–3 (B) 10–11 (C) 10–5 (D) 10–9


0999DJA110319008


LTS-23/46

ALLEN3. Which has least pKa value ?

fdldk pKa eku U;wure gksrk gS&

(A) HCO2H (B) CH3CO2H (C) - -3 2

3

CH CH CO H|CH

(D) CO H2

4. Which is most basic in aqueous solution ?

fuEu esa ls dkSulk ;kSfxd tyh; foy;u esa lokZf/kd {kkjh; gS&

(A) CH3—NH2 (B) (CH3)2NH (C) (CH3)3 N (D) NH2


0999DJA110319008


LTS-24/46














� 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 feysax s 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 (fodYiksa ) dks pquk x;k gks ;k u pquk x;k gksA

0999DJA110319008


LTS-25/46


iz'u 5 ,oa 6 ds fy;s vuqPNsnRadioactive tracers provide a convenient mean of measuring the small concentrationencountered in determining ksp values. Exactly 30 ml of 0.001M AgNO3 solutioncontaining radioactive silver, with an activity of 40000 counts per ml was mixed with 10 ml of0.002 M KIO3 solution and the mixture is diluted to exactly 400 ml. After equilibrium wasreached, a portion of sample was filtered to remove any solid and was found to have anactivity of 2000 counts per ml.ksp ekuksa dks Kkr djus esa jsfM;kslfØ; Vªslj ,d mi;qDr fof/k gksrh gS tks U;wu lkUnzrk dh x.kuk }kjk dh tkrh gS tSls

30 ml 0.001M AgNO3 foy;u esa mifLFkr jsfM;kslfØ; flYoj ftldh lfØ;rk 40000 izfr ml gS] dks 10 ml]0.002 M KIO3 foy;u eas feyk;k x;k gS vkSj feJ.k dks 400 ml rd ruq fd;k x;k gSA lkE; izkIr gksus ds ckn uewus

ds ,d Hkkx dks Nkudj fdlh Hkh Bksl dks gVk;k x;k vkSj bldh lfØ;rk 2000 counts izfr ml ik;h x;hA5. The ksp of AgIO3 is -

AgIO3 dk ksp gS-(A) 6 × 10–10 (B) 1.25 × 10–9 (C) 6 × 10–12 (D) 4 × 10–10

6. Number of moles of solid AgIO3 formed -cuus okys Bksl AgIO3 ds eksyksa dh la[;k gS -(A) 10–5 (B) 10–4 (C) 10–2 (D) 0.4 × 10–3


0999DJA110319008


LTS-26/46


iz'u 7 ,oa 8 ds fy;s vuqPNsn

NaHCO3 when treated with acids it may evolve CO2. Acids which are stronger & comparableacidic with H2CO3 may evolve CO2.NaHCO3 dks tc vEyksa ds lkFk mipkfjr fd;k tkrk gS rks ;g CO2 mRl£tr dj ldrk gSA vEy tks H2CO3 ds

leku ;k blls vf/kd vEyh; gksrs gSa , CO2 mRlftZr dj ldrs gSaA7. Gas liberated during reaction.

vfHkfØ;k ds nkSjku mRl£tr xSl gS&

SO H3

COOH

NaHCO3

14

(Xs)

(A) CO2 (B) SO2 (C) 14

2CO (D) SO3

8. Compound which is soluble in NaOH & NaHCO3 both.;kSfxd tks NaOH rFkk NaHCO3 nksuksa esa foys; gS&

(A) (B) (C)

OHNO2

(D)


0999DJA110319008


LTS-27/46







lgh gSA




½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA9. Note : log2 = 0.3 : All data are at 25ºC

uksV : log2 = 0.3 : lHkh vkadMs 25ºC ij gS

List-I/lwph-I List-II/lwph-II(P) 1 M XOH(aq.) (1) [H+] > [OH–]

[(pKb (XOH) = 6)](Q) 0.1M C6H5NH2(aq.) (2) pH = 2.4

[pKb(C6H5NH2) = 7](R) At equivalence point when (3) pOH = 3

20 ml of 0.5 M BOH is titrated with 2M HCl[Kb(BOH) = 1/4 × 10–9M]tc 20 ml 0.5 M BOH , 2M HCl ds lkFk

ds lkFk vuqekfir fd;k tkrk gS] rks rqY;kad fcUnq ij

[Kb(BOH) = 1/4 × 10–9M](S) 20 ml 0.125M, C5H5NHCl + 5 ml, 5 × 10–3 M HCl (4) pH = 10

[Kb (C5H5N) = 5 × 10–10M]The CORRECT option islgh fodYi gS(A) P ® 3,4 ; Q ® 4 ; R ® 1,2 ; S ® 1 (B) P ® 1,4 ; Q ® 3 ; R ® 1,3 ; S ® 3(C) P ® 1,3 ; Q ® 2, 4 ; R ® 1,2 ; S ® 4 (D) P ® 3,4 ; Q ® 1 ; R ® 1,2 ; S ® 2


0999DJA110319008


LTS-28/46

ALLEN

10. List–I/lwph-I List–II/lwph-II

(P)CH=CH–CO H2

(1) Reacts with NaHCO3

NaHCO3 ds lkFk fØ;k djrk gSA

(Q)

CH=CH–CH3

HO(2) Soluble in NaOH

NaOH esa foys; gSA

(R) (3) Reacts with HBr to give product containing chiral centre

HBr ds lkFk fdjSy dsUnz j[kus okyk mRikn nsus ds fy;s fØ;k djrk gSA(S) CH = CH–CH–CO H2 2

CH3

(4) Decolorise Br2 solution.

Br2 foy;u dks jaxfgu djrk gSAThe CORRECT option islgh fodYi gS

(A) P® 1,2,3,4 ; Q®2, 3, 4 ; R ® 4 ; S ®1, 2, 3(B) P® 1,2,3,4 ; Q®2, 3, 4 ; R ® 4 ; S ®1, 2, 3, 4(C) P® 1,2,3,4 ; Q®2, 3, ; R ® 4 ; S ®1, 2, 4(D) P® 1,2; Q® 3, 4 ; R ® 4 ; S ®1, 2, 3, 4

SECTION-II : Numerical Value Type (Up to second decimal place)No question will be asked in section II


0999DJA110319008


LTS-29/46

ALLENSECTION–III : (Maximum Marks : 24)

[k.M–III : (vf/kdre vad : 24)� This section contains SIX questions.� The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both





1. The activity of a radioactive sample decreases to 1/3 of the original activity (A0) in a periodof 9 years. After 9 years more, its activity is A0/x. Find the value of x.,d jsfM;ks,fDVo uewus dh lfØ;rk 9 o"kksZ ds le; dky esa ?kV dj ewy lfØ;rk (A0) dh 1/3 gks tkrh gSA vkSj

vxys 9 o"kks Z ds ckn bldh lfØ;rk A0/x gks tkrh gSA x dk eku Kkr dhft,A

2. The number of neutrons emitted when 23592 U undergoes controlled nuclear fission to 141

54 Xe

and 8938Sr is ?

tc 23592 U dk fu;af=r ukfHkdh; fo[k.Mu 141

54 Xe rFkk 8938Sr esa gksrk gS] rc mRlftZr U;wVªkuks a dh la[;k D;k gS?

3.M 8M

Z Z 4A(g) B(g)

-

-¾¾® + (a. particles)

(a. particles are helium nuclei, so will form He gas by trapping electron). The radioactivedisintergration follows 1st order kinetics starting with 1 moles of A in a 16.4 litre closed flaskat 127ºC. Pressure (atm) devoloped after two half life is (R = 0.082 litre atm/mol-K)

M 8MZ Z 4

A(g) B(g)-

-¾¾® + (a. d.k)

(a. d.k] ghfy;e ds ukfHkd gksrs gS] blfy;s bySDVªkWu dks xzg.k dj He xSl cukrs gS) jsfM;ks lfØ; fo?kVu] 127ºCij 16.4 yhVj ds can ¶ykLd esa A ds 1 eksy ds lkFk izkjEHk djrs gq, izFke dksfV cy xfrdh dk vuqlj.k djrk

gS nks v¼Zvk;q ds ckn fodflr gksus okyk nkc (atm) gSSA (R = 0.082 litre atm/mol-K)Space for Rough Work / dPps dk;Z ds fy, LFkku

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ALLEN


4. The solubility of Pb(OH)2 in water is 5 × 10–6 M. Calculate the pH of buffer solution in whichsolubility of Pb(OH)2 is 0.5 × 10–3 mol litre–1.ty esa Pb(OH)2 dh foys;rk 5 × 10–6 M gSA cQj foy;u] ftlesa Pb(OH)2 dh foys;rk 0.5 × 10–3 mol litre–1 gS]

dh pH Kkr dhft;sA5. Total number of esters and acids possible for the compound having molecular formula C4H8O2.

vkf.od lw= C4H8O2 j[kus okys ;kSfxd ds fy;s , sLVj rFkk vEyksa dh lEHkkfor dqy la[;k gSA6. How many of the following pair(s) can be distinguished by aq. NaHCO3 extraction method

(i) Phenol & Salicyclic acid (ii) Benzene sulphonic acid & Cyclohexanol(iii) Benzoic acid & Squaric acid (iv) Picric acid & Carbolic acid(v) Fumeric acid & o-methyl phenol (vi) p-nitrophenol & EthynefuEu esa ls fdrus ; qXe tfy; NaHCO3 fu"d"kZ.k fof/k }kjk foHksfnr fd;s tk ldrs gSA

(i) fQukWy rFkk lsfyflfyd vEy (ii) csUthu lYQksfud vEy rFkk lkbDyksgsDlsukWy

(iii) csatksbd vEy rFkk LDok;fjd vEy (iv) fiØhd vEy rFkk dkcks Zfyd vEy

(v) ¶;wesfjd vEy rFkk o-esfFky fQukWy (vi) p-ukbVªksfQukWy rFkk , sFkkbu

0999DJA110319008


LTS-31/46



(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)





(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

� dkWye-I dh izR;sd izfof"V ds fy, lHkh lqesfyr izfof"V;ks a ds cqycqyksa dks dkyk djsaA mnkgj.k Lo:i] ;fndkWye-I dh izfof"V (A) izfof"V;ksa (Q), (R) rFkk (T) ls lqesfyr gks] rks bu rhuksa cqycqyksa dks vks-vkj-,l- esa dkykdjsaA blh izdkj izfof"V;ksa (B), (C) rFkk (D) ds fy;s Hkh djsa



0999DJA110319008


LTS-32/46

ALLEN

1. Column-I/dkWYke–I Column-II/dkWYke–II(A) Addition of HCl can make the buffer solution (P) CH3COONa(aq.)

HCl dk ;ksx cQj foy;u cuk ldrk gSA(B) Addition of NaOH can make a buffer solution (Q) NH3(aq.)

NaOH dk ;ksx cQj foy;u cuk ldrk gSA(C) Increase solubility when added to a saturated solution of (R) NH4Cl(aq.)

Fe(OH)3 (Assuming no complex formation)tc Fe(OH)3 ds lar`Ir foy;u dks feyk;k tkrk gS] rc

foys;rk c<rh gS (eku yhft, ladqy dk fuekZ.k ugha gksrk gS)(D) Form buffer on mixing with solution of their conjugate acid (S) HCl(aq.)

or basebuds l;qXeh vEy ;k {kkj ds foy;u dkss ijLij feykus ij cQj curk gS (T) AgCl(s)

2. Match Column-I with Column-II.Column - I (Facts) Column - II (Reasons)

(A) Guanidine H N–C–NH2 2

NH

is example (P) Resonance stabilisation of conjugate acid

of strong base of strong base.(B) Carbanion stability 3 3CCl CF> (Q) Due to s-character of central atoms(C) Alkyne is more acidic than alkene (R) Due to d-orbital resonance

(D) Acidity : < (S) Due to formation of aromatic anion

(T) Stability of conjugate base due to morenumber of identical resonatingstructure

dkWye-I dk dkWye-II ds lkFk lqesfyr dhft,A

dkWye-I (rF;) dkWye-II (dkj.k)(A) XokfuMhu H N–C–NH2 2

NH

, izcy {kkj dk mnkjg.k gSA (P) izcy {kkj ds la;qXeh vEy dk vuqukn LFkk;hdj.k ds

dkj.k gSA(B) dkcZ½.kk;u LFkkf;Ro 3 3CCl CF> (Q) dsUæh; ijek.kqvksa ds s-y{k.k ds dkj.k

(C) ,Ydhu dh rqyuk esa ,Ydkbu vf/kd vEyh; gSA (R) d-d{kd ds vuqukn ds dkj.k

(D) vEyh;rk < (S) ,sjksesfVd ½.kk;u cukus ds dkj.k

(T) la;qXeh {kkj dk LFkkf;Ro] leku vuquknh lajpukvksa

dh vf/kd la[;k ds dkj.kSpace for Rough Work / dPps dk;Z ds fy, LFkku

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ALLEN

PART-3 : MATHEMATICS

Hkkx-3 : xf.krSECTION–I(i) : (Maximum Marks : 12)

[k.M–I(i) : (vf/kdre vad : 12)� This section contains FOUR questions.� Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options










1. If x1, x2,........, xn are the angles lying in (0, p) ; arithmetic mean of sinx1, sinx2,.....,sinxn is |seca|,then G.M. of sinx1, sinx2,........, sinxn is -;fn dks.k x1, x2,........,xn, (0, p) esa fLFkr gS rFkk sinx1, sinx2,....., sinxn dk lekUrj ek/; |seca|gS] rks

sinx1, sinx2,........, sinxn dk xq.kksÙkj ek/; gksxk &(A) 0 (B) 1 (C) 21/n (D) (1 + |seca|)

2. Let a function ƒ (x) be defined as ƒ (x) =1 2cos (µ) x , 0 x 1

4x, x 1

-é + < <ê ³ë

, ƒ (x) can have a local minimum

at x = 1, if the value of µ lies in the interval -

ekuk Qyu ƒ (x) =1 2cos (µ) x , 0 x 1

4x, x 1

-é + < <ê ³ë

ls ifjHkkf"kr gksrk gSA ƒ (x) dk LFkkuh; fufEu"B x = 1 ij gksxk ;fn µ dk

eku fuEu vUrjky esa fLFkr gks &(A) [–1, cos3) (B) (cos3, 1] (C) (cos3, cos1) (D) (cos3, cos2)


0999DJA110319008


LTS-34/46

ALLEN3. Let ƒ (x) = x3 + ax + b with a ¹ b and suppose the tangent lines to the graph of ƒ at x = a and

x = b have the same gradient. Then the value of ƒ (1) is equal to -ekuk ƒ (x) = x3 + ax + b tgk¡ a ¹ b rFkk ekuk ƒ ds vkjs[k esa x = a rFkk x = b ij Li'kZ js[kkvksa dh izo.krk leku gSA

rc ƒ (1) dk eku gksxk -

(A) 0 (B) 1 (C) – 13

(D) 23

4. Consider the function ƒ (x) = 2x 2x 2 x 0

x 2 x 0

ì + + <í

+ ³î , and identify the correct statement -

(A) Mean value theorem is applicable in [–2, 2](B) Mean value theorem is not applicable in [–5, –1]

(C) Mean value theorem is applicable in [–4, –1] and its c is 52

-

(D) Mean value theorem is applicable in [–4, –1] and its c is –1

ekuk Qyu ƒ (x) = 2x 2x 2 x 0

x 2 x 0

ì + + <í

+ ³î gS] rks lR; dFku dk p;u dhft, &

(A) ek/; eku izes; [–2, 2] esa mi;qDr gSA

(B) ek/; eku izes; [–5, –1] esa mi;qDr ugha gSA

(C) ek/; eku izes; [–4, –1] esa mi;qDr gS rFkk bldk c, 52

- gksxkA

(D) ek/; eku izes; [–4, –1] esa mi;qDr gS rFkk bldk c, –1 gksxkA


0999DJA110319008


LTS-35/46














� 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 feysax s 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 (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

0999DJA110319008


LTS-36/46


iz'u 5 ,oa 6 ds fy;s vuqPNsnConsider an AP 65, 62, 59, 56, .......ekuk lekUrj Js.kh 65, 62, 59, 56, .......

5. Maximum value of sum of A.P. is -lekUrj Js.kh ds ;ksxQy dk vf/kdre eku gksxk -(A) 507 (B) 701 (C) 633 (D) 737

6. The value of first negative term of the series is -Js.kh dk izFke J.kkRed in dk eku gksxk -(A) –1 (B) –2 (C) –3 (D) –4

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

Let ƒ (x) = 2

11 x+

and m be the slope, a and b are x-intercept & y intercept respectively of a

tangent to y = ƒ (x).On the basis of above information, answer the following :

ekuk ƒ (x) = 2

11 x+

rFkk Li'kZ js[kk y = ƒ (x) dh izo.krk m ,oa a rFkk b Øe'k% x vUr%[k.M rFkk y vr%[k.M gSaA

mijksDr tkudkjh ds vk/kkj ij] fuEu iz'uksa ds mÙkj nhft, :7. Absissa of the point of contact of the tangent for which m is greatest is -

Li'kZ js[kk ds Li'kZ fcUnq dk Hkqt ftlds fy, m vf/kdre gks] gksxk -

(A) 1

3(B) 1 (C) –1 (D)

1

3-

8. The greatest value of 'b' is -b dk vf/kdre eku gS &

(A) 98

(B) 38

(C) 18

(D) 58


0999DJA110319008


LTS-37/46







lgh gSA




½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA9. Consider an arrangement of numbers

1 2

3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18 19 20

List-I List-II(P) 1st term in the 20th row is (1) 16020(Q) 20th term from the end in the 20th row is (2) 401(R) sum of the terms in the 20th row is (3) 27000(S) sum of odd positioned terms in 30th row is (4) 381(A) P ® 4; Q ® 2; R ® 1; S ® 3 (B) P ® 4; Q ® 2; R ® 3; S ® 1(C) P ® 4; Q ® 1; R ® 2; S ® 3 (D) P ® 3; Q ® 2; R ® 1; S ® 4


0999DJA110319008


LTS-38/46

ALLENekuk la[;kvks a ,d foU;kl fuEu gS %

1 2

3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18 19 20

lwph-I lwph-II(P) 20 oha iafDr dk izFke in gksxk (1) 16020(Q) 20 oha iafDr esa vUr ls 20 ok¡ in gksxk (2) 401(R) 20 oha iafDr esa inks a dks ;ksxQy gksxk (3) 27000(S) 30 oha iafDr esa fo"ke LFkku ds inks a dk ;ksxQy gksxk (4) 381(A) P ® 4; Q ® 2; R ® 1; S ® 3 (B) P ® 4; Q ® 2; R ® 3; S ® 1(C) P ® 4; Q ® 1; R ® 2; S ® 3 (D) P ® 3; Q ® 2; R ® 1; S ® 4

10. List-I List-II(P) ƒ (x) = x5 – 5x4 + 5x3 – 10 has local maxima and minima at (1) 4

x = l and x = m respectively then 2l +m is(Q) Ladder of length 5m leaning against a wall is being pulled along the (2) 5

ground at 2cm/s. When the foot of the ladder is 4m away from the wallif the top of the ladder slides down on the wall at 8/l cm/s, then l =

(R) If ƒ (x) = x – 2 sinx, 0 £ x £ 2p is increasing in the interval [ap, bp] (3) 3then a + b is

(S) If equation of the tangent to the curve y = –e–x/2, where it crosses the (4) 2

y-axis is x y

1p q

+ = , then p – 2q is

(A) P ® 2; Q ® 3; R ® 1; S ® 4 (B) P ® 2; Q ® 3; R ® 4; S ® 1(C) P ® 2; Q ® 4; R ® 3; S ® 1 (D) P ® 2; Q ® 4; R ® 1; S ® 3


0999DJA110319008


LTS-39/46

ALLEN


lwph-I lwph-II(P) ƒ (x) = x5 – 5x4 + 5x3 – 10 dk LFkkuh; mfPp"B rFkk fufEu"B Øe'k% (1) 4

x = l rFkk x = m ij gS] rks 2l + m gS

(Q) nhokj ds lgkjs [kM+h gqbZ 5m yEch lh<+h dks ry ds vuqfn'k 2 lseh@ls- ls [khapk (2) 5tkrk gSA tc lh<+h dk fupyk fljk nhokj ls 4 ehVj nwj gS] ;fn lh<+h dk

mGijh fljk 8/l lseh/ls- dh nj ls uhps fQlyrk gS] rks l =(R) ;fn ƒ (x) = x – 2 sinx, 0 £ x £ 2p, vUrjky [ap, bp] esa o/kZeku gS] rks (3) 3

a + b gksxk

(S) ;fn oØ y = –e–x/2 dh Li'kZ js[kk dk lehdj.k] tgk¡ ;g y-v{k dks dkVrk gS (4) 2

x y1

p q+ = gS] rks p – 2q gksxk &

(A) P ® 2; Q ® 3; R ® 1; S ® 4 (B) P ® 2; Q ® 3; R ® 4; S ® 1(C) P ® 2; Q ® 4; R ® 3; S ® 1 (D) P ® 2; Q ® 4; R ® 1; S ® 3

SECTION-II : Numerical Value Type (Up to second decimal place)No question will be asked in section II

0999DJA110319008


LTS-40/46

ALLEN


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

� This section contains SIX questions.� The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both





1. Given that a, g are roots of equation Ax2 – 4x + 1 = 0 and b, d the roots of the equationBx2 – 6x + 1 = 0. Find value of B – A such that a, b, g, d are in H.P.fn;k x;k gS fd a, g lehdj.k Ax2 – 4x + 1 = 0 rFkk b, d lehdj.k Bx2 – 6x + 1 = 0 ds ewy gSA B – A dk eku Kkr

dhft, rkfd a, b, g, d gjkRed Js.kh esa gksA

2. If 2

x

kx xƒ(x) x

2e= - + is decreasing for every x < 0, then least value of k2 is-

;fn 2

x

kx xƒ(x) x

2e= - + lHkh x < 0 ds fy, ákleku gS] rks k2 dk U;wure eku gksxk

3. A curve passes through (1, 0) and the slope of the tangent at any point (x, y) is x2 – 2x for allvalues of x. If the point of minimum ordinate, when x Î [–2, 2] is (a, b), then the value of a – b is,d oØ (1, 0) ls xqtjrk gS rFkk fdlh fcUnq (x, y) ij lHkh x ds fy, Li'kZ js[kk dh izo.krk x2 – 2x gSA ;fn U;wure

dksfV dk fcUnq] tc x Î [–2, 2] gks] (a, b) gS] rks a – b dk eku gksxk

0999DJA110319008


LTS-41/46

ALLEN


4. Maximum value of ƒ(x) = 8x3 – x4 – 22x2 + 24x is equal toƒ(x) = 8x3 – x4 – 22x2 + 24x dk vf/kdre eku gksxk

5. Let ƒ(x) = ( )1p3log x log x x x 0

0 x 0

ìï + ¹íï =î

. If Rolle's theorem is applicable to ƒ(x) in x Î [0,1] then least

integral value of p is

ekuk ƒ(x) = ( )1p3log x log x x x 0

0 x 0

ìï + ¹íï =î

gSA ;fn x Î [0,1] esa ƒ(x) ij jkWy izes; lR;kfir gksrh gS] rks p dk

U;wure iw.kk±d eku gksxk

6. Let 2

1 x 0

x a 0 x 1ƒ(x)

bx 1 x 2

6 c x 2

=ìï - < <ï= í

£ <ïï - =î

be such that Lagrange's mean value theorem is applicable to ƒ(x) for

x Î [0,2]. Let P(p,ƒ(p)), p Î (0,2) be a point on ƒ(x) such that tangent at P is parallel to chordjoining points with x = 0 & x = 2 on y = ƒ(x). Then a + b + c + 4p is equal to

ekuk 2

1 x 0

x a 0 x 1ƒ(x)

bx 1 x 2

6 c x 2

=ìï - < <ï= í

£ <ïï - =î

bl izdkj gS fd x Î [0,2] esa ƒ (x) ij ykaxzjkt e/; eku izes; ykxq gksrh gSA ekuk

P(p,ƒ(p)), p Î (0, 2), ƒ(x) ij ,d fcUnq bl izdkj gS fd P ij Li'kZ js[kk] oØ y = ƒ(x) ds fcUnq x = 0 rFkk x = 2dks tksM+us okyh thok ds lekUrj gS] rks a + b + c + 4p dk eku gksxk

0999DJA110319008


LTS-42/46



(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)





(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

� dkWye-I dh izR;sd izfof"V ds fy, lHkh lqesfyr izfof"V;ks a ds cqycqyksa dks dkyk djsaA mnkgj.k Lo:i ] ;fndkWye-I dh izfof"V (A) izfof"V;ksa (Q), (R) rFkk (T) ls lqesfyr gks] rks bu rhuksa cqycqyksa dks vks-vkj-,l- esa dkykdjsaA blh izdkj izfof"V;ksa (B), (C) rFkk (D) ds fy;s Hkh djsa



0999DJA110319008


LTS-43/46

ALLEN


1. If 1 2 1

, ,4 n2 n 2 n-+ -

......... is given series, then

Column-I Column-II(A) The given series is (P) 1(B) If sum of first 5 terms of given series is equal to 5, (Q) 2

then n can be(C) If given series is increasing, then n may be (R) 3

(D) If n

n 2 n n nT

4 n- +

=-

, then value of n is (S) G.P.

(T) A.P.

;fn 1 2 1

, ,4 n2 n 2 n-+ -

......... ,d fn;k x;k vuqØe gS] rks

dkWye-I dkWye-II(A) fn;k x;k vuqØe gksxk (P) 1(B) ;fn fn, x, vuqØe ds izFke 5 inksa dk ;ksxQy 5 gks] rks n gks ldrk gS (Q) 2(C) ;fn fn;k x;k vuqØe o/kZeku gS] rks n gks ldrk gS (R) 3

(D) ;fn n

n 2 n n nT

4 n- +

=-

gks] rks n dk eku gksxk (S) xq.kksÙkj Js.kh

(T) lekUrj Js.kh

0999DJA110319008


LTS-44/46

ALLEN


2. Column-I Column-II(A) Number of points of local maximas of the function (P) 4

ƒ (x) = x2 + 4cosx + 5 is(B) Integral value of 'k' for which ƒ '(x) = 0 where (Q) 2

ƒ(x) = (k2 – 5k + 6) tanx + (3k –6)x(C) Length of subtangent drawn to function (R) 3

ƒ(x) = 7sinx + tan2xæ ö

ç ÷pè ø at point ,8

2pæ ö

ç ÷è ø

is

(D) If ƒ (x) = 2|x|3 + 3x2 – 12|x| + 3, where x Î [–1, 1], (S) 1then greatest value of ƒ (x) is (T) 0

dkWye-I dkWye-II(A) Qyu ƒ (x) = x2 + 4cosx + 5 ds LFkkuh; mfPp"B fcUnqvksa dh (P) 4

la[;k gksxh

(B) 'k' dk iw.kk±d eku ftlds fy, ƒ '(x) = 0 gks] tgk¡ (Q) 2ƒ(x) = (k2 – 5k + 6) tanx + (3k –6)x gS] gksxk

(C) fcUnq ,82pæ ö

ç ÷è ø

ij Qyu ƒ(x) = 7sinx + tan2xæ ö

ç ÷pè ø dh v/kksLi'kZ js[kk (R) 3

dh yEckbZ gksxh

(D) ;fn ƒ (x) = 2|x|3 + 3x2 – 12|x| + 3 tgk¡ x Î [–1, 1] (S) 1

gks] rks ƒ (x) dk vf/kdre eku gksxk (T) 0


LTS-45/460999DJA110319008

ALLEN

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

+3 0 –1Single If only the bubble If none In all

correct option corresponding to of the other,dy lgh the correct option bubbles is cases

I(i) fodYi 4 is darkened — darkened vU; lHkh 12;fn flQZ lgh fodYi ds ;fn fdlh Hkh ifjfLFkfr;ksa

vuq:Ik cqycqys dks cqycqys dks dkyk esadkyk fd;k gS ugha fd;k g S

Paragraph +4 +1 0 –2Based If only the bubble(s) For darkening a bubble If none In all

(One or more corresponding to all corresponding to each of the othercorrect the correct option(s) correct option, provided bubbles is cases

option(s)) is(are) darkened NO incorrect option darkened vU; lHkh I(ii) vuqPNsn ij 4 ;fn flQZ lkjs lg h fodYi darkened ;fn fdlh Hkh ifjfLFkfr;ksa 16

vk/kkfjr (fodYiksa) ds vuq:Ik izR;sd lgh fodYi ds vuq:i cqycqys dks esa(,dy ;k ,d ls cqycqys (cqycqyksa) dks cqycqys dks dkyk djus ij] dkyk ugha

vf/kd lgh dkyk fd;k x;k gS ;fn dksbZ xyr fodYi dkyk fd;k gSfodYi) ugha fd;k g S

Matching +3 0 –1Lists Type If only the bubble If none In all

(Single corresponding to of the othercorrect option) the correct option bubbles is cases

I(iii) lqpha lqesyu 2 is darkened — darkened vU; lHkh 6izdkj ;fn flQZ lgh fodYi ds ;fn fdlh Hkh ifjfLFkfr;ksa

(,dy lgh vuq:Ik cqycqys dks cqycqys dks dkyk esafodYi) dkyk fd;k gS ugha fd;k g S

+4 0Single digit If only the bubble In all

Integer corresponding other(0-9) to correct answer cases

III ,dy vadh; 6 is darkened — vU; lHkh — 24iw.kk±d ;fn flQZ lgh mÙkj ds ifjfLFkfr;ksa(0-9) vuq:Ik cqycqys dks esa

dkyk fd;k gS+8 +2 0

Matrix If only the bubble(s) For darkening a bubble In allMatch Type corresponding to corresponding to each otherlqesfyr izdkj all the correct correct match is cases

match(es) is(are) darkened vU; lHkh IV 2 darkened izR;sd lgh feyku ifjfLFkfr;ksa — 16

;fn flQZ lkjs lgh feyku ds vuq:i cqycqys esa(feykuksa) ds vuq:Ik dks dkyk djus ijcqycqys (cqycqyksa) dks

dkyk fd;k x;k gS

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

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

Your Target is to secure Good Rank in JEE 2020 0999DJA110319008LTS-46/46


I have read all the instructionsand shall abide by them.

eSaus lHkh funs Z'kks a dks i<+ fy;k g S vkSj eS mudkvo'; ikyu d:¡xk@d:¡xhA

____________________________Signature of the Candidate / ijh{kkFkhZ ds gLrk{kj

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 g S ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks a leku g SaA

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

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


ALLEN

Dear Student,

We request you to provide feedback for the test series till you have appeared. Kindly answer the questions

provided 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.

If you have not prepared well for today's test and if you are not feeling good today, then do not blame test

series for it.

We strive to prepare you for all kinds of situations and facing variations in paper, as this can also happen in

Main exam. It is important for you to concentrate on your rank.

Go through the feedback form thoroughly and answer with complete loyalty. Darken your response (2, 1, 0) in

OMR sheet corresponding to :

Distance Learning Programmes(Session-2019-20)

ALL INDIA TEST SERIES

ABOUT FEEDBACK SYSTEM

Questions

1. How convenient it was for you to enroll in our Distance Learning Course through online mode?

[2] Very Convenient [1] Average [0] Difficult

2. How do you find location of Test Center ?

[2] Approachable from all part of city [1] Average Approachable [0] Difficult to reach

3. Test Timing :

[2] Comfortable [1] Average [0] Need to be change

4. Do you feel Test starts on time :

[2] Yes Always [1] Some time delayed [0] Always delay

5. The level of test paper [meet all the requirement of competitive examination]

[2] Good standard [1] Average [0] Below average

6. Number of mistake in test papers :

[2] Negligible [1] Are very less [0] Too High

7. Are you satisfied with result analysis ?

[2] Outstanding [1] Average [0] Below average

8. Do you feel our Test Series is able to improve speed, accuracy & developing examination temperament?

[2] Yes I feel [1] Partly [0] Not at all

9. Response from Allen on email / telephonically

[2] Always good and prompt [1] Some time delay [0] Not satisfactory

10. Response at test center

[2] Satisfactory [1] Partly Satisfactory [0] Not Good

(0999dja110319008) )) test pattern distance learning ... · vks- vkj- ,l- esa gsj&qsj@foÏfr u...

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