TARGET : JEE (Main + Advanced) 2019 · LT-2/38 Tae : JEE (Ma + Adaced) 2019/14-04-2019/Pae-1...

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 gksA

QUESTION 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

Time : 3 Hours Maximum Marks : 183

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PAPER – 1

Test Type : MAJOR Test Pattern : JEE-Advanced

TEST # 10 TEST DATE : 14 - 04 - 2019

)0999DJA110318018)Paper Code

(0999DJA110318018)

LTS-2/38

Target : JEE (Main + Advanced) 2019/14-04-2019/Paper-1

0999DJA110318018

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

0999DJA110318018

Leader Test Series/Joint Package Course/14-04-2019/Paper-1

LTS-3/38

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

PART-1 : PHYSICS

-1 : SECTION–I(i) : (Maximum Marks : 28)

–I(i) : ( : 28)

This section contains SEVEN 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.

() ()

: +4 : +3 : +2

: +1

: 0 : –2

+4 +2 +1 , –2

0999DJA110318018


LTS-4/38

1. A stationary singly ionized helium atom emits a photon corresponding to the first line ofLyman series. The emitted photon is then absorbed by a stationary hydrogen atom in theground state. Point out INCORRECT statement/s from the following :-(A) The emitted photon from He+ ion excites the hydrogen atom to sixth excited state.(B) The kinetic energy of recoiling of He+ of mass m when it emits a photon of frequency v is

given by n2 2

2h2mc

(C) The emitted photon from He+ ion ionizes the hydrogen atom with liberated electroncarrying kinetic energy of about 20 eV.

(D) The transition energy during the emission of photon from He+ ion is about 40.8 eV.

,d fLFkj ,dy vk;fur ghfy;e ijek.kq] ykbeu Js.kh dh izFke js[kk ds laxr ,d QksVksu mRlftZr djrk gSA ;g

mRlftZr QksVksu] ewy voLFkk esa fLFkr fLFkj gkbMªkstu ijek.kq }kjk vo'kksf"kr dj fy;k tkrk gSA xyr dFku pqfu;s %&

(A) He+ vk;u ls mRlftZr QksVksu] gkbMªkstu ijek.kq dks NBh mÙksftr voLFkk esa mÙksftr djrk gSA

(B) æO;eku m okys He+ }kjk v vkofÙk dk ,d QksVksu mRlftZr djus ij bldh izfrf{kIr xfrt ÅtkZ n2 2

2h2mc }kjk nh

tkrh gSA

(C) He+ vk;u ls mRlftZr QksVksu gkbMªkstu ijek.kq dks vk;fur djrk gS rFkk yxHkx 20 eV xfrt ÅtkZ ds lkFk ,d

bysDVªkWu mRlftZr gksrk gSA

(D) He+ vk;u ls QksVksu ds mRltZu ds nkSjku laØe.k ÅtkZ dk eku yxHkx 40.8 eV gSA2. A projectile is thrown with velocity U = 20m/s ± 5% at an angle 60°. If the projectile falls back

on the ground at the same level then which of following can be possible answers for range?Consider g = 10m/s2.fdlh iz{ksI; dks osx U = 20m/s ± 5% ls dks.k 60° ij QSadk tkrk gSA ;fn iz{ksI; iqu% /kjkry ij leku Lrj ij gh

fxjrk gS rks ijkl ds fy;s laHkkfor eku gks ldrs gS (g = 10m/s2 )(A) 39.0 m (B) 37.5 m (C) 34.6 m (D) 32.0 m


0999DJA110318018


LTS-5/38


3. A parallel beam of light (l = 5000 Å) is incident at an angle q = 30° with the normal to the slitplane in a Young's double slit experiment. The intensity due to each slit is I0. Point O isequidistant from S1 and S2. The distance between slits is 1 mm, then :-

(A) The intensity at O is 4 I0.(B) The intensity at O is zero(C) The intensity at a point on the screen 4 mm above O is 4I0.(D) The intensity at a point on the screen 4 mm above O is zero.

S1

q

2m

O

,d lekUrj izdk'k iqat (l = 5000 Å) dks ;ax f}fLyV iz;ksx esa fLyV ry ij vfHkyEc ls q = 30° dks.k ij vkifrrfd;k tkrk gSA izR;sd fLyV ds dkj.k rhozrk I0 gSA fcUnq O, S1 o S2 ls lenwjLFk gSA fLyVksa ds e/; nwjh 1mm gSA rc%&

(A) O ij rhozrk 4 I0 gSA

(B) O ij rhozrk 'kwU; gSA

(C) insZ ij O ls 4 mm Åij fLFkr fcUnq ij rhozrk 4I0 gSA

(D) insZ ij O ls 4 mm Åij fLFkr fcUnq ij rhozrk 'kwU; gSA

0999DJA110318018


LTS-6/38

4. Oil enters the bend of a pipe in the horizontal plane with velocity 4 ms–1 and pressure280 × 103 Nm–2 as shown in figure. (Take specific gravity of oil as 0.9 and sin 37° = 0.6)(A) The pressure of oil at the point Q, (P2) = 172 × 103 N/m2

(B) The pressure of oil at the point Q, (P2) = 388 × 103 N/m2

(C) The force required to hold the bend in the place is 63 × 103 N.(D) The force required to hold the bend in the place is 76 × 103 N.

A =0.2m12

V =4m/s1

37°

QV2

A = —2A4

1

P

A = —2A4

1

iznf'kZr fp= esa rsy {kSfrt ry esa ,d ikbi ds eksM + esa 4 ms–1 osx o 280 × 103 Nm–2 nkc ls izos'k djrk gSA

(rsy dk fof'k"V xq:Ro 0.9 o sin 37° = 0.6 ysa)

(A) fcUnq Q ij rsy dk nkc, (P2) = 172 × 103 N/m2 gSA

(B) fcUnq Q ij rsy dk nkc, (P2) = 388 × 103 N/m2 gSA

(C) eksM+ dks blds LFkku ij cuk;s j[kus ds fy;s vko';d cy 63 × 103 N gSA

(D) eksM+ dks blds LFkku ij cuk;s j[kus ds fy;s vko';d cy 76 × 103 N gSA5. The indicator diagram for two processes 1 and 2 carrying on an ideal gas is shown in figure.

W denotes work done by gas, DQ is heat given to gas and T is temperature of gas. Then :fdlh vkn'kZ xSl ij yxk;s x;s nks izØeksa 1 rFkk 2 ds fy, lwpd vkjs[k fp= esa n'kkZ;k x;k gSA ;fn xSl }kjk fd;k x;k

dk;Z W, xSl dks nh x;h Å"ek DQ ,oa xSl dk rkieku T gks rks %&

VO

Process 1Process 2

P isothermal

(A) Wprocess 1 > Wprocess 2 (B) Tfinal process 2 > Tfinal process 1

(C) DQprocess 1 > DQprocess 2 (D) DQprocess 2 > DQprocess 1


0999DJA110318018


LTS-7/38

6. A uniform circular loop of mass 'm' and radius R is lying in space where a uniform magneticfield 0

ˆ ˆB B (i 3 j)= +r

. The loop carries constant current I in clockwise direction and lies in agravity free region at rest in xy–plane with its centre at origin. Then choose the CORRECTstatement(s) :-

(A) The initial angular acceleration of the loop has magnitude 04 Bm

Ip

(B) The initial angular acceleration of the loop has magnitude 02 Bm

Ip

(C) The unit vector in the direction of initial angular acceleration is 3 1ˆ ˆi j

2 2æ ö

+ -ç ÷ç ÷è ø

(D) The initial acceleration of the point with coordinate 3 RR,

2 2æ ö-ç ÷ç ÷è ø

is zero

æO;eku 'm' o f=T;k R okyk ,d le:i o`Ùkkdkj ywi fdlh LFkku ij j[kk gqvk gS] tgk¡ ,d le:i pqEcdh; {ks=

0ˆ ˆB B (i 3 j)= +

r fo|eku gSA bl ywi esa fu;r /kkjk I nf{k.kkorZ fn'kk esa izokfgr gks jgh gS rFkk ;g x-y ry esa xq:Ro

eqä {ks= esa bl izdkj fLFkr gS fd bldk dsUæ ewy fcUnq ij gksrk gSA lgh dFku@dFkuksa dks pqfu, %&

(A) bl ywi ds izkjfEHkd dks.kh; Roj.k dk ifjek.k 04 Bm

Ip gSA

(B) bl ywi ds izkjfEHkd dks.kh; Roj.k dk ifjek.k 02 Bm

IpgSA

(C) izkjfEHkd dks.kh; Roj.k dh fn'kk esa bdkbZ lfn'k 3 1ˆ ˆi j

2 2æ ö

+ -ç ÷ç ÷è ø

gSA

(D) funsZ'kkad 3 RR,

2 2æ ö-ç ÷ç ÷è ø

okys fcUnq dk izkjfEHkd Roj.k 'kwU; gSA


0999DJA110318018


LTS-8/38

7. A load W is suspended from a self-propelled crane by a cable of length d (figure). The craneand load are moving at a constant speed v0 (assume v0 < 2gd ). The crane is stopped by abumper and the load on the cable swings out. Mark the correct statement(s) :fp= esa ,d Hkkj W dks d yEckbZ dh dscy }kjk ,d Lopkfyr Øsu ls yVdk;k x;k gSA Øsu rFkk Hkkj fu;r pky v0

(ekuk v0 < 2gd ) ls xfr'khy gSA bl Øsu dks ,d cEij }kjk jksdk tkrk gS rFkk dscy ij yVdk Hkkj >wyus yxrk gSAlgh dFku@dFkuks dks pqfu;s%&

W

d

v0

(A) The angle through which the load swings is 0v2arcsin2 gd

æ öq = ç ÷ç ÷

è ø.

(B) Tension in the cable just before impact has magnitude W.

(C) Tension in the cable just after impact has magnitude 20vWW

g d+ ´ .

(D) Tension in the cable just after impact has magnitude 20vWW

g d- ´ .

(A) Hkkj 0v2arcsin2 gd

æ öq = ç ÷ç ÷

è ø dks.k ij xfr djrk gSA

(B) VDdj ds Bhd iwoZ dscy esa ruko dk ifjek.k W gSA

(C) VDdj ds Bhd i'pkr~ dscy esa ruko dk ifjek.k 20vWW

g d+ ´ gSA

(D) VDdj ds Bhd i'pkr~ dscy esa ruko dk ifjek.k 20vWW

g d- ´ gSA


0999DJA110318018


LTS-9/38

SECTION–I(ii) : (Maximum Marks : 18) [k.M–I(ii) : (vf/kdre vad : 18)

� This section contains SIX questions of matching type.� This section contains Two tables (each having 3 columns and 4 rows)� Based on table, there are THREE 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 lqesy izdkj ds N% iz'u gSaA� bl [k.M es a nks Vscy gSa (izR;sd Vscy esa 3 dkye vkSj 4 iafDr;k¡ gSa)� Vscy ij vk/kkfjr rhu iz'u gSaA� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa flQZ ,d fodYi lgh gSaA� izR;sd iz'u ds fy, vks- vkj- ,l- ij lgh mÙkj 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 esaAAnswer Q.8, Q.9 and Q.10 by appropriately matching the information given in thethree columns of the following table.uhps nh x;h Vscy ds rhu dkyeksa esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10ds mÙkj nhft;sA

Column-1 Column-2 Column-3Equation of wave Energy in one Average power

wavelength transmitted atpoint X

dkWye-1 dkWye-2 dkWye-3rjax lehdj.k ,d rjaxnS/; Z es a ÅtkZ fcUnq X ls lapfjr vkSlr

'kfä

(I)2 5y A sin 2 t xv 6p pæ ö= pn - n +ç ÷

è ø(i) 2p2µA2n2l (P) 0

(II) ( )5y A sin x cos 2 t2pæ ö= pnç ÷

è øl(ii) 4p2µA2n2l (Q) 2p2µvA2n2

(III) ( )4y A sin x cos 2 tpæ ö= pnç ÷è øl

(iii) p2µA2n2l (R)2 2 2vA

2p m n

(IV)2y A sin x 2 tv 6p pæ ö= n - pn +ç ÷

è ø(iv)

2 2 2A2

p m n l (S)2 2vA

4pm n


0999DJA110318018


LTS-10/38

8. A progressive wave is moving in the positive x direction with speed v. The amplitude of the

wave is A and frequency of the source is n. At t = 0, the particle at x = 3l

is located at Ay2

= +

and moving away from the mean position. Point X is located at a distance 34l

from the source.

,d izxkeh rjax /kukRed x fn'kk esa v pky ls xfr'khy gSA rjax dk vk;ke A rFkk L=ksr dh vkofÙk n gSA t = 0 ij

x = 3l

ij fLFkr d.k Ay2

= + ij gksrk gS rFkk ek/; fLFkfr ls nwj tk jgk gSA fcUnq X L=ksr ls 34l

nwjh ij fLFkr gSA

(A) (II) (iv) (P) (B) (I) (i) (Q) (C) (III) (ii) (R) (D) (IV) (iii) (S)9. A string of length l fixed at both ends and vibrating in its 3rd overtone. The maximum amplitude

of the particle at anti-node is A. At t = 0, the particle at x = 6l

is located at its positive

extreme. Point X is located at a distance 2l

from the source.

nksuksa fljksa ls ca/kh gqbZ l yEckbZ dh jLlh blds r`rh; vf/kLojd esa dEiUu dj jgh gSA izLiUn ij d.k dk vf/kdre

vk;ke A gSA t = 0 ij x = 6l

ij fLFkr d.k bldh /kukRed lhekUr fLFkfr ij ik;k tkrk gSA fcUnq X L=ksr ls 2l

nwjh

ij fLFkr gS %&(A) (II) (iii) (Q) (B) (III) (iv) (P) (C) (III) (iii) (P) (D) (II) (iv) (R)

10. A string is vibrating in its 5th harmonic with fixed end at x = 0 and free end at x = l. The

maximum amplitude of the particle at antinode is A. At t = 0, the particle at x = 4l

is at the

positive extreme. Point X is located at a distance 5l

from the source.

,d jLlh bldh 5 oha xq.kko`fr esa dEiUu dj jgh gS rFkk bldk fLFkj fljk x = 0 ij ,oa eqä fljk x = l ij gSA izLian

ij d.k dk vf/kdre vk;ke A gSA t = 0 ij x = 4l

ij fLFkr d.k /kukRed lhekUr fLFkfr ij gksrk gSA fcUnq X L=ksr

ls 5l

nwjh ij fLFkr gS %&

(A) (III) (ii) (P) (B) (II) (ii) (R) (C) (II) (iv) (P) (D) (II) (iii) (P)Space for Rough Work / dPps dk;Z ds fy, LFkku

0999DJA110318018


LTS-11/38

Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following table.uhps nh x;h Vscy ds rhu dkyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uk s a Q.11, Q.12 vkSjQ.13 ds mÙkj nhft;sA

Column–1 Column-2 Column-3Mass of the Object is 'm' Kinetic energy at the time Time till slipping ceases.situation is given at t = 0 when slipping ceases.dkWye–1 dkWye-2 dkWye-3fi.M dk æO;eku 'm' gS rFkk fQlyu lekIr gksus ds le; le; tc fQlyu lekIr gksrh gSAnh xbZ fLFkfr t = 0 ij gSA xfrt ÅtkZ

(I) v0

v /R0

ringµ R (i) Zero (P) 0vgm

(II) v0

v /R0

Discµ R (ii) 20

1 mv12 (Q) 02v

3 gm

(III) v

v /R0solid sphere µ R (iii) 2

09 mv

70 (R) 04v7 gm

(IV) v

v /R0

spherical shell

µ R (iv)20mv

30 (S) 04v5 gm


0999DJA110318018


LTS-12/38

11. Name the object that has maximum loss in kinetic energy, its kinetic energy after slippingceases and the time till slipping ceases.fdl oLrq dh xfrt ÅtkZ esa vf/kdre âkl gksrk gS\ fQlyu lekIr gksus ds i'pkr~ bldh xfrt ÅtkZ rFkk og le;tc fQlyu lekIr gks tkrh gS] Kkr dhft;sA

(A)(II) (iii) (Q) (B) (I) (i) (P) (C) (III) (iii) (R) (D) (IV) (iv) (S)12. Name the object has minimum loss in kinetic energy during slipping, its kinetic energy after

slipping ceases and the time till slipping ceases :-

fQlyu ds nkSjku fdl oLrq dh xfrt ÅtkZ esa U;wure âkl gksrk gS\ fQlyu lekIr gksus ds i'pkr~ bldh xfrt ÅtkZ

rFkk og le; tc fQlyu lekIr gks tkrh gS] Kkr dhft;sA

(A)(II) (ii) (P) (B) (I) (i) (S) (C) (III) (iii) (R) (D) (IV) (iv) (Q)13. Name the object, which stops when slipping ceases its kinetic energy at the time of slipping

ceases and time till slipping ceases :-

dkSulh oLrq fQlyu lekIr gksus ij :d tkrh gS \ fQlyu lekIr gksus ds le; bldh xfrt ÅtkZ rFkk og le; tc

fQlyu lekIr gks tkrh gS] Kkr dhft;sA

(A)(II) (ii) (R) (B) (III) (iii) (S) (C) (IV) (iv) (Q) (D) (I) (i) (P)

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

[k.M–II : la[;kRed eku izdkj (n'keyo ds nks LFkku rd)[k.M II es a dksb Z iz'u ugha gSA


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LTS-13/38

SECTION–III : (Maximum Marks : 15)[k.M–III : (vf/kdre vad : 15)

� This section contains FIVE 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 : +3 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 In all other cases.

� bl [k.M esa ik¡p 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 : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

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

1. In a Coolidge tube the atomic number of target material is 41. A potential difference of 20 kVis applied across the tube. Let lK be Ka line produced by tube and lmin be cut off wavelength.Calculate ( )1

K min2 l - l in the order of 10–12 m.

dwyht uyh esa iz;qDr y{; inkFkZ dk ijek.kq Øekad 41 gSA uyh ds fljksa ij 20kV foHkokUrj vkjksfir fd;k tkrk gSA

ekuk lK ] uyh }kjk mRiUu Ka js[kk rFkk vard rjaxnS/;Z lmin gSA ( )1K min2 l - l dh x.kuk 10-12 m dksfV esa dhft,A

2. An open organ pipe containing air resonates in fundamental mode due to a tuning fork. Themeasured values of length l(in cm) of the pipe and radius r (in cm) of the pipe are l= 94 ± 0.1,r = 5 ± 0.05. The velocity of the sound in air is accurately known. The maximum percentageerror in the measurement of the frequency of that tuning fork by this experiment is given bya2 %. Find the value of 10a.

ok;q ls Hkjk gqvk ,d [kqyk vkWxZu ikbi ,d Lofj= ds lkFk ewy fo/kk esa vuqukn djrk gSA ikbi dh ekih xbZ yEckbZ

l(cm esa) rFkk f=T;k r (lseh esa) Øe'k% l = 94 ± 0.1, r = 5 ± 0.05 gSA ok;q esa /ofu dk osx ;FkkFkZ :i ls Kkr gSA

bl iz;ksx }kjk Lofj= dh vko`fÙk ds ekiu esa vf/kdre izfr'kr = qfV a2 % gSA 10a dk eku Kkr dhft;sA


0999DJA110318018


LTS-14/38

3. The concave spherical mirror of radius of R = 9 cm forms the image of a source which coincideswith a source. When a little liquid is poured in the mirror, a second image is formed betweenthe source and the mirror at a distance l = 4 cm from the source. Find the refractive index µof the liquid. Write '5µ' in OMR sheet.f=T;k R = 9 cm okyk vory xksyh; niZ.k fdlh L=ksr dk izfrfcEc cukrk gS tks L=ksr ds laikrh gSA tc niZ.k ij dqN

æo Mkyk tkrk gS rks L=ksr ls l = 4 cm nwjh ij L=ksr o niZ.k ds e/; ,d f}rh; izfrfcEc curk gSA ;fn æo dk

viorZukad µ gks rks '5µ' dk eku Kkr dhft;sA

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

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

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

O

4. Due to the excess pressure of surface tension find the change in the radius (in Å) of a liquiddrop of surface tension 0.075 N/m and Bulk modulus 1.25 × 108 N/m2.i`"B ruko ds nkc vkf/kD; ds dkj.k ,d æo cwan] ftldk i`"B ruko 0.075 N/m rFkk vk;ru izR;kLFkrk xq.kkad

1.25 × 108 N/m2 gS] dh f=T;k esa ifjorZu (Å esa) Kkr dhft,A


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LTS-15/38

5. A ring of mass m and radius 3R is rotating with constant angular speed w around a planet ofmass M and radius R. Center of ring and planet concide with each other. Tension in the ring

is given as T = 2GMm3n Rp

. Find value of n 3GM9R

æ öw =ç ÷

è øæO;eku m rFkk f=T;k 3R okyh ,d oy; æO;eku M rFkk f=T;k R okys xzg ds pkjksa vksj fu;r dks.kh; pky w ls ?kw.kZu

dj jgh gSA oy; rFkk xzg dk dsUæ ,d&nwljs ds lEikrh gSA oy; esa ruko T = 2GMm3n Rp

gks rks n dk eku Kkr dhft;sA

3GM9R

æ öw =ç ÷

è ø

w


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LTS-16/38

PART-2 : CHEMISTRY

-2 : SECTION–I(i) : (Maximum Marks : 28)

–I(i) : ( : 28)

This section contains SEVEN 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,


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 with

second 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), without

selecting 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 correct

option(s) will result in –2 marks.

() ()

: +4 : +3 : +2

: +1

: 0 : –2

+4 +2 +1 , –2

0999DJA110318018


LTS-17/38

1. Which statement about the behaviour of a catalyst is/are incorrect-(A) A catalyst reacts with the reactant and shifts the equilibrium to the right(B) A catalyst lowers the activation energy of the original reaction pathway(C) A catalyst provides additional energy to a reactant so it can achieve the activation energy(D) A catalyst provides an alternative reaction pathway with a lower activation energy

,d mRizsjd ds O;ogkj ds lUnHkZ esa fuEu esa ls dkSulk dFku xyr gS&

(A) ,d mRizsjd] fØ;kdkjdksa ls vfHkfØ;k djrk gS vkSj lkE; dks nk;sa vksj foLFkkfir dj nsrk gSA

(B) ,d mRizsjd] okLrfod vfHkfØ;k iFk dh lfØ;.k ÅtkZ dks de dj nsrk gSA

(C) ,d mRizsjd] vfHkdkjdksa dks vfrfjDr ÅtkZ iznku djrk gS ftlls ;g lfØ;.k ÅtkZ izkIr dj ldsA

(D) ,d mRizsjd] U;wu lfØ;.k ÅtkZ ds lkFk ,d oSdfYid vfHkfØ;k iFk miyC/k djkrk gSA

2. Which of the following are correct in acidic medium(A) The oxidising power of KMnO4 will increase with increase in pH(B) The oxidising power of KMnO4 will decrease with increase in pH(C) The oxidising power of K2Cr2O7 will increase with increase in pH(D) The oxidising power of K2Cr2O7 will decrease with increase in pH

vEyh; ek/;e esa fuEu esa ls dkSuls fodYi lgh gS&

(A) KMnO4 dh vkWDlhdkjd lkeF;Z pH c<kus ds lkFk c<sxh

(B) KMnO4 dh vkWDlhdkjd lkeF;Z pH c<kus ds lkFk ?kVsxh

(C) K2Cr2O7 dh vkWDlhdkjd lkeF;Z pH c<kus ds lkFk c<sxh

(D) K2Cr2O7 dh vkWDlhdkjd lkeF;Z pH c<kus ds lkFk ?kVsxh


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LTS-18/38


3. The diatomic molecule(s) that have two p-bonds.f}ijekf.o; v.kq ftlesa nks p-ca/k mifLFkr gSa&(A) B2 (B) C2 (C) N2 (D) O2

4. Which of the following statements are NOT correct ?(A) Copper is extracted by self reduction method.(B) Cast iron is the purest form of iron.(C) The composition of malachite ore is Cu(OH)2.2CuCO3(D) Cupellation process is used for the refining of AgfuEu esa ls dkSuls dFku lgh ugha gS\

(A) dkWij dks Lovip;u fof/k }kjk fu"df"kZr fd;k tkrk gSA

(B) <yok¡ yksgk (Cast iron)] yksgs dk 'kq¼re :i gSA

(C) esysdkbZV v;Ld dk la?kVu Cu(OH)2.2CuCO3 gSA

(D) D;wisyhdj.k izØe dk iz;ksx Ag ds 'kqf¼dj.k ds fy, fd;k tkrk gSA5. Which of the following complex/ion is/are paramagnetic.

fuEu esa ls dkSuls ladqy@vk;u vuqpqEcdh; gS@gSa\(A) Ni(CO)4 (B) [MnCl4]

2– (C) K4[Fe(CN)6] (D) [CoCl4]2–

0999DJA110318018


LTS-19/38


6. Identify correct order of heat of combustion ?ngu dh Å"ek dk lgh Øe crkb;s\

(A) > (B) >

(C) > (D) <

7. Identify compound(s) which contains centre of symmetry ?,sls ;kSfxd crkb;s ftuesa lefefr dk dsUnz mifLFkr gksrk gS&

(A)

COOHOH

HHO

H

COOH

(B)

Cl

Cl

(C) O

OCl

Cl

(D)

Cl

Cl

0999DJA110318018


LTS-20/38






� bl [k.M esa lqesy izdkj ds N% iz'u gSaA

� bl [k.M es a nks Vscy gSa (izR;sd Vscy esa 3 dkye vkSj 4 iafDr;k¡ gSa)

� Vscy ij vk/kkfjr rhu iz'u gSaA

� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa flQZ ,d fodYi lgh gSaA

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

Answer Q.8, Q.9 and Q.10 by appropriately matching the information given in thethree columns of the following table. Column-I Column-II Column-III(A) Combustion of butane (P) Entropy change of (I) Low temperature

gas at S.T.P. universe constant conditions favourable(B) Dissolution of sodium (Q) Entropy change of (II) More spontaneous at

chloride in water at STP system positive high temperature(C) Adding carbon dioxide (R) Enthalpy change of (III) High pressure condition

gas to water to form system negative favourablecarbonated water at highpressure and 298K

(D) Vaporisation of liquid (S) Entropy change of (IV) More spontaneous atwater at 1 atm and 373K universe positive low pressure

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LTS-21/38


8. Which of following option is incorrect -(A) A – S – I (B) B – Q – IV (C) C – R – I (D) D – P – IV

9. In column-I (A) if the reaction was carried at 400 K which option would be correct -(A) A – P – II (B) A – Q – III (C) A – S – IV (D) A – R – III

10. In column-I (D), if vaporisation is carried out at 15 atm which option is correct -(A) D – P – IV (B) D – S – II (C) D – R – I (D) D – Q – II

0999DJA110318018


LTS-22/38


uhps nh x;h Vscy ds rhu dkyeksa esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10ds mÙkj nhft;sA

dkWye-I dkWye-II dkWye-III(A) S.T.P. ij C;wVsu xSl dk (P) czãk.M dh , aVªkWih (I) fuEu rki dh vuqdwy

ngu ifjorZu fu;r (fLFkj) gS ifjfLFkfr gSA(B) STP ij ty esa (Q) ra= dh ,aVªkWih ifjorZu (II) mPprki ij vf/kd

lksfM;e DyksjkbM dk ?kqyuk èkukRed gSA Lor% izofrZr gSA(C) mPp nkc o 298 K ij ty esa (R) ra= dh , aFkSYih ifjorZu (III) mPp nkc dh vuqdwy

CO2 feykdj dkcksZfuÏr ty ½.kkRed gSA ifjfLFkfr gSAdk cuuk

(D) 1 atm rFkk 373K ij æo ty (S) czãkaM dh ,aVªkWih ifjorZu (IV) fuEu nkc ij vf/kddk ok"iu èkukRed gSA Lor% izofrZr gSA

8. fuEu fodYi esa dkSulk xyr gS\(A) A – S – I (B) B – Q – IV (C) C – R – I (D) D – P – IV

9. dkWye-I (A) esa ;fn vfHkfØ;k 400 K ij djkbZ tk, rks dkSulk fodYi lgh gksxk\(A) A – P – II (B) A – Q – III (C) A – S – IV (D) A – R – III

10. dkWye-I (D) esa] ;fn ok"iu 15 atm ij djk;k tk, rks dkSulk fodYi lgh gksxk\(A) D – P – IV (B) D – S – II (C) D – R – I (D) D – Q – II

0999DJA110318018


LTS-23/38


Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following table.

Column-I Column-II

(A)

(B)

(C)

(D)

(P)

(Q)

(R)

(S)

Column-III

(i)

(ii)

(iii)

(iv)

Carbocation

Benzyne

Carbanion

NaEther

Zn Free radical

Cl

CH – CH 2 2

Cl Cl

Cl NaNH2NH3

HCH – C – OH 3

CH3

CH3

D

+

CH – CH 3 2

NH 2

11. Correct set for A :(A) P,Q, ii, iii (B) Q,R, ii, iii (C) S, iii, ii, iv (D) P, i, iii, iv

12. Correct set for B :(A) Q, R, ii, iii (B) P, Q, ii, iii (C) S, iii, ii, iv (D) P, i, iii, iv

13. Correct set for D(A) P, i (B) Q, i (C) R, iii (D) S, iv

0999DJA110318018


LTS-24/38

uhps nh x;h Vscy ds rhu dkyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uk s a Q.11, Q.12 vkSjQ.13 ds mÙkj nhft;sA

(A)

(B)

(C)

(D)

(P)

(Q)

(R)

(S)

(i)

(ii)

(iii)

(iv)

dkcZ/kuk;u

csUtkbZu

dkcZ½.kk;u

NaEther

Zn eqDr ewyd

Cl

CH – CH 2 2

Cl Cl

Cl NaNH2NH3

HCH – C – OH 3

CH3

CH3

D

+

CH – CH 3 2

NH 2

11. A dk lgh leqPp; gS:(A) P,Q, ii, iii (B) Q,R, ii, iii (C) S, iii, ii, iv (D) P, i, iii, iv

12. B dk lgh leqPp; gS :(A) Q, R, ii, iii (B) P, Q, ii, iii (C) S, iii, ii, iv (D) P, i, iii, iv

13. D dk lgh leqPp; gS %¥(A) P, i (B) Q, i (C) R, iii (D) S, iv


0999DJA110318018


LTS-25/38








� bl [k.M esa ik¡p 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 : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. An impure sample of CaCO3 contains 38% of Ca. The percentage of impurity present in thesample is :

CaCO3 ds ,d v'kq¼ uewus esa 38% Ca gSA uewus esa mifLFkr v'kq¼rk dk izfr'kr gS

2. Find out the number of planar speciesClF3, I3

+, I3–, XeF4, XeF5

–, I2Cl6, SO2Cl2, SF4

fuEu esa ls leryh; Lih'kht dh la[;k crkb;sA

ClF3, I3+, I3

–, XeF4, XeF5–, I2Cl6, SO2Cl2, SF4

0999DJA110318018


LTS-26/38

3. Among the following, total number of possible metals present in anode mud, which is obtainedby electrolytic refining of copper.Fe, Ag, Au, Pt, Pb, ZnfuEu esa ls] ,uksM eM esa mifLFkr , slh lEHkkfor /kkrqvksa dh dqy la[;k crkb;s ftUgsa dkWij ds oS|qr vi?kVuh; 'kqf¼dj.k

}kjk izkIr fd;k tkrk gSAFe, Ag, Au, Pt, Pb, Zn

4. Find the number of chemical species which are paramagnetic in nature.O3, NO2, KO2, [PtCl4]2–, N2, Na2[Fe(CN)5NO]fuEu esa ls ,slh jklk;fud Lih'kht dh la[;k crkb;s tks vuqpqEcdh; izd`fr dh gSaAO3, NO2, KO2, [PtCl4]2–, N2, Na2[Fe(CN)5NO]

5.

OH

Conc. H SO2 4

Identify total number of 1,2-shift for above reaction ?OH

Conc. H SO2 4

mijksDr vfHkfØ;k ds fy;s 1,2-f'k¶V dh dqy la[;k crkb;sA


0999DJA110318018


LTS-27/38

PART-3 : MATHEMATICS Hkkx-3 : xf.krSECTION–I(i) : (Maximum Marks : 28)

[k.M–I(i) : (vf/kdre vad : 28)� This section contains SEVEN 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,


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 lkr 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 (fodYiksa ) dks pquk x;k gks ;k u pquk x;k gksA

0999DJA110318018


LTS-28/38

1. Consider a twice derivable function ƒ (x) such that ƒ (1) = 3, ƒ (2) = 6 & ƒ (3) = 9, then(A) ƒ (x) = 8 atleast once in (1, 3) (B) ƒ '(x) = 3 atleast twice in (1, 3)(C) ƒ ''(x) = 0 atleast once in (1, 3) (D) ƒ ''(x) = 0 nowhere in (1, 3)ekuk nks ckj vodyuh; Qyu ƒ (x) bl izdkj gS fd ƒ (1) = 3, ƒ (2) = 6 rFkk ƒ (3) = 9 gS] rks&

(A) (1,3) esa de ls de ,d ckj ƒ (x) = 8 gksxk (B) (1, 3) esa de ls de nks ckj ƒ '(x) = 3 gksxk

(C) (1, 3) esa de ls de ,d ckj ƒ ''(x) = 0 gksxk (D) (1, 3) esa dgha Hkh ƒ ''(x) = 0 ugha gksxk2. The complex number z, which satisfies the equations az3 + bz2 + g = 3, where |a|£ 1, |b| £ 1,

|g|£ 1 and |z| ¹ 1(A) Always lies inside the circle |z| = 1 (B) Lies either outside or on the circle |z| = 1(C) Must lie outside the circle |z| = 1 (D) Can lies insdie the circle |z| = 2lfEeJ la[;k z tks lehdj.k az3 + bz2 + g = 3 dks lUrq"V djrh gS ] tgk¡ |a|£ 1, |b| £ 1, |g|£ 1 rFkk|z| ¹ 1(A) lnSo o`Ùk |z| = 1 ds vUnj fLFkr gksxh (B) ;k rks o`Ùk|z| = 1 ds ckgj ;k mGij fLFkr gksxh

(C) o`Ùk |z| = 1 ds ckgj fLFkr gksuh pkfg, (D) o`Ùk|z| = 2 ds vUnj fLFkr gks ldrh gS

3. f(x) is continuous in [0, 1] and satisfies f(0) = 1

f2æ öç ÷è ø

= f(1). Then identify the correct statement

(A) f(x) = 1f x

4æ ö+ç ÷è ø

has atleast 2 solutions in [0, 1]

(B) f(x) = 1f x

4æ ö+ç ÷è ø

has atleast 1 solutions in 1

0 ,4

é ùê úë û

(C) f(x) = 1f x

4æ ö+ç ÷è ø

has atleast 1 solutions in 1 3,

2 4é ùê úë û

(D) f(x) = x has atleast 1 solution in [0, 1]Space for Rough Work / dPps dk;Z ds fy, LFkku

0999DJA110318018


LTS-29/38


Qyu f(x) vUrjky [0, 1] esa larr~ gS] rFkk f(0) = 1f2æ öç ÷è ø

= f(1) dks laUrq"V djrk gSA rc lR; dFku igpkfu,

(A) f(x) = 1f x

4æ ö+ç ÷è ø

dk vUrjky [0, 1] esa de ls de nks gy gSA

(B) f(x) = 1f x

4æ ö+ç ÷è ø

dk vUrjky 1

0 ,4

é ùê úë û

esa de ls de ,d gy gSA

(C) f(x) = 1f x

4æ ö+ç ÷è ø

dk vUrjky 1 3,

2 4é ùê úë û

esa de ls de ,d gy gSA

(D) f(x) = x dk vUrjky [0, 1] esa de ls de ,d gy gSA

4. If x

3

2

ƒ(x) (sin t )dt= ò & 4x

3

5

(sin t )dtl = ò , then -

;fn x

3

2

ƒ(x) (sin t )dt= ò rFkk 4x

3

5

(sin t )dtl = ò , rks -

(A) l = ƒ (x4) + ƒ (5) (B) l = ƒ (x4) – ƒ (5) (C) l = x4 – ƒ (5) (D) l = ƒ (5) – ƒ (x4)5. If sides of a triangle ABC satisfy a relation 4a2 + 9b2 + c2 = 2a(a + 3b + c), then -

(A) the triangle is isosceles (B) the triangle is obtuse angled.

(C) 1 17B cos18

- æ öÐ = ç ÷è ø

(D) 1 35A sin

6-æ ö

Ð = ç ÷ç ÷è ø

;fn f=Hkqt ABC dh Hkqtk;sa] lEcU/k 4a2 + 9b2 + c2 = 2a(a + 3b + c) dks lUrq"V djrh gS] rks -(A) f=Hkqt lef}ckgq gksxk (B) f=Hkqt vf/kd dks.k dk gksxk

(C) 1 17B cos18

- æ öÐ = ç ÷è ø

(D) 1 35A sin

6-æ ö

Ð = ç ÷ç ÷è ø

0999DJA110318018


LTS-30/38

6. If 1/ x

3

x 0

ƒ(x)lim 1 x e

x+®

é ù+ + =ê úë û

, then -

;fn 1/ x

3

x 0

ƒ(x)lim 1 x e

x+®

é ù+ + =ê úë û

gS] rks &

(A) 2x 0

ƒ(x)lim 2

x+®= (B)

x

2x 0

ƒ(x)lim 1 0

tan x+®

é ù+ =ê úë û

(C) x

2

x

1lim 1 xƒ e

x®¥

é ùæ ö+ =ê úç ÷è øë û(D) +®

é ùê ú+ + =ê úë û

ƒ (x)x

x 0lim 1 ( 1) 2

7. If one vertex of an equilateral triangle of side 'a' lies at the origin and the other lies on theline x - 3 y = 0 then the co-ordinates of the third vertex are :

;fn Hkqtk a okys leckgq f=Hkqt dk ,d 'kh"kZ ewyfcUnq ij gS rFkk nwljk 'kh"kZ js[kk x - 3 y = 0 ij gks] rks rhljs 'kh"kZ dsfunsZ'kkad gksaxs :

(A) (0, a) (B) 3 a a,

2 2

æ ö-ç ÷ç ÷

è ø(C) (0, - a) (D)

3a a,

2 2

æ ö-ç ÷ç ÷è ø


0999DJA110318018


LTS-31/38






� bl [k.M esa lqesy izdkj ds N% iz'u gSaA� bl [k.M es a nks Vscy gSa (izR;sd Vscy esa 3 dkye vkSj 4 iafDr;k¡ gSa)� Vscy ij vk/kkfjr rhu iz'u gSaA� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa flQZ ,d fodYi lgh gSaA� izR;sd iz'u ds fy, vks- vkj- ,l- ij lgh mÙkj 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 esaAAnswer Q.8, Q.9 and Q.10 by appropriately matching the information given in thethree columns of the following table.uhps nh x;h Vscy ds rhu dkyeksa esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10ds mÙkj nhft;sAz1, z2, z3, ....., zn are roots of equation zn – 1 = 0 ...(1)Column-1 Value of 'n' for equation (1)Column-2 Minimum value of |zi – zj| : i, j Î {(1, 2), ....., n}Column-3 Area of polygon formed by z1, z2, ....., zn

z1, z2, z3, ....., zn lehdj.k zn – 1 = 0 ds ewy gSaA ...(1)dkWye-1 esa lehdj.k (1) ds fy, n ds eku gSaA

dkWye-2, esa |zi – zj| : i, j Î {(1, 2), ....., n} ds U;wure eku gSaA

dkWye-3, esa z1, z2, ....., zn }kjk fufeZr cgqHkqt ds {ks=Qy gSaA

Column-1/dkWye-1 Column-2/dkWye-2 Column-3/dkWye-3

(I) 4 (i) -2 2 (P) 3 32

(II) 6 (ii) 2 (Q) 2

(III) 18 (iii) -2 3 (R) 2 2(IV) 12 (iv) 1 (S) 3

0999DJA110318018


LTS-32/38


8. Which of the following options is the only CORRECT combination ?

fuEu esa ls dkSu lk fodYi dsoy lgh la;kstu gS ?

(A) (I) (i) (R) (B) (I) (ii) (Q) (C) (I) (iv) (S) (D) (II) (ii) (Q)9. Which of the following options is the only CORRECT combination ?


(A) (II) (iv) (P) (B) (II) (iv) (S) (C) (III) (i) (P) (D) (III) (ii) (R)10. Which of the following options is the only CORRECT combination ?


(A) (III) (iv) (Q) (B) (IV) (iv) (S) (C) (IV) (iii) (S) (D) (IV) (iv) (P)

0999DJA110318018


LTS-33/38


Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following table.uhps nh x;h Vscy ds rhu dkyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uk s a Q.11, Q.12 vkSjQ.13 ds mÙkj nhft;sAColumn-1 contains same differential equations with boundary conditionsColumn-2 contains values of area enclosed by function represented by them between ordinates

0 and 1 (in sq. units)Column-3 represents value of derivative of function represented by these equation at x = 2.dkWye-1 esa izfrcU/k ds lkFk leku vodyuh; lehdj.k gSA

dkWye-2 esa iznf'kZr Qyuksa ds }kjk dksfV 0 rFkk 1 ds e/; ifjc¼ {ks=Qy dk eku (oxZ bdkbZ esa) dkWye-2 esa gSA

dkWye-3 esa iznf'kZr Qyuksa ds x = 2 ij vodyt dk eku dkWye-3 esa gSA

Column-1/dkWye-1 Column-2/dkWye-2 Column-3/dkWye-3

(I) xdy + 2ydx = x(x – 3)dx (i)5

12 (P) –4

y(0) = 4

(II)2

22

d y dy(x 1) 2xdxdx

+ = (ii) ln2 (Q)625

-

y'(0) = 3, y(0) = 1(III) y(1 – x2)dx = x(x2 + 1)dy (iii) 2ln2 (R) 0

y(1) = 1, (x Î (0,¥))

(IV) ( )2dy y y , y 1 1

dx x xæ ö æ ö= - =ç ÷ç ÷

è øè ø(iv)

114 (S) 15

11. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSu lk fodYi dsoy lgh la;kstu gS ?(A) (I) (i) (P) (B) (II) (iii) (P) (C) (II) (iv) (S) (D) (I) (ii) (S)

12. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSu lk fodYi dsoy lgh la;kstu gS ?(A) (III) (ii) (Q) (B) (IV) (ii) (R) (C) (I) (iv) (S) (D) (III) (i) (P)

13. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSu lk fodYi dsoy lgh la;kstu gS ?(A) (II) (i) (S) (B) (I) (iv) (P) (C) (II) (iii) (R) (D) (IV) (ii) (Q)

0999DJA110318018


LTS-34/38








� bl [k.M esa ik¡p 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 : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

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

1. If 2

2

x x 1y (x R)

x x 1- +

= Î+ +

& z = y2 + 2y + 3. Let l1 be the maximum value of z & l2 be the minimum

value of z, then |2l1 – 9l2|is equal to

;fn 2

2

x x 1y (x R)

x x 1- +

= Î+ +

rFkk z = y2 + 2y + 3 gSA ekuk l1, z dk vf/kdre eku rFkk l2, z dk U;wure eku gks] rks

|2l1 – 9l2| dk eku gksxk

2. If 1 2 3

A 5

6 7 9

é ùê ú= a bê úê ú-ë û

satisfies x3 – 24x – 72 = 0, then 3a + b – 425 is

;fn 1 2 3

A 5

6 7 9

é ùê ú= a bê úê ú-ë û

lehdj.k x3 – 24x – 72 = 0 dks lUrq"V djrk gks] rks 3a + b – 425 dk eku gksxk

0999DJA110318018


LTS-35/38


3. The three planes 9x – 3y + 4z = 3, x + 2y + 2z = 5 and 21x + ay + bz = 3 intersect at a point Aon XOY plane and at a point B on XOZ plane, where O is the origin. If the area of DAOB is

k, then a bk

2+æ ö+ ç ÷

è ø is equal to

rhu lery 9x – 3y + 4z = 3, x + 2y + 2z = 5 rFkk 21x + ay + bz = 3, XOY lery dks fcUnq A ij] XOZ

lery dks fcUnq B ij izfrPNsn djrs gS] tgk¡ O ewy fcUnq gSA ;fn DAOB dk {ks=Qy k gS] rks a bk

2+æ ö+ ç ÷

è ø dk

eku gksxk

4. Three normals are drawn from P(a, b) to y2 = 4x & their equation are y = ax, y + bx = 3 & y = x – 3,then a + b is equal to

P(a, b) ls y2 = 4x ij rhu vfHkyEc [khaps tkrs gaS rFkk mudh lehdj.k y = ax, y + bx = 3 rFkk y = x – 3 gS] rks

a + b gksxk

0999DJA110318018


LTS-36/38

5. f(x) is a function defined in [–1, 1] such that +®

=x 0lim f(x) 2 and

-®=

x 0lim f(x) 3 . Find the value of l

for which ® ®

é ù é ù- = l -ë û ë û4 2 4 6

x 0 x 02 lim f(x x ) lim f(x x ) .

f(x), vUrjky [–1, 1] esa bl izdkj ifjHkkf"kr gS fd +®

=x 0lim f(x) 2 rFkk

-®=

x 0lim f(x) 3 gSA rc l dk eku gksxk ftlds

fy, ® ®

é ù é ù- = l -ë û ë û4 2 4 6

x 0 x 02 lim f(x x ) lim f(x x ) gksA



LTS-37/380999DJA110318018

NURTURE PH-I English 1001 03/04/2019 ASAT/TALLENTEX

NURTURE PH-II Eng/Hindi 1002 15/04/2019 ASAT/TALLENTEX

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ENTHUSIAST PH-II Eng/Hindi 1022 01/04/2019 Direct

ENTHUSIAST PH-III Eng/Hindi 1023 08/05/2019 Direct

LEADER PH-I Eng/Hindi 1031 24/04/2019 Through ASAT

LEADER PH-II Eng/Hindi 1032 08/05/2019 Through ASAT

LEADER PH-III Eng/Hindi 1033 29/05/2019 Through ASAT

NURTURE PH-I Eng/Hindi 2001 01/04/2019 Direct

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ENTHUSIAST PH-I Eng/Hindi 2021 01/04/2019 Direct

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NURTURE PH-I English 3001 07/04/2019 Direct

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LEADER PH-I English/Hindi 3031 08/04/2019 Direct

LEADER PH-II English/Hindi 3032 29/04/2019 Direct

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ACHIEVER PH-I English/Hindi 3051 15/05/2019 Direct

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For JEE (Adv.) JEE (Main) NEET AIIMS NTSE OLYMPIADS PNCF Class 6 to 12th th

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Your Target is to secure Good Rank in JEE 2019 0999DJA110318018LTS-38/38

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

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eSaus lHkh funs Z'kks a dks i<+ fy;k gS 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 rjg tk¡pfy;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 .............................................


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, provided bubbles is cases

I(i) ,dy ;k ,d ls 7 option(s) is(are) NO incorrect option darkened vU; lHkh 28vf/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 gSMatching +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(ii) lqpha lqesyu 6 is darkened — darkened vU; lHkh 18izdkj ;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 gS

+3 0Single digit If only the bubble In all

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

III ,dy vadh; 5 is darkened — vU; lHkh — 15iw.kk±d ;fn flQZ lgh mÙkj ds ifjfLFkfr;ksa(0-9) vuq:Ik cqycqys dks esa

dkyk fd;k gS

Dear Student,

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