(acade ses : 2015 - 2016) leader test series / joint

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 instructed to do so.

;g eksgjcU/k iqfLrdk 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 vkWIVhdy 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 and form number in the space provided on the back cover of this booklet.

,d iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke rFkk QkWeZ uEcj fyf[k,A5. After breaking the seal of the booklet, verify that the booklet contains 40 pages and all the 28 questions in each

subject and along with the options are legible.bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ys fd blesa 40 i`"B gSa vkSj vkSj izR;sd fo"k; ds lHkh 28 iz'u vkSj muds mÙkj fodYi

Bhd ls i<+ s tk ldrs gSA

QUESTION PAPER FORMAT AND MARKING SCHEME / iz'ui= dk izk:i vkSj vadu ;kstuk :6. The question paper has three parts : Physics, Chemistry and Mathematics. Each part has two sections.

bl iz'ui= esa rhu Hkkx gSa % HkkSfrd foKku] jlk;u foKku vkSj xf.krA gj Hkkx esa nks [k.M gSaA7. Carefully read the instructions given at the beginning of each section.

izR;sd [k.M ds izkjEHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+ sA

8. Section-I / [k.M-I :(i) Section-I(i) contains 8 multiple choice questions with only one correct option.

Marking scheme : +3 for correct answer, 0 if not attempted and –1 in all other cases.[k.M-I(i) esa 8 cgqfodYih; iz'u gSA ftuds dsoy ,d fodYi lgh gSaAvad ;kstuk : +3 lgh mÙkj ds fy,] 0 iz;kl ugha djus ij rFkk –1 vU; lHkh voLFkkvksa esaA

(ii) Section-I(ii) contains 5 multiple choice questions with one or more than one correct option.Marking scheme : +3 for correct answer and 0 in all other cases.[k.M-I(ii) esa 5 cgqfodYih; iz'u gSA ftuds ,d ;k ,d ls vf/kd fodYi lgh gSaA

vad ;kstuk : +3 lgh mÙkj ds fy, rFkk 0 vU; lHkh voLFkkvksa esaA(iii) Section-I(iii) contains 2 ‘paragraph’ type questions. Each paragraph describes an experiment, a situation or a

problem. Three multiple choice questions will be asked based on first paragraph and Two multiple choicequestions will be asked based on second paragraph. only one correct option.Marking scheme : +3 for correct answer, 0 if not attempted and –1 in all other cases.[k.M-I(iii) esa 2 ‘vuqPNsn’ izk:i iz'u gSA izR;sd vuqPNsn ,d iz;ksx] ,d n'kk vFkok ,d leL;k dks n'kkZrk gSA izFke vuqPNsn ij rhu

cgqfodfYi; iz'u iwNs tk;xs rFkk nwljs vuqPNsn ij nks cgqfodfYi; iz'u iwNs tk;xsA dsoy ,d fodYi lgh gSaA

vad ;kstuk : +3 lgh mÙkj ds fy,] 0 iz;kl ugha djus ij rFkk –1 vU; lHkh voLFkkvksa esaA

9. There is no questions in SECTION-II & III / [k.M–II o III esa ,d Hkh iz'u ugha gS10. Section-IV contains 10 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive)Marking scheme : +3 for correct answer and 0 in all other cases.[k.M-IV esa 10 iz'u gSaA izR;sd iz'u dk 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSAvad ;kstuk : +3 lgh mÙkj ds fy, rFkk 0 vU; lHkh voLFkkvksa esaA

Test Type : UNIT TEST TEST # 03 Test Pattern : JEE-AdvancedTEST DATE : 09 - 08 - 2015

LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : JEE (Main + Advanced) 2016

Paper Code : 0999DT110315003

DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)

Please see the last page of this booklet for rest of the instructions / Ïi;k 'k s"k funs Z'kk s a ds fy, bl iq fLrdk ds vfUre i`"B dks i<+ s a A

Time : 3 Hours Maximum Marks : 252

DO

NOT

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THE

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LS W

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UT

BEIN

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STRU

CTED

TO

DO

SO B

Y TH

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LATO

R / f

ujh{

kd d

s vuqn

s'kksa d

s fcuk

eqgj

sa u r

ksM+s

LTS-2/40

Target : JEE (Main + Advanced) 2016/09-08-2015

0999DT110315003

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

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17, Br = 35,

Xe = 54, Ce = 58,

Atomic masse s : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

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

· Coulomb's law constant pe9

0

1 = 9×104

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

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

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

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

· Permittivity of vacuum Î0 = 20

1cm

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

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)

Leader Test Series/Joint Package Course/09-08-2015

PHYS

ICS

LTS-3/400999DT110315003

PART-1 : PHYSICS Hkkx-1 : HkkSfrd foKku

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

� This section contains EIGHT questions.� Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option(s) is correct.� For each question, darken the bubble corresponding to the correct option in the ORS� Marking scheme :

+3 If the bubble corresponding to the correct option is darkened0 If none of the bubbles is darkened–1 In all other cases

� bl [k.M esa vkB iz'u gSa� izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSaA� izR;sd iz'u ds fy,] lgh fodYi ds vuq:i cqycqys dks vks-vkj-,l- esa dkyk djsaA� vadu ;kstuk :

+3 ;fn lgh fodYi ds vuq:i cqycqys dks dkyk fd;k tk;0 ;fn dksbZ Hkh cqycqyk dkyk u fd;k gks–1 vU; lHkh voLFkkvksa esa

1. Electric field at the center of a charged parallel plate capacitor having uniform surface charge density s onfacing surface is and –s on the other facing surface is

,d vkosf'kr lekUrj iê la/kkfj= dh ,d lrg ij le:i i`"Bh; vkos'k ?kuRo s rFkk nwljh lEeq[k lrg ij –s gSAblds dsUæ ij fo|qr {ks= Kkr dhft;s %&

(A) 04

se (B)

0es

(C) 02e

s(D) zero ('kwU;)

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


LTS-4/40

PHYS

ICS


0999DT110315003


2. The block shown in the figure is equilibrium. Find acceleration of the block just after the string burns.

(A) 35g

(B) 45g

(C) 43g

(D) None of these

53°

string

fp= esa iznf'kZr CykWd lkE;koLFkk esa gSA jLlh dks tykus ds ckn CykWd dk Roj.k gksxk

(A) 35g

(B) 45g

(C) 43g

(D) buesa ls dksbZ ugha

3. Two identical capacitors 1 and 2 are connected in series. The capacitor 2 contains a dielectric slab of constantK as shown. They are connected to a battery of emf V0 volts. The dielectric slab is then removed. Let Q1 andQ2 be the charge stored in the capacitors before removing the slab and Q1¢, and Q2¢ be the values afterremoving the slab. Thennks ,dleku la/kkfj= 1 o 2 Js.khØe esa tqM+s gSaA la/kkfj= 2 esa ijkoS|qrkad K okyh ,d ifV~Vdk dks fp=kuqlkj j[kk x;k

gSA la/kkfj=ksa dks V0 fo|qr okgd cy okyh cSVjh ls tksM+rs gSaA vc ijkoS|qr ifV~Vdk dks fudky ysrs gSaA ekuk ifV~Vdk

dks fudkyus ls igys la/kkfj=ksa esa lafpr vkos'k Q1 o Q2 o ifV~Vdk dks fudkyus ds ckn vkos'k Q1¢ o Q2¢ gS rks

(A) ÷øö

çèæ +

=K

1KQ

'Q

1

1(B) 2

)1K(Q

'Q

2

2 += (C) K2

1KQ

'Q

2

2 += (D) 2

KQ

'Q

1

1 =


PHYS

ICS

LTS-5/400999DT110315003


4. A particle of mass 1 kg is acted upon by a force 'F' which varies as shown in the figure. If initial velocity ofthe particle is 10 ms–1, the maximum velocity attained by the particle during the period isæO;eku 1 kg okys d.k ij ,d cy 'F' yxk;k tkrk gS] tks fp=kuqlkj ifjofrZr gksrk gSA ;fn d.k dk izkjfEHkd osx

10 ms–1 gks rks bl nkSjku d.k }kjk izkIr vfèkdre osx gksxk %&

(A) 210 ms–1 (B) 110 ms–1 (C) 100 ms–1 (D) 90 ms–1

5. In the circuit shown, the energy stored in 1mF capacitor isiznf'kZr ifjiFk esa 1mF la/kkfj= esa lafpr ÅtkZ gksxh %&

(A) 40 mJ (B) 64 mJ (C) 32 mJ (D) none (dksbZ ugha)6. For the circuit shown, which of the following statements is true?

(A) with S1 closed, V1 = 15 V, V2 = 20 V (B) with S3 closed, V1 = V2 = 25 V

(C) with S1 and S2 closed, V1=V2 = 0 (D) with S1 and S2 closed, V1 = 30V, V2 = 20V

iznf'kZr ifjiFk ds fy;s fuEu esa ls dkSuls dFku lR; gSa\

(A) S1 cUn djus ij V1 = 15 V, V2 = 20 V (B) S3 cUn djus ij V1 = V2 = 25 V(C) S1 rFkk S2 cUn djus ij V1=V2 = 0 (D) S1 rFkk S2 cUn djus ij V1 = 30V, V2 = 20V

LTS-6/40

PHYS

ICS


0999DT110315003


7. A child of mass 30 kg was sitting on a snow cart of mass 10 kg. The friction between the cart and thesnow was negligible. If a force of 120 N was applied to the child, as shown in the figure. The minimumcoefficient of friction required between the child and cart to keep the child from slipping off isæO;eku 30 kg okyk ,d yM+dk 10 kg æO;eku okyh cQZ dh ,d xkM+h ij cSBk gqvk gSA xkM+h rFkk Q'kZ ds e/; ?k"kZ.k

ux.; gSA bl yM+ds ij 120 N dk cy fp=kuqlkj yxk;k tkrk gSA yM+dk fQlyus ls cp tk;s blds fy;s yM+ds rFkk

xkM+h ds e/; vko';d U;wure ?k"kZ.k xq.kkad gksxk%&

120N

ICE

Cart

(A) 0.1 (B) 0.2 (C) 0.4 (D) 0.38. Find the potential difference across point a and b in fig.:-

fp= esa fcUnq a o b ds e/; foHkokUrj dk eku Kkr dhft, :-e1

e2

C1 C2

a

b

(A) 1 2

1 2C Cæ öe + eç ÷+è ø

C1 (B) 1 2

1 2C Cæ öe + eç ÷+è ø

C2 (C) 1 2

1 2C Cæ öe - eç ÷+è ø

C1 (D) 1 2

1 2C Cæ öe - eç ÷+è ø

C2


PHYS

ICS

LTS-7/400999DT110315003

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

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

option(s) is (are) correct.� For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS� Marking scheme :

+3 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened0 In all other cases

� bl [k.M esa ik¡p iz'u gSa� izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi

lgh gSaA� izR;sd iz'u esa] lHkh lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks vks-vkj-,l- esa dkyk djsaA� vadu ;kstuk :

+3 ;fn flQZ lHkh lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k tk;0 vU; lHkh voLFkkvksa esa

9. A parallel plate capacitor of area A and separation d is charged to potential difference V and removedfrom the charging source. A dielectric slab of constant K = 2, thickness d and area A/2 is inserted, asshown in the figure. Let s1 be free charge density at the conductor-dielectric surface and s2 be thecharge density at the conductor-vacuum surface.(A) The electric field has the same value inside the dielectric as that in the free space between the plates.

(B) The ratio 1

2

ss is equal to

21 .

(C) The new capacitance is 03 A2dÎ

.

(D) The new potential difference between the plates is 23 V..

d

A

Ks2

– s2– s1

s1

{ks=Qy A rFkk d nwjh okys ,d lekUrj iV~V la/kkfj= dks V foHkokUrj rd vkosf'kr dj] vc vkos'ku lzksr dks gVk fy;k

tkrk gSA vc ,d ijkoS|qr ifV~Vdk ftldk ijkoS|qrkad K = 2, eksVkbZ d rFkk {ks=Qy A/2 gS] dks fp=kuqlkj bldh IysVksa

ds e/; izfo"V djk;k tkrk gSA ekuk pkyd&ijkoS|qr lrg ij eqDr vkos'k ?kuRo s1 rFkk pkyd&fuokZr~ lrg ij vkos'k

?kuRo s2 gSA

(A) ijkoS|qr esa fo|qr {ks= dk eku] IysVksa ds e/; fjDr LFkku esa fo|qr {ks= ds cjkcj gh gSA

(B) vuqikr 1

2

ss dk eku

21 gSA

(C) ubZ /kkfjrk 03 A2dÎ

gSA

(D) IysVksa ds e/; u;k foHkokUrj 23 V gSA

LTS-8/40

PHYS

ICS


0999DT110315003


10. Two blocks having same mass are placed on rough incline plane and the coefficient of friction between

A and incline is µ1 = 1.0 and between block B and incline is µ2 = 34 . As the inclination of the plane ‘q’

with respect to horizontal increases, choose the correct answer (s).(A) There is no contact force between block A and B if q £ 37°(B) There is no contact force between block A and B if q £ 45°(C) As µ1 > µ2 both the blocks move together

(D) They starts moving at an angle q = 1 7tan8

-

B A

q

leku æO;eku ds nks CykWd ,d [kqjnjs urry ij j[ks gSaA A o urry ds e/; ?k"kZ.k xq.kkad µ1 = 1.0 o B rFkk urry ds

e/; ?k"kZ.k xq.kkad µ2 =34 gSA tc {kSfrt ds lkis{k vkur dks.k ‘q’ dks c<+krs gSa rks lgh dFku@dFkuksa dks pqfu;sA

(A) q £ 37° gksus ij CykWd A o B ds e/; dksbZ laidZ cy ugha yxsxkA(B) q £ 45° gksus ij CykWd A o B ds e/; dksbZ laidZ cy ugha yxsxkA(C) pwafd µ1 > µ2 gS vr% nksuksa CykWd lkFk&lkFk xfr djrs gSaA

(D) ;s q = 1 7tan8

- ij xfr djuk izkjEHk dj nsxsaA

11. In the connection shown in the figure the switch K is open and the capacitor is uncharged. Then we close theswitch and let the capacitor charge upto the maximum and open the switch again. The values indicated by thegalvanometer.(A) increases after closing the switch and then decreases after opening the switch(B) direction of current is same through the galvanometer just before and after opening the switch again(C) after opening the switch again energy dissipation in R1 is zero(D) after opening the switch again energy dissipation in R2 is zero

iznf'kZr ifjiFk esa fLop K [kqyk gS rFkk la/kkfj= vukosf'kr gSA vc fLop dks can dj nsrs gSa rFkk la/kkfj= vf/kdre ekurd vkosf'kr gksrk gS vc fLop dks iqu% [kksy nsrs gSaA xsYosuksehVj esa iznf'kZr eku %&(A) fLop dks cUn djus ds ckn c<+rk gS rFkk [kksyus ds i'pkr~ ?kVrk gSA(B) fLop dks [kksyus ds Bhd igys rFkk ckn esa xsYosuksehVj ls izokfgr /kkjk dh fn'kk leku jgrh gSA(C) iqu% fLop [kksyus ds ckn R1 esa O;f;r ÅtkZ 'kwU; gSA(D) iqu% fLop [kksyus ds ckn R2 esa O;f;r ÅtkZ 'kwU; gSA


PHYS

ICS

LTS-9/400999DT110315003


12. A block of mass m is placed on inclined smooth surface of wedge of mass M as shown. Wedge isaccelerating horizontally with an acceleration a such that block is relatively at rest on the inclined surface.(A) the value of a is gcotq(B) the value of a is g tanq(C) normal force on the wedge due to surface is (m+M)g(D) normal force on the surface due to wedge is Mg.

nzO;eku m ds ,d CykWd dks fp=kuqlkj M nzO;eku ds ost dh vkur fpduh lrg ij j[kk x;k gSA ost Roj.k als {kSfrt :i ls bl izdkj Rofjr gksrk gS fd CykWd vkur lrg ij fojke esa jgrk gSA(A) a dk eku gcotq gSA(B) a dk eku g tanq gSA(C) lrg ds dkj.k ost ij yEcor~ cy (m + M)g gSA(D) ost ds dkj.k lrg ij yEcor~ cy Mg gSA

13. The circuit shown in the figure consists of a battery of emf e = 10 V ; a capacitor of capacitance C = 1.0 mFand three resistor of values R1 = 2W, R2 = 2W and R3 = 1W. Initially the capacitor is completely unchargedand the switch S is open. The switch S is closed at t = 0.(A) The current through resistor R3 at the moment the switch closed is zero.(B) The current through resistor R3 a long time after the switch closed is 5A.(C) The ratio of current through R1 and R2 is always constant.(D) The maximum charge on the capacitor during the operation is 5mC.

fp= esa iznf'kZr ifjiFk esa e = 10 V fo|qr okgd cy okyh ,d cSVjh] C = 1.0 mF /kkfjrk okyk la/kkfj= rFkk R1 = 2W,R2 = 2W rFkk R3 = 1W okys rhu izfrjks/k yxs gq, gSaA izkjEHk esa la/kkfj= iw.kZr;k vukosf'kr gS rFkk fLop S [kqyk gSAt = 0 ij fLop S dks can dj nsrs gSa:-(A) fLop dks can djus ds {k.k ij R3 ls fuxZr /kkjk 'kwU; gSA

(B) fLop dks can djus ds yEcs le; i'pkr~ R3 ls fuxZr /kkjk 5A gSA(C) R1 rFkk R2 ls izokfgr /kkjk dk vuqikr lnSo fu;r jgrk gSA

(D) bl izfØ;k ds nkSjku la/kkfj= ij vf/kdre vkos'k 5mC gSA

LTS-10/40

PHYS

ICS


0999DT110315003

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

� This section contains TWO paragraphs.� Based on first paragraph will be THREE questions and second paragraph will be TWO questions� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four option(s) is correct.� For each question, darken the bubble corresponding to the correct option in the ORS� Marking scheme :


� bl [k.M esa nks vuqPNsn gSa� izFke vuqPNsn ij rhu iz'u gSa rFkk nwljs vuqPNsn ij nks iz'u gSa� izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa dsoy ,d fodYi lgh gSaA� izR;sd iz'u ds fy,] lgh fodYi ds vuq:i cqycqys dks vks-vkj-,l- esa dkyk djsaA� vadu ;kstuk :


Paragraph for Questions 14 to 16iz'u 14 ls 16 ds fy;s vuqPNsn

In the circuit shown in figure, the battery is an ideal one with emf V. The capacitor is initially uncharged. Theswitch S is closed at time t = 0.fn[kk;s x;s ifjiFk fp= esa] cSVjh vkn'kZ gS ftldk fo|qr okgd cy V gSA la/kkfj= izkjEHk esa vukosf'kr gSA le;t = 0 ij fLop dks cUn fd;k tkrk gSA

R/2 A 5R/2

R

R/2

V

B

S

C

14. The charge Q on the capacitor at time t is–le; t esa la/kkfj= ij vkos'k Q gksxk&

(A) t

RCCV 1 e2

-æ ö-ç ÷

è ø(B)

t3RCCV 1 e

2-æ ö

-ç ÷è ø

(C) 2t

5RCCV 1 e2

-æ ö-ç ÷

è ø(D)

2t9RCCV 1 e

2-æ ö

-ç ÷è ø

15. The current in AB at time t is–le; t ij AB esa /kkjk gksxh&

(A) t

3RCV 1 e2R

-æ ö-ç ÷

è ø(B)

t3RC2V 1 e

R-æ ö

-ç ÷è ø

(C)

t3RC2V e1

R 6

-æ öç ÷-ç ÷ç ÷è ø

(D)

t3RCV e1

2R 6

-æ öç ÷-ç ÷ç ÷è ø

16. What is its limiting value iAB at t ® ¥ ?t ® ¥ ij iAB dk lhekUr eku D;k gksxk \

(A) V

2R (B) VR (C)

2VR (D)

V3R


PHYS

ICS

LTS-11/400999DT110315003


Paragraph for Questions 17 and 18iz'u 17 ,oa 18 ds fy;s vuqPNsn

The following two questions refer to the circuit shown. Assume that the capacitors are initially uncharged.iznf'kZr ifjiFk esa la/kkfj= izkjEHk esa vukosf'kr gSaA

17. At time t = 0, the switch S in the circuit is closed. At that instant, what is the equivalent resistance of thecircuit?le; t = 0 ij fLop S dks can dj fn;k tkrk gSA bl {k.k ifjiFk dk rqY;kadh izfrjks/k gksxk %&

(A) 9.00 W (B) 1.20 W (C) 1.80 W (D) 0.92 W18. After the switch has been closed for a long time, how much charge is on the positive plate of the 3 mF

capacitor ?fLop dks yEcs le; rd can j[kus ds i'pkr~ 3 mF la/kkfj= dh /kukRed IysV ij fdruk vkos'k gksxk %&

(A) 24 µC (B) 32 µC (C) 56 µC (D) 72 µC

SECTION –II / [k.M – II & SECTION –III / [k.M – IIIMatrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj

No question will be asked in section II and III / [k.M II ,oa III esa dksbZ iz'u ugha gSA

LTS-12/40

PHYS

ICS


0999DT110315003


SECTION–IV : (Maximum Marks : 30)

[k.M–IV : (vf/kdre vad : 30)� This section contains TEN 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� Marking scheme :

+3 If the bubble corresponding to the answer is darkened0 In all other cases

� bl [k.M esa nl iz'u gSa

� izR;sd iz'u dk mÙkj 0 ls 9 rd] nksuksa 'kkfey] ds chp dk ,d ,dy vadh; iw.kk ±d gS

� izR;sd iz'u esa] vks-vkj-,l- ij lgh iw.kk±d ds vuq:i cqycqys dks dkyk djsa

� vadu ;kstuk :

+3 ;fn mÙkj ds vuq:i cqycqys dks dkyk fd;k tk;

0 vU; lHkh voLFkkvksa esa

1. A 10µF capacitor is connected to a 12V battery and charged. Then it is disconnected from battery and filledwith dielectric of constant k = 2? What is the change in potential difference (in volt) across the capacitor ?10µF dk la/kkfj= 12V dh cSVjh ls tksM+k tkrk gS rFkk vkosf'kr fd;k tkrk gSA fQj bls cSVjh ls foPNsn dj

fn;k tkrk gS rFkk k = 2 fu;rkad ds ijkoS|qr ls Hkj fn;k tkrk gS ? la/kkfj= ds fljksa ij foHkokUrj esa ifjorZu

(oksYV esa) D;k gS \2. In the circuit (Figure), E = 5 V, r = 1 ohm, R2 = 4 ohms, R3 = 3 ohms and C = 3 µF. Find the charge on the

plates of capacitor A (in µC).ifjiFk esa (fp=), E = 5 V, r = 1 vkse, R2 = 4 vkse, R3 = 3 vkse] rFkk C = 3 µF gSA la/kkfj= A dh IysVksa ij vkos'k

Kkr dhft;sA (µC esa).

C C

C C

R1

R2

R3

e r

A


PHYS

ICS

LTS-13/400999DT110315003

3. The figure shows two circuits with a charged capacitor that is to be discharged through a resistor as

shown. The initial charge on capacitors is1

2qq

= 2. If both switches are closed at t = 0, the charges

become equal at 10–4 ln2 sec. Find the resistance R (in W).fp= esa nks ifjiFkksa dks ,d vkosf'kr la/kkfj= ds lkFk fn[kk;k x;k gS ftls n'kkZ;s vuqlkj ,d izfrjks/kd ls

fujkosf'kr djuk gSA la/kkfj=ksa ij izkjfEHkd vkos'k 1

2qq

= 2 gSA ;fn nksuksa fLop t = 0 ij cUn gksa rks vkos'k

10–4 ln2 sec. ij cjkcj gks tkrs gSaA izfrjks/k R (W esa) Kkr dhft;sA

q1

–q1 5µF5W

q2

–q2 10/3µFR

4. A block of mass 45 kg resting on a horizontal surface is acted upon by a force F which varies as shown in thefigure. If the coefficient of friction between the block and surface is 0.2, find the displacement(in m) when the block will come to rest.nzO;eku 45 kg dk ,d CykWd] ,d {kSfrt lrg ij fojkekoLFkk esa j[kk gqvk gSA bl ij ,d cy F dk;Zjr~ gS] tksfp=kuqlkj ifjofrZr gksrk gSA ;fn CykWd rFkk lrg ds e/; ?k"kZ.k xq.kkad 0.2 gks rks CykWd ds fojkekoLFkk esa vkus rd bldkfoLFkkiu (ehVj esa) Kkr dhft,A

0

240

F(N)

s(m)21


LTS-14/40

PHYS

ICS


0999DT110315003

5. In given figure, find the acceleration (in m/s2) of the block B. Assuming pulley is ideal andrope is massless.iznf'kZr fp= esa CykWd B dk Roj.k (m/s2 esa) Kkr dhft,A f?kjuh dks vkn'kZ rFkk jLlh dks nzO;ekughu ekfu;sA

10kg

10kgA

B

100N

\\\\\\\\\\\\\\\\\\

6. Initially the spring is in natural length. At t = 0, F = 20 N force starts acting on the block B along the line joiningthe masses. At some instant, acceleration of the block A is 5 m/s2 left side. Find the acceleration (in m/s2) ofthe block B at that instant.fp= esa iznf'kr fLizax izkjEHk esa viuh ewy yEckbZ esa gSA t = 0 le; ij nksuksa nzO;ekuksa dks tksM+us okyh js[kk ds vuqfn'k

F = 20 N dk ,d cy] CykWd B ij dk;Z djuk izkjEHk dj nsrk gSA fdlh {k.k CykWd A dk cka;h vksj Roj.k 5 m/s2 gSA

bl {k.k ij CykWd B dk Roj.k (m/s2 esa) Kkr dhft,A

A B F=20Nm =2kgA m =5kgB



PHYS

ICS

LTS-15/400999DT110315003

7. Two blocks of weights 10 N and 20 N are connected by a string and pulleys as shown. Assuming that

the string and pulley are massless and frictionless, Find the net force acting on the block on the left.

nks CykWdksa ds Hkkj 10 N rFkk 20 N gS] bUgsa fp=kuqlkj jLlh rFkk f?kjfu;ksa dh lgk;rk ls tksM+k x;k gSA eku yhft, fd jLlh

rFkk f?kjuh nzO;ekughu rFkk ?k"kZ.kjfgr gSA cka;h vksj fLFkr CykWd ij dk;Zjr~ dqy cy dk eku Kkr dhft,A

A B 20 N 10 N

8. A cart of mass 3 kg is pulled by a 5kg object as shown. The cart, whose length is 40 cm moves along the tablewithout friction. There is a block of mass 2kg on the cart, which falls from it 0.8s after the start of the motion.Find the coefficient of kinetic friction between the cart and the brick. Use g = 10 m/s2. Write the value of 10µ.,d 3 kg æO;eku dh xkM+h dks 5kg æO;eku ds fi.M }kjk [khapk tkrk gSA bl xkM+h dh yEckbZ 40 cm gS rFkk ;g est

ij fcuk ?k"kZ.k xfr djrh gSA bl xkM+h ij 2kg æO;eku dk ,d CykWd j[kk gS tks xkM+h dh xfr izkjEHk gksus ds 0.8s i'pkr~

bl ij ls fxj tkrk gSA xkM+h rFkk CykWd ds e/; xfrt ?k"kZ.k xq.kkad m gks rks 10µ dk eku Kkr dhft,sA g = 10 m/s2

3kg2kg

5kg


LTS-16/40

PHYS

ICS


0999DT110315003

9. A climber of mass 60 kg sits between two parallel rock walls so that he presses his boots to one andhis back to the other wall. The friction coefficient between the boots and the wall is 1.2 and betweenthe wall and the climber’s back is 0.8. The normal force (in hundred Newton) does he need to applyon the wall is given by 100 aN. Fill a in OMR sheet.,d 60 kg æO;eku dk ioZrkjksgh nks lekUrj pêkuksa ds chp bl izdkj cSBk gS fd og vius twrksa ls ,d pêku dks

rFkk viuh ihB ls nwljh pêku dks nckrk gSA twrksa rFkk pêku ds e /; ?k"kZ.k xq.kkad 1.2 vkSj ihB rFkk pêku ds eè;

?k"kZ.k xq.kkad 0.8 gSA mls pêku ij U;wure 100 aN vfHkyEc cy yxkuk gksxkA a dk eku Kkr djksA

10. Three forces acting on a body are shown in figure. To have the resultant force only along the x-direction,find the magnitude of the minimum additional force (in newton) needed.fp= esa fdlh oLrq ij rhu cy dk;Zjr~ gSaA buds ifj.kkeh cy dks dsoy x-fn'kk ds vuqfn'k izkIr djus ds fy;s

vko';d vfrfjDr tksM+s tkus okys U;wure cy dk ifjek.k (U;wVu esa) Kkr dhft,Ay

5NF1

x37°

4N

53°

5N

F3

F2



CHEM

ISTR

Y

LTS-17/400999DT110315003

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






1. Which of the following is O, P directing ?

fuEu esa ls dkSulk lewg O, P funsZ'k; gS&

(A)ClClCl

–C (B) –NH3+

(C) Cl

Cl–CH (D)

H

H–O

+

2. Select the correct statement :(A) Anti-aromatic species have negligible tendency to form dimer.(B) –Cl is having lesser negative inductive effect than –F but has more +M effect(C) The one which has higher resonance energy is always more stable than the other which has

lower resonance energy(D) –NH2 is having lesser negative inductive effect than –OH but has more +M effect.lgh dFku dk p;u dhft;s&

(A) ,sUVh , sjkseSfVd Lih'kht dh f}yd cukus dh izo`fr ux.; gksrh gS(B) –F dh rqyuk esa –Cl dk ½.kkRed izsjf.kd izHkko de gksrk gSA ijUrq +M izHkko vf/kd gksrk gS

(C) ,d ;kSfxd ftldh vuqukn ÅtkZ vf/kd gks og de vuqukn ÅtkZ j[kus okys vU; ;kSfxd dh rqyuk esa ges'kk

vf/kd LFkk;h gksrh gSA

(D) –OH dh rqyuk esa –NH2 dk ½.kkRed izsjf.kd izHkko de gksrk gS ijUrq +M izHkko vf/kd gksxk


LTS-18/40

CHEM

ISTR

Y


0999DT110315003


3. Which of the following is / are +I group(s).

(A) –SiMe3 (B) –CMe3 (C) COO- 1 (D) All of these

fuEu esa ls +I lewg gS&

(A) –SiMe3 (B) –CMe3 (C) COO- 1 (D) mijksDr lHkh

4. Which of the following will have highest dissociaton constant (Ka)?

fuEu esa ls fdldk fo;kstu fLFkjkad (Ka) mPpre gksxk\

(A) COOHCH3

(B) COOHNO2

(C) COOHF

(D) COOHCl

5. Select the one which is most basic among the following:

fuEu esa ls lokZf/kd {kkjh; gS&

(A) N–H (B) NÅ Me

Me(C) N–HO (D) O

6. Correct order of Rotational energy barrier will be

?kw.kZu ÅtkZ vojks/k dk lgh Øe gksxk&

(I) (II) OMe (III) OMe

O N2

(A) I > II > III (B) II > I > III (C) III > II > I (D) III > I > II


CHEM

ISTR

Y

LTS-19/400999DT110315003


7. Which of the following is incorrect order of basic nature.

fuEu esa ls dkSulk {kkjh; izd`fr dk xyr Øe gS&

(A) in aqueous solution (tyh; ek/;e esa) Me2NH > Me – NH2 > Me3N

(B) CH – C – NH > CH – CH – NH >3 2 3 2 2 NH2

NH

(C)

NH2

CH3

>

NH2NH2

>

Me

(D)

NMe2

CH3

>

NMe2 NMe2

>

CCl3

8. Which of the given statement is incorrect ?

(A) CH2=CH–CH2–CH=CH2 is less acidic than

(B) CH3– NÅ

Me3 is less acidic than CH3– PÅ

Me3

(C) H H

CNNC

is more acidic than H H

(D) OO

OHHO is less acidic then

O

OHHOfn;s x;s dFku esa ls dkSulk xyr gS&

(A) dh rqyuk esa CH2=CH–CH2–CH=CH2 de vEyh; gS

(B) CH3– PÅ

Me3 dh rqyuk esa CH3– NÅ

Me3 de vEyh; gS

(C) H H

dh rqyuk esa

H H

CNNC

vf/kd vEyh; gS

(D) O

OHHO dh rqyuk esa

OO

OHHO de vEyh; gS

LTS-20/40

CHEM

ISTR

Y


0999DT110315003



� This section contains FIVE questions.� Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.� For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS� Marking scheme :





9. Correct order of basic strength in aq. solution is/are :

tyh; foy;u esa {kkjh; lkeF;Z dk lgh Øe gSa&

(A)N

> >

NH2

CH –C–NH3 2

O(B)

N> >

NH2N

H

(C) CH3–CH = N

:

H > CH3CH2 N

:

H2 > NH3 (D) OH > NH > F2–– –

10. Which of the following is/are electrophile.

fuEu esa ls dkSu bysDVªksuLusgh gS@gS&

(A) AlCl3 (B) BH3 (C) CO2 (D) ROH

11. Which of the following is / are aromatic species.

fuEu esa ls dkSu ,sjkseSfVd Lih'kht gS@gS&

(A) NH

(B) O

Å(C) (D)


CHEM

ISTR

Y

LTS-21/400999DT110315003


12. Correct statement(s) is/are regarding given compounds

(A) Basic strength order I > II > III

(B) Basic strength order I < II < III

(C) III compound having all inductive, mesomeric & hyperconjugation effects.

(D) All I, II & III are heterocyclic compounds

fn;s x;s ;kSfxdksa ds lUnHkZ es lgh dFku gS@gS&

(A) {kkjh; lkeF;Z Øe I > II > III

(B) {kkjh; lkeF;Z Øe I < II < III

(C) III ;kSfxd esa lHkh izsjf.kd] ehlksesfjd rFkk vfrla;qXed izHkko gksrs gSa

(D) I, II rFkk III lHkh fo"kepØh; ;kSfxd gS

13. Identify compounds which are soluble in aqueous NaOH ?

(A) Phenol (B) Picric acid (C) Salicylic acid (D) Asprin

;kSfxd igpkfu;s tks tyh; NaOH eas ?kqyu'khy gS&

(A) QhukWy (B) fifØd vEy (C) lsfylhfyd vEy (D) ,Lizhfju

LTS-22/40

CHEM

ISTR

Y


0999DT110315003







OHH

H N2

H

H

C CHº

CH –H2

(a)

(b)

(c)

(d)

(g)

(e)

(f)

14. Which of the following represent most acidic site in the above moleculemijksDr v.kq esa fuEu esa ls dkSu lokZf/kd vEyh; fLFkfr dks iznf'kZr djrk gS&(A) a (B) b (C) c (D) d

15. Which of the following represent most basic site in the moleculefuEu esa ls dkSu v.kq esa lokZf/kd {kkjh; fLFkfr dks iznf'kZr djrk gS&(A) d (B) e (C) f (D) g

16. Which of the following represent the correct order C – H bond energyfuEu esa ls dkSu C—H ca/k ÅtkZ dk lgh Øe iznf'kZr djrs gSa&

(A) f > g > c (B) c > f > g (C) g > f > c (D) c > g > fSpace for Rough Work / dPps dk;Z ds fy, LFkku


CHEM

ISTR

Y

LTS-23/400999DT110315003



Planar monocyclic ring containing (4n + 2)p electron (where n = 0, 1, 2, 3.... ¥) are aromatic on applying

this rule benzene and its derivative are aromatic.Non benzenoid aromatic compound are those which contain ring that is not six membered such as

–

. Cyclic planar conjugated compound which posses 4np electron are antiaromatic. In a reaction

that reactant would tend to react with reagent that can result into aromatic product salt.

leryh; ,dypfØ; oy; ftlesa (4n + 2)p bysDVªkWu (tgk¡ n = 0, 1, 2, 3.... ¥) gks , sjksesfVd ;kSfxd dgykrs gSaA blfu;e ds vuqlkj csfUtu oy; ,sjksesfVd gSA

;kSfxd ftldh oy; 6 lnL; ugha gksrs mUgsa ukWu csUthuksbM ,sjksesfVd dgrs gSa tSls& –

leryh; pØh; ;kSfxd

ftuesa la;qXeh 4np bysDVªkWu gksrs gSa ,sUVh,sjkesfVd dgykrs gSaA ;fn fdlh vfHkfØ;k esa ,sjksesfVd mRIkkn curk gks rks

vfHkfØ;k rhozrk ls lEiUu gks tkrh gSA

17. Which of the following is correct ?

(A) Br

reacts immediately with AgNO3 (B) Br reacts immediately with AgNO3

(C) Br reacts immediately with AgNO3 (D) Both (A) and (B)

fuEu esa ls dkSulk lgh dFku lgh gS&

(A) Br

, AgNO3 ds lkFk rqjUr fØ;k djrk gS (B) Br , AgNO3 ds lkFk rqjUr fØ;k djrk gS

(C) Br , AgNO3 ds lkFk rqjUr fØ;k djrk gS (D) (A) rFkk (B) nksuksa

LTS-24/40

CHEM

ISTR

Y


0999DT110315003


18. Which of the given statement is/are not correct ?

(A) H H

is less acidic than CH2=CH–CH2–CH=CH2

(B) CH3– NÅ

Me3 is less acidic than CH3– PÅ

Me3

(C) H H

CNNC is more acidic than

H H

(D) OO

OHHO is less acidic then

O

OHHO

fn;s x;s dFku esa ls dkSuls lgh ugha gS&

(A) CH2=CH–CH2–CH=CH2 dh rqyuk eas H H

de vEyh; gS

(B) CH3– PÅ

Me3 dh rqyuk esa CH3– NÅ

Me3 de vEyh; gS

(C)

H H

dh rqyuk eas

H H

CNNC vf/kd vEyh; gS

(D) O

OHHO dh rqyuk esa

OO

OHHO de vEyh; gS




CHEM

ISTR

Y

LTS-25/400999DT110315003








� vadu ;kstuk :



1. Number of reactions correctly matched with their major product :

vfHkfØ;kvksa dh la[;k crkb;sa] tks muds eq[; mRikn ds lkFk lgh lqesfyr gSA

(i) HCOONa 4KHSO¾¾¾® HCOOH

(ii) 2Ph–OH 2 3Na CO¾¾¾¾® 2PhONa + H2O + CO2

(iii) R–SO3H 3NaHCO¾¾¾¾® RSO3Na + H2O + CO2

(iv) R–CO2H14

3NaHC O¾¾¾¾®14

2R C O Na- + H2O + CO2

(v) Ph–SO3H 3CH COONa¾¾¾¾® Ph–SO3Na + CH3COOH

(vi) R–CºCH NaOH¾¾¾® R–CºCNa + H2O

(vii) R–CºCH Na¾¾® RCºCNa + 12 H2

LTS-26/40

CHEM

ISTR

Y


0999DT110315003

2. How many moles of a gas A released when 244 gm of PhCOOH is treated with sufficient amount of NaHCO3 ?tc 244 gm PhCOOH , NaHCO3 dh i;kZIr ek=k ds lkFk fØ;k djrs gSa rks A xSl ds fdrus eksy eqDr gksrs gSaA

3. In how many pairs Ist group or atom has more –I effect than IInd group.Ist ;qXe esa fdrus lewg ;k ijek.kq dk IInd lewg dh rqyuk esa vf/kd –I izHkko gksrk gSA

(i) –F, –Cl (ii) –OH, –F (iii) –NH2, –OH (iv) –OH, –CH3 (v) –F, –NMe3+

4. How many lone pairs are present in resorcinol ?fjlksflukWy esa fdrus ,dkadh bysDVªksu ; qXe mifLFkr gSA

5. Number of reaction which is occur in forward direction.,slh vfHkfØ;kvksa dh la[;k tks vxz fn'kk esa gksrh gSA(a) PhSO3Na + MeCOOH�MeCOONa + Ph – SO3H

(b) CH Na Et – OH3 Å + � 4EtONa + CH

(c) Ph – ONa + MeCOOH�MeCOONa + PhOH

(d) NaNH2 + CH3 – H�CH31NaÅ + NH3

(e) CH3CºCH + NaOH ¾® CH C CNa + H O3 2– +

6. Number of compounds which is/are more acidic than phenol.

(i) o-nitrophenol (ii) p-nitrophenol (iii) Benzyl alcohol(iv) Pyruvic acid (v) p-methoxyphenol (vi) m-methoxyphenol(vii) Propan-2-ol (viii) Ethanol

,sls ;kfxdksa dh la[;k tks QhukWy dh rqyuk esa vf/kd vEyh; gS@gSA

(i) o-ukbVªksfQukWy (ii) p-ukbVªksfQukWy (iii) csathy ,sYdksgkWy

(iv) ikbfod vEy (v) p-esFkksDlhQhukWy (vi) m-esFkksDlhQhukWy

(vii) izksisu-2-vkWy (viii) ,sFksukWy



CHEM

ISTR

Y

LTS-27/400999DT110315003

7. How many atoms or groups has higher priority than –OH group according CIP rule in following groups.–CN, –Cl, –D, –H, –SO3H, –SH, –F,–Br, –I, –NH2

fuEu lewgksa esa CIP fu;e ds vuqlkj –OH lewg dh rqyuk esa fdrus ijek.kqvksa ;k lewgksa dh mPp izkFkfedrk gksrh gSA

–CN, –Cl, –D, –H, –SO3H, –SH, –F,–Br, –I, –NH2

8. How many compounds are quasi aromatic in given compound :

fn;s x;s ;kSfxdksa esa fdrus Dokth ;kSfxd , sjkseSfVd gksrs gSaA

(1) (2) (3)

O

(4) O

(5) t-Bu

t-Bu

t-Bu

N(6) Azulene (, stqfyu) (7) (8)

O

O

9. Degree of unsaturation in given compound is :

fn;s x;s ;kSfxd esa vlar`Irrk dh dksfV gSA

Ph

O

10. The total number of contributing structures showing hyperconjugation (involving C–H bonds) for the

following carbocation is.

fuEu dkcZ/kuk;u ds fy;s vfrla;qXeu (C–H ca/k dks lfEefyr djrs gq;s) iznf'kZr djus okyh vuquknh lajpukvksa

dh dqy la[;k gSA

H C3 H C2CH CH2 3Å


LTS-28/40

MAT

HEM

ATIC

S


0999DT110315003

PART-3 : MATHEMATICS Hkkx-3 : xf.kr






1. If (x, y, z) º (a, b, c) satisfies the system of equation 2x + y = 2z, 9z – 7x = 6y & x3 + y3 + z3 = 216 andacosq + bsinq = c, then number of values of [0, 3 ]qÎ p is

(A) 0 (B) 1 (C) 2 (D) more than 2

;fn (x, y, z) º (a, b, c) lehdj.k fudk; 2x + y = 2z, 9z – 7x = 6y o x3 + y3 + z3 = 216 rFkk acosq + bsinq = c

dks larq"V djrk gS] rks [0, 3 ]qÎ p ds ekuksa dh la[;k gksxh&

(A) 0 (B) 1 (C) 2 (D) 2 ls vf/kd

2. If x Î [0,100p), is the solution of the equation cos2x – sin2014x = 1, then sum of all possible solutions is-

;fn x Î [0,100p), lehdj.k cos2x – sin2014x = 1 dk gy gks] rks lHkh laHko gyksa dk ;ksxQy gksxk&

(A) 4950p (B) 5000p (C) 5050p (D) 11000p



MAT

HEM

ATIC

S

LTS-29/400999DT110315003


3. If in DABC, with usual notations sinA sinB = sinC and 3b – 5c = 0, then the value of Acot2

is-

(A) 2 (B) 3 (C) 4 (D) No such triangle exist.

;fn f=Hkqt ABC esa] lkekU; ladsrksa ds lkFk sinA sinB = sinC rFkk 3b – 5c = 0 gks] rks Acot2

dk eku gksxk -

(A) 2 (B) 3 (C) 4 (D) ,slk f=Hkqt fo|eku ugha gS

4. Let ( )n

n n1 2 A B 2+ = + , where AAn and Bn are rational numbers and n Î N, then the value of n

nn

AlimB®¥

is equal to -

(A) x 0

sin xlimx®

é ùê úë û

(B) x 0

sin xlimx®

é ùê úë û

(C) x

4

| cos x sin x |limx

4p

®

-p

-(D) Does not exist

(where [.] denotes greatest integer function)

ekuk ( )n

n n1 2 A B 2+ = + , tgk¡ An rFkk Bn ifjes; la[;k;sa rFkk n Î N gS] rks n

nn

AlimB®¥

dk eku gksxk&

(A) x 0

sin xlimx®

é ùê úë û

(B) x 0

sin xlimx®

é ùê úë û

(C) x

4

| cos x sin x |limx

4p

®

-p

-(D) fo|eku ugha gSA

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

5. If 2

xlim ax x bx 3

®¥+ + = , then-

;fn 2

xlim ax x bx 3

®¥+ + = gks] rks&

(A) a = 1, b = 6 (B) a = 1, b Î R (C) a = –1, b Î R (D) a = –1, b = 6

LTS-30/40

MAT

HEM

ATIC

S


0999DT110315003


6.1

3 2

2x tan x b x 0ƒ(x)

x x ax c x 0

-ì + + <= í

+ + + ³î. If ƒ(x) is differentiable at x = 0, then

2

2

b ac

+ is equal to (bc ¹ 0) -

ekuk 1

3 2

2x tan x b x 0ƒ(x)

x x ax c x 0

-ì + + <= í

+ + + ³î gSA ;fn ƒ(x), x = 0 ij vodyuh; gks] rks

2

2

b ac

+ dk eku gksxk (bc ¹ 0) -

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

7. Let ƒ(x) is a continuous function in [a, b] such that ƒ(a) ¹ ƒ(b) and ƒ(a), ƒ(b) > 0, then which of the followingis INCORRECT ?

(A) If ƒ(c) =ƒ(a) ƒ(b)

2+

, then there always exist at least one 'c' in (a, b)

(B) If ƒ(c) ƒ(a)ƒ(b)= then there always exist at least one 'c' in (a, b)

(C) If 5ƒ(c) 3ƒ(a) 2ƒ(b)= + , then there always exist at least one 'c' in (a, b)

(D) If 2ƒ(c) 3ƒ(a) ƒ(b)= - , then there always exists at least one 'c' in (a, b)ekuk vUrjky [a, b] esa larr~ Qyu ƒ(x) bl izdkj gS fd ƒ(a) ¹ ƒ(b) rFkk ƒ(a), ƒ(b) > 0 gks] rks fuEu esa ls dkSulk dFkulgh ugha gksxk \

(A) ;fn ƒ(c) =ƒ(a) ƒ(b)

2+

gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(B) ;fn ƒ(c) ƒ(a)ƒ(b)= gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(C) ;fn5ƒ(c) 3ƒ(a) 2ƒ(b)= + gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

(D) ;fn 2ƒ(c) 3ƒ(a) ƒ(b)= - gks] rks (a,b) esa de ls de ,d 'c' lnSo fo|eku gksxkA

8. Number of points where ƒ(x) = |x – sgn(x)| is non differentiable, is (sgn(.) denotes signum function)fcUnqvksa dh la[;k tgk¡ ƒ(x) = |x – sgn(x)| vodyuh; ugha gS] gksxh (sgn(.) flXue Qyu dks n'kkZrk gS)(A) 0 (B) 1 (C) 2 (D) 3


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

option(s) is (are) correct.� For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS� Marking scheme :





9. If sin x cos x 1

ƒ(x) cos x sin x 11 sin x cos x

= , then identify the correct statement(s) -

(A) Number of solutions of ƒ(x) = 0 is six in [0, 2p](B) Number of solutions of ƒ(x) = 0 is three in [0, 2p](C) ƒ'(0) = 1(D) ƒ'(0) = 0

;fn

sin x cos x 1ƒ(x) cos x sin x 1

1 sin x cos x= gks] rks lgh dFku@dFkuksa dk p;u dhft;s&

(A) [0, 2p] esa ƒ(x) = 0 ds gyksa dh la[;k N% gS(B) [0, 2p] esa ƒ(x) = 0 ds gyksa dh la[;k rhu gS(C) ƒ'(0) = 1(D) ƒ'(0) = 0

LTS-32/40

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10. If 0,2pæ öa Îç ÷

è ø, then expression cosa(tan2a + cosec2a) + tana(cos2a + cosec2a) + coseca(tan2a + cos2a)

cannot take the value/s-

;fn 0,2pæ öa Îç ÷

è ø gks] rks O;atd cosa(tan2a + cosec2a) + tana(cos2a + cosec2a) + coseca(tan2a + cos2a) dk

eku fuEu ugha gks ldrk -(A) p (B) e (C) 6 (D) 5

11. In DABC, a = 2, b = 3, C3p

Ð = , then -

(A) length of internal angle bisector through vertex C is 6 3

5

(B) length of internal angle bisector through vertex C is 3 3

5

(C) length of median through vertex C is 192

(D) length of median through vertex C is 194

f=Hkqt ABC esa a = 2, b = 3, C3p

Ð = gks] rks -

(A) 'kh"kZ C ls xqtjus okys vUr% dks.k v¼Zd dh yEckbZ 6 3

5 gksxhA

(B) 'kh"kZ C ls xqtjus okys vUr% dks.k v¼Zd dh yEckbZ 3 3

5 gksxhA

(C) 'kh"kZ C ls xqtjus okyh ekf/;dk dh yECkkbZ 192

gksxhA

(D) 'kh"kZ C ls xqtjus okyh ekf/;dk dh yECkkbZ 194

gksxhA


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LTS-33/400999DT110315003


12. If y cos x cos x cos x ......= + + + ¥ (where cosx > 0), then which of the following can be the value

of dydx ?

;fn y cos x cos x cos x ......= + + + ¥ (tgk¡ cosx > 0) gks] rks fuEu esa ls dkSu dydx

dk eku gks ldrk gS-

(A) sin x

1 2y- (B) sin x

1 4cos x-+

(C) 2

ysin xy cos x-

+ (D) 2

ysin xy cos x-

13. Given [ ]( )

( )

33

3x

sin x xlim

x®a

é ù -ë û

- aexists finitely, then the value of a can be (where a Î I & [.] denotes greatest

integer function)-

fn;k x;k gS [ ]( )

( )

33

3x

sin x xlim

x®a

é ù -ë û

- a ifjfer :i ls fo|eku gks] rks a dk eku gks ldrk gS (tgk¡ a Î I rFkk [.] egÙke

iw.kk±d Qyu dks n'kkZrk gS) -(A) 0 (B) 1 (C) –1 (D) 2

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0999DT110315003







If ƒ (x) be real valued & differentiable function in its domain & ƒ (x) ƒ (y)ƒ(x y)1 ƒ (x)ƒ (y)

++ =

- & ƒ '(0) = l.

;fn ƒ (x) okLrfod eku rFkk vius izkUr esa vodyuh; Qyu gS rFkk ƒ (x) ƒ (y)ƒ(x y)

1 ƒ (x)ƒ (y)+

+ =-

rFkk ƒ '(0) = l gSA

14. If l = 3, then x 0

ƒ(x)limxƒ3

® æ öç ÷è ø

is -

;fn l = 3 gks] rks x 0

ƒ(x)limxƒ3

® æ öç ÷è ø

gksxk -

(A) 13 (B) 3 (C) – 1

3(D) –3

15. If l = 1, then ƒ (tan–1x) is -(A) one-one (B) periodic (C) even (D) bounded;fn l = 1 gks] rks ƒ (tan–1x) gksxk -(A) ,dSdh (B) vkorhZ (C) le (D) ifjc¼

16. If l = 2, then -;fn l = 2 gks] rks -

(A) ( )1 1ƒ ƒ ƒ 12 4

æ ö æ ö> >ç ÷ ç ÷è ø è ø

(B) ( )1 1ƒ ƒ 1 ƒ2 4


(C) ( ) 1 1ƒ 1 ƒ ƒ4 2


(D) ( )1 1ƒ ƒ ƒ 14 2



MAT

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LTS-35/400999DT110315003



In the figure. It is given that ÐC = 90°, AD = DB; ED is perpendicular to AB, AB = 20 and AC = 12.fn;s x;s fp= esa] ;g fn;k x;k gS fd ÐC = 90°, AD = DB gSA ED,AB ds yEcor~ gS] AB = 20 bdkbZ rFkkAC = 12 bdkbZ gSA C

A Bg

dq

ab

E

D17. Area of triangle AEC is -f=Hkqt AEC dk {ks=Qy gksxk (oxZ bdkbZ esa) -

(A) 24 (B) 21 (C) 42 (D) 212

18. The value of tang tan(60° – g)tan(60° + g) is -tang tan(60° – g)tan(60° + g) dk eku gksxk -

(A) 11744

- (B) 174

- (C) 34 (D)

54



LTS-36/40

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� vadu ;kstuk :



1. Number of value(s) of x [0, 2 ]Î p which satisfies the equation sin2x + cosec2x = cos2x + sec2x isx [0, 2 ]Î p ds ekuksa dh la[;k] tks lehdj.k sin2x + cosec2x = cos2x + sec2x dks lUrq"V djrh gS] gksxh

2. If P is a point on the altitude AD of triangle ABC such that ÐCBP = B3

and if AP = l Bcsin3

æ öç ÷è ø

(where symbols

have their usual meanings), then l is -

;fn P, f=Hkqt ABC ds 'kh"kZ yEc AD ij ,d fcUnq bl izdkj gS fd ÐCBP = B3

rFkk AP = l Bcsin3

æ öç ÷è ø

(tgk¡ ladsrksa ds lkekU; vFkZ gSaA) gS] rks l gksxk -

3. If the value of 2x 0

sin(2 sec x)limn (x 1)®

p+l

is k.p, then k is equal to

;fn 2x 0

sin(2 sec x)limn (x 1)®

p+l

dk eku k.p gks] rks k dk eku gksxk


MAT

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LTS-37/400999DT110315003


4. Number of values in [0,6p] where ƒ(x) = {x}sin2x + x3 – {x} [sin2x] + [x]sin2x – [x][sin2x] is non-derivable(where [.] denotes greatest integer function & {.} denotes fractional part function)vUrjky [0,6p] esa ekuksa dh la[;k] tgk¡ ƒ(x) = {x}sin2x + x3 – {x} [sin2x] + [x]sin2x – [x][sin2x] vodyuh;

ugha gS (tgk¡ [.] egÙke iw.kk±d Qyu dks rFkk {.} fHkUukRed Hkkx Qyu dks n'kkZrk gS), gksxh

5. Value of ƒ8pæ ö

ç ÷è ø

so that 2tan 4xƒ(x) (sin 4x)= becomes continuous at x

8p

= is of the form e–a/b, where a and

b are coprime numbers, then (a + b) is

ƒ8pæ ö

ç ÷è ø

dk eku rkfd 2tan 4xƒ(x) (sin 4x)= , x

8p

= ij larr~ gS] e–a/b (tgk¡ a rFkk b lgvHkkT; la[;k;s gS) :i

dk gks] rks (a + b) dk eku gksxk

6. Let ƒ(x) = x3 + x and g(x) be the inverse of ƒ(x). The value of g''(–2) is of the form ab ,

(where a and b are coprime numbers), then (b – 9a) is

ekuk ƒ(x) = x3 + x rFkk g(x), ƒ(x) dk izfrykse gSA g''(–2) dk eku ab (tgk¡ a rFkk b lgvHkkT; la[;k gS) gks] rks

(b – 9a) dk eku gksxk

LTS-38/40

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0999DT110315003

7. If tan( cos x) cot( sin x)p = p , then value of 8 2 cos x4pæ ö-ç ÷

è ø is

;fn tan( cos x) cot( sin x)p = p gks] rks 8 2 cos x4pæ ö-ç ÷

è ø dk eku gksxk

8. The number of solutions between 0 and 2p of sin2q = cos3q is0 rFkk 2p ds e/; sin2q = cos3q ds gyksa dh la[;k gksxh

9. ( )7

7 72xlim x x 1 x 1

®¥+ - - is equal to

( )7

7 72xlim x x 1 x 1

®¥+ - - dk eku gksxk

10. If ƒ(x) is a continuous and differentiable function satisfying x ƒ(x) + ƒ(x + 1) = x, then the value ofƒ'(1) + ƒ'(2) is;fn ƒ(x), Qyu x ƒ(x) + ƒ(x + 1) = x dks lUrq"V djus okyk larr rFkk vodyuh; Qyu gks] rks ƒ'(1) + ƒ'(2)dk eku gksxk



LTS-39/400999DT110315003

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Stream Course Name (Eligibility) Admission Classes FromMode Phase-I Phase-II

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(Main+Advanced) (X to XI Moving) ASAT 05-04-15 19-04-15 03 (XI to XII Moving) Direct 01-04-15 04-05-15

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(Ex-ALLEN / XII Passed before 2015)

ASAT 08-04-15 08-07-15 08-10-15 NA NA NA NA NA NA NA


Your Target is to secure Good Rank in JEE 2016 0999DT110315003LTS-40/40

OPTICAL RESPONSE SHEET / vkWfIVdy fjLikal 'khV :11. The ORS is machine-gradable and will be collected by the invigilator at the end of the examination.

vks- vkj- ,l- e'khu&tk¡P; gS rFkk ;g ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk;sxkA12. Do not tamper with or mutilate the ORS. / vks- vkj- ,l- dks gsj&Qsj@foÏfr u djsaA13. Write your name, form number and sign with pen in the space provided for this purpose on the original. Do not write

any of these details anywhere else. Darken the appropriate bubble under each digit of your form number.viuk uke] QkWeZ uEcj vkSj vks- vkj- ,l- esa fn, x, [kkuksa esa dye ls Hkjsa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh tkudkjh dghavkSj 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 cqycqyksa dks dkyk djus dh fof/k :14. Use a BLACK BALL POINT PEN to darken the bubbles in the upper sheet.

Åijh ewy i`"B ds cqycqyksa dks dkys ckWy IokbUV dye ls dkyk djsaA15. Darken the bubble COMPLETELY / cqycqys dks iw.k ± :i ls dkyk djsaA16. Darken the bubbles ONLY if you are sure of the answer / cqycqyksa dks rHkh dkyk djsa tc vkidk mÙkj fuf'pr gksA

17. The correct way of darkening a bubble is as shown here :

cqycqys dks dkyk djus dk mi;qDr rjhdk ;gk¡ n'kkZ;k x;k gS : 18. There is NO way to erase or "un-darken" a darkened bubble

dkys fd;s gq;s cqycqys dks feVkus dk dksbZ rjhdk ugha gSA19. The marking scheme given at the beginning of each section gives details of how darkened and not darkened

bubbles are evaluated.gj [k.M ds izkjEHk esa nh x;h vadu ;kstuk esa dkys fd;s x;s rFkk dkys u fd;s x, cqycqyksa dks ewY;kafdr djus dk rjhdk fn;kx;k gSA

20. 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

I HAVE READ ALL THE INSTRUCTIONS

AND SHALL ABIDE BY THEM

eSaus lHkh funs Z'kk s a dks i<+ fy;k gS vkSj es a mudk

vo'; ikyu d:¡xk@d:¡xhA

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

I have verified the identity, name and rollnumber 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 fd iz'u i= rFkk 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 ..............................................

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