Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk...

45
1 Answers & Solutions for JEE (Advanced)-2018 Time : 3 hrs. Max. Marks: 180 Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472 PAPER - 2 DATE : 20/05/2018 PART-I : PHYSICS [kaM -1( vf/ dr e vad : 24) bl [ ka Mes a Ng (06) iz ' u gS a A çR ;s d iz 'u ds l ghmÙk j (mÙk jks a ) ds fy, pk j fodY i fn, x, gS a A bu pk j fodY iks a es a ls ,d ;k ,d ls vf/ d fodY i lgh gS (gS a )A iz R ;s d iz 'u ds fy, ] iz ' u dk(ds ) mÙk j nsus gs rq l ghfodY i (fodY iksa ) dks pq usa A iz R ;s d iz 'u ds mÙk j dk ew Y ;ka du fuEu va du ; ks t uk ds vuq lk j gks xk% iw .kZ va d : +4 ; fn ds oy(l k js ) l gh fodY i (fodY iks a ) dks pq uk x; k gS A vka f' kd va d : +3 ; fn pk jks a fodY i l gh gS a ijU rq ds oy r hu fodY iks a dks pq uk x; k gS A vka f' kd va d : +2 ; fn rhu ; k r hu l svf/ d fodY i lgh gS a ijU rq ds oy nksfodY iks a dkspq uk x; k gS vkS j pq us gq , nks uks a fodY i l gh fodY i gS a A vka f' kd va d : +1 ;fn nks ;k nks ls vf/ d fodY i lgh gS a ijU rq ds oy , d fodY i dks pq uk x; k gS vkS j pq uk gq vk fodY i l gh fodY i gS A 'kw U ; va d : 0 ; fn fdl hHkhfodY i dks ugha pq uk x; k gS (vFkkZ r~ iz ' u vuq Ùk fjr gS )A ½.kkRed va d : –2 vU ; l HkhifjfLFk fr;ks a es a A mnkgj. k Lo:i % ; fn fdl h iz 'u ds fy, dsoy i gyk ] r h ljk , oa pkS Fkk l gh fodY i gS a vkS j nw ljk fodY i xyr gS a ; rks ds oy l Hkh rhu l gh fodY iksa dk p; u djus i j gh +4 va d feysa sxs A fcuk dks bZ xyr fodY i pq us (mnkgj.k es a nw ljk fodY i)] rhu l gh fodY iks a esa ls fl i Q Z nks dks pq uus ij (mnkgja r% i gyk r Fkk pkS Fkk fodY i) +2 va d feys a xs A fcuk dks bZ xyr fodY i pq us (bl mnkgj.k es a nw ljk fodY i )] r hu l ghfodY iks a es a ls fliQZ , d dks pq uus ij (igyk ;k rh ljk ;k pkS Fkk fodY i) +1 va d feys a xs A dks bZ Hkhxyr fodY i pq uus i j (bl mnkgj.k es a nw ljk fodY i )] –2 va d feya s xs ] pkgs l gh fodY i (fodY iks a ) dks pq uk x; k gks ;k u pq uk x; k gks A

Transcript of Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk...

Page 1: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

1

Answers & Solutionsfor

JEE (Advanced)-2018

Time : 3 hrs. Max. Marks: 180

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005Ph.: 011-47623456 Fax : 011-47623472

PAPER - 2

DATE : 20/05/2018

PART-I : PHYSICS[ kaM-1(vf/ dre vad : 24)

bl [ kaM esa Ng (06) i z' u gSaA

çR;sd i z' u ds l gh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa l s , d ; k ,d l s vf/ d fodYi l ghgS(gSa)A

i zR;sd i z'u ds fy, ] i z'u dk(ds) mÙkj nsus gsrq l gh fodYi (fodYi ksa) dks pqusaA

i zR; sd i z'u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk%

i w.kZ vad : +4 ; fn dsoy(l kjs) l gh fodYi (fodYi ksa) dks pquk x; k gSA

vkaf' kd vad : +3 ; fn pkjksa fodYi l gh gSa i jUrq dsoy rhu fodYi ksa dks pquk x; k gSA

vkaf' kd vad : +2 ; fn rhu ; k rhu l s vf/ d fodYi l gh gSa i jUrq dsoy nks fodYi ksa dks pquk x; k gS vkSjpqus gq, nksuksa fodYi l gh fodYi gSaA

vkaf' kd vad : +1 ; fn nks ; k nks l s vf/ d fodYi l gh gSa i jUrq dsoy ,d fodYi dks pquk x; k gS vkSj pqukgqvk fodYi l gh fodYi gSA

' kwU; vad : 0 ; fn fdl h Hkh fodYi dks ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)A

½.kkRed vad : –2 vU; l Hkh i fjfLFkfr ; ksa esaA

mnkgj.k Lo: i % ; fn fdl h i z' u ds fy, dsoy i gyk] rhl jk , oa pkSFkk l gh fodYi gSa vkSj nwl jk fodYi xyr gSa; r ks dsoyl Hkh rhu l gh fodYi ksa dk p; u djus i j gh +4 vad feysasxsA fcuk dksbZ xyr fodYi pqus (mnkgj.k esa nwl jk fodYi )] rhul gh fodYi ksa esa l s fl i QZ nks dks pquus i j (mnkgjar% i gyk rFkk pkSFkk fodYi ) +2 vad feysaxsA fcuk dksbZ xyr fodYi pqus(blmnkgj.k esa nwl jk fodYi )] rhu l gh fodYi ksa esa l s fl i QZ ,d dks pquus i j (i gyk ; k rhl jk ; k pkSFkk fodYi ) +1 vad feysaxsAdksbZ Hkh xyr fodYi pquus i j (bl mnkgj.k esa nwl jk fodYi )] –2 vad feyasxs] pkgs l gh fodYi (fodYi ksa) dks pquk x; k gks; k u pquk x; k gksA

Page 2: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

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JEE (ADVANCED)-2018 (PAPER-2)

1. nzO;eku (mass) m dk ,d d.k 'kq:vkr esa ewy fcUnq (origin) ij fojkekoLFkk esa gSA d.k ij ,d cy yxkus ls og x-v{k ij

pyus yxrk gS vkSj d.k dh xfrt ÅtkZ (kinetic energy) K, le; ds lkFk dKt

dt ds vuqlkj ifjo£rr gksrh gS] tgk¡

,d mfpr foekvksa okyk /ukRed fu;rkad (positive constant) gSA fuEufyf[kr dFkuksa esa ls dkSu lk (ls) lgh gS (gSa)\

(A) d.k ij yxk;k x;k cy fu;r (constant) gS

(B) d.k dh pky le; ds lekuqikfrd (proportional) gS

(C) d.k dh ewy fcUnq ls r; dh x;h nwjh] le; ds lkFk js[kh; rjhds ls (linearly) c<+rh gS

(D) cy laj{kh (conservative) gS

mÙkjmÙkjmÙkjmÙkjmÙkj (A, B, D)

gygygygygy ⇒ 21 dK dvK mv mv

2 dt dt

fn;k gS] ⇒ dK dvt mv t

dt dt

⇒ ∫ ∫v t 2 2

0 0

v tvdv tdt

m 2 m 2

v t

m

dv

adt m

F ma m fu;r

⇒ 2

dS tt S

dt m m 2

2. eku yhft, fd ,d ';ku (viscous) nzo ds ,d cM+s VSad (tank) esa ,d iryh oxkZdkj IysV (thin square plate) rSj jgh

gSA VSad esa nzo dh Å¡pkbZ h, VSad dh pkSM+kbZ ls cgqr de gSA rSjrh gqbZ IysV dks ,d fu;r (constant) osx u0 ls {kSfrt

fn'kk esa [khapk tkrk gSA fuEufyf[kr dFkuksa esa ls dkSu lk (ls) lgh gS (gSa)\

(A) nzo ds }kjk IysV ij yxk;k x;k çfrjks/d cy (resistive force) h ds O;qRØekuqikfrd (inversely poroportional) gS

(B) nzo ds }kjk IysV ij yxk;k x;k çfrjks/d cy IysV ds {ks=kiQy ij fuHkZj ugha djrk gS

(C) VSad dh iQ'kZ (floor) ij yxrk gqvk Li'kZjs[kh; çfrcy (tangential/shear stress) u0 ds lkFk c<+rk gS

(D) IysV ij yxus okys Li'kZjs[kh; çfrcy nzo dh ';kurk (viscosity) ds lkFk js[kh; rjhds ls (linearly) cnyrh gS

mÙkjmÙkjmÙkjmÙkjmÙkj (A, C, D)

gygygygygy u0

Fv

A

Plate

⎛ ⎞ ⎜ ⎟⎝ ⎠

v

dvF A

dz

pw¡fd VSad esa nzo dh Å¡pkbZ h vR;Ur de gS ⎛ ⎞⇒ ⎜ ⎟ ⎝ ⎠

0udv v

dz z h

⎛ ⎞ ⎜ ⎟⎝ ⎠

0v

uF ( )A

h

⎛ ⎞ ⎜ ⎟⎝ ⎠

v v 0 v v

1F ,F u ,F A,F

h

Page 3: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

3

JEE (ADVANCED)-2018 (PAPER-2)

3. z-v{k ds l ekUr j ,d vuUr yEckbZ dh i ryh vpkyd (non-conducting) rkj i j ,dl eku js[ kh; vkos' k ?kuRo (uniform

line charge density) gSA] ; g r kj R f=kT; k okys ,d i r ys vpkyd xksyh; dks' k (spherical shell) dks bl çdkjHksnrk gS fd vkdZ (arc) PQ, xksyh; dks' k ds dsUnz O i j 120° dk dks.k cukrh gS] t Sl k fd fp=k esa n' kkZ; k x; k gSA eqDrvkdk' k dk i jkoS| qrkad (permittivity of free space)

0 gSA fuEufyf[ kr dFkuksa esa l s dkSu l k (l s) l gh gS (gSa)\

P

R

O

Q

120°

z

(A) dks' k l s xqt jus okyk oS| qr ÝyDl (electric flux) 0

3R gS

(B) oS| qr {ks=k (electric field) dk z-?kVd (z-component) dks' k ds i "B (surface) ds l Hkh fcUnqvksa i j ' kwU; gS

(C) dks'k l s xqt jus okyk oS| qr ÝyDl (electric flux) 0

2R gS

(D) oS| qr {ks=k (electric field) dks'k ds i "B ds l Hkh fcUnqvksa i j yEcor (normal) gS

mÙkj (A, B)

gy PQ = (2) R sin 60°

3

(2R) 3R2

P

R

O

Q

120°

z

R

Q 3 R i fjc¼

0

q

i fjc¼

0

3 R

rFkk fo| qr {ks=k rkj ds yEcor gS] bl fy, Z-?kVd ' kwU; gksxkA

4. , d r kj dks ,d l edks.k f=kHkqt ds vkdkj esa eksM+ dj f i Qksdl nwjh (focal length) okys ,d vory ni Z.k (concave

mirror) ds l keus j[ kk x; k gS] t Sl k fd fp=k esa n' kkZ; k x; k gSA pkj fodYi fp=kksa esa l s dkSu l k (l s) fp=k eqM+s gq, r kj dsçfr fcEc dk l gh vkdkj xq.kkRed r jhds l s n' kkZrk gS (n' kkZrs gSa)\ (; s fp=k Ldsy (scale) ds vuql kj ugha gSaA)

45°ff

2

Page 4: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

4

JEE (ADVANCED)-2018 (PAPER-2)

(A)

> 45°

(B)

(C) 0 < < 45°

(D)

mÙkj (D)

gy fcUnq A dk i zfr fcEc

f/2

B

A

Q

(f – x)

x CPO

1 1 1v f

fv f2

ABAB

I fI 2AB

fAB2

For Height of PQ

1 1 1 1 1 1 f(f x)

vv (f x) f v (f x) f x

PQI f(f x) fPQ x{ (f x)} x

PQf f 2(AB)x

I PQx x f

2(AB)x

PQf

IPQ = 2AB

rFkk 2(AB)x

PQf

5. , d jsfM; ks, fDVo {k; Ja[ kyk (decay chain) esa 23290 Th ukfHkd] 212

82 Pb ukfHkd esa {kf; r gksrk gSA bl {k; i zØe (process)

esa mRl ft Zr gq, (emitted) vkSj – d.kksa dh l a[ ; k Øe' k% N vkSj N gSaA fuEufyf[ kr dFkuksa esa l s dkSu l k (l s) l gh gS (gSa)\

(A) N 5 (B) N 6

(C) N 2 (D) N 4

mÙkj (A, C)

Page 5: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

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JEE (ADVANCED)-2018 (PAPER-2)

gy A A 4Z Z 2X Y

{k;

A AZ Z 1 1X X

{k;

232 21290 82Th Pb

mRl ft Zr -d.kksa dh l a[ ; k

232 2125

4

pw¡fd Z, (90 – 82) rd ?kV t krk gS = 8 dssoy

bl fy, – {k; dh l a[ ; k = 2

6. vuquknh ok; q&LraHk (resonating air column) ds ,d i z; ksx esa èofu dh pky eki us ds fy, 500 Hz dh vkofÙk okys ,dLofj=k f}Hkqt (tuning fork) dk mi ; ksx fd; k t krk gSA vuqukn uyh esa t y dk Lr j cnydj ok; q&LraHk dh yEckbZ cnyht krh gSA nks mÙkjksÙkj (successive) vuqukn] ok; q LrEHk dh yEckbZ 50.7 cm vkSj 83.9 cm i j l qus t krs gSaA fuEufyf[ krdFkuksa esa l s dkSu l k (l s) l gh gS (gSa)\

(A) bl i z; ksx l s fuèkkZfjr èofu dh pky 332 m s–1 gS

(B) bl i z; ksx esa vaR; l a' kksèku (end correction) 0.9 cm gS

(C) èofu r jax dh r jaxnSè; Z (wavelength) 66.4 cm gS

(D) 50.7 cm i j l qus t kus okyk vuqukn] ewy xq.kkofÙk (fundamental harmonic) gS

mÙkj (A, B, C)

gy

(2n – 1) 50.7 e4

2 (n 1) – 1 83.9 e4

83.9 – 50.7 33.2 cm2

= 66.4 cm

v = f = 66.4 × 500 × 10–2 m/s = 332 m/s

For, n = 2, e = – 0.9 cm

[ kaM-2(vf/ dre vad : 24)

bl [ kaM esa vkB (08) i z' u gSaA i zR;sd i z'u dk mÙkj , d l a[ ; kRed eku (NUMERICAL VALUE) gSA

çR; sd i z' u ds mÙkj ds l gh l a[ ; kRed eku(n' keyo vadu esa] n' keyo ds f}rh; LFkku rd : f.Mr@fudfVr ; mnkgj.kr%6.25, 7.00, -0.33, -.30, 30.27, -127.30) dks ekmt + (MOUSE) vkSj vkWu LØhu (ON-SCREEN) opqZvy U; wesfjd dhi SM(VIRTUAL NUMERIC KEYPAD) ds i z; ksx l s mÙkj ds fy, fufnZ"V LFkku i j nt Z djsaA

i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk&

i w.kZ vad : +3 ; fn fl i QZ l gh l a[ ; kRed eku (Numerical value) gh mÙkj Lo: i nt Z fd; k x; k gSA

' kwU; vad : 0 vU; l Hkh i fjfLFkfr ; ksa esaA

Page 6: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

6

JEE (ADVANCED)-2018 (PAPER-2)

7. , d Bksl {kSfr t r y (solid horizontal surface) r sy dh ,d i r yh i jr (thin layer) l s <dk (covered) gqvk gSAnzO; eku (mass) m = 0.4 kg dk , d vk; r kdkj xqVdk (rectangular block) bl r y i j fojkekoLFkk esa gSA1.0 N s i fjek.k dk ,d vkosx (impulse) xqVds i j t = 0 l e; i j yxk; k t kr k gS ft l ds i QyLo: i xqVdk x-v{k(x-axis) i j (t) = 0 e–t/ osx l s pyus yxrk gS] t gk¡ 0 , d fLFkj jkf' k gS vkSj = 4 s gSA l e; t = i j] xqVds dkfoLFkki u (displacement)_____ ehVj gSA e–1 = 0.37 ysaA

mÙkj (6.30)

gy m/s0

1v = = 2.5

0.4

–t /ds= 2.5e

dt

s t–t /

0 0

ds = 2.5e dt

t–t /

0

2.5s = e

–1 = –t /2.5 1– e

t = i j] s = 10[1 – e–1] = 10 × 0.63 = 6.30 m

8. , d xsan dks Hkwfe (ground) i j {kSfr t r y (horizontal surface) l s 45º ds dks.k i j i z{ksfi r (projected) fd; k t krkgSA xsan 120 m dh vfèkdre Å¡pkbZ i j i gq¡p dj Hkwfe i j oki l ykSV vkrh gSA Hkwfe l s i gyh ckj Vdjkus ds mi jkar xsan dhxfr t Åt kZ (kinetic energy) vkèkh gks t krh gSA Vdjkus ds rqjar ckn xsan dk osx {kSfr t r y l s 30º dk dks.k cukrk gSAVdjkus ds ckn xsan_____ ehVj dh vfèkdre Å¡pkbZ i j i gq¡prh gSA

mÙkj (30.00)

gy 2 2u sin 45°

= 1202g

2u

= 1204g

; fn i gyh VDdj ds ckn pky v gS] rc

2v= 60

4g pw¡fd u

v2

hvfèkdre = 2 2v sin 30°

2g = 2v

8g = 30 m = 30.00 m

9. , d d.k] ft l dk nzO; eku (mass) 10–3 kg vkSj vkos' k (charge) 1.0 C gS] ' kq#vkr esa fojkekoLFkk esa gSA l e; t = 0 i j

; g d.k , d fo| qr ~ {ks=k (electric field) 0ˆE (t) E sin t I

ds i zHkko esa vkr k gS] t gk¡ 1

0E 1.0 N C gS vkSj

3 110 rad s gSA d.k i j dsoy fo| qr~ cy (electrical force) dk gh i zHkko ekfu; sA rc i jorhZ (subsequent) l e;

i j d.k dh vfèkdre pky_____ m s–1 gSA

mÙkj (2.00)

gy ˆ 3F = qE = 1.0 N sin (10 t) i

3 3Fa = = 10 sin (10 t)

m

Page 7: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

7

JEE (ADVANCED)-2018 (PAPER-2)

3 3dv= 10 sin (10 t)

dt

V t

3 3

0 0

dv = 10 sin(10 t)dt 3

33

10v = 1 – cos(10 t)

10

vmax = 2 m/s = 2.00 m/s

10. , d py dqaMyh xSYosuksehVj (moving coil galvanometer) esa 50 i sQjs (turns) gSa vkSj gj i sQjs dk {ks=ki Qy (area)

2 × 10–4 m2 gSA xSYosuksehVj esa mi fLFkr pqEcd l s 0.02 T dk pqEcdh; {ks=k (magnetic field) mRi Uu gksrk gSA fuyacurkj (suspension wire) dk , saBu fu; r kad (torsional constant) 10–4 N m rad–1 gSA xSYosuksehVj esa èkkjk cgus dsl e; ] ; fn dqaMyh 0.2 rad ?kwerh gS rks xSYosuksehVj esa i w.kZ i Sekuk fo{ksi (full scale deflection) gksrk gSA xSYosuksehVj dhdqaMyh dk i zfr jksèk 50 gSA bl xSYosuksehVj dks 0 – 1.0 A dh jsUt (range) esa èkkjk ds eki u djus ; ksX; ,d , sehVj(ammeter) ds : i esa i fjofrZr djuk gSA bl ds fy, , d ' kaV (shunt) i zfr jksèkd dks xSYosuksehVj l s i kÜoZØe (parallel) esal a;ksft r djuk i M+rk gSA bl ' kaV i zfr jksèkd dk eku_____vkse (ohms) gSA

mÙkj (5.56)

gy = BANim = K

–4

m –4

K 10 × 0.2i = =

BAN 0.02 × 2 × 10 × 50

0.2= = 0.1 A

2

0.1 × 50 = 0.9 S 50

S = = 5.56 9

11. , d bLi kr ds r kj] ft l dk O; kl (diameter) 0.5 mm gS vkSj ; ax xq.kkad (Young's modulus) 2 × 1011 Nm–2 gS] l s M

nzO; eku (mass) dk ,d Hkkj yVdk; k t krk gSA Hkkj yVdkus ds ckn r kj dh yackbZ 1.0 m gSA bl r kj ds var esa 10

Hkkxksa okyk ,d ofuZ; j i Sekuk (vernier scale) yxk; k t krk gSA bLi kr ds rkj ds i kl , d vkSj l anHkZ (reference) rkj gSft l i j 1.0 mm vYi rekad (least count) okyk ,d eq[ ; i Sekuk (main scale) yxk gqvk gSA ofuZ; j i Sekus ds 10 Hkkxeq[ ; i Sekus ds 9 Hkkxksa ds cjkcj gSaA ' kq#vkr esa] ofuZ; j i Sekus dk ' kwU; eq[ ; i Sekus ds ' kwU; l s l ai krh (coincident) gSA ; fnbLi kr ds r kj i j yVdk; k x; k Hkkj 1.2 kg l s c<+k;k t krk gS] r ks eq[ ; i Sekus ds Hkkx l s l ai krh gksus okyk ofuZ; j i Sekus dkHkkx_____gSA g = 10 ms–2 vkSj = 3.2 ysaA

mÙkj (3.00)

gyW

LYAL

11 2 –6

1.2 10 4 1

2 10 (0.5) (10 )= 0.3 mm

ofuZ; j dk vYi rekad

91–

10 mm = 0.1 mm

ofuZ; j dk i kB~;kad = 3.

Page 8: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

8

JEE (ADVANCED)-2018 (PAPER-2)

12. , di jekf.od vkn' kZ xSl (monatomic ideal gas) ds , d eksy dk vk; r u (volume), #¼ks"e i zl kj (adiabatic

expansion) l s] vi us vkjafHkd eku dk vkB xquk c<+ t krk gSA l koZf=kd xSl fu; r kad (universal gas constant) R dkeku 8.0 J mol–1 K–1 ysaA ; fn xSl dk vkjafHkd rki eku 100 K gks] r ks bl i zfØ; k esa xSl dh vkar fjd Åt kZ (internal

energy) _____t wy (Joule) l s de gks t krh gSA

mÙkj (900.00)

gy 5

n 1,3

T1 V1–1 = T2V2

–1

2–131

2 12

V 1T T 100

V 8

T2 = 25 K

2 13R

U n T – T2

3

1 8 (25 – 100)2

= – 900 J

vkUrfjd Åt kZ esa deh = 900 J

13. ,d çdk'k fo| qr (photoeletric) ç;ksx esa 200 W ' kfDr (power) okyk ,d l ekUr j ,do.khZ çdk' k fdj.k i qat (a parallel

beam of monochromatic light) i w.kZ : i l s vo' kksf"kr djus okys ,d mRl t Zd (perfectly absrobing cathode)

i j fxjrk gSA mRl t Zd ds i nkFkZ dk dk;Z&i Qyu (work function) 6.25 eV gSA çdk'k dh vkofÙk] nsgyh vkofÙk (threshold

frequency) l s FkksM+h gh vf/ d gS] ft l l s mRl £t r gksus okyh çdkf' kd bysDVªkWuksa (photoelectrons) dh xfr t Åt kZ (kinetic

energy) ux.; gSA eku yhft ,s fd çdk'k fo| qr mRl t Zu n{krk (photoelectron emission efficiency) 100% gSA mRl t ZdvkSj l axzkgd (anode) ds chp 500 V dk foHkokUr j (potential difference) yxk; k t krk gSA mRl £t r gksus okys l HkhbysDVªkWu l axzkgd i j vfHkyEc vki fr r (normal incidence) gksdj vo' kksf"kr gks t krs gSaA bysDVªkWuksa dh l axzkgd i j VDdjl s f = n × 10–4 N dk cy yxrk gSA n dk eku ________ gSA bysDVªku dk nzO; eku (mass) me = 9 × 10–31 kg gS vkSj1.0 eV = 1.6 × 10–19 J gSA

mÙkj (24.00)

gy P = 200 J/s,

i zfr l sd.M i QksVkWu dh l a[ ; k (N) = –19

200

(6.25 1.6 10 )

= 2 × 1020

P 2m (KE) 2m(eV)

–31 –192 9 10 1.6 10 500

= 12 × 10–24

F = P × N = 12 × 10–24 × 2 × 1020

= 24 × 10–4N

Page 9: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

9

JEE (ADVANCED)-2018 (PAPER-2)

14. , d gkbMªkst u&t Sl k vk; fur (hydrogen-like ionized) i jek.kq dk i jek.kq Øekad (atomic nunber) Z gSA bl i jek.kq esa,d gh bysDVªku gSA bl i jek.kq ds mRl t Zu&Li sDVªe (emission spectrum) esa] n = 2 l s n = 1 l aØe.k (transition) l smRi Uu gksus okys i QksVkWu (photon) dh Åt kZ] n = 3 l s n = 2 l aØe.k (transition) l s mRi Uu gksus okys i QksVkWu (photon)

dh Åt kZ l s 74.8 eV vf/ d gSA gkbMªkst u i jek.kq dh vk; uu Åt kZ (ionization energy) 13.6 eV gSA Z dk eku___________ gSA

mÙkj (3.00)

gy

2 21 1 1 113.6 – Z 74.8 13.6 – Z

1 4 4 9

2 3 513.6 Z – 74.8

4 36 Z2 = 9 Z = 3

[ kaM 3 (vf/ dre vad % 12)

• bl [ kaM esa pkj (04) i z'u gSaA i zR;sd i z' u esa nks (02) l qesyu l wfp;k¡ (matching lists) gSa% l wph–I vkSj l wph–II |

• l wph–I vkSj l wph–II ds rÙoksa ds l qesykuks dks n'kkZrs gq, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa fl iQZ ,d fodYi gh l gh

l qesyu i znf' kZr djr k gSA

• i zR; sd i z' u ds fy, l gh l qesyu i znf' kZr djr s okys fodYi dks pqusaA

• i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk

i w.kZ vad % +3 ; fn fl i QZ l gh fodYi gh pquk x; k gSA

' kwU; vad % 0 ; fn dksbZ Hkh fodYi ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)A

Í .k vad % –1 vU; l Hkh i fjfLFfr ; ksa esaA

15. fofHkUu vkos' k for j.kksa (charge distributions) l s mRi Uu gksus okys fo| qr {ks=k (electric field) E dk ,d fcUnq P(0, 0, d)

i j eki u fd; k t krk gS vkSj bl fo| qr {ks=k E dh d i j fuHkZjr k vyx&vyx i k; h t krh gSA l wph-I esa E vkSj d ds chp esa vyx&vyx l EcU/ (relations) fn; s x; s gSaA l wph-II fofHkUu çdkj ds vkos' k for j.kksa vkSj muds LFkkuksa dks crkrh gSaA l wph-I ds i Qyuksadk l wph-II l s l acaf/ r vkos' k for j.kksa l s l qesy dhft ,A

l wph-I l wph-II

P. E, d i j fuHkZj ugha djr k gS 1. ewy fcUnq (origin) i j fcUnq vkos'k (point charge) Q

Q. 1

Ed

2. , d y?kq f}/ qzo (small dipole) ft l dk fcUnq vkos' k Q t ks

(0, 0, l) i j gS vkSj –Q t ks (0, 0, –l) i j gSA ekfu; s 2l <<d

R. 2

1E

d3. vuUr (infinite) yEckbZ dk ,dl eku js[ kh; vkos'k ?kuRo

(uniform linear charge density) okyk r kj t ks x-v{kl s l Ei krh (coincident) gS

S. 3

1E

d4. vuUr yEckbZ ds ,dl eku js[ kh; vkos'k ?kuRo okys nks r kj t ks

x-v{k ds l ekUr j gSaA (y = 0, z = l) okys r kj i j + vkos'k?kuRo gS rFkk (y = 0, z = –l) okys r kj i j – vkos'k ?kuRogSA ekfu, 2l <<d

5. ,dl eku vkos'k ?kuRo (uniform surface charge density)

dk vuUr l ery pknj (infinite plane sheet) t ks xy- ryl s l Ei krh gS

Page 10: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

10

JEE (ADVANCED)-2018 (PAPER-2)

(A) P 5; Q 3, 4; R 1; S 2

(B) P 5; Q 3; R 1, 4; S 2

(C) P 5; Q 3; R 1, 2; S 4

(D) P 4; Q 2, 3; R 1; S 5

mÙkj (B)

gy l wph-II

(1) 2

0

1 QE

4 d

2

1E

d

(2) 30

1 2Q(2 )E

4 d

v{k 3

1E

d

(3)

0

E2 d

1E

d

(4)

2

0 0 0

(2 )E –

2 (d – ) 2 (d ) 2 d 2

1E

d

(5)

0

E2 E, d i j fuHkZj ugha djrk gSA

16. M nzO;eku (mass) okys ,d xzg (planet) ds nks çkÑfrd mi xzg (natural satellites) nks oÙkh; d{kksa esa i fjØe.k (revolve)

dj jgs gSaA mi xzgksa ds chp xq: Rokd"kZ.k cy (gravitational attraction) dh mi s{kk dhft ; sA i gyk mi xzg] ft l dk nzO; eku

m1, d{kh; pky 1, dks.kh; l aosx (angular momentum) L1, xfr t Åt kZ (kinetic energy) K1 vkSj vkorZdky (period

of revolution) T1 gS] R1 f=kT; k okyh d{kk esa LFkkfi r gSA nwl jk mi xzg] ft l dk nzO; eku m2, d{kh; pky v2, dks.kh; l aosx

L2, xfr t Åt kZ K2 vkSj vkorZdky T2 gS] R2 f=kT; k okyh d{kk esa LFkkfi r gSA ; fn m1/m2 = 2 vkSj R1/R2 = 1/4 gks] r ks l wph-

I esa fn, x, vuqi krksa dk l qesy l wph-II esa nh x; h l a[ ; kvksa ds l kFk djsaA

l wph-I l wph-II

P.1

2

v

v 1.18

Q.1

2

L

L 2. 1

R.1

2

K

K 3. 2

S.1

2

T

T 4. 8

(A) P 4; Q 2; R 1; S 3

(B) P 3; Q 2; R 4; S 1

(C) P 2; Q 3; R 1; S 4

(D) P 2; Q 3; R 4; S 1

mÙkj (B)

Page 11: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

11

JEE (ADVANCED)-2018 (PAPER-2)

gy 2

2

GMm mv GMv

R RR

ekuk R1 = R R2 = 4R

m2 = m m1 = 2m

l wpht-I

(P) 1 2

2 1

v R 4R2 : 1

v R R

(Q) L = mvR

1 1

2 2

L R(2m)v 1(2) 1: 1

L 4R(m)v 2

m2

v2

R1

m1

v1

M

R2

(R)

21

1

222

1(2m)vK 2 2(4) 8 : 1

1K (m)v2

(S)

3 32 2

1 1

2 2

T R 11: 8

T R 4

17. , di jek.fod vkn' kZ xSl (monatomic ideal gas) dk ,d eksy (one mole), pkj Å"ekxrh; i zØeksa (thermodynamic

processes) l s xqt jr k gS] t Sl k fd uhps PV O; oLFkk fp=k (schematic diagram) eas n' kkZ; k x; k gSA ; gk¡ fn; s x; s i zØeksa

esa ,d l enkch; (isobaric)] , d l evk; r fud (isochoric)] , d l erki h; (isothemal) vkSj , d : ¼ks"e (adiabatic)

gSaA l wph&I esa fn, x, i zØeksa dk l wph&II esa fn, x, l axr dFkuksa l s l qesy djsaA

3P0

P

V0 3V0

I

II

IIIIV

V

P0

l wph-I l wph-II

P. i zØe-I esa 1. xSl }kjk fd; k x; k dk; Z ' kwU; gS

Q. i zØe-II esa 2. xSl dk rki eku ugha cnyrk gS

R. i zØe-III esa 3. xSl vkSj i fjos' k ds chp Å"ek i zokg ugha gksrk gS

S. i zØe-IV esa 4. xSl }kjk fd; k x; k dk;Z 6P0V0 gS

(A) P 4; Q 3; R 1; S 2

(B) P 1; Q 3; R 2; S 4

(C) P 3; Q 4; R 1; S 2

(D) P 3; Q 4; R 2; S 1

mÙkj (C)

Page 12: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

12

JEE (ADVANCED)-2018 (PAPER-2)

gy adiabatic Isothermal

dP dPdV dV

l wph-1

(P) i zØe I #¼ks"e Q = 0

(Q) i zØe II l enkch;

W = PV = 3P0 [3V0 – V0] = 6P0V0

(R) i zØe III l evk; r fud W = 0

(S) i zØe (IV) l er ki h; rki = fu; rA

18. uhps nh x; h l wph-I esa] , d d.k ds pkj fofHkUu i Fk] l e; ds fofHkuu i Qyuksa (functions) ds : i esa fn; s x; s gSaA bu i Qyuksaeas vkSj mfpr foekvksa okys / ukRed fu; r kad (positive constants) gSa] t gk¡ | i zR; sd i Fk eas d.k i j yxus okyk

cy ; k r ks ' kwU; gS ; k l aj{kh (conservative) gSA l wph-II eas d.k dh i k¡p HkkSfrd jkf' k; ksa dk fooj.k fn; k x; k gS%p

js[ kh;

l aosx (linear momentum) gS] L ewyfcanq (origin) ds l ki s{k dks.kh; l aosx (angular momentum) gS] K xfr t Åt kZ

(Kinetic energy) gS] U fLFkfr t Åt kZ (potential energy) gS vkSj E dqy mt kZ (total energy) gSA l wph-I ds i zR; sdi Fk dk l wph-II esa fn; s x; s mu jkf' k; ksa l s l qesy dhft , ] t ks ml iFk ds fy, l aj{kh (conserved) gSaA

l wph-I l wph-II

P.

ˆ ˆr(t) t i t j 1.p

Q.

ˆ ˆr(t) cos t i sin t j 2.L

R.

ˆ ˆr(t) (cos t i sin t j) 3. K

S.

2ˆ ˆr(t) t i t j2

4. U

5. E

(A) P 1, 2, 3, 4, 5; Q 2, 5; R 2, 3, 4, 5; S 5

(B) P 1, 2, 3, 4, 5; Q 3, 5; R 2, 3, 4, 5; S 2, 5

(C) P 2, 3, 4; Q 5; R 1, 2, 4; S 2, 5

(D) P 1, 2, 3, 5; Q 2, 5; R 2, 3, 4, 5; S 2, 5

mÙkj (A)

gy t c cy F = 0 rc fLFkfr t Åt kZ U = fu; r

F 0 cy l aj{kh gS dqy Åt kZ E = fu; r

l wph-I

(P) ˆ ˆr(t) t i tj

dr ˆ ˆv i j

dt = fu; r kad p

fu; rkda

2 2|v| constant K = fu; r kad

Page 13: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

13

JEE (ADVANCED)-2018 (PAPER-2)

dv

a 0 F 0 U constantdt

E = U + K = fu; r kad

L m(r v) 0

L constant

P 1, 2, 3, 4, 5

(Q) ˆ ˆr(t) cos ti sin tj

dr ˆ ˆv sin t( i) cos tjdt

fu; r kda p

fu; r kda

2 2|v| ( sin t) ( cos t) fu; rkad K fu; rkad

2dv

a r 0dt

E = fu; rkad = K + U

But K fu; rkad U fu; rkad

ˆL m(r v) m (k)

fu; rkda

Q 2, 5

(R) ˆ ˆr(t) (cos ti sin tj)

dr ˆ ˆv [sin t( i) cos tj]dt

fu; rkda p

fu; r kda

|v|

fu; rkda K = fu; rkad

2dva r 0 E , U

dt

fu; rkda fu; rkda

2 ˆL m(r v) m k

fu; rkda

R 2, 3, 4, 5

(S)

2ˆ ˆr(t) t i t j

2

dr ˆ ˆv i tj pdt

fu; r kda fu; r kda

2 2|v| ( t) K

fu; rkda fu; rkda

dv ˆa j 0 E K Udt

fu; rkda

But K fu; rkda

U fu; rkda

21 ˆL m(r v) t k2

fu; r kda

S 5

Page 14: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

14

JEE (ADVANCED)-2018 (PAPER-2)

PART-II : CHEMISTRY[ kaM-1(vf/ dre vad : 24)

bl [ kaM esa Ng (06) i z' u gSaA

çR;sd i z' u ds l gh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa l s , d ; k ,d l s vf/ d fodYi l ghgS(gSa)A

i zR;sd i z'u ds fy, ] i z'u dk(ds) mÙkj nsus gsrq l gh fodYi (fodYi ksa) dks pqusaA

i zR; sd i z'u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk%

i w.kZ vad : +4 ; fn dsoy(l kjs) l gh fodYi (fodYi ksa) dks pquk x; k gSA

vkaf' kd vad : +3 ; fn pkjksa fodYi l gh gSa i jUrq dsoy rhu fodYi ksa dks pquk x; k gSA

vkaf' kd vad : +2 ; fn rhu ; k rhu l s vf/ d fodYi l gh gSa i jUrq dsoy nks fodYi ksa dks pquk x; k gS vkSjpqus gq, nksuksa fodYi l gh fodYi gSaA

vkaf' kd vad : +1 ; fn nks ; k nks l s vf/ d fodYi l gh gSa i jUrq dsoy ,d fodYi dks pquk x; k gS vkSj pqukgqvk fodYi l gh fodYi gSA

' kwU; vad : 0 ; fn fdl h Hkh fodYi dks ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)A

½.kkRed vad : –2 vU; l Hkh i fjfLFkfr ; ksa esaA

mnkgj.k Lo: i % ; fn fdl h i z' u ds fy, dsoy i gyk] rhl jk , oa pkSFkk l gh fodYi gSa vkSj nwl jk fodYi xyr gSa; r ks dsoyl Hkh rhu l gh fodYi ksa dk p; u djus i j gh +4 vad feysasxsA fcuk dksbZ xyr fodYi pqus (mnkgj.k esa nwl jk fodYi )] rhul gh fodYi ksa esa l s fl i QZ nks dks pquus i j (mnkgjar% i gyk rFkk pkSFkk fodYi ) +2 vad feysaxsA fcuk dksbZ xyr fodYi pqus(blmnkgj.k esa nwl jk fodYi )] rhu l gh fodYi ksa esa l s fl i QZ ,d dks pquus i j (i gyk ; k rhl jk ; k pkSFkk fodYi ) +1 vad feysaxsAdksbZ Hkh xyr fodYi pquus i j (bl mnkgj.k esa nwl jk fodYi )] –2 vad feyasxs] pkgs l gh fodYi (fodYi ksa) dks pquk x; k gks; k u pquk x; k gksA

1. l adqy [Co(en)(NH3)3(H2O)]3+ (en = H2NCH2CH2NH2) ds fo"k; esa l gh fodYi gS (gSa)

(A) bl ds nks T; kferh; l eko; o (geometrical isomers) gksrs gSa

(B) bl ds rhu T; kferh; l eko; o gksaxs ; fn f}narqj (bidentate) 'en' dks nks l k; ukbM fyxUMksa (cyanide ligands) l s cnykt k,

(C) ; g vuqpqEcdh; (paramagnetic) gS

(D) ; g [Co(en)(NH3)4]3+ dh rqyuk esa yach r jax&nSè; Z (wavelength) dk i zdk'k vo' kksf"kr djr k gS

mÙkj (A, B, D)

gy (A) [Co(en)(NH3)3(H2O)]3+

Co

OH2

NH3

NH3

NH3

NH2CH2

H N2CH2

Co

NH3

OH2

NH3

NH3

NH2CH2

NH2CH2

3+ 3+

Page 15: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

15

JEE (ADVANCED)-2018 (PAPER-2)

(B) [Co(CN)2(NH

3)3(H

2O)]1+

Co

OH2

NH3

NH3

NH3

NC

NC

NH3

OH2

NH3

NH3

NC

NC

CN

OH2

NH3

CN

H N3

H N3

Co Co

1+ 1+ 1+

(C) Co3+ = [Ar]3d6

en rFkk NH3 dh mifLFkfr esa ;g fuEu pØ.k ladqy cukrk gS

(D) [Co(en)(NH3)

4]3+ e s a e

g rFk k t

2g ds eè; vUrj ky [Co(en)(NH

3)

3(H

2O)]3+ l s vfè kd g SA vr%

[Co(en)(NH3)

4]3+ dh vis{kk [Co(en)(NH

3)

3(H

2O)3+ nh?kZ rjaxnSè;Z vo'kksf"kr djrk gS

2. vyx ls fy, x, Mn2+ vkSj Cu2+ ds ukbVªsV yo.kksa ds foHksnu ds fy, lgh fodYi gS (gSa)

(A) Tokyk ijh{k.k (flame test) esa Mn2+ vfHky{kf.kd (characteristic) gjk jax fn[kkrk gS

(B) vEyh; ekè;e esa H2S izokfgr djus ij dsoy Cu2+ vo{ksi dk cuuk fn[kkrk gS

(C) gYds {kkjdh; ekè;e esa H2S izokfgr djus ij dsoy Mn2+ vo{ksi dk cuuk fn[kkrk gS

(D) Cu2+/Cu dk vip;u foHko (reduction potential) Mn2+/Mn ls mPprj gS (le:i voLFkk ij ekik x;k)

mÙkjmÙkjmÙkjmÙkjmÙkj (B, D)

gygygygygy (A) eSaxuht Tokyk ijh{k.k esa gYdk cSaxuh jax n'kkZrk gS

(B) H S2 2

HClCu CuS

dkyk vo{kis

(C) Cu+2 rFkk Mn+2 nksuksa {kkjh; ekè;e esa H2S ds lkFk vo{ksi cukrs gS

(D) 2Cu /Cu

E 0.34

2Mn /Mn

E 1.18 V

3. ,sfufyu feJ.k vEy (lkUnz HNO3 rFkk lkUnz H

2SO

4 ) ds lkFk 288 K ij vfHkfØ;k djds P (51%), Q (47%) vkSj R(2%)

nsrk gSA fuEufyf[kr vfHkfØ;k vuqØeksa dk (ds) eq[; mRikn (major products(s)) gS (gSa)

1) Ac O, pyridine

2) Br , CH CO H2

2 3 2

3) H O

4) NaNO , HCl/273-278K

5) EtOH,

3

2

+

R

1) Sn/HCl2) Br /H O (excess)

2 2

3) NaNO , HCl/273-278K

4) H PO2

3 2

Smajor product(s)

(A)

BrBr

Br

Br

(B)

Br

Br

Br

Br

(C)

BrBr

Br

(D)

Br

Br

Br Br

mÙkjmÙkjmÙkjmÙkjmÙkj ( D )

Page 16: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

16

JEE (ADVANCED)-2018 (PAPER-2)

gy NH2

Conc. HNO 3

& Conc. H SO2 4

NH2

NO2

+

NH2

NO2

+

NH2

NO2

51%(P)

47%(Q)

2%(R)

NH2

NO2 Ac O, Pyridine2

NH – C – CH3

NO2

O

Br2

CH COOH3

NO2

NH – C – CH3

O

Br

H O3+

NO2

NH2

Br

NaNO , HCl2

273-278 K

NO2

N Cl2+ –

Br

EtOH

NO2

Br

(S)

(R)

Sn/HClNO2

Br

(S)

NH2

Br

Br2

H O (excess)2

NH2

Br

Br

Br Br

NaNO , HCl2

Br

Br

Br Br

N Cl2

+ –H PO3 2

Br

Br

Br Br

4. D-Xywdksl dk fi Q' kj i zLrqrhdj.k (Fischer presentation) uhps fn; k x; k gSA

CHO

HHO

HH

OHHOHOH

CH OH2

D-glucose

-L-Xywdksi kbjSuksl (-L-glucopyranose) dh l gh l ajpuk (l ajpuk, ¡) gS (gSa)

(A)

CH OH2

OH

HO OH

H

H

HOH H

OH

(B)

CH OH2

OH

HO OH

H

OH

HH H

OH

Page 17: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

17

JEE (ADVANCED)-2018 (PAPER-2)

(C)

CH OH2

O

HO

OH

H

H

HO

H H

H

OH (D)

CH OH2

OHO

OH

H

H

HOH

H

H

OH

mÙkj (D)

gy -L-Xywdksi k; jsukst dh l ajpuk fuEu gS

OH

OH

H

OH

H

CH OH2

H

HO

H

OH5. fLFkj vk; ru ,oa 300 K i j , d i zFke dksfV dh vfHkfØ; k A(g) 2B(g) + C(g) ds fy, ] i zkjaHk (t = 0) vkSj l e; t i j

l ai w.kZ nkc Øe' k% P0 vkSj Pt gSaA ' kq: esa fl i QZ A, [A]0 l kanzrk ds l kFk mi fLFkr gS vkSj A ds vkaf' kd nkc (partial pressure)dksi zkjafHkd ewY; (initial value) ds 1/3 rd i gqapus dk l e; t1/3 gSA l gh fodYi gS (gSa)(eku ys fd ; s l kjh xSl sa vkn' kZ xSl ksa t Sl k O; ogkj djrh gSa)

(A)

Time

In(3

P–P

)0

t

(B)

[A]0

t 1/3

(C)

Time

In(P

– P

)0

t

(D)

[A]0

Rat

e c

on

stan

t

mÙkj (A, D)

gy A 2B + C

P0 – –

P0 – P 2P P

Pt = P0 + 2P

P = t 0P P2

0 0

t 0 0 t0

P 2PKt ln ln

P P 3P PP

2

0 0 tKt ln 2P ln(3P P ) Time

ln (

3P

– P

)0

t

vfHkfØ; k dk osx fu; rakd i zkjfEHkd l aknzrk i j fuHkZj ugh djr k

Page 18: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

18

JEE (ADVANCED)-2018 (PAPER-2)

6. vfHkfØ; k A P ds fy, ] [A] vkSj [P] ds l e; ds l kFk r ki eku T1 vkSj T2 i j vkys[ k uhps fn, x, gSa

Time

T1

T2

5

10[A

] / (

mo

l L)

–1

Time

T1

T25

10

[P]

/ (m

ol L

)–1

; fn T2 > T1 ] rks l gh i zdFku gS (gSa)

( H vkSj S dks rki eku fuHkZjrk l s Lora=k ekfu; s vkSj T1 i j ln(K) rFkk T2 i j ln(K) dk vuqi kr 2

1

T

T l s vf/ d gSA ; gk¡

H, S, G vkSj K, Øe' k% ,UFkSYi h] ,UVªkWi h] fxCt (Gibbs) Åt kZ vkSj l kE; koLFkk fLFkjkad gSa)

(A) H 0, S 0 (B) G 0, H 0

(C) G 0, S 0 (D) G 0, S 0

mÙkj (A, C)

gy 1 2

2 1

ln K Tln K T

r ki c<kus i j K ?kVrk gS

H° < 0

xzki Q l s K > 1 G° < 0

1 1 2

2 1

2

– H Sln K T R R T

H Sln K TT R R

1 2 2

2 1 1

( H T S ) T T( H T S ) T T

–H° + T1S° > –H° + T2S°

S° < 0

[ kaM-2(vf/ dre vad : 24)

bl [ kaM esa vkB (08) i z' u gSaA i zR;sd i z'u dk mÙkj , d l a[ ; kRed eku (NUMERICAL VALUE) gSA

çR; sd i z' u ds mÙkj ds l gh l a[ ; kRed eku(n' keyo vadu esa] n' keyo ds f}rh; LFkku rd : f.Mr@fudfVr ; mnkgj.kr%6.25, 7.00, -0.33, -.30, 30.27, -127.30) dks ekmt + (MOUSE) vkSj vkWu LØhu (ON-SCREEN) opqZvy U; wesfjd dhi SM(VIRTUAL NUMERIC KEYPAD) ds i z; ksx l s mÙkj ds fy, fufnZ"V LFkku i j nt Z djsaA

i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk&

i w.kZ vad : +3 ; fn fl i QZ l gh l a[ ; kRed eku (Numerical value) gh mÙkj Lo: i nt Z fd; k x; k gSA

' kwU; vad : 0 vU; l Hkh i fjfLFkfr ; ksa esaA

Page 19: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

19

JEE (ADVANCED)-2018 (PAPER-2)

7. uhps fn, x, v.kqvksa esa l s] de l s de ,d l srqca/ (bridging) vkWDl ks l ewg okys ; kSfxdksa dh dqy l a[ ; k____gSA

N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5

mÙkj (5.00)

gy

N—NO

N O2 3

O ON N

O

N O2 5

O OO

O

P O4 6

PP O

P O P

O OO O

P O4 7 H P O4 2 5

P O P

PP

O O

O

O O

O

P

O

HOH

O POH

O

H

H P O5 3 10

P

O

OH HO O

P

OH

O

O P

OH

O

OH

H S O2 2 3

S

S

OH HO O

H S O2 2 5

S

O

O HO

S

O

OH

8. mPp rki eku i j gok ds i zokg l s xysuk (galena) (, d v; Ld) dk vkaf' kd vkWDl hdj.k gksrk gSA dqN l e; ckn gok dki zokg can dj fn; k x; k] fdUrq can HkV~Vh dks xje djuk pkyw j[ kk x; k r kfd varoZLrqvksa (contents) dk Lo; a&vi p; u(self-reduction) gksA O2 ds i zfr kg xzg.k i j mRi kfnr Pb dk (kg esa) Hkkj gS____A

(i jek.kq Hkkj g mol–1 esa % O = 16, S = 32, Pb = 207)

mÙkj (6.47)

gy 2 2

2

2PbS 3O 2PbO 2SO

2PbO PbS 3Pb SO

3 eksy O2 l s 3 eksy ysM i zkIr gksrk gS

96 kg vkWDl ht u l s i zkIr ysM = 621 kg

1 kg vkWDl ht u l s i zkIr ysM = 621

6.468 6.47 kg96

9. , d t yh; foy; u esa ?kqfyr MnCl2 dh ek=kk ds eki u ds fy, ] bl s vfHkfØ; k MnCl2 + K2S2O8 + H2O KMnO4 +

H2SO4 + HCl (l ehdj.k l arqfyr ugha gS) ds vuql kj i w.kZr ; k KMnO4 esa i fjofrZr fd; k x; kA l kUnz HCl dh dqN cw¡nsa blfoy; u esa Mkyh x; h vkSj ml s gYds l s xje fd; k x; kA vkxs] i jeSaxusV vk; u dk jax xk;c gksus rd vksDl kfyd vEy (225

mg) dks va'kksa esa Mkyk x; kA i zkjafHkd foy;u esa MnCl2 dh ek=kk (mg esa) _____ gSA

(i jek.kq Hkkj g mol–1 esa : Mn = 55, Cl = 35.5)

mÙkj (126.00)

gy POAC l s,

MnCl2 ds m moles = KMnO4 ds m moles = x((ekuk)

rFkk KMnO4 ds meq = vkWDl sfyd vEy ds meq

225x 5 2

90x = 1

MnCl2 ds m moles = 1

MnCl2 ds mg = (55 + 71) = 126 mg

Page 20: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

20

JEE (ADVANCED)-2018 (PAPER-2)

10. fn, x, ; kSfxd X ds fy, / zqo.k ?kw.kZd f=kfoe l eko; oh; ksa (optically active steroisomers) dh l ai w.kZ l a[ ; k ____ gSA

mÙkj (7.00)

gy

HO

HO

OH

OH

dsoy rhu f=kfo; dsUnz mi fLFkr gS

dqy l eko; o = 23 = 8

ysfdu ,d l eko;o i zdkf' kd fuf"Ø; gS

HO

OH

OH

OH

11. fuEufyf[ kr vfHkfØ; k vuqØe esa] ,sl hVksi Q+hukWu ds 10 eksy l s i zkIr D dh cuh ek=kk (xzke esa)_____ gSA

(fn; k x; k gS] i jek.kq Hkkj gmol–1 esa % H = 1, C = 12, N = 14, O = 16, Br = 80. i zR; sd pj.k esa mRi kn dh mi t (%)

dks"Bd esa nh x; h gS)

O

NaOBr

H O3+ A

(60%)

NH , 3 B

(50%)

Br /KOH2 C(50%)

Br (3 equiv)2

AcOHD

(100%)

mÙkj (495.00)

gy

O

NaOBr

H O3+

Acetophenone10 moles

COOH

A(60%)

NH 3

CONH2

B(50%)

Br KOH2

NH2

C(50%)

NH2

D(100%)

Br

Br

Br

D dh yfCèk = 60 50 50

10 1.5 moles100 100 100

D dh ek=kk = 1.5 × 330 = 495.00

Page 21: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

21

JEE (ADVANCED)-2018 (PAPER-2)

12. dkWij dk i`"B] dkWij vkWDlkbM ds cuus ls efyu gksrk gSA dkWij dks 1250 K ij xje djrs le; vkWDlkbM cuus ls jksdusds fy, ukbVªkstu xSl dk izokg fd;k x;kA fdUrq ukbVªkstu xSl esa 1 eksy % tyok"i dk vinzO; gSA tyok"i dkWij dkuhps fn, x, vfHkfØ;k ds vuqlkj vkWDlhdj.k djrk gS%

2Cu(s) + H2O(g) Cu

2O(s) + H

2(g)

1250 K ij vkWDlhdj.k jksdus ds fy, H2 dk U;wure vkaf'kd nkc (bar esa) pH

2 pkfg,A In(pH

2) dk eku ____ gSA

(fn;k x;k gS] iw.kZ nkc = 1 bar, R (lkoZtfud xSl fu;rakd) = 8 J K–1mol–1, In(10) = 2.3; Cu(s) vkSj Cu2O(s)

ijLij vfeJ.kh; gSA

Cu(s) vkSj Cu2O(s) ijLij vfeJ.kh; gSA

1250 K: ij% 2Cu(s) + 1

2O

2(g) Cu

2O(s); G– = –78,000 J mol–1

H2(g) +

1

2O

2(g) H

2O(g); G– = –1,78,000 J mol–1; G fxCt ÅtkZ gS)

mÙk jmÙ k jmÙ k jmÙ k jmÙ k j (–14.60)

gygygygygy (i) 2 2

12Cu(s) O (g) Cu O(s), G –78 kJ/mole

2

(ii) 2 2 2

1H (g) O (g) H O(g), G –178 kJ/mole

2

(i) – (ii)

2 2 22Cu(s) H O(g) Cu O(s) H (g), G 100 kJ

G G RT ln K 0

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

H5 2

H O2

P10 8 1250 ln 0

P

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

H4 52

H O2

P10 ln 10 0

P

H H O2 2ln P – ln P –10

vc, –2H O H O Total2 2

P X P 0.01 1 10

H2ln P 2 ln10 –10

H2ln P 4.6 –10

H2ln P –14.60

13. fuEufyf[kr mRØe.kh; vfHkfØ;k (reversible reaction) ij fopkj djsaA

A(g) + B(g) ��⇀↽��

AB(g)

izrhi vfHkfØ;k (backward reaction) dh lfØ;.k ÅtkZ (activation energy)

vxz vfHkfØ;k (forward reaction)

dh

lfØ;.k ÅtkZ ls 2 RT (J mol–1 esa) vfèkd gSA ;fn vxz vfHkfØ;k dk iwoZ pj?kkrkadh xq.kd (pre-exponetial factor)

izrhi vfHkfØ;k ds iwoZ pj?kkrkadh xq.kd ls 4 xq.kk gS] rks 300 K ij vfHkfØ;k ds G (J mol–1 esa) dk fujis{k (absolute)

eku ____ gSA

(fn;k x;k gS] In(2) = 0.7, 300 K ij RT = 2500 J mol–1, G fxCt ÅtkZ gS)

mÙk jmÙ k jmÙ k jmÙ k jmÙ k j (8500.00)

Page 22: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

22

JEE (ADVANCED)-2018 (PAPER-2)

gygygygygy A(g) + B(g)� AB(g)

b f

a aE –E 2RT

f

b

A4

A

f

b

KK

K

af–E /RT

f fK A e

ab–E /RT

b bK A e

a ab f(E –E )/RTf f

b b

K Ae

K A

K = 4e2RT/RT

K = 4e2

G° = – RT In K

= – RT (2 + ln 4)

= – 2500 (2 + 2 × 0.7)

= – 8500 J mol–1

fujis{k eku 8500.00 gS

14. ,d oS|qrjklk;fud lsy% A(s) | An+(aq, 2 M) || B2n+(aq, 1 M) | B(s) ij fopkj dhft,A 300 K ij lsy vfHkfØ;k ds

H dk ewY; mlds G ls nqxuk gSA ;fn lsy dk emf 'kwU; gS] rks 300 K ij lsy vfHkfØ;k esa B ds izfr eksy cuus ds

fy;s S (J K–1 mol–1 esa) dk eku _____ gSA

(fn;k x;k gS] In(2) = 0.7, R ( lkoZtfud xSl fu;rkad) = 8.3 J K–1mol–1 | H, S vkSj G, Øe'k% ,UFkSYih] ,UVªkih vkSj fxCt

(Gibbs) ÅtkZ gSaA)

mÙk jmÙ k jmÙ k jmÙ k jmÙ k j (–11.62)

gy n –

2n –

2n n

A A ne

B 2ne B

2A B 2A B

H° = 2G°, Ecell

= 0

G° = H° – TS°

G° = TS° G

ST

n 2

2n

– RTlnK [A ]S – R ln

T [B ]

2

2– 8.3 ln

1

S° = – 11.62 JK–1 mol–1

Page 23: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

23

JEE (ADVANCED)-2018 (PAPER-2)

[ kaM 3 (vf/ dre vad % 12)• bl [ kaM esa pkj (04) i z'u gSaA i zR;sd i z' u esa nks (02) l qesyu l wfp;k¡ (matching lists) gSa% l wph–I vkSj l wph–II |• l wph–I vkSj l wph–II ds rÙoksa ds l qesykuks dks n'kkZrs gq, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa fl iQZ ,d fodYi gh l gh

l qesyu i znf' kZr djr k gSA• i zR; sd i z' u ds fy, l gh l qesyu i znf' kZr djus okys fodYi dks pqusaA• i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk

i w.kZ vad %+3 ; fn fl i QZ l gh fodYi gh pquk x; k gSA' kwU; vad %0 ; fn dksbZ Hkh fodYi ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)AÍ .kkRed vad %–1 vU; l Hkh i fjfLFfr ; ksa esaA

15. l wph-I (List-I) ds i zR; sd l adj d{kd (hybrid orbitals) ds l sV dks l wph-II(List-II) esa fn, x, l adqy (l adqyksa) ds l kFkl qesy djsaA

l wph-I l wph-II

P. dsp2 1. [FeF6]4–

Q. sp3 2. [Ti(H2O)

3Cl

3

R. sp3d2 3. [Cr(NH3)

6]3+

S. d2sp3 4. [FeCl4]2–

5. Ni(CO)4

6. [Ni(CN)4]2–

l gh fodYi gS(A) P 5; Q 4,6; R 2,3; S 1

(B) P 5,6; Q 4; R 3, 4; S 1,2

(C) P 6; Q 4,5; R 1; S 2,3

(D) P 4,6; Q 5,6; R 1,2; S 3

mÙkj (C)

gy 1. [FeF6]4–

Fe+2

4s 4p3d

mPp pØ.k l adqy gS D;ksafd F– , d nqcZy {ks=k fyxs.M gSA vr% l adj.k sp3d2 gS

2. [Ti(H2O)3Cl3]

Ti3+

d sp2 3

d sp2 3

3. [Cr(NH3)6]3+

Cr3+

d sp2 3

d sp2 3

4. [FeCl4]2–

Fe+2 : 3d6, Cl– nqcZy {ks=k fyxs.M gSA

sp3

3d 4s 4p

Page 24: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

24

JEE (ADVANCED)-2018 (PAPER-2)

5. Ni(CO)4

Ni0 – 3d84s2, CO i zcy {ks=k fyxs.M gSA

Ni

sp3

6. [Ni(CN)4]2–

Ni+2

dsp2

CN– i zcy {ks=k fyxs.M gSA

16. l wph-I (List-I) dh vfHkfØ;kvksa ds eq[ ; mRi kn dk l wph-II(List-II) esa fn, x, , d ; k vusd l q; ksX; vfHkdkjdksa ds l kFk vfHkfØ;kdjus i j bfPNr mRi kn X cuk;k t k l drk gSA

(fn; k x; k] vfHkxkeh vfHkofÙk(migratory aptitude) dk Øe% ,sfjy > , sfYdy > gkbMªkst u)

O

Ph

Ph

OH

Me

X

l wph-I l wph-II

P.

PhHO

PhOH

Me

Me

H SO2 4+1. I

2, NaOH

Q.

PhH N2

PhOH

H

Me

HNO2+2. [Ag(NH

3)

2]OH

R.

PhHO

MeOH

Ph

Me

H SO2 4+3. i sQgfyax foy; u

S.

PhBr

PhOH

H

Me

AgNO3+4. HCHO, NaOH

5. NaOBr

Page 25: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

25

JEE (ADVANCED)-2018 (PAPER-2)

l gh fodYi gS

(A) P 1; Q 2,3; R 1,4; S 2,4

(B) P 1,5; Q 3,4; R 4,5; S 3

(C) P 1,5; Q 3,4; R 5; S 2,4

(D) P 1,5; Q 2,3; R 1,5; S 2,3

mÙkj (D)

gy (P)Me

MeHOPhHO

Ph

+ H SO2 4

O

MePhMe

PhI /NaOH2

O

PhMe

PhC – OH

NaOBr

O

PhMe

PhC – OH

P 1, 5

(Q)

H

MeH N2

Ph

+ HNO2

O

HPhMe

Ph[Ag(NH ) ]OH3 2

PhMe

PhCOOH

Fehling

Ph

OH

SolutionPh

Me

PhCOOH

Q 2, 3

(R)Ph

MeHO

Ph

+ H SO2 4

O

MeMePh

Ph

Me

OH

NaOBr I /NaOH2

PhMePh

COOHPh

MePh

COOH

R 1, 5

(S)Ph

MePh

Ph

+ AgNO3

MePh

Ph

OH

Fehlingsolution

[Ag(NH ) ]OH3 2

Me

Ph

Ph

COOH

Br

CHO

Me

Ph

Ph

COOH

S 2, 3

Page 26: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

26

JEE (ADVANCED)-2018 (PAPER-2)

17. l wph-I (List-I) esa vfHkfØ;k; sa gSa vkSj l wph-II (List-II) esa eq[ ; mRi kn gSaA

l wph-I l wph-II

P.ONa Br

+ 1.OH

Q.OMe

HBr+ 2.Br

R.Br

NaOMe+ 3.OMe

S.ONa

MeBr+ 4.

5.O

l wph-I dh i zR; sd vfHkfØ; k dk l wph-II ds ,d ; k vusd mRi knksa ds l kFk l qesy djsa vkSj l gh fodYi pqusaA

(A) P 1,5; Q 2; R 3; S 4

(B) P 1,4; Q 2; R 4; S 3

(C) P 1,4; Q 1,2; R 3,4; S 4

(D) P 4,5; Q 4; R 4; S 3,4

mÙkj (B)

gy (P) ONa + Br + OH

3° gSykbM ds l kFk i zHkkoh : i l s foyksi u gksrk gSA

(Q) OMe + HBr Br + MeOH

(R) Br + NaOMe

(S) ONa + MeBr OMe

P 1, 4; Q 2; R 4; S 3.

Page 27: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

27

JEE (ADVANCED)-2018 (PAPER-2)

18. l wph-I (List-I) esa vyx vyx t yh; foy; uksa dk t y ds l kFk ruqdj.k djus ds i zØe fn, x, gSA foy; u ds ruqdj.k l s[H+] i j gq, i zHkko l wph-II(List-II) esa fn, x, gSaA

(è; ku nsa] nqcZy vEy vkSj nqcZy {kkj dh fo; kst u ek=kk ()(degree of dissociation)<<1 gS; yo.k ds t y&vi ?kVu dhek=kk(degree of hydrolysis of salt)<<1 gS; [H+], H+ vk; uksa dh l kanzrk dks fu: fi r djr k gS)

l wph-I l wph-II

P. (0.1 M NaOH dk 10 mL + 0.1 M , fl fVd 1. ruqdj.k djus i j [H+] ds eku esa dksbZ cnyko ugha gksrkgS

vEy dk 20 mL) dk 60 mL rd ruqdj.k

Q. (0.1 M NaOH dk 20 mL + 0.1 M , fl fVd 2. ruqdj.k djus i j [H+] dk eku cnydj bl ds i zkjafHkdvEy dk 20 mL) dk 80 mL rd ruqdj.k eku dk vkèkk gksrk gS

R. (0.1 M HCI dk 20 mL + 0.1 M veksfu;k 3. ruqdj.k djus i j [H+] dk eku cnydj bl ds i zkjafHkdfoy; u dk 20 mL) dk 80 mL rd ruqdj.k eku dk nks xq.kk gksrk gS

S. 10 ml Ni(OH)2 dk l ar Ir foy;u (saturated 4. ruqdj.k djus i j [H+] dk eku cnydj bl ds i zkjafHkd

solution) t ks vkfèkD; Bksl Ni(OH)2 ds l kFk eku dk 1

2 xq.kk gksrk gS

l kE;koLFkk esa gS] ml dk 20 mL rd ruqdj.k 5. ruqdj.k djus [H+] dk eku cnydj bl ds i zkjafHkd eku

fd;k x; k (Bksl Ni(OH)2 ruqdj.k ds i ' pkr dk 2 xq.kk gksrk gSA

Hkh mi fLFkr gSA)

l wph-I esa fn, x, i zR;sd i zØe dks l wph-II esa fn, x, ,d ; k vusd i zHkko (i zHkkoksa) ds l kFk l qesy djsaA l gh fodYi gS

(A) P 4; Q 2; R 3; S 1

(B) P 4; Q 3; R 2; S 3

(C) P 1; Q 4; R 5; S 3

(D) P 1; Q 5; R 4; S 1

mÙkj (D)

gy (P) 3 old

20 0.1– 10 0.1 1CH COOH

30 30

3old

1CH COO

30

[yo.k] = [vEy] okyk ci Qj

ruqrk i j pH i fjofrZr ugha gksrh (P) (1)

(Q) [ ] CH COO 3

–old

=

20 0.1 2

40 40

[ ] CH COO 3

–new

= 2

80

Page 28: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

28

JEE (ADVANCED)-2018 (PAPER-2)

CH COO 3

– + H O2

c

–3

xxCH COOH OH

– 2 – 22old new

h[OH ] [OH ]x

Kc 2/40 2/80

[OH ] – 2

new =

– 2old[OH ]

2

– oldnew

[OH ][OH ]

2

new old[H ] 2[H ]

(Q) (5)

(R) 4 old

20 0.1 2[NH ]

40 40

4 new2

[NH ]80

4 2 4yc y

NH H O NH OH H

2 22old new[H ] [H ]y

Khc 2/40 2/80

22 old

new[H ]

[H ]2

oldnew

[H ][H ]

2

(R) (4)

(S) , d l ar Ir foy;u ds fy; s – 3 sp[OH ] 2K

foy;u dk vk; ru dqN Hkh gks] [H+] fu; r jgrk gSA

(S) (1)

END OF CHEMISTRY

Page 29: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

29

JEE (ADVANCED)-2018 (PAPER-2)

MATHEMATICS[ kaM-1(vf/ dre vad : 24)

bl [ kaM esa Ng (06) i z' u gSaA

çR;sd i z' u ds l gh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa l s , d ; k ,d l s vf/ d fodYi l ghgS(gSa)A

i zR;sd i z'u ds fy, ] i z'u dk(ds) mÙkj nsus gsrq l gh fodYi (fodYi ksa) dks pqusaA

i zR; sd i z'u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk%

i w.kZ vad : +4 ; fn dsoy(l kjs) l gh fodYi (fodYi ksa) dks pquk x; k gSA

vkaf' kd vad : +3 ; fn pkjksa fodYi l gh gSa i jUrq dsoy rhu fodYi ksa dks pquk x; k gSA

vkaf' kd vad : +2 ; fn rhu ; k rhu l s vf/ d fodYi l gh gSa i jUrq dsoy nks fodYi ksa dks pquk x; k gS vkSjpqus gq, nksuksa fodYi l gh fodYi gSaA

vkaf' kd vad : +1 ; fn nks ; k nks l s vf/ d fodYi l gh gSa i jUrq dsoy ,d fodYi dks pquk x; k gS vkSj pqukgqvk fodYi l gh fodYi gSA

' kwU; vad : 0 ; fn fdl h Hkh fodYi dks ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)A

½.kkRed vad : –2 vU; l Hkh i fjfLFkfr ; ksa esaA

mnkgj.k Lo: i % ; fn fdl h i z' u ds fy, dsoy i gyk] rhl jk , oa pkSFkk l gh fodYi gSa vkSj nwl jk fodYi xyr gSa; r ks dsoyl Hkh rhu l gh fodYi ksa dk p; u djus i j gh +4 vad feysasxsA fcuk dksbZ xyr fodYi pqus (mnkgj.k esa nwl jk fodYi )] rhul gh fodYi ksa esa l s fl i QZ nks dks pquus i j (mnkgjar% i gyk rFkk pkSFkk fodYi ) +2 vad feysaxsA fcuk dksbZ xyr fodYi pqus(blmnkgj.k esa nwl jk fodYi )] rhu l gh fodYi ksa esa l s fl i QZ ,d dks pquus i j (i gyk ; k rhl jk ; k pkSFkk fodYi ) +1 vad feysaxsAdksbZ Hkh xyr fodYi pquus i j (bl mnkgj.k esa nwl jk fodYi )] –2 vad feyasxs] pkgs l gh fodYi (fodYi ksa) dks pquk x; k gks; k u pquk x; k gksA

1. fdl h Hkh / ukRed i w.kk±d (positive integer) n ds fy, , fn : (0, ) ,

n 1

n j 11

f (x) tan1 (x j)(x j 1)

l Hkh x (0, ) ds fy, ] ds }kjk i fjHkkf"kr gSA(; gk¡ i zfrykse f=kdks.kferh; i Qyu (inverse trigonometric function)

tan–1x,

,

2 2 esa eku / kj.k djr k gSA) rc fuEufyf[ kr esa l s dkSul k (l s) dFku l R; gS (gSa)\

(A) 5 2j 1 jtan (f (0)) 55

(B) 10 2j 1 j j(1 f (0))sec (f (0)) 10

(C) fdl h Hkh fu; r (fixed) / ukRed i w.kk±d n ds fy, ]

nx

1lim tan(f (x))

n

(D) fdl h Hkh fu; r (fixed) / ukRed i w.kk±d n ds fy, ]

2n

xlim sec (f (x)) 1

mÙkj (D)

Page 30: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

30

JEE (ADVANCED)-2018 (PAPER-2)

gygygygygy ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

1 1 1

n

nf (x) tan (n x) tan (x) tan

1 (n x)x

⇒ 1

n nf (0) tan (n) tan(f (0)) n

2

n n2 2 2 2

1 1 1 nf (x) f (0) 1

1 (n x) 1 x 1 n 1 n

(A)

∑5

2 2 2 2 2 2j

j 1

5 6 11tan (f (0)) 1 2 3 4 5 55

6

(B)

⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦

∑ ∑210 10

2 2j j 2

j 1 j 1

j(1 f (0))sec (f (0)) 1 1 j 10

1 j

(C)

nx x

nlim tan(f (x)) lim 0

1 (n x)x

(D)

2

nx

lim (1 tan (f (x)) 1 0 1

2. ekuk fd T, fcanqvksa P(-2, 7) vkSj Q(2, -5) ls xqtjus okyh js[kk (line) gSA ekuk fd F1 mu lHkh o`Ùk ;qXeksa (pairs

of circles) (S1, S

2) dk leqPp; (set) gS fd js[kk T, S

1 ds fcanq P ij vkSj S

2 ds fcanq Q ij Li'khZ (tangent)

gS rFkk o`Ùk S1 o S

2 ,d nwljs dks fcanq] ekuk fd M, ij Li'kZ djrs gSaA tc ;qXe (S

1, S

2), F

1 esa fopfjr (varies)

djrk gS rks ekuk fd leqPp; (set) E1, fcanq M ds fcanqiFk (locus) dks n'kkZrk gSA ekuk fd F

2 mu ljy js[kk&[k.Mksa

(straight line segments) dk leqPp; gS] tks fcanq R(1, 1) ls xqtjrha gSa rFkk E1 ds nks fHkUu fcanqvksa ds ;qXe

(pair of distinct points) dks tksM+rha gSaA ekuk fd E2, leqPp; F

2 ds js[kk[k.Mksa ds eè; fcanqvksa dk leqPp; gSA rc

fuEufyf[kr esa ls dkSulk (ls) dFku lR; lR; lR; lR; lR; gS (gSa)\

(A) fcanq (–2, 7) leqPp; E1 esa fLFkr gS

(B) fcanq ⎛ ⎞⎜ ⎟⎝ ⎠

4 7,

5 5 leqPp; E

2 esa fLFkr ugha ugha ugha ugha ugha gS

(C) fcanq ⎛ ⎞⎜ ⎟⎝ ⎠

1,1

2 leqPp; E

2 esa fLFkr gS

(D) fcanq ⎛ ⎞⎜ ⎟⎝ ⎠

30,

2 leqPp; E

1 esa fLFkr ugha ugha ugha ugha ugha gS

mÙkjmÙkjmÙkjmÙkjmÙkj (B, D)

gygygygygyP

Q

T

M

Page 31: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

31

JEE (ADVANCED)-2018 (PAPER-2)

PMQ = 90°

5 7

12 2

M dk fcanqiFk

x2 + y2 – 2y – 39 = 0 ...(1)

oÙk dh ml thok dk lehdj.k ftldk eè; fcnq (, ) gS

S1 = T

x + y – (y + ) – 39 = 2 + 2 – 2 – 39

R(1, 1)

( , )

x + y – 2y – 39 = 02 2

;g (1, 1) ls xqtjrh gS

2 + 2 – 2 – + 1 = 0

fcUnqiFk : x2 + y2 – x – 2y + 1 = 0 ...(2)

fodYi (A) xyr gS ;|fi ;g lehdj.k (1) dks larq"V djrk gS vU;Fkk js[kk T f}rh; o`Ùk dks nks fcanqvksa ij Li'kZ djsxhA fcanq

(4/5, 7/5) lehdj.k (2) dks larq"V djrs gS ysfdu iqu% bl fLFkfr esa thok dk ,d fljk (–2, 7) gksxk tks E1 esa lfEefyr

ugha gSA vr% ⎛ ⎞⎜ ⎟⎝ ⎠

4 7,

5 5, E

2 esa fLFkr ugha gSaA (1/2, 1), lehdj.k (2) dks larq"V ugha djrk gS] vr% ;g E

2 esa fLFkr ugha gSaA

(0, 3/2), lehdj.k (1) dks larq"V ugha djrk gS] vr% E1 esa fLFkr ugha gSA

3. ekuk fd S mu lHkh LrEHk vkO;wgksa (column matrices)

1

2

3

b

b

b

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

dk leqPp; (set) gS ftuds fy, b1, b

2, b

3 � vkSj

okLrfod pjksa (real variables) okys lehdj.k fudk; (system of equations)

–x + 2y + 5z = b1

2x – 4y + 3z = b2

x – 2y + 2z = b3

dk de ls de ,d gy (solution) gSA rc fuEufyf[kr okLrfod pjksa okys fudk;ksa esa ls fdl (dkSuls) fudk; (fudk;ksa)

dk Hkh izR;sd 1

2

3

b

b

b

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

S ds fy, de ls de ,d gy gS\

(A) x + 2y + 3z = b1, 4y + 5z = b

2 and x + 2y + 6z = b

3

(A) x + 2y + 3z = b1, 4y + 5z = b

2 vkSj x + 2y + 6z = b

3

(B) x + y + 3z = b1, 5x + 2y + 6z = b

2 vkSj –2x – y – 3z = b

3

(C) –x + 2y – 5z = b1, 2x – 4y + 10z = b

2 vkSj x – 2y + 5z = b

3

(D) x + 2y + 5z = b1, 2x + 3z = b

2 vkSj x + 4y – 5z = b

3

mÙkjmÙkjmÙkjmÙkjmÙkj (A, D)

Sol. nh x;h lehdj.kksa ds fudk; dks fuEukuqlkj fy[kk tk ldrk gS

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1

2

3

1 2 5 x b

2 4 3 y b AX B

1 2 2 z b

Page 32: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

32

JEE (ADVANCED)-2018 (PAPER-2)

⎡ ⎤

⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∼ ∼

11 1

2 1 2 3 1 2 3

3 1 31 3

b1 2 5 b 1 2 5 b 1 2 5

1 13A |B 2 4 3 b 0 0 6 b b b 0 0 0 b b b

7 71 2 2 b 0 0 7 b b 0 0 7

b b

bl fudk; ds gy ds fy,, 1 3

2

b 13bb 0

7 7

b1 + 7b

2 – 13b

3 = 0 (b

1, b

2, b

3) = (–7K

2 + 13K

3, K

2, K

3) tgk¡ K

2, K

3 R...(i)

(A) 0 (b1, b

2, b

3) dk dksbZ laHkkfor leqPp; ,d gy nsrk gSA

(i) ds ekuksa dk çR;sd leqPp; fodYi (A) ds fudk; dk de ls de ,d gy çnku djrk gS

(B )

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∼ ∼

1 1 1

2 1 2 1 2 3

3 1 3 1 3

1 1 3 b 1 1 3 b 1 1 3 b

5 2 6 b 3 0 0 2b b 0 0 0 b b 3b

2 1 3 b 1 0 0 b b 1 0 0 b b

bl fudk; ds gy ds fy,] b1 + b

2 + 3b

3 = 0

(b1, b

2, b

3) = (–K

2 –3K

3, K

2, K

3), K

2, K

3 R ...(ii)

Li"Vr% (i) }kjk çnf'kZr leqPp;] (ii) esa fLFkr ugha gS

bl fudk; esa (i) ds (b1, b

2, b

3) ds çR;sd leqPp; ds fy, dksbZ gy ugha gSA

(C) lehdj.ksa lekUrj leryksa ;k le:i leryksa dh gSA gy ds fy,] bu leÙkyksa dk le:i gksuk vko';d gS] ftlds fy,(b

1, b

2, b

3) = (–K

3, 2K

3, K

3) ...(iii)

(b1, b

2, b

3) ds ,sls çR;sd eku (i) esa fLFkr gS ysfdu (i) esa (b

1, b

2, b

3) dk çR;sd eku (iii) esa mifLFkr ugha gS

(i) esa çR;sd (b1, b

2, b

3), fodYi (C) ds lehdj.k fudk; dk gy çnku ugha djrk gSA

(D) 0 (i) ds ekuksa dk çR;sd leqPP; fodYi (D) ds fudk; dk de ls de ,d gy çnku djrk gSA

4. ,slh nks ljy js[kkvksa (straight lines) ij fopkj dhft,] ftuesa ls izR;sd] o`Ùk (circle) 2 2 1x y

2 vkSj ijoy;

(parabola) y2 = 4x nksuksa ij gh Li'khZ (tangent) gSA eku fd ;s js[kk,a fcanq Q ij izfrPNsn (intersect) djrh gSA ,d ,sls

nh?kZo`Ùk (ellipse) ij fopkj dhft, ftldk dsanz (centre) ewyfcanw (origin) 0(0,0) ij gS vkSj ftldk v/Z&nh?kkZ{k (semi-

major axis) OQ gSA ;fn bl nh?kZo`Ùk ds y?kq v{k (minor axis) dh yEckbZ 2 gS] rc fuEufyf[kr esa ls dkSulk (ls) dFku

lR; lR; lR; lR; lR; gS (gSa)\

(A) nh?kZo`Ùk dh mRdsUnzrk (eccentricity)

1

2 gS vkSj ukfHkyEc thok (latus rectum) dh yEckbZ 1 gS

(B) nh?kZoÙk dh mRdsUnzrk 1

2 gS vkSj ukfHkyEc thok dh yEckbZ

1

2 gS

(C) js[kkvksa 1x

2 o x = 1 ds chp nh?kZo`Ùk }kjk ifjc¼ (bounded) {ks=k (region) dk {ks=kiQy (area) 1

( 2)4 2

gS

(D) js[kkvksa 1x

2 o x = 1 ds chp nh?kZoÙk }kjk ifjc¼ {ks=k dk {ks=kiQy 1

( 2)16

gS

Page 33: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

33

JEE (ADVANCED)-2018 (PAPER-2)

mÙkj (A, C)

gy 1

y mxm

, 2 2 1x y

2 i j Li ' kZ js[ kk Hkh gSA

2

11m

m 121 m

mHk; fu"B Li ' kZ js[ kk; sa y = x + 1 o y = –x – 1 Q(–1, 0)

nh?kZoÙk dk l ehdj.k

2 2

2 2

x y1

1 1

2

b2 = a2(1 – e2)

21 11(1 e ) e

2 2

ukfHkyEc dh yEckbZ =

2

122b 2 1

a 1

y

xQ(–1, 0) (0, 0)

x = 11

x2

O

21y 1 x

2

1 2

1

2

1A 2 1 x dx

2

12 1

1

2

1 12 x 1 x sin x

2 2

2

4 2 oxZ bdkbZ

5. ekuk fd s, t, r ' kwU; sÙkj (non-zero) l fEeJ l a[ ; k; sa (complex numbers) gSa vkSj L l ehdj.k (equation) sz tz r 0 ds

gyksa (solutions) z = x + iy (x, y ,i 1) dk l eqPp; gS] t gk¡ z x iy A rc fuEufyf[ kr esa l s dkSul k (l s) dFku

l R; gS (gSa)\

(A) ; fn L esa Bhd ,d vo; o (element) gS] r c |s| |t|

(B) ; fn |s| = |t|, r c L esa vuUr (infinitely many) vo; o gSa

(C) L {z : |z – 1 + i| = 5} esa vo;oksa dh vf/ dre l a[ ; k 2 gS

(D) ; fn L esa ,d l s T;knk vo; o gS] r c L esa vuUr vo; o gSa

mÙkj (A, C, D)

Page 34: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

34

JEE (ADVANCED)-2018 (PAPER-2)

gy sz t z r 0 ...(i)

(i) dk l a; qXeh ysus i j

s z t z r 0 ...(ii)

l ehdj.k (i) o (ii) esa l s z dk foyksi u djus i j

(ss z t s z sr) (t s z t t z t r) 0

2 2z(| s | | t | ) tr rs

(A) ; fn | s | | t | , r c z dk vf}rh; eku gS

(B) ; fn | s | | t | vkSj r t r s 0 , r c z ds vuUr eku gSa

; fn | s | | t | rFkk r t r s 0 , rc z dk dksbZ eku ugha gS

L fjDr l eqPp; ; k vuUr l eqPp; gks l drk gS

(C) z dk fcanqi Fk l Hkh fLFkfr ; ksa esa fjDr l eqPp; ; k , dy l eqPp; ; k ,d js[ kk gSA ; g fn, x, oÙk dks vf/ d l s vf/ d nksfcanqvksa i j çfrPNsn djsxkA

(D) ; fn L esa ,d l s vf/ d vo;o gSa] r c L ds vuUr vo;o gSaA

6. ekuk fd f : (0, ) , d , sl k f}vodyuh; (twice differentiable) i Qyu (function) gS fd

2

t x

f(x)sint f(t)sinxlim sin x

t x l Hkh x (0, ) ds fy, A

; fn

f ,

6 12 r c fuEufyf[ kr esa l s dkSul k (l s) dFku l R; gS (gSa)\

(A)

f

4 4 2

(B) 4

2xf(x) x

6 l Hkh x (0, ) ds fy,

(C) , d , sl s (0, ) vfLrRo (existence) gS ft l ds fy, f() = 0

(D)

f f 0

2 2

mÙkj (B, C, D)

gy

2

t x

f(x) sint f(t) sinxlim sin x

t x

L.H. fu;e ,oa N.L. çes; dk ç; ksx djus i j,

f(x) cosx – f(x) sinx = sin2x

1

f(x) x csinx

...(i)

Page 35: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

35

JEE (ADVANCED)-2018 (PAPER-2)

f c 0 f(x) x sinx6 12

(A)

f

4 4 2

(B) pw¡fd 3x

sinx x6

4

2 xx sinx x

6

4

2 xf(x) x

6

(C) [0, ] esa f(x) l r r gS rFkk (0, ) esa vodyuh; gS rFkk f(0) = f() = 0

f() = 0, (0, )

(D) f(x) = –x cosx – sinx, f(x) = x sinx – 2 cosx

f f 02 2 2 2

[ kaM-2(vf/ dre vad : 24)

bl [ kaM esa vkB (08) i z' u gSaA i zR;sd i z'u dk mÙkj , d l a[ ; kRed eku (NUMERICAL VALUE) gSA

çR; sd i z' u ds mÙkj ds l gh l a[ ; kRed eku(n' keyo vadu esa] n' keyo ds f}rh; LFkku rd : f.Mr@fudfVr ; mnkgj.kr%6.25, 7.00, -0.33, -.30, 30.27, -127.30) dks ekmt + (MOUSE) vkSj vkWu LØhu (ON-SCREEN) opqZvy U; wesfjd dhi SM(VIRTUAL NUMERIC KEYPAD) ds i z; ksx l s mÙkj ds fy, fufnZ"V LFkku i j nt Z djsaA

i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk&

i w.kZ vad : +3 ; fn fl i QZ l gh l a[ ; kRed eku (Numerical value) gh mÙkj Lo: i nt Z fd; k x; k gSA

' kwU; vad : 0 vU; l Hkh i fjfLFkfr ; ksa esaA

7. l ekdy (integral)

12

10 2 6 4

1 3dx

(x 1) (1 x)

dk eku gS ____A

mÙkj (2.00)

gy

1/2

1/42 60

1 3I dx

(x 1) (1 x)

1/2

1/460 8

1 3dx

1 x(x 1)

1 x

1/2

3/20 2

1 3dx

1 x(1 x)

1 x

2

1 x 2dxt dt

1 x (1 x)j[ kus i j

1/31/3 1/2

3/21

1

1 1 3 1 tI (1 3) (1 3)( 3 1) 2

12 2t2

Page 36: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

36

JEE (ADVANCED)-2018 (PAPER-2)

8. ekuk fd P, 3 × 3 dksfV (order) dk ,d , sl k vkO; wg (matrix) gS fd P dh l Hkh i zfof"V; k¡ (entries) l eqPp; (set) {–1,

0, 1} esa l s gSA rc P ds l kjf.kd (determinant) dk vf/ dre l aHkkfor eku (maximum possible value) gS ____ A

mÙkj (4.00)

gy

11 12 13

21 22 23

31 32 33

a a a

P a a a

a a a

|P| dh vf/ dre l aHkkouk 6 gks l drh gS ; fn

21 22 22 23 21 23

31 32 32 33 31 33

a a a a a a2

a a a a a a

i jUrq] eSfVªDl 21 22

31 32

a a

a a , 2 (ekuk) ds : i esa l eqPp; gS rFkk 22 23

32 33

a a

a a , 2 ; k –2 ds : i esa l eqPp; gS] 21 23

31 33

a a

a a

Lor% ' kwU; eku ysrk gSA

vr%, |P| 6. vxyh l aHkkouk 4 gSA

1 1 1

P 1 1 1

1 1 1

, sl h ,d l aHkkouk gSA

9. ekuk fd l eqPp; (set) X esa Bhd 5 vo; o (elements) gS vkSj l eqPp; Y esa Bhd 7 vo; o gSaA ; fn X l s Y esa ,dSdh

i Qyuksa (one-one functions) dh l a[ ; k gS vkSj Y l s X esa vkPNknd (onto) i Qyuksa dh l a[ ; k gS] rc 1

( )5!

dk eku

gS ____ A

mÙkj (119.00)

gy = X l s Y esa ,dSdh i Qyuksa dh l a[ ; k

= Y l s X esa vkPNknd i Qyuksa dh l a[ ; k

= Y ds 7 vo; oksa esa l s 5 l ewg cukb, rFkk bu l ewgksa dk X ds 5 vo; oksa esa Øe i fjorZu dhft ,

4 2 3

7 75 5

3. 1 . 4 2 1 2. 3

3

1 7 7 75 3. 4 5. 22 3

= 7 [5 + 15 – 3] = 119

10. ekuk fd f : ,d l sl k vodyuh; i Qyu (differentiable function) gS ft l ds fy; s f(0) = 0A ; fn y = f(x), vody

l ehdj.k (differential equation) dy

(2 5y)(5y – 2)dx

dks l arq"V djr k gS] rc x –lim f(x) dk eku gS___A

mÙkj (0.40)

Page 37: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

37

JEE (ADVANCED)-2018 (PAPER-2)

gy dy

(2 5y)(5y 2)dx

dydx

(5y 2)(5y 2)

1 1 1dy dx

4 5y 2 5y 2

1 5y 2ln x C

20 5y 2

5y 2

f(0) 0 ln x5y 2

5y 2

x ln5y 2

5y 2 20 y

5y 2 5

x

2lim f(x) 0.40

5

11. ekuk fd f : , d , sl k vodyuh; i Qyu (differentiable function) gS ft l ds fy; s f(0) = 1, vkSj t ks l Hkhx, y ds fy, l ehdj.k f(x + y) = f(x)f '(y) + f '(x)f(y) dks l arq"V djr k gSA rc loge(f(4)) dk eku gS _____A

mÙkj (2.00)

gy

f(0) = 1, f : R R

f(x + y) = f(x) f(y) + f(x) f(y) x, y R

x = y = 0 j[ kus i j f(0) = 2f(0) f(0) f(0) = 1/2

y = 0 j[ kus i j ,

f(x) = f(x) f(0) + f(x) f(0)

f(x) = 12

f(x) + f(x)

2f (x) 1

=f(x) ln(f(x)) =

x+ C

2

f(0) = 1 C = 0

ln(f(x)) =x2

ln(f(4)) = 4

= 22

Page 38: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

38

JEE (ADVANCED)-2018 (PAPER-2)

12. ekuk fd P çFke v"Bka' k (first octant) esa ,d fcanq gS] ft l dk l ery (plane) x + y = 3 esa çfr fcEc (image) Q (vFkkZrjs[ kk[ k.M PQ l er y x + y = 3 ds yEcor gS vkSj PQ dk eè; fcanq l ery x + y = 3 esa fLFkr gS) z-v{k (axis) i j fLFkrgSA ekuk fd P dh x-v{k l s nwjh 5 gSA ; fn P dk xy-l er y esa çfr fcEc R gS] r c PR dh yEckbZ gS ______A

mÙkj (8.00)

gy P = (,, ), Q = (0, 0, K) pw¡fd ;g z-v{k i j fLFkr gS

PQ dk eè; fcanq vFkkZr~, , ,

K2 2 2

fuEu dks l arq"V djrk gS

x + y = 3+ = 6 ...(1)

PQ ds fnd~ vuqi kr = (– 0, – 0, – K) = (p, p, 0)

= ,oa = K ...(2)

(1) , oa (2) = = 3

P = (, , K) = (3, 3, K)

x-v{k l s P dh nwjh = 5

2 + 2 = 25 2 + K2 = 25 32 + K2 = 25|K| = 4

PR dh yEckbZ = 2|K| = 8

13. çFke v"Bkax (first octant) esa ,d , sl s ?ku (cube) i j fopkj dhft ; s] ft l dh Hkqt kvksa (sides) OP, OQ vkSj OR yEckbZ 1

gS vkSj t ks Øe' k% x-v{k (axis), y-v{k vkSj z-v{k ds vuqfn' k (along) gSa] t gk¡ O(0, 0, 0) ewyfcanq (origin) gSA ekuk fd

?ku dk dsaæ (centre)

1 1 1S , ,

2 2 2 gS] vkSj ' kh"kZ (vertex) T ewyfcanq O ds l Eeq[ k (Opposite) okyk og ' kh"kZ gS fd

fcanq S fod.kZ (diagonal) OT i j fLFkr gSA ; fn

p SP, q SQ, r SR t ST, | (p q) (r t) | vkjS rc dk ekugS_______A

mÙkj (0.50)

gy 1 1 1 1ˆ ˆˆ ˆ ˆ ˆp SP i j k = (i j k)

2 2 2 2

Y(0,1,0)Q

T (1,1,1)

O

S1 1 1

, ,2 2 2

P(1,0,0)X

ZR(0, 0, 1)

1 ˆˆ ˆq SQ ( i + j k)

2

1 ˆˆ ˆr SR ( i j + k)

2

1 ˆˆ ˆt ST (i + j + k)

2

ˆˆ ˆi j k

1 1 ˆ ˆp q 1 1 1 (i + j)4 2

1 1 1

ˆ ˆˆ ˆ ˆ ˆi j k i j k 1 1 1 ˆ ˆr × t 1 1 1 1 1 1 ( i + j)4 4 2

1 1 1 0 0 2

1 1 1ˆ ˆˆ ˆ ˆ ˆ(p q) (r t) (i + j) ( i + j) 2k k

4 4 2

1

(p q) (r t) 0.502

Page 39: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

39

JEE (ADVANCED)-2018 (PAPER-2)

14. ekuk fd X = (10C1)2 + 2(10C2)2 + 3(10C3)2 + ... + 10(10C10)2, t gk¡ 10Cr, r{1, 2, ...,10}, f}i n xq.kkadksa (binomial

coefficients) dks n'kkZrs gSaA rc 1

X1430

dk eku gS____A

mÙkj (646.00)

gy (1 + x)10 = 10C0 + 10C1x + 10C2x2 + ... + 10C10x10

nksuksa vksj x ds l ki s{k vodyu djus i j

10(1 + x)9 = 10C1 + 2·10C2x + 310C3x2 + ...+ 1010C10x9 ...(1)

(1 + x)10 = 10C0x10 + 10C1x9 + ... + 10C10 ...(2)

10(1 + x)9(1 + x)10 esa x9 dk xq.kkad l eku gS

pw¡fd (10C1)2 + 2(10C2)2 + 3(10C3)2 + ... + 10(10C10)2

= 10(1 + x)19 esa x9 dk xq.kkad = 10 × 19C9 = X

1X

1430 =

1 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11× 10 ×

1430 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 646

[ kaM 3 (vf/ dre vad % 12)

• bl [ kaM esa pkj (04) i z'u gSaA i zR;sd i z' u esa nks (02) l qesyu l wfp;k¡ (matching lists) gSa% l wph–I vkSj l wph–II |

• l wph–I vkSj l wph–II ds rÙoksa ds l qesykuks dks n'kkZrs gq, pkj fodYi fn, x, gSaA bu pkj fodYi ksa esa fl iQZ ,d fodYi gh l ghl qesyu i znf' kZr djr k gSA

• i zR; sd i z' u ds fy, l gh l qesyu i znf' kZr djus okys fodYi dks pqusaA

• i zR; sd i z' u ds mÙkj dk ewY; kadu fuEu vadu ; kst uk ds vuql kj gksxk

i w.kZ vad %+3 ; fn fl i QZ l gh fodYi gh pquk x; k gSA

' kwU; vad %0 ; fn dksbZ Hkh fodYi ugha pquk x; k gS (vFkkZr~ i z' u vuqÙkfjr gS)A

Í .kkRed vad %–1 vU; l Hkh i fjfLFfr ; ksa esaA

15. ekuk fd

1

xE x : x 1 0

x – 1vkSj

vkSj

–1

2 1 e

xE x E : sin log , (real number)

x –1,d okLrfod l a[ Õkk gS

(; gk¡ çfr ykse f=kdks.kferh; i Qyu (inverse trigonometric function) sin–1x, – ,

2 2 esa eku / kj.k djr k gS)

ekuk fd i Qyu

1 e

xf : E , f(x) log

x – 1 ds }kjk i fjHkkf"kr gS vkSj i Qyu

–12 2

xg : E , g(x) sin log

x – 1

ds }kjk i fjHkkf"kr gSA

Page 40: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

40

JEE (ADVANCED)-2018 (PAPER-2)

l wph-I l wph-II

P. f dk i fj l j (range) gS 1.

1 e, ,

1 e e 1

Q. g dk i fj l j esa l ekfgr (contained) gS 2. (0, 1)

R. f ds çkUr (donain) esa l ekfgr gS 3.

1 1,

2 2

S. g dk çkUr gS 4. ( , 0) (0, )

5.

e,

e 1

6.

1 e( , 0) ,

2 e 1

fn, gq, fodYi ksa esa l s l gh fodYi gS:

(A) P 4; Q 2; R 1; S 1

(B) P 3; Q 3; R 6; S 5

(C) P 4; Q 2; R 1; S 6

(D) P 4; Q 3; R 6; S 5

mÙkj (A)

gy f(x) ds çk¡r ds fy, ;

x

0 x 0 x 1x 1

Õkk

f

1 eD : ( , 0) (1, ) , ,

1 e e 1

f(x) ds i jkl ds fy, ;

ekuk

y

e y

x ey log x

x 1 e 1

vc x < 0 ; k x > 1

y

y

e0

e 1;k

y

y

e1

e 1

0 < ey < 1 ;k y

10

e 1

– < y < 0 ;k ey > 0

f(x) dk i jkl ( , 0) (0, ) gS

g(x) ds çk¡r ds fy, ;

e

x1 log 1

x 1rFkk

x

0x 1

Page 41: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

41

JEE (ADVANCED)-2018 (PAPER-2)

1 xe

e x 1rFkk x < 0 or x > 1

x(e 1) 10

x 1rFkk

(e 1)x e

0x 1

;k x < 0 ;k x > 1

⎛ ⎤ ⎡ ⎞ ⎜ ⎟⎥ ⎢ ⎝ ⎦ ⎣ ⎠

1 ex , ,

e 1 e 1

g(x) ds ijkl ds fy,;

⎛ ⎞ ⎜ ⎟⎝ ⎠e

x1 log 0

x 1 ;k ⎛ ⎞ ⎜ ⎟⎝ ⎠

x0 log 1

x 1

g(x) 0 0 g(x)2 2

Õkk

g(x) dk ijkl = ⎡ ⎞ ⎛ ⎤ ⎟ ⎜⎢ ⎥

⎣ ⎠ ⎝ ⎦, 0 0,

2 2

R 1, P 4, S 1, Q 2

16. ,d gkbZ Ldwy (high school) esa] 6 ckydksa M1, M

2, M

3, M

4, M

5, M

6 vkSj 5 ckfydkvksa G

1, G

2, G

3, G

4, G

5 ds lewg (group)

esa ls ,d lfefr (committee) cukbZ tkuh gSSA

(i) ekuk fd 1 lfefr dks bl çdkj ls cukus ds rjhdksa (ways) dh dqy la[;k gS fd lfefr esa 5 lnL; gSa] ftuesa ls Bhd

(exactly) 3 ckyd vkSj 2 ckfydk,a gSaA

(ii) ekuk fd 2 lfefr dks bl çdkj ls cukusa ds rjhdksa dh dqy la[;k gS fd lfefr esa de ls de (at least) 2 lnL; gSa]

vkSj ckydksa vkSj ckfydkvksa dh la[;k cjkcj (equal) gSA

(iii)ekuk fd 3 lfefr dks bl çdkj ls cukusa ds rjhdksa dh dqy la[;k gS fd lfefr esa 5 lnL; gSa] ftuesa ls de ls de 2

ckfydk,a gSA

(iv) ekuk fd 4 lfefr dks bl çdkj ls cukusa ds rjhdksa dh dqy la[;k gS fd lfefr esa 4 lnL; gSa] ftuesa ls de ls de 2

ckfydk,a gSa vkSj M1 o G

1 lfefr esa ,d lkFk ugha gSaA

lwphlwphlwphlwphlwph-I lwphlwphlwphlwphlwph-II

P. 1 dk eku gS 1. 136

Q. 2 dk eku gS 2. 189

R. 3 dk eku gS 3. 192

S. 4 dk eku gS 4. 200

5. 381

6. 461

fn, fodYiksa esa ls lgh fodYi gS:

(A) P 4; Q 6; R 2; S 1

(B) P 1; Q 4; R 2; S 3

(C) P 4; Q 6; R 5; S 2

(D) P 4; Q 2; R 3; S 1

mÙkjmÙkjmÙkjmÙkjmÙkj (C)

Page 42: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

42

JEE (ADVANCED)-2018 (PAPER-2)

gygygygygy

(i) 1 = 6C

3 × 6C

2 = 20 × 10 = 200

(ii) 2 = 6C

1 × 5C

1 + 6C

2 × 5C

2 + 6C

3 × 5C

3 + 6C

4 × 5C

4 + 6C

5 × 5C

5

= 30 + 150 + 200 + 75 + 6 = 461

(iii) 3 = 5C

2 × 6C

3 + 5C

3 × 6C

2 + 5C

4 × 6C

1 + 5C

5

200 + 150 + 30 + 1 = 381

(iv) 4 = (5C

2 × 6C

3 – 4C

1 × 5C

1) + (5C

3 × 6C

1 – 4C

2 × 5C

0)

= (150 – 20) + (60 – 6) + 5

= 130 + 60 – 1 = 190 – 1 = 189

17. ekuk fd 2 2

2 2

x yH: 1

a b, tgk¡ a > b > 0, xy-lery (plane) esa ,d ,slk vfrijoy; (hyperbola) gS ftldk la;qXeh v{k

(conjugate axis) LM mlds ,d 'kh"kZ (vertex) N ij 60º dk dks.k (angle) varfjr (subtend) djrk gSA ekuk fd

f=kHkqt (triangle) LMN dk {ks=kiQy (area) 4 3 gSA

lwphlwphlwphlwphlwph-I lwphlwphlwphlwphlwph-II

P. H ds la;qXeh v{k dh yEckbZ gS 1. 8

Q. H dh mRdsUærk (eccentricity) gS 2.4

3

R. H dh ukfHk;ksa (foci) ds chp dh nwjh gS 3.2

3

S. H ds ukfHkyEc thok (latus rectum) dh 4. 4

yEckbZ gS

fn, gq, fodYiksa esa lgh foDyi gS :

(A) P 4; Q 2; R 1; S 3

(B) P 4; Q 3; R 1; S 2

(C) P 4; Q 1; R 3; S 2

(D) P 3; Q 4; R 2; S 1

mÙkj (B)

Sol.30°

L

b

M

O a

y

xN

btan30

a

a

b3

...(i)

Page 43: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

43

JEE (ADVANCED)-2018 (PAPER-2)

vc OLN dk {ks=ki Qy = 1

ab2

1

ab 2 32

ab 4 3 ...(ii)

(i) rFkk (ii) l s a 2 3, b = 2

vc 2

2

b 4 2e 1 1

12a 3

ukfHk; ksa ds eè; nwjh = 2ae

2

2 2 3 83

ukfHkyEc dh yEckbZ = 22b 2 4 4

a 2 3 3

18. ekuk fd i Qyu f1 : , f

2: – ,

2 2 , f

3 :

2– 1, e – 2 rFkk f4 : bl çdkj i fjHkkf"kr gSa fd

(i) f1(x) = 2– x1 – e ,sin

(ii) f2(x) =

–1

sinx

tan xx 0

1 x 0

Õkfn

Õkfn, t gk¡ çfrykse f=kdks.kferh; i Qyu (inverse trigonometric function) tan–1x,

– ,2 2 esa

eku / kj.k djrk gS]

(iii) f3(x ) = [sin(log

e(x + 2))]; t gk¡ t , [t ], t l s NksVk t ds cjkcj egÙke i w.kk±d (greatest integer) dks n'kkZrk gS

(iv) f4(x) =

2 1x sin x 0x

0 x 0

Õkfn

Õkfn

l wph-I l wph-II

P. i Qyu f1

1. x = 0 i j l arr (continuous) ugha gS

Q. i Qyu f2

2. x = 0 i j l ar r gS vkSj x = 0 i j vodyuh;

(differentiable) ugha gS

R. i Qyu f3

3. x = 0 i j vodyuh; gS vkSj x = 0 i j bl dk vodyt(derivative) l arr ugha gS

S. i Qyu f4

4. x = 0 i j vodyuh; gS vkSj x = 0 i j bl dk vodytl ar r gS

fn, gq, fodYi ksa esa l s l gh fodYi gS :(A) P 2; Q 3; R 1; S 4(B) P 4; Q 1; R 2; S 3(C) P 4; Q 2; R 1; S 3

(D) P 2; Q 1; R 4; S 3

mÙkj (D)

Page 44: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

44

JEE (ADVANCED)-2018 (PAPER-2)

gy ( i ) f1 : R R

2x1f (x) sin 1 e

= 2x

1sin 1

e

f1(x) l oZ=k l r r~ gS] i jUrq f1(0) fo| eku ugha gS] i QyLo: i x = 0 i j f1(x) vodyuh; i Qyu ugha gS f1(x) vodyuh;i Qyu ugha gSA

(ii) 2f : , R2 2

1

2

1

sinx; x 0

tan xf (x) 1 ; x 0

sinx; x 0

tan x

x = 0 i j LHL dk eku –1 gS

x = 0 i j RHL dk eku 1 gS

x = 0 i j f2(x) vl r r~ gS,

(iii) 23f : 1, e 2 R

3 ef (x) sin(log (x 2)

21 x e 2

21 x 2 e

e0 log (x 2)2

e0 sin(log (x 2)) 1

[sin(loge(x + 2)] = 0

f3(x) = 0

x = 0 i j f3(x) l r r~ ,oa vodyuh; gS

(iv) f4 : R R

2

4

1x sin ; x 0

f (x) x

0 ; x 0

Page 45: Time : 3 hrs. Max. Marks: 180 Answers & Solutions for JEE ... · (D) 50.7 cm ij l u st ku sokyk vuuqkn] ey x.kkofÙk (fundamental harmonic) gS mÙkj (A, B, C) gy (2n–1) 50.7 e 4

45

JEE (ADVANCED)-2018 (PAPER-2)

24

x 0 x 0

1lim f (x) lim x sin 0

x

= x = 0 i j f4(x) dk eku

x = 0 i j f4(x) l r r~ gS

vc, h 0 h 0

1sin

f(0 h) f(0) hf (0 ) lim lim 01hh

h 0 h 0

1sin

f(0 h) f(0) hf (0 ) lim lim 01hh

x = 0 i j f(x) vodyuh; gS

vc, 24 2

1 1 1f (x) 2 sin x . cos

x xx

= 1 1

2 sin cosx x

fujar j gS

x = 0 i j vl r r~ i Qyu gS