JEE ADVANCED MOCK TESTS - IITPK...JEE ADVANCED MOCK-TEST SOLUTION - PRACTICE TEST - 1 1 1 JEE ADVANCED MOCK TESTS SOLUTION OF PRACTICE TEST - 1 PAPER – 1 MATHEMATICS Sol.1 …

JEE ADVANCED MOCK-TEST

SOLUTION - PRACTICE TEST - 1

1

1 JEE ADVANCED MOCK TESTS SOLUTION OF PRACTICE TEST - 1PAPER – 1

MATHEMATICS

Sol.1 (a) Let14612234 222 zyxzyxzyxf ),,(

= 113321 222 )()()( zyxFor the least value of , we have),,( zyxf

0103201 zandyx ,

1231 zyx ,,

Hence, the least value of 11231

,,),,( fiszyxf

Sol.2 (c) Let be a positive integer for which m 22 96 mn or 969622 )()( nmnmornmor 962 })({)( nnmnmHence must be both evennmandnm As , the number of solution is 4.12816624448296 ororor

Sol.3 (a) We have,

xy

yx

xyyx

21

2

22

Now, xy

yx2

222 sin

02

22

xyyx ]sin[ 02

Therefore, and have the same sign. Now, x y

)122

12

2222

xy

yxxy

yx

xyyx ....( MGMA

But Therefore, .sin 12 yxxy

yx

12

22


The IITian’s Prashikshan Kendra Pvt. Ltd.

2

Sol.4 (b) We have,

77

76

75

74

72

7 coscoscoscoscoscos

= coscoscoscoscoscoscos

74

73

75

72

76

7

= coscoscoscoscoscoscos

73

73

72

72

77= 1cos

Sol.5 (a) )sincos( xBxAey x

]sincos[]cossin[ xBxAexBxAedxdy xx

…… (i)yxBxAedxdy x ]cossin[

Again ,differentiating w.r.t. x, we get

dxdyxBxAexBxAe

dx

yd xx ]sincos{]cossin[22

[…..using (i) ]dxdyyy

dxdy

dx

yd

2

2

02222

ydxdy

dx

yd

Sol.6 (c) kjiOBOCBC ˆˆˆ 424 kiAB ˆˆ 33

kjiAC ˆˆˆ 72

541836 222 ACABBC ,,,

Clearly, 222 ABBCAC 90B

Sol.7 (a) We have , whereGDGDGCGB 211 )(D is the midpoint of BC. Therefore,

02 GAGAGDGAGCGBGA

G divides AC in the ratio 2 : 1, )( GAGD 2

Sol.8 (a) If there are more than one rational points on the circumference of the circle [as ( ) is the center), then will be a rational multiple of , 02222 ceyxyx e, e

which is not possible. Thus, the number of rational points on the circumference of the circle is atmost one.

Sol.9 (a) The point of intersection of diagonals lies on the circumcircle, i.e. (1, 1), since 01212 )()( xyxy

722 sinRl



3

72

72236

cossin

sinR

(1,1)72

R

l

Therefore, the locus is : 7211 222 cos)()( yx

or 027212222 sinyxyx

Sol.10 (c) Let the equation of the chord OA of the circle …. (i)04222 yxyx

be ….(ii)mxy

OX

AC

YO

B

Solving (i) and (ii), we get042222 mxxxmx

or 0421 22 xmxm )()(

or 21420m

mxandx

Hence, the points of intersection are

(0, 0) and

22 1

42

142

m

mm

m

mA)(

,

or 22

22

22

1

421

142

m

mm

m

mOA

)()(

Since OAB is an isosceles right-angled triangle,22

21 ABOA

Where AB is a diameter of the given circle. Hence,

101

4210 2

22

m

morOA

)(

or )( 22 11016164 mmm or 0383 2 mm



4

i.e.313 orm

Hence, the required equations are :033 yxandxy

Sol.11 (a, c, d)12

12

ydx

dyandyxy

Also,yx

ydxdyor

yxy

21

Also, 02 xyy

or (as y > 0)2

411 xy

orxx

y41

141

441

'

Sol.12 (a, b, c)

214214

728

34165312728

][ 12233 RRandRRR = 0

abccabcbabca

abcabccacbbbcaa

333

222

111

1

111

///

],,[ 332211 cRRbRRaRR

= = 0 (taking common from 111111

333

cba

abcabc abc ]3C

aaaaaa

bababa

bababababababababa

222

32

65443232

],[ 122233 RRRRRR = 0

233477612

217343576432

][ 322 7CCC

= 2230470611

][ 211 CCC

Sol.13 (a, b, c) Let 221 tan)(tan oror ),(),/( 0202 or



5

53

11202 2

2

tantancos)(cos

or53

532 11

coscos

or54

532 11 tancos

= 43

243 11 tancot

Sol.14 (a, b) We have ).(|||||| bababa 2222 or 22222 cos|||||||||| bababa or cos|| 222 ba )||||( 1 ba

= 24 sinor |sin||| 2 ba

Now, 21121 |sin||sin||| orba

],/()/,[ 6560 or

Sol.15 (a, b, d)xxxx 311 1111 tan)(tan)(tan)(tan

or )(tantan)(tan)(tan 131 1111 xxxx

or

)()(

tan)()(

)(tan

13113

111 11

xxxx

xxxx

orxx

x

xx

x

33112

112

22

or )()()()( 12331121 22 xxxxxx

or210 ,x

Sol.16 (2) are extremities of the latus rectum having positive ordinates. Then,)/,( abae 2

…..(i)

22

222

abea

But, …..(ii))( 222 1 eab Therefore, from (i) and (ii), we get

0422 222 aaeea

or 02222 )()( aaae 0222 ))(( aaeHence, 2a



6

Sol.17 (1) Let occurs in term, then2x 1rT

rrr

rr

rrn

rn

r xb

aCbx

axCT

222

11112

11 .)(

For 573227 rorrx ,

Hence, coefficients of is 7x 56

511

b

aC

Let occur in term, then7x 1rTr

rrr

bxaxCT

2

11111

1)(

= rrr

r xb

aC 31111

11

)(

For 73117 rx 6r

Hence, co-efficient of 65

6117

b

aCisx

Now, 56

611

6

55

11b

aCb

aC

6

56

115

11b

aCaC

b

CaC 161111

511

b

CaC 1511

511

1ab

Sol.18 (1) Interior angle of regular polygon of side is n

n360180

Hence, 150144120180 ;;;

41518108 sincoscos

2120 secsec

41536144coscos

2150 ecec coscos

124

1524

15

)()(

Sol.19 (4) Given, 41

23 23 xxxxf )(

= )( 146441 23 xxx



7

= )( 2146441 23 xxx

= 421

41 44 ])([ xx

421

411 44 ])([)( xxxf

…. (i) 142

421 )()( xfxf

Replacing by , we have x )(xf…..(ii)11 )]([])([ xffxff

Now, …..(iii)43

41

.

/))(( dxxffI

Also, ……(iv) 43

41

43

4111

/

/

/

/))(())(( dxxffdxxffI

[ using (i) ]Adding (iii) and (iv), we get

43

41

43

412112

/

/

/

/))](())(([ dxdxxffxffI

Or I = 41

41 I

Sol.20 (9) The centre of the given circle is O ),( 34

(4,-3)

B

A

P(2,3)

The circumcircle of will circumscribe the quadrilateral also. Hence, one of the PAB PBOAdiameters must be OP.So, the equation of circumcircle of PAB will be:

….(i)03342 )()()(),( yyxxor 01622 xyxDirector circle of the given ellipse will be

222 935 byx )()(

or …..(ii)025610 222 byxyxSo, from (i) and (ii), by applying the condition of orthogonality, we get,

225130532 b )]()([

or 22430 bHence, 542 b



8

PHYSICS

Sol.21.(a) Distance traveled = 50 +40 + 20 = 110 mmDCABBFABAF 302050

His displacement, 22 DFAFAD

= m5025004030 22

In 43

4030

AEDEAED tan,

431tan

His displacement from his house to the field is 50 m, EtoN

431tan

Sol.22 (b) Given, sec,,, 65010 TcmntAtcmr

So, 1sec36

22 T

At cmxt 50 ,So, [Y= ])(sin 0105 tr sin

= 10 sin

6

21

sin

Equation of displacement

6310 tcmx sin)(

(ii) At sec4t

6910

6810

64

310 sinsinsinx

= 102

102

102

310

sinsinsin

Acceleration, xa 2

= 22

11910109

sec/.)( cm

E

N

S

W



9

Sol.23 (c) smvsmvair /,/ 5200330 Here, s = 7 m.

So, sec. 321 1075252001

3301

ttt

= 2.75 ms

Sol.24 (a) The variation of temperature is given by :…. (i)

dTT

TT)( 12

1

We know that, TV

273273TVVT

TV

TT

TVdu

VdxdtT

273

d

xd

TTT

dxV

t0 121

273)(

= )()( 1212

2273 TTTT

dV

= )( 1212

2732 TTTTV

d

12

2732TTV

dT

[Since ]1212 TTTT (Putting the given values, we get

ms96310280

273330

332

Sol.25 (d) We know that 2

1

1

2vv

so,v

310314721

.

sec/. mv 8400 10042

sec]/,,[ mvandairfor 81031

Again,760

810314521

v

.

sec/. mv 8760 10072

Sol.26 (c) The time period at temperature is

glgl

T o /)( 122



10

=

21112 21 o

o Tgl /)(

Thus,

)( CTT o 402

1120

and,

)( CTT o 402

1140

or, 12040 101201 ])([])([ CC

TT

= 1101201 ])([])[ CC= )( C 101

or ….. (i)420

2040 102110

.)( CT

TT

This is fractional loss of time. As the temperature increases, the time period also increases. Thus, the clock goes slow. The time lost in 24 ours is, by (i)

shourst 410102124 4 .).()(

Sol.27 (a) 262 10111 mxmmAAi ,3/9000 mkgcu

Molecular mass has atoms.

m

NhasmkgatomsN oo

= 3105639000

.

ANo

No. of atoms = No. of electrons

MpN

lmPNn oVolumeUnit

electronsofNumber

...

= 323

10563

9000106

.AenVi d /

1963

23106110

10563

90001061

.

.xVd

19626

3

1061109106

10563

.

.

19623

3

1061109000106

10563

.

.

= smmsm /./.. 07301007301696

10563 33



11

Sol.28 (b) VKVV 31 101212 qEF

16101 msv

aSuv 222

or 310122 mqv

or 326 10122101 mq)(

or mq312 102410

or qBmvrand

qm

91024

and,20

101024 69

.

r

cmmr 121012 2

Sol.29 (c) The side of the square is:mml 0501052 23 ..

As it is uniformly pulled out in 1.0 s, the speed of the loop is :1050 msv .

The emf induced in the left arm of the loop is NvBl

= VmTms 10050400050100 1 .).().().(

The current in the loop is AVi 31001100

10

..

The force on the left arm due to the magnetic field is ).().().( TmANilBF 4000501001100 3

= N31002 .This force is towards left in the figure. To pull the loop uniformly, an external force of

towards right number be applied. The work force by this force is N31002 .JmNW 43 10010501002 .).().(

Sol.30 (a) Force on the block are :

mg

F

N

N



12

(i) its weight Mg, (ii) the normal force N (iii) the applied force F and(iv) the kinetic friction NThe forces are shown in above figure. As the block moves with a uniform velocity, the forces add up to zero.Taking horizontal and vertical components,

MgNFandNF sincosEliminating N from these equations,

sincos

,)sin(cos

Mg

ForFMgF

The work done by this force during a displacement is

sincos

coscos

MgdFdw

Sol.31 (a, b)

2 kg 3 kg 2 kg 3 kg

36 km hr-1 Stationary v

Before After

The initial velocity of 2 kg ball.smhrkm // 10

18536

Let the common velocity of 2 kg and 3 kg after collision = vUsing conservation of momentum,

vkgkgmskgmskg )()()()()( 3203102 11

11

45

20

mskg

kgmsv

Hence, Initial kinetic energy JmskgKi 10010221 21 )()(

[option a ]

And final kinetic energy = 2143221 )()( mskgkg

= 40 J [option b ]

Sol.32 (b) Let 210 smg /

53

5030

sin,Here

30 m

50 m

50 KJ

Nmg cos

mg sin

54

cos

SmgSNWfriction )cos()(



13

= J80005054105040 )().(

JmnkghmgU 15000301050 2 )()sec/()(Using work-energy theorem.

KUWW frictionex Here, 0K

)()( JJWUW frictionex 800015000 = 23000 J = 23 KJ

(b) is correct

Sol.33 (a, d)r

GMvo

oooop vvvvvv 451

45

oe vv 2

rr’

m

v’

m

v

Earth

Since, , it means particle will go around the earth in an elliptical orbit. epo vvv Now, from conservation of angular momentum.

'' rmvmvr or …..(i)'.'. rvrv From conservation of energy

….(ii)'

'r

GMmmvr

GMmmv 2221

21

Substituting for from (i) in (ii),'v

'' rGMm

r

rmvr

GMmmv 2

222

21

21

As,r

GMv45

'' rGM

r

rr

GMr

GMr

GM

2

2

45

21

45

21

'' rr

r

r

2

2

851

85

Put 083

85 2 xxx

rr'

0385 2 xx03355 2 xxx

53101315 orxxxx )()(



14

rrrrorrr35

53

'''

Hence, the perigee and apogee distances for the particle are respectively.35randr

Hence option (a) and option (d) are correct.

Sol.34 (b, c) cos.. rmsrmsAV IVP In LCR,

IZIR

VVR cos

IVg

VC

V

Vg

VL-VC

2

2

Z

RVZR

ZV

VP rmsrmsrmsAV

Here, ;VVrms 12LfLXR L ., 250

= 21020502 3 LX

and 6101005021

211

CfCXc .

=

100

5622 )( XCXLRZ

Energy used in 1000 s is tZ

RVtP rmsAV .

.)()(

2

2

= J322

1032100056

5012

.

)(

option (b) is correct ][

For Energy used is = ,st 200 20056

50122

2

= J3104600 .option (c) is correct ][



15

Sol.35 (a, b) uoruvvum 333

cmf 15

Here, fuv111

Case I :1511

313

uuuv

151

34

ucmu 20

Case II :1511

313

uuuv

151

331

u

cmuoru 103

152

)(

Options (a) and (b) are correct.

Sol.36 (a) Let the ball strike the stepnth210

2120 tn )(.

or nnt 04010

0222 ..

2120 /. nt Horizontally the ball travels

2121 902054 // ).(sec).(sec)/.(. nmnmtux Also, horizontal distance of nth step = (0.3 m)n

(0.9 m) nnth 30.

9321 nn ;/

Sol.37 (0) The maximum contribution may come from the charge forming pairs with others. q8To reduce its effect, it should be placed at a corner and the smallest change in the middle. qThis arrangement shown in figures ensures that the charges in the strongest pair are qq 82 ,at the largest separation.

x 9cm-x

2qx

q 8q

The potential energy is ,

xcmcmxqU

98

9162

4 0

2

This will be minimum if :is minimum.

xcmxA

982



16

For this, …… (i)09

8222

)( xcmxdx

dA

or, cmxorxxcm 329 ,The electric field at the position of charge is :q

….from (i)09

824 220

2

)( xcmxq

Sol.38 (7) The average temperature of the liquid in the first case is:C

CC

65

26070

1

The average temperature difference from the surrounding is:CCC 35306501

The rate of fall of temperature is:11 2

56070 min

minCCC

dtd

From Newton’s law of cooling,)(min CbAC 352 1

or ….(i)min352

bA

In the second case, the average temperature of the liquid is

CCC

552

50602

So that, CCC 25305502 If it takes a time to cool down from to , the rate of fall in temperature ist C60 C50

tC

tCC

dtd

1050602

From Newton’s law of cooling and (i), or c

tC

25

35210min

min7t

Sol.39 (1) We have, qBmK

qBmvr

2

Where, energykineticmvk 221

Thus,Bq

Kmr

Bq

Kmr

d

dd

p

pp

22 ,

andBq

Kmr

a

aa

2

We get, 21

2

p

p

p

p

d

p

p

d

d

pm

m

q

q

m

m

qq

r

r

and 14

2

p

p

p

p

a

p

p

a

d

pm

m

q

q

m

m

qq

r

r



17

Sol.40.(2) Consider water as the system. At the top of the circle its acceleration towards the centre is

vertically downward with magnitude . The force on water are :r

v 2

(a) weight downward and Mg(b) normal force by the bucket, also downward.So, from Newton’s second law,

rMvNMg /2For water not to fall out from the bucket, .0N

Hence, rgvorMgrMv 22 ,/The minimum speed at the top must be rgIf the bucket continues on the circle with this minimum speed , the forces at the bottom of the rgpath are :(a) weight Mg downward and (b) normal contact force N’ by the bucket upward.The acceleration is towards the centre which is vertically upward, so

or rMvMgN /' 2 MgrvgMN 22 )/('

CHEMISTRY

Sol.41 (b) Molar mass of Glouber’s salt,1

242 32210 molgOHSONa .

Molar mass of Na2SO4 = 142 g mol-1

Mass of Na2SO4 in solution =

1

1

322

1422540

molg

gmolg ).(

= 17.75 gMass of solution = ).()( 107751500 molgmL

= 538.75 gMass of water = (538.75 – 17.75 )g = 521 g

Amount of Na2SO4 = molmolg

g1250

142

55171 .

.

Molality of Na2SO4 = 124052101250 kgmol

kgmol

..

.



18

Sol.42 (a) Higher value of standard reduction potential for shows thatAgAg /

reduction would take place on silver electrode.Hence, cell would be AgAgCuCu |||).(| 0102

VEEE oanodeocathode

ocell 462033707990 ...

The cell reaction is:222 neiAgAgCu .,.

Applying Nernst Equation,

2

205910][

][log.

Ag

Cun

EE ocellcell

)(givenEcell 0

2010

2059104620

][.log..

Ag

0591020462010 2

2

..][loglog Ag

263152 .][log Ag

815826317 ..]log[ Ag

Taking antilog MAg 9105231 .][

Sol .43(a) We have

Choice (a) representsDCHDCHCHCHHCRR 235232 )()(),(

Sol.44 (d) is:butadienemethoxy 311 ,

32 CHOCHHCHCH

Its resonating structure may be visualized as follows,



19

32 CHOCHCHCHCH

(Choice b)32 CHOCHCHHCCHCH

32 CHOCHHCCHCH

(Choice a)32 CHOCHCHCHCHCH

32 CHOCHCHCHCH

(Choice c)22 CHOCHCHHCCH

The structure 32 CHOHCHCCHCH

(choice d) is expected to be least stable as the movement of electrons is in the opposite direction to that of electrons from oxygen of methoxy group.

Sol.45 (b) Bond angle in is near to . The bonding of P with H atoms HPH 3PH 90involves its orbitals. The lone pair on P is in atomic orbital. Bond angle is p s HNH near to . The bonding of atoms involves orbitals. The fourth orbital 109 HwithN 3sp 3spcontains lone pair, which has directional characteristics and more available for bonding. Thus is a stronger base than .3NH 3PH

is stronger acid than and thus is weaker than .SH2 OH2 HS OHis stronger acid than and thus is a weaker base than COOHCH3 OHCH3 COOCH3

.OCH3The S and Se involve bonds with the three oxygen atoms, which bear some of the dp dispersed negative charge. Because of larger size of and involvement of orbital in Se d4the overlap with orbital of oxygen, there is less overlap and lesser dispersal of negative p2charge and hence negative charge is more available in and thus acts as a 33SeOCHstronger base than .33SOCH

Sol.46 (b)



20

CH3

Dipole moment is smaller than chlorobenzene

Cl

Cl

Dipole moment is identical to monohclorobenzene

Cl

Cl

Dipole moment is 1.732 times that of monohclorobenzene

Cl

Cl

Zero Dipole moment

Hence, the increasing order is IV < I < II < III

Sol.47 (c) The given reaction proceeds as follows,

So.l48 (b) The reaction proceeds via elimination-addition mechanism.



21

Sol.49 (d) The half-cell reaction is ][

][ln )(

nx

x

A

AnFRTEE

][][

log.

formoxidizedformreduced

nV

E o

059150

Substituting the given values, we get

6754240591501010..log..

nVEV

= …..(i)).(. 4910059150

nVE

2518480591501150

.

.log..

nVEV

= ….(ii)).(. 0210059150

nVE

Subtracting Eq (i) from Eq.(ii), we get

)..(.. 021049100591500140

nVV

or 29810140

470059150 .

).().().(

VV

n

Sol.50 (a)



22

Sol.51 (a, b, c)(a) The reaction is :

(b) The reaction is :

(c) The general reaction for the formation of a Schiff base is:'' RNRCHNHRRCHO OH 22

(d) It is the trans R’ group which migrates

Sol.52 (a, b)reacts with the faster than 4NaBH OC CC

(b) The reaction is

(c) The products are and .COOHCH3 HCOOH

(d) The product is adipic acid



23

Sol.53 (a, b, d)Reactants (a, b, d) contain asymmetric C atom so in reaction they would give racemised 1SNand retention product. Reactant (c) does not contain any asymmetric C atom.

Sol.54 (a, b)

Also (c) and (d) would give more strained three-and four-membered ring, respectively.

Sol.55 (a, b, d)HF is more polar than HBr as EN of F is more than that of Br.

is more covalent than NaCl, as 18-electron shell of is more EN because CuCl )]([ 103dArCu

inner electrons have poor shielding effect on nucleus increasing the polarizing power of nucleus.

Sol.56 (5) 22 4224 ])([ SONaCNCuNaCNCuSO

22 2 )()( CNCuCNCNCu

23 43 ])([[ CNCuNaNaCNCuCN

422434 22102 SONaCNCNCuNaNaCNCuSO )(])([1 mole NaCNmoleCuSO 54

Sol.57 (5) Hydration energy of = Hydration energy of + Hydration energy of NaCl Na Cl= 11 382389 molkJmolkJ= 1771 molkJ

Heat of solution ( ) Hsol n

= Hydration energy – Lattice energy= 1776771 molkJ]([= 15 molkJ

Sol.58 (5) Given, LVatmP 10561 ;.0820317 .; RKT

Total moles molRTPVn 60

317082010561

..

.)(



24

Let be a mol, therefore moles of (0.6 – a) mol ; mass of C in a mole of 8HCx 12HCx.; mass of C in (0.6 – a) mol of 12 gaxHCx 1212 12HCx ga).( 60

Total mass of C in mixture = gaxax ).( 601212= 14.4 g

% of C in mixture = 100441

27

.. x

Given, % of C = 87 %

or 587441

720 xor

x.

Sol.59 (4)

This shows that the generation of one bond in cyclohexene requires of )( CC 1119 molkJenthalpy. To calculate RE resonance energy.

REHmolkJHHH 41

321 119 ,

)()( ecyclohexenHbenzeneHH ff 5

= 120515646 molkJ)( From Hess’ law :

)( 32154 HHHHH

= 11521193205 molkJ15238 x

4x

Sol.60 (3) 43234 3412 CHOHAlOHCAl )(3 moles of are produced.4CH



25

PAPER – 2

MATHEMATICS

Sol.1 (b) Let , where qp

184log Iqp ,

qp

qp

2132

2129 244 logloglog

)(log saynm

qp

2132

Where, and Inm , mnnm orn 23230 /)((possible only when which is not true)0 nmHence, is an irrational number.184log

Sol.2 (a) Let , then or ),,(log 10 andyxxyt 21 t

t 01 2 )(t

, we get yxeixt y ..,log 1

340122 ,, xxx)( rejectedonlyx 43

Sol.3 (c)

B

C

A

D 1B

1A

)( db

)(b)(0

)(d

Let P.V. of A, B and D be respectively.,, dandb0

Then P.V. of C, dbc

Also P.V. of 21d

bA

and P.V. of 21b

dB

ACdbABAA23

23

11 )(

Sol.4 (b) Let mxy be a chordThen the points of intersections are given by

04431 22 )()( mxmx



26

221221 14

143

mxxand

m

mxx

Since (0, 0) divides chord in the ratio 1 : 4, we have 12 4xx

22121 1

441

433m

xandm

mx

mmm 2416999 22

i.e.7

240 ,m

Therefore, the lines are y = 0 and 024 xy

Sol.5 (a) is a tangent12 yx

O

A (2, 5)012 yx

42 xx

Slope of line 21

OA

The equation of OA is

)()( 2215 xy

or 122 yxTherefore, intersection with will give the coordinates of center as (8, 2). Hence, 42 yx

535228 22 )()(OAr

Sol.6 (b)

21

221

11

nnnnS

nn

n.......limlim

nn

nn

nnnnnn

122

111

11 ........lim

= 1

0 1)( xxdx

Put, dzdxx

orzx 2

1

10

1

012

12

)(loglim

zzdz

Snn

= )log(log 122 = 422 loglog



27

Sol.7 (c)

n

rrn nrrn4

1243 )(

lim

243

1

nrn

nr

Tr

n

n

nr

nrn

S4

12

43

11lim

=

4

0243 )( xx

dx

Put dtdxx

ortx 12343

10

4

4

102 1011

32

32

ttdtS

Sol.8 (d) . Putting , we get

cb

cadxxfI )( dtdxorctx

b

a

b

adxcxfdtctfI )()(

= bc

acdxxf )(

Putting we get,dtcdxortcx

b

a

b

adxcxfcdtctfcI )()(

Sol.9 (c) Sol.10 (d)Sol.11 (b) Hint : In given determinant, applying and , we get122 CCC 233 CCC

bcbxcabcbx

cacxxf

3

22

11

0

0)(

= bcb

cabcbcac

bccabc

cax

3

22

11

3

22

1

0

0

011

01

So, is linear, Let . Then)(xf ,)( QPxxf QbPbfQaPaf )(,)(

Then,

….. (i))(

)()()(

abbafabf

QQPf

00



28

Also,acabab

acabac

af

3

2

1000

)(

= )()()( acacac 321Similarly,

)()()()( bcbcbcbf 321)()()()( xcxcxcxg 321

)()()()( bfbgandafag Now, from (i), we get

)()()(

)(ab

bagabgf

0

Sol.12. (d)Sol.13 (d)

Sol.14 (c) Hint : xcbcca

bcxbbaacabaxa

a

22

22

31

Applying and taking 3211 cCbCCC xcba 222

Common, we get

xcbccbcxbbacaba

xcbaa

2

22221 )(

Applying and , we get 122 bCCC 133 cCCC

xcxb

axcba

a0

000

1 222 )(

= )()( 22221 axxcbaa

= )( xcbax 2222

Thus, is divisible by . Also, graph of is 2xandx )(xf



29

Sol.15 (a,c) bbacbaca ).()(),(

= cxbyx )()sin( 124 2

or bbacbabca ).().().(

= cxbyx )()sin( 124 2

Now, , Therefore,cacc ).(

caorcccacc .).().().(

).(,sin. baxyxba 1241 2

or 1241 2 xyx sin

or 1122 22 )(sin xxxyBut 111 yxy sin,sin

Inny ,)(2

14

Sol.16 (a, b, c, d) Since are unit vectors inclined at an angle , we have candba ,

1 |||| ba

and cbca ..cos

Now, …. (i))( babac

})(.{).().(. baabaaaca

))(.,.(||cos 002 baabaa

Similarly, by taking dot product on both sides of (i) by , we get b cos

Again, )( babac 22 ])(||| babac

= 222222 |||||| baba

)}{.(})(.{).( babbaaba 22222221 || ba

=

22 22222 sin|||| ba

212

2222 or

But cos2221

221 22 coscos



30

Sol.17 (a, b) 222 coscossin 212 cossin

2

21212 cos)sin(cos

42

42 22 cossin

Sol.18 (a, c, d)Let be a real root. Then,z

01233 )()( aii

02133 )()( ai

2

0133 aand

012

38

3

aa

08123 aa

Let 8123 aaaf )( and 01020001 )(,)(,)(,)( ffff 03 )(f

Hence, ),(),(),( 321201 aoraora

Sol.19 (a, b, c)For , given equation becomes 0a

2

00

211 aoraordxax )(

For ,20 a

2

0 0

211

a

adxaxdxxaordxax )()(||

or 01212

222

222

aaoraaa

or 01 2 )(aFor 2a

2

0

2

011 dxxaordxax )(||

or 223122 aoraora

Sol.20 (a, c) The point from which the tangents drawn are at right angle lies on the director circle.Equation of director circle is 3216222 yxPutting , we get 2x



31

282 y

or 72y

So, the points can be ),(),( 722722 or

PHYSICS

Sol.21 (c) ijvvgvvgv grrr ˆˆ)(/ 34

112 5916 kmhkmhvvv rgrgr /

Sol.22 (b) cos)(

mgrR

mv

2

h

2221

21 Imvmgh

222107

51

211 mvmvmvrRmg )cos()(

cos)cos()( mgrRmg 17

10

)cos()( 17

102 rRmgmv

1710

717

710

coscos or

)(cos)( rRgrRgv 1710

and 21710

r

rRgrv )(

Sol.23 (b) Initially, when the switch is closed on position 1, the capacitor C isconnected in series with batteries and . From KVL, we have1E 2E

012 EECiQ

or ….. (i)CEEQi )( 12 Depending upon the sign of , charge on the left plate may be positive )( 12 EE iQ

, or negative ; charge on right plate would be equal and opposite. )( 12 EEif )( 12 EEif When the switch is moved to position 2, the left plate (earlier having charge +Qi ) will not have charge

…….(ii)CEQf 1



32

The net charge flow through the circuit is :CECEEEQQQ if 2121 ])([

We can say that a net positive charge equal to is pulled by the battery of emf from CE2 1Ethe left plate of the capacitor, which flows through battery and is transferred to the right 1Eplate of the capacitor. Work done by battery in the process of charge transfer is:1E

…… (iii)CEEW 21A part of this work changes the energy of the capacitor.

CEECEC

Q

C

QW ifC

212

21

22

21

21

22)(

= CEEE )( 2221221

and the remaining part is lost as Joule heat :CEWWH c

222

1

Sol.24 (c) This work done by us is stored in the capacitor in the volume where new electric field is created. If you calculate the work done by using the )( 12 ddA

expression

0112 12

20

ddAV

dUEW elext

you may get confused. Here, battery is also doing work, so from energyconservation principle, we get

0 batteryelext WWW

Sol.25 (a) From Snell’s law,

ri

risinsin

sinsin 12

21

From the graph,

12360

ri

ri

sinsin

sinsin

tan

12

12

vv

vc

21212

1 3 vvvv

Sol.26 (a)uvf111

20 cm 40 cm 20 cm v = ?

340

201

4011

ff )(

Focal length for the combination,



33

320

eqf

uvfeq

111

2011

2030

v

201

2031

v

(to the right)cmv 10

Sol.27 (c) The neutron following straight path. This is because of zero force on neutron as it is neutral.

Sol.28 (c)mqVvorqVmv 2

21 2

Centripetal force, qvBR

mv

2

RmqBv

Hence, RmqB

mqV

2

orBq

mVR 1221

/

Here, and are constant. Hence, qV , B 2Rm

So,2

21

21

RR

mm

Sol.29 (b)Sol.30 (c)

Hint : AR

I 21224

1kx 1kx kx kx

mg mg BI /

Magnetic force = 10

10522

2 BBBI sin

andxmgkmgkx

11 22 ;

BImgxxk )( 212BImgkxkx 21 22



34

BIxx

mg21

105030101010 3 B

.

.

TB 6010600 3 .

Sol.31 (c)

Sol.32 (a) Hint : For the first case : BvqF k̂31025

= )ˆˆˆ([)ˆˆ( kBjBiBji zyx 21010

65

= ]ˆ)(ˆˆ[ kBBjBiB xyzz

210

…… (i)TBBB xyz 3100 ,Similarly, for the second case :

])ˆˆˆ()ˆ()(ˆ kBjBiBkjF zyx 65

2 1010

)ˆˆ(ˆ jBjBjF yx 102…… (ii)0102 yx BBF ,

Using eqs. (i) and (ii), we get TBx 310Thus, iTB ˆ)( 310Also, NBF x 22 1010

Sol.33(a, b) Hint : As the magnetic field is along the axis, the magnetic force will be along x axisz )( 1at . So, the particle will move in helical path along (1). At , the direction of field 0t 0Tt changes, so force becomes along direction and now the particle will move in helical path zalong (2). It will be moving along axis, so that resulting path will be helical. x

, particle will be at 20TtAt 1E

0

0V

00 cosV

00 sinV

y

12

y

zx E1

E2

(half of pitch)20Pcoordinatex

0 (from fig.) coordinateyand (from fig.)02Rcoordinatez Hence, (a) is correct.



35

Similarly, at particle will be at .2

3 0Tt 2E

The coordinates are

02023

RPo ,,

Hence, (b) is correct.From fib. We can see that distance between two extremes : 021 4RisEandE

Sol.34 (b)

Sol.35 (d) is parallel to and is perpendicular to both. Therefore, path of the particle is E B va helix with increasing pitch. Speed of particle at any time is t

….. (i)222 zyx vvvv

Here, 2022 vvv zy

and 00 32 vvvv x

qEmv

ttmqEvtav xx

00

33

Sol.36 (a, b, d)

Since , the Wheatstone bridge is balanced. Hence, . No current passes 1

2

2

1CC

RR

DC VV

through the galvanometer. Hence, option (a) is correct.Potential difference across potential difference across C1 = 4 V1RPotential difference across potential difference across C2 = 5 V2RPotential difference across is the potential difference across capacitor is 5 F8

. Hence, option (b) is correct and so is option (d).CVFCVQ 4058

Sol.37 (b, c) Net force towards the centre of the earth.

33

2

3 Rmgx

RxmgR

RmGMxmg

)('

Normal force, sin'mgN Thus, pressing force,

xR

RmgxN

2

is constant and independent of x.2

mgN

Hence, option (b) is correct.Tangential force, cos'mgmaF



36

x

xR

Rgxga

22

4

cos'

22 4xRRgxa

Curve is parabolic and at 02

aRX ,

Hence, option (c) is correctSol.38 (a, b, c, d)

(i) If and the body is projected vertically upwards, the body will rise up to that height evv where its velocity becomes zero. After that it will fall freely due to gravity following a straight line path.

(ii) If and the body is projected at some angle w.r.t. vertical direction, the body will reach evv up to a certain height (where vertical component of the velocity becomes zero) and then fall down following a parabolic path (this will be a case of projectile motion).

(iii) If , then the orbit will be circular. But if (escape velocity), then RGMvv o / eo vvv the orbit will be an ellipse.So, all the options are correct.

Sol.39 (b, c, d)If they collide, their vertical component of velocities should be same, i.e.

5430160100 sinsinsin

x1x2x

Their horizontal components will always be same. Horizontal components.138030160 mscos

and 16053100100 mscos

They are not same, hence their velocities will not be same at any time. So (b) is correct.ttxxx coscos 1003016021

tx )( 60380

Time of flight : (same for both)sg

T 16301602

sin

Now to collide in air xTt

960312801660380

xx

Since their times of flight are same, they will simultaneously reach their maximum height. So it is possible to collide at the highest point for certain values of .x



37

Sol.40 (b, c)Initially, accelerations are opposite to velocities. Hence, motion will be retarded. But after sometimes velocity will become zero and then velocity will be in the direction of acceleration. Now, the motion will be acceleration. As the particle is blown over by a wind with constant velocity along horizontal direction, the particle has a horizontal component of velocity. Let this component be . 0vThen it may be assumed that the particle is projected horizontally from the top of the tower with velocity .0v

Hence, for the particle, initial velocity and angle of projection .0vu o0

We know equation of trajectory is

22

2

2 costan

ugxxy

Here, (putting 20

2

2vgxy )o0

The slope of the trajectory of the particle is:

xvg

v

gxdxdy

20

202

2

Hence, the curve between slope and x will be a straight line passing through the origin and will have negative slope. It means that option (b) is correct.Since horizontal velocity of the particle remains constant, .tvx 0

We get0v

gtdxdy

So the graph between and time will have the same shape as the graph between andm t m. Hence, option (a) is wrong.x

The vertical component of velocity of the particle at time is equal to . Hence, at time t gt ,t

])()[( 202

21 vgmKE

It means, the graph between KE and time should be a parabola having value att 2021 mv

. Therefore, option (c) is correct.0tAs the particle falls, its height decreases and KE increases.

, where H is the initial height. The KE increases linearly with)( hHmgmvKE 2021

height of its fall or the graph between KE and height of the particle will be a straight linehaving negative slope. Hence, option (d) is wrong.

CHEMISTRY

Sol.41 (b) (mole fraction of 2O )22 OtotalO PP

222

22

CONO

OO nnn

n

347760

1008650076010760

20

20

2 ..

)...(.

O



38

347760734202 ..

totalO PP

= Hgofmm076.

Sol.42 (a) , where and are moles present initially.2

2

2

21

SO

OSXn

n

OnOn

f '

')(

2OSn 2On

or

2

2

2

21 '

'loglog

SO

O

O

SO

n

n

n

nfX

2

2

2

2

2

2

SO

O

O

SO

SO

O

n

n

n

n

M

MX

'

'loglog

161

11

6432

loglogX

707064328

2

1

2

1 .; rr

nn

alsoX

If X = 6, then

2

2

2

264326

SO

O

O

SO

n

n

n

n

'

'loglog

=

161

2

2

O

SO

n

n

'

'log

122

2 :'

'

O

SO

n

n

Rate of diffusion is i.e. in each operation.,1

2

2

1MM

rr

7070.

Sol.43 (a) net dipole moment is zero.4CH resultant dipole moment towards nucleus of nitrogen.3NFresultant dipole moment towards lone pair3NHresultant dipole moment 2p cos .OH2 5104.

Sol.44 (b) Hybridisation of with one it is angular.231521 spNOF )( ,lp

Hybridisation of no l, it is planar.22 31521 spFNO )( lp

Sol.45 (d) )()()( aqSOaqCuaqCuSO isElectrolys 242

4



39

At cathode : )()( reductionCueaqCu 22

The blue colour of disappears due to the deposition of on 4CuSO Cu PtElectrode.

At anode : . Since oxidation potential of )(gOeHOH 22 2122 OH2

oxidation potential of , so oxidation of occurs and is evolved at anode.24SO OH2 )(gO2The colourless solution is due to the formation of H2SO4 as follows :

42242 SOHSOanodefromH

)(

Sol.46 (a) )( molCHMgBrHMeMgBrHH

14

Sol.47 (a) Since the reactant is 1 alkyl halide, so in the presence of NaCN, it will follow E2 3path rather than so path (II) is not feasible.,2SNThe possible product by path (II) is :

Sol.48 (a) Here (a) is stable because it would not change to other stable carbocation. It can only change to C2 C2

On the other hand, (b) can change to two structuresC2



40

Furthermore, (c) is stabilized by 1, 2-Me shift and (d) is stabilized by shift. H21,

So (a) is most stable.Sol.49 (a)

Sol.50 (b)

Sol.51 (d)

COOH

COOH COOH

COOHPhthalic acid(D) or (I)

Terephtalic acid(E) or (II)

Sol.52 (a)

COOH

COOH

O

O

O

Sol.53 (b) At anode : )()()( oxidationeaqHgH 222

At cathode : )(Re

)()()(duction

aqBrsAgesAgBr 2222

Cell reaction :)()()()()( aqBraqHsAggHsAgBr 2222 2

cellnFECatG 15

= = VmolC 230965002 1 . 144390 molJ cellnFECatG 35

= = VmolC 210965002 1 . 140530 molJ



41

Now, STHCatG 15

….. (i) SKHmolJ )(28844390 1

and SKHCatG )(30835

…..(ii) SKHmolJ )(30840530 1

(Assuming and at 298 K are temperature independent, i.e. H S AT and at )H Hc 15 H H35

Solving for and from equation (i) and (ii) we get, H S….(iii) HS28844390

….(iv) HS30840530Equating equations (iii) and (iv), we get

11193 molKJS

Sol.54 (d) The value of at is calculated as follows :G C25= SKHG )(298 )()( 11 19329899974 KJKmolJ

142460 molJ

The value of at will be:AgAgBrBrE || C25( at equilibrium = 0)cellE

VmolC

molJnFC

E cellcell 22096500242460

1

1.

)()()(

2HHEAgAgBrClEEcell |||

0220 AgAgBrClEV

||.

VEAgAgBrCl

220.||

The cell representation for the calculation of Ksp is : AgAgBrBrAgAg |||||

At anode : BrAgAg

At cathode : BrAgeAgBr

Cell reaction : AgBr BrAgAt equilibrium, 0cellE

][][log. BrAgEE cellcell 1060

0 spAgAgAgAgBrBr

KEE log.)(||| 1

060

spAgAgAgAgBrBr

KEE log.|||

060

( spKVV log.).. 060800220



42

010060

800220 ..

)..(log V

VK sp1010spK

Sol.55 (a, b, c)X is ][])([: 42326233 2 BHNHBHHBNHNH

][])([; 42232622333 2 BHNHCHCHHBNHCHNHCH ][])([)(;)( 422326242323 2 BHNHCHBHHBNHCHNHCH

With forms an adduct :6233 HBNCH ,)(

])[()( 3336233 2 BHNCHHBNCH

Sol.56 (a, d)

When ABBAB

B

A

A MTMTorMT

MT

andAA

AA RTM

WornRTPV )( B

B

BB RTM

WPV )(

When BA WW BA PVPV )()(

MRT

rms3

A

AA M

RT3

A

BB M

RT3

BA

B

B

A

AMT

MT

Sol.57 (b, c)

Hybridisation Geometry Shape(a) 2

3 33321 spBF )(

Planar Planar

33 4352

1 spPCl )(TH Pyramidal

(b)dspFXe 32 5282

1 )(

Tbp Linear

spCO 20421

2 )(Linear Linear

(c) 34 4442

1 spCF )(TH TH

34 4442

1 spSiF )(TH TH

(d)dspPF 35 5552

1 )(

Tbp Tbp

235 6572

1 dspIF )(OH Square pyramid



43

Sol.58 (a, c)

ClHNCHHClNHCH 3323Initial 0.1 0.08 Moles final (0.1 – 0.08) (0.008 – 0.008) 0.008

= 0.02 = 0Since is left, so it forms basic buffer solutionBW

44 10251080020105

.

.

.][][][

SaltBaseKHO b

MHO

KH w 114

14108

1025110

.][][

= = )(log 11108 pH 1124 log 8911430 .. pOH = 14 – 9.8 = 4.2

Sol.59 (b, c, d)(a) is wrong.For reaction of AS with , basic buffer is formed is maximum. When BW bpKpOH At 50% neutralization (25 mL of 0.1 N Hcl)Slope of the given graph will be least and the buffer will have maximum buffer capacity.

Sol.60 (d) For precipitation, ion should be minimum in the solution.][ Ag

For MCl

AgClKAgAgCl sp 91

1010

10511051

..

][][: min

For AgBr : MBr

AgBrKAg sp 94

1310

10051005

..

][][ min

For MCrO

CrOAgKAgCrOAg sp 52

12

24

4242 10

10911091

..

][][: min

Therefore, min in solution in and AgBr , so both will be precipitated][ Ag AgCl

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