of iit-jee - 2011momentumacademy.net/new_website/downloads/iitjee2011-2.pdf · iit-jee - 2011 of...

MOMENTUM Also at

GWALIOR : (0751) 3205976, / REWA : (07662) 421444 / BALAGHAT : (07632) 324766 / AMRAVATI :(0721) 2572602

www.momentumacademy.com

1

Our Top class IITian faculty team promises to give you an authentic solution which will be fastest in the whole country.

TIME : 3 hrs. Max. Marks : 240

JABALPUR : 1525, Wright town, Near Stadium & Vechile Turn , Ranjhi Ph :(0761) 4005358, 2400022

NAGPUR : 5, E.H.C.Road,New Ramdaspeth & 24 Pragati Colony,Opp.Sai Mandir,Wardha Rd.Ph :(0712) 3221105

FEEL THE POWER OF OUR KNOWLEDGE & EXPERIENCE

SOLUTIONS

IIT-JEE - 2011

OF

INSTRUCTIONS

1. The question paper consists of 3 parts (Chemistry, Physics and Mathematics). Each

part consists of four sections.

2. In Section I (Total Marks : 24), for each question you will be awarded 3 marks if you

darken ONLY the bubble corresponding to the correct answer and zero marks if no

bubble is darkened. In all other cases, minus one (–1) mark will be awarded.

3. In Section II (Total Marks : 16), for each question you will be awarded 4 marks if you

darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero

marks otherwise. There are no negative marks in this section.

4. In Section III (Total Marks : 24), for each question you will be awarded 4 marks if you

darken ONLY the bubble corresponding to the correct answer and zero marks other-

wise. There are no no negative mark in this section.

5. In Section IV (Total Marks : 16), for each question you will be awarded 2 marks for

each row in which you have darken ALL the bubble(s) corresponding to the correct

answer(s) ONLY And zero marks otherwise. Thus, each question in this section car-

ries a maximum of 8 marks. There are no negative marks in this section.

For IIT-JEE / AIEEE / MEDICAL ENTRANCE EXAMS

R

PAPER - II

[CHEMISTRY/PHYSICS/ MATHEMATICS]

MOMENTUM Also at



2

For IIT-JEE / AIEEE / MEDICAL ENTRANCE EXAMS

SOLUTIONS OF IIT-JEE 2011

FEEL THE POWER OF OUR KNOWLEDGE & EXPERIENCE

Our Top class IITian faculty team promises to give you an authentic solution which will be fastest in the whole country.

PAPER - II

[ CHEMISTRY]

R

JABALPUR : 1525, Wright town, Near Stadium & Vechile Turn , Ranjhi Ph :(0761) 4005358, 2400022

NAGPUR : 5, E.H.C.Road,New Ramdaspeth & 24 Pragati Colony,Opp.Sai Mandir,Wardha Rd.Ph :(0712) 3221105

Questions are given in short here and answer corresponding to each question is mentioned. So Answers for each paper

set is covered. Detailed solutions are available at our center and website www. momentumacademy.com

SECTION - ISingle Correct Choice Type

1. The following carbohydrate is

HO

HO

H

HHO

H

OH

OHH

H

(a) a ketohexose (b) an aldohexose (c) an α − furanose (d) an α −pyranose

1.(b) an aldohexose

H C OH− −

HO C−

HO C H− −

H C OH− −

H C OH− −

HO C H− −

2HO C CH OH− − −

H C OH− −

2HO H C C− −

H H

C H−

O

O

[An aldohexose]

H

or HOH

2. Oxidation states of the metal in the minerals haematite and magnetite, respectively are

(a) II, III in haematite and III in magnetite (b) II, III in haematite and II in magnetite(c) II in haematite and II, III in magnetite (d) III in haematite and II, III in magnetite

2(d) Haematite → 2 3Fe O in +2

Magnetite → 3 4Fe O in +2 & +3

3. Among the following complex ( )K - P , 3 6 3 6 3 3 3[ ( ) ]( ),[ ( ) ] ( ), [ ( ) ]K Fe CN Co NH Cl Na Co oxalateK L

2 6 2[ ( ) ] ( )Ni H O Cl N , 2 4[ ( ) ]( )K Pt CN O and 2 6 3 2[ ( ) ]( ) ( )Zn H O NO P the diamagnetic complexes are

(a) K,L,M,N (b) K,M,O,P (c) L,M,O,P (d) L,M,N,O

3.(c) L,M,O,P

Code (8)

MOMENTUM Also at



3

4. Passing 2H S gas into a mixture of 2 2 2, ,Mn Ni Cu

+ + + and 2Hg

+ ions in an acidified aqueous solution pre-

cipitates :

(a) CuS and HgS (b) MnS and CuS (c) MnS and NiS (d) NiS and HgS

4.(a) ,CuS HgS

5. Consider the following cell reaction

2

( ) 2( ) ( ) ( .) 22 4 2 2 ( )s g aq aqFe O H Fe H O I+ ++ + → + 1.67E V° =

At 2 3

2[ ] 10 , ( ) 0.1Fe M P O+ −= = atm and 3pH = , the cell potential at 25°C is :

(a) 1.47 V (b) 1.77 V (c) 1.87 V (d) 1.57 V

5.(d)

2 20

4

2

0.0591 [ ]log

[ ]

FeE E

n H PO

+

+= −

⋅

3 2

3 4

.0591 [10 ]1.67 log

4 [10 ] 0.1

−

−= −

×

1.57E V=

6. The freezing point (in C°) of a solution containing 0.1 g of 3 6[ ( ) ]K Fe CN (Mol.w.t. 329) in 100g of water

(1

1.86fK K kg mol−= ) is

(a) 22.3 10−− × (b) 25.7 10−− × (c) 35.7 10−− × (d) 21.2 10−− ×

6.(a) f fT K mi∆ = ×

[ 1 ( 1)i x y= + −

y - for 3 6[ ) ] 4K FeCN = ]

0.11.86 1000 4 ( 4)

329 100fT i∆ = × × × =

×

22.3 10T −= − ×

7. Amongst the compounds given, the one that would form a brilliant colored dye on treatment with 2NaNO

dil HCl followed by addition to an alkaline solution of β − naphthol is

(a)

3 2( )N CH

(b)

3NHCH

(c)

2NH

3H C

(d)

2 2CH NH

MOMENTUM Also at



4

7.(c)

2NH

3H C

8. The major product of the following reaction is

2RCH OH

H+(anhydrous)

(a) a hemiacetal (b) an acetal (c) an ether (d) an ester

8.(b)

5H ⊕+

⊕ ⊕

H H

H

2O CH R− −

H

NuΘ2O CH R− −

2O CH R− −

An acetal

SECTION - IIMore than one correct

9. The correct functional group X and the reagent/reactioin conditions Y in the following scheme are

2 4( )X CH X− − ( )i Y

( )ii

HO2 4

( )C CH C- -O O

OHheat

condensation polymer..

(a) 3 2, / /X COOCH Y H Ni heat= = (b) 2 2, / /X CONH Y H Ni heat= =

(c) 2 2, /X CONH Y Br NaOH= = (d) 2, / /X CN Y H Ni heat= =

9.(b,d)2 4( )

YX CH X− →

2 | |

2 2 4 2( )H Ni Heat

YH N C CH C NH− − − →

O O

Tetra methylene diamide

2 2 2 4 2 2( ) 6Adipic

acidH N CH CH CH NH Nylon− − − → −

Hexa methylene tetramine

2X CONH=

MOMENTUM Also at



5

10. For the first order reaction2 5 2 22 ( ) 4 ( ) ( )N O g NO g O g→ +

(a) the concentration of the reactant decreases exponentially with time

(b) the half-life of the reaction decreases with increasing temperature(c) the half-life of the reaction depends on the initial concentration of the reactant

(d) the reaction proceeds to 99.6% completion in eight half-life duration

10.(a,b,d)11. Reduction of the metal centre in aqueous permanganate ion involves :

(a) 3 electrons in neutral medium (b) 5 electrons in neutral medium

(c) 3 electrons in alkaline medium (d) 5 electrons in acidic medium

11.(a)(c)(d)

Basic 4 2 22 2 2 3[0]

OHKMnO H O MnO KOH

−

+ → + +

number of e− transfered = 3

Neutral

4 2 22 3 40MnO H O e MnO H− − −+ + → +


Acidic

2

4 28 5 4MnO H e Mn H O− + − ++ + → +


12. The equilibrium 02 I IICu Cu Cu+ in aqueous medium at 25°C shifts towards the left in the presence

of :

(a) 3NO−

(b) Cl − (c) SCN − (d) CN −

12.(c,d) 2 2

3 2 2 42 2 2Cu SO SCN H O CuSCN H SO+ − −+ + + → +

2

22 4 2 ( )Cu CN Cu CN CN+ − −+ → + +

SECTION - III

Integer Answer type

13. The volume (in mL) of 0.1 M 3AgNO required for complete precipitatioin of chloride ions present in 30 mL of

0.01M solution of 2 5 2[ ( ) ]Cr H O Cl Cl , as silver chloride is close to

13.(6)2 5 2 3

[ ( ) ] 2 2Cr H O Cl Cl AgNO AgCl+ → ↓

30 0.01×0.3 0.6milimole

0.1 0.6V × =6V = ml

MOMENTUM Also at



6

14. The number of hexagonal faces that are present in a truncated octahedron is

14.(8) On truncated octahedral no. of hexagonal faces.

15. The total number of contributing structres showing hyperconjugation (involving C-H bonds) for the following

carbocation is :

3H C

2 3CH CH

15.(6)

3 2 3H C C CH CH- - - α α

α

6 H.C. structures ∵ 6 Hα16. The maximum number of isomers (including) stereoisomers) that are possible on mono-chlorination of the

following compound is

C

2 3CH CH

3CH

H3 2CH CH

16.(8)3 2 2 3CH CH CH CH CH− − − −

1P

4P

2S 3t 2S 1P

3CH

/Cl Cl hv−Total 4-mono-Chlorinated products

2 2 2 3 3 2 3CH CH CH CH CH CH CH CH CH CH− − − − + − − − −

Cl Cl

3CH 3CH

* *

*

(Two enantiomers)

(4 enantiomers)

MOMENTUM Also at



7

3 2 2 3 3 2 2 3CH CH C CH CH CH CH CH CH CH− − − − + − − − −

Cl

3CH 2CH Cl−

17. Among the following, the number of compounds than can react with 5PCl to give 3POCl is

2 2 2 2 2 4 4 10, , , , ,O CO SO H O H SO PO

17.(4)4 10 5 36 10PO PCl POCl+ →

2 4 5 2 2 32 2 2H SO PCl SO Cl POCl HCl+ → + +

5 2 3 2PCl H O POCl HCl+ → +

5 2 2 3PCl SO SOCl POCl+ → +

18. In 1 L saturated solution of 10

[ ( ) 1.6 10 ]spAgCl K AgCl−= × , 0.1 mol of CuCl

6[ ( ) 1.0 10 ]spK CuCl

−= × is

added. The resultant concentration of Ag + in the solution is 1.6 10 x−× . The value of ‘’X’’ is

18.(7) AgCl Ag Cl+ −→ +

x x y+

CuCl Cu Cl+ −→ + y x y+

( )x x y+ = 101.6 10−× ........(i)

( )y x y+ = 61 10−× ........(ii)

41.6 10x

y

−= ×

4101.6

xy = ×

Put y in (i) eqn.

4 10( 10 ) 1.6 101.6

xx x −× + × = ×

102

4

1.6 1.6 10

10x

−× ×=

71.6 10x −= ×SECTION - IV

Matrix - Match Type

19. Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in thesereactions as given in Column II

Column I Column II

MOMENTUM Also at



8

(A)

OO3H C

.aq NaOH

O

(p) Necleophillic substitution

(B)

O

2 2 2CH CH CH Cl3

CH MgI

3CH(q) Electrophilic substitution

(C)

O

2 2 2CH CH CH OH 2 4H SO18

18

(r) Dehydration

(D)

2 2 2 3 2( )CH CH CH C CH 2 4H SO

OH

3H C 3CH

(s) Nucleophilic addition

(t) Carbanion

19.(A)

OO2H C

.aq NaOH

O

Aldol condensation

O

Strong base

OHO

H

2H O-

O

∴ (r,s,t)

MOMENTUM Also at



9

(B)

O

2 2 2CH CH CH Cl- - -3CH MgCl-

Nu addition

C

OMgX

3CH

Cl

C

3CH

Nu Substitution

∴ (p,s,t)

(C)

O

2 2 2CH CH CH OH- - -2 4H SO18

C

O

2 2 2CH CH CH OH- - -

C

OH

2 2 2CH CH CH OH- - -

18Nu

C

OH

2 2 2CH CH CH OH- - -

α

β18

Nu

Addition

MOMENTUM Also at



10

18

2H O-

∴ (r,s)

(d)

2 2 2CH CH CH C- - -

OH

3CH

3CH

2 4H SO2H O-

2 2 2CH CH CH C- - -3CH

3CHEASR

3CH3H C

∴ (q,r)

20. Match the transformations in Column I with appropriate options in Column II

Column I Column II

(A) 2 2( ) ( )CO s CO g→ (p) phase transition

(B) 3 2( ) ( ) ( )CaCO S CaO S CO g→ + (q) allotropic change

(C) 22 ( ).H H g→ (r) H∆ is positive

(D) ( , ) ( , )white solid red solidP P→ (s) S∆ is positive

(t) S∆ is negative

20. A → prs , B→ rs , C → t , D → tq

[ PHYSICS ]

Single Choice Questions :

21. A satellite is moving with a constant speed ' 'V in a circular orbit about the earth. An object of mass ' 'm is

ejected the satellite such that it just escapes from gravitational pull of the earth. At the time of its ejection,

the kinetic energy of the object is

(a) 21

2mV (b) 2mV (c)

23

2mV (d) 22mV

MOMENTUM Also at



11

Sol.(b) ( )2 212

2m V mV=

22. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, ( )1 sinx t A tω=

and ( )2

2sin

3x t A t

πω = +

. Adding a third sinusoidal displacement ( ) ( )3 sinx t B tω= +∅ brings the

mass to a complete rest. The values of B and ∅ ARE.

(a) 3

2 ,4

Aπ

(b) 4

,3

Aπ

(c) 5

3 ,6

Aπ

(d) ,3

Aπ

Sol.(b)4

;3

πφ = B A=

23. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a

screw gauge, Whose pitch is 0.5 mm and there are 50 divisions on the circle scale. The reading on the mainscale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative

error of 2%, the relative percentage error in the density is

(a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2%

Sol.(c)0.5

0.0150

LC mm= =

2.5 20 0.01 2.70d mm= + × =

34

3 2

m

dρ

π

=

100 100 3 100m d

m d

ρρ∆ ∆ ∆

× = × + ×

0.01

2 3 100 3.1%2.70

= + × × =

24. A long insulated copper wire is closely would as a spiral of ‘N’ turn. The spiral has inner radius ' 'a and outer

radius ' 'b . The spiral lies in the X-Y plane and a steady current ' 'I flows through the wire. The Z-component

of the magnetic field at the center of the spiral

MOMENTUM Also at



12

Y

X

I a

b

(a) ( )2

NI bIn

b a a

µ −

0

(b) ( )2

N I b aIn

b a b a

µ + − −

0

(c) 2

N I bIn

b a

µ

0

(d) 2

N I b aIn

b b a

µ + −

0

Sol.(a)( )

0

0 ln2 2

b

a

Ndx I

NI bb aB

x b a a

µµ

− = = −∫

25. A light ray treveling in glass medium is incident on glass-air interface at an angle of incidence θ . The

reflected (R) and transmitted (T) intensities, both as function of θ , are plotted. The correct sketch is

(a)

100%

Intensity

T

R

0 900θ

(b)

100%

Intensity

T

R

0 900θ

MOMENTUM Also at



13

(c)

100%Intensity

T

R

0 900θ

(d)

100%

Intensity

T

R

0 900θ

Sol.(c)

26. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity

V m/s in a horizontal direction, hits the centre of the ball, After the collision, the ball and bullet travelindependently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the

foot of the post. The initial velocity V of the bullet is

0 20 100

V m/s

(a) 250 /m s (b) 250 2 /m s (c) 400 /m s (d) 500 /m s

Sol.(d)20 1

100 5

ball

bullet

V

V= =

ballV u=

5bulletV u=

1T s=

20 /u m s∴ =

5 100 /u m s=

0.01 0.2 20 100 0.01V∴ = × + ×

0.01 4 1 5V = + =

500 /V m s=27. Which of the field pattern given below is valid for elcetric field as well as for magenetic field ?

MOMENTUM Also at



14

(a) (b)

(c) (d)

Sol.(c)

28. A wooden block performs SHM on a frictionless surface with frequency, 0v . The block carries a chagre Q+

on its surface. If now a uniform electric field E

is switched-on as shown, then the SHM of the block will be

E

+Q

(a) of the same frequency and with shifted mean position.

(b) of the same frequency and with the same mean position.

(c) of changed frequency and with shifted mean position.

(d) of changed frequency and with the same mean position.

Sol.(a)

29. A series R-C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a

dielectric medium and (B) when C is filled with dielectric of constant 4. The current RI through the resistor

MOMENTUM Also at



15

and voltage CV across the capacitor are compared in the two cases. Which of the following is/are true?

(a) A B

R RI I> (b)

A B

R RI I< (c)

A B

C CV V> (d)

A B

C CV V<

Sol.(b,d)

2

2 2

1A

VI

RCω

=

+

2

2 2

1

16

B

VI

RCω

=

+

B AI I> (b)

A B

R RV V>

A B

C CV V< (d)

30. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/

s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the 10 m/s. Immediately after the collision.

0.75m

20m/s

10m/s

1m/s

(a) the ring has pure rotation about its stationary CM.(b) the ring comes to a complete stop.

(c) friction between the ring and the ground is to the left.

(d) these is no friction between the ring and the ground.

Sol.(a,c)

0System

iP =

0Final

xP∴ =

∴ Ring comes to rest.

31. Which of the following statement(s) is/are correct?

(a) If the electric field due to a point charge varies as 2.5r− instead of 2r− , then the gauss law will still be

valid.

(b) The gauss law can be used to calculate the field distribution around an electric dipole.

(c) If the electric field between two point charges is zero somewhere, then the sign of the two chargesis the same.

MOMENTUM Also at



16

(d) The work done by the external force in moving a unit positive charge from point A at potential AV to point

B at potential BV is ( )B AV V− .

Sol.(c,d)

32. Two solid spheres A and B of equal volume but of different densities Ad and Bd are connected by a string.

They are fully immersed in a fluid of density Fd . They get arranged into an equilibrium state as shown in the

figure with a tension in the string. The arrangement is possible only if

A

B

(a) A Fd d< (b) B Fd d> (c) A Fd d> (d) 2A B Fd d d+ =Sol.(a,b,d)

F Ad d>

Vd gB

Vd gF

VdB

TT

Vd gA

VdA

Vd gF

2 F A BVd g Vd g Vd g= +

2 F A Bd d d= +

B Fd d>

33. Water (with refractive index =4

3) in a tank is 18 cm deep. Oil of refractive index

7

4lies on water making a

convex surface of radius of curvature ' 6 'R cm= as shown. Consider oil to act as a thin lens. An object ' 'S

is placed 24 cm above water surface. The location of its image is at ' 'X cm above the bottom of the tank.

Then ' 'X is

MOMENTUM Also at



17

S

µ = 1.0

µ = 7/4

µ = 4/3

R = 6 cm

Sol.(2) S

cm18

cm24

Ist Refraction :

1

71

7 1 4

4 24 6V

−− =−

1

7 1 3

4 24 24V+ =

1

7 2 1

4 24 12V= =

1 21V cm⇒ =

IInd Refraction :

4

4 43721 3 7

4

h= = ×

16h cm=

1 7 11 0

4 6Lf

= − −

3

24= 8Lf cm∴ =

∴ Ans 2cm

MOMENTUM Also at



18

34. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from and insulating thread in free in

free-spce. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the

sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the

sphere is 10ZA× (where 1 < A < 10). The value of ‘Z’ is

Sol.(7)1242

6.21200

hceV eV

λ= =

max 1.5K eV∴ =

maxeV K=

1.5V V=

1.5q

KR=

2

9

1.5 10

9 10q

−×=

×

2

9 19

1.5 10

9 10 1.6 10n

−

−

×=

× × ×

8 8 71.5

10 0.10 10 1 109 1.6

= × = × = ××

35. A train is moving along a straight line with a constant acceleration ' 'a . A boy standing in the train through a

ball forward with a speed of 10 m/s, at angle of 600 to the horizontal. The boy has to move forward by 1.15 m

inside the train to catch the ball back at initial height. The acceleration of the train, in 2/m s , is

Sol.(5)

60o

a

10

1.15 = Range relative to train

32 102 2 3

10

yuT

g

× ×= = =

Relative acceleration in horizontal direction = ( )a ←

211.15

2xu T aT∴ − =

1 110 3 3 1.15

2 2a× × − × =

5 3 1.5 1.15a− =

5.0a =

MOMENTUM Also at



19

36. Two batteries of different emfs and diffrent internal resistances are connected as shown. The voltage across

AB in volts is 6 V

3 V 2 Ω

1 Ω

A B

Sol.(5)3

13

i A= =

6 1A BV V+ − =

5B AV V− =37. A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between

the block and the floor is 0.1. Intially the block is at rest and the spring is un-stretched. An impulse is given

to the block in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The intial

velocity of the block in m/s is V=N/10. Then N is

Sol.(4)21

2 (0.06) 0.1 0.18 10 0.062× × + × × ×

210.18

2V= ×

4 4 236 10 108 10 0.09V− −× + × =2 2(4 12) 10V −= + ×

1 44 10

10V −= × =

38. A series R-C combination is connected to an AC voltage of angular frequency 500ω = radian/s. If the

impedance of the R-C circuit is 1.25R , the time constant (in millisecond) of the circuit is

Sol.(4)2 2 2

CZ R X= +

2 2 21.25C

R R X= +

2 20.25CX R=

0.5CX R=

MOMENTUM Also at



20

10.5R

Cω=

1 1

0.5 500 0.5RCτ

ω= = =

× ×

1

4250

s ms= =

(MATRIX-MATCH TYPE)This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column -

I and five statements (p, q, r, s and t) in Column - II. Any given statement in Column - I can have correct

matching with ONE or MORE statement(s) given in Column - II. For example, if for a given question, statement

B matches with the statements given in q and r, then for the particular question, against statement B, darken

the bubbles corresponding to q and r in the ORS.

39. Column - I shows four systems, each of the same length L , for producing standing waves. The lowest

possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as

fλ . Match each system with statements given in Column -II describing the nature and wavelength of the

standing waves.

Column - I Column - II

(A) Pipe closed at one end (p) Longitudinal waves

0 L

(B) Pipe open at both ends (q) Transverse waves

0 L

(C) Stretched wire clamped at both ends (r) f Lλ =

0 L

(D) Stretched wire clamped at both ends and at mid-point (s) 2f Lλ = 4f Lλ =

0 LL/2

(t) 4f Lλ =

Sol. (A) p, t (B) p, s (C) q, s (D) q, r

40. One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P V− diagram.

Column - II gives the characteristics involved in the cycle. Match them with each of the processes given inColumn - I.

MOMENTUM Also at



21

3P

1P

1V 3V 9V V0

C

B A

D

P

Column - I Column - II

(A) Process A B→ (p) Internal energy decreases

(B) Process B C→ (q) Internal energy increases

(C) Process C D→ (r) Heat is lost

(D) Process D A→ (s) Heat is gained

(t) Work is done on the gas

Sol. (A) p, r, t

Isobaric compression

temperature decreases ( )P∴

0U∆ < ,

0W <

0Q U W∆ = ∆ + <(B) p, r

Isobaric Process

temperature decreases ( )P∴

0U∆ < ,

0W =

0Q∆ <(C) q, s

Isobaric Expansion

temperature Increases

0U∆ > ( )q

0W > ,

0Q∆ >(D) r, t

Isothermal Process

0U∆ = , 0W <

0Q∆ <

MOMENTUM Also at



22

[ MATHEMATICS ]

SECTION - I

Single Correct Choice Type

41. Let ( )6,3P be a point on the hyperbola

2 2

2 21

x y

a b− = . If the normal at the point P intersects the x - axis at

( )9,0 , then eccentricity of the hyperbola is

(a) 5

2(b)

3

2(c) 2 (d) 3

41.(b) Given ( sec , tan ) (6,3)P a b Pθ θ = . Now equation of normal is 2 2

sec tan

ax bya b

θ θ+ = +

i.e.

2 22 2

6 3

a x b ya b+ = + . Since it passes through (9, 0) therefore

22 2 2 29

26

aa b a b

×= + ⇒ = . So,

1 31

2 2e = + =

42. Let ( ),x y be any point on the parabola 24y x= . Let P be the point that divides the line segment from

( )0,0 to ( ),x y in the ratio 1:3 . Then the locus of P is

(a) 2x y= (b) 2

2y x= (c) 2y x= (d) 2

2x y=

42.(c) Let Q be

2

,4

yy

(0,0)

2

,4

yy

Q

For ,P

2

2

1 3 04' 16 '1 3

y

x x y

× + ×= ⇒ =

+

1 3 0' 4 '

1 3

yy y y

× + ×= ⇒ =

+

Eliminating y , we get 2 216 ' (4 ') ( ') 'x y y x= ⇒ = then locus is 2y x=

43. Let ( ) 2f x x= and ( ) sing x x= for all x R∈ . Then the set of all x satisfying

( )( ) ( )( )f g g f x g g f x= , where ( )( ) ( )( )f g x f g x= , is

(a) , 0,1,2......n nπ± ∈ (b) , 1,2,.....n nπ± ∈

(c) 2 , ..., 2, 1,0,1,2,....2

n nπ

π+ ∈ − − (d) 2 , ..., 2, 1,0,1,2,....n nπ ∈ − −

43.(a) ( ) 2f x x= , ( ) sing x x= then ( )( ) ( )( )f g g f x g g f x= ( )( )( )( ) ( )( )( )f g g f x g g f x⇒ =

( )( )( ) ( )( )2 2f g g x g g x⇒ = ( )( ) ( )2 2sin sinf g x g x⇒ = ( )( ) ( )2

2 2sin sin sin sinx x⇒ =

( )2sin sin 0,1x⇒ = x nπ⇒ = ±

MOMENTUM Also at



23

44. Let [ ] [ ): 1,2 0,f − → ∞ be a continuous function such that ( ) ( )1f x f x= − for all [ ]1,2x∈ − . Let

( )2

1

1

R x f x dx−

= ∫ and 2R be the area of the region bounded by ( ) , 1, 2y f x x x= = − = and the x -axis.

Then

(a) 1 22R R= (b)

1 23R R= (c) 1 22R R= (d)

1 23R R=

44.(c) ( )2

1

1

R x f x dx−

= ∫ ...(i)

( ) ( )2

1

1 1x f x dx−

= − −∫ .....(ii) ( ) ( )

a a

b b

f x dx f a b x dx

= + − ∫ ∫∵

By (i) +(ii) we get ( )2

1

1

2R f x dx−

= ∫ . Here ( )2

2

1

R f x dx−

= ∫ 1 22R R⇒ =

45. If ( )1

2 2

0lim 1 ln 1 2 sin , 0x

xx b b bθ

→ + + = > and ( ],θ π π∈ − , then the value of θ is

(a) 4

π± (b)

3

π± (c)

6

π± (d)

2

π±

45.(d) ( )1/

2 2

0lim 1 ln 1 2 sin

x

xx b b θ

→ + + = ( )2

0

1lim 1 ln 1 1

22 sinxx b

xe b θ→ + + − ⇒ = ( )2ln 1 22 sin

be b θ+⇒ =

21 2 sinb b θ⇒ + = 2 2 sin 1 0b b θ⇒ − + = b R∈∵

So, 24sin 4 0θ − ≥ sin 1θ⇒ ≤ − or 1≥ 2

πθ⇒ = ±

46. The circle passing through the point ( )1,0− and touching the y -axis at ( )0,2 also passes through the point

(a) 3,0

2

−

(b) 5,2

2

−

(c) 3 5,

2 2

−

(d) ( )4,0−

46.(d) Let circle be 2 2 2 2 0x y gx fy c+ + + + =

(ο,2)Here 2f− = as it touches y axis− at ( )0,2 . Also 2c f= 4c∴ =

∴ circle 2 22 4 4 0x y gx y+ + − + =

Since it passes through ( )1,0− i.e. 1 0 2 4 0 4 0g+ − − × + = 5

2g⇒ =

∴ circle is 2 2 5 4 4 0x y x y+ + − + = . For 0y = , 2 5 4 0x x+ + = 1, 4x⇒ = − −

MOMENTUM Also at



24

47. Let 1ω ≠ be a cube root of unity and S be the set of all non-singular matrices of the form

2

1

1

1

a b

cωω ω

,

where each of ,a b and c is either ω or 2ω . Then the number of distinct matrices in the set S is

(a) 2 (b) 6 (c) 4 (d) 8

47.(a)2

1

1

1

a b

cωω ω

21 ( ) 0c a c bω ω ω= − + − + × 2

( ) 1ac a cω ω= + − − + ( 1)( 1)a cω ω= − −

For 0∆ ≠ we get ( , ) ( , )a c ω ω= and for all cases b can be ω or 2ω

48. A value of b for which the equations 2 1 0x bx+ − = , 20x x b+ + = have one root in common is

(a) 2− (b) 3i− (c) 5i (d) 2

48. (b) 2 1 0x bx+ − = , 2 0x x b+ + = have one root common ⇒ 2

1 2 2 1 1 2 2 1 1 2 2 1( ) ( )( )c a c a b c b c a b a b− = − −

∴ 2 2( 1 ) ( 1)(1 )b b b− − = + − ⇒ 2 2 32 1 1b b b b b+ + = − + − ⇒ 3 3 0b b+ = ⇒ 0, 3 , 3b i i= −

SECTION - II

Multiple Correct Answer(s) Type

49. If ( )

,2 2

cos , 02

1, 0 1

ln , 1,

x x

x xf x

x x

x x

π π

π

− − ≤ −− − < ≤=

− < ≤ >

then

(a) ( )f x is continuous at 2

xπ

= − (b) ( )f x is not differentiable at 0x =

(c) ( )f x is differentiable at 1x = (d) ( )f x is differentiable at 3

2x = −

49.(a,b,c,d) ( )

,2 2

cos , 02

1, 0 1

ln , 1,

x x

x xf x

x x

x x

π π

π

− − ≤ − − − < ≤=

− < ≤ >

-1

1

1−π/2

O

( ) ( ) ( )0

1 1' 1 lim

h

f h ff

h

−

→

− −=

−

( )0

1 1 0lim 1h

h

h→

− − −= =

−

( ) ( ) ( )0

1 1' 1 lim

h

f h ff

h

+

→

+ −=

( )0

ln 1 0lim 1h

h

h→

+ −= =

MOMENTUM Also at



25

50. Let L be a normal to the parabola 24y x= . If L passes through the point ( )9,6 , then L is given by

(a) 3 0y x− + = (b) 3 33 0y x+ − = (c) 15 0y x+ − = (d) 2 12 0y x− + =

50.(abd) 2 4 1y x a= ⇒ =

Normal at 2( ,2 )at at is 32y tx at at+ = + ⇒ 32y tx t t+ = +

Since it passes through (9,6) therefore 36 9 2t t t+ = + 3 7 6 0t t⇒ − − =1t = − satisfies

therefore

3 7 60

1

t t

t

− −=

+

2 ( 1) ( 1) 6( 1)0

1

t t t t t

t

+ − + − +⇒ =

+ 2 6 0t t⇒ − − = 3, 2t⇒ = −

∴ Normals are 3y x− = − , 2 12y x− = − , 3 33y x+ =

51. Let E and F be two independent events. The probability that exactly one of them occurs is 11

25 and the

probability of none of them occurring is 2

25. If ( )P T denotes the probability of occurrence of the event

T , then

(a) ( ) ( )4 3,

5 5P E P F= = (b) ( ) ( )1 2

,5 5

P E P F= =

(c) ( ) ( )2 1,

5 5P E P F= = (d) ( ) ( )3 4

,5 5

P E P F= =

51.(a,d) ( ) ( ) 11

25P E F P E F∩ + ∩ =

( ) ( ) ( ) ( ) ( ) ( ) 11

25P E P E P F P F P E P F− + − =

Given ( ) 2

25P E F∩ = ...(i) ( ) 2

125

P E F⇒ − ∪ = ( ) 23

25P E F⇒ ∪ =

( ) ( ) ( ) ( ) 23

25P E P F P E P F⇒ + − = .......(ii)

From (i) & (ii) ( ) ( ) 12

25P E P F =

52. Let ( ): 0,1f R→ be defined by

( )1

b xf x

bx

−=

−, where b is a constant such that 0 1b< < . Then

(a) f is not invertible on ( )0,1 (b) 1f f

−≠ on ( )0,1 and ( )( )1

0f b

f′ =

′

(c) 1f f −− on ( )0,1 and ( )( )1

0f b

f′ =

′ (d) 1f − is differentiable on ( )0,1

MOMENTUM Also at



26

52. (a,b) ( )1

b xf x

bx

−=

−

( ) ( )( ) ( )( )( )2

1 1

1

bx b x bf x

bx

− − − − −′ =

− ( )

2

2

1

1

bx b bx

bx

− + + −=

− ( )

2

2

1

1

b

bx

−=

−

'( ) 0 (0,1)f x x< ∀ ∈

Function is decreasing in (0,1) hence function is into and 1f − cannot be found

SECTION - III

Integer Answer Type

53. Let ( ) ( ) ( ) ( ) ( )y x y x g x g x g x′ ′ ′+ = , ( )0 0y = , x R∈ , where ( )'f x denotes ( )df x

dx and ( )g x is a

given non-constant differentiable function on R with ( ) ( )0 2 0g g= = . Then the value of ( )2y is

53. (Ans : 0) '( ) ( ) '( ) ( ) '( )y x y x g x g x g x+ = . Let ( )y x u= & ( )g x v=

then du dv dv

u vdx dx dx

+ = du udv v dv⇒ + = du

u vdv

⇒ + =

solution ( )v vu e v e dv c⇒ = +∫ v v vue ve e c⇒ = − + ( ) ( ) ( )( ) ( )g x g x g xy x e g x e e c⇒ = − +

put 0x = 0 0 1 1c c⇒ = − + ⇒ =

put 2x = ⇒ 0 0(2) 0 1y e e= − + ⇒ (2) 0y =

54. Let ˆˆ ,a i k= − − ˆ ˆ,b i j= − +

and ˆˆ ˆ2 3c i j k= + +

be three given vectors. If r

is a vector such that

r b c b× = × and 0r a⋅ =

, then the value of r b⋅

is

54. (Ans : 9) r b c b× = ×

( ) 0r c b− × =

i.e. ( ) ||r c b−

r c bλ⇒ = +

. 0r a=

( ) ( )( ) ( )2 3 . 0i j k i j i kλ⇒ + + + − + − − = 1 3 0 4λ λ⇒− + − = ⇒ =

3 6 3r i j k∴ = − + +

and ( ) ( ). 3 6 3 .r b i j k i j= − + + − +

3 6 9= + =

55. Let 3ie πω = , and , , , , ,a b c x y z be non-zero complex numbers such that

a b c x+ + =

2a b c yω ω+ + =

2a b c zω ω+ + =

Then the value of

2 2 2

2 2 2

x y z

a b c

+ +

+ + is

55. (Ans : Question is wrong)

56. Let M be a 3 3× matrix satisfying

MOMENTUM Also at



27

0 1 1 1

1 2 , 1 1

0 3 0 1

M M

− = − = −

and

1 0

1 0

1 12

M

=

Then the sum of the diagonal entries of M is

56. (Ans : 9)

11 12 13

21 22 23

31 32 33

0 1

1 2

0 3

a a a

a a a

a a a

− =

12 1a = − , 22 2a = ,

32 3a =

Also, 11 12 1a a− = ,

21 22 1a a− = ,31 32 1a a− = − ⇒

11 0a = , 21 3a = ,

31 2a =

again 11 12 13 0a a a+ + = ,

21 22 23 0a a a+ + = , 31 32 33 12a a a+ + = then we get

33 7a =

So sum 11 22 33a a a= + + 0 2 7 9= + + =

57. The number of distinct real roots of 4 3 24 12 1 0x x x x− + + − = is

57. (Ans : 2) Let ( ) 4 3 24 12 1f x x x x x= − + + − ⇒ ( ) 3 24 12 24 1f x x x x′ = − + +

and ( ) 212 24 24f x x x′′ = − + ( )212 2 2x x= − + 0 x R> ∀ ∈ ( ) is increasingf x x R′⇒ ∀ ∈

then ( ) ( ),f f′ ′−∞ = −∞ ∞ =∞

So, ( ) 0f x′ = for only one ' 'x So, ( )f x has only two real roots

58. The straight line 2 3 1x y− = divides the circular region 2 2 6x y+ ≤ into two parts. If

3 5 3 1 1 1 12, , , , , , ,4 2 4 4 4 8 4

S − =

,

then the number of point(s) in S lying inside the smaller part is

58. (Ans : 2) 2 3 1x y− = , 2 26x y+ ≤

If ( )', 'x y is inside the smaller region then 2 ' 3 ' 1 0x y− − > & 2 2' ' 6 0x y+ − ≤

If 3

'4

y = 8 ' 13x >5

' 18

x⇒ >

1(0, )

3- 1

( ,0)2

32,4

is inside but

5 3,

2 4

is outside the circle

If 1

'4

x = then 1

'6

y < − i.e. 1 1,

4 4

− is inside

If 1

'4

y = then 7

'8

x > 1 1,

8 4

∴ is outside ∴ 2 points are inside.

MOMENTUM Also at



28

SECTION - IV

Matrix - Match Type

59. Match the statements given in Column I with the values given in Column IIColumn I Column II

(A) If ˆ ˆˆ ˆ3 , 3a j k b j k= + = − +

and 2 3c k=

form (p) 6

π

a triangle, then the internal angle of the triangle between

a and b

is

(B) If ( )( ) 2 23b

af x x dx a b− = −∫ , then the value of

6f

π

is (q) 2

3

π

(C) The value of ( )5/62

7 / 6

secln3

x dxπ

π∫ is (r) 3

π

(D) The maximum value of 1

1Arg

z

−

for 1, 1z z= ≠ is given by (s) π

(t) 2

π

59. (A)→q; (B) →p; (C) →s; (D) → t

(A) 0a b c+ − =a

bc

( ) ˆˆcos a bπ θ− = ⋅ ( ) ( )3 3

2 2

j k j k+ ⋅ − +=

⋅

1

2 3

ππ θ= ⇒ − = ,

2

3

πθ =

(B) ( )( ) 2 23

b

a

f x x dx a b− = −∫ ( ) ( )2 2 2 23

2

b

a

f x dx b a a b⇒ − − = −∫ ( )2 2

2 2

b

a

b af x dx⇒ = −∫

( )2 2

2

b

a

b af x dx

−⇒ =∫ ( )f x x⇒ = / 6

6f

ππ ⇒ =

(C) ( )2

5/6

7 / 6sec

ln3x dx

ππ∫

5/62

7/6

1ln tan

ln3 2 4

xπ π ππ = +

2 5ln tan ln tan

ln3 3 6

π π π = −

1ln 3 ln

ln3 3

π = −

π=

(D) ( )1

1 cos siniθ θ− + 2

1

ˆ2sin / 2 2sin / 2cos / 2iθ θ θ=

− ⋅ ( )

1 1

2sin / 2 sin / 2 cos / 2iθ θ θ= ×

−

1

1cos sin

2sin / 2 2 2 2 2i

θ π θ πθ

− = − + −

1cos sin

2sin / 2 2 2 2 2i

π θ π θθ

= ⋅ − + −

1 1cot

2 2 2i

θ= + is a complex number moving on line x=1/2, and its max(|arg|) = / 2π in the limiting case.

MOMENTUM Also at



29

60. Match the statements given in Column I with the intervals/union of intervals given in Column II

Column I Column II

(A) The set 2

2Re :

1

izz

z

−

is a complex number, 1, 1z z= ≠ ± is (p) ( ) ( ), 1 1,−∞ − ∪ ∞

(B) The domain of the function ( ) ( )( )

2

1

2 1

8 3sin

1 3

x

xf x

−

−−

=

− is (q) ( ) ( ),0 0,−∞ ∪ ∞

(C) If ( )

1 tan 1

tan 1 tan

1 tan 1

f

θθ θ θ

θ= −

− −, then the set (r) [ )2,∞

( ) : 02

fπ

θ θ ≤ <

is

(D) If ( ) ( )3/ 2 3 10 , 0f x x x x= − ≥ , then ( )f x (s) ( ] [ ), 1 1,−∞ − ∪ ∞

is increasing in

(t) ( ] [ ),0 2,−∞ ∪ ∞

60. (A)→s; (B) → t; (C) → r; (D) → r

(A) 2

2Re , 1

1

i zzz

z

= −

Now, 2 2

2 2

1 1

2

i z iz

z z

−+

− − ( ) ( )( )

( )( )

2 2

22 2

1 1

1

i z z z z

z z zz

− − −=

− − +

( )( )( )( )2

1

2 2

i z z z z

z z zz

− +=

− + −

( )( )2

2

4

i z z

z z

−=

− + ( )24 1

4 1

y

yx

− −= =

− but 1 1y− ≤ ≤ So Re( )z is ( ] [ ), 1 1,−∞ − ∪ ∞

(B)2

8.31 1

9 3

x

x− ≤ ≤

−

So, 2 2

8.3 8.31 & 1

9 3 9 3

x

x x

x≥ − ≤

− −for 1x <

28.3 9 3x x≥ − + ⇒ 23 8.3 9 0x x− − ≤ ⇒ ( )( )3 9 3 1 0x x− + ≤ ⇒ 2x ≤

and ( ) 28 3 9 3x x≤ − ⇒ ( ) ( )2

3 8 3 9 0x x+ − ≤ ⇒ ( )( )3 9 3 1 0x x+ − ≤

So, 0x ≤Now for 1x >

MOMENTUM Also at



30

28.3 9 3x x≤ − + ⇒ ( )23 8.3 9 0xx − − ≥ ⇒ ( )( )3 9 3 1 0x x− + ≥ ⇒ 2x ≥

and ( ) 28 3 9 3

x x≥ − ⇒ ( )23 8(3 ) 9 0x x+ − ≥ ⇒ ( )( )3 9 3 1 0x x+ − ≥ ⇒ 0x ≥

So, 2x ≥

(C) ( ) 22.secf θ θ=

(D) ( ) ( )3/ 2 3 10f x x x= −

( ) ( ) ( )3/ 2 1/ 233 3 10

2f x x x x′ = + −

1/ 2 3 103

2

xx x

− = + ( )

1/ 23

5 102

xx= −

( ) 2f x is x↑∀ ≥

of iit-jee - 2011momentumacademy.net/new_website/downloads/iitjee2011-2.pdf · iit-jee - 2011 of...

Documents