momentumacademy.netmomentumacademy.net/new_website/downloads/iitjee2012paper1keys.pdf · SECTION I : Single Correct Answer Type This section contains 10 multiple choice questions

PPPPARARARART I :T I :T I :T I : PHY PHY PHY PHYSICSSICSSICSSICS

SECTION I : Single Correct Answer Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.

1. A thin uniform rod, pivoted at O , is rotating in the horizontal plane with z

O

vω

constant angular speed ω , as shown in the figure. At time 0t = , a small

insect starts from O and moves with constant speed v with respect to the

rod towards the other end. It reaches the end of the rod at t T= and stops.

The angular speed of the system remains ω

throughout. The magnitude of the torque (| |)τ

on the system about O , as

a function of time is best represented by which plot ?

(A) t

TO

τ (B)

tTO

τ(C)

tTO

τ (D)

tTO

τ

1.(B) ( )2

rodL I mx ω= +

22 2dL

m xv m v tdt

τ ω ω= = =

2. Three very large plates of same area are kept parallel and close to each other. They are considered asideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at

temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady

state condition is:

(A)

1

465

2T

(B)

1

497

4T

(C)

1

497

2T

(D) ( )1

497 T

2.(C) ( ) ( )4 44 4

0 03 2T T T T− = − 2T 3TT0

4 4

02 97T T⇒ =

JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,

HISAR - (01662) 296006, REWA - (07662) 421444 , BALAGHAT - 324766

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1/ 4

0

97

2T T

=

3. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface

charge density. The variation of the magnitude of the electric field | ( ) |E r

and the electric potential ( )V r

with the distance r from the centre, is best represented by which graph?

(A)

E ( )r

O R r

V r( )

(B)

E ( )r

O R r

V r( )

(C)

E ( )r

O R r

V r( )

(D)

E ( )r

O R r

V r( )

3.(D)

4. In the determination of Young’s modulus 2

4MLgY

ldπ

= by using Searle’s method, a wire of length 2L m=

and diameter 0.5d mm= is used. For a load 2.5M kg= , an extension 0.25l mm= in the length of

the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively..

They have the same pitch of 0.5mm . The number of divisions on their circular scale is 100 . The

contributions to the maximum probable error of the Y measurement :

(A) due to the errors in the measurements of d and l are the same

(B) due to the error in the measurement of d is twice that due to the error in the measurement of l

(C) due to the error in the measurement of l is twice that due to the error in the measurement of d

(D) due to the error in the measurement of d is four times that due to the error in the measurement of l

4.(A) 2

4MLgY

ldπ=

2Y M L l d

Y M L l d

∆ ∆ ∆ ∆ ∆= + + +

0.5d mm= , 0.25l mm=

L.C for d = L.C. for l

l d∴∆ = ∆0.25

l l

l

∆ ∆∴ =

22

0.5 0.25

d l l

d

∆ ⋅ ∆ ∆⋅ = =




5. A small block is connected to one end of a massless spring of un-stretched length 4.9m . The other end of

the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched

by 0.2m and released from rest at 0t = . It then executes simple harmonic motion with angular frequency

rad/sec3

πω = . Simultaneously at 0t = , a small pebble is projected with speed v from point P at an

angle of 45o as shown in the figure. Point P is at a horizontal distance of 10m from O . If the pebble hitss

the block at 1t s= , the value of v is (take 210 /g m s= )

45o

v

Px

10 mO

z

(A) 50 /m s (B) 51 /m s (C) 52 /m s (D) 53 /m s

5.(A) Time of flight 1sec=

21

2

v

g

⋅= 50 /

2

gv m s⇒ = =

6. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The

fringe widths recorded are ,G Rβ β and Bβ , respectively. Then,

(A) G B RBβ β> > (B) B G RBβ β> >

(C) R B GBβ β> > (D) R G BBβ β> >

6.(D) R G Bλ λ λ> > R G Bβ β β∴ > >

7. A small mass m is attached to a massless string whose other end is P

O

fixed at P as shown in the figure. The mass is under going circular

motion in the

x y− plane with centre at O and constant angular speed ω . If the

angular momentum of the system, calculated about O and P are

denoted by 0L

and P

L

respectively, then

(A) O

L

and P

L

do not vary with time.

(B) O

L

varies with time while P

L

remains constant.

(C) O

L

remains constant while P

L

varies with time.

(D) O

L

and P

L

both vary with time.

7.(C) L r p= ×

0L is always along z-axis




pL rotates about z-axis

8. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40

amu) is kept at 300 K in a container. The ratio of the r m s speeds ( )

(arg )

rms

rms

v helium

v on

is

(A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

8.(D)0

rms

RTV

M

γ=

4010

4

He

Ar

V

V= =

9. Two large vertical and paralle metal plates having a separation of 1 cm are connected to a DC voltage

source of potential difference X . A proton is released at rest midway between the two plates. It is found to

move at 045 to the vertical JUST after release. Then X is nearly

(A) 51 10 V

−× (B) 71 10 V

−× (C) 91 10 V

−× (D) 101 10 V

−×

9.(C) tan 45oeE

mg=

45o

mg

d

mgE

e=

910mgd

V Ed Ve

−= = =

10. A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of

the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of

curvature 14 .R cm= For this bi-convex lens, for an object distance of 40 cm, the image distance will be

1.5n = 1.2n =

14R cm=

(A) 280.0 cm− (B) 40.0cm (C) 21.5cm (D) 13.3cm

10.(B) Focal length of plane-convex lens

( )1

1 1 11.5 1

14 28f

= − =

128f cm=




( )2

1 1 11.2 1

14 70f

= − =

270f cm=

Focal length of combination

1 1 1 7 1

28 70 140 20f= + = =

1 1 1

40 20V∴ + =

1 1

40V= , 40V cm=

SECTION II : Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.

11. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q− at

(0, / 4,0), 3a q− + at (0,0,0) and q− at (0, / 4, 0)a+ . Choose the correct option(s)

3q

(A) The net electric flux crossing the plane / 2x a= + is equal to the net electric flux crossing the plane

/ 2x a= −

(B) The net electric flux crossing the plane / 2y a= + is more than the net electric flux crossing the plane

/ 2y a= −

(C) The net electric flux crossing the entire region is 0

q

ε .

(D) The net electric flux crossing the plane / 2z a= + is equal to the net electric flux crossing the plane

/ 2x a= +11.(A,C,D)

A → because both planes are symmetrically located with respect to charge B → both planes are

once again symmetrical

C → Gauss law

D → because both are symmetrical




12. For the resistance network shown in the figure, choose the correct option(s)

2Ω 2Ω

2Ω

1Ω1Ω

4Ω4Ω4Ω

Q T

12V1I

SP

(A) The current through PQ is zero (B) 1

3I A=

(C) The potential at S is less than that at Q (D) 2

2I A=

12.(A,B,C,D)

From symmetry current through PQ & ST is zero.

∴ circuit is

I1

O

2

4

P

Q T

SΩ 2Ω 2Ω

Ω 4 Ω 4 Ω

12 V

I2

4eq

R = Ω1 3I A∴ =

2

123 2

18I A= × =

0 2 04 8SV V I V= − = − , 0 01 4 4

QV V V= − × = −

13. A small block of mass of 0.1 Kg lies on a fixed inclined plane PQ which makes an angle θ with the

horizontal. A horizontal force of 1N acts on the

1N

O P

Q

θ

block through its center of mass as shown in the figure. The block remainsstationary if

(take 210 /g m s= )

(A) 045θ =

(B) 045θ > and a frictional force acts on the block towards P .

(C) 045θ > and a frictional force acts on the block towards Q .

(D) 045θ < and a frictional force acts on the block towards Q .

13.(A,C)

For 45oθ =




45o

1N

1N

N

2R N=

There is no net force along the incline

∴ friction will be zero & block at rest

If 45oθ > then resultant of 1N and 1N will have a component down the incline.

1N

R

N

14. Consider the motion of a positive point charge in aregion where there are simultaneous uniform electric

and magnetic fields 0ˆE E j=

and 0ˆB B j=

. At time 0t = , this charge has velocity v

in the x y− plane,

making an angle θ with the x-axis. Which of the following option(s) is (are) correct for time 0t > ?

(A) If 0oθ = , the charge moves in a circular path in the x z− plane

(B) If 0oθ = , the charge undergoes helical motion with constant pitch along the y-axis

(C) If 10oθ = , the charge undergoes helical motion with its pitch increasing with time, along the y-axis

(D) If 90oθ = , the charge undergoes linear but accelerated motion along the y-axis

14.(C,D)

15. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe.When this pulse reaches the other end of the pipe :

(A) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open

(B) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open

(C) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed (D) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed

15.(B,D)

At open end reflected pressure wave suffers a phase change of π .

At closed end there is no phase change.

SECTION III : Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to9 (both inclusive).

16. An infinitely long solid cylinder of radius R has a uniform volume charge density ρ .It has a spherical

cavity of radius / 2R with its centre on the axis of the cylinder, as shown in the figure.The magnitude of the

electric field at the point P , which is at a distance 2 R from the axis of the cylinder, is given by the

expression 0

23

16

R

k

ρ

ε . The value of k is :




P/2R

x

16.(6) P Cylinder SphereE E E= −

2

0 02 (2 ) 4 (2 )

Q

R R

λ

πε πε= −

3

2

2

0 0

4

3 2

4 16

R

R

R R

π ρρ

ε πε

⋅

= − 0 0 0

23

4 96 96

R R Rρ ρ ρ

ε ε ε= − =

17. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the

cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude

of the magnetic field at the point P is given by 012

NaJµ , then the value of N is :

P

O

2a

a

17.(5) P Cylinder CavityB B B= −

2

2 00 4

322

2

aJ

J a

aa

µ πµ π

π π

⋅ ⋅ = −

⋅0 0

1 1 5

2 12 12aJ aJµ µ

= − =

18. A lamina is made by removing a small disc of diameter 2 R from a bigger disc of uniform mass density and

radius 2 R , as shown in the figure. The moment of inertia of this lamina about axes passing through O

and P is 0

I and pI , respectively. Both these axes are perpendicular to the plane of the lamina. The ratio

/p O

I I to the nearest integer is :




2R

2RO

18.(3)( )

2 2

2

0

4 2

2 2

M R MRI MR

= − +

213

2MR= [ M = mass of cavity]

( )2

2 23(4 )(2 ) 5

2 2p

MRI M R M R

= − +

237

2Rπ=

0

373

13

pI

I∴ = ≈

19. A circular wire loop of radius R is placed in the x y− plane centered at the origin O . A square loop of side

( )a a R<< having two turns is placed with its center at 3z R= along the axis of the circular wire loop,

as shown in figure. The plane of the square loop makes an angle of 45o with respect to the z-axis. If the

mutual inductance between the loops is given by

2

0

/ 22 p

a

R

µ, then the value of p is :

R

O

x

y

45o

z

a

3 R

19.(7) ( )

2

0

3 / 22 22 3

IRB

R R

µ=

+

[ I = current in circular loop] 0

16

I

R

µ=

Flux through square loop

cos45oNABφ =

2 0 12

16 2

Ia

R

µ= ⋅

2

0

7 / 22

aI

R

µ= ⋅

2

0

7 / 22

aM

R

µ∴ =




20. A proton is fired from very far away towards a nucleus with charge 120Q e= , where e is the electronic

charge. It makes a closest approach of 10 fm to the nucleus. The de-Broglie wavelength (in units of fm )

of the proton at its start is : (take the proton mass, 27(5 /3) 10

pm kg

−= × ,

15/ 4.2 10 /h e J s C−= × ⋅ ,

9 2 2

01/ 4 9 10 . /N m Cπε = × ,

151 10fm m−= )

20.(7) K KE= at start

( )

2 9 2

14140

1 120 9 10 120

4 1010

e eK

πε −−

× × ×= =

24 2108 10 e= × ×

2P mK=

27 24 252 10 108 10

3

h h

pe

λ−

= =

× × × × ×

15

15

1

4.2 107 10 7

6 10fm

−−

−

×= = × =

×

PPPPARARARART II :T II :T II :T II : CHEMISTR CHEMISTR CHEMISTR CHEMISTRYYYY



21. In allene 3 4( )C H , the type(s) of hybridisation of the carbon atoms is (are)

(A) sp and 3sp (B) sp and 2

sp (C) Only 2sp (D) 2

sp and 3sp

21.(B) H C C C H- = = -sp

2

H H

sp sp2

22. For one mole of a van der Waals gas when 0b = and 300 ,T K= the PV vs 1/V plot is shown below..

The value of the van der Waals constant a 2 -2( . ) atm liter mol is

[Graph not to scale]24.6

23.1

21.6

20.1

3.02.0

(Mol liter )-11

V

0

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0




22.(C) ( )2

aP V RT

V

+ =

aPV RT

V+ =

aPV RT

V= −

Slope = a−2 1

2 1

y y

x x

−=

−

21.6 20.1

3 2

−=

−1.5=

23. The number of optically active products obtained from the complete ozonolysis of the given compound is

CH -CH = CH - C - CH = CH - C - CH = CH - CH 3 3

HCH3

H CH3

(A) 0 (B) 1 (C) 2 (D) 423.(A)

3 3CH CH CH C CH CH C CH CH CH- = - - = - - = -

O O- O O- O O-

3CH

H3

CH

H

3 2|O H O

3 3CH C H O C C C H O C C C H O C CH- - + = - - - + = - - - + = -

O

H

3CH

H H3CH

HO HO

(A ) is correct i.e. none is optically acitive

24. A compound p qM X has cubic close packing ( ) ccp arrangement of X . Its unit cell structure is shown

below. The empirical formula of the compound is

M =

X =

(A) MX (B) 2MX (C) 2M X (D) 5 14M X

24.(B)1

4 1 24

M = × + =

1 18 6 4

8 2X = × + × =

2MX




25. The number of aldol reaction ( )s that occurs in the given transformation is

. .

3 4 conc aq NaOHCH CHO HCHO+ →

OH OH

HO OH

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

25.(C) 2H C H CH C H- + - - -

H

O

αO

2H CH CH C H- - - -

OH

αO

H C H CH C H- + - - -

H

α

OO

2CH OH-

2CH CH C H- - -

OH

2CH OH-

O

H C H C C H- + - - -

H

OO

( )2 2CH OH-

α

2CH C C H- - -

OH

( )2 2CH OH-

O

Then after

C H- -

O

Is convert in to

Through cannizzaros rxn.2CH OH

∴ (C) is correct

26. The colour of light absorbed by an aqueous solution of 4CuSO is

(A) orange-red (B) blue-green (C) yellow (D) violet26.(A) Orange Red

27. The carboxyl functional group (- )COOH is present in

(A) picric acid (B) barbituric acid (C) ascorbic acid (D) aspirin

27.(D)

OH

2O N2NO

2NO

Picric acid

(2,4,6- Trinitrophenol)

O O

O

Barbituric acid

HO OH

O

H

HO

HO




Ascorbic acid

C OH-

3O C CH- -

O

OAspirin ( O-Acetyl salicylic acid)

28. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ 0a is Bohr radius]

(A)

2

2 2

04

h

maπ(B)

2

2 2

016

h

maπ(C)

2

2 2

032

h

maπ(D)

2

2 2

064

h

maπ

28.(C) 22

hm rν

π=

22

2 2

1 1

2 2

hm

m rν

π=

04r a=

2

2 2

0

1. .

2 16

hK E

m aπ= ×

29. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

(A) 3 4 2, , ,HNO NO NH CI N (B) 3 2 4, , ,HNO NO N NH CI

(C) 3 4 2, , ,HNO NH CI NO N (D) 3 4 2, , ,NO HNO NH CI N

29.(B)3 2 4; ; ;HNO NO N NH Cl

+5 +2 0 -3

30. As per IUPAC nomenclature, the name of the complex [ ]2 4 3 2( ) ( )Co H O NH 3CI is

(A) Tetraaquadiaminecobalt (III) chloride (B) Tetraaquadiamminecobalt (III) chloride(C) Diaminetetraaquacobalt (III) chloride (D) Diamminetetraaquacobalt (III) chloride

30.(D) ( ) ( )2 3 34 2Co H O NH Cl

Diamminetetraaquacobalt ( )III chloride



31. Identify the binary mixture ( )s that can be separated into individual compounds, by differential extraction,

as shown in the given scheme.

Binary mixture containing

Compound 1 and Compound 2

Compound 1 Compound 2

Compound 1 Compound 2

NaHCO (aq)3

NaOH(aq)

(A) 6 5C H OH and 6 5C H COOH




(B) 6 5C H COOH and 6 5 2C H CH OH

(C) 6 5 2C H CH OH and 6 5C H OH

(D) 6 5 2C H CH OH and 6 5 2C H CH COOH

31.(B,D)In option (A)

Both 6 5C H OH & 6 5C H COOH both can react with NaOH & 3NaHCO

hence cannot be seperated.

32. Choose the correct reason ( )s for the stability of the lyophobic colloidal particles.

(A) Preferential adsorption of ions on their surface from the solution(B) Preferential adsorption of solvent on their surface from the solution(C) Attraction between different particles having opposite charges on their surface(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the

colloidal particles32. (A,C,D)

33. Which of the following molecules, in pure form, is ( )are unstable at room temperature?

(A) (B) (C)

O

(D)

O

33. (B,C) stable ∵of conjugated system

(B) Unstable ∵ of conjugated system but antiaromatic

(C) O ↔ O

Unstable ∵ 4 &eπ − i.e. antiaromatic polarisation

(D)

O

↔

O

stable ∵ 6 eπ − aromatic conjugated system

∴ (b,c) is correct

34. Which of the following hydrogen halides react ( )s with 3 AgNO ( )aq to give a precipitate that dissolves

in 2 2 3( )Na S O aq ?

(A) HCl (B) HF (C) HBr (D) HI34.(A,C,D)

( )2 2 3 3 2 3 22AgBr Na S O Na Ag S O NaBr + → +




( )2 2 3 3 2 3 22AgCl Na S O Na Ag S O NaCl + → +

( )2 2 3 3 2 3 22AgI Na S O Na Ag S O NaI + → +

35. For an ideal gas, consider only P V− work in going from an initial state X to the final state Z . The final

state Z can be reached by either of the two paths shown in the figure. Which of the following choice ( )s is

( )are correct ? [take S∆ as change in entropy and w as work done]

X Y

Z

V (liter)

(A) x z x y y zS S S→ → →∆ = ∆ + ∆ (B) x z x y y zw w w→ → →= +

(C) x y z x yw w→ → →= (D) x y z x yS S→ → →∆ = ∆

35.(AC) (A) Entropy is state function.(C) Work done is area under the P-V curve.



36. The substituents 1R and 2R for nine peptides are listed in the table given below. How many of these

peptides are positively charged at pH = 7.0 ?

H N - CH - CO - NH - CH - CO - NH - CH - CO - NH - CH - COO3

H R1 R2 H

Peptide R1

R2

I H H

II H CH3

III CH2COOH H

IV CH2CONH

2(CH

2)

4NH

2

V CH2CONH

2CH

2CONH

2

VI (CH2)

4NH

2(CH

2)

4NH

2

VII CH2COOH CH

2CONH

2

VIII CH2OH (CH

2)

4NH

2

IX (CH2)

4NH

2 CH

3

36.(4) Peptide (I) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O OH H O O

H




Peptide (II) 3 2

H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O OH 3CH O O

H

Peptide (III) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

2CH C OH- - H O O

H

O

Peptide (IV) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

2 2CH C NH- - ( )2 24CH NH-

O

O

H

O

Peptide (V) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

2 2CH C NH- -

O

O

H

O

2 2CH C NH- -

O

Peptide (VI) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

( )2 24CH NH-

O

O

H

( )2 24CH NH-

Peptide (VII) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

2CH C OH- - 2 2CH CONH- O

H

O

O

Peptide (VIII) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O

( )2 24CH NH-

O

O

H

2CH OH-

Peptide (IX) 3 2H N CH C NH CH C NH CH C NH CH C O- - - - - - - - - - - -

O

O O

O

H

( )2 24CH NH- 3CH

Ans . 3 Peptide IV /Peptide VI / Peptide VIII/Peptide IX because of lone pair of electron present on nitrogen whichat pH = 7 undergoes protonation i.e. positive in nature.




37. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes

a nuclear reaction yielding element X as shown below. To which group, element X belongs in the

periodic table?

63 1 1 1

29 1 0 16 2Cu H n H Xα+ → + + +

37.(8) 63 1 1 4 1 52

29 1 0 2 1 266 2Cu H n He H Fe+ → + + +

Fe in 8th group

38. When the following aldohexose exists in its D -configuration, the total number of stereoisomers in itss

pyranose form is

CHO

CH

CHOH

CHOH

CHOH

CH OH

2

2

38.(8)

H

C O=

2CH

H C OH- -

H C OH- -

H C OH- -

2CH OH-

D-Configuration

*

or

OH

H C-

2CH

H C OH- -

H C OH- -

H C-

2CH OH-6

5

4

3

2

1

or

H

2CH OH-

HO

H

H

OH

OH

H

*

*

*

In Pyranose form total 3 chiral or stereocentre hence total 32 8= stereo isomers.

39. 29.2% ( / ) w w HCI stock solution has a density of 1.25 1g mL

− . The molecular weight of HCI is 36.5

1g mol

− . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCI is

39.(8) 29.2 % (w/w) HCl ; d = 1.25 g/L , 1 29.2w g= , 100

1.25 1000

wV L

d= =

×

100W g= , 1

1

WM

M V=

× , 1 36.5 /M g mol= , 1.25 /d g ml=

1 1 2 2M V M V= , 2

11000 10 0.4 200V−× × = × , 110 80V× = , 1 8V ml=

40. An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8

and 1/10 of its initial concentration are 1/8t and 1/10t respectively.. What is the value of

[ ][ ]

( )1/8

10

1/10

x10? log 2 0.3t

taket

=




40.(9) 1/8 0

1

8t C C→ = , 1/10 0

1

10t C C→ = ,

01/8

0

01/10

0

82.303log

2.303log 10

Ct

K C

Ct

K C

×=

= ×,

1/8

1/10

log8

1

t

t=

1/8

1/10

3 0.3 0.9t

t= × = ,

1/8

1/10

10 9t

t× =

PPPPARARARART III :T III :T III :T III : MA MA MA MATHEAMTICSTHEAMTICSTHEAMTICSTHEAMTICS



41. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that

each person gets at least one ball is

(A) 75 (B) 150 (C) 210 (D) 243

41.(B) 5 balls ⇒ 3 pesons

1,1,3

2, 2,1

number of ways 5! 5!

3!1!1!3!2! 2!2!1!2!

= + ×

( )10 15 6 150= + × =

42. Let

2cos , 0

( ) ,

0 , 0

x xf x x IRx

x

π≠

= ∈ =

then f is

(A) differentiable both at 0x = and at 2x =

(B) differentiable at 0x = but not differentiable at 2x =

(C) not differentiable at 0x = but differentiable at 2x =

(D) differentiable neither at 0x = nor at 2x =

42.(B)

2cos , 0

( )

0 , 0

x xf x x

x

π≠

= =

At 0x =

L.H.D.

2

0

(0 ) cos 0

lim( )h

hh

h

π

+→

− =

=−

0

lim ( ) cos 0h

hh

π+→

= − =

R.H.D.

2

0

(0 ) cos 0

lim 0( )h

hh

h

π

+→

− −

= =

L.H.D. = R.H.D. ( )f x⇒ is differentiable at 0x =




but ( )f x⇒ is not differentiable at 2x =

43. The function :[0,3] [1,29]f → defined by 3 2( ) 2 15 36 1f x x x x= − + + , is

(A) one- one and onto (B) onto but not one - one

(C) one - one but not onto (D) neither one - one nor onto

43.(B) 3 2( ) 2 15 36 1f x x x x= − + +

ve

0 2 3

2'( ) 6 30 36f x x x= − +

26( 5 6)x x⇒ − + 6( 2)( 3)x x⇒ − −

for max 0,2,3⇒

(0) 1f =

(2) 16 60 72 1f = − + + 29=

(3) 28f =

( )f x∴ is onto as range [1,29]= & Co- domain [1,29]= (B) onto but not one - one

44. If

2 1lim 4

1x

x xax b

x→∞

+ +− − = +

then

(A) 1, 4a b= = (B) 1, 4a b= = − (C) 2, 3a b= = − (D) 2, 3a b= =

44. (B)

2 1lim 4

1x

x xax b

x→∞

+ +− − = +

2 21lim 4

1x

x x ax ax bx b

x→∞

+ + − − − −⇒ =

+

2(1 ) (1 ) (1 )lim 4

1x

a x x a b b

x→∞

− + − − + −⇒ =

+. Since limit existe

therefore 1 0 1a a− = ⇒ =

( ) (1 )lim 4

1x

x b b

x→∞

− + −⇒ =

+4 4

1

bb

−⇒ = ⇒ = −

45. Let z be a complex number such that the imaginary part of z is nonzero and 2 1a z z= + + is real. Then

a cannot take the value

(A) 1− (B) 1

3(C)

1

2(D)

3

4

45.(D)

21 3

2 4a z

= + +

when 1 3

,2 4

z a= − = but z must have imaginary part non - zero 1 3

2 4z a∴ ≠ − ∴ ≠

46. The ellipse

2 2

1 : 19 4

x yE + = is inscribed in a rectangle R whose sides are parallel to the coordinate axes.

Another ellipse 2E passing through the point (0,4) circumscribes the rectangle R . The eccentricity of the

ellipse 2E is

(A) 2

2(B)

3

2(C)

1

2(D)

3

4




46.(C)

2 2

2 2 2:

x yE

a b+ .......(i)

equation (i) passes through (3,2)

2 2

9 41

a b∴ + =

(3,2)(0,2)

(0, 4)

(3,0)(0,4) passes through equation (i)

2 2

0 161

a b∴ + =

2 2 9 416, 12

3b a

×= = =

2 212 4 11

16 16 2e e e= − ⇒ = ∴ =

47. Let [ ]ijP a be a 3 3× matrix and let [ ]ijQ b= , where 2i jij ijb a+= for 1 , 3i j≤ ≤ . If the determinant of P is

2 , then the determinant of the matrix Q is

(A) 102 (B) 112 (C) 122 (D) 132

47.(D)

11 12 13

21 22 23

31 32 33

| |

a a a

P a a a

a a a

=

2 3 411 12 13

3 4 521 22 23

4 5 631 32 33

2 2 2

| | 2 2 2

2 2 2

a a a

Q a a a

a a a

=

211 12 13

2 3 4 221 22 23

231 32 33

2 2

2 .2 .2 2 2

2 2

a a a

a a a

a a a

=2 3 4 22 .2 .2 .2.2 | |P= 122 | |p=

48. The integral

2

9 / 2

sec

(sec tan )

xdx

x x+∫ equals (for some arbitrary constant K )

(A) 2

11/ 2

1 1 1(sec tan )

11 7(sec tan )x x K

x x

− − + +

+(B)

2

11/ 2

1 1 1(sec tan )

11 7(sec tan )x x K

x x

− + +

+

(C) 2

11/ 2

1 1 1(sec tan )

11 7(sec tan )x x K

x x

− + + +

+(D)

2

11/ 2

1 1 1(sec tan )

11 7(sec tan )x x K

x x

+ + +

+

48.(C) Put 1

(sec tan ) ,sec tanx x t xt

+ = − =21 1

2sec sec2

tx t x

t t

+⇒ = + ⇒ =

sec(sec tan )x x dx dt+ =




∴

2

9 / 2

sec (sec tan )

(sec tan )(sec tan )

x x xdx

x xx x

+

++∫

11

2

(sec )(sec tan )sec

(sec tan )

x x x x dx

x x

+

+

∫2

11/ 2 11/ 2

1

(sec ) 2

( )

t

x dt t dtt t

+

⇒ =∫ ∫

2

13/ 2 13/ 2

1 1

2 2

dt tdt

t t= +∫ ∫

11/ 2(2 13/ 2)1 1

2 11/ 2 2

tt dt

−−

= + −

∫9 / 2 1

11/ 2

1 1 1

11 2 7 / 2

tC

t

− + − = + + −

11/ 2 7 / 2

1 1 1 1

11 7C

t t

− = + +

2

11/ 2

1 1 1(sec tan )

11 7(sec tan )x x C

x x

= − + + + +

49. The point P is the intersection of the straight line joining the point (2,3,5)Q and (1, 1,4)R − with the plane

5 4 1x y z− − = . If S is the foot of the perpendicular drawn from the point (2,1,4)T to QR then the length

of the line segment PS is

(A) 1

2(B) 2 (C) 2 (D) 2 2

49.(A) Direction ratios of (1,4,1)QR

P divides QR in the ratio :1k where (5 2 4 3 5 1) 2

1 (5 1 4 ( 1) 4 1) 1

k − × − × − −= =

× − × − − −

P∴ is

2 1 1 2 2 ( 1) 1 3 2 4 1 5, ,

2 1 2 1 2 1

× + × × − + × × + ×

+ + +

4 1 13, ,

4 3 3

equation of PQ

1 1 4

1 4 1

x y zk

− + −= = =

S∴ can be ( 1,4 1, 4)k k k+ − +

TS PQ⊥

(2,1, 4)T

S

P

R (1, 1,4)−

(2,3,5)




( 1 2) 1 (4 1 1) 4 ( 4 4) 1 0k k k∴ + − × + − − × + + − × =

1 16 8 0k k k− + − + =

118 9

2k k⇒ = ⇒ =

3 9,1,

2 2S

∴ =

2 2 23 4 1 9 13

12 3 3 2 3

PS

= − + − + −

1 4 1 1

36 9 36 2= + + =

50. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line

4 5 20x y− = to the circle 2 2 9x y+ = is

(A) 2 220( ) 36 45 0x y x y+ − + = (B) 2 220( ) 36 45 0x y x y+ + − =

(C) 2 236( ) 20 45 0x y x y+ − + = (D) 2 236( ) 20 45 0x y x y+ + − =

50.(A) 4 5 20x y− =

Let a point on the line be

5 20,

4

tP t

+

let the mid point of chord of contact be ( ', ')x y

equation of chord 1T S=

2 2' ' 9 ' ' 9xx yy x y+ − = + − ........(i)

2 2' ' ' 'xx yy x y+ = + & equation of chord of contact of tangents from P be 0T = 5 20

9 04

tx yt

+ + − =

......(ii)(i) & (ii) are identical

2 2' ' ' '

5 20 9

4

x y x y

t t

+∴ = =

+ 5 20

' '4

ttx y

+ ⇒ =

2 2

20 ' 9 '

4 ' 5 ' ' '

y yt

x y x y∴ = =

− +

2 220( ' ' ) 36 ' 45 'x y x y+ − +



51. Let , [0,2 ]θ φ π∈ be such that 22cos (1 sin ) sin tan cot cos 1

2 2

θ θθ φ θ φ

− = + −

, tan(2 ) 0π θ− > and




31 sin

2θ− < < − . Then φ cannot satisfy

(A) 02

πφ< < (B)

4

2 3

π πφ< < (C)

4 3

3 2

π πφ< < (D)

32

2

πφ π< <

51.(A,C,D)3

1 sin2

θ− < < − ...(i)

4

3

π 5

3

π

22sin

2cos (1 sin ) cos 12cos / 2.sin / 2

θθ φ φ

θ θ− = −

2cos (1 sin ) 2sin cos 1θ φ θ φ⇒ − = −

2sin cos 2cos sinθ φ θ φ⇒ + 1 2cosθ= +

1sin cos cos sin cos

2φ φ θ φ θ+ = +

1sin( ) 1

2θ φ< + < from (i)

13 17

6 6

π πθ φ⇒ < + < and

5 3

3 2

π πθ

− −< − <

4

2 3

π πφ⇒ < <

tan(2 ) 0π θ− > ...(ii)

/ 2π

2π

3

2

π

∴ (i) & (ii) 3 5

,2 3

π πφ

⇒ ∈

3 5

2 3

π πθ< < , 0 2φ π< <

3 11

2 3

π πθ φ⇒ < + <

52. Let S be the area of the region enclosed by 2

, 0, 0x

y e y x−= = = and 1x = . Then

(A) 1

Se

≥ (B) 1

1Se

≥ − (C) 1 1

14

Se

≤ +

(D)

1 1 11

2 2S

e

≤ + −

52.(A,B,D)2xy e−=

( ) ( )21 2x

y e x−= −

∴ function is decreasing

1/ 2 1

1

0

1

e

1/ e

1x 1S

e

≥

and 1 1 1

1 12 2 e

≤ ⋅ + −

∵ 2

[0,1]x xe e x

− −≥ ∀ ∈1

0

11

xS e dx S

e

−⇒ ≥ ⇒ ≥ −∫53. A ship is fitted with three engines

1 2,E E and 3E . The engines function independently of each other with




respective probabilities 1 1

,2 4

and 1

4. For the ship to be operational at least two of its engines must function.

Let X denote the event that the ship is operational and let 1 2,X X and

3X denote respectively the eventss

that the engines 1 2,E E and

3E are functioning. Which of the following is (are) true?

(A) 1

3

16

cP X X =

(B) P [Exactly two engines of the ship are functioning X ]7

8=

(C) 2

5

16P X X =

(D) 1

7

16P X X =

53.(B,D) 1x 2x

3x

shaded region represent X

1 2 3

1 1 1( ) ( ) ( )

2 4 4P x P x P x= = =

1 2 3 1 2 3( ) ( ) ( )c cP X P X X X P X X X= ∩ ∩ + ∩ ∩

1 2 3 1 2 3( ) ( )cP X X X P X X X+ ∩ ∩ + ∩ ∩

1 1 3 1 3 1

2 4 4 2 4 4

= +

1 1 1 1 1 1

2 4 4 2 4 4

+ +

( )11

( )

c P X XXP

X P X

∩=

2 3 1 2 3( ) ( ) 1

( ) 8

P X X P X X X

P X

∩ − ∩ ∩= =

Let T ⇒ exactly two engines are functioning

then ( )

( )

T P T XP

X P X

∩ =

1 2 2 3( ) ( )

( )

P X X P X X

P X

∩ + ∩=

1 3 1 2 3( ) 3( )

( )

P X X X X X

P X

+ ∩ − ∩ ∩

1 1 1 3

8 16 8 321

4

+ + −=

7

7321 8

4

= =

2

2 2

( )

( )

P X XXP

X P X

∩=

1 2 2 3 1 2 3

2

( ) ( ) ( )

( )

P X X P X X P X X X

P X

∩ + ∩ − ∩ ∩=

1 2 2 3 1 2 3

2

( ) ( ) ( ) ( ) ( ) ( ) ( )

( )

P X P X P X P X P X P X P X

P X

+ −=




1 1 1 1 1 1 1

2 4 4 4 2 4 4

1

4

+ −

=

1 1 1 5

58 16 32 321 1 8

4 4

+ −= = =

1

1 1

( )

( )

P X XXP

X P X

∩= =

( )1 2 1 3 1 2 3

1

( ) ( ) ( )P X X P X X P X X X

P X

∩ + ∩ − ∩ ∩ 7

16

=

54. Tangents are drawn to the hyperbola

2 2

19 4

x y− = ,parallel to the straight line 2 1x y− = . The points of

contact of the tangents on the hyperbola are

(A) 9 1

,2 2 2

(B) 9 1

,2 2 2

− −

(C) ( )3 3, 2 2− (D) ( )3 3, 2 2−

54.(A,B)

Let (3sec ,2 tan )P θ θ be a point on hyperbola tangent at p

sec tan1

3 2

x yθ θ− =

given sec 2

23 tan

θ

θ

−− × =

1sec 3tan sin

3θ θ θ= ⇒ =

3 1sec tan

2 2 2 2θ θ∴ = ± = ±

point of contact will be is 1st and 3rd quadrant and will be

9 1 9 1( sec , tan ) , , ,

2 2 2 2 2 2a bθ θ

− − =

55. If ( )y x satisfies the differential equation ' tan 2 secy y x x x− = and ( )0 0y = , then

(A)

2

4 8 2y

π π =

(B)

2

'4 18

yπ π

=

(C)

2

3 9y

π π =

(D)

24 2'

3 3 3 3y

π π π = +

55.(A,D) Differentiable equation

tan 2 secdy

y x x xdx

− = it is form dy

py qdx

+ =

tan. . 1/ sec

pdx x dxI F e e x

−∫ ∫= = =




( . . ) ( . .)y I F Q I F dx C= +∫ 1

2sec

y x dx Cx

⇒ = +

∫

22/(sec )

2

xy x C⇒ = + .

Since (0) 0 0y c= ⇒ =

2 secy x x= ⇒2 2

24 16 8 2

yπ π π

= =

2' 2 sec sec tany x x x x x= +

( )22

' (2) (2) 33 3 9

yπ π π

= +

24 2 3

3 9

π π= +

24 2

3 3 3

π π= +



56. Let :f IR IR→ be defined as ( ) 2 1f x x x= + − . The total number of points at which f attains either a

local maximum or a local minimum is

56.(5)

2| | | 1 |x x+ −O

1

5 / 4

1− 11/ 2− 1/ 2

1x < −

2( ) 1f x x x= − + −

21 5

2 4x

= − −

1 0x− ≤ <

2( ) 1f x x x= − − +

25 1

4 2x

= − +

20 1, ( ) 1x f x x x≤ < = − + ( )251/ 2

4x= − −

21, ( ) 1x f x x x≥ = + −

21 5

2 4x

= + −

57. The value of 3

2

1 1 1 16 log 4 4 4 .....

3 2 3 2 3 2 3 2

+ − − −

is




57.(4)1 1 1

6 log3/ 2 4 43 2 3 2 3 2

+ − −

Now 1 1 1

4 4 .....3 2 3 2 3 2

x− − =

21 1 14 4 ....

18 3 2 3 2x

− − =

24 1

18 18x x− =

24 18x x− =

218 4 0x x+ − =

1 4( 4)(18) 289

2(18) 36x

± − − ±= =

1 17 16 4

36 36 9

+= = =

2(2 / 3)3/ 2 (3/ 2)

46 log 6 log

9

⇒ + = +

2(2 / 3)(3 / 2)6 log

−

= + 6 2 4= − =

58. Let ( )p x be a real polynomial of least degree which has a local maximum at 1x = and a local minimum

at 3x = . If (1) 6p = and (3) 2p = , then '(0)p is

58.(9) ( ) ( )1( ) 1 3P x a x x= − −

32( ) 2 3

3

xP x a x x C

= − + +

1(1) 6 2 3

3P a C

= ⇒ − + +

46

3a C⇒ = +

( ) ( )3 2 2 9 18 9p a c= ⇒ = − + + 2C⇒ =

4, 6 2

3So a= + 3a⇒ = ( )(0) 3 (0 1) 0 3P⇒ = − − 9=

59. If ,a b

and c

are unit vectors satisfying 2 2 2

9a b b c c a− + − + − =

, then 2 5 5a b c+ +

is

59.(3) 2 2 2 22 . 2 .a b a b b c b c+ − + + − 2 2 2 . 9c a c a+ + − =

6 2( . . . ) 9a b b c c a− + + =

( . . . ) ( 3/ 2)a b b c c a∴ + + = −

only when ( 0)a b c+ + =

Now, (2 5 5 )a b c+ +

2(2 5( ))a b c⇒ + +

( )2

2 5a a⇒ − 2

( 3 )a= − 29 3a= =




60. Let S be the focus of the parabola 2 8y x= and let PQ be the common chord of the circle

2 2 2 4 0x y x y+ − − = and the given parabola. The area of the triangle PQS is

60.(4) 2 8y x= & 2 2 2 4 0x y x y+ − − =

Q

SP

42

(2, 2)

(2,0)

Focus (2,0)S = ends of latus rectum are (2,4) & (2 4)− circle & parabola meet at (0,0) say P & (2,4) ,

say Q PQS∆1

2 4 42

= × × =

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