Part – I : (CHEMISTRY) - JEE Main, JEE Advanced, NEET ... · PAPER-1 IITIIIITTIIT ----JEE 201...

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PAPER-1 CAREER POINT IITIITIITIIT----JEE 201JEE 201JEE 201JEE 2011111 EXAMINATION EXAMINATION EXAMINATION EXAMINATION

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Part – I : (CHEMISTRY) SECTION – I (Total Marks : 21)

(Single Correct Answer Type) 10/04/2011

This section contains 7 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which

ONLY ONE is correct.

1. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN– and H2O, respectively, are

(A) octahedral, tetrahedral and square planar

(B) tetrahedral, square planar and octahedral

(C) square planar, tetrahedral and octahedral

(D) octahedral, square planar and octahedral

Ans. [B]

Sol. Complexes are : [NiCl4]–2, [Ni(CN)4]

–2 & [Ni(H 2O)6]+2

Ni+2 = 3d84s0

[NiCl4]–2 : Now Since, Cl– is a weak legand so no pairing of electron take place and geometry is tetrahedral

[Ni(CN)4]–2 : Since, CN– is a strong legand so pairing of electron will take place & geometry is square

planar.

[Ni(H2O)6]+2 : It will formed octahedral complex since C.N. = 6

2. AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was

measured. The plot of conductance (Λ ) versus the volume of AgNO3 is -

Λ

volume

(P)

Λ

volume

(Q)

Λ

volume

(R)

Λ

volume

(S)

(A) (P) (B) (Q) (C) (R) (D) (S)

Ans. [D]

Sol. Because in the beginning of the reaction no of ions remain constant so conductivity remains constant but

after complete precipitation of Cl– the no. of ions increases in solution. So conductivity increases.

CODE - 9

2222 / / / / 33333333



3. Bombardment of aluminum by α-particle leads to its artificial disintegration in two ways, (i) and (ii) as

shown. Products X, Y and Z respectively are -

Al2713 YP30

15 +

ZSi3014 +XSi30

14 +

(ii)

(i)

(A) proton, neutron, positron (B) neutron, positron, proton

(C) proton, positron, neutron (D) positron, proton, neutron

Ans. [A]

Sol. nPAl 10

3015

He2713

42 + →

He42

Si3014 + H1

1 Si3014 + e0

1+

4. Extra pure N2 can be obtained by heating -

(A) NH3 with CuO (B) NH4NO3 (C) (NH4)2Cr2O7 (D) Ba(N3)2

Ans. [D]

Sol. Ba(N3)2 →∆ Ba + 3N2 ↑

5. Among the following compounds, the most acidic is -

(A) p-nitrophenol (B) p-hydroxybenzoic acid (C) o-hydroxybenzoic acid (D) p-toluic acid Ans. [C]

Sol. o-hydroxy benzoic acid is stronger acid due to ortho effect.

6. The major product of the following reaction is -

C

O

NHC

O

(i) KOH

(ii) Br CH2Cl

(A)

C

O

N–CH2C

O

Br (B)

C

O

NC

O

CH2Cl

(C)

C

O

N

O–CH2 Br

(D)

C

O

N

O CH2Cl

3333 / / / / 33333333



Ans. [A]

Sol.

C

O

NH C

O

KOH

C

O

NK C

O

Cl–CH2– – Br C

O

N – CH2– –Br C

O

Θ ⊕

7. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity

of the solution is -

(A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M

Ans. [C]

Sol. M = wtmol

10dx ×× =

60

1015.17.10 ×× = 2.05 M x = percentage by weight

x = 1001000120

120 ×+

SECTION – II (Total Marks : 16) (Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out of which

ONE OR MORE is/are correct.

8. Extraction of metal from the ore cassiterite involves -

(A) carbon reduction of an oxide ore (B) self-reduction of a sulphide ore

(C) removal of copper impurity (D) removal of iron impurity

Ans. [A, D]

Sol. Cassiterite is SnO2.

To reduce SnO2 into Sn, carbon reduction process is used.

Sn has iron impurity.

SnO2 + C → Sn + CO2

* The most appropriate answer to this question is (A,C,D), but because of ambiguity in language, IIT

has declared (A, D) as correct answer

9. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible

conformations (if any), is (are)

(A)

C–CH

CH2H2C

H (B)

H–C≡C–C

H

CH2 (C) H2C=C=O (D) H2C=C=CH2

Ans. [B,C]

Sol. Factual.

4444 / / / / 33333333



10. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are)

(A) Adsorption is always exothermic

(B) Physiosorption may transform into chemisorption at high temperature

(C) Physiosorption increases with increasing temperature but chemisorption decreases with increasing

temperature

(D) Chemisorption is more exothermic than physiosorption, however it is very slow due to higher energy of

activation

Ans. [A, B, D]

Sol. Factual.

11. According to kinetic theory of gases -

(A) collisions are always elastic

(B) heavier molecules transfer more momentum to the wall of the container

(C) only a small number of molecules have very high velocity

(D) between collisions, the molecules move in straight lines with constant velocities

Ans. [A, C, D]

Sol. Factual.

* The most appropriate answer to this question is (A, D), but because of ambiguity in language, IIT has

declared (A,C,D) as correct answer

SECTION – III (Total Marks : 15)

(Paragraph Type) This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the

second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A),

(B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions Nos. 12 to 14

When a metal rod M is dipped into an aqueous colourless concentrated solution of compound N, the solution

turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aq.

NH3 dissolves O and gives an intense blue solution.

12. The metal rod M is -

(A) Fe (B) Cu (C) Ni (D) Co

Ans. [B]

Sol. Metal rod M is Cu

13. The compounds N is -

(A) AgNO3 (B) Zn(NO3)2 (C) Al(NO3)3 (D) Pb(NO3)2

Ans. [A]

Sol. Cu + AgNO3(conc.) —→ Cu(NO3)2 + Ag

light blue

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14. The final solution contains -

(A) [Pb(NH3)4]2+ and [CoCl4]

2– (B) [Al(NH3)4]3+ and [Cu(NH3)4]

2+

(C) [Ag(NH3)2]+ and [Cu(NH3)4]

2+ (D) [Ag(NH3)2]+ and [Ni(NH3)6]

2+

Ans. [C]

Sol. AgCl + NH2(aq) → [Ag(NH3)2]+

Cu+2 + NH3(aq) → [Cu(NH3)4]+2

Intense blue

Paragraph for Questions Nos. 15 to 16

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product

through the following sequence of reactions, in which Q is an intermediate organic compound.

P

(C6H10)

(i) dil. H2SO4/HgSO4

(ii) NaBH4/ethanol (iii) dil. acid

Q

(i) conc. H2SO4 (catalytic amount) (–H2O)

(ii) O3 (iii) Zn/H2O

2 C

H3C CH3

O

15. The structure of compound P is -

(A) CH3CH2CH2CH2–C≡C–H (B) H3CH2C–C≡C–CH2CH3

(C) H–C–C≡C–CH3

H3C

H3C

(D) H3C–C–C ≡C–H

H3C

H3C

Ans. [D]

Sol.

CH3–C–C≡C–H (i) dil. H2SO4/HgSO4

(ii) NaBH4

(iii) dil. HCl

CH3

CH3

(P)

CH3–C–CH–CH3

CH3

H3C OH

(Q)

–H2O

(i) conc. H2SO4

CH3–C=C CH3

CH3

(ii) O3 (iii) Zn/H2O

C=O CH3

CH3

H3C

2

16. The structure of compound Q is -

(A)

H–C–C–CH2CH3

H3C

H3C

OH

H

(B)

H3C–C–C–CH3

H3C

H3C

OH

H

(C)

H–C–CH2CHCH3

H3C

H3C

OH

(D)

CH3CH2CH2CHCH2CH3

OH

Ans. [B]

Sol. Factual.

6666 / / / / 33333333



SECTION – IV (Total Marks : 28)

(Integer Answer Type) This section contains. 7 questions. The answer to each questions is a single digit integer ranging from 0 to 9. The

bubble corresponding to the correct answer is to be darkened in the ORS.

17. Reaction of Br2 with Na2CO3 in aq. solution gives sodium bromide and sodium bromate with evolution of

CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is.

Ans. [5]

Sol. 3 Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2

18. The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is.

Ans. [5]

Sol.

NaO–S–S–S–S–ONa

O#

O

⊕ Θ Θ ⊕ ∗ ∗ O

O

#

O.N. S* = 0

O.N. S# = + 5

∴ Difference = 5

19. The maximum number of electrons that can have principal quantum number, n = 3 and spin quantum

number, ms = – 1/2, is.

Ans. [9]

Sol. For n = 3, max e– = 2n2 = 18

Half of them can have ms = – 1/2

20. A decapeptide (mol. wt. 796) on complete hydrolysis gives glycine (mol. wt. 75), alanine and phenylalanine.

Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units

present in the decapeptide is.

Ans. [6]

Sol. Decapeptide water

molecule9 → (x) glycine + (y) alanine + (z) phenylalanine

Mass of hydrolysed product = 796 + 18 × 9

mass of glycine = 958 × 100

47 = 450.26

No. of glycine unit = 675

26.450 =

7777 / / / / 33333333



21. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1 mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm, at 0ºC) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0ºC is close to.

Ans. [7] Sol.

He + compounds

Pext = 1 atm

Vapour pressure of compound = 0.68

∴ PHe = 1 – 0.68 = 0.32 ∴ By PV = nRT, for He

V = 32.0

2730821.01.0

P

RTn

He

He ××=

V ~ 7L

22. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is.

Ans. [5]

Sol.

C–C–C–C–C–C 1 2 3 4 5 6

Br

alc.KOH

C–C–C–C= C– C

+ G.I

C–C–C=C– C– C

+ G.I

+

C–C–C–C– C– C

23. The work function (φ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is.

Metal Li Na K Mg Cu Ag Fe Pt W

φ φ φ φ (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

Ans. [4]

Sol. Efalling = 199

834

106.110300

1031062.6hc−−

−

××××××=

λ = 4.137 eV

The metals having less work function will show photoelectric effect Hence Li, Na, K, Mg

8888 / / / / 33333333



Part –II (PHYSICS)

SECTION – I

Single correct answer type This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

24. 5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1,

the work done in the process is-

(A) 1RT8

9 (B) 1RT

2

3 (C) 1RT

8

15 (D) 1RT

2

9

Ans. [A]

Sol. T1V1γ–1 = T2V2

γ–1

T2 = T1

1

2

1

V

V−γ

= T1

135

7.0

7.5−

= T1(8)2/3 = 4T1

no. of mole = 4

1

W = 1

)TT(nR 21

−γ−

= 3/2

)TT4(R4

111 −×

= 8

9RT1

25. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a

horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The

maximum possible value of angular velocity of ball (in radian/s) is-

m

L

(A) 9 (B) 18 (C) 27 (D) 36

Ans. [D]

Sol.

L

Tθ

θ

mg T cos θ = mg

Code : 9

9999 / / / / 33333333



T sin θ = mω2L sin θ

T = mω2L

ω2max =

mL

Tmax

ωmax = mL

Tmax = 5.05.0

324

× = 4324×

= 36 rad/s

26. Consider an electric field xEE 0=r

, where E0 is a constant. The flux through the shaded area (as shown in

the figure) due to this field is-

(a,a,a)

(0,0,0) (0,a,0)

(a,0,a)

z

x

y

(A) 2E0a2 (B) 2

0aE2 (C) E0a2 (D)

2

aE 20

Ans. [C] Sol.

(a,a,a)

(0,0,0) (0,a,0)

(a,0,a)

z

x

y

H(a,a,a)

(0,a,0)

E(a,0,a)

x

y

(a,a,0)D(a,0,0)

G (0,a,a)F(0,0,a)

AB

C

z

flux through EHBA

= flux through EHDC = E0a

2

10101010 / / / / 33333333



27. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall

building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren

heard by the car driver is-

(A) 8.50 kHz (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

Ans. [A]

Sol. Vs = 3600

1036 3× m/s = 10 m/s, ν = 8 KHz

V0 = 320 m/s

V0 = Vs = 10 m/s

Vs

Observer

ν′ =

′−+

s

0

VV

VV ν

=

−+

10320

10320 8 KHz

= 310

330 × 8 = 8.51 KHz

28. A meter bridge is set-up as shown, to determine an unknown resistance 'X' using a standard 10 ohm resistor.

The galvanometer show null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2

cm respectively for the ends A and B. The determine value of 'X' is-

X 10Ω

A B

(A) 10.2 ohm (B) 10.6 ohm (C) 10.8 ohm (D) 11.1 ohm

Ans. [B]

Sol. 10

x =

248

152

++

= 50

1053×= 10.6

11111111 / / / / 33333333



29. A 2 µF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the

switch S is turned to position 2 is-

1 2

S

2µF 8µF V

(A) 0 % (B) 20 % (C) 75 % (D) 80 %

Ans. [D]

Sol.

2µF 8µF

∆U = 2

1 ×

82

82

+×

[V – 0]2

= 2

1×

10

16 × V2 =

10

V8 2

Ui′ =

2

1 × 2 × V2

= V2

% dissipated = iU

U∆ =

10

8 × 100

= 80%

30. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength

of the second spectral line in the Balmer series of singly-ionized helium atom is-

(A) 1215 Å (B) 1640 Å (C) 2430 Å (D) 4687 Å

Ans. [A]

Sol. 6561

hc = 13.6

−22 3

1

2

1

λhc

= 13.6 × 4

−22 4

1

2

1

6561

hc = 13.6 ×

36

5

λhc

= 13.6 × 4 × 163

12121212 / / / / 33333333



6561

hc =

36

5×

3

4

λ = 27

5

λ = 27

56561× = 243 × 5

= 1215 Å

SECTION – II

Multiple Correct Answers Type This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or More may be correct.

31. A spherical metal shall A of radius RA and a solid metal sphere B of radius RB(<RA) are kept far apart and

each is given charge '+Q'. Now they are connected by a thin metal wire. Then-

(A) insideAE = 0 (B) QA > QB (C)

B

A

σσ

= A

B

R

R (D) surfaceon

Bsurfaceon

A EE <

Ans. [A, B, C, D]

Sol.

RB

Q

RA

Q

S

Ans.(A)

A

A

R

kQ =

B

B

R

kQ [Final potential will be same]

B

A

Q

Q =

B

A

R

R

as RA > RB

∴ QA > QB Ans. [B]

2B

B

2A

A

R4

QR4

Q

π

π =

2B

B

2A

A

R

RR

R

= B

A

σσ

= A

B

R

R Ans [C]

as RA > RB

∴ σB > σA

∴ EB > EA Ans. [D]

13131313 / / / / 33333333



32. A metal rod of length 'L' and mass 'm' is pivoted at one end. A thin disk of mass 'M' and radius 'R' (<L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached :(case A) The disc is not free to rotate about its center and (case B) the disc is free to rotate about its center. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true ?

(A) Restoring torque in case A = Restoring torque in case B (B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A > Angular frequency for case B (D) Angular frequency for case A < Angular frequency for case B

Ans. [A,D] 33. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite

region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true ?

(A) They will never come out of the magnetic field region (B) They will come out traveling along parallel paths (C) They will come out at the same time (D) They will come out at different times Ans. [B, D], [B, C], [B, C, D] Sol.

e–

p

v

v

Rp

Re

× × ×

× ×

× ×

× ×

× × ×

×

× × × × ×

×

× × × × ×

eB

vmR

R

vmevB

eB

VmR

R

vmevB

pp

p

2p

ee

e

2e

=

=

=

=

Re Rp>

eB

m

v

ReT e

eπ=π= ⇒ Te Tp>

eB

mT

pp

π=

But as it's not mentioned that whether they entered in field together or not (C) and (D) could be right depending on data.

* The most appropriate answer to this question is (B,D), but because of ambiguity in language, IIT has declared [(B, C), (B, D), (B, C, D)] as correct answer

14141414 / / / / 33333333



34. A composite book is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a

constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat

'Q' flow only from left to right through the blocks. Then in steady state-

heat 0 1L 5L 6L

A B

C

D

4K

3K

5K

E

6K

3L

4L

1L

(A) heat flow through A and E slabs are same

(B) heat flow through slab E is maximum

(C) temperature difference across slab E is smallest

(D) heat flow through C = heat flow through B + heat flow through D

Ans. [A, C, D]

Sol.

A B

C

D

4K

3K

5K

E

6K

2A

A

4A

A

2K

4A

L 4L L

H

16

H3

2

H

16

H5 H

All the three system shown are in series hence rate of heat flow will be same through both A & E.

)KA(8

LRA = ;

KA3

L4RB = ;

KA8

L4RC = ;

KA5

L4RD = ;

KA24

LRE =

Using parallel combination rate of heat flow across C = rate of heat flow through B+ rate of heat flow

through D.

KA8

HLHRAA ==θ∆

KA4

HL

KA3

L4

16

H3R

16

H3BB =

==θ∆

( )KA4

HL

KA8

L4

2

HR

2

HCC =

==θ∆

KA4

HL

KA5

L4

16

H5D =

=θ∆

KA24

HL

KA24

LHE =

=θ∆

15151515 / / / / 33333333



SECTION – III

(Paragraph type) This section contains 2 paragraphs.. Based upon the first paragraph 3 multiple choice question and based upon on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 35 to 37

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially

useful in studying the changes in motion as initial position and momentum are changed. Here we consider

some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which

position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space

diagram is x(t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the

phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We

use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or

to left) is negative.

Mo

men

tum

Position

35. The phase space diagram for a ball thrown vertically up from ground is-

(A)

Position

Momentum

(B)

Position

Momentum

(C)

Position

Momentum

(D)

Position

Momentum

Ans. [D]

Sol. From conservation of mechanical energy

22 mu2

1mgxmv

2

1 =+

16161616 / / / / 33333333



gxm2umvm 22222 =−

p2 – p02 = 2m2gx

p2 = p02 + 2m2gx

36. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two

circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical

energies respectively. Then-

Momentum

E1E2

a

2a

Position

(A) E1 = 2 E2 (B) E1 = 2E2 (C) E1 = 4E2 (D) E1 = 16 E2

Ans. [C]

Sol.

K' EMax 1

K' Emax 2

K.E1= 0

K.E2 = 0

maximal position

maximal position

4)a(k

21

)a2(k21

E

E

2

2

2

1 ==

E1 = 4E2

37. Consider the spring-mass system, with the mass submerged in water, as shown in figure. The phase space

diagram for one cycle of this system is-

17171717 / / / / 33333333



(A)

Position

Momentum

(B)

Position

Momentum

(C)

Position

Momentum

(D)

Position

Momentum

Ans. [B]

Paragraph for Questions 38 to 39

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids

containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the

number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they

are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons

begin to oscillate about the positive ions with a natural angular frequency 'ωp', which is called the plasma

frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular

frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp, all the

free electrons are set to resonance together and all the energy is reflected. This is the explanation of high

reflectivity of metals.

38. Taking the electronic charge as 'e' and the permittivity as 'ε0', use dimensional analysis to determine the

correct expression for ωp.

(A) 0mε

Ne (B)

Ne

mε0 (C) 0

2

mε

Ne (D)

20

Ne

mε

Ans. [C]

Sol. F = mω2l ≡

20

2

4

e

lπε

ω2 ≡ 3

0

2

4

e

lπε ≡

ε0

2

m

e

3

3N

l

l

ω = 0

2

m

Ne

ε

18181818 / / / / 33333333



39. Estimate the wavelength at which plasma reflection will occur for metal having the density of electrons

N ≈ 4 × 1027 m–3. Take ε0 = 10–11 and m ≈ 10–30, where these quantities are in proper SI unit-

(A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm

Ans. [B] c = λf

ωp = ω = λπc2

= 0

2

εm

Ne

20

Ne

mc2

επ=λ N

m

e

c2 0επ= = 27

1130

19

8

104

)10)(10(

106.1

10314.32

××××× −−

−

= 589 × 10–9 m ≈ 600 nm

SECTION – IV

Integer Type

This section contains 7 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The

bubble corresponding to the correct answer is to be darkened in the ORS.

40. A block is moving on an inclined plane making an angle 45º with the horizontal and the coefficient of

friction is µ. The force required to just push it up the inclined plane is 3 times the force required to just

prevent it from sliding down. If we define N = 10 µ, then N is.

Ans. [5]

Sol.

45ºN=mgcosθ

mgsinθ + µmgcosθ = 3(mgsinθ – µmgcosθ)

⇒ 2

µ

2

1 + = 2

µ3

2

3 −

⇒ 2

µ4 =

2

2

µ = 2

1

N = 10µ =

2

110 = 5

41. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a

force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of

friction between the ground and the ring is large enough that rolling always occurs and the coefficient of

friction between the stick and the ring is (P/10). The value of P is.

19191919 / / / / 33333333



Stick

Ground Ans. [4]

Sol.

2N2N

f2

f1= µ×2 = 5

P

10

2P =×

mg

2 – f2 = Macm …….(1) f2 = 2 – 2 × 0.3 = 1 .4 N (f2 – f1) R = Icm α

(f2 – f1) R = R

aMR cm2 ×

f2 – f1 = Macm f1 = f2 – macm = 1.4 – 2 × 0.3 = 0.8 N

5

P8.0 = ⇒ P = 4

Note : It has been assumed that the stick applies horizontal force of 2N (only normal reaction)

42. Four point charge, each of +q, are rigidly fixed at the four corners of a square planar soap film of side 'a'.

The surface tension of the soap film is γ. The system of charges and planar film are in equilibrium, and

a = k

1/N2

γ

q

, where 'k' is a constant. Then N is.

Ans. [3]

Sol.

FAB FAC

FAD A D

B C

+q

+q +q

+q

20202020 / / / / 33333333



FAC = 2

0

2

a8

q

πε

FAD = FAB = 2

0

2

a4

q

πε

FR =

+πε 2

1º45cos2

a4

q2

0

2

= r (2) (BD) = 2 r( 2 a)

⇒ a3 = r28

21

2q

0

2

πε

+

a =

3/12

r

qk

⇒ N = 3

where k =

3/1

2821

2

π

+

43. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a

square of side 4 cm. The momentum of inertia of the system about the diagonal of the square is N × 10–4kg-m2,

then N is.

Ans. [9]

Sol.

m

m m

m

l

r = 2

5cm =

2

5 × 10–2 m

m = 2

1 kg

l = 4 × 10–2 m

Using parallel axis theorem

Itotal =

×××× −4104

5

2

1

5

24 +

××× −41082

12

⇒ 10–4 + 8 × 10–4 ⇒ 9 × 10–4 kg m2

21212121 / / / / 33333333



44. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is

109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is.

Ans. [1]

Sol. A = λN

1010 = 910

1× N

N = 1019

mass of sample = 1019 × 10–25 × 1 × 106 = 1 mg

45. A long circular tube of length 10 m and radius 0.3 carries a current I along its curved surface as shown. A

wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with

the axis of the tube. The current varies as I = I0cos(300 t) where I0 is constant. If the magnetic moment of

the loop is N µ0 I0 sin(300 t), then 'N' is.

I

Ans. [6]

Sol. B = 10

t300cosIµ 00

φ = 10

Iµ 00 × 3.14 × 0.01 cos 300 t

φ = 3.14 × µ0I0 cos 300 t × 10–3

e = dt

dφ− = 3.14 × 300 µ0I0 sin 300t × 10–3

i = R

e =

005.0

10t300sinIµ30014.3 300

−××

i = 3.14 × 60 µ0I0 sin 300 t

Magnetic moment = 3.14 × 60 × 3.14 × 100

01.0 × µ0I0 sin 300 t

= 5.9 µ0I0sin 300 t

= 6 µ0 I0 sin 300 t

46. Steel wire of length 'L' at 40ºC is suspended from the ceiling and then a mass 'm' is hung from its free end.

The wire is cooled down from 40ºC to 30ºC to region its original length 'L'. The coefficient of linear thermal

expansion of the steel is 10–5/ºC, Young's modulus of steel is 1011 N/m2 and radius of the wire is 1 mm.

Assume that L >> diameter of the wire. Then the value of 'm' in kg is nearly.

Ans. [3]

22222222 / / / / 33333333



Sol.

m

∆L = AY

mgL = Lα(∆θ)

⇒ m = g

)(AY θ∆α =

10

10101010 5116 ××××π −−

m = 3.14 kg ⇒ 3 kg

Part – III : (MATHEMATICS)

SECTION – I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

47. The value of ∫ +

3

222

2

)–6sin(sin

sinn

nxnx

xxl

ll

dx is

(A) 4

1ln

2

3 (B)

2

1 ln

2

3 (C) ln

2

3 (D)

6

1 ln

2

3

Ans. [A]

Sol. Let x2 = t xdx = 2

dt

I = ∫ −+

3

2)6sin(sin

sin

2

1n

n

dttnt

tl

ll

…… (1)

I = ∫ −+−3

2)6sin(sin

)6(sin

2

1n

n

dttnt

tnl

ll

l …… (2)

Add (1) & (2)

2I = ∫3

22

1n

n

dtl

l

I = 4

1(ln3 – ln2) =

2

3

4

1nl

Code-9 10/04/2011

23232323 / / / / 33333333



48. Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0, and x = 0 into two parts R1(0 ≤ x ≤ b) and

R2(b ≤ x ≤ 1) such that R1 – R2 = 4

1. Then b equals

(A) 4

3 (B)

2

1 (C)

3

1 (D)

4

1

Ans. [B]

Sol.

0 b 1

R2 R1

R1 – R2 = 4

1

∫ −b

dxx0

2)1( – ∫ −1

2)1(b

dxx = 4

1

bx

0

3

3

)1(

− –

13

3

)1(

b

x

− =

4

1

3

)1( 3−b +

3

1 – 0 +

3

)1( 3−b =

4

1

3

)1(2 3−b =

4

1 –

3

1 =

12

1−

(b – 1)3 = – 8

1

b – 1 = 2

1− ⇒ b = 2

1

49. Let →a = i + j + k ,

→b = i – j + k and

→c = i – j – k be three vectors. A vector

→v in the plane of

→a

and →b , whose projection on

→c is

3

1, is given by

(A) i – 3 j + 3k (B) – 3i – 3 j – k (C) 3i – j + 3k (D) i + 3 j – 3k

Ans. [C]

Sol. Let →ν = x i + y j + zk

Q [→a

→b

→ν ] = 0

zyx

111

111

− = 0

On solving x = z ….(1)

24242424 / / / / 33333333



Q projection of →ν on

→c is

3

1

So, 3

1=

||

.→

→→ν

c

c ⇒

3

zyx −−=

3

1

⇒ x – y – z = 1 ….(2) So solving (1) & (2) y = –1 & x = z

50. Let (x0, y0) be the solution of the following equations

(2x)ln 2 = (3y)ln 3

3ln x = 2ln y

Then x0 is

(A) 6

1 (B)

3

1 (C)

2

1 (D) 6

Ans. [C]

Sol. (2x)ln 2 = ( 3y)ln 3

ln 2 (ln 2 + ln x) = ln 3 (ln 3 + ln y)

ln 2 . ln x – ln 3 ln y = (ln 3)2 – (ln 2)2 .....(1)

3ln x = 2ln y

ln x . ln 3 = lny . ln 2

ln y = ln x 2

3

n

n

l

l .....(2)

Solving (1) & (2)

ln x = – ln 2 ⇒ x = 2

1

51. Let α and β be the roots of x2 – 6x – 2 = 0, with α > β. If an = αn – βn for n ≥ 1, then the value of 9

810

2

2–

a

aa is

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

Ans. [C]

Sol. ∴ x2 – 6x – 2 = 0 has roots α, β

So, α2 – 2 = 6α & β2 – 2 = 6β

an = αn – βn

So, 9

810

2

2

a

aa − =

)(2

)(2)(99

881010

β−αβ−α−β−α

= )(2

)2()2(99

2828

β−α−ββ−−αα

= )(2

)6()6(99

88

β−αββ−αα

= 3.

52. A straight line L through the point (3, –2) is inclined at an angle 60° to the line 3 x + y = 1. If L also

intersects the x-axis, then the equation of L is

(A) y + 3 x + 2 – 3 3 = 0 (B) y – 3 x + 2 + 3 3 = 0

(C) 3 y – x + 3 + 2 3 = 0 (D) 3 y + x – 3 + 2 3 = 0

Ans. [B]

25252525 / / / / 33333333



Sol. Let the slope of the line is m

tan 60º = m

m

31

3

−+

3 = m

m

31

3

−+

So, m + 3 = ± 3 (1 – 3 m)

m + 3 = 3 – 3m m + 3 = – 3 + 3m

m = 0 m = 3

hence line hence line

y = – 2 y + 2 = 3 (x – 3)

y – 3 x + 2 + 33 = 0

as line intersects x-axis

So line will be, y – 3 x + 2 + 33 = 0

53. Let P = θ : sin θ – cos θ = 2 cos θ and Q = θ : sin θ + cos θ = 2 sin θ be two sets. Then

(A) P ⊂ Q and Q – P ≠ ∅ (B) Q ⊄ P

(C) P ⊄ Q (D) P = Q Ans. [D]

Sol. P : sin θ – cos θ = 2 cos θ

sin θ = ( 2 + 1) cos θ

tan θ = 2 + 1

tan θ = tan 672

1°

θ = nπ + 8

3π, n ∈ I ……(1)

Q : sin θ + cos θ = 2 sin θ

cos θ = ( 2 – 1) sin θ

tanθ = 12

1

− = 2 + 1

θ = nπ + 8

3π, n ∈ I ……(2)

∴ P = Q

26262626 / / / / 33333333



SECTION – II (Total Marks : 16)

(Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

54. The vector(s) which is/are coplanar with vectors i + j + 2k and i + 2 j + k , and perpendicular to the

vector i + j + k is /are

(A) j – k (B) –i + j (C) i – j (D) – j + k

Ans. [A,D]

Sol. kzjyixr ˆˆˆ ++= is coplanar with the given vector so

∴

121

211

zyx

= 0

So, 3x = y + z .....(1)

∴ kjir ˆˆˆ ++⊥→

So, 0)ˆˆˆ(. =++→

kjir

So, x + y + z = 0 .....(2) On solving (1) & (2)

So, x = 0 ∴ y + z = 0 ∴ (A) & (D) Satisfy

55. Let f : R → R be a function such that f(x + y) = f(x) + f(y), ∀x, y ∈ R If f(x) is differentiable at x = 0, then (A) f(x) is differentiable only in a finite interval containing zero (B) f(x) is continuous ∀ x ∈ R (C) f '(x) is constant ∀ x ∈ R (D) f(x) is differentiable except at finitely many points Ans. [B,C] Sol. f(x + y) = f(x) + f(y) By Partial differentiation with respect to x f ' (x + y) = f ' (x) f ' (y) = f '(0) f(y) = (f '(0))y + c f(y) = ky +c ∴ f(y) = ky as f(0) = 0 ∴ f(x) = kx Alternate

f '(x) = 0

lim→h

h

xfhxf )()( −+

= 0

lim→h

h

xfhfxf )()()( −+ =

h

hfh

)(lim

0→

27272727 / / / / 33333333



= λ (let) f(x) = λx + c As f(0) = 0 ⇒ c = 0 f(x) = λx * The most appropriate answer to this question is (B, C), but because of ambiguity in language, IIT has

declared (BC,BCD) as correct answer.

56. Let M and N be two 3 × 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M2N2 (MTN)–1(MN–1)T is equal to

(A) M2 (B) –N2 (C) –M2 (D) MN Ans. [C] Sol. Q MN = NM

M2 N2 = MN MN Q (MT)–1 = (–M)–1 = –M–1 Given, M2N2(MTN)–1. (MN–1)T = – MN MN N–1 M–1 N–1

M I

M

= –M NN–1 M = – M2 Although, the most suitable answer is (C), But given information is contradictory as Skew symmetric

matrix of odd order cannot be non singular * The most appropriate answer to this question is (C), but because of ambiguity in language, IIT has

declared this question as bonus (marks to all students)

57. Let the eccentricity of the hyperbola 2

2

a

x –

2

2

b

y = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the

hyperbola passes through a focus of the ellipse, then

(A) the equation of the hyperbola is 3

2x–

2

2y = 1 (B) a focus of the hyperbola is (2, 0)

(C) the eccentricity of the hyperbola is 3

5 (D) the equation of the hyperbola is x2 – 3y2 = 3

Ans. [B,D] Sol. Let e1 = eccentricity of hyperbola e2 = eccentricity of ellipse

∴ e1 = 2

1

e

so eccentricity of ellipse = 2

3= e2

eccentricity of ellipse = 3

2= e1

Now focus of ellipse is (± ae2, 0) ≡ (± 3 , 0)

Hyperbola passes through it

So, 2

2)3(

a – 0 = 1 ⇒ a2 = 3

also b2 = a2 (e12 – 1)

b2 = 3

1–

3

4 = 1

and hyperbola

28282828 / / / / 33333333



113

22

=−yx

also focus (± ae1, 0) ≡ (± 2, 0)

SECTION – III ( Total Marks : 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple choice questions and based on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question Nos. 58 to 60 Let a, b and c be three real numbers satisfying

[a b c]

737

728

791

= [0 0 0] …….(E)

58. If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (A) 0 (B) 12 (C) 7 (D) 6

Ans. [D]

Sol. [a b c]

737

728

791

= [0 0 0]

a + 8b + 7c = 0

9a + 2b + 3c = 0

7a + 7b + 7c = 0

On solving above equation

(a, b, c) ≡

λλ−λ− ,7

6,

7

∴ (a, b, c) lies on the plane 2x + y + z = 1

So 7

2λ− – 7

6λ + λ = 0

on solving λ = – 7

So 7a + b + c = 6

59. Let ω be a solution of x3 – 1 = 0 with Im(ω) > 0. If a = 2 with b and c satisfying (E), then the value of

aω

3 +

bω1

+ cω

3 is equal to

(A) –2 (B) 2 (C) 3 (D) –3

Ans. [A]

Sol. ∴ (a, b, c) ≡

λλ−λ− ,7

6,

7

∴ a = 2 is given so λ = – 14

29292929 / / / / 33333333



So (a, b, c) ≡ (2, 12, – 14)

So cba ω

+ω

+ω

313 = –2

60. Let b = 6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then

n

n∑

∞

=

β+

α0

11 is

(A) 6 (B) 7 (C) 7

6 (D) ∞

Ans. [B]

Sol. ∴ (a, b, c) ≡

λλ−λ− ,7

6,

7

∴ b = 6 so λ = – 7.

So (a, b, c) ≡ (1, 6, –7)

So the equation ax2 + bx + c = 0

x2 + 6x – 7 = 0

So α = 1, B = – 7

S = ∑∞

=

β+

α0

11

n

n

= n

∑

−7

1

1

1

= n

∑

7

6 = 1 +

7

6 +

2

7

6

+ .... ∞

=

7

61

1

− = 7

Paragraph for Question Nos. 61 and 62 Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball.

A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2.

61. The probability of the drawn ball from U2 being white is

(A) 30

13 (B)

30

23 (C)

30

19 (D)

30

11

Ans. [B]

Sol. RW

23 W1

U1 U2

Required probability = P(H)[P(W/H) × P(W2) + P(R/H)P(W2)] + P(T) [P

T

Wboth P(W2) + P

T

Rboth

P(W2) +

T

WRP 11 &

P(W2)]

=

××+×+×+

×+×3

2

3

11

2

1

2

1

5

21

5

3

2

1

25

12

13

25

22

25

23

C

CC

C

C

C

C

30303030 / / / / 33333333



=

+++

+5

2

30

1

10

3

2

1

5

1

5

3

2

1 =

30

23

30

11

5

2 =+

62. Given that the drawn ball from U2 is white, the probability that head appeared on the coin is

(A) 23

17 (B)

23

11 (C)

23

15 (D)

23

12

Ans. [D] Sol. Required probability

=

+

+

+

+

+

)(&

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

)()()(

211

2221

21

21

21

WPT

WRPWP

T

RbothPWP

T

WbothPTPWP

H

RPWP

H

WPHP

WPH

RPWP

H

WPHP

= 13

12

30

232

1

5

21

5

3

2

1

=

×+×

SECTION – IV(Total marks : 28)

(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.

63. Let a1, a2, a3, ….., a100 be an arithmetic progression with a1 = 3 and Sp = ∑=

p

iia

1

, 1 ≤ p ≤ 100. For any integer

n with 1 ≤ n ≤ 20, let m = 5n. If n

m

S

Sdoes not depend on n, then a2 is

Ans. [9]

Sol. a1 = 3

n

m

S

S=

n

n

S

S5 = ])1(2[

2

])15(2[2

5

1

1

dnan

dnan

−+

−+

= ndd

ndd

+−+−

)6(

]5)6[(5

Q n

n5

S

S is independent of n so d = 6

So a2 = a1 + d = 3 + 6 = 9 * The most appropriate answer to this question is (9), but because of ambiguity in language, IIT has

declared (3, 9 ; 3 & 9 both) as correct answer. 64. Consider the parabola y2 = 8x. Let ∆1 be the area of the triangle formed by the end points of its latus rectum

and the point P

2,

2

1 on the parabola, and ∆2 be the area of the triangle formed by drawing tangents at P

and at the end points of the latus rectum. Then 2

1

∆∆

is

31313131 / / / / 33333333



Ans. [2] Sol. It is a property that area of triangle formed by joining three points lying on parabola is twice the area of

triangle formed by tangents at these points Alternate : y2 = 8x

P

2,

2

1

•

• • A

P

B(2, –4)

3/2

(2, 4)

∆1 = 2

1|Base × Height|

= 2

1×

2

3× 8 = 6

Also

Equation of tangent at P

2,

2

1

• • P

•

•

y (2) = 4.

+2

1x

y = 2x + 1 ….(1) Tangent at A : y = x + 2

Tangent at B : – y = + x + 2 ⇒ y = – x – 2 Point of intersection L(–2, 0), M (1, 3), N (–1, –1)

∆2 =

111

131

102

2

1

−−

−

= |2

1[–2(4) + (–1 + 3)]|

= ]138[2

1 −+− = 3

So, 2

1

∆∆

= 3

6= 2

32323232 / / / / 33333333



65. The positive integer value of n > 3 satisfying the equation

πn

sin

1 =

πn

2sin

1 +

πn

3sin

1 is

Ans. [7]

Sol. Let n

π = θ

θsin

1 =

θ2sin

1 +

θ3sin

1

θsin

1 –

θ3sin

1 =

θ2sin

1

[sin 3θ – sin θ] sin 2θ = sin θ sin 3θ

2 sin θ cos 2θ sin 2θ = sin θ sin 3θ

Q sin θ ≠ 0

2 cos 2θ sin 2θ = sin 3θ

sin 4θ = sin 3θ

so either 4θ = 3θ or 4θ = π – 3θ

so θ = 0 or θ = 7

π so n = 7

66. Let f(θ) = sin

θθ2cos

sintan 1– , where –

4

π < θ <

4

π. Then the value of

)(tanθd

d (f (θ)) is

Ans. [1]

Sol. Q tan–1

θθ2cos

sin = sin–1 tan θ

so f (θ) = sin (sin–1 tan θ) = tan θ

Q )(tan

))((

θθ

d

fd =

)(tan

)(tan

θθ

d

d = 1

67. If z is any complex number satisfying | z – 3 – 2i| ≤ 2, then the minimum value of | 2z – 6 + 5i| is

Ans. [5]

Sol.

A

P

2

5,–3

Min value

(3, 2) B

So, Min of |2z – 6 + 5i| = PA

= Min 22

53–

iz + = 2 ×

2

5 = 5

68. The minimum value of the sum of real numbers a–5, a–4, 3a–3, 1, a8 and a10 with a > 0 is Ans. [8] Sol. A.M. ≥ G.M.

33333333 / / / / 33333333



8

1 1083–3–3–4–5– aaaaaaa +++++++ ≥ (a–5. a–4. a–3. a–3. a–3. 1. a8 .a10)1/8

a–5 + a–4 + a–3 + a–3 + a–3 + 1 + a8 + a10 ≥ 8 so minimum value is 8

69. Let f : [1, ∞) → [2, ∞) be a differentiable function such that f(1) = 2. If 6∫x

dttf1

)( = 3x f(x) – x3 for all x ≥ 1,

then the value of f (2) is

Ans. [6]

Sol. 6∫x

dttf1

)( = 3xf(x) – x3

6 f(x) = 3f(x) + 3xf '(x) –3x2 3f(x) = 3x f '(x) – 3x2

3y = 3xdx

dy– 3x2

xdx

dy – y = x2

dx

dy –

x

y = x

I.F. = ∫ dx

xe1–

= e–ln x = x

1

y .x

1 = ∫ dx

xx

1. ⇒

x

y = x + c ⇒ y = x2 + cx

Q f(1) = 2 ⇒ c = 1

y = x2 + x f(2) = 4 + 2 = 6 * The most appropriate answer to this question is (6), but because of ambiguity in language, IIT has

declared this question as bonus (marks to all students)

Part – I : (CHEMISTRY) - JEE Main, JEE Advanced, NEET ... · PAPER-1 IITIIIITTIIT ----JEE 201...

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