IIT JEE 2011 Paper1 Solutions

8/7/2019 IIT JEE 2011 Paper1 Solutions

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Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Date : 11-04-2011 Duration : 3 Hours Max. Marks : 240

PAPER - 1

QUESTIONS AND SOLUTIONS OF IIT-JEE 2011

INSTRUCTIONSA. General :

1. The question paper CODE is printed on the right hand top corner of this sheet and on the back

page of this booklet.

2. No additional sheets will be provided for rough work.

3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and

electronic gadgets are NOT allowed.

4. Write your name and registration number in the space provided on the back page of this booklet.

5. The answer sheet, a machine-gradable Optical Response Sheet (ORS), is provided separately.

6. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.

7. Do not break the seals of the question-paper booklet before being instructed to do so by the

invigilators.

8. This Question Paper contains having 69 questions.

9. On breaking the seals, please check that all the questions are legible.

B. Filling the Right Part of the ORS:

10. The ORS also has a CODE printed on its Left and Right parts.

11. Make sure the CODE on the ORS is the same as that on this booklet. If the codes do not

match, ask for a change of the booklet.

12. Write your Name, Registration No. and the name of centre and sign with pen in the boxes

provided. Do not write them anywhere else. Darken the appropriate bubble UNDER each digit

of your Registration No. with a good quality HB pencil.

C. Question paper format and Marking Scheme:

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

consists of four sections.

14. In Section I (Total Marks: 21), 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.

15. In Section Il (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.

16. In Section III (Total Marks: 15), 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.

17. In Section IV (Total Marks: 28), for each question you will be awarded 4 marks if you darken

ONLY the bubble corresponding to the correct answer and zero marks otherwise. There are no

negative marks in this section.

Write your Name registration number and sign in the space provided on the back

page of this booklet.

DONOTBREAKTHESEALS

WITHOUTBEING

INSTRUCTEDTODOSOBYT

HEINVIGILATOR


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Useful Data :

R = 8.314 JK1 mol1 or 8.206 102 L atm K1 mol1

1 F = 96500 C mol1

h = 6.626 1034 Js

1 eV = 1.602 1019 J

c = 3.0 108 m s1

NA = 6.022 1023


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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.

1. Extra pure N2

can be obtained by heating

(A) NH3 with CuO (B) NH4NO3(C) (NH

4)

2Cr

2O

7(D) Ba(N

3)

2

Ans. (D)

Sol. Ba (N3)2 Ba + 3N2

2. 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. Ni+2 + 4Cl [NiCl4]2

SP3

[NiCl4]2 = 3d8 confiuration with Nickel in + 2 Oxidation state. Clbeing weak field ligand does not compel

for pairing of electrons.

So,

Hence, complex has tetrahedral geometry

Ni+2 + 4Cl [Ni(CN)4]2

[Ni(CN)4]2

= 3d8 configuration with nickel in + 2 oxidation state. CN

being strong field ligand with complesfor pairing of electrons.

So,

Hence, complex has square planar gemetry.

CHEMISTRY


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Ni+2 + 6H2O [NI(H2O)6]+2[Ni(H2O)6] = 3d

8 configuration with nickel in + 2 oxidation state. As with 3d8 configuration two d-orbitals are

not available for d2sp3 hybridisation. So, hybridisation of Ni (II) is sp3d2. As Ni (II) with six co-ordination will

have octahedral geometry.

Note : With water as ligand Ni (II) forms octahedral complexes.

3. Bombardment of aluminum by -particle leads to its artificial disintegration in two ways, (I) and (ii) asshown. Products X, Y and Z respectively are,

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

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

Ans. (A)

Sol. +

4. 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)

CHEMISTRY


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Sol. Mole =60

120= 2

mass of solution = 1120 g

V =100015.1

1120

=

115

112L

M = 112

1152

= 2.05 mol/litre

5. AgNO3

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

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

is :

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

Ans. (D)

Sol.

6. 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)

CHEMISTRY


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Sol.

Due to intramolecular hydrogen bonding in conjugate base of o-Hydroxybenzoic acid, it is strongest acid.

7. The major product of the following reaction is :

(A) (B)

(C) (D)

Ans. (A)

Sol.

CHEMISTRY


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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.

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. Important ore of tin is cassiterite (SnO2). SnO2 is reduced to metal using carbon at 1200 1300C in an

electric furnace. The product often contains traces of Fe, which is removed by blowing air through the

molten mixture to oxidise FeO. Which then floats to the surface.

SnO2 + 2C Sn + 2COFe + O2 FeO

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

(A) Adsorption is always exothermic

(B) Physisorption may transform into chemisorption at high temperature

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

temperature

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

activation.

Ans. (A, B, D)

Sol. (A) H = ve for adsorption(B) fact

(D) chemical bonds are stronger than vander waals forces so chemical adsorption is more exothermic.

10. According to kinetic theory 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, B, C, D)

Sol. (A) Fact

(B) P = MV =M

RT3M = MRT3

(C) Max well distribution

(D) Fact

CHEMISTRY


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11. 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) (B) HC

(C) H2C=C=O (D) H

2C=C=CH

2

Ans. (B, C)

Sol. In (B) and (C) CH2

== C == O all atoms are always in same plane.

SECTION Ill (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. 12 and 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

aqueous NH3dissolves O and gives an intense blue solution.

12. The metal rod M is :

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

Ans. (B)

13. The compound N is :

(A) AgNO3

(B) Zn(NO3)

2(C) Al(NO

3)

3(D) Pb(NO

3)

2

Ans. (A)

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(NH

3)

4]2+ (D) [Ag(NH

3)

2]+ and [Ni(NH

3)

6]2+

Ans. (C)

Sol. (12), (13) & (14)

Cu(M) + AgNO3 (aqueous colorless solution) Resultant solution (Cu(NO3)2 + AgNO3(N) (blue solution)

Note : Here it is considered that complete AgNO3 is not utilized in the reaction.

AgNO3 + NaCl AgCl + NaNO3(O) (white ppt.)

Solution containing white ppt. of AgCl also contains Cu(NO3)2.

CHEMISTRY


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Which developed deep blue colouration with aqueous NH3 solution

AgCl (white) + 2NH3 (aq.) [Ag(NH3)2]+

Cu(NO3)2 (aq.) + 4NH3 (aq.) [Cu(NH3)4] (NO3)2(deep blue coloration)

So, Metal rod M is Cu.

The compound N is AgNO3 and the final solution contains [Ag(NH3)2]+ and [Cu(NH3)4

2+


An acyclic hydrocarbon P, having molecular formula C6H

10, gave acetone as the only organic product

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

15. The structure of compound P is

(A) CH3CH2CH2CH2

CC

H (B) H3CH2C

CC

CH2CH3

(C) (D)

Ans. (D)

16. The structure of the compound Q is

(A) (B)

(C) (D)

Ans. (B)

Sol. (15 & 16)

CHEMISTRY


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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.

17. The difference in the oxidation numbers of the two types of sulphur atoms in Na2S

4O

6is

Ans. 5

Sol.

18. Reaction of Br2with Na

2CO

3in aqueous solution gives sodium bromide and sodium bromate with evolution

of CO2gas. The number of sodium bromide molecules involved in the balanced chemical equation is

Ans. 5

Sol. 3Br2

+ 3Na2CO

3 5NaBr + NaBrO

3+ CO

2

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.

So, electrons with spin quantum number = 2

1will be 1 + 3 + 5 = 9.

20. The work function () of some metas is listed below. The number of metals which will show photoelectriceffect when light of 300 nm wavelength falls on the metal is

Metal Li Na K Mg Cu Ag Fe Pt W

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

Ans. 4

Sol. Ephoton

=3000

12400= 4.13 ev

Photoelectric effect can take place only if Ephoton

Thus,

Li, Na, K, Mg can show photoectric effect.

CHEMISTRY


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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 0C) are introduced. Considering the ideal gas

behaviour, the total volume (in litre) of the gases at 0C is close to

Ans. 7

Sol. PHe

= 1 0.68 = 0.32 atm

V = ?

n = 0.1 ; V =P

nRT=

32.0

2730821.01.0 = 7

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

KOH is

Ans. 5

Sol.

Total 5 products.

23. 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. Molecular weight of decapeptide = 796 g/mol

Total bonds to be hydrolysed = (10 1) = 9 per molecule

Total weight of H2O added = 9 18 = 162 g/mol

Total weight of hydrolysis product = 796 + 162 = 958 g

Total weight % of glycine (given) = 47%

Total weight of glycine in product = g100

47958 = 450 g

Molecular weight of glycine = 75 g/mol

Number of glycine molecule =75

450 = 6.

CHEMISTRY


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PART - II


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. 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.

finisident

= freflected

=

10320

320

8 kHz

fobserved

=320

10320 freflected

= 8 310

330

= 8.51 kHz 8.5 kHz

25. 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. Number of moles of He =4

1

Now T1

(5.6) 1 = T2

(0.7) 1

T1 = T2

3/2

8

1

PHYSICS


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4T1

= T2

Work done = 1

]TT[nR 12

=

32

]T3[R4

11

= 1RT8

9

26. Consider an electric field xEE 0

, where E0

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

the figure) due to this field is :

(A) 2E0a2 (B) 2 E0a

2 (C) E0a2 (D)

2

aE 20

Ans. (C)

Sol.

flux = (E0

cos 45) area)

= a2a2

E0

= E0a2

PHYSICS


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

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

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

Ans. (A)

Sol.

9

1

4

1RZ

1 2H

2H= R(1)2

36

5

161

4

1RZ

1 2He

He= R(4)

16

3

27

5

36

5

3

16

4

1

2H

He

He = 27

5

6561 = 1215

28. 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 :

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

Ans. (D)

Sol. T sin = mLsin2

324 = 0.5 0.5 2

2 =5.05.0

324

=5.05.0

324

= 5.018

= 36 rad/sec.

PHYSICS


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MATHEMATICS29. A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10 ohm

resistor. The galvanometer shows 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 determined value of X is

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

Ans. (B)

Sol. 1 = 52 + 1 = 53 cm

2

= 48 + 2 = 50 cm

R

x

2

1

10

x

50

53

x = 10.6

30. A 2F capacitor is charged as shown in figure. The percentage of its

stored energy dissipated after the switch S is turned to position 2 is

(A) 0%

(B) 20%

(C) 75%

(D) 80%

Ans. (D)

Sol. Ui=

2V)2(

2

1, V

common=

5

V

Uf=

2

1(2 + 8)

2

5

V

100U

UU

i

fi

= 100V

5

VV

2

22

1005

4

= 80% Ans.


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MATHEMATICS

SECTION II (Total Marks : 16)


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


31. A spherical metal shell 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) 0EinsideA (B) QA > QB (C)A

B

B

A

R

R

(D) surfaceonB

surfaceonA EE

Ans. (A), (B), (C), (D)

Sol.

QA

+ QB

= 2Q ...(i)

B

B

A

A

R

KQ

R

KQ ...(ii)

(i) and (ii) QA = QB

B

A

RR

& QB

B

A

R

R1 = 2Q Q

B=

B

A

R

R1

Q2=

BA

B

RR

RQ2

& QA

=BA

A

RR

RQ2

QA > QB

2BB

2AA

B

A

R4/Q

R4/Q

=

A

B

R

Rusing (ii)

EA

=0

A

& EB

=0

B

A

< B

EA

< EB (at surface)


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PHYSICS32. 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 travelling along parallel paths.

(C) They will come out at the same time.

(D) They will come out at different times.

Ans. (B), (D)

Sol.

tp

=eB

m2

eBv

vm2

v

R2 ppP

te

=v

R)22( e=

eB

m)22(

eBv

vm)22( ee

te t

p


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PHYSICS

33. A composite block 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 flows only from left to right through the blocks. Then in steady state

(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 : At steady state, heat flow through A and E are same.

C : T = i R

i is same for A and E but R is smallest for E.

D : iB

=BR

T

iC

=CR

T

iD

=DR

T

if ic

= iB

+ iD

HenceDBC R

1

R

1

R

1

kA5KA3KA8


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PHYSICS

34. A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R ( B

A

< B

SECTIONIII (Total Marks :15)

(Paragraph Type)



has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


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PHYSICSParagraph for Question Nos. 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 down-

wards (or to left) is negative.

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)


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PHYSICS

Sol.

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

(A)21 E2E (B) 21 E2E (C) 21 E4E (D) 21 E16E

Ans. (C)

Sol. In 1st case amplitude of SHM is a.

In 2nd case amplitude of SHM is 2a

Total energy =2

1k(amplitude)2

E1

=2

1k(2a)2

E2

=

2

1k(a)2

21 E4E


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PHYSICSAlternative :

Linear momentum P = mv

= m 22 xA

P2 = m22 (A2 x2)

P2 + (m)2x2 = m22A2 ...(i)

Equation of circle (bigger)

P2 + x2 = (2a)2

P2 + x2 = 4a2 ...(ii)

Equation of circle (smaller)

P2 + x2 = a2 ...(iii)

Comparing (i) and (ii)

Amplitude A = 2a

and (m)2 = 1 m2 =m

1

22

)A(m2

1

So energy E1

=22 )a2(m

2

1

= )a4(m

1

2

1 2

=m

a2 2

Comparing (i) and (iii)

A = a

(m)2 = 1 m2 =m

1

So E2

=22Am

2

1 = 2a

m

1

2

1 =

2

2

m

a

2

1

m

a

2

1 2

So 4E

E

2

1 E1

= 4E2


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PHYSICS

37. Consider the spring-mass system, with the mass submerged in water,

as shown in the figure. The phase space diagram for one cycle of this

system is :

(A)Position

Momentum

(B)

(C) (D)

Ans. (B)

Sol. Linear momentum

P = mv

=22 xAm

P2 + m22x2 = m22A2

represents a circle on Px diagram with radius of circle R = A ( m22 = 1)

of spring mass system remains constant and equal tom

k


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PHYSICSAmplitude of oscillation inside liquid will decrease due to viscous force

So radius of circular arcs will decrease as position change

Correctly shown in option B


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 p, which is called the

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

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

pall 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 eand the permitlivity as 0, use dimensional analysis to determine the

correct expression for p.

(A)0m

Ne

(B) Ne

m 0(C)

0

2

m

Ne

(D) 20

Ne

m

Ans. (C)

Sol. T

1

FL

QM

QL

1

m

Ne

2

2

2

3

0

2

So only (C) is dimensionally correct


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PHYSICS

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

N 4 x 1027 m3. Take 0 10111 and m 1030, where these quantities are in proper SI units.s.

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

Ans. (B)

Sol. For resonance

= P

= 1130

21927

0

2

1010

)106.1(104

m

Ne

= 3.2 1015

f =14.32

102.3

2

15

2

1 1015

=f

c= 15

8

102

1103

600 nm

SECTION IV (Total Marks : 28)

(Integer Answers Type)

This sections 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.

40. 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


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PHYSICS

Sol.

Note : If net force applied by the rod is considered to be 2 N.

2Ff 22' ...(i)

FR f 'R = 2mR2R

a

F f ' = 2ma = 1.2 . ..(i i)From (i) & (ii)

(1.2 + f')2 + f '2 = 22

2f '2 + 2.4f ' + 1.44 = 4

f '2 + 1.2f' + 0.72 2 = 0

f '2 + 1.2f' 1.28 = 0

f ' =2

)28.1(444.12.1

= 0.6 28.136.0

= 0.6 64.1

= 0.68

From eq. (2)

F = 1.88

=88.1

68.0=

10

P P = 3.61 4 Ans.

Note : In Hindi friction force is aksed, so the answer is P = 6.8. (for Hindi)

Note : But if only normal reaction applied by the rod is considered to be 2 N.


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PHYSICS

Law 2 f = 2 [0.3]

f = 2 0.6

f - 1.4 Nx ...(i)

a = R

0.3 = [0.5]

= 53

rad/s ....(ii)

c

= c

fR 2R = mR2

f 2 = mR

1.4 2 =2

2

5

3

1.4 0.6 = 2

0.8 = 2 = 0.4 = 10

P

P = 4 Ans.

Note : In Hindi friction force is aksed, so the answer is P = 8. (for Hindi)

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

is . 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 , then N is

Sol.

F1

=2

mg

2

mg


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PHYSICS

F2

=2

mg

2

mg

F1

= 3F2

1 + = 3 3

4 = 2

=21

N = 10

N = 5 Ans.

42. Four point charges, 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 . The system of charges and planar film are in equilibrium, and

a = k

N/12q

, wherekis a constant. Then Nis

Sol. Surface Tension =length

force

2

2

2

2

a2

kq

a

kq22 = 2a

a = (Some constant)

3

12q

So N = 3 Ans.

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

The wire is cooled down from 40C to 30C to regain its original length L. The coefficient of linear thermal

expansion of the steel is 105 /C, Youngs 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.

Sol.A

F=y

L

L

)(yA

mg

m =g

)(yr

g

)(Ay 2

=

10

101010)10( 51123 = 3

Ans. 3


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PHYSICS44. 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 1025 kg. The mass (in mg) of the radioactive sample is

Sol. N = N0

et

dt

dN= 1010 = N

0() t10 9e

at (t = 0)

1010 = N0

109

N0

= 1019

mass of sample = N0

1025

= N0

(mass of the atom)

= 106 kgm

= 106 103 gm

= 103 gm

= 1 mg

Ans. 1

45. A long circular tube of length 10 m and radius 0.3 m carries a current 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 = 0

cos (300 t) where 0 is constant. If the magnetic

moment of the loop is N 0

0sin (300 t), then N is

Sol. Flux through circular ring

= (0

ni) r2

=20 r

L

0cos 300 t

i =Rdt

d

i =RL

r 02

0 . sin 300 t 300

= 00 sin 300 t

RL

300.r2


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PHYSICSM = . r2

= 0

0sin 300 t

RL

300.r42

(Take 2 = 10)

=10100

3001010 4

N = 6 Ans.

46. 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 4cm. The moment of inertia of the system about the diagonal of the square is

N 104 kg-m2, then N is

Sol.

=

2MR

5

22 + 2MxMR

5

2 22

= 2MR5

2 2

+ 2MR

5

2 2

+ (Mx2) 2

= 4

2MR5

2+ 2mx2

=2

MR5

8

+ 2mx2

=4

2

10)24()5.0(22

55.0

5

8

=

8

5

5 104

= 9 104= N 104

So, N = 9 Ans.


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MATHEMATICS

PART - III


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

(2x)n2 = (3y)n3

3nx = 2ny .

Then x0 is

(A)6

1(B)

3

1(C)

2

1(D) 6

Ans. (C)

Sol. (2x)n2 = (3y)n3

n2 n(2x) = n3 n(3y) = n3 (n3 + ny) ......... (1)

also 3nx = 2ny

nx n3 = ny n2 ......... (2)

by (1) n2 n(2x) = n3 (n3 + ny) n 2 . n (2x) = n3

2n

3nnx3n

n22 n2x = n23 (n2 + nx)

3n2n 22 (n2x) = 0

n2x = 0 x =2

1

48. The value of 3n

2n

22

2

dx)x6nsin(xsin

xsinx

is

(A)4

1n

2

3(B)

2

1n

2

3(C) n

2

3(D)

6

1n

2

3

Ans. (A)

Sol. Put x2 =t

x dx =2

dt


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I = 3n

2n)t6n(sintsin

tsin

.2

dt......(1)

apply b

adx)x(f =

b

adx)xba(f

I =2

1

3n

2ntsin)t6nsin(

)t6nsin(

dt .......(2)

adding (1) and (2)

2I =2

1

3n

2n

dt.1

I = 23

n4

1

49. Let kjia

, kjib

and kjic

be three vectors. A vector

in the plane of a

and b

, whose

projection on c

is3

1, is given by

(A) k3j3i (B) kj3i3 (C) k3ji3 (D) k3j3i

Ans. (C)

Sol. Let

= ba

= )( i + )( j + k)(

Now c. =3

1

3

)()()( =

3

1

=

= (2+ 1) i j + (2) k

For 1,

= k3jl3

MATHEMATICS


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50. Let P = { : sin cos = 2 cos } and Q = { : sin + cos = 2 sin } be two sets. Then

(A) P Q and Q P (B) Q P

(C) P Q (D) P = Q

Ans. (D)

Sol. P = {: sin cos = 2 cos }

sin = ( 2 + 1) cos

tan = 2 + 1

= n+8

3; n I

Q = {: sin + cos = 2 sin }

cos = ( 2

1) sin

tan =12

1= 2 + 1

= n+8

3; n I

P = Q

51. 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 = 41

. Then b equals

(A)4

3(B)

2

1(C)

3

1(D)

4

1

Ans. (B)

Sol. R1

= b

0

2dx)1x( =

b

0

3

3

)1x(=

3

1)1b( 3

also R2

= 1

b

2dx)1x( =

1

b

3

3

)1x(=

3

)1b( 3

R1 R

2=

3

1

3

)1b(2 3

4

1=

3

1

3

)1b(2 3 (b 1)3 =

8

1

b =2

1

MATHEMATICS


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52. Let and be the roots of x2 6x 2 = 0, with > . If an

= nn for n 1, then the value of9

810

a2

a2a

is

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

Ans. (C)

Sol. x2 6x 2 = 0 having roots and

2 6 2 = 0

10 69 28 = 0

10 28 = 69 .... (i)

similarly 10 28 = 69 .... (ii)

by (i) and (ii)

(10 10) 2(8 8) = 6 (9 9)

a10

2a8

= 6a9

9810

a2

a2a

= 3

Aliter

)(2

)(299

881010

=

)(2

)(99

881010

=

)(2

)()(99

99

=2

=

2

6= 3

53. A straight line L through the point (3, 2) is inclined at an angle 60 to the line 1yx3 . 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)

Sol. Let slope of line L = m

)3(m1

)3(m

= tan 60 = 3 m31

3m

= 3

taking positive sign, m + 3 = 3 3m

m = 0

taking negative sign

m + 3 + 3 3m = 0

m = 3

As L cuts x-axis m = 3

so L is y + 2 = 3 (x 3)

MATHEMATICS


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SECTION - II (Total Marks : 16)


The section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)


54. The vector(s) which is/are coplanar with vectors k2ji and kj2i , and perpendicular to the vector

kji is/are

(A) kj (B) ji (C) ji (D) kj

Ans. (A, D)

Sol. a = i + j + 2 k

b = i+2 j

+ k

c = i + j + k

Required vector is c

( a

b

)

[( c

.b

) a

( c

. a

) b

]

[(1+2+1) ( i + j +2 k ) (1+1+2) ( i +2 j + k ) ]

[4 j +4 k ]

so our vector in parallel j + k

55. 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 M2 N2 (MT N)1 (MN1)T is equal to

(A) M2 (B) N2 (C) M2 (D) MN

Ans. (C)

Sol. Data inconsistent

A 3 3 non-singular matrix cannot be skew-symmetricHowever considering M, N matrices as even order, we obtain correct answer.

M2 N2 (MT N)1 (MN1)T = M2N2 N1 (MT)1 (N1)T MT

M2 N2 N1 M1 N1 M

M2 NM1 N1 M

MNN1 M M2

(So the question is wrong)

MATHEMATICS


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56. 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

x2

2

y2= 1

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

(C) the eccentricity of the hyperbola is3

5

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

Ans. (B, D)

Sol. Eccentricity of ellipse =4

11 =

2

3

2

2

a

b1 =

3

2

a

b=

3

1

focus of ellipse 0,3

2

2

a

)3(= 1 a = 3

b = 1 & focus of hyperbola (2, 0)

Hence equation of hyperbola

1

y

3

x 22 = 1

57. Let f : RR 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) = kx

Hence f(x) is continuous & differentiable at x R & f(x) = k (constant)

MATHEMATICS


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J10411Page # 37

SECTION - III (Total Marks : 15)

(Paragraph Type)



has four choice (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 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)

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

a

3

+ b

1

+ c

3

is equal to

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

Ans. (A)

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

then

n

0n

11

is

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

6(D)

Ans. (B)

Sol. (Q.No. 58 to 60)

a + 8b + 7c = 0 . .. ... .. ... ( i)

9a + 2b + 3c = 0 ........... (ii)

a + b + c = 0 ........... (iii)

=111329 781 = 1.(1) 8(6) + 7(7) = 0

MATHEMATICS


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Let C =

a + 8b = 7

a + b =

b =7

6 & a =

7

(a, b, c) ,

76,

7where R

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

7

6

7

2= 1

7

= 1 = 7

7a + b +c = 7 + 6 7 = 6

59. a = 2 = 14

b = 12 & c = 14

Now cba313

=

14

122.3

13

= 3 + 1 + 32 = 3(+2) + 1 = 2

60. b = 6 = 7

a = 1 & c = 7

now ax2 + bx + c = 0 x2 + 6x 7 = 0

x = 7 , 1

n

0n

11

=

n

0n76

= 1 +7

6+

2

7

6

+ ....... =

7

61

1

= 7


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. P(white) = P (H white) + P(T white)

=

32

CCC

31

CC1

CC

21

21

521

53

21

25

1213

25

22

25

23

MATHEMATICS


39/43

J10411Page # 39

=

30

12

30

1

10

3

2

1

10

8

2

1

=30

22

2

1

10

4

=30

23

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. WhiteHead

P = )white(P

)whiteHead(P

=

30

23

2

1

5

21

5

3

2

1

=

30

2310

4

=23

12

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

63. 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

2,

2

1P

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. Then2

1

is

Ans. (2)

Sol. 2 = 21

(by property)

2

1

= 2

MATHEMATICS


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

p

1i

ia , 1 p 100.

For any integer n with 1 n 20, let m = 5n. Ifn

m

S

Sdoes not depend on n, then a2 is

Ans. (9)

Sol.n

m

S

S=

n

n5

S

S=

]d)1n(6[2

n

]d)1n5(6[2

n5

=]nd)d6[(

]nd5)d6[(5

d = 6 or d = 0 Now if d = 0 then a2 = 3 else a2 = 9

for single choice more appropriate choice is 9, but in principal, question seems to have an error.

a2 = 3 + 6 = 9

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

n

3sin

1

n

2sin

1

nsin

1is

Ans. (n = 7)

Sol.

nsin

1

n

3sin

1

=

n

2sin

1

n

3sin

nsin

nsin

n

2cos2

=n

2sin

1

sinn

4= sin

n

3

n

4= (1)k

n

3+ k , k

If k = 2m n

= 2m

n

1= 2m , not possible

If k = 2m + 1 n

7= (2m + 1)

n = 7, m = 0

Ans. n = 7

MATHEMATICS


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66. Let f : [1, ) [2, ) be a differentiable function such that f(1) = 2. If

x

1

3x)x(xf3dt)t(f6

for all x 1, then the value of f(2) is

Ans. (6)

Sol. Data inconsistent.

Putting x = 1 , in given integral equation f(1) = 1/3 , a contradiction (given that f(1) = 2).

However if considering integral equation as x

1

3x)x(xf3dt)t(f6 5

we obtain correct answer.

Differentiating the integral equation

6f(x) = 3f(x) + 3xf(x)

3x2

f(x) x

1f(x) = x

put y = f(x)

dx

dy

x

1y = x

I.F. =x

1

General solution is yx

1= x + c

Put x = 1, y = 2 c = 1

y = x2 + x

f(x) = x2 + x

f(2) = 4 + 2 = 6

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

Ans. (5)

Sol. |2z 6 + 5i| = 2

2

i53z

for minimum = 2 2

5= 5

MATHEMATICS


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68. The minimum value of the sum of real numbers a5, a4, 3a3, 1, a8 and a10 where a > 0 is

Ans. (8)

Sol. A.M. G.M.

8/1

108

33345

108

33345

a.a.1.a

1.

a

1.

a

1.

a

1.

a

1

8

aa1a

1

a

1

a

1

a

1

a

1

8/1108

345)1(8aa1

a

3

a

1

a

1

minimum value of103

345aa1

a

3

a

1

a

1 = 8, at a = 1

69. Let f() =

2cos

sintansin 1 , where

4

IIT JEE 2011 Paper1 Solutions

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