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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.
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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)
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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)
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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
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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.
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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+
Paragraph for Question Nos. 15 and 16
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)
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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.
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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.
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PART - II
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.
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)
(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 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)
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.
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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
Paragraph for Question Nos. 38 and 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 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
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. 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)
(Multiple Correct Answers Type)
The 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 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)
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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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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 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
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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
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. P(white) = P (H white) + P(T white)
=
32
CCC
31
CC1
CC
21
21
521
53
21
25
1213
25
22
25
23
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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
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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
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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
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