Paper - I
Code-7
Time: 3 Hours
Maximum Marks: 240
Please read the instructions carefully. You are alloted 5 minutes specifically for this purpose.
INSTRUCTIONS
A. General:
1. The question paper CODE is printed on the right hand top corner of this sheet and also on the
back page (page no. 36) 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 pages 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 THIS BOOKLET.
7. Do not break the seals of the question-paper booklet before instructed to do so by the invigilators.
8. This Question Paper contains 36 pages 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 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 mark if no bubbles are
darkened. In all other cases, minus one (-1) mark will be awarded.
15. In Section II (Total Marks : 16), for each question you will be awarded 4 marks if you darken
ALL the bubbles 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.
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-1-
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
s
n
o
i
t
a
PART I : CHEMISTRY
SECTION- I (Total Marks - 21)
n
i
m(B), (C) and (D) out
This section contains 7 multiple choice questions. Each question has four choices (A),
a
x
of which ONLY ONE is correct
E
e
v
1.
The major product of the following reaction is
i
t
ti
O
e
p
(i) KOH
C
m
NH
o
C
(ii)C
Br
CH Cl
O r
o
F
O
O
e
t
C
C
N CHitu
Br
N
CH Cl
(A)
(B)
t
C
C
s
On
O
I
YO
O
T
I C
C
N
N
AM
(C)
(D)
Single Correct Answer Type
2
2
2
O CH2
Br
CH2Cl
O
Sol: Ans [A]
CO
NH
CO
KOH
N– K+
CO
CO
CH2 – Cl
Br
e
v
i
t
i
t
e
p
m
o
C
C
N – CHr–
–Br
o
C F
Oe
t
tu by heating
i
2.
Extra pure N can be obtained
t
s (B) NH NO
n
(A) NH with CuO
(C) (NH ) Cr O
I
Y
Sol: Ans [D]T
I
M )  Ba N + 8N
A3Ba(N
n
i
am
x
E
s
n
o
i
t
a
O
2
2
3
3 2
4
3
2
3
4 2
2
7
(D) Ba(N3)2
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-2-
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
s
n
o
i
t
(A) octahedral, tetrahedral and square planar
(B) tetrahedral, square planar and octahedral
a
n
i
(C) square planar, tetrahedral and octahedral
(D) octahedral, square planar and octahedral
m
a
x
Sol: Ans [B]
E
e
[NiCl ] – sp tetrahderal (weak field ligand)
v
i
t
ti
[NiCN] – dsp square planer (strong field ligand)
e
p
[Ni(H O) ] – sp d octahedral
m
o
C
4.
Bombardment of aluminum by -particle
rleads to its artificial disintegration in two ways, (i) and (ii) as
o
shown. Products X, Y and Z respectively
(ii)
F are,
Al
P + Y
e
(A) proton, neutron, positront
tu
i
t
(B) neutron, positron,
proton
s
(i)
n
I
(C) proton, positron, neutron
Y proton, neutron
(D) positron,
T
I
Si + X
Si + Z
Sol: M
Ans [A]
A
3.
Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN– and H2O, respectively,
are
–2
3
4
–2
4
2
+2
2
3 2
6
27
13
30
14
27
13
Al + 42He
30
15
30
14
30
14
1
Si + 1H (x)
30
14
P + 0n1 (y)
s
n
o
i
t
a
(Neutron)
4
He
2
30
15
0
Si + +1e (z)
n
i
m 1.15 g/mL. The
5.
Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution ofa
density
x
molarity of the solution is
E
e
(A) 1.78 M
(B) 2.00 M
(C) 2.05 Miv
(D) 2.22 M
t
ti
Sol: Ans [C]
e
p
W  120g W
 1000g 
m
 D  1.15g ml o
M  60
C

r
o
W  1000 120  1000
F
M

M V
60  V
e
t
u 1000  120(g)
M(g)
W it  W
t
D
 1.15 s

V(ml)
V
V(ml)
n
I
1120 Y
V 
I1.15T
AMM 120 1.15 1000  2.05
(Proton)
urea
(Positron)
water
1
urea
solution
Water
Urea
s
60  1120
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-3-
Chemistry, Physics & Mathematics (Paper-I) Code-7
6.
IIT-JEE-2011
s
n
o
i
t
a
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
n ××
i
×
×
m ××
× ×
a
× × × ×××
×
×
x
×
× × ×
E
e
volume
volume
volume
v
volume
i
t
(P)
(Q)
(R)
(S)
i
t
e(R)
(A) (P)
(B) (Q)
(C)
(D) (S)
p
m
Sol: Ans [D]
o
rC
7.
Among the following compounds, theo
most
acidic is
F
(A) p-nitrophenol
(B) p-hydroxybenzoic acid
e
t
(C) o-hydroxybenzoiciacid
(D) p-toluic acid
tu
t
s
Sol: Ans [C]
In
Y
H
O
OHIT
O
COOH
C
O
AM
×
×
×
×
×
×
×
Intramoleculer H-Bonding & ortho effect
SECTION- II (Total Marks - 16)
Multiple Correct Answer Type
n
i
am
x
E
s
n
o
i
t
a
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.
e
v
i
t
i
t
e
According to kinetic theory of gases
(A) collisions are always elastic
p
m
(C) only a small number of molecules have very
ohigh velocity
C
(D) between collisions, the molecules move
r in straight lines with constant velocities
o
F
Sol: Ans [A,B,C,D]
e
t
u
t
9.
Amongst the given options,
i the compound(s) in which all the atoms are in one plane in all the possible
t
s
conformations (if n
any), is (are)
I
H
H Y H
(A)
IH TC C C CH (B) H C C – C CH (C) H C C O (D) H C C CH
AM
(B) heavier molecules transfer more momentum to the wall of the container
2
2
2
2
2
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-4-
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
Sol: Ans [A, B, C]
H
C=C=C
H
s
n
o
i
t
a
H
n
i
am
x
10. The correct statements(s) pertaining to the adsorption of a gas on a solid
surface is (are)
E
e
(A) Adsorption is always exothermic
v
i
t
i
t
(B) Physisorption may transform into chemisorption at high
temperature
e
p
(C) Physisorption increases with increasing temperature but chemisorption decreases with increasing
m
o
temperature
C
r
(D) Chemisorption is more exothermic
than
physisorption, however it is very slow due to higher energy
o
F
of activation
e
t
Sol: Ans [A,B,D]: Factual itu
t
s
11. Extraction of metal
In from the ore cassiterite involves
Yreduction of an oxide ore
(A) carbon
(B) self-reduction of a sulphide ore
T
I
(C) removal of copper impurity
(D) removal of iron impurity
M
A
Sol: Ans [A, D]: Factual
H
All the atoms are not in same plane.
s
n
o and
i
This Section contains 2 paragraphs. Based upon one of the paragraph 3 multiple choice questions
t
a
based paragraph 2 multiple choice questions have to be answered. Each of these questionsin
has four choices
(A), (B), (C) and (D) out of which ONLY ONE is correct.
am
x
E
Paragraph for Question 12 to 14 e
v of compound N, the solution
isolution
When a metal rod M is dipped into an aqueous colourless concentrated
t
i
ta white precipitate O. Addition of aqueous
turns light blue. Addition of aqueous NaCl to the blue solution gives
e
p
NH dissolves O and gives an intense blue solution.
m
o
C
12. The metal rod M is
r
(A) Fe
(B) Cu Fo
(C) Ni
(D) Co
e
t
Sol: Ans [B]
u
t
ti
13. The compound N is s
n
I
(A) AgNO
(B) Zn(NO )
(C) Al(NO )
(D) Pb(NO )
Y
IT
Sol: Ans [A]
AM
SECTION- III (Total Marks - 15)
Paragraph Type
3
3
3 2
3 3
3 2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-5-
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
s
n
(A) [Pb(NH ) ] and [CoCl ]
(B) [Al(NH ) ] and [Cu(NH ) ]
o
i
t
(C) [Ag(NH ) ] and [Cu(NH ) ]
(D) [Ag(NH ) ] and [Ni(NH ) n
] a
i
Sol: Ans [C]
am
x
E
M is dissolved in solution to give Cu Blue
e
v) ] soluble and also Cu give Blue
i
O is AgCl white ppt dissolved in aquous NH to give [Ag(NH
t
ti
e
[Cu(NH ) ] .
p
m
Paragraph for Question
Nos. 15 to 16
o
C C H , gave acetone as the only organic product through
An acyclic hydrocarbon P, having molecular formula
r
oQ is an intermediate organic compound.
the following sequence of reactions, in which
F
e
t
O
tu
i
t
C
P
Q
2
s
n
(C H
)
H
C
CH
I
Y of compound P is
T
15. The structure
I
(A) CH CH CH CH  C  C  H
(B) CH CH C  C  C  CH CH
AM
HC
HC
14.
The final solution contains
2+
2–
3 4
3+
4
3 4
+
2+
3 2
2+
3 4
+
3 4
2+
3 2
3 6
++
+
3
++
3 2
2+
3 4
6
6
3
10
2
10
(i) dil. H2SO4/HgSO4
(i) conc. H2SO4
(catalytic amount)
(–H2O)
(ii) NaBH4/ethanol
(iii) dil. acid
(ii) O3
(iii) Zn/H2O
2
3
2
3
2
2
3
3
(C) H C C C CH3
H3C
3
(D) H 3C C C
H3 C
3
C H
s
n
o
i
t
aCH
Sol: Ans [D]
CH3
CH3
(i) dil. H2SO4/HgSO4
CH3 (P)
3
CH3
(ii) NaBH 4/C2H5OH
+
dil H
e
v
i
t
i
t
e
conc. H2SO4/H2O
p
m
o
C
r
CH3 – C – CH – CH 3
(Q)
CH3 OH
o
F
e
n
i
amCH – C –C – CH
x
CH O
E
tautomerisation
CH3 – C –C = CH2
rearrangment
CH3 OH
3
2
3
CH3
CH3 – C – CH – CH 3
CH3 +OH2
CH3
CH3
+
rearrangment
+
CH3 – C – CH – CH3
CH3 – C – CH – CH3
t
u
t
i
CHst
In
3
AM
I
TY
CH3
CH3
(i) O3
CH3 – C = C – CH3
(ii) Zn/H2O
CH3
CH3
2CH3 – C = O
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-6-
Chemistry, Physics & Mathematics (Paper-I) Code-7
16.
IIT-JEE-2011
The structure of the compound Q is
H 3C
(A) H C
H 3C
H 3C
(C) H C
H 3C
OH
H3 C
H
C
3C
(B)
H3 C
C CH 2CH3
H
OH
H
OH
e
v
i
t
i
t
e
CH2CHCH3
n
i
m
OHxa
E
C CH3
s
n
o
i
t
a
(D) CH3CH2CH2CHCH 2CH 3
Sol: Ans [B]
p
m
o
IntegerC
Answer Type
r to each of the question is a single digit integer, ranging
This Section contains 7 questions. The answer
o
Fthe correct answer is to be darkened in the ORS.
from 0 to 9. The bubble corresponding to
e
t
u
t
i 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and
17. A decapeptide (Mol.tWt.
s
n contributes 47.0% to the total weight of the hydrolysed products. The number of
phenylalanine. I
Glycine
glycine units
present in the decapeptide is
Y
T
Sol. M = I
796

AM
SECTION- IV (Total Marks - 28)
hydrolysis
decapeptide 
 glycine(M  75)  ala min e  pheny
9 H2 O
glycine contribute = 47%
mass of product 796 + 9 × 18 = 958
18.
weight of glycine =
958  47
= 450
100
Moles of glycine =
450
 6.00 Ans.: 6
75
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is
O
O
+5 0 0
+5
Sol. O – S – S – S – S – O Ans.: 5
O
O
p
m
o
C
19. The total number of alkenes possible byrdehydrobromination of 3-bromo-3-cyclopentylhexane using
o
alcoholic KOH is
F
e
Sol. CH –CH –C–CH –CH –CHut
t
i
t
s
n
CH –CH=C–CHI–CH –CH
Y
T
I
M–CH –C=CH–CH –CH
ACH
3
2
3
3
2
2
2
2
3
2
3
2
3
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-7-
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
s
n
o
i
t
a
CH3–CH2–C–CH2–CH2–CH3
n
i
am
x
20. Reaction of Br with Na CO in aqueous solution gives sodium bromide
E and sodium bromate with
e
vinvolved in the balanced chemical
evolution of CO gas. The number of sodium bromide molecules
i
t
ti
equation is
e
p
Sol. Balanced equation is
m
o
3Br + 3Na CO  5NaBr + NaBrO + 3OC Ans.: 5
r
21. To an evacuated vessel with movableo
piston under external pressure of 1 atm., 0.1 mol of He and 1.0
e F pressure 0.68 atm. at 0° C) are introduced. Considering the ideal
mol of an unknown compoundt(vapour
tu (in litre) of the gases at 0°C is close to
gas behaviour, the totalivolume
t
s
Sol. Let X be the unknown
In compound.
YAt equilibrium

T
I
P  P 1 P
AMLet V be the volume at equilibium
CH3–CH2–CH–CH2–CH2–CH3
Ans.: 5
2
2
3
2
2
He

2
3
X
3
2
total
Ptotal V  (n He  n X )RT
...(i)
V  (n He  n X )RT
Now, PX V  n X RT

nX 
Px V
RT
Substituting the value of nX in (i)
P V

V   0.1  X  RT
RT 

p
m
o
C
r
V = 0.1 R + 0.68 V
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
o
F
0.1  0.082  273
e= 7L Ans.: 7
V

0.32 ut
telectrons that can have principal quantum number, n = 3, and spin quantum
i
t
22. The maximum number
of
s
number, m – / I
, isn
Yof electron in 3rd shell = 2n = 2 × 3 = 18
Sol. Total number
T
I
Total number of electron have m = – / is 9
Ans: 9
M
A

0.32 V = 0.1 RT
1
s
2
2
2
1
s
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-8-
Chemistry, Physics & Mathematics (Paper-I) Code-7
23.
IIT-JEE-2011
s
n
io
Fe
Pt at W
n 4.75
4.7
6.3
i
am
x
E
The work function (  ) 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
 (eV)
Li
2.4
Na
2.3
K
2.2
Sol. Energy of incident light in eV 
Mg
3.7
Cu
4.8
Ag
4.3
12400
12400

 4.13eV
 (in Å) 300  10
Li, Na, K and Mg will show photoelectric effect.
e
v
i
Ans.:
t4
i
t
e
p
m
o Marks: 21)
SECTION- I (Total
C
r Answer Type
Single Correct
o
This section contains 7 multiple choice F
questions. Each question has 4 choices (A), (B), (C) and (D) out of
e
t
which ONLY ONE is correct
u
t
ti
s
1
2
24. A 2 µF capacitor is
ncharged as shown in figure. The percentage
I
of its stored
energy dissipated after the switch S is turned to
s
Y
T
position
I 2 is
(A) 0 %
(B) 20 %
V
AM
PART II : PHYSICS
(C) 75%
(D) 80%
Sol: Ans [D]
C1C2 (V  0) 2 2(C1  C2 )
 100
% Energy dissipated =
1
2
CV
2 1
where C1 = 2F
C2 = 8F
= 80%
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
p
m
building which reflects the sound waves. The speed
of sound in air is 320 m/s. The frequency of the
o
C
siren heard by the car driver is
r
o
(A) 8.50 kHz
(B) 8.25F
kHz
(C) 7.75 kHz
(D) 7.50 kHz
e
t
Sol: Ans [A]
u
t
ti
Frequency heard by s
driver
n 320  10 330
vI u
f  Y
f.
 8
 8.5 kHz
= 8
vu
320  10
310
T
I
AM
25.
A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
-9-
Chemistry, Physics & Mathematics (Paper-I) Code-7
26.
IIT-JEE-2011
T1, the work done in the process is
(A)
9
RT1
8
(B)
3
RT1
2
(C)
15
RT1
8


TV–1 = const.
V 
T2  T1  1 
 V2 
 1
 5.6 
= T1  
 0.7 
1
R  4T1  T1 
9
W4
 RT1
1  5/ 3
8
5
1
3
 T1 (8)2 / 3  4T1
o
F
e
p
m
o
C
r
e
v
i
t
i
t
e
t
u
t
i area (as shown in the figure)
tshaded
The flux through the
s
due to this fieldIisn
(A) 2E aY
(B) 2E a
T
I
Ea
M
(C) E a
(D)
2
A
9
nRT
i
2
am
x
E
(D)
Sol: Ans [A]
27.
s
n
o
i
t
a
5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be

Consider an electric field E  E0 xˆ , where E0 is a constant.
2
1
z
(a,0,a)
(a,a,a)
2
0
0
2
(0,0,0)
0
2
0
(0,a,0)
x
y
Sol: Ans [C]
Angle of normal to plane with x axis = 45º
 1 ˆ 1 ˆ
i
k
Area of plane = a2 2 
2 
 2

 1 ˆ 1 ˆ
 
2
i
k  = E a2
flux through plane = E .ds = E0iˆ.(a 2) 
0
2 
 2
n
i
am
x
E
s
n
o
i
t
a
e
v
i
thelium atom is
i
of the second spectral line in the Balmer series of singly-ionized
t
e
(A) 1215 Å
(B) 1640 Å
(C)p 2430 Å
(D) 4687 Å
m
o
Sol: Ans [A]
C
r
 1
1
1  5R
o
 R(1) 



F

 (2) (3)  36 e
t
1
1  i3tRu
1
 R(2)    t


2 4 s
 4
n
I

5
Dividing, Y
IT  27
AM   275  6561Å  1215 Å
28.
The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength
2
2
2
1
2
2
2
2
2
1
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 10 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
29.
IIT-JEE-2011
s
n
o
i
t
a
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
Sol: Ans [D]
e
v
i
t
i
t
e
nL
i
am
x
E
m
p
m
o
C
30. A meter bridge is set-up as shown, to determine
r an unknown resistance ‘X’ using a standard 10 ohm
o
resistor. The galvanometer shows null
F point when tapping key is at 52 cm mark. The end-corrections are
e
t the ends A and B. The determined value of ‘X’ is
1 cm and 2 cm respectively for
u
t
ti
s
X
In
Y
T
I
A
B
M
A (A) 10.2 ohm
(B) 10.6 ohm
(C) 10.8 ohm
(D) 11.1 ohm
mr2 = 324
2
  
324
 = 36 rad/sec
0.5  0.5
Sol: Ans [B]
X
10

(52  1) (48  2)
 X 
10  53
 10.6 
50
SECTION- II (Total Marks: 16)
Multiple Correct Answers Type
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of
which ONE OR MORE may be correct.
p
m
o heat
E of different thermal conductivities (given in C
terms
rlength,
of a constant K) and sizes (given in terms
of
o
Fare of same
L) as shown in the figure. All slabs
e
tleft to right through
width. Heat ‘Q’ flows only from
u
t
ti state
the blocks. Then in steady
s
In A and E slabs are same
(A) heat flow through
Y
(B) heat
flow through slab E is maximum
T
I
(C)
temperature
difference across slab E is smallest
M
A
31.
A composite block is made of slabs A, B, C, D and
0
1L
3L
4L
1L
5L
6L
A
B
3K
E
2K
C
4K
6K
D
5K
(D) heat flow through C = heat flow through B + heat flow through D
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 11 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
Sol: Ans [A,C,D]
Let heat Q passes through slab A

Heat through E is also Q

L
RTh =
kA
RA 
1
8kt
RB 
4
3kt
RC 
1
2kt
o
F
e
p
m
o
C
r
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
4
5kt
where t is thickness of slabs
t
u
t
ti
s
T n
Using Q  I,
R
Y
ITQ = 3Q/16
AM Q = Q/2
RD 
Th
B
C
QD = 5Q/16

T = heat through slab × thermal resistance, which is minimum for slab E.
Also, QC = QB + QD
s
n
32. A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end. A thin disk of mass ‘M’ and
oradius
i
t
‘R’ (< L) is attached at its centre to the free end to the rod. Consider two ways the discais attached:
n
i
(case A) The disc is not free to rotate about its centre and (case B) the disc is free to rotate about its
mthe same displaced
afrom
centre. The rod-disc system performs SHM in vertical plane after being released
x
E
position. Which of the following statement (s) is (are) true?
e
v
(A) Restoring torque in case A = Restoring torque in case B ti
i
t
e
(B) Restoring torque in case A < Restoring torque in case
B
p
m
(C) Angular frequency for case A > Angular frequency for case B
o
C
(D) Angular frequency for case A < Angular frequency for case B
r
o
F
Sol: Ans [A,D]
e
tA writing the mechanical energy in the general form we get
For the case
u
t
ti

1s mL MR
L

 ML    mg cos   MgL cos   E
n

I 2 3 2  2
Y
IT Differentiating the above equation with respect to time we get
AM
2
2
2
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 12 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
=
IIT-JEE-2011
d
mgL / 2  MgL


dt
 mL2 MR 2
2
 3  2  ML 


n
i
am
x
E
s
n
o
i
t
a
Comparing with the standard equation of SHM we get angular frequency of the above SHM as
mgL / 2  MgL
1 =  mL2 MR 2
2
 3  2  ML 


e
v
i
t
i
t
e
p
m
L
1 mL
1
 MgL cos   E '
  M ( L ) mg cos o
2 C
2 3
2
r
now differentiating the above equationowith respect to time
F
e
d
mgL / 2 
MgL
t 

 =
u
mL
1
dt
t
ti6  2 ML
s
n
now comparingIwith the standard equation of SHM we get angular frequency of the above SHM as
Y mgL / 2  MgL
T
I  mL 1 
AM  =  6  2 ML 
now writing the mechanical energy for case B in the general equation we get
2
2
2
2
2
2
2
2
Clearly 2 > 1
s
n
o
i
infinite region of uniform magnetic field perpendicular to the velocity. Which of the followingtstatement
na
(s) is/are true?
i
(A) They will never come out of the magnetic field region
am
x
(B) They will come out travelling along parallel paths
E
e
(C) They will come out at the same time
iv
t
i
t
(D) They will come out at different times.
e
p
m
Sol: Ans [B,D]
o
C
 mv 
r
r 

  rm
o
 qB 
× × × ×
F
e
× × × ×
t

r
<r
u
t electron and proton both
× × × ×
Also radius is very smallifor
t
s as shown in the figure
× × × ×

Path should
be
n
I
T
m
Y
Also, t  
IT 2 qB
AorM t  m
33.
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-
electron
proton
Proton
Electron
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 13 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
34.
IIT-JEE-2011
s
n
o
i
t
a
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
n
i
(D) E
E
am
x
E
e
Sol: Ans [A,B,C,D]
vQ and Q respectively
i
t
Let final charges on spherical shell A and solid metal sphere B
are
i
t
e
kQ
kQ
p

R
R
m
o
C
Q
R
r


...(i)
o
Q
R
F

Q >Q
e
t
u
E
 0 (inside uniformly
charged spherical shell electric field is zero)
t
i
t
s
Q
  In

4R
Y
T  Q / 4R = Q   R  = R /R
 I

Q / 4R
Q R 
M
A E = kQ
(A) E Ainside  0
(B) QA  QB
 A RB
(C)   R
B
A
on surface
A
on surface
B
A
A
B
B
A
B
A
A
B
B
A
B
inside
A
2
surface

A
A
B
B
2
A
2
B
2
A
B
B
A
B
A
R2
EA
Q
(on surface)  A
EB
QB
s
n
o
i
t
a
2
R 
R
 B   B
RA
 RA 
n
i
m
Paragraph Type
a
x
This Section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple
choice questions and
E
e Each of these questions has
based on the other paragraph 2 multiple choice questions have to be answered.
v
i
four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
t
i
t
e
p
Paragraph for Question
35 to 37
m
o
Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially
C
useful in studying the changes in motion
as initial position and momentum are changed. Here we
r
o in one-dimension. For such systems, phase space is a plane in
consider some simple dynamical systems
F
e axis and momentum is plotted along vertical axis. The phase
which position is plotted alongthorizontal
space diagram is x(t) vs.ip(t)
tucurve in this plane. The arrow on the curve indicates the time flow.
t
s space diagram for a
For example, the phase
n
particle movingIwith constant velocity is a
straight lineY
as shown in the figure. We use the
T
I
sign convention in which position or momentum
M (or to right) is positive and downwards
Aupwards
(or to left) is negative.
Momentum
SECTION- III (Total Marks: 15)
Position
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 14 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
35.
IIT-JEE-2011
The phase space diagram for a ball thrown vertically up from ground is
Momentum
n
i
am
x
E
s
n
o
i
t
a
Momentum
(A)
(B)
Position
Momentum
o
F
e
Position
t
tu
i
t
s
In
(C)
Y
T
I
p
m
o
C
r
e
v
i
t
i
t
e
Position
Momentum
(D)
Position
Sol: Ans [D]
v = u – 2g s
M
A m v = m u – 2gm s
2
2
2 2
or
2 2
2
p2 = p02 – k s
2
where k = 2m2s
Momentum
2
p = –k(s – p0 /k)
36.
The phase space diagram for simple harmonic
E1
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
p
m
o
(C) E = 4 E
(D) E = 16 E C
r
o
Sol. [C]
F
e
t
P
1
u
E  ME  mv 
t
2
2tmi
s
n
E
I

4

E Y
T
I
AM
(A) E1 = 2 E2
1
E2
e
v
i
t
i
t
e
2a
a
Position
(B) E1 = 2E2
2
1
2
max
n
i
am
x
E
s
n
o
i
t
a
2
2
max
1
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 15 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
37.
IIT-JEE-2011
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
n
i
am
x
E
Momentum
(A)
Momentum
Position
Momentum
(C)
Y
T
I
s
n
o
i
t
a
o
F
e
t
u
t
ti Position
s
In
e
v
i
t
(B) ti
e
p
m
o
Momentum
rC
Position
Position
(D)
AM
Taking equilibrium position to be reference for displacement
Sol. [B]
Paragraph for Question Nos. 38 to 39
s
n
o
i
t
a
A dense collection of equal number of electron and positive ions is called neutral plasma. Certain solids
containing fixed positive ions surrounded by free electron 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 p, all the
free electrons are set to resonance together and all the energy is reflected. This is the explanation of high
reflectivity of metals.
e
v
i
t
i
t
e
n
i
am
x
E
p
m
o as  , use dimensional analysis to determine the
38. Taking the electronic charge as ‘e’ and the permittivity
C
correct expression for  .
r
o
Ne
Ne
mF

m
e
(A)
(B)
(C)
(D)
t
m
m
Ne
u Ne
t
i
Sol. [C]
t
s
n
(N) = [L ]
I
(e ) = [A TY
]
T
(m) =I[M]
1
AM
 ML T A

0
p
2
o
o
o
2
o
–3
2
2 2
3
4
2
0
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 16 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
39.
IIT-JEE-2011
s
n
o
i
t
(D) 200a
n nm
i
am
x
E
Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons
N  4  1027 m3 . Take  o  1011 and m  1030 , where these quantities are in proper SI units.
(A) 800 nm
(B) 600 nm
(C) 300 nm
Sol: Ans [B]
wP 
4  1027  (1.6  1019 ) 2
 3.2  1015 rad / sec
1030  1011
3  108  2
 600 nm
3.2  1015
e
v
i
t
i
t
e
p
m
o
SECTION-C
IV (Total Marks: 28)
r ANSWER TYPE
o
INTEGER
F
This Section contains 7 questions. The
answer to each of the question is a single digit integer ranging from
e
t
u
0 to 9. The bubble corresponding
to the correct answer is to be darkened in the ORS.
t
ti
s
Ineach of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners
40. Four solid spheres
Yof side 4 cm. The moment of inertia of the system about the diagonal of the square is
of a square
T
N ×I
10 kg-m , then N is
M
A 2
 d 
 2 = 9  10 kg  m
Sol.: I = mr  4  m

–4
2
2
2

4


 2
5
2
s
n
o
i
t
a
N=9
Ans. N  9
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
e
v
i
and the ring is large enough that rolling always occurs andtit
the coefficient of friction between the stick and the ringp
ise
m
(P/10). The value of P is
o
C
Sol. Free body diagram of ring
r
o
Given, a = 0.3 m/sec
F
e
2 – f = 2 × 0.3  f = 1.4 N
t
tu
i
I .a
t
Also, f R – f R = I s=
In R
Y Ia  f = 0.8 N
f f 

T
I
R
4
AM
nStick
i
am
x
E
of 0.3 m/s2. The coefficient of friction between the ground
Ground
2
1
1
1
2
1
2
2
F = 0.8   = 0.4 =
2
10
 P=4
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 17 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
42.
IIT-JEE-2011
A long circular tube of length 10 m and radius 0.3 m 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  I 0 cos(300t ) where I0 is constant. If the magnetic moment of
the loop is N  o I o sin(300t ) , then N is
Sol. Magnetic field due to cylinderical tube inside it
B
0 Ni  0 I

L
L
where I is total surface current
 = (B.A)
d
dB
=A
dt
dt
AM

dB
dt
Y
T
I
|emf| = A
t
u
t
ti
s
In
o
F
e
A. A
Magnetic moment =
p
m
o
C
r
e
v
i
t
i
t
e
n
i
am I
x
E
s
n
o
i
t
a
dB
dt
R0
R0  resistance of coil
 2 r 4  0 I 0  300
.
=
sin (300 t)
R0
L
n
i
am
x
E
= 6 µ0 i0 sin (300 t)
s
n
o
i
t
a
e
v
i
43. A block is moving on an inclined plane making an angle 45° i
with
t the horizontal and the coefficient of
t
e plane is 3 times the force required to just
friction is . The force required to just push it up the inclined
p
prevent it from sliding down. If we define N = 10 ,
mthen N is
o
Sol. mg sin 45° +  mg cos 45° = 3[mg sin 45° –C
r  mg cos 45°]
o

 = 0.5
F
e

N = 10  = 10 × 0.5u
=t
5
t
i
t
Ans.  5
s
n
I
44. The activityY
of a freshly prepared radioactive sample is 10 disintegration per second, whose mean life
is 10 I
s.T
The mass of an atom of this radioisotope is 10 kg. The mass (in mg) of the radioactive sample
AisM
Ans.: 6
10
9
–25
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 18 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
Sol. N0 = 1010 dp s
Average life =
1
 109 s

–9
10
19

10 N0 = 10
 N0 = 10

Mass of radioactive sample = 10–25 × 1019 kg = 1 mg
n
i
am
x
E
s
n
o
i
t
a
e
v
i
t
i
t
e corners of a square planar soap film of side
45. Four point charges, each of +q, are rigidly fixed at the four
p
‘a’. The surface tension of the soap film is . The m
system of charges and planar film are in equilibrium,
o
C N is
q 
r
and a  k   , where k is a constant.
Then
o
 
F
e
Sol. Net electrostatic force on onetcharge
tu
i
t
1 q s
1

. n  2 
=
4 Ia  2

Y
Considering
IT MN as system,
AM 41 . qa  12  2   2cos 45  2a
Ans.  1
2
1/ N
2
2
0
2
2
0

 q2 
ak 
r 
s
n
o
i
t
a
1/ 3
n
i
m from its free end.
46. Steel wire of length L at 40°C is suspended from the ceiling and then a mass m is
hung
a
x
The wire is cooled down from 40°C to 30°C to regain its original length L. The coefficient of linear
EN/m and radius of the wire is
e
thermal expansion of the steel is 10 /°C, Young’s modulus of steel is 10
imvin kg is nearly
t
1 mm. Assume that L >> diameter of the wire. Then the valueiof
t
e
p
 1  10   10
Sol. mg 
 l 
m
L
o
C
r
mgL

l =
....(i)Fo
  10
e
t
Also l = L × 10 × 10 tu ....(ii)
i
t
s
Comparing (i) andn
(ii)
I

m =Y
 = 3.14
IT
Ans. 3
AM

N =3
–5
3 2
11
2
11
5
–5
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 19 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
PART III : MATHEMATICS
SECTION- I (Total Marks : 21)
Single Correct Choice Type
n
i
am
x
E
s
n
o
i
t
a
This section contains 7 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct
47.
e
v
i
t
i
t
e
A straight line L through the point (3, –2) is inclined at an angle 60° to the line
intersects the x-axis, then the equation of L is
p
(B)
m
o (D)
C
r
(A) y  3x  2  3 3  0
(C)
3y  x  3  2 3  0
y  3x  2  3 3  0
3y  x  3  2 3  0
o
F
e3
Slope of 3x  y  1 is m = t
u
So slope of required lineit
is tan (–120°) = 3
t
Equation is y + 2n
= s 3 (x –3)
I
 y  3x  3 3  2  0
Y
T
I , y ) be the solution of the following equations (2x)
48. Let (x
AM
1
1
1
(A)
(B)
(C)
Sol: Ans [B]
ln2
0
0
6
3
3x  y  1. If L also
= (3y)ln3 3lnx = 2lny. Then x0 is
(D) 6
2
Sol: Ans [C]
ln 2
(2x)
ln 3
= (3y)
....(i)
3ln x = 2ln y
....(ii)
taking log :
ln 2 ln 2 x = ln 3 ln 3y ....(iii)
ln 3 ln x = ln y ln 2
....(iv)
1
solving (iii) and (iv) we get x0 =
2
ln 3
49.
The value of

ln 2
(A)
1 3
ln
4 2
p
m
o
C
r
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
x sin x 2
dx is
sin x 2  sin(ln 6  x 2 )
(B)
o
F
e
1 3
ln
2 2
t
u
t
Sol: Ans [A]
ti
s
n dt we get
Puting x = t andIx dx =
2
Y
T sin t dt  1  (ln 3  ln 2)
1 I
 sin t  sin(ln 6  t ) 4
M
2
A
(C) ln
3
2
(D)
1 3
ln
6 2
2
ln 3
ln 2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 20 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
50.
IIT-JEE-2011
       
   



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,

1
, is given by
whose projection on c is
3






 
 


(A) i  3 j  3k
(B) 3i  3 j  k
(C) 3i  j  3k
(D) i  3 j  3k
Sol: Ans [C]
e
v
i
t
i
t
e
[iˆ  ˆj  kˆ ] 1
[(iˆ  ˆj  kˆ )   (i  j  k )].

3
3
 = 2
n
i
am
x
E
s
n
o
i
t
a
p
m

o
C
51. Let P  { : sin   cos   2 cos } andrQ  { : sin   cos   2 sin } be two sets. Then
o
F
(A) P  Q and Q  P   e
(B) Q  P
t
(C) P  Q
(D) P = Q
tu
i
t
s
Sol: Ans [D]
In
sin  – cosY
 = 2 cos 
T
 Itan  = 2 + 1
and sin  + cos  = 2 sin 
AM
The vector is 3iˆ  ˆj  3kˆ
 tan  =
2 +1
s
n
o parts
i
52. Let the straight line x = b divide the area enclosed by y = (1 – x) , y = 0, and x = 0 intottwo
a
n
1
i
R (0  x  b) and R (b  x  1) such that R  R  . Then b equals
m
4
a
x
3
1
1
1
E
(A)
(B)
(C)
(D)
4
2
3
e
4
v
i
t
i
Sol: Ans [B]
t
e
p
According to the question
m
o
1
C
 (1  x) dx   (1  x ) dx  4
r
o
F
1
1e
1
t
 (1 – b) =  (1 – b)u=
b=
t
8
2
2
i
t
s of x – 6x – 2 = 0, with  > . If a =  –  for n  1, then the value of
nroots
53. Let  and  be I
the
a  2a Y
is
2aIT
(B) 2
(C) 3
(D) 4
A(A)M1

P = Q hence (D)
2
1
1
2
b
2
1
2
2
0
b
3
2
n
n
n
10
8
9
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 21 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
Sol: Ans [C]
(9 – 9) ( + ) = 10 – 10 +  (8 – 8)

a9 () = a10 +  . a8

6a9 = a10 –2a8
n
i
am
x
E
a10  2a8
3
2a9
s
n
o
i
t
a
e
v
i
SECTION- II (Total Marks i: t
16)
t
e Type
Multiple Correct Answers
p
mquestion has 4 choices (A), (B), (C) and (D) out of
This section contains 4 multiple choice questions. Each
o
C
which ONE OR MORE may be correct.
r
o
F
e x  y  1 be reciprocal to that of the ellipse x + 4y = 4. If the
t
54. Let the eccentricity of the hyperbola
tu a b
i
t
s a focus of the ellipse, then
hyperbola passes through
In
x
y
Y
(A) the equation
of the hyperbola is

 1 (B) a focus of the hyperbola is (2, 0)
3
2
T
I
5
(C) the eccentricity of the hyperbola is
(D) the equation of the hyperbola is x – 3y = 3
AM
3

2
2
2
2
2
2
2
2
2
Sol: Ans [B, D]
e=

b2
2
1 2 
a
3
3b2 = a2 and also hyperbola passes through
( 3,0) 

55.
3
 1  a2 = 3 and b2 =1.
a2
e
v
i
t
i
t
e
n
i
am
x
E
2
s
n
o
i
t
a
x2 y2

 1 is equation of the hyperbola and its focus is (2, 0)
3
1
p
m
o
C
r
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
o
F
(B) f (x) is continuous x  R e
uR t
(C) f (x) is constant xi
t
t except at finitely many points
s
(D) f (x) is differentiable
In
Sol: Ans [B, C]Y
IT
f (x + y) = f (x) + f (y)
AMf (x) = kx
(A) f (x) is differentiable only in a finite interval containing zero
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 22 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
56.
IIT-JEE-2011
s
n
o
i
t
(D) MNa
n
i
am
x
E
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 M 2 N 2 ( M T N ) 1 ( MN 1 )T is equal to
(A) M2
(B) –N2
(C) –M2
Sol: Ans [C]
M2N2 (MTN)–1 (MN–1)T

M2N2N–1 (MT)–1 (N–1)T MT

M2N(N–1 M–1)T MT

M2N(M–1 N–1)T MT

M2N(N–1 )T(M–1)T MT

M2 (–N)T(N–1 )T
o
F
e
p
m
o
C
r
e
v
i
t
i
t
e
t
u
t
ti coplanar with vectors i  j  2k and i  2 j  k, and perpendicular to the
57. The vector(s) whichsis/are
  n
vector i  j  kI is/are
 
 
 Y
 
(A) IjT
(B) i  j
(C) i  j
(D)  j  k
k
Sol:
Ans [A,D]
AM

–M2
[(iˆ  ˆj  2kˆ )  (iˆ  2 ˆj  kˆ )] . [i + 2j + k] = 0

 = –1

± (  ˆj  kˆ ) are vectors
s
n
o
i
t
a
n
i
Paragraph Type
am
x
This Section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple
choice questions and
E
e Each of these questions has
based on the other paragraph 2 multiple choice questions have to be answered.
v
i
t
four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
i
t
e
p
Paragraph for Question
m 58 to 60
o
Let a, b and c be three real numbers satisfying C
r
1 9 o7 
8 F
[ a b c]e
2 7   [0 0 0]
(E)

t
u 7 3 7
t
i
t
s
58. If the point P(a,
Inb, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of
7a + b + c Y
is
T
(A) 0I
(B) 12
(C) 7
(D) 6
M
A
SECTION- III (Total Marks : 15)
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 23 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
s
n
o
i
t
a
Sol: Ans [D]

a+b+c=0
n
i
 7a + b + c = 6
am
x
59. Let  be a solution of x – 1 = 0 with Im() > 0. If a = 2 with b and c satisfying
(E), then the value of
E
e
v
i
3
1
3
t
i


is equaltto
  
e
p
(A) –2
(B) 2
(D) –3
m(C) 3
o
C
Sol: Ans [A]
r
o
a + 8 b + 7c = 0
....(i)
F
e ....(ii)
t
9a + 2 b + 3c = 0
tu ....(iii)
a + b + c = 0 ti
s
Equation (i), Equation
In (iii) gives 7b + 6c = 0
Y
Equation
(ii), Equation (iii) gives 7a+ c = 0
T
Ia b c
 

AM
1 6 7
2a + b + c = 1  a = 1
3
a
60.

b = 12 and c = –14

3
1
3
 b  c  2
a
  
b
c
n
i
am
x
E
s
n
o
i
t
a
Let b = 6, with a and c satisfying (E). If  and  are the roots of the quadratic equation

2
ax + bx + c = 0, then

n0
(A) 6
 1 1
  
 
n
is
(B) 7
Sol: Ans [B]

n

 1 1
 b 

       c 
n 0 
 n 0  
t
u
t
ti
s
In
=
c
7

c  b 7  6
=
7
I
TY
n
o
F
e
e
v
i
t
i
t
e
6
(C)
7
p
m
o
C
r
(D) 
{If b = 6 and c = –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
M
A
appears then 2 balls are drawn at random from U and put into U . Now 1 ball is drawn at random from U .
fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail
1
2
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 24 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
61.
IIT-JEE-2011
The probability of the drawn ball from U2 being white is
(A)
13
30
(B)
23
30
19
30
(C)
Sol: Ans [B]
Prob. =
=
3
1  3 C2
C12C1 2 2 C2 1 
1 3
2 1

1

 5  

1



×
5C2
2  5 C2
3
C2 3 
2  5
5 2 
e
v
i
t
i
t
e
23
30
s
n
11
io
(D)
30 at
n
i
am
x
E
p
m
62. Given that the drawn ball from U is white, theo
probability that head appeared on the coin is
C
17
11
15
12
r
(A)
(B)
(C)
(D)
o
23
23F
23
23
e
t
Sol: Ans [D]
u
t
i
2 st 2
n 12
Prob.
= I5  5 
2 11 23 23
Y

5 30 30
T
I
SECTION- IV (Total Marks : 28)
AM
2
Integer Answer Type
This Section contains 7 questions. The answer to each of the 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.
63.
n
i
am
x
E
 1  sin   


Let f ()  sin  tan 
  , where  4    4 . Then the value of
 cos 2  

Sol. f ()
 1 sin  
= sin  tan

cos2 

p
m
o
C
r
 1  sin   
= sin  sin 
  = tan 
 cos   

d
(tan ) = 1
d (tan )
e
v
i
t
i
t
e
s
n
o
i
t
a
d
( f ()) is
d (tan )
o
F
e  be the area of the triangle formed by the end points of its latus
64. Consider the parabola y = 8x.tLet
u
i1t 
t
rectum and the pointsP  , 2  on the parabola, and  be the area of the triangle formed by drawing
In  2 

tangentsT
atY
P and at the end points of the latus rectum. Then
is

I
AM
2
1
2
1
2
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 25 -
Chemistry, Physics & Mathematics (Paper-I) Code-7
IIT-JEE-2011
1

8 2  
2  83
Sol. 1 = 

6
2
4
2 = 3
1
so   2.
2
e
v
i
t
i
t
e
n
i
am
x
E
s
n
o
i
t
a
65.
The minimum value of the sum of real numbers a–5, a–4, 3a–3, 1, a8 and a10 with a > 0 is
Sol.
a 5  a 4  3  a 3  1  a 8  a10
1
8
 ‘8’ is answer
66.
o
F
e
p
m
o
C
r
p
Let a1, a2, a3, …, a100 be an arithmetic progression with a1 = 3 and S p   ai , 1  p  100. For any
t
u
t
S
i
integer n with 1  nst20, let m = 5n. If
S
In
m Y
[6  ( m  1)d ]
S IT
2
Sol. M
n
S
A 2 [6  (n  1)d ]
i 1
m
does not depend on n, then a2 is
n
m
n
5[6  (5n  1)d ]
[6  ( n  1) d ]
s
n
o
 d = 6 and a = 3 + 6 = 9
i
t
67. If z is any complex number satisfying |z – 3 – 2i|  2, then the minimum value of |2z –n
6 +a5i| is
i
m
5i 

a
Sol. |2z – 6 + 5i| = 2 |z –  3   |
x
2

E
e
It is nearest to the point (3, 0) on the circle |z –3 – 2i| = 2
v
i
t
i
5i
t
so |z| = 2  3  3 
=5
e
2
p
m
o
1
1
1


68. The positive integer value of n > 3 satisfyingC
the equation
is
 
 2 
 3 
r
sin
sin
sin
 
 
 
o
n
 n 
 n 
F
e
t
1
1
1
u
t


Sol.

2
3 i
sin
sin
sin st
n
n
In n
Y

3

2
2
3
sin

T
I n sin n  sin n sin n  sin n sin n
AM 3 4
=
2
min

sin
n
 sin
n
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 26 -
Chemistry, Physics & Mathematics (Paper-I) Code-7

3
4
= r + (–1)r
n
n

n=7
IIT-JEE-2011
s
n
o
i
t
a
n
i
m
69. Let f : [1, )  [2, ) be a differentiable function such that f (1) = 2. If 6af (t )dt  3xf ( x)  x
x
E
for all x  1, then the value of f (2) is
e
v
i
t
ti
Sol. 6  f (t )dt  3xf ( x )  x
e
p
m
differentiating w.r.t. ‘x’
o
C
6 f (x) = 3f (x) + 3x f ´(x) –3 x
r

f (x) = x f ´(x) – x
o
F
1te

Integrating factor = e
u x
t
i
st
y n
Solution is
I= x + c [ f (1) = 2]  c = 1
x
Y
T
I xy = x + 1

AM

f (2) = 6
x
3
1
x
3
1
2
2
1
 dx
x

s
n
o
i
t
1.
[A]
2.
[D]
3.
[B]
4.
[A]
5.
[C]
a
n
i
6.
[D]
7.
[C]
8.
[A,B,C,D] 9.
[A, B, C]
10. [A,B,D]
m
a 15. [D]
11. [A, D]
12. [B]
13. [A]
14. [C]
x
16. [B]
17. [6]
18. [5]
19. [5] E
20. [5]
e
21. [7]
22. [9]
23. [4]
iv
t
i
t
Physics
e
p 27. [D]
24. [D]
25. [A]
26. [A]
28. [A]
m
o
29. [D]
30. [B]
31. [A,C,D]
32. [A,D]
33. [B,D]
C
34. [A,B,C,D] 35. [D]
36.
37. [B]
38. [C]
r [C]
o
F 41. [4]
39. [B]
40. [9]
42. [6]
43. [5]
e
44. [1]
45. u
[3]t
46. [3]
t
i
Mathematics
t
s
47. [B] In 48. [C]
49. [A]
50. [C]
51. [D]
52. [B]
Y 53. [C] 54. [B, D] 55. [B, C] 56. [C]
T
57.I [A,D]
58. [D]
59. [A]
60. [B]
61. [B]
[D]
63. [1]
64. [2]
65. [8]
66. [9]
AM62.
67. [5]
68. [7]
69. [6]
Answers
Chemistry
Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42
- 27 -