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Sri Chaitanya IIT Academy., India
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
INSTRUCTIONS
A. General
1. This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by the
invigilators.
2. The question paper CODE is printed on the left hand top corner of this sheet and on the back cover page of this booklet.
3. Blank spaces and black pages are provided in the question paper for your rough work. No additional sheets will be
provided for rough work.
4. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadget of
any kind are NOT allowed inside the examination hall.
5. Write your Name and Roll number in the space provided on the back cover of this booklet.
6. Answers to the questions and personal details are to be filled on an Optical Response Sheet, which is provided
separately. The ORS is a doublet of two sheets-upper and lower, having identical layout. The upper sheet is a machine-
gradable Objective Response Sheet (ORS) which will be collected by the invigilator at the end of the examination. The
upper sheet is designed in such a way that darkening the bubble with a ball point pen will leave an identical impression at
the corresponding place on the lower sheet. You will be allowed to take away the lower sheet at the end of the
examination. (see Figure-1 on the back cover page for the correct way of darkening the bubbles for valid answers).
7. Use a black ball point pen only to darken the bubbles on the upper original sheet. Apply sufficient pressure so that
the impression is created on the lower sheet. See Figure-1 on the back cover page for appropriate way of darkening the
bubbles for valid answers.
8. DO NOT TAMPER WITH /MUTILATE THE ORS OR THIS BOOKLET.
9. On breaking the seal of the booklet check that it contains 28 pages and all the 60 questions and corresponding answer
choices are legible. Read carefully the instruction printed at the beginning of each section.
B. Filling the right part of the ORS
10. The ORS also has a CODE printed on its left and right parts.
11. Verify that the CODE printed on the ORS (on both the left and right parts) is the same as that on this booklet and put
your signature in the Box designated as R4.
12. IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKET / ORS AS APPLICABLE.
13. Write your Name, Roll No. and the name of centre and sign with pen in the boxes provided on the upper sheet of ORS.
Do not write any of this anywhere else. Darken the appropriate bubble UNDER each digit of your Roll No. in such
way that the impression is created on the bottom sheet. (see example in Figure 2 on the back cover)
C. Question Paper Format
The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections.
14. Section 1contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which
ONE is correct.
15. Section 2 contains 3 paragraphs each describing theory, experiment and data etc. Six questions relate to three paragraphs with two questions on each paragraph. Each question pertaining to a particular passage should have only one correct answer among the four given choices (A), (B), (C) and (D).
16. Section 3 contains 4 multiple choice questions. Each question has two lists (List-1: P,Q,R and S; List-2: 1,2,3 and
4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE is correct.
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PART I : PHYSICS SECTION – 1
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A),(B),(C),and
(D) out of which ONLY ONE option is correct
1. If Cu is the wavelength of K X ray line of copper (atomic number 29) and Mo is
the wavelength of the K X ray line of molybdenum (atomic number 42), then the
ratio /Cu Mo is close to
A) 1.99
B) 2.14
C) 0.50
D) 0.48
Key: B
Sol: 2
1(z 1)
2 2
Cu MO
Mo Cu
Z 1 41 2.14Z 1 28
2. A metal surface is illuminated by light of two different wavelengths 248 nm and 310
nm . The maximum speeds of the photoelectrons corresponding to these wavelengths
are 1u and 2 ,u respectively. If the ratio 1 2: 2 :1u u and 1240hc eV ,nm the work
function of the metal is nearly
A) 3.7eV
B) 3.2eV
C) 2.8eV
D) 2.5eV
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Key: A
Sol: 21 0
1
1 cmu ..................(1)2
22 0
2
1 Cmu ..................(2)2
2 0
0 0
0
1240(1) 2 248 (4 )4 51240(2) 1
310
0 3.7eV
3. Parallel rays of light of intensity 2912I Wm are incident on a spherical black body
kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant 8 2 45.7 10 Wm K and assume that the energy exchange with the surroundings is
only through radiation. The final steady state temperature of the black body is close to
A) 330 K
B) 660 K
C) 990 K
D) 1550 K
Key: A
Sol:
RR
2 4 40I( R ) Ae(T T )
2 2 4 4
0I( R ) (4 R )(1)(T T )
4 40
IT T4
8
8 8912 10 81 10 121 104 5.67
1/4 2T (121) 10 330K
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4. During an experiment with a metre bridge, the galvanometer shows a null point when
the jockey is pressed at 40.0 cm using a standard resistance of 90 , as shown in the
figure. The least count of the scale used in the metre bridge is 1 mm. The unknown
resistance is
R 90
40.0cm
A) 60 0.15
B) 135 0.56
C) 60 0.25
D) 135 0.23
Key: C
Sol: R 40R 90 6090 100 60
C) Again, (90)R nR n90 n n(100 )100
dR d d0R 100
1 1 1 1 1dR (R)(d ) 60 0.1 0.25100 40 60 4
R 60 0.25
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5. A wire, which passes through the hole in a small bead, is bent in the form of quarter of
a circle. The wire is fixed vertically on ground as shown in the figure. The bead is
released from near the top of the wire and it slides along the wire without friction. As
the bead moves from A to B, the force it applies on the wire is
090
A
B
A) always radially outwards
B) always radially inwards
C) radially outwards initially and radially inwards later
D) radially inwards initially and radially outwards later
Key: D
Sol:
00 mg
mgcos
2mvrP
At P, let 2
0mvmg cos
r
2
0v g cos ...................(1)r
And 20
1mgr(1 cos ) mv2
2
0v 2g(1 cos ).....................(2)r
02(1) (2) cos3
At this angle 0 , force between wire and bead is zero
2
0mvmg cos
r force by the wire on the bead will be outwards and by the bead on the
wire will be radially inwards.
2
0mvmg cos
r , force by the bead on the wire will be radially outwards.
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6. A planet of radius 110
R (radius of Earth) has the same mass density as Earth.
Scientists dig a well of depth 5R on it and lower a wire of the same length and of
linear mass density 3 110 kgm into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth
66 10 m and the acceleration due to gravity on Earth is 210ms ) A) 96 N B) 108 N C) 120 N D) 150 N Key: B
Sol:
Ry
dy
C
gp
planet
On the surface, 4g GR3
p p p 1p e
e e e
g R R 1g g 10 1msg R R 10
All a depth ‘y’ from the surface of planet,
1p
yg g 1R
Gravitational force on element of length ‘dy’ is 1p
ydF g (dm) (dy) 1 gR
R /5R/5 2
p p0 0
y yF dF g 1 dy g yR 2R
2
pR 1 R
g5 2R 5
3 6
2p p
R 1 R 9 10 6 10 5.4g 1 g 0.9 10 108N5 10 5 10 5 10 5
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7. Charges , 2Q Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii / 2R , R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are 1 2,E E and 3E respectively, then
R
2RQ
Sphere 1 Sphere 2 Sphere 3
2Q
R4Q
PR
2R
P P
A) 1 2 3E E E B) 3 1 2E E E C) 2 1 3E E E D) 3 2 1E E E Key: C Sol:
1 2 12 2
0 0
1 Q 1 2QE , E 2E4 R 4 R
3 130
1 4Q 1E R E4 (2R) 2
2 1 3E E E
8. A glass – capillary tube is of the shape of a truncated cone with an apex angle so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height ,h where the radius of its cross section is .b If the surface tension of water is ,S its density is , and its contact angle with glass is , the value of h will be (g is the acceleration due to gravity)
h
A) 2 cosS
b g
B) 2 cosSb g
C) 2 cos / 2Sb g
D) 2 cos / 2Sb g
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Key: D
Sol:
R
hb
/2
/2
/2
bcos
2 R
bR
cos2
2Sh gR
2ShR g
cos
2S 2hg b
9. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is
BLOCK
LIQUID
S A) 1.21
B) 1.30
C) 1.36
D) 1.42
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Key: C
Sol:
2 2g
22rel
2h 4hD 1
D1
22.72 4 10 2.721 4 2 1.36
11.54 2
10. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its
original position after hitting the surface. The force on the ball during the collision is
proportional to the length of compression of the ball. Which one of the following
sketches describes the variation of its kinetic energy K with time t most
appropriately? The figures are only illustrative and not to the scale
A) K
t
B) K
t
C) K
t
D) K
t
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Key: B
Sol: Before it collides, 2 2 21 1KE mv m(gt) KE t2 2
Contact time is very small and in that time, KE drops to zero very quickly and increases and after
that it rises with decreasing KE under gravity. This is best described by graph (B)
SECTION – 2 Comprehension Type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has ONLY ONE correct answer among the four given options (A), (B), (C), and (D).
Paragraph For Questions 11 & 12
In the figure a container is shown to have a movable (without friction) piston on top.
The container and the piston are all made of perfectly insulating material allowing no
heat transfer between outside and inside the container. The container is divided into
two compartments by a rigid partition made of a thermally conducting material that
allows slow transfer of heat. The lower compartment of the container is filled with 2
moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2
moles of an ideal diatomic gas at 400 K . The heat capacities per mole of an ideal
monatomic gas are 3 5, ,2 2V PC R C R and those for an ideal diatomic gas are
5 7,2 2V PC R C R .
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11. Consider the partition to be rigidly fixed so that it does not move. When equilibrium
is achieved, the final temperature of the gases will be
A) 550 K
B) 525 K
C) 513 K
D) 490 K
Key: D
Sol: 1 21 v 2 pn C R(700 T) n C (T 400) T 490K
12. Now consider the partition to be free to move without friction so that the pressure of
gases in both compartments is the same. Then total work done by the gases till the
time they achieve equilibrium will be
A) 250R
B) 200R
C) 100 R
D) 100 R
Key: D
Sol:
1 1 2 2
1 2 1 2 initial final
( U W ) ( U W ) 0( W W ) ( U U ) U U
Calculation for Tf
1 2
1 2 1 2
1 P f 2 P f f
gases 1 v 2 v 1 V 2 V f
gas
6300n C (700 T ) n C (T 400) T K12
( W) n C (700) n C (400) (n C n C )T
so, W 100R
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Paragraph For Questions 13&14
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin
tube of uniform cross section is connected to the nozzle. The other end of the tube is
in a small liquid container. As the piston pushes air through the nozzle, the liquid from
the container rises into the nozzle and is sprayed out. For the spray gun shown, the
radii of the piston and the nozzle ar 20mm and 1mm respectively. The upper end of the
container is open to the atmosphere.
13. If the piston is pushed at a speed of 15mms the air comes out of the nozzle with a
speed of
A) 10.1ms
B) 11ms
C) 12ms
D) 18ms Key: C
Sol: 1 1 2 2 2A V A V V 2m / s
14. If the density of air is a and that of the liquid ,l then for give piston speed the rate
(volume per unit time) at which the liquid is sprayed will be proportional to
A) a
l
B) a l
C) l
a
D) l
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Key: A
Sol:
P
Consider a fluid element at point P due to flow in horizontal pipe pressure energy of
liquid in vertical pipe at ‘P’ reduces = 20 a 0
1P V2
But net energy of fluid element remains same
22 1 1 a0 0 a 0 0
1 1P P V V V V2 2
Paragraph For Questions 15 & 16
The figure shows a circular loop of radius a with two long parallel wires (numbered 1
and 2) all in the plane of the paper. The distance of each wire from the centre of the
loop is d . The loop and the wires are carrying the same current I . The current in the
loop is in the counterclockwise direction if seen from above.
a
P R
Q S
Wire1 Wire2
d d
15. When d a but wires are not touching the loop, it is found that the net magnetic field
on the axis of the loop is zero at a height h above the loop. In that case
A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h a
B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h a
C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and 1.2h a
D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and 1.2h a
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Key: C
Sol:
a
P R
Q S
Wire1 Wire2
d d
Bwire
Bclosed loop
Bwire
h
a a
2 2h a
2
0 0wire closed loop 1/2 2 2 3/22 2
2 i ia2B cos B cos2 2(a h )2 d h
2 2
2 2 1/2 2 2 3/2 2 2
2 1cos a 2 acos(a h ) [a h ] [a h ]
2 2
2 2 2 2 2 2
2 h a aha h (a h ) 2 a h
2 2 2 2 44h (a h ) a
22h x
On solving we get
2ax (2 11 1)
2
2
2 a2h 2 11 1 h 1.2a2
16. Consider ,d a and the loop is rotated about its diameter parallel to the wires by 030
from the position shown in the figure. If the currents in the wires are in the opposite
directions, the torque on the loop at its new position will be (assume that the net field
due to the wires is constant over the loop)
A) 2 2
oI ad
B) 2 2
2oI a
d
C) 2 23 oI a
d
D) 2 23
2oI ad
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Key: B
Sol:
2 00net
2 i| | M B (i a ) sin 302 d
2 2
0i a 1d 2
SECTION-3
Matching List Type (Only One Option Correct)
This section contains four questions, each having two matching lists. Choices for the correct
combination of elements from list-I and List-II are given as options (A), (B),(C) and (D), out of
which ONE is correct.
17. Four charges 1 2, 3,Q Q Q and 4Q of same magnitude are fixed along the x axis at x =–2a,
-a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at a
distance b >0. Four options of the signs of these charges are given in List I. The
direction of the forces on the charge q is given in List II. Match list I with List II and
select the correct answer using the code given below the lists.
Q1 Q3Q2 Q4
q(0,b)
(-2a,0) (-a,0) (+a,0) (+2a,0) List I List II
P. 1 2 3 4, , ,Q Q Q Q all positive 1. x
Q. 1 2,Q Q positive; 3 4,Q Q negative 2. x
R. 1, 4Q Q positive; 2, 3Q Q negative 3. y S. 1 3,Q Q positive; 2 4,Q Q negative 4. y Code : A) P-3, Q-1, R-4 S-2 B) P-4, Q-2, R-3, S-1 C) P-3, Q-1, R-2, S-4 D) P-4, Q-2, R-1, S-3
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Key: A
Sol: Q1 Q3Q2 Q4
q(0,b)
(-2a,0) (-a,0) (+a,0) (+2a,0)
1 2 3 4 y
1 2 3 4 x
1 4 2 3 y
1 3 2 4 x
ˆ ˆP) Q Q Q Q ve; E 0i E jˆ ˆQ) Q Q ve; Q Q ve; E E i 0 j
ˆ ˆR) Q Q ve; Q Q ve; E 0i E jˆ ˆS) Q Q ve; Q Q ve; E E i 0j
P 3;Q 1; R 4;S 2
18. Four combinations of two thin lines are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in list I with their focal length in List II and select the correct answer using the code given below the lists.
List I List II
P. 1. 2r
Q. 2. 2r
R. 3. r
S. 4. r Code: A) P-1, Q-2, R-3,S-4 B) P-2, Q-4, R-3, S-1
C)P-4, Q-1, R-2, S-3
D) P-2, Q-1, R-3, S-4
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Key: B
Sol: for convex lens 1 1 1 1(1.5 1) f rf r r r
For plano-convex it is f1 = 2r
For concave lens 1 1 1 1(1.5 1) f rf r r r
For plao-concave it is 11f 2r
P. eqeq
1 1 1 2 rff r r r 2
Q. eqeq
1 1 1 1 f rf 2r 2r r
R. eqeq
1 1 1 1 f rf 2r 2r r
S. eqeq
1 1 1 f 2rf r 2r
P 2;Q 4; R 3;S 1
19. A block of mass 1 1m kg another mass 2 2 ,m kg are placed together (see figure) on an inclined plane with angle of inclination . Various values of are given in List I. The coefficient of friction between the block 1m and the plane is always zero. The coefficient of static and dynamic friction between the block 2m and the plane are equal to 0.3. In List II expressions for the friction on block 2m are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g.
[Useful information : tan 5.5 0.1; tan 11.5 0.2; tan 16.5 0.3; ]
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1m2m
List I List II
P. 5 1. 2 sinm g
Q. 10 2. 1 2 sinm m g
R. 15 3. 2 cosm g
S. 20 4. 1 2 cosm m g
Code :
A) P-1, Q-1, R-1, S-3
B) P-2, Q-2, R-2, S-3
C) P-2, Q-2, R-2, S-4
D) P-2, Q-2, R-3, S-3
Key: D
Sol: L 1 2f (m m )gsin
22 1 2
1 2
mm g cos (m m )gsin tan 0.2(m m )
Where is angle of repose = 11.50
0 0
2
15 & 20 ; blocks are in motionHence friction is m g cos
0 0
1 2
5 & 10 ; blocks are at rest.Hence friction (m m )g sin
P 2;Q 2;R 3;S 3
20. A person in a lift is holding a water jar, which has a small hole at the lower end of its
side. When the lift is at rest, the water jet coming out of the hole hits the floor of the
lift at a distance d of 1.2 m from the person. In the following, state of the lift’s
motion is given in List I and the distance where the water jet hits the floor of the lift is
given in List II. Match the statements from List I with those in List II and select the
correct answer using the code given below the lists.
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List I List-II
P. Lift is accelerating vertically up. 1. 1.2d m
Q. Lift is accelerating vertically down with an 2. 1.2d m
acceleration less than the gravitational acceleration
R. Lift is moving vertically up with constant speed 3. 1.2d m
S. Lift is falling freely 4. No water leaks out of the jar
Code:
A) P-2, Q-3, R-2, S-4
B) P-2, Q-3, R-1, S-4
C) P-1, Q-1, R-1, S-4
D) P-2, Q-3, R-1, S-1
Key: C
Sol: Of the lift is accelerations upward are downward with acceleration less than
acceleration due to gravity d = 1.2
Lift is moving vertically up with constant speed
d = 1.2
When lift is falling freely, water does not leak.
Hence ‘C’ is the correct option
P 1;Q 1;R 1;S 4
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PART II: CHEMISTRY SECTION – 1
(Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct. 21. The product formed in the reaction of 2SOCl with white phosphorous is A) 3PCl
B) 2 2SO Cl
C) 2SCl D) 3POCl Key: A
Sol: 4 2 3 2 2 28 4 4 2 P SOCl PCl SO S Cl
22. The major product in the following reaction is
[Figure]
Cl
3CH
O1 dry ether, 03
2 aq. acid.CH MgBr, C
.
A) 3H C
3CH
O
B) 2H C
3CH
OH
3CH
C) O
2CH
D) O
3CH
3CH
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Key: D
Sol:
ClO
3CH MgBr
ClO-
o
23. Hydrogen peroxide in its reaction with 4KIO and 2NH OH respectively, is acting as a
A) reducing agent, oxidising agent
B) reducing agent, reducing agent
C) oxidising agent, oxidising agent
D) oxidising agent, reducing agent
Key: A
Sol: 2 2 2 2 22 4 NH OH H O N H O
24. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is
CompleteHydrolysis
6 other productXeF P
2OH / H O
Q
2slow disproportionation in OH / H O
products A) 0 B) 1 C) 2 D) 3 Key: C
Sol: 3 2 2 4 6 24 6 .8 XeO NaOH H O Xe O Na XeO H O
Gaseous products are Xe and 2O
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25. The acidic hydrolysis of ether (X) shown below is fastest when
[Figure]
acid OH ROH
X
OR
A) one phenyl group is replaced by a methyl group.
B) one phenyl group is replaced by a para-methoxyphenyl group.
C) two phenyl groups are replaced by two para-methoxyphenyl groups.
D) no structural change is made to X.
Key: D
Sol: Additional stability of carbocation is attained by 4 methoxy phenyl substitution than
simple phenyl substitution.
26. Isomers of hexane, based on their branching, can be divided into three distinct classes
as shown in the figure.
[Figure]
andI andII
III
The correct order of their boiling point is
A) I > II > III
B) III > II > I
C) II > III > I
D) III > I > II
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Key: B
Sol: Boiling points of isomeric alkanes are dependent on surface area [branching decreases
surfaces area increases ] with increasing branching bp decreases
So order is III>II>I
27. For the identification of -naphthol using dye test, it is necessary to use
A) dichloromethane solution of -naphthol.
B) acidic solution of -naphthol.
C) neutral solution of -naphthol.
D) alkaline solution of -naphthol.
Key: D
Sol: Alkaline solution of -naphthol increases the electron density at position and
thereby it enhances the rate of reaction .
28. Assuming 2 2s p mixing is NOT operative, the paramagnetic species among the
following is
A) 2Be
B) 2B
C) 2C
D) 2N Key: C
Sol: If there is no s-p mixing M.O electro configuration of 2C is
2 * 2 2 2 1 11 1 2 2 2z x ys s s sp p p
So 2C is paramagnetic
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29. For the elementary reaction M N , the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is
A) 4 B) 3 C) 2 D) 1 Key: B
Sol: nr kM 30. For the process
2 2H O l H O g at 0100T C and 1atmosphere pressure, the correct choice is
A) 0systemS and 0surroundingsS
B) 0systemS and 0surroundingsS
C) 0systemS and 0surroundingsS
D) 0systemS and 0surroundingsS
Key: B
Sol: qST
SECTION – 2
Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc.Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only ONE correct answer among the four given options (A), (B), (C)and(D)
Paragraph For Questions 31 and 32 Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider
only the major products formed in each step for both the schemes.
HO
H1 (excess)22.CH (1 equivalent)3 2
3 (1 equivalent)34 Lindlar's catalyst2
Scheme-1.NaNH
CH I.CH I.H .
X
M
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H
21 2equivalent.NaNH
OHBr2.
33 mild.H O.
24.H ,Pd / C
35.CrO
Y Scheme-2
N
31. The product X is
A) 3H CO
H H
B)
3H CO
H
H
C) 3 2CH CH O
H H
D) 3 2CH CH O
H
H
Key: A
Sol: Since acetalyde carbanion is more nucleophilic in nature it firstly react with ethyl
iodide and give the product accordingly
32. The correct statement with respect to product Y is
A) It gives a positive Tollens test and is a functional isomer of X.
B) It gives a positive Tollens test and is a geometrical isomer of X.
C) It gives a positive iodoform test and is a functional isomer of X.
D) It gives a positive iodoform test and is a geometrical isomer of X.
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Key: C
Sol: As 2 /H Pd C reduces triple bond to single bond in absence of poison catalyst like
Barium sulphate and 3CrO oxidises secondary alcohol to ketone finally we get the
functional isomer of alcohol which gives iodoform test
Paragraph For Questions 33 and 34
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in
excess to give tetrahedral and square planar complexes, respectively. An aqueous
solution of another metal ion M2 always forms tetrahedral complexes with these
reagents.
Aqueous solution of M2 on reaction with reagent S gives white precipitate which
dissolves in excess of S. The reactions are summarized in the scheme given below:
SCHEME:
Tetrahedral Q Rexcess excessM1 Squareplanar
Tetrahedral Q Rexcess excessM2 Tetrahedral
S, stoichiometric amount
White precipitate Sexcess precipitatedissolves
33. M1, Q and R, respectively are
A) 2Zn , KCN and HCl B) 2 ,Ni HCl and KCN C) 2 ,Cd KCN and HCl
D) 2 ,Co HCl and KCN
Key: B
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34. Reagent S is
A) 4 6K Fe CN
B) 2 4Na HPO
C) 2 4K CrO
D) KOH Key:-D
Sol: 33 & 34
2Ni with Cl gives tetrahedral 24NiCl and with CN gives square planar
24
Ni CN
2Zn gives tetrahedral complexes with both Cl and CN
2Zn gives white ppt with KOH but dissolve in excess
Paragraph For Questions 35 and 36
X and Y are two volatile liquids with molar weights of -110g mol and 140g mol
respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are
simultaneously placed at the ends of a tube of length 24L cm , as shown in the
figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature
of 300K. Vapours of X and Y react to form a product which is first observed at a
distance d cm from the plug soaked in X. Take X and Y to have equal molecular
diameters and assume ideal behaviour for the inert gas and the two vapours.
Initial formation ofthe product
Cotton woolsoaked in Y
Cotton woolsoaked in X d
24L cm
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35. The value of d in cm (shown in the figure), as estimated from Graham’s law is
A) 8
B) 12
C) 16
D) 20
Key: C
Sol: 4024 10
d
d
36. The experimental value of d is found to be smaller than the estimate obtained using
Graham’s law. This is due to
A) larger mean free path for X as compared to that of Y.
B) larger mean free path for Y as compared to that of X.
C) increased collision frequency of Y with the inert gas as compared to that of X
D) increased collision frequency of X with the inert gas as compared to that of Y with
the inert gas.
Key: D
Sol: Collision frequency increases then d decreases
SECTION -3
Matching List Type (Only One Options Correct)
This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which ONE is correct 37. Different possible thermal decomposition pathways for peroxyesters are shown below.
Match each pathway from List -1 with an appropriate structure from List-II and select the
correct answer using the code given below the lists.
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OO
R'R
O
Peroxyester
P
2-CO R R' O
Q
2-CO R R' O carbonyl compoundR X
R2RCO R' O
2-CO carbonyl compoundR X '
S2RCO R' O
2-CO R R' O
List-I List-II
P. Pathway P 1. 6 5 2C H CH O O3CH
O
Q. Pathway Q 2. 6 5C H O O
3CH
O
R. Pathway R 3.
6 5 2C H CH O O 3CHO
3CH2 6 5CH C H
S. Pathway S 4.
6 5C H O O 3CHO
3CH6 5C H
Codes:
P Q R S
A) 1 3 4 2
B) 2 4 3 1
C) 4 1 2 3
D) 3 2 1 4
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Key: A
Sol: P
6 5 2 3 2 3 21C H CH C O O CH Ph CH CH O CO
O
Q
6 5 2 3 2 3 23CH CH C O O C CH Ph CH O C CH CO
O CH3
CH2Ph
CH3
CH2Ph
2Ph CH
O
R
6 5 2 3 34CH CH C O O C CH Ph C O O C CH
O CH3
Ph
CH3
Ph
Ph
O
OH
+
S
6 5 3 6 5 32C H C O O CH C H C O CH O
O
Ph
O2CO
38. Match the four starting materials (P, Q, R, S) given in List-1 with the corresponding reaction
schemes (I, II, II, IV) provided in List-II and select the correct answer using the code given
below the lists.
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List-I List-II
P. H H 1. Scheme I
4 22 3
( ) , , ,
7 6 2 3?i KMnO HO heat ii H H Oiii SOCl iv NH C H N O
Q.
OH
OH 2. Scheme II
3 2 4
3 2 4
( ) / .. ,
6 6 2 2?i Sn HCl ii CH COCl iii conc H SOiv HNO v dil H SO heat vi HO C H N O
R.
NO2
3. Scheme III
3 2 4
2 3 2 2 4
,873 fuming HNO , ,. ,
6 5 3?i red hot iron K ii H SO heatiii H S NH iv NaNO H SO v hydrolysis C H NO
S.
NO2
CH3 4. Scheme IV
2 4
3 2 4 2 4
. ,60. , . .
6 5 4?i conc H SO Cii conc HNO conc H SO iii dil H SO heat C H NO
Codes:
P Q R S
A) 1 4 2 3
B) 3 1 4 2
C) 3 4 2 1
D) 4 1 3 2
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Key: C
Sol: 3P
3/873
redHotFe K
min/ /3 2 4
Fu gHNO H SO
2NO
2NO
/3 2NH H S2NO
2NH
2 /NaNO HCl Hydrolysis
2NO
OH
6 5 3C H NO
4Q
OH
oy
2 4060
ConH SOC
OH
3SOH
2 43
ConH SOHNO
2NO
3SOH
OH
2 4
dilH SO
OH
OH
2NO
6 5 4C H NO
OH OH
R
2NO
(2) /Sn HCl2NH
3 CH C Cl
O NH
O
3CH
2 4 /ConH SO
NH
O
3CH
3SOH
3HNO
NH
O
3CH
3SOH
2 4dilH SO2NO
3SOH
2NH2NO
OH
2NH2NO
6 6 2 2C H N O
2
S 1
2NO
3CH
/ /4
KMnO OH Heat
2NO
2CO H
2SOCl
2NO
COCl
3NH
2NO
2CONH
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39. Match each coordination compound in List-I with an appropriate pair of characteristics from
List-II and select the correct answer using the code given below the lists.
2 2 2 2en=H NCH CH NH ;atomicnumbers:Ti=22; 24;Co=27;Pt=78Cr
List-I List-II
P. 3 24Cr NH Cl Cl 1. Paramagnetic and exhibits ionisation
isomerism
Q. 2 35 2Ti H O Cl NO 2. Diamagnetic and exhibits cis-trans
isomerism
R. 3 3Pt en NH Cl NO 3. Paramagnetic and exhibits cis-trans
isomerism
S. 3 3 34 2Co NH NO NO 4. Diamagnetic and exhibits ionisation
isomerism
Codes:
P Q R S
A) 4 2 3 1
B) 3 1 4 2
C) 2 1 3 4
D) 1 3 4 2
Key: B
Sol: 3 24 Cr NH Cl Cl exhibit paramagnetic and geometrical isomerism.
2 35 2 Ti H O Cl NO exhibit paramagnetism and ionisation isomerism
3 3 Pt en NH Cl NO exhibit diamagnetism and ionsation isomers
3 3 34 2 Co NH NO NO exhibit diamagnetism and as-trans isomers
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40. Match the orbital overlap figures shown in List-I with the description given in List-II and
select the correct answer using the code given below the lists.
List-I List-II
P. 1. p d antibonding
Q. 2. d d bonding
R. 3. p d bonding
S. 4. d d antibonding
Codes:
P Q R S
A) 2 1 3 4
B) 4 3 1 2
C) 2 3 1 4
D) 4 1 3 2
Key: C
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List –I List-II
Sol: P. d-d (possible ovelap)
Q. p-d (possible-side wise where so p-d bond )
R. p-d (Negative side wise overlap)
S. d-d (Negative overlap) antibonds d-d anti bond
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PART –III_MATHEMATICS SECTION-1
(Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices A,B,C and D out of which ONLY ONE is correct. 41. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in
envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is
A) 264 B) 265 C) 53 D) 67 Key: C
Sol: Req. no. of ways = 41
1 1 1 34 2 1 2 532 3 4 4
C
42. In a triangle the sum of two sides is ‘x’ and the product of the same two sides is ‘y’. If 2 2 , x c y where ‘c’ is the third side of the triangle, then the ratio of the in-radius to
the circum-radius of the triangle is
A) 3
2 y
x x c
B) 3
2 y
c x c
C) 3
4 y
x x c
D) 3
4 y
c x c
Key: B
Sol: 2 2 2 1 22 2 2 cos4 3
a b c ab s s c ab c c
1 sin 321 22 3
ab Cr ycR sR c x cc x
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43. The common tangents to the circle 2 2 2 x y and the parabola 2 8y x touch the
circle at the points P, Q and the parabola at the points R, S. Then the area of the
quadrilateral PQRS is
A) 3
B) 6
C) 9
D) 15
Key: D
Sol: 2 2 22 tan 8 2 Let y mx be a gent to y x if also touches x ym
1 m
2, 4 , 1, 1 R P
Req. area = (1 + 4) (1 + 2) = 15
44. Three boys and two girls stand in a queue. The probability, that the number of boys
ahead of every girl is at least one more than the number of girls ahead of her, is
A) 12
B) 13
C) 23
D) 34
Key: A
Sol: Required probability = 5 3! 2!5!
45. The quadratic equation 0p x with real coefficients has purely imaginary roots.
Then the equation 0p p x has
A) Only purely imaginary roots B) All real roots C) two real and two purely imaginary roots D) neither real nor purely imaginary roots Key: D Sol: Let p(x) = 2 4 21 2 2 x p p x x x
102
iIf p p x then x
Hence 0p p x has neither real nor purely imaginary roots
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46. For 0, x , the equation sin 2sin 2 sin 3 3 x x x has
A) infinitely many solutions
B) three solutions
C) one solution
D) no solution
Key: D
Sol: Apply boundary conditions
47. The following integral
2
17
4
2cos
ec x dx is equal to
A) log 1 2 16
02
u ue e du
B) log 1 2 17
0
u ue e du
C) log 1 2 17
0
u ue e du
D) log 1 2 16
02
u ue e du
Key: A
Sol: Let cos2
u ue eecx
cos cot2
u ue eecx x dx du
2
u ududx
e e
021717
log 2 14
22cos
u uu u
duec x dx e ee e
log 2 1
16
0
2
u ue e du
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48. Coefficient of 11x in the expansion of 4 7 122 3 41 1 1 x x x is
A) 1051
B) 1106
C) 1113
D) 1120
Key: C
Sol: 4 7 124 7 122 3 4 4 2 7 3 12 4
0 0 0
1 1 1
s r ts r t
s r t
x x x C x C x C x
2 3 4 11 Hence s r t
Req. value is 4 7 12 4 7 12 4 7 12 4 7 122 1 1 0 1 2 4 1 0 1 3 0. . . . . . . . 1113 C C C C C C C C C C C C
49. Let : 0, 2 f be a function which is continuous on [0, 2] and is differentiable on
(0, 2) with f(0) = 1. Let 2
0
x
F x f t dt
0, 2 . ' ' 0, 2 , 2 for x If F x f x for all x then F equals
A) 2 1e
B) 4 1e
C) 1e
D) 4e
Key: B
Sol: ' .2 ' F x f x x f x
2'2 log 0 0 1
f xx f x x c c f
f x
4 ' ' 1 0 0 f x e and F x f x F x f x C F x f x F
42 2 1 1 F f e
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50. The function y f x is the solution of the differential equation
4
2 2
21 1
dy xy x xdx x x
in (-1, 1) satisfying f(0) = 0. Then
32
32
f x dx is
A) 33 2
B) 33 4
C) 36 4
D) 36 2
Key: B
Sol: 2 21. 1 . 1, 1 x dx
xI F e x x
5
2 4 21 2 , 0 0 05
xy x x x dx x c but c t
5 2
2 25 1 1
x xf xx x
3 3 322 2 2
1 2 1
2003
2
1 1 12 2 sin 1 sin2 21
x xf x dx dx x x xx
3 323 8 6 3 4
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SECTION – 2
Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc .Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only ONE correct answer among the four given options (A), (B), (C)and(D)
Paragraph For Questions 51 and 52
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing
numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6,
7. A card is drawn from each of the boxes. Let ix be the number on the card drawn
from the thi box, i = 1, 2, 3
51. The probability that 1 2 3 x x x is odd, is
A) 29105
B) 53105
C) 57105
D) 12
Key: B
Sol: Required probability 2 3 4 1 2 4 1 3 3 2 2 3 533 5 7 3 5 7 3 5 7 3 5 7 105
52. The probability that 1 2 3, ,x x x are in an arithmetic progression, is
A) 9105
B) 10105
C) 11105
D) 7105
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Key: C
Sol: 2 1 32 x x x
Combinations are 1,1,1 2,2,2 3,3,3 1,2,3 3,2,1 2,3,4 3,4,5 3,5,7 2,4,6 1,3,5 1,4,7
Required probability 11105
Paragraph For Questions 53 and 54
Let a, r, s, t be non-zero real numbers. Let 2 2 2, 2 , , , 2 , 2P at at Q R ar ar and S as as
be distinct points on the parabola 2 4y ax . Suppose that PQ is the focal chord and lines QR and PK are parallel, where ‘K’ is the point (2a, 0).
53. The value of ‘r’ is
A) 1
t
B) 2 1t
t
C) 1t
D) 2 1t
t
Key: D
Sol: Given that 1 1 t and 2
121
2 2 2 1 12 2
at tt r t r t
t at a t t t
54. If st = 1, then the tangent at ‘P’ and the normal at S to the parabola meet at a point whose ordinate is
A) 22
3
12tt
B) 22
3
12a tt
C) 22
3
1a tt
D) 22
3
2a tt
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Key: B
Sol: Equation of tangent at P is 2 xyt x at y att
&
Equation of normal at S is 3 32 2 xy sx as as at sax as ast
But 23 2
2 21 2 x a a ast at a att t t t
222
2
112 2 2 3
a tat ay a att t
Paragraph For Questions 55 and 56
Given that for each 0, 1a 1
1
0lim 1
haa
hh
t t dt
exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0, 1).
55. The value of 12
g is
A) B) 2
C) 2
D) 4
Key: A
Sol: 1
1
0
1 aag a t t dt
1
0
12 1
dtg
t t
56. The value of 1'2
g is
A) 2
B)
C) 2
D) 0
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Key: D
Sol: 1
1
0
1 aag a t t dt
1' 02
g
SECTION-3
Matching List Type (Only one Option Correct)
This section contains four questions, each having two matching lists. Choices for the correct
combination of elements from List-II are given as options (A), (B), (C) and (D), out of which ONE
is correct.
57. List-I List-II
P. The number of polynomials f x with non-negative integer 1. 8
Coefficients of degree 2 , satisfying 0 0f and
1
01, f x dx is
Q. The number of points in the interval 13, 13 at which 2. 2
2 2sin cos f x x x attains its maximum value, is
R.
22
2
31 x
x dxe
equals 3. 4
S.
121212
0
1cos 2 log1
1cos 2 log1
xx dxx
xx dxx
equals 4. 0
P Q R S
(A) 3 2 4 1
(B) 2 3 4 1
(C) 3 2 1 4
(D) 2 3 1 4
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
Key: D Sol: P) Let 2 f x a x bx
1 2 3 6 0, 2 3,03 2
a b a b
No. of 2f x
Q) 22 sin4
f x x from the graph of ,f x no. of max is 4
R) 2 2 22 2
2
2 2 2
3 3 2 3 81 1
x x
x xI dx dx I x dx Ie e
S)
12
12
1cos 2 log 01
xx dxx
58. List-I List-II
P. Let 1 3cos 3cos , 1,1 ,2
y x x x x .Then 1. 1
22
2
1 1
d y x dy xx x
y x dx dx equals
Q. Let 1 2, ,... 2nA A A n be the vertices of a regular polygon 2. 2
of n sides with its centre at the origin. Let
ka be the position
vector of the point , 1,2,...,kA k n . If
1 11 11 1
. ,
n nk k k kk k
a a a a then the minimum value of n is
R. If the normal from the point ,1P h on the ellipse 2 2
16 3
x y 3. 8
is perpendicular to the line 8, x y then the value of h is
S. Number of positive solutions satisfying the equation 4. 9
1 1 12
1 1 2tan tan tan2 1 4 1
x x x is
P Q R S (A) 4 3 2 1 (B) 2 4 3 1 (C) 4 3 1 2 (D) 2 4 1 3
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
Key: A
Sol: P) 1 1 2 2 2112 2
3cos 3cos 1 9 11 1
yy x y x yy x
2 21 2 1 12 1 11 2 18 y y xy yy
22 11 9 y x xy y
Q) 1 2 2 3 1 1 2 2 3 1...... . . ...... . n n n na a a a a a a a a a a a
2 21 sin 1 cos
n nn n
2tan 1 n
2tan 1 n
2 3,4 4
or
n
2 1 34 4 4 or
8n
R) Equation of normal 6 3 31
x y
h
Slope 6 / 23
h
h but 2 1 2 h
h
S) 2
1 112 1 4 1
1 112 1 4 1
x xx
x x
2 2
2 3 1 28 6
x
x x x
3 1 18 6
xx x
23 7 6 0 x x
3x only one solution
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
59. Let 1 2 3: , :[0, ) , : f f f and 4 : [0, ) f be defined by
10,0;
x
x if xf x
e if x
22 ;f x x
3
sin 0,0
x if xf x
x if x
And
2 1
42 1
0,
1 0.
f f x if xf x
f f x if x
List-I List-II
P. 4f is 1. Onto but not one-one
Q. 3f is 2. Neither continuous nor one-one
R. 2 1f of is 3. Differentiable but not one-one
S. 2f is 4. Continuous and one-one
P Q R S
(A) 3 1 4 2
(B) 1 3 4 2
(C) 3 1 2 4
(D) 1 3 2 4
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
Key: D
Sol :
1
, 0, 0
x
x xf x
e x
y = -xy = ex
2 f x
y
x
2y x
3f x
siny x y x
2
4 2
, 01, 0
x
x xf xe x
2y x 2 1 xy e
p) From the graph range of 4 [0, ) f , but not 1 – 1 q) from graph range is not R, but not 1 – 1 and continuous, differentiable at ‘0’ r) 2
2 1 , 0 f f x x 2 , 0xe x 0at x L. L = 0, R.L = 1 Disc at x = 0, Clearly from graph not 1 – 1 s) Clearly from graph one one and continuous
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2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS
60. Let 2 2cos sin ; 1, 2,...,910 10
kk kZ i k .
List-I List-II
P. For each kZ there exists a jZ such that . 1k jZ Z 1. True
Q. There exists a 1,2,....,9k such that 1 . kZ Z Z has no 2. False
solution Z in the set of complex numbers.
R. 1 2 91 1 ... 110
z z z equals 3. 1
S. 9
1
21 cos10
k
k equals 4. 2
P Q R S A) 1 2 4 3
B) 2 1 3 4
C) 1 2 3 4
D) 2 1 4 3
Key: C Sol: 1, 1 2 3 9, , ,....z z z z are 10th roots of units
p) 10
2 10 2 1c s10 10
k k
k
k kz i cis zz
, . 1 k j k jfor each z a z such that z z
q) If 11
. kk
zz z z zz
19 1 2 9 10
. , .... ,1
kk
z z z z z z z z
r) 1 2 3 91 , 1 , 1 ,... 1 z z z z are the roots of 101 1 0 x
10 10 91 1 10 ..... 10 0 x x x x 1 2 91 1 ...... 1 z z z = product of roots for 9 910 .... 10 0 x x s) sum of 10th roots of units = 0
9
1
21 010
k
kcis
9
1
2 110
k
kcos
Ans = 2
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